Piskunov -Differential And Integral Calculus - N Piskunov

N.PISKUNOV DIFFERENTIAL and INTEGRAL CALCULUS Mik PUBLISHERS Moscow 1980 TRANSLATED FROM THE RUSSIAN BY 0. YANKOVSKY H.C. Mnexynon AN@SEPEHUMARBHOE HMHHTETPANBHOE VCuHCREHHS Ha anzautcnon sone CONTENTS Preface oo eeeeeeeeeeee (Chapter 1.NUMBER. VARIABLE. FUNCTION 1.Real Numbers, Real Numbers asPoints on aNumber Scale... -19 2 The Absolute Value ofaRealNumber. vvsweslll 3VariablesandConstants. see TTI Till 6 4TheRangeofaVariable cs a6 &Ordered Varabes. increasing ‘aid’Becreasing ‘Variabie. “Bonded Bina Sona ngmppeoocouoooocededes Ly 7,Ways ofRepresenting Funétigns <0 000 2D DIDI LIf 9 8Basie Elementary Functions, Elementary’ Functions’ <2... 21. 39.AlgebraicFunctions=Ssounwsevents ©:ears 10.PolarCoordinate Sysiem32D LillwsExercisesonChapter|veee ees (Chapter Hl LIMIT, CONTINUITY OF AFUNCTION 1.TheLimit ofaVariable. AnInfinitely Large Variable... .... 322TheLimitofaFunctionoswesc eeeeee lll 3.AFunction thet Approaches lafinily: Bounded Funclions <2222 33 4Infinitesimals and Their Basic Properties vs. sess sss ss a25BasieTheoremsonLimitss.rvss0201222 lllB®6.TheLimitoftheFunction“2% asx—e0.0.20.20... 6.807.TheNumber vee teeeeeeeeeeeeeeeBLSNaturalLogarithms <2 DTDDID DIIDIIIITT bs9Continuity ofFunctions 220 LDL DIDlfgr 10.Certain Properties ofContinuous Functions’ 220020222222 61MW;Comparing fntnitesimals svnsesssSlt 8ExercisesonChapterowesvtcles 88 (ChapterI.DERIVATIVEANDDIFFERENTIAL 1.VelocityofMotionov.eeeeeeeeeeeee %DefinitionofDerivative ©oli iiiiiin 5.Geometric Meaning of the Defivative 2211202222 I1014Ditterentiability ofFunctions ene 22ST te 5.Finding the Derivatives ofElementary Functions! The ‘Derivative oftheFactiony=,WherentsPositiveandintegrals. vivswv+ 6,DerivativesoftheFunctionsymin;yaiconn oS 18 7.Derivatives of:aConstant, the Produet ofaConstant byaFunction,2Sum,aProduct,and-aQuotient. sv.vsrssee ss80 8.The Derivative of«Logarithmic Function’ 222220220022 at 9.The Derivative ofaComposite Function. <2. 2. 212 10.Derivatives oftheFunctions y=tanx, y=cotx, y=Injx] +. ++. 88 1" 11Aatlt Function an WsDiferentiton ge ee 8 2. The Gesoe siitancoftieDineen 222222:02:00 withRespecttothePolarAnglessseeeceeeeeeee1DExercises onChapter Hl26200ee ee ee ee 4AHeworPeetetndanetrastceenee SE 6.TheLimitolaRsloo Twiii LargeQuantities (vsantin Contents 5 7.Applying the Theory ofMaxima and Minima ofFunctions tothe SolutionofProblems wwe wet renee eeHD 8,Testing aFunction for’Maxisuin and Minimim byMeans 'ofTaylor's 9,Convesity” and’ Coneaviiy'of @Curve. Poinis’of taiection’ <> 2: +16310.Asymploles rneeneeee eee18811;GeneralPlanforinvestigaiing ‘Functions’ andCoasirdeting ‘Graphs19412;Investigating Curves Represented Parametrically sr se ces199 ExercisesonChapler Vivevecevecere reres208 Chapter Vic THE CURVATURE OF ACURVE 1.The Length ofanArc and ItsDerivative... 2... e208 3.CalculationotCurvature72TIIIT Ili lae 4Caleulation ofthe CurvatureofaLineRepresenied arameirically ||215 &Calculation of‘the Curvature ofa'Line Given byan’Equation ofPolarCoordinates’. cnn eeeeneenee,5 6.The Radius and Circie ofCurvature. Centee ofCurvature’ Evotite and 7,The Properties ofanEyoluie21 TTDLLLitat &Approximating the Real Roots ofanEquation’ |11.121 1235 ExercisesonChapterVoweeeeee eevee vO Chapter VIL. COMPLEX NUMBERS. POLYNOMIALS 1.ComplexNumbers.BasicDefinitions... 6...ceeeee28ZBasie Operations onComplex Numbers 2212212222212 age 3.Powers and Roots ofComplex Numbers <2 111001011 lagr 4Exponential Function witt Complex Exponent anditsProperties °|240,5.Euler's Formula, ‘TheExponential FormofaComplex Number =©|243, &Factoring #Polynomial nsneeeeecsate 7.TheMultipleRootsofaPolynomial: 2212121 L1DLar &Factorisation ofaPolynomial inthe Caie ofComipiex Roots 2! 2248 8,Interpolation. Lagrange's Interpolation Formula na. 2.2:29 10;OntheBestApproximation ofFunctions byPolyrioruiats, Chebyshev's ExercisesonChapter VILLE DDDDDDass Chapler VIII. FUNCTIONS OF SEVERAL VARIABLES 1,Definition ofaFunction ofSeveral Variables... ow... «258 %Geometric Representation ofaFunction ofTwo Variables’ *) <> |238 4.Partial and Total Increment of'a Function ss sess =1.2 22288 4Continuity ofaFunction ofSeveral Variables ©22222222 12 360 5,Partial Derivatives ofaFunction ofSeveral Variables <2 <1 21263 6.The Geometric’ Interpretation ofthe Parlisl Derivatives of&Func:tionofTwoVariable veswseeses eea6 7,Total Increment and Total Dittrentiais 2222222222522 365 &Approximation byTotal Differentias 222212222212 1368 8.Error Approximation byDifferentials |<2222022222120 10,The DerWvative of2Composite Function. The Toial Derivative <> |273 Ii:The Derivative ofaFunction Defined Implicitly «ss ss ss 7218 12,Partial Derivatives ofDifferent Orders es). 2000221209 6 Contents 1B.LevelSurfaces eeee ee288 1H.Divectional Derivatives’ 2222222002000 002tae 16.Taylor's Forindla fora Function ofTwo Variables <2 211 1390 17. Maximum and-Minimum of aFunction ofSeveral Variables” <° 202 18. Maximum and Minimum ofaFunction of Several Variables Relaied tyGivenEauatigns (Conditional MasiaandMinima)=«»«'s»300 19.SingularPointsofaCuvevssvrstent ceceesOBExercisesonChapterVIII... ttt tect lit. l310 Chapler 1X. APPLICATIONS OFDIFFERENTIAL CALCULUS T0SOLID GEOMETRY 1.The Equations ofaCurveinSpace oeeee esOM 2The Limits and. Derivative of the Vector” Function ‘of ‘aScalar Argument, The Equation ofaTangent to#Curve. The Equation of@ Normal Plane cee eset ee cere tenesSIT 3.RulesforDifferentiating Vectors (Véctor’ Functions) ¢2111 1San 4.The First and ‘Second “Derivatives ofaVector with Respect” toihe ‘ArcLength. TheCurvature ofaCurve. ThePrincipal Normal... =324 5.Osculating Plane. Binormal. Torsion swvss sc=022381 6A'Tangent PlaneandNormaltoaSurdace’22222222211 2338ExercisesonChapterIXovevvcece cee eee MO Chapter XINDEFINITE INTEGRALS 1,Antiderivative and theIndefinite Integral... 00.0. 342 2Table ofIntegralsene estle 3.SomeProperties ofanindefinitednteprei gS 4Inlegration bySubstitution (ChangeofVarisbley" 222S2Saas 5.Integrals ofPunetions Containing @Quadratic Trinomial |<.111351 &IntegrationbyParisceneeesee eS.BB 7Rational Fractions. Partial Rational Fractions and “Their ‘Integration 357 &Decomposition ofaRational Fraction into Partial Fractions =~ 361 9,Integration ofRational Fractions vv wsvee sen =02 236510.Ostrogeadsky's Methodvase sscvtect rset: 88M1Integrals ofIrrational Funcdions 1212222222221 iiian 12,Integrals oftheForm(R(x,VarFoxFOdr ...2.5... .372 13.Integration ofBinomial Differentials... ewe ee SS 14.Integration ofCertain Classes ofTrigonometrie “Functions” <<2>.378 15,Integration ofCertain Irrational Functions byMeans ofTrigonomelricSUDBMIMUHON. ee es teecnn 238 16.Funetions” Whose"Thtegials“Cannot"Be"Expressed’ in’Terms“ofElementary Functions sscerscee neneneeoe885 ExercisesonChapterXeveecece cee 6B86 Chapter Xi.THE DEFINITE. INTEGRAL 1.Statement ofthe Problem. The Lower and, Upper Integral Sums ...9962TheDefiniteIntegralcecwseeteete eeeee638 8:BasieProperties ofthe’Definite fotegral ©021222211211 240k 4.Evaluating aDefiniteIntegral.Newton-Leibniz “Formula><<.><407 8:ChangingtieVariableintheDefiniteintegral. ssso.22412 Contents 7 §,IntegrationbyParteeeeeeeeeee etd improperIntegrates TILT LITILLilitae &Appromimating Definite integrats 2222222222220 20Dae 9.Chebyshev's “Formula. DEDEDLED PDbiti0 10; Integrals Dependent on’aParameter 2222222222221 288 ExercisesonChapterXPoeoevvcecece eee AB [Chapter XI. GEOMETRIC AND MECHANICAL APPLICATIONS OF THE DEFINITE INTEGRAL 1.Computing Areas inRectangular Coordinates cee MMe2!The‘Area‘ofaCurvilinear SectorinPolarCoordinaies ©1212!1445,3,The Are Length ofa Curve ce we ee LAAT 4.Computing the Volume of Soiid from ‘theAress ofParaliei Sections (WalletsbySic) imTees ttttes2a 5,The VolumeofaSolidofRevoiuiion” 2222222222222. 2458 6.The Surface ofaSolidofRevolution. 20222222201 2458 7.Computing WorkbytheDefinite Integral 2222222222222 457 8Coordinates ofthe Centre ofGravity ne. 2112 LLTD 1112 489 ExercisesonChapter XII...vvvveveevee sess5462 ‘Chapter XIII. DIPERENTIAL EQUATIONS 1.Statement ofthe Problem. The Equation ofMotion ofaBody with Resistance of“the Medium Proportional to. the Velocity. ‘The Equationof@Catenaryossvn en eeee469 2,Delinitions Plilii i ie 3.First-Order Differential Equations (General Notions) oer 4.Equations with Separated and Separable Variables. The’ Protiem ‘of theDisintegration ofRadium eee ooaT8 5,Homogeneos FisOrdeEqusiiong PEDDLE Da &.Equations Reducible toHomogeneous Equations’ <2) 211): 4at 7,FirstOrder Linear Equations. ss sss 21111 Laer&Bernoulli's Equation 222222 DDTTLILIla 9.ExactDifferential Equations ©2222 222221221212 l4m 10,Integrating FactorOOONCoDDD LITT las iTheEnvelopeof FémilyofCurves DLbitlfer 12!SinguiarSolutionsofaFirst-OrderDifferentialEquation|2<2{‘S04 13:GatautsEquation 9 NI,ASE LDis U4Lagrange’ ‘Eauation¢2UDI LSDD 15.Orthogonal andIsogonaiTrajectories |2022222 LID22gg 16.Higher-Order Differentia! Equations(Fundamentats) |1...111Sia 17.AntEquationoftheFormyf=f(a) bu 518 18:Some Mtyper ofSecond-Order Dire’ Gaiations‘Redusbie' toFirst-Order ‘Equations. si 19,Graphical Metiod ofIniegrating Second-Order’ Differential "Equations 527 29,Homogeneous Linear Equation Defitions andGeneral Properties508 21;Second-Order Homogeneous Linear Equation withContant Coelic 22,Homogeneous’ Litiear’ Equations ‘ofthe ‘nth Order “with “Constant 23,Nonhomogeneous Second-Grier Linear’ Equations |) 2° >: 3a 2.Nonhomogeneous“Secnd-Order “Linear “Equatins) withCnstai 8 Contents 35,Higher-Order Nonhomogeneous Linear Equations... ss... S51 38.The Diflerential Equation ofMechanical Vibrations ©<<. << 2:855ZrFreeOstilations tsetse seeeens colle228S Be:ForeedOscillations 222022202 DTTLIIII LILIDD 29.SystemsofOrdinaryDifferential Equations©712712 °221865 55,SystemsofLinearbiterential Equations withContantCocticients869 BI:Ohyapunen's Theory atStabI eepincecnig2 32. Euler's Method ofApproximate Solution of First-Order Differential Equations ...teens we eeeee OBL 28,A‘Difference Method forApprosimaie’Solition ot“Differential” Equa: tinyBasedonTaylors Formula Adams: Method vn uy:504 34,An Approximate Method for" Integrating Systems ‘ofFirsi-OrderDisrerential Equation. vssteswees sterneeSOLExercisesonChapterXIvee ee888 (Chapter XIV. MULTIPLE INTEGRALS 1.Double Integrals. oe eeeeesOB 2.Calculating Double Integrals... ee ee BO 3.Calculating DoubleIntegrals(Continued) eee eeBIT4Gleuating AtesandVolumes byMeansofDobe inteats 2<J623 5:The Double Integral inPolar Coordinates = DLT eee6.Changing Variabfes in'aDouble Integral (General Case)|.<211633, 7Computing theAreaofaSurfacessnner 2111 88 8.TheDensity ofDistribution ofMatier andtheDouble Integral <<1642 9.TheMoment ofInertia oftheAreaofaPlaneFigure... ||643 10;TheCoordinates oftheCentreofGravityoftheAreaol«PianeFigure648 MH:TripleIntegrals snenssescwesnetreesee85 12Evaluating TripleUnlegial«ooo ah 13.ChangeofVariablesinaTripleIntegralee 656 14.uALboatofInertiaandtheCoordinatesoftheCentreofGravity0 15.Computing Integrals Dependent’ on ‘a’Parameter 2.21.2 1662EnercissonChapterXIVveveceene eecccee5668 (Chapter XV. LINE INTEGRALS AND SURFACE INTEGRALS [LineIntegrals eee eeeeeee+670 2Evaluating aLineIntegral2222222222 2D2I1IIiI lee 3GrenFormula ce ILD 4.Conditions for a”Line’ Integral Being Tadependent of”theBai’ of Imegratio cree cee er weceecee sencetesOL BSutonInteginis” LLLDETDID DEED &Evaluating Surface Integrals 2222222222222 222120 1689 &Ostrogeadsky's Formula2222222 DLILIES DILLleer 9.TheHamiltonian Operator anidCértain ‘Applications ofit’:.21700 ExerdsesonChapterXVvvcecevceveceseeveee518 ChapterXV1seats 1Series.SumofaSerieseee TO 2Necessary Condition forGoivergene ota‘Series 211222122. 73 3.Comparing Series with Positive Terms ese's22222222218 Contents 9 4. DiAlembert’sTestooeee eeeeeeeeTB S.Cauchys Tete ILILIILiiliiiiiiiial 6;TheIntegral TestforConvergence of‘a’Series 2212222221703 7Alternating Series. Lelbnie Theorem ene 2221 liae&Plusand-Minus Series,AbsoluteandContitional’ Convergence ©<<729 9: FunctionalSeries vsssrwtesteense Dis 10;MajorisedSeries DEDDDEDDDD DDDlie IL.TheContinuity oftheSumof«Series 2222222222220 738 12;Integration and Differentiation ofSeries 122212222112 28 15,Power Series. Interval ofConvergence. 221222212012 h2 If; Differentiation of Power Series ss2222222222112 0816.SeriesinPowersofsays 2.222222 l12 222018 16.Taylor's Series andMaclaurin’s ‘Series 0222222212211 7 17.ExamplesofExpansionofFunctionsin’Series|2212122111 751 18EulersFormulas sssveessessessclslelllBS 19.TheBinomialSeries2222) 22222 LiLLiiitpe 20, Expansion ofthe Function in’(1-f2) in’aPower’ Series. ‘ComputingLogarithms. « See eeeieeeeT 186 21,Integration byUse‘ofSeries(Calculating Definite’tntegraisy|2||788 22°Integrating Differential Equations byMeans ofSeries‘. "<2||780 2Beseel’sEquations ssn senescence2BB ExercisesonChapterXVIoweeee eee ll368 (Chapter XVI1. FOURIER SERIES 1,Definition. Statement ofthe Problem... os... eee ee 762Expansions ofFunctionsinFourierSeries©222°22122222 0 3.A‘Remark ontheExpansion ofPerlodic Function in’&FourierSeles ceetentcieeegnctne neseennTS 4,Pourier Series for'Even’ and Odd’ Functions |) 22222112 Ber 5.The Fourier Series for aFunction with Period 2 2). ¢2!!! m96.OntheExpansion ofaNonperiodie FunctioninaFourierSeries||7917.Roproximation by a.Trigonometric Polynomial of aFunction Represented inthe Mean.esassseeeeeceeeeeTB 8.TheDirichletIntegrals. 2122S LLLLiililme 8:TheConvergence ofaFourierSeriesataGiven’Boini:<|!2°:aor 10;Certain ‘Sufficient Conditions TortheConvergence ofa.Fourier’ Series 802 ULPractical Harmonie Analysis vss ces fee ewes ne ews80812.FourierIntegral eeee STILL LILlila 13.The Fourler‘integrai “taComplex’ Form 2222222222222 180ExercisesonChapterXVI... esvvsveveeeves ess8R Chapter XVIII. EQUATIONS OF MATHEMATICAL PHYSICS 1.BasicTypesofEquations ofMathematical Physics0... ..8152Derivation ofthe Equation ofOscillations of"aString.” FormulationoftheBoundary-VaiueProblem.Derivation ofEquations ofElectric Oscillations in Wires sense ee vans 816 3.Solution oftheEquation ‘ofOscillations of String’ Sy’the MethodofSeparationofVariables(TheFourierMelbod)2<7."°°.620 4,The Equation forPropagation ofHeat Ina Rod: Formuiaiion ofiheBoundary-Value Probien®ssssnes mtaa 5.HeatPropagation inSpace|22010 Bs 6:Solution "oltheFirst” Boundary:Value_ Problem forthe’Heat=Conductivity Equation bytheMethod ofFinite Differences... .=»829 10 Contents 1.Propagation ofHeat inan Unbounded Rod... gt 8Problems “ThatReduce fo.Investigating. Siuiions ofiheLaplaceEquation. Stating Boundary-Value Problems esos os886 9.The Laplace Equation inCylindrical Coordinates.” Solution ‘ofihe Dirichlet Probiem for aRing with Constant. Values of the DesiredFunctionontheInnerandOuterCircumlerences ssssssSAL 10.TheSolutionofDirichlet's Problemforacircle©.||”.*1|g43 11;Solution oftheDirichlet Problem bytheMethod ofFinile Differences 847 ExercisesonChapterXVIvovvew ee eeBO (Chapter XIX. OPERATIONAL CALCULUS AND CERTAIN OFITS APPLICATIONS 1.The Initial Function and Its Transform... oe es BH 2Transforms ofthe Functions.og (0), sin, cosf°0272? 855 8.The Transform ofaFunction with Changed Scaie ofthe IndependentVariable.Transforms oftheFunctionssinaf,cosafowen=.856 4,TheLinearityPropertyof@Transform pbogoooagoat 4 5.The Shift Theorem... 5 Polls8 6TransformsoftheFunctions¢~,sinhat,coshat,¢°*!sinat,e~*!cosat858 7.Differentiation ofTransforms ‘weeveeeneeeeteee=880 8The TranslormsofDerivatives 2002002202200 012aor 9.TableofTransforms ee TILL ase 10:AnAuxiliaryEquationforaGiven’Differential’ Equation |!||g64 1,Decomposition Theorem i gin“sigSystemsofOH 12Examples ofSolutions otDiferentia’ Equations “and.Systems ofDifferential EquationsbytheOperational Method... +..869 13. The Convolution Theorem we ee a 14.TheDiferential Equations afMechanical Oavttatins’ ine‘Ditferen:lialEquations of“Electrie-Cireult Theory Sinus oidbibiot .) 15,Solution oftheDifferential Oscillation Equation ©1211.1! |am 16.Investigating FreeOscillations eeneneennes IT;Investigating Mechanical and Electrical Oscilations iniheCase’ of@ Periodic External Force owe ny nee 816 18.Solving theOscillation Equation’ iniheCase’ of’Resonance << !|87819.TheDelayTheorem Seennne 89ExercisesonChapterXIXooeecence cen esB80Subject Index PREFACE, This text isdesigned asacourse ofmathematics for higher technical schools. Itcontains many worked examples that illustrate the theoretical material and serve asmodels forsolving problems. ‘The first two chapters “Number. Variable. Function” and “Limit. Conti« nuity ofaFunction” have been made asshort aspossible. Some oftheques- tions that areusually discussed inthese chapters have been put inthethird and subsequent chapters without loss ofcontinuity. This has made itpossible totake upvery early thebasic concept ofdifferential calculus—the deriva. tive—which isrequired inthe study oftechnical subjects. Experience has shown this arrangement ofthe material tobethe best and most convenient for the student Alarge number ofproblems have been included, many ofwhich illust- rate the interrelationships ofmathematics and other disciplines. The problems arespecially selected (and insufficient number) foreach section ofthecourse thushelpingthestudenttomasterthetheoretical material. Toalargeextent, this makes theuseofaseparate book ofproblems unnecessary and extends the usefulness ofthis text as acourse of mathematics for self-instruction. N.S. Piskunoo CHAPTER I NUMBER. VARIABLE. FUNCTION SEC, 1,REAL NUMBERS. REAL NUMBERS AS POINTS ON A NUMBER SCALE ~Number isone ofthebasic concepts ofmathematics. Itoriginated inancient times and has undergone expansion and generalisation over the centuries. ‘Whole numbers andfractions, both positive andnegative, together with thenumber zero arecalled rafional numbers. Every rational number mayberepresented intheformofaratio,2,oftwo integers pand q;forexample, 5 5 5, 125=4. Inparticular, the integer pmay beregarded asaratio ofthe integers+;forexample, 6° 6-8, 0-9. Rational numbers may berepresented intheform ofperiodic terminating ornonterminating fractions. Numbers represented by nonterminating, but nonperiodic, decimal fractions are called irrational numbers; such arethenumbers V2, V3, 5—V2, etc. The collection ofallrational and irrational numbers makes up thesetofrealnumbers. The realnumbers areordered inmagnitude; that istosay, foreach pair ofreal numbers xandythereisone, and only one, ofthefollowing relations: x<y, x=y *>y. Real numbers may bedepicted aspoints onanumber scale, Anumber scale isaninfinite straight line onwhich arechosen: 1)acertain point Ocalled theorigin, 2)apositive direction indicated byanarrow, and 3)asuitable unit oflength. Weshall’ usually make the number scale horizontal and take the positive direction tobefrom leittoright Ifthenumber x,ispositive, itisdepicted asapoint M,at adistance OM, =x, totheright oftheorigin Q;ifthenumber x, isnegative, it'is represented byapoint M,totheleftofOata “ Number. Variable. Function distance OM,=—x, (Fig. 1).The point Orepresents thenumber zero. Itisobvious that every real number isrepresented bya definite point onthenumber scale. Two different real numbers are represented bydifferent points onthenumber scale. ‘The following assertion isalso true: each point onthenumber scale represents only one real number (rational orirrational), Tosummarise, allreal numbers and allpoints onthenumber scale are inone-to-one correspondence: toeach number there cor- responds only one point, and conversely, toeach point there cor- responds only onenumber. This frequentiy enables ustoregard “the number x"and “the point x”as, inacertain sense, equivalent expressions. Weshallmakewideuse Me 2 My ofthiscircumstance inourcourse.STs" Westate without proof thefollow- Fig.t ingimportant property oftheset; ofreal numbers: both rational and irrational numbers maybefound betweenanytwoarbitrary realnumbers. Ingeometrical terms, thisproposition readsthus:bothrationalandirrational pointsmay befound between any two arbitrary points onthenumber scale. Inconclusion wegive thefollowing theorem, which, inacertain sense, represents abridge between theory and ‘practice. Theorem. Every irrational number amay beexpressed, toany degree ofprecision, with theaid ofrational numbers. Indeed, letthe irrational number a>0 and letitberequired toevaluate@withanaccuracyof$(torexample,#5»qpg»andso forth). Nomatter what ais,itlies between two integral numbers N and N+1. We divide the segment between Nand N-+1 into n parts; then awill liesomewhere between therational numbers N+ andN+"#!, Sincetheirdifference isequalto+,each ofthem expresses atothegiven degree ofaccuracy, theformer being smaller and thelatter greater. Example.TheirrationalnumberV7isexpressed byrationalnumbers:Wedad15,t0onedecimal lace, 141"andi:42 totwo decimal places, U4l4'and 1-415 tothree decimal places, etc. SEC, 2,THE ABSOLUTE VALUE OF AREAL NUMBER Let usintroduce aconcept which weshall need later on: the absolute value ofareal number. The Absolute Value ofaReal Number 15 Definition. The absolute value (or modulus) ofareal number « (written |x|) isanonnegative real number that satisfies thecon- ditions Jzl=x ite 0; [x|=—x ifx<0, Examples. |2/=2; |—5|=5; |0|=0. From thedefinition itfollows’ that therelationship x<|x| holds forany x. Let _usexamine some oftheproperties ofabsolute values. 1.The absolute value ofanalgebraic sum ofseveral realnumbers isnogreater than the sum ofthe absolute values oftheterms lxt+yls}x|+lyl- Proof. Let x-+y>0, then lxt+yl|=x+y<]x]+/y] (since x<|x] and y<ly)). Let x+y<0, then let+yl=—@+9)=(—9)+(—9) </¥1+19) This completes the proof. The foregoing proof isreadily extended toany number ofterms. Examples. |—243] <|—2/413]=243—5 of1<5i |=3-5 |=|—3|+1—51=345=8 or88. 2.The absolute value ofadifference isno less than the difference oftheabsolute values ofthe minuend and subtrahend: le—yl>le1—lyl Proof. Let x—y=z, then x=y+2z and from what has been proved lel=ly+2]<[yl +lzl=lyl+le—yh, whence IeI—lyl<ix—yl, thus completing the proof. 3.The absolute value ofaproduct isequal totheproduct of theabsolute values ofthe factors: lxyz|=12] |!lek 4.The absolute value ofaquotient isequal tothe quotient oftheabsolute values ofthedividend and thedivisor: lE|-4ty yl” The latter two properties follow directly from thedefinition of absolute value, 6 Number. Variable. Function SEC. 3,VARIABLES AND CONSTANTS The numerical values ofsuch physical quantities astime, length, area, volume, mass, velocity, pressure, temperature, etc., aredeter- mined bymeasurement. Mathematics deals with quantities divested ofany specific content. From now on,when speakingofquantities, ‘weshall have inview their numerical values. Invarious phenomena, thenumerical values ofcertain quantities vary, while thenumerical values ofothers remain fixed. For instance, inuniform motion ofapoint,timeanddistancechange,whilethevelocityremainsconstant.Avariable isaquantity thattakesonvariousnumerical values. Aconstant isaquantity whose numerical values remain fixed. We shall use the letters x,y,2,u,...,et., todesignate variables, and the letters a,6,c,...,etc., todesignate constants. Note. Inmathematics, aconstant isfrequently regarded asa special case ofvariable whose numerical values are thesame. Itshould benoted that when considering specific physical pheno- mena itmay happen that one and thesame quantity inonepheno- menon isaconstant while inanother itisavariable. Forexample, thevelocity ofuniform motion isaconstant, while thevelocity of uniformly accelerated motion isavariable.” Quantities that have the some value under allcircumstances are called absolute constants. Forexample, theratio ofthecircumference ofacircle toitsdia- meter isanabsolute constant: «=3.14159. Asweshall seethroughout this course, theconcept ofavariable quantity isthe basic concept ofdifferential and integral calculus. In*Dialectics ofNature”, Friedrich Engels wrote: “The turning point inmathematics was Descartes’ variable magnitude. With that came motion and hence dialectics inmathematics, and at once, too, ofnecessity thedifferential and integral calculus.” SEC, 4,THE RANGE OF AVARIABLE Avariable takes on aseries ofnumerical values. The collection ofthese values may differ depending onthecharacter oftheprob- lem. For example, thetemperature ofwater heated under ordinary conditions will vary from room temperature (15-18°C) tothe boiling point, 100°C. The variable quantity x—cosa can take on all_values from—1 to-+1. The values ofavariable aregeometrically depicted aspoints on anumber scale. For instance, the values ofthe variable x=cosa forallpossible values ofaaredepicted asthesetofpoints ofan interval onthe number scale, from —1 to|,including thepoints =I and |(Fig. 2). The Range ofaVariable "7 Definition. The set ofallnumerical values ofavariablequantity iscalled therange ofthevariable. We shall now define the following ranges ofavariable that will befrequently used later on. ‘Anopen interval isthe collection of allnumbers xlying between andexcluding thegiven numbers aand6(a<6); it 4 isdenoted (a,6)orbymeans ofthe = |inequalities a<x<b. U ape ‘Aclosed interval is the set of all numbers xlying between and including the two given numbers aand 6;itis = denoted [a,6]or,bymeans ofinequali- Fig.2 ties, ax. Ifone ofthe numbers aor6(say, a)belongs tothe interval, while theother does not, wehave apartly closed interval, which may begiven bythe inequalities a<x<b and isdenoted [a,6).Ifthe number 6belongs tothesetand a does not, wehave ‘thepartly closed interval (a,6,which may be given by’the inequalities a<x<b. Ifthevariable xassumes allpossible values greater than a,such aninterval isdenoted (a,co) and isrepresented bytheconditional inequalities acrco, Inthesamewayweregardtheinfiniteintervals andpartlyclosedinfi-nite intervals represented bytheconditional inequalities a<x< 00;<x <j;00<1GOK <0. Example. Therangeofthevariable x<copa forallpossible values of istheinterval{—1,1]andisdefined‘bytheinequalities —I-<ix< | The foregoing definitions may beformulated fora“point” in place ofa“number”, An interval isthesetofallpointsxlyingbetweenthegivenpoints aand 6(the end points) and iscalled closed oropen accordingly asitdoes ordoes not include itsend points. The neighbourhood ofagiven point x,isanarbitrary interval (a,6)containing this point within it;that is,theinterval (a,6) whose end points satisfy the condition a<x,<6, One often 18 Number. Variable. Function considers theneighbourhood (a,6) ht Et ofthepointx,forwhichx,isthe e€ midpoint. Then x,iscalled the Fig.3. centre ofthe neighbourhood and thequantity 2%, theradius of theneighbourhood. Fig. 3shows theneighbourhood (x,—e, x,-+#) ofthe point x,with radius e. SEC. 5.ORDERED VARIABLES. INCREASING AND DECREASING VARIABLES. BOUNDED VARIABLES We shall say that the variable xisanordered variable quantity ifitsrange isknown and ifabout each ofany two ofitsvalues itmay besaid which value isthe precedingoneandwhichisthe following one. Here, the notions “preceding” and “following” are notconnected with time, butserve asaway to“order” thevalues ofthevariable, i.e.,toestablish theorder oftherespective values of the variable. Definition 1.Avariable iscalled increasing ifeach subsequent value ofitisgreater than thepreceding value. Avariable iscalled decreasing ifeach subsequent value isless than thepreceding value. Increasing variable quantities and decreasing variable quantities arecalled monotonically varying variables orsimply monotonic quantities. 7 Example. When the number ofsides ofategular polygon inscribed ina circle1doubled, theareas'ofthepolygon isanincreasing variable, The trea of2regular polygon citeumscribed about acircle, when the number of Sides isdoubled, fsadecreasing variable. Itmay benoted that not every Variable quantity isnecessarily” increasing ordecreasing. ‘Thus, ifais.an increasing! variable ‘over‘theinterval (0,En),thevariable z—slaa &nota monotonie quantity. Itfrst increases from 0'to1,then decreases from 1to eiPand then increases from —~1 to0. Definition 2.The variable xiscalled bounded ifthere exists a constant M>0 such that allsubsequent values ofthevariable, after acertain one, satisfy thecondition —M<x<M, thatis,|x|)<M. Inother words, avariable iscalled bounded ifitispossible to indicate aninterval [—M, MJsuch that allsubsequent. values of thevariable, afteracerfain one,willbelong tothisinterval However, oneshould notthink that the variable will necessarily assume allvalues ofthe interval [—M, Ml]. For example, the variable that assumes all possible rational values onthe interval [—2, 2]isbounded, and nevertheless itdoes notassume allvalues on[—2, 2},namely, the irrational values, Function 19 SEC, 6, FUNCTION Inthestudy ofnatural phenomena and thesolution oftechnical and mathematical problems, one finds itnecessary toconsider the variation ofone quantity asdependent onthevariation ofanother. For instance, instudies ofmotion, the path traversed isregarded asavariable which varies with time. Here, the path traversed is afunction ofthe time. Let usconsider another example. We know that thearea ofa circle, interms ofthe radius, isQ—AR*. Iftheradius Rtakes ona’variety ofnumerical values, the area Qwill also assume various numerical values. Thus, thevariation ofone variable brings about avariation inthe other. Here, thearea ofacircle Qisa function ofthe radius R.Let usformulate adefinition ofthe con- cept “function”. Definition 1.Iftoeach value ofthe variable x(within acertain +range) there corresponds one definite value ofanother variable y, then yisafunction ofxof,infunctional notation, y=f(x), y=@(x), and so forth. The variable xiscalled the independent variable orargument. The relation between the variables xand yiscalled afunctional relation. The letter finthefunctional notation y=/(x) indicates that some kind ofoperations must beperformed onthe value of xinorder toobtain the value ofy.Inplace ofthe notation y=[(x), u=@(), ete., one occasionally finds y=y(x), u=u(x), etc., theletters y,udesignating both thedependent variable and the symbol ofthetotality ofoperations tobeperformed onx. The notation y=C, where Cisaconstant, denotes afunction whose value forany ‘value ofxisthe same and isequal toC. Definition 2,The set ofvalues ofxfor which the values ofthe function yaredetermined byvirtueoftherulef(x)iscalledthedomainCldefinition ofthejunctionExample1.Thefunctiony=sinxisdefinedforallvaluesofx.Therefore, its domain ofdefinition isthe infinite interval —oo<x< o. Note 1.Ifwehave afunctional relation oftwo variable quan- tities xand y=f(x) and ifxand y=f(x) areregarded asordered variables, then ofthe two values ofthefunctiony*=/(x*)and y**=f(x**)corresponding totwovaluesoftheargument’x*and xt, thesubsequent value ofthe function will bethat one which corresponds tothesubsequent value oftheargument. Thefollowing definition is,therefore, natural. Definition’ 3.Ifthe function y=f(x) issuch that toagreater value oftheargument, xtherecorresponds agreatervalueofthe 20 Number.Variable. Function function, then thefunction y=f(x) iscalled increasing. Adecreas- ing function issimilarly defined. Example 2.The function QAR forO-<R<aisanIncreasingfuntion because toa greater value ofRthere corresponds agreater value ofQ. Note 2.The definition of function issometimes broadened so that toeach value ofx,within acertain range, there corresponds notone but several values ofyoreven aninfinitude ofvalues ofy.Inthis case wehave amultiple-valued function incontrast totheone defined above, which iscalled asingle-valued function. Henceforward, when speaking ofafunction, weshall have inview only single-valued functions. Ifitbecomes necessary todeal with multiple-valued functions weshall specify this fact. SEC. 7,WAYS OF REPRESENTING FUNCTIONS I.Tabular representation ofafunction Here, the values oftheargument x,,x,,...,%, and thecor- responding values ofthe function y,,"yy. s¥, are written out inadefinite order. Examples are tables of trigonometric functions, tables of logarithms, and soon. ‘Anexperimental study ofphenomena can result intables that express afunctional relation between themeasured quantities. For example, temperature measurements oftheairatameteorological station onadefinite day yield atable like the following. Thetemperature T(indegrees)isdependent onthetime 1(inhours). ‘Lofaftele] sede els r|of- =[=|-as[«Jsfos[ This table defines Tasafunction of¢. Ways ofRepresenting Functions 2 Il.Graphical representation ofafunction Ifinarectangular coordinate system onaplanewehaveaset ofpointsM(x,4),andnotwo)pointslieonastraight lineparallel tothey-axis,thissetofpointsdefines Fi acertain single-valued function y= yf =F(x); the abscissas ofthe points arethe values ofthe argument, the corresponding ordinates arethevalues y ofthefunction (Fig. 4). The collection ofpoints inthe q * * ay-plane whose abscissas are the Fig.4values ofthe independent variable and whose ordinates arethe corresponding values ofthefunction iscalled agraph ofthegiven function. IIL. Analytical representation ofafunction Let usfirst explain what “analytical expression” means. Byana- lytical expression wewill understand aseries ofsymbols denoting atotality ofknown mathematical operations that areperformed in adefinite sequence onnumbers and letters which designate constant orvariable quantities. Bytotality ofknown mathematical operations wemean notonly themathematical operations familiar from thecourse ofsecondary school (addition, subtraction, extraction ofroots, etc.) but also those which will bedefined asweproceed inthis course. The following areexamples ofanalytical expressions: sg, logemsine. ox ETH.xa; REET, 2xVira, ele. Ifthe functional relation y=f(x) issuch that fdenotes an analytical expression, wesaythat thefunction yofxisrepresented analytically. Examples offunctions represented analytically are: 1)y=x*—2; 2)y= 3)y=VI—x% 4)y=sinx; 5)Q=aR?, andsoforth. Here, thefunctions arerepresented analytically bymeans ofa single formula (aformula isunderstood tobetheequality oftwo analytical expressions). Insuch cases onemay speak ofthenatural domain ofdefinition ofthe function. The setofvalues ofxforwhich theanalytical expression on theright-hand side hasafully definite value isthenatural domain 2 Number. Variable. Function ‘ofdefinition ofafunction represented analytically. Thus, thenatu- ral domain ofdefinition ofthe function y=x'—2 isthe infiniteinterval —co<x<oo, because thefunction isdefinedforallvalues ofx.Thefunction y= isdefined forallvalues ofx, with thtexception ofx=1, because forthis value ofxthe deno- minator vanishes. Forthefunction y=V/1—*, thenaturaldomain _ofdefinition istheclosed interval—I<x<l, Yhyx and soon. Note. Itissometimes necessary toconsider only apart ofthenatural domain ofafunction, | and not the whole domain. For instance, the dependence oftheareaQofacircleuponthe radius Risdefined bythe function Q=aR* The domain ofthis function, when considering agiven geometrical problem, isthe infinite .<interval O0<R<-+ oo.Butthenatural domain Uof this function isthe infinite interval—co< Fig. 5. <R<to. Ifthe function y=f(x) isrepresented analy- tically, itmay beshown graphically onacoordinate xy-plane. Thus, thegraph ofthe function y=x* isaparabola asshown in Fig. 5 SEC. 8,BASIC ELEMENTARY FUNCTIONS. ELEMENTARY FUNCTIONS The basic elementary functions are the following analytically represented functions. 1.Power function: y=x*, where aisareal number. *) Il.Exponential function: y=a*, where aisapositive number notequal tounity. IIL. Logarithmic function: y=log, x,where thebase oflogarithms aisapositive number notequal tounity. IV. Trigonometric functions: y=sinx, y=cosx, y=tanx, y=cotx, y=secx, y=cscx. V.Inverse trigonometric functions: y=aresinx, y=arccosx, y=arctanx, y=arceotx, y=arcsecx, y=arcescx. Let usconsider thedomains ofdefinition and thegraphs ofthe basic elementary functions. *)Iaisirrational, thisfunctionIsevaluated bytakinglogarithms andantilogarithms: logy=alogx.Itisassumedthalx>, Basic Elementary Functions, Elementary Functions 23 Power function y=x". 1.@isapositive integer. The function isdefined inthe infi- nite interval —co<x<-+ co,Inthis case, thegraphs ofthe func- tion for certain values ofahave the form shown inFigs. 6 and 7. b yyet Shoes 7 x Fig. 6. Fig. 7. 2.@isanegative integer. Inthis case, thefunction isdefined forallvalues ‘ofxwith theexception ofx=0. The graphs ofthe functions for certain values ofa have theform shown inFigs. 8 v and 9, . oha \s Fig. &. Fig. 9. Figs. 10,11,and 12show graphs ofapower function with fractional rational values ofa, 0y 4 = oe Shs ae x an roar Fig.10. Fig.11. Fig.12. Py Number.Variable. Function Exponential function, y=a*, a>0 anda+1. This function is defined forallvalues ofx.ItsgraphisshowninFigs.13and14. y BHOLPose oh" yet 3 2 } 2 x QF 7 2-* Fig, 13. Fig. 14. Logarithmic function, y=log,x,a>0anda+1.Thisfunction isdefined forx>0. Itsgraph’is shown inFig. 15, Trigonometric functions. Inthe formulas y—sinx, etc., the independent variable xisexpressed inradians. Alltheenumerated 7 trigonometric functions areperiodic.Let usgive ageneral definition ofa periodicfunction. y=l0gax Definition 1.The function y=/(x) iscalled periodic ifthere exists acon- stant C,which, when added to(orsub- *tracted from) the argument x,does not change. the value of the” function: F(x-+C)=f(x).Theleastsuchnumber iscalled the period ofthe function; it Fig. 16 will henceforward bedesignated as'2/. Fromthedefinition itfollows directly that y=sinx isaperiodic function with aperiod 2n: sinx— =sin (x-+2n). The period ofcosx islikewise 2a, The functions y=tanx and y=cotx have aperiod equal tox. The functions y=sinx, y=cosx aredefined forallvalues ofx;thefunctions y=tanx andy=secx aredefinedeverywhere except thepoints x=(2e+1)$(e=0, 1,2, ...);thefunctions y=cotx and y=cscx are defined forallvalues ofxexcept the points x=ka(k=0,1,2,...). Graphs oftrigonometric functions are shown inFigs.’ 16, 17,18, and 19. The inverse trigonometric functions will bediscussed inmore detail later on, ¥ nnn (Sandie a E (i. ix* Fig. 16. w cos De [a ox* Pig. 7 1 ee a hag\ in i} ivi, Net \G toWg iF| iINOING |- Tad i xm/e 7 NEY Fig. 18. Fig 1. Letusnowintroduce theconcept ofafunction ofafunction. Ifyisafunction ofu,and u(inturn) isdependent onthevar- iable x,then yisalso dependent onx.Let y=F(u) u=@(x). We getyasafunction ofx y=F[oe(x)}- Thisfunction iscalledafunction ofafunction oracomposite function. Example 1.Lety=sinu, u=x?, The function y=sin(x*) isacomposite Note. The domain ofdefinition ofthefunction y=F[@(x)] is either theentire domain ofthefunction, u=@(x), orthat part 6 Number. Variable. Function ofitinwhich those values ofuaredefined that donotgobeyond thedomain ofthefunction F(u). Example 2,Thedomainofdefinition olthefuntion y=VThy=Vi, ata) isibeclosed interval [1,1], beeause when|x['>lw<Oand,conse! uenty, thefunction Vitenotdefined although tefuntion wana" Uetined torallvalues of=).‘The graph ofthis function isthe upper hail of 2°ehelewithcentre attheorigina ihecordnate system andithradstally The operation “function ofafunction” may beperformed any number oftimes. For instance, the function y=In (sin(x*+-1)] is obtained asaresult ofthe following operations (defining the following functions): Daatl, using, y=Inu, Let usnow define anelementary function. y Definition 2.Anelementary function is afunction which may berepresented by asingle formula ofthe type -y—f(t), where the expression onthe right-hand side ismade upofbasic elementary func- tions and constants bymeans ofafinite of 7 number of operations of addition, Fie2. subtraction, multiplication, division and takingthefunction ofafunction, From the’ definition itfollows that elementary functions are functions represented analytically. Examples ofelementary fonctions maVipmR yaleett7etotane gavetmRalent Veazine andthe Tike. Examples ofnon-elementary functions:Tneunconyset237..cnyer(a)].isnotelementary becausetheumber ofoperations that musi beperionmed toobtain y-ineteases with ny that isto sayr itWsnol bounded 2.The function given inFig.” 20isnot elementary either because itis represented bymeats oftwo formule: fwe=x, if0cxel, Hare-1, eres SEC. 8,ALGEBRAIC FUNCTIONS Algebraic functions include elementary functions ofthe following kind: Algebraic Functions a 1.The rational integral function, orpolynomial y=ax"+ax""4...44,, where a,,a,,...,d, areconstants called coefficients, and nisa nonnegative’ integer called the degree ofthe polynomial. Itis obvious that this function isdefined forallvalues ofx,that is, itisdefined inan infinite interval. Examples: 1.y=ax+6 isalinear Yh as Yh axefunctionWhen”60,telinerfunetion wo weYaar expressesyabeingdirectlypro-feet eaTO 2 ymaxttbxte is a quadraticfunctionTheengiotagstaeaicTune: irfunctions"are.considered*indetailin@ ” analytic geometry. Fig. 21. N.Fractional rational function. This function isdefined asthe ratio oftwo polynomials: atfaye. tnelaee For example, thefollowing isafractional rational function: y=t) ittexpresses inverse variation, Itsgraph isshown inFig. 22.Itis obvious that afractional rational function isdefined forallvalues ‘ofxwiththeexception yy ofthose for which the a0 aco denominator becomes zero. oi Hl.Irrational func- "77 tion, Ifinthe formula y=f(x), operations of addition, subtraction, multiplication, division@ iO) andraising toapowerwith rational non-inte- Fig.22. gral exponents areper- formed ontheright- handside,thefunction9=J)jpcalledirrational, Examples,epee DEVE 1 1 ye AVE, VG; ete. ofirrational functions are:y=EEE yVx;ete. 8 Number. Variable, Function : Note1.Theabove-mentioned threetypesofalgebraic functions donotexhaust allalgebraic functions. Analgebraic function isany function y=f(x) which satisfies anequation oftheform Py(xy +P,(x)y+... +P,(x)=0, a where P,(x), P,(x), «..,P(x) arecertain polynomials inx. Itmay beproved that ‘each oftheenumerated three types of function satisfies acertain equation oftype (1),butnotevery func- tion that satisfies anequation like (I)isafunction ofone of the three types given above. Note 2.Afunction which isnot algebraic iscalled transcendental. Examples oftranscendental functions are: y=cosx;y=10% and the like. SEC. 10. POLAR COORDINATE SYSTEM The position of@point ina plane may bedetermined bymeans ofaso-called polar coordinate system. Wechoose apoint Oinaplane and call itthepole; thehalf- line issuing from this point iscalled thepolar axis. The position ofthepoint Mintheplane may bespecified bytwo numbers: the number @,which expresses the distance ofMfrom the pole, andthenumber @,which istheangleformed /bytheline segment OM and the polar axis. 0 The positive direction oftheangle @isreckoned counterclockwise. The numbers @ 920 and@arecalled thepolar coordinates ofthe ° xpoint M(Fig. 23). Wewillalways consider theradius vector Fig.28. @nonnegative. Ifthepolar angle @istakenwithinthelimits0<@<2z, thentoeach point ofthe plane (with the exception ofthe pole) there corre- sponds adefinite number pair gand y.For thepole, g=0 and @ isarbitrary. Let usnow seehow the polar and rectangular Cartesian coordi- nates arerelated. Let the origin ofthe rectangular coordinate system coincide with thepole, and the positive direction ofthe a-axis, with the polar axis. We establish arelationship between therectangular and polar coordinates ofone and the same point. From Fig. 24itfollows directly that x= csp, y=esing and, conversely, that e=VEFH, tang=4, Note. Tofind@, itisnecessary totake into account thequad- rant inwhich thepoint islocated and then take the correspond- y, eneLZaZ* sj X=QCOSP ingvalue [email protected] equation 9=F(g) inpolar coordinates defines acertain line. Example 1.Equation g=a, where a=const, defines inpolar coordinates ViFoma ortpyma? ED n x 3 3me ielel-lel=[=[ cfoaare[ist]ese|xoine|ware]eeewoes Thecorresponding curveisshowninFig.26.ItiscalledthespiralofArchi-neErample 3. rT ay, [AKY ‘ » Number. Variable. Function angularcoordinates. Substituting a=V"+y?, cosp=——A— intothegi- e ee Bhcos=r, « venequation,wegetVFFFam Vere xttyt—2ar=0. Exercises onChapter 1 1.Giventhefunction/(a)=x*4+6x—4. Verilytheequalities1(1)=3, 10)=23. = 2fla=xt4l. Evaluate: 3)£(4). Ans. 17.b)1(V.Ans.3.©)atl). ansafShakar a)flayleAneerede)fadeUnsack:8Fare fins.4201.@)1Bal.Ansdat.a:: 3.9)=Fe-Writetheexpressions +)andgi.Ans.o(2)=mx, 13m45 (=) “37s! G@)__I-x *4.piy=VaFFE.Writetheexpressions (2x)end(0).Ans.(24)= =2VFTy=25.TeVeritytheequality/29)=Trea . 8en=logFe.Verilytheequality9(a)+90)=—9 (FER). 7.f(x)=logx;@(x)=a*. Writetheexpressions: 8)1p(2)]-Ans.3log2. b)/[@(a)]. Ans. 3loga.c)@{f(a).Ans.{loga}. &.Find thenatural domain ofdefinition ofthe function y=2st+1 Ans. -e<r<te. 8.Findthenatural domains ofdefinition ofthefunctions: =),YT=w.Ans. —leoxc4l. b)VOFR+/7—«. Ans. —3cxa7.9 Vrta— =VIR. Ans,<0<xcto. &SEE. Ans.x0.©)aresints,Ans. —lex. f)y=logs. Ans. x>0. g)y=a*(a>0). Ans. <x<+o.Construct thegraphs ofthefunctions” 1ye$5.Meyeyatthe12yeBHO18.yestOe, MaysLy.15,yensinde,16.ymreosSe.17.yort—ae$6.18.ym x 1 i 19,yasin(242).20,pcos(2—F).21.yotangex.2yoeotje.23,yedF24,yd"25,yologyt 28.yoattle Meyd28,ym Bo yest80.ye Bhgees 82.yar288.yet yeleh 35.yelogy|x|. 98.yorlogsa).97.yadain(2e4-). 98ym Exercises onChapter 1 a =4cos(++). 99.Thefunction/(4)isdefinedontheinterval{—1,1)afollows farits poderse Gala foroeeci 40, The function /(2)isdefined onthe interval (0,2}a8follows: Jayex forocrci: laos forlcre?. Plotthecurvesgivenbythepolarequations: 41.e=-=(hyperbolic spl-ral),42.g=a¥(logarithmic spiral).48.o—aV'GO52H(lemniscate). 4.Q= =a(1—cos q)(cardioid). 45.g=asin3p. CHAPTER I LIMIT. CONTINUITY OF AFUNCTION SEC. 1,THE LIMIT OF AVARIABLE. ‘AN INFINITELY LARGE VARIABLE Inthis section weshall consider ordered variables that vary in aspecial way defined asfollows: “the variable approaches a limit”. Throughout theremainder ofthecourse, the concept of limit ofavariable will play afundamental role, foritisintimate- lybound upwith the basic concepts ofmathematical analysis, such asderivative, integral, etc. Definition 1.Aconstant number aissaid tobe the limit ofa variable x,ifforevery preassigned arbitrarily small positive num- bereitispossible toindicate avalue ofthe variable xsuch that allsubsequent values ofthevariable will satisfy theinequality |x—a]<e. Ifthe number aisthe limit ofthe variable x,one says that x approaches thelimit a;insymbols wehave x—a orlimx=a. Ingeometric terms, limit may bedefined asfollows. a Theconstant number aisthelimitofthevariable xifforany preassigned o ae" arbitrarily small neighbourhood with ea centreinthepointaandwithradius Fig. 28. ethere isavalue ofxsuch that all points corresponding tosubsequent values ‘ofthe variable will bewithin this neighbourhood (Fig. 28). Let us consider several cases ofvariables approaching limits. Example 1.The variable xtakes onsuccessive values: u Le gaged be wat gelgswelgiital bie ‘Weshall prove that this variable hasunity asitslimit. Wehave Foranye,allsubsequent values ofthevariable begin with n,where 4ce,or.n>+willsatisfytheinequality[xy=1|<eandtheproofis complete.- The Limit ofaVariable 33 Itwill benoted here that the variable quantity decreases asit,approaches the limit. Example 2Thevarlable xtakesonsuccessive values: xy=l—ys x= 1 1 to jelsled yelog aetops ostel Hoga This variable has alimit ofunity. Indeed, 1 beaH=|(1-4(0" ge)—![=p For any e,beginning with n,which satisfies the relation 1 Bs from which itfollows that >t, 1 nlog2> log or 1 lose">gt allsubsequent values ofxwill satisly the relation [xt]<e. Itwill benoted here that thevalues ofthevariable are greater than or tessthanthelimit,andthevariable: approaches itslimit“by“oscilating Note 1.Aswas pointed out inSec. 3(see Ch. 1),aconstant quantity’c isfrequently regarded asavariable whose values all coincide: x=c, Obviously, the limit ofaconstant isequal tothe constant itself, since wealways have the inequality |x—c|=|c—c|=0<e forany e. Note 2.From the definition ofalimit itfollows that avari- able cannot have two limits. Indeed, iflimx=a and limz= =(a<6), then xmust satisfy, atone and thesame time, two inequalities: |x—a|<eand|x—b|<e ° foranarbitrarily smalle;butthisisimpossible ife<tt(Fig. 29). 2088 u Limit, Continuity of@Function Note 3,One should not think that every variable has alimit,LettheVariablextakeonthefollowingsuccessive values: 1 1 Lye,nediaelodi ged: weld i 1 1Suslope)fan=BET (Fig. 30), For&sufficiently lange,thevaluex,andallsubsequent values with even labels will differ from unity byassmall a ad ks yeewyyo f ecleg % Fig. 29. Fig. 90 rurmber asweplese, whilethenextvaluege,and.llsubse quent values ofxwith odd labels will differ irom zero byas small anumber asweplease. Consequently, the variable xdoes not approach a.limit. Inthe definition ofalimit itisstated that ifthe variable approaches thelimit a,then @isa constant. But theword “appro- caches” isused also todescribe another type ofvariation ofa variable, aswill beseen from thefollowing definition, Definition 2.Avariable xapproaches infinity ifforevery preassigned positive number Mitispossible toindicate avalue Ofxsuch that, beginning with this value, allsubsequent values ofthevariable will satisly theinequality’ |x|>M. Iithevariable xapproaches infinity, itiscalled aninfinitely large variable and wewrite x0. Example &The variable +takes onthe values ech eB EO es (DA oe “This isaninfinitely large variable quantity, sine for anarbitrary M>0allaluesothe variebler “beginning. wilt8certain "one,"area aboolute Tnagaitude, greater than ‘The variable x“approaches plus infinity", x—--+00, iffor an arbitrary M>0 allsubsequent values ofthe variable, beginning with acertain one, satisfy the inequality M<x. ‘Anexampleofavariablequantityapproaching plusInfinityisthevariablethal fakes onthevalues sys, abs BM oo The Limit ofaFunction 38 Avariable approaches minus infinity, x—-—co, ifforanarbi- trary M>0, allsubsequent values ofthevariable, beginning with acertain one, satisfy the inequality x<—M. For example, avariable xthat assumes the values x=—l, y=—2, soos Agency vees approaches minus infinity. SEC, 2,THE LIMIT OF AFUNCTION Inthis section we shall consider certain cases of the variation ofafunction when the argument xapproaches acertain limit a orinfinity. Definition 1.Letthefunction y=/(x) bedefined inacertain neighbourhood ofthepoint aoratcertain points ofthis neigh- bourhood. The function y=f(x) approaches thelimit b(y—b) asx approaches a(x—-a), itfoteverypositive number e,nomatter how small, itispossible toindicate apositive number 8such that forailx,different from aand satisfying the inequality *) |x—a|<6, wehave theinequality [Fab] <e. If6isthe Limit ofthefunction f(x) asx—+a, wewrite limf(x)=6 v1 y-tto aedbey 6 YY, oFf(x)—+basx—a. og BOY,hapeabsxe, thisisyeZZ. illustrated onthe graph ofthe function y=f(x) a8 follows (Fig. 31). ‘Since from the inequality |x—a|<6 there follows the aeinequality |f(x)—b|<e, this U CORTICES means that for all points x Fig.$1. *)Here. wemean the values ofxthat satisy the inequality|x—a| <6andbelong tothedomain ofdefinition ofthefunction. WeshallEncounter similar elreumstances inthe future. For instance, when considering thebehaviour ofafunction asx", itmay happen that"thefunction'sdefined only forpositive integral values of"x."And sointhis case x. 0 Sssuming only positive integral values. We ‘shall not specify’ this. when i comes uplater on. 2 % Limit, Continuity ofaFunction that are not more distant from the point athan 8,the points Mofthe graph ofthe function y=f(x) liewithin aband of width 2ebounded bythe lines y=b—e and y=b-+e. Note {.Wemay define thelimit ofthefunction f(x) asx—a asfollows. Let variable xassume values such (that is,ordered insuch fashion) that if |—a|>|s"*—al, then x" isthe subsequent value and 2*isthe preceding value; but if |#—a|=|2"—a| andPH, then X* isthe subsequent value and 3*isthe preceding value. Inother words, oftwo points onanumber scale, thesubsequent one isthat which iscloser tothepoint a;atequal distances, the subsequent one isthat which istotheright ofthepoint a. Letavariable quantity xordered inthis fashion approach the limit alx—a orlimz=a). Letusfurther consider thevariable y=/(x). Weshall here and henceforward consider that ofthe two values ofafunction, the subsequent one isthat which corresponds tothe subsequent value oftheargument. If,asx—~a, avariable ythus defined approaches acertain limit 6,we shall write limf(x)=6 and weshall say that thefunction y=f(s) approaches the limitfi basx—a.yo Itiseasy toprove that bothdefinitions ofthe limit of afunction are equivalent. Note 2.Iff(x) approaches thelimit b,asxapproaches acertain number bea,sothat xtakes ononly values less | than a,wewrite lim f(x)=, and call5,thelimitofthefunction f(x) 7a % ontheleft ofthepoint a,Ifxtakes on_onlyvalues greater than a,we eewritelimf()=6, andcall6.the limitofthefunction onthe“gtofthe,pointa(Fis.32).Tt-can beproved that ifthelimit ontheight and thelimit on theleftexist and areequal, that is,6,=0,—0, then 6will be The Limit ofaFunction 37 the limit inthesense ofthe foregoing definition ofalimit atthe point a.And conversely, ifthere exists alimit 6ofafunction at thepoint c,then there exist limits ofthefunction atthepoint a both ontheright and ontheleftand they areequal Example 1.Let usprove that lim(3x+1)=7. Indeed, let anarbitrary 0begiven;forthe inequality|(3r-+1)—7| <etobefulfilleditis. arytehvetheTolowinginequalities futaniegs see RS ee ‘ |8x—6|<e, |x—2| <v7T <1-2 <z: Thus,givenanye,forallvaluesofxsatisfying theInequality |x—2|<3 =6,the value ofthe function Gx-+1 will differ from 7byless than e.And ihis' means that 7isthe limit ofthe function asx—+2. Note 3,For afunction tohave alimit asx—+a, itisnot_ne> cessary that thefunction bedefined atthepoint xa. When find- ing the limit we consider the values ofthe function inthe neighbourhood ofthe point athat are different from a;this is clearly illustrated inthefollowing case. Example2.Weshallprovethattin=}=4,Here,thefunction =4 isnot defined for x=2. Itisnecessary toprove that foranarbitrary e,there will bea&such that theTollowing inequality will befulfilled: $4[ES <e o W[x—2 <0, But when x#2 inequality (1)isequivalent totheinequality e=2)+2)|SPS? atetenace lx-21<e. Cy Thus, foranarbitrary e,inequality (1)will befulfilled ifinequality (2) isfulfilled (here, 5=e), which means that the given function has the number 4-as its limit as'x—»2. Let us now consider certain cases of variation ofafunction as x—> 00. Definition 2.The function f(x) approaches thelimit 6af&— co ifforeach arbitrarily small positive number ¢itispossible to indicate apositive number Nsuch that for allvalues ofxthat satisfy the inequality |x|>WN the inequality |f(x)—b|<e will be fulfilled. 8 Limit. Continuity ofaFunction Example 8.Toprove that tim®)=I ua(144)=1, 1isnecessary to prove that, foranarbitrary e,the following inequalitwillbetuinited * Ila) provided |x|> ,where Nisdetermined bythechoice ofe.Inequality (3) isequivalent tothefollowing inequality: |El<« whichwillbefulfilled ifi>den, Andtismenetattig(142)=timEtat (ig.29, y ~-------| sa SS aece x Fig. 33 Knowing the meanings of the symbols + co and x—+—a the meaningofthefollewing expressions areobvious ands canes “Hay approaches’ bas4 fo" and +fO5 approaches 8asxe", or, insymbols, pe edges lim feb. SEC. 3,AFUNCTION THAT APPROACHES INFINITY. BOUNDED FUNCTIONS Wehave considered cases when the function f(x) approaches a certain limit 6asx» aor asx 00. Letusnow take thecase when thefunction y=f(x) approaches infinity when theargument varies insome way. AFunction that Approaches Infinity. Bounded Functions 39 Definition 1,Thefunctionf(x)approaches infinityasx~a,ie., itisan infinitely large quantity asxa, ifforeach positive number M, nomatter how large, itispossible tofind a6>0 such that forallvalues ofxdifferent from aand satisfying the condition [x—a|<8, wehave the inequality |f(x)|>M. Iff(x) approaches infinity asxa,wewrite limf(2)=00 orf(x)+00asxa. IF[(@)approaches infinityasx—aand,intheprocess, assumes onlypositive oronlynegative values, the’appropriate notation is Timf(x)=-+90oflimf{(x)=—o0. Example WestallpvethatttyaInesforny M>0 we will have 1 fiestemM, provided (2<p,Nal<Gao. m va Thetution gyasumes onypestve vasa.8 trample 2Wesatpovhattn(—)meIndeed, trap M>0 we will have provided Irlete=o cobeat, tse(—1)30or20and(4)<0or130,8. UEthefunction 1(x)approaches infinity asx» oo,wewrite Um /(x)=, and wemay have the particular cases: .Um1)=0, lim[(x)=, limf(x)=—. Forexample,Im#=4+e, |lintoe 0 Limit. Continuity of@Function Note 1.The function y=f(x) asxa orasx-+ comay not approach afinite limit orinfinity. 4 y 7 /" a*Hl mfI a3 a Fig. 34, Fig. 35: Example 3. The function y=sinx defined onthe infinite interval move tm, asz+t2,doesnotapproacheithera”finitelimitor infinity (Fig. 36) A yas bi io bg— Fig. 96. Example 4.Thefunction y=sin+definedforallvaluesofx,except £=0,doesnotapproach citherafinitelimitorinfinity asx»0.The Graph ofthis function isshown inFig. 37. y wannannapagthpget iT fc ¥ <M. Fig. 97. Definition 2.’The function y—f(x) iscalled bounded inagiven range ofthe argument xifthere exists apositive number M_such that forall values ofxinthe range under consideration the ‘AFunction that Approaches Infinity. Bounded Functions a inequality |f(x)|< Mwill befulfilled. Ifthere isnosuch num- berM,the function f(x) iscalled unbounded inthe given range. Example 5.The function y=sinx, defined in the infinite interval ect <+e, isbounded, since forall values of Isinxl<t=M. Definition 3.The function f(x) iscalled bounded asx—-a if there exists aneighbourhood with centre atthe point a,inwhich thegiven function isbounded. Definition 4.The function y—f(x) iscalled bounded asx—-co ifthere exists anumber N>O such that for all values ofx satisfying the inequality |x|>N, thefunction f(x) isbounded Theboundedness ofafunction approaching alimitisdecided bythefollowing theorem. Theorem 1.Iflim f(x)=b, where &isafinite number, the Junction f(x)isbounded asx—+a. Proof. From theequality lim f(x)—6 itfollows that forany e>0 there willbea6such that intheneighbourhood a—6<x<a+6 theinequality IFe)—bl<e or IF)|<lb|+e will befulfilled, which means that thefunction f(x) Isbounded as x—-a, Note 2.From the definition ofabounded function f(x) it follows that if lim f(x)=00 or limf(x) 00, that is,iff(&) isan infinitely large function, itisunbounded. The converse isnot true: anunbounded function may not be infinitely large. For example, the function y=xsinx asx—soo isunbounded because, for any M>0, values ofxcan befound such that |xsinx|>M. But the function y=xsinx isnot infinitely large because itbecomes zero when x=0, x,2m,... The graph ofthe unetion y=xsinx isshown inFig. 38. Theorem 2.Iflimf(x)=0 40,thenthefunction y=zr> isa bounded function” asx—a. Proof. From the statement ofthe theorem itfollows that for an arbitrary e>0 inacertain neighbourhood ofthepoint x=a we 2 Limit. Continutty of@Function willhave|f(xJ—6|<e, oF[If(x)|—[bll<e, of—e<|/()|——lb|<e, or[b|—e</f(x)|<[b|+e. 4 fi t{yaxsink|} HNNoe AE \! '' Hte\aePAA se]te 1--- Wb Fig. 38. From thelatter inequality itfollows that 1 1 1. o=e> Ten >ere Forexample, taking e=75161, weget 10 1 0 Tei> Weal>Wey? whichmeansthatthefunction rt;isbounded. SEC. 4,INFINITESIMALS AND THEIR BASIC PROPERTIES Inthis seétion weshall consider functions approaching zero as the argument varies inacertain manner. Definition. The function a=a(x) iscalled infinitesimal asx—+a orasx—roo iflim a(x)=0 or lim a(x)=0. From the definition ofalimit itfollows that if,forexample, lima(x)=0, this means that forany preassigned arbitrarily small positive etherewillbea8>0 suchthatforallxsatisfying thecondition |x—a|<6, the condition |a(x)|<e will besatisfied. Example 1.The function a=(r—1}* isaninfinitesimal asx—p1because ia=" lim(e—I'=0 (Fig. 39). Infinitesimals and Their Basic Properties 43 Example 2.Thefunction a=isaninfinitesimal asxr@(Pig.40) (see Example 3,See. 2) 4 iy ee ee | W x Fig. 38. Fig. 40. Let usestablish arelationship that will beimportant later on. Theorem 1.Ifthefunction y=f(x) isinthe form ofasum of aconstant band aninfinitesimal a: y=b+a, @ then limy=6 (asx—+a orx—+00). Conversely, iflimy=6, wemay write y=b-+a, where aisan infinitesimal. Proof. From equality (1)itfollows that |y—6|=Ja|. But for anarbitrary e,all values ofa,from acertain value onwards, satisly the relationship |a|<e; consequently, the inequality |y—6|<e will befulfilled for‘allvalues ofyfrom acertain value onwards. And this means that limy=6, Conversely: iflimy=6, then given anarbitrary e,for all values ofy,from acertain value onwards, wewill have |y—|<e. But ifwe denote y—b=a, then itfollows that for allvalues ofa,from acertain one onwards, wewill have |a|<e; and this means that @isan infinitesimal. Example 3.Let afunction begiven (Fig. 41) gattt, then timy=1, and, conversely, if um y=1 “ Limtt, Continuity of@Function . thevariable ymayberepresented inthe 7form of asum of the limit 1and the ontit infinitesimal a=;thatis(Fig.41), alte 1 a] Theorem 2.Ifa=a(x) approaches+ zero as x—~a (or asx—+oo) and doesnotbecome zero,thengat 7 "%approaches infinity. big.at Proof. ForanyM>0, nomattera howlarge,theinequality ;2)>M will befulfilled provided theinequality |a|<-j7 isfulfilled. Thelatter inequality will befulfilled forallvalues ofa,from acertain one onwards, since a(x)—0. Theorem 3.The algebraic sum oftwo, three and, ingeneral, a definite number ofinfinitesimals isaninfinitesimal function. Proof. We shall prove the theorem fortwo terms, since the proof issimilar forany number ofterms. Let u(x)=a(x)+B(x), where lima(x)=0, limB(x)=0. We shallprovethatforany¢>0,nomatterhow’small,therewill bea8>0 such that when the inequality |x—a|<6 issatisfied, theinequality Mens willbefulfilled. Sincea(x)isaninfinites- imal, a6will befound such that ina neighbourhood with centre atthepoint aand radius 6,,wewill have lewl<z- SinceB(x)isaninfinitesimal, wewillhave[B(x)|<3 inthe neighbourhood ofthepoint awith radius 6. Letustake 5equal tothesmaller ofthetwo quantities 6,and 8,thentheinequalities Ja|<- and|p|<4willbefulfilledin theneighbourhood ofthepoint awith radius 3.Hence, inthis neighbourhood wewill have le]=la)+8|SJ) +/BI<F+55% and so|u|<e, asrequired. The proof is’similar forthecase when lima(x)=0, limB(x) =0. Basic Theorems on Limits 45 Note. Later on we shall have toconsider sums ofinfinitesimals such thatthe number ofterms increases with adecrease ineach term. Inthis case, thetheorem may nothold. Totake anexample, consider u=L4t4...4-+ where xtakesononlypositive Tie integral values (x=1, 2,3,..., 1,...). Itisobvious that as x—rco each term isaninfinitesimal, but the sum «=I isnot an infinitesimal. Theorem 4.The product ofthe function ofan infinitesimala=a(x) byafunction bounded byz=z(a), asx—+a(orx00)isaninfinitesimal quantity (function). Proof. Let usprove the theorem’ for the case x—+a. For a certain M>0 there will beaneighbourhood ofthepoint x=a inwhich ‘the inequality |z|<M will besatisfied. For any e>0 there willbeaneighbourhood inwhich theinequality |a|<-f will befulfilled. The following inequality will befulfilled inthe least ofthese two neighbourhoods: laz|<apM=e which means that azisaninfinitesimal. The proof issimilar for the case x—+oo. Two corollaries follow from this theorem. Corollary 1.Tflimas0, limB=0, thenlimap-—0 because B(x) isabounded quantity. This holds forany finite number offactors. Corollary 2.Iflima=0 and c=const, then limoa=0. Theorem 5.Thequotient aobtained bydividing theinfini- tesimal a(x) byafunction whose limit differs from zero isan infinitesimal. .Proof. Letlima(x)=0, lim2(x)=6 #0.ByTheorem 2,Sec.3,itfollowsthatzaisabounded quantity. Forthisreason,thefractionsHB=0) 75areaproductofaninfinitesimal byabounded quantity, that is,aninfinitesimal. SEC. 5,BASIC THEOREMS ON LIMITS Inthis section, asinthe preceding one, we shall consider sets offunctions that depend onthesame argument x,where x—+a or x—+00, We shall carry out. the proof forone ofthese cases, since the other isproved analogously. Sometimes we will not ‘even write x—a orx—+0o, butwill take them forgranted, 6 Limit. Continuity ofaFunction Theorem 1.The limit ofanalgebraic sum oftwo, three and, ingeneral, any definite number ofvariables isequal tothe algebraic sum ofthelimits ofthese variables: lim,++... uy)lima,+lima,+...+limay, Proof. Weshall carry out the proof fortwo terms, since itis thesame forany number ofterms. Let limu,=a,, limu,—a,.Then onthebasis ofTheorem 1,Sec. 4,wecanwrite 4,=4,+4,u,=4,+¢, where a,and a,areinfinitesimals. Consequently, 4,+4, =(a,+4,) +(a,+4). - Since (a,+a,) isaconstant and (a,+a,) isaninfinitesimal,again byTheorem 1,Sec.4,weconclude that lim(4,+4,)=a,+4,=limu,+lima. Example 1. tin257 in(142)—tim1+tim214 timZ=140=1. Theorem 2.The limit ofaproduct oftwo, three and, ingeneral, any definite number ofvariables isequal’ totheproduct ofthe limits ofthese variables: limu-u, ..dy=tima,-lima, ...timuy. Proof. Tosave space weshall carry out the proof fortwo factors. Letlimu,=a,, limu,=a,. Therefore, WH ata, Weta, 4,4,=(a,+4,)(a,+0,)=0,0,+4,4,+4,0,+4,4,. The product a,a, isaconstant. Bythe theorems ofSec. 4,the quantity a,a,a,,-+0,a, isaninfinitesimal, Hence, limu,u,= =a, =limd, -limd,. Corollary. Aconstant factor may betaken outside thelimit sign. Indeed, iflimu,—a,, ¢isaconstant and, consequently, lime=c, then lim(cu,)=lime-limu,=c-limu,, asrequired. Example2. lim5e*=5limx*=5-8=40, Theorem 3..The limit ofaquotient oftwo variables isequal tothequotient ofthelimits ofthese variables ifthelimit ofthe denominator isnot zero: lim4=E24iflime#0, Basic Theorems on Limits a Proof. Letlimu=a, limo=b #0. Then u=a+a, v=b+8, where aand 6areinfinitesimals. We write the identities Haotela,(ata_2)a,ab—fa crite i+(Sp-3)-t tia or Hoy bBooot bOHB) a, F ab—fa Thefraction%isaconstantnumber,whilethefractionrem; isaninfinitesimal variable byvirtueofTheorems 4and5(Sec.4),since ab—a isaninfinitesimal, while the denominator 6(6+) A Fo in 8 lin hasthelimit 6*40. Thus, lim == 14, Example 3. q Slimsmn25 ETD AHS gts sy20G2=TimGe4im2219 3= Here, wemade use ofthe already proved theorem forthe limit ofafraction because the limit ofthe denominator differs from zero asx—+ 1.ifthe limit ofthe denominator iszer0,. the theorem for the limit of fraction isnot pplicable, and. special considerations have tobeinvoked. Example4.Findtim=}. Herethenominator, anaumerstor approach seroax+2.andsconsequently, Theorem 3.isinapplicable. Perform the following Identical transformation: POA 99),ir aie This transformation holds for allvalues ofxdifferent from 2.And so, having inview thedefinition of limit, weeanwrite tn,PhtigC=DEEN (2) Example 5.Findtim17. Asx-+1thedenominator approaches zero but the numerator does not (itapproaches unity), Thus, the limit ofthe Teciprocal quantity iszero: lim (21)eon ° faim TO 48 Limit. Continuity ofaFunction Whence, byTheorem 2ofthe preceding section, wehave ain Theorem 4. Jfthe inequalities u<z<v are fulfilled between the corresponding values ofthree functions u=u(x), 2=2(x), and v=v(x), where u(x) and v(x), asx—+a (orasx—+00), approach one and the same limit 6,then z=z(x) asx—+a (or asx—+00) approaches thesame limit. Proof. For definiteness we shall consider variations of the functions asx—+a. From the inequalities u<z<v follow the inequalities u—b<z—b<v—b; itisgiven that limu=6, limo=6, Consequently, forany e>0 there will beacertain neighbourhoodwithcentreatthepointa,inwhichtheinequality Tuoblce will befulfilled; likewise, there will beacertain neighbourhood with centre atthe point ainwhich theinequality |v—6|<e will befulfilled. The following inequalities will befulfilled inthe smaller ofthese neighbourhoods: —e<u—b<e and —e<vu—b<e, and thus the inequalities —ecz—b<e will befulfilled; that is, lim z= 6. Theorem 5.Ifasx—+a (orasx—+00) the function ytakes on nonnegative values y>=0 and, althesame time, approaches the limit 6,then 6isanonnegative number b>0. Proof. Assume that 6<0, then |y—b|>6; that is, the difference modulus |y—6| isgreater than the positive number |6| and, hence, does notapproach zero asx—+a. But then ydoes not approach 6asx—+a; this contradicts thestatement ofthe theorem. Thus, theassumption that 6<0 leads toacontradiction. Consequently, 6>0. Insimilar fashion wecan prove that ify<0, then limy<0. Theorem 6./fthe inequality v>u isfulfilled between corre- sponding values oftwo functions u=u(x) and v=v(x) which approach limits asx—+a (orasx—+00), then limvy>limu. Basie Theorem onLimits 0 Proof. Itisgiven that v—u>0. Hence, by Theorem5,lim(o—u)0orHim»—limus0, andsolimo >limu. A Example6.Provethatlimsinx=0. From Fig,42itfollowstnatifOA=1,x>0,then AC=sins,AB=x,sine<x.Obviously, whenx<0 wewillhave[sin|<[x].ByTheorems 5and6,itg 8{ollows, from these inequalities, that lim sinx=0, ; fo Fig 42. Example7Provethattimsinon0.Indeed,[sin]<{snx1,Conse quently,limsin$=0. Example 8.Prove that lim cosx=1; note that coss=1tant, therefore, x 1 Jimcoee—tim(1—2stot5)—1—2timsat$101, Insome investigations concerning the limits ofvariables, one has tosolve two independent problems: 1)toprove that the limit ofthe variable exists ‘and to establish the boundaries within which the limit under consideration exists; 2)tocalculate the limit tothe necessary degree ofaccuracy. The first problem issometimes solved bymeans ofthefollowing theorem whichwillbeimportant lateron. Theorem 7.Ifavariable visanincreasing variable, that is, each subsequent value isgreater than thepreceding value, and if itisbounded, that is, v<M, then this variable has the limit limv=a, where a< M. Asimilar assertion may bemade with respect toadecreasing bounded variable quantity. Wedonotgive the proof ofthis theorem here since itisbased onthe theory ofreal numbers, which we shall not consider in this text. Inthefollowing two sections weshall derive the limits oftwo functions that find wide application inmathematics. %0 Limit,Continuity ofaFunction SEC,6,THELIMITOFTHEFUNCTION “22asx0 This function isnot defined for x=0 since the numerator and denominator ofthe fraction cbecome zero. Letusfind the limit ofthis function asx—-0. Let us consider acircle ofradius 1(Fig. 43); denote thecentral angle MOB byx;O<x<$. From Fig.43it follows instraightforward fashion that °@A area AMOA<area ofsector MOA < Fig. 43. area ACOA. Ww TheareaAMOA=40A-MB=4-1-sinx= 4sinx. TheareaofsectorMOA=40A-AM=4-1-x= bx, TheareaofACOA=40A-AC=4.1-tanx=4 tanx. Altercancelling +},inequality (1)isrewritten sinx<x<tanx. Divide allterms bysin.x: xt '<ing<tore or 1>E>cosx. We derived this inequality onthe assumption that x>0; noting that“52-824 andcos(—x)=cosx, weconclude thatit holds forx<O aswell But lim cosr=1, lim 1=1, Hence, thevariable “2tiesbetween twoquantities thathave the same limit (unity). Thus byTheorem 4ofthe preceding section, im 0 eid Thegraphofthefunction y=" isshown inFig.44, The Number € 5 y lea B a 7 fe ard Fig. 44 Examples. tansjysine sitim1oy! OeTe cee Aeee 2)tnSEEimaS timSED tk(const). iss ast gin 3)timA=S08tim—?=tim—?sing1-0-0. z sinar imHae imMOE gySO rcsOLsafe 2B” SBE Bq BOB Beet BE a 1aSBT Tpmconst, B=const). SEC. 7.THE NUMBER £ Let usconsider the variable (+3), where nisanincreasing variable that takes onthe values 1, 2,3,... Theorem 1.Thevariable (1+ 1)",asn—voo, hasalimit between the numbers 2and 3. Proof. ByNewton's binomial formula wehave 1)"pyLymot)/1)*a(n)(n—2)(1)* (1+2)a4p42. (FE)uesgena(tyten(n—1(n—2)...fa—(a—1)) /4)" EG O) 52 Limit.Continuity of@Function Carrying out the obvious algebraic manipulations in(1), weget 1 1 2 n=lsotmmca('-a) (1-8). @ Fromthelatterequality itfollows thatthevariable (1+2) isanincreasing variable asnincreases. Indeed, when passing from the value atothe value n-+1, each term inthe latter sum increases, 1 1) 4 1ta(1—%)<7a(1— spa)and50forth, and another. term isadded. (All terms ofthe expansion arepositive.) .Weshallshowthatthevariable (1+) isbounded. Noting that(12)<1(1-2)(1-2)<1,ete,weobtainfromexpression (2)the inequality (144) <l4lt htt tie: Further noting that tor. a oi 1Taser tesserae wecan write the inequality (14h)<1+l4ptyt ten:eae a ‘The grouped terms ontheright-hand side ofthis inequality form ageometric progression withthecommon ratioq=5 andthe first term a=1, and so * lit 1 (1+)<1+[l+ptat eetee]= 18ree nipteer+ =1+[2(3)|<3 The Number e 53 Consequently, forallnweget (14+4)"<s. From equality (2)itfollows that (1+t)'>2 Thus, wegettheinequality : 2<(1+4)'<3 @) Thisprovesthatthevariable (1+iyisbounded. Thus,thevariable (144) isanincreasing andbounded variable; therefore, by‘Theorem 7,Sec. 5,ithas alimit, This limitisdenoted bytheletter e. .Definition. Thelimitofthevariable (1+)" asn—roo isthe number e: e=lim(14gy9 ByTheorem 6,Sec.5.itfollows frominequality (2)thatthe number esatisfies the inequality 2<e<3. Thetheorem isthus proved. The number eisanirrational number. Later on, amethod will beshown that permits calculating etoany degree ofaccuracy. Its value toten significant decimal places is e=2.7182818284... Theorem 2.Thefunction (1+1)approaches thelimiteasx approaches infinity, lim(1+t)'me Proof.Ithasbeenshownthat(1-+2)"—se asn—voo, ifm takes onpositive integral values. Now letxapproach infinity while taking onfractional and negative values. 9)maybeshownthat(1-4;2)"Feasavoevenifisnotan increasing variable quantity. ot Limit. Continuity ofaFunction 1)Let x—+-+00. Each ofitsvalues lies between two positive integral numbers, n<x<cnt+l. The following inequalities will befulfilled: eee aT 1 1 1 l4d>1¢i>14+5h, (14¢)"'> (144) >(14ch)- Ifx—+00, itisobvious thatn—+co. Letusfindthelimits ofthe variables between whichthevariable (1-+4)" lies: lim(1+t)"=lim(i+4)(144)- =tim(144)"+ tim(1+4)<e1=6, 1 yen. '4+245) lim(t+)=lim(4a) note atl note =ea Lye watt.(43)afne, im—_ uu tte('4a41) Hence, byTheorem 4,See. 5, lim(1+4)'=e @) 2)Letx—+—oo. Weintroduce anew variable ¢=—(x-+1) or xo—(4+1), When f—--+00 then x——oo. We can write Wt Lyset pgyetet slim,(14g)=lim(1a) =in(Ga)RLV Lye =a(GAY=sim(14-7) n 1y\t 1=jim(1+) (1+q)sels0 The Number ¢ 55 Thetheorem isproved. Thegraphofthefunction, y=(1+4)° isshown inFig. 45, y, le aI " Fig. 45. Ifinequality (4)weputL=a, thenasx00 wehavea—+0 (but a0) and weget lin(140)=e. Examples: mite(142) te,(144) ata,(4b) encenes ©ae(FE) in(SER in(1) =win(14h)asin,(14-4)= 56 Limit.Continuity of@Function SEC. 8,NATURAL LOGARITHMS InSee,8ofChapter Twedefined thelogarithmic. function y=log,x. The number aiscalled the base ofthe logarithms. Ifa=T0, then yisthedecimal (common) logarithm ofthenum- berxand isdenoted y—logz. Inschool courses ofmathematics wehave tables ofcommon logarithms, which arecalled Briggs’ logarithms after the English mathematician Briggs (1556-1630). Logarithms tothe base e=2.71828... arecalled natural or Napierian logarithms after one ofthefirst inventors oflogarithmic yi yor | a 2 eee Cy (A Fig. 4. Uybles, themathematician Napier (1650-1617).2) Therefore, i =x, then yiscalled thenatural logarithm ofthe number x.In writing wehave y=Inx (after the initial letters oflogarithmus naturalis) inplace ofy=log,x. Graphs ofthefunction y=Inx and y=logx areplotted inFig. 46. Let usnow establish arelationship between decimal and natural logarithms ofone and thesame number x. Let y=logx orx=10". We take logarithms ofthe leftand right sides ofthelaiter equality tothebase eandgetInx—y|n 10. Wedetermine y=_-45Inx, of,substituting thevalueofy,we havelogx=,hyInx. Thus, ifweknow the natural logarithm ofanumber x,thecom- mon (decimal) logarithm ofthis number isfound bymultiplying bythefactor M=5=0.434294, which factor isindependent ofx.The number Misthemodulus ofcommon logarithms with respect tonatural logarithms: logx=MIn-x. *)ThefirstJogarithmic tableswereconstructed bytheSwissmathemati- cian’ Bargi (1852-1632) toabase close tothe number Continuity ofFunctions 87 Ifinthis identity we put x=e, weobtain anexpression ofthe number Minterms ofcommon logarithms: loge=M (Ine=1). Natural logarithms areexpressed interms ofcommon logarithms asfollows: Inx=Htloge where 1#7=2-302585. SEC. §CONTINUITY OF FUNCTIONS Letthefunction y=f(x) bedefined forsome value x,andin some neighbourhood with centre atx,.Lety,=f(x,). Ifxreceives some positive ornegative (it'is immaterial which) increment Axandassumes thevalue yx=x,+Ax, then thefunction ytoowill " receive an’increment Ay.Thenewin- aycreased value ofthefunction will be iy 4+ Ay=f(%,+Ax)(Fig.47).Theincre- mentofthefunctionAywillbeexpressed bytheformula 4 Ay=f (x,+A) (x,). a Definition 1.Thefunction y=f(x) is4| 4 ote x called continuous forthevalue x-=x, Fig.47. (oratthe point x,)ifitisdefined in some neighbourhood ofthe point x,(obviously, atthe point x, aswell) and if lim Ay=0 w aes or,which isthe same thing, dimUG+42)—f 1=0. (2) Indescriptive geometrical terms, the continuity ofafunction ata given point signifies that the difference ofthe ordinates ofthe graph ofthefunction y=/(x) atthepoints x,+Ax and x,will, inabsolute magnitude, bearbitrarily small, ‘provided |Ax| is sufficiently small. 58 Limit.Continuity ofa-Function Example 1.We shall prove that the function y=x* iscontinuous atanarbitrary pointx.Indeed) , Wah etAVRO EAR, Ay (ay+Astana tae, limAy=lim(Qe,Ax-+Ax)=2e limAx+limax-limax=0are ge weeeTrygee forany way that 4xmay approach zero (Figs. 48,a and 48,b). +gp4x20,4490 yyA<0,Ay<0,, @ ly = ax i a a me Fig. #8. Example 2.Weshallprove thatthefunction y=sinx iscontinuous at anarbittary point x,-Indeed, Wainy yet Aysin (2+ AX), Aysin(x44)—siny=2staSFcos(+447).1kwasshownthatmsn£0(Example7,See.5).Thefunctionx44)isbounded.Therefore, timay=0. cas(ny)isbounded.Therefore, limay=0. Insimilar fashion, itispossible toprove thefollowing theorem by considering each basic elementary function and each elementary function. Theorem. Every elementary function iscontinuous ateach point atwhich itisdefined. The condition ofcontinuity (2)may bewritten thus: limf(x, +Ax)=/(x,) or limf(x)=F(x), but x= limx, ih Continuity ofFunctions 89 Consequently, limf(x)=f(lim2). @) Inother words, inorder tofind the limit ofacontinuous functionasx—x,itissufficient tosubstitute intotheexpression ofthefunction thevalue oftheargument, x,,inplaceoftheargument x. Example 3.The function y-=+" iscontinuous at.every point. x,and therelore limx=33, lim st=3"=9, ncExample 4.Thefunction yosiax icontinuous ateverypointand therelore x_V3 erteattateat Example 5.The function y=e* iscontinuous atevery point and therefoimeet. ’ eae eneey Eeample 6.ty2 unLinge tnnla)? since lim(+x)*=eandthefunction Inziscontinuous forz>0,and,Consequently, for=e, limIn(1-4) =in{lim(1-+2)*]=Ine =1. Definition 2.Ifthefunction y=f(x) iscontinuous ateachpoint ofacertain interval (a,6), where a<o, then itissaid that the function iscontinuous in this interval. Ifthefunction isalso defined for x=a and lim f(x)=f(a), itissaid that f(x) atthepoint x—a iscontinuous ontheright.Iilimf(x)=/(6),itissaidthatthefunctionf(x)éscontinuous sobre ontheleft ofthe point x=. Ifthe function f(x) iscontinuous ateach point ofthe interval (a,6)and iscontinuous attheend points ofthe interval, onthe Tight and left, respectively, itissaid that the function f(x) is continuous over theclosed interval (a,6}. Example 7.The function y=" iscontinuous inanyclosed interval [a,0) This follows from Example 1. C Limit. Continuity ofaFunction Ifatsome point x=x,, atleast one oftheconditions ofconti-nuityisnotfulfilledforthefunctiony=/(x), thatis,ifforx=x,theTunction isnotdefined orthere does notexist alimit limf(x) orlimf(x)#f(x,) inthearbitrary approach ofx—+x,, although the’expressions ontherightandleftexist, thenatx—x, the function y=f(x) isdiscontinuous. Inthis case, thepoint x=x, iscalled thepoint ofdiscontinuity ofthefunction. Example 8.Thefunction y=isdiscontinuous atx=0.Indeed, the function isnot defined atx=0. tyLape: tinbane (ieFig991 iseasytoshowthatthisfunction iscontinuous forany Value x0. Example 9.Thefunction y=2* sdiscontinuous atx=0. Indeed, lim2*=o, lim2*=0. Thefunction isnotdefined atx=0(Fig.49). y| 2hy yptt)1 ee I + rio Fig. 49. Fig. 50. Example 10.Consider thefunction f(2)=75p. Atx<0,ack atx50,7Epa Henes, afs Meaaa ‘the function isnot defined atx=0.We have thus established the fact that thefuetiona)=27isdiscontinuous atx=0(Fig.5) Certain Properties ofConttnuous Functions 6 Example 11.Theearlierexamined function y=sin+is discontinuous at2=0. Definition 3.Ifthefunction f(x)issuchthatthereexistfinite limits lim F(x)=f(x,+0) and lim f(x)=/(%,—0), but either lim¥()% limf(x)orthevalue ofthefunction f(x)atx=x, isnotdefined, thenx=,iscalledapointofdiscontinuity ofthefirst kind. (For example, forthefunction considered inExample 10, the point x=0 isapoint ofdiscontinuity ofthefirst kind). SEC. 10, CERTAIN PROPERTIES OF CONTINUOUS FUNCTIONS Inthis section we shall consider anumber ofproperties of functions that are continuous onaninterval. These properties will bestated inthe form oftheorems given without proof. Theorem 1.Ifafunction y=f(x) iscontinuous onsome inter- val [a,6](a<x<b), there will be,onthis interval atleast one point x=x, such that thevalue ofthefunction afthis point will satisfy therelation Fa) =F), where xisany other point ofthe interval, and there will beat least one point x,such that the value ofthefunction atthis point will satisfy therelation Fe) <F(#). Weshallcallthevalueofthefunction f(x,)thegreatest value ofthefunction y=f(x) onthe interval (a,6],and thevalue of the’ function f(x,) the smallest (least) value ofthe function on 4 the interval (a,6]. This theorem isbriefly stated as follows: Afunction continuous on the interval a=x<b attains on this interval (atleast once) agreatest oy & oe value Mandasmallestvaluem. Pan The meaning ofthis theorem is ig. clearly illustrated inFig. 51. Note. The assertion that there exists agreatest value ofthe function may prove incorrect ifone considers the values ofthe function inthe interval a<x<b. For instance, ifwe consider the function y=x inthe interval O<x<1, there will beno o Limit. Continuity ofaFunction greatest and noleast (smallest) values among them. Indeed, there isnoleast value orgreatest value ofxinthe interval. (There isnoextreme left point, since nomatter what point x*wetake there willbeapoint leftofit,forinstance, thepoint >;like- wise, there isnoextreme right point; consequently, there isnoleastandnogreatestvalueofthefunction y=x.)Theorem2.Letthefunctiony=[(x)becontinuous ontheintr. val {a,6]and aftheend point ofthis interval let ittake on valuesofdifferentsign;thenbetweenthepointsaand6therewill beatleast one point x=c, atwhich thefunction becomes zero: f@)=0, a<c<b. This theorem has asimple geometrical meaning. The graph ofa continuous function y=f(x) joining the points M,[a, f(a)] and M,{6,F(6)|, where f(a)<0 andf(b)>0 gy 1orf(a) >0and f(6)<0, cuts thex-axis 1 atleast atonepoint (Pig. 52). te yi mlo.tcoy to gy 7 a7 ”ne LE Milanjla.ftal] Y Fig. 52. Fig. 53. Example. Given the function y=x*—2. Yen=—Iy Yeny=6. It1sconti rnuous inthe interval [1,2].Henge,inthis(nterval_there™'s apointwhere y=x*—2 becomes zero. Indeed, y=0 when x=j/(Fig.53). Theorem 3.Let thefunction y=f(x) bedefined and continuous inthe interval (a,6].Ifatthe end points ofthis interval the Junction takes onunequal values f(a)=A,[(6)=B, thennomat- terwhat thenumber wbetween numbers Aand B,there will bea point x=c between aand 6such that f(c)=w. The meaning ofthis theorem isclearly illustrated inFig. 54. Inthegiven case, any straight line y=p cuts the graph ofthe function y=f(x). Comparing Infinitesimats 6 Note. Itwill benoted that Theorem 2isaparticular case of this theorem, forifAand Bhave different signs, then for#one can take 0,and then »=0 will liebetween thenumbers AandB. a ara“ +fem}ya oF ae Fig. 54. Fg. 65 Corollary ofTheorem 3./fafunction y=f(x) iscontinuous in some interval and takes onagreatest value and aleast value, then inthis interoal ittakes on, atleast once, any value lying between thegreatest and least values. Indeed, letf(x,)=M, f(%,)=m. Consider theinterval [x,,x1ByTheorem 3,inthis inferval the function y=f(x) takes on any value plying between Mand m.But theinterval. [x,,x] lies inside the interval under consideration inwhich the function F(x) isdefined (Fig. 55). SEC, 11.COMPARING INFINITESIMALS Let several infinitesimal quantities %BLY ee beatthesame time functions ofone and the same argument x and letthem approach zero asxapproaches some limit aor infinity. We shall describe the approach ofthese variables tozero when weconsider their ratios. *) Weshall, infuture, make useofthefollowing definitions. Definition 1.Iftheratio&hasafinitenonzero limit,that is,iftim=440,andtherefore,lim=-1-40,theinfinites- Imalsfandaarecalledinfinitesimats ofthesameorder. |1)Weassume that theinfinitesimal inthedenominator does not vanish insme neighbourhood ofthe point a. ot Limit. Continuity ofaFunction Example 1.Leta—x, B=sin2z, where x—+0. The infinitesimals @and 6. are-ol the same order because timBW timS0249 Example 2.When x-+0, the infinitesimals x,sin3x, tan2x, 7In(14x) are infinitesimals’ ofthe ‘same’ order” The proof issimilar to’that given in Example I. Definition 2.Iftheratiooftwoinfinitesimals ®approaches zero,thatis,ifim—0 (andlim-#-—oo),thentheinfinitesi-malBiscalled aninfinitesimal ofhigher order than a,andthe infinitesimal @iscalled, aninfinitesimal oflower order than B. Example3.Leta=s,B=x",n>1,20.Theinfinitesimal fisaninfinitesimal ofhigher order than the infinitesimal a,since tim.alimx*-1=0. Here, theininitesimal aisaninfinitesimal oflower order than p. Definition 3.Aninfinitesimal Biscalled aninfinitesimal ofthe kth order relative toaninfinitesimal a,ifBanda*areinfinitesimalsofthesameorder, thatis,iftim=A#0. Example 4.If=x, Box’, then asx-+0 the infinitesimal pisan infinitesimal ofthe third order relative tothe infinitesimal @since timFotimat. co eee) Definition 4.Iftheratiooftwoinfinitesimals &approaches unity,thatis,iftim£=1, theinfinitesimals Bandaarecalled equivalent infinitesimals and wewrite a~p. Example 8.Let a=x and Basins, where x-+0. The infinitesimals a and: areequivalent, since tnME, Example 6.Leta=x, B=In(l+x), where x+0. The infinitesimals a and Bate equivalent, since tim12+) (eee Example 6,Sec. 9). Comparing Infinitesimats 6 Theorem 1.Ifaand Bareequivalent infinitesimals, their difler- ence a—B isaninfinitesimal ofhigher order than aand than B. Proof. Indeed, a 6 mB nlltim$2=tim(1—£)=1—tim£=1-1=0. Theorem 2./fthedifference oftwo infinitesimals a—B isan infinitesimal ofhigher order than cand than ®,then aand Bare equivalent infinitesimals. Proof.LetlimS=®—0, thenlim(1-8)=0,or1—lim8=o, =lim®, i im2b 1) orL=lim£, i.e. amB.Iflim P=,thenlim($1)0, lim=1,thatis,awB. Example 7.Let ams, Bux-+s9, where x—+0 The infnitesimals aand Pateequivalent, since their diference Ba—s* isanintinitesimal ofhigher order than aand than B.Indeed, um22 im2umxt=0, jm28timtim oe een Ma’ Example 8.Asx» theinitesinalsamtt! andpxareequivatent infinitesimals, sincetheirdiference a—B=*4!— 1-1isaninfiaitesimal olhigher orderthanaandthanB.Thelimitoftheratioof@andBis unity 4Bom eoumEth 1y_ YinFminemtinSEEum(14+-b)=1 oe Note.Iftheratiooftwoinfinitesimals &hasnolimitand does not approach infinity, then Band aarenot comparable in the above sense. Example8.Letams,Bexsin-L,wherex—+0.Theinfniesimats 4andBcannotbecompared because theirratioP=sintasx—~0doesnot approach either afinite limit orinfinity (see Example 4,Sec. 3). aeaaee 6 Limit, Continuity ofaFunction Exercises onChapter tt Find the indicated limits: htmERZEES. ane42m(Ramemconeeatsl Aan2 x—2 14 4-28 $1 3tim32.Ans.0.4tim(2—L44).Ans.2.6im=28th| ARR is, (2-3+3) te,ae mo 241 Ane VAD bn 1 AnsA.6.imEEEans11timPEREERans,1, 8WnPEPE EEOAnsFe Hint.Writetheformula (k+1)*—k* =3k*+43k+41 fork=0,1,2... a. Pat Poraa.t$o14ts Pa 9.49-241; (n+1)'—n? =3n?+3n +1. ‘Adding the left and right sides, wegot OF IRBPEMSMDESLED EMEOED,atath (otatet. pay—3EDsty, whence Ey pateAetna th| imP42—1 Ans,oo, 0,timSBN gs, oi ee Ae ar jimS220 angte,um2X4,Ans.4,13.tim2":Ans.3 MeMS aerpae AnpeI Se Aha 4" imP56 agg I im243—10 Ang “(ecie AgS gare At tnPEW Ans 2. 7, tim MAW Ans 0,ener ri) 3 ann Qu AN imELAN2pngget 13 a.— WeagENHaneoetn[Lt]. Ane= te.fin2.Ans.(aiepoitivetegen,inTEESad Exercises onChapter It er Vitis 2V3 VETE—» ¢ 22,timVEFINB gs,2VE95,timVBEHPns,2, iNVr-v3 a i”Vee 5ayy Ra "5 timLEEans2a, tmARV ysVO sot Va=T 3 Pr aa 26, timViFEEE—I Ans 1. a timVER3 Ans. 1, homer z ne Vaal 28,inVET. ans,Laseetee,1asxem.2.in(VFFTA =VFRD,Ans0.80.ime(VEFIMw. Ans.Laseebe, —aat sax sings shy some,ahtimMEAns1.a2timAEAns.498,tmTS, 1 z 2 An.La timans, 2,35, timxeots.Ans.z aeVisca va ners 36.imiaieese .Ans.V3.37.im(t—2ytan®. ans.2,v=Ban(e—3) 98,tim28H pgs,2timEEA Ane,Deas, tanx—sinx gg, 1 m (42). Aneomer “SF atm(143) am , Ly 1 x\ 1wim (\-LYF. ans. La tim (YF. an,2.('-7) o as,(Tez) 7 tm (LY aneeM im{nttn(et)—Innl). An. 46.timpcos AnsAAT, tmOEE sg, CN) ee 80,in(eos£)".Ans1.sttimBOE,an,1asare,O08 sina a. mlest. Am ae. a ee PD 68 Limit.Continuity of@Function weve,Ossrs—e. 4time(6?1),AneIne.a8timSH, Ans.a—B. 56.lim _@=—e Ans. 1. 6iSinarmsinpe 4" Determinethepointsofdiscontinuity ofthefunctions: 57.y=——* =". Ans.Discontinuities of iowxe—2;<1; I= ZEEE AAMDiscontinuities ofsecondKindfo21; 0:2.68poten. AnsDisontnutien ofsecond kindfor#0and 2,2, .,2 59.Find thepoints ofdiscontinuity ofthefunctions y=1+2* andcon- struct the graph ofthis function. Ans, Discontinuity ofsecond kind atx=0 Gate ae040, yori az —0-0). 60.From among thefollowing infinitesimals (asx-»0); x,Vx(I=ah sin3s, 2xosx 9/tanFx, xe, select infinitesimals ofthe same order as1, and also ofhigher and lover order than x.Ans. Infinitesimals ofthe same order are sin3xand xe**; infinitesimals ofhigher orde, x#and2xcosx$/Tana, infinitesimalsoflowerorder,V3x3, 61.Choose from among the same infinitesimals (asx-+0)suchthatare equivalent totheinfinitesimal <:Qsinx, Jtan2e, x30, VEER, Ina)A434Ans.Jtan2x,2-3,In(1+9), 62.Check tosee that asx-+1,theinfinitesimal 1—xand1—j/Fare ofthesameorderofsmallness.Aretheyequivalent?Ans.limayehence, these infinitesimals areofthesame order, butthey arenotequivalent. CHAPTER ML DERIVATIVE AND DIFFERENTIAL SEC. 1.VELOCITY MOTION Let usconsider the rectilinear motion ofsome solid, sayastone, thrown vertically upwards, orthemotion ofapiston ‘inthecylin: der ofanengine, etc. Idealising the situation and disregarding dimensions and shapes, weshall always represent such abody in the form ofamoving ‘point M. The distance softhe moving point reckoned from some initial position M, will depend onthe time ¢;inother words, swill bea function oftime #: ast» s=f(0. i) at Atsome instant oftime*) f,letthemoving point M $l.” beatadistance sfrom the initial position M,, and at Yw, some later instant ¢-+A¢ letthe point beatM,, a distance s+Asfrom theinitial position (Fig. 56). Thus, Fig. 56 during theinterval oftime Afthe distance schanged bythequantity As.Insuch cases, one says that during thetime Atthequantity sreceived anincrement As. Letusconsider theratio$8;itgivesustheaverage velocity of motion ofthepoint during thetime Af: bsYao=Ri" 2) The average velocity cannot inallcases give anexact picture oftherate oftranslation ofthepointMattime¢.If,forexample, thebody moved very fast atthebeginning oftheinterval Afand very slow attheend, theaverage velocity obviously cannot reflect these peculiaritiesinthemotionofthepointandgiveusacorrect idea ofthetrue velocity ofmotion attime ¢.Inorder toexpress more precisely this true velocity interms oftheaverage velocity, one has totake asmall interval oftime Af.The most complete description oftherate ofmotion ofthepoint attime ¢isgivenbythelimitwhichtheaverage velocity approaches asAf—0, *)Here and henceforward weshall denote thespecific value ofavariable and the variable itself bythe same letter. 70 Derivative andDifferential This limit iscalled the rate ofmotion atagiven instant: v=li = (3) dinat ® Thus, therate (velocity) ofmotion atagiven instant isthelimit ofthe ratio ofincrement ofpath Astoincrement oftime At,as the time increment approaches zero. Let uswrite equality (3)infull. Since As= f(t+At)—f(t), imLao—hO 1) arrr ” This isthe velocity ofvariable motion. Itisthus obvious that the notion ofvelocity ofvariable motion isintimately related to ‘the concept ofalimit. Itisonly with theaidofthelimit concept that wecan determine the velocity ofvariable motion. From formula (3°)itfollows that oisindependent oftheincrement intime Af, but depends on the value of¢andthetypeof function f(t). Laesod Spee pata yg(UE2 EMM) as 1bsETE ga as 1 cota,Smt(ertea)met Definition ofDertuative a SEC. 2,DEFINITION OF DERIVATIVE Let there be function y=f) ) defined inacertain interval. The function y=f(x) hasadefinite value foreach value ofthe argument xinthis interval. Let theargument xreceive acertain increment Ax(itisimma- terial whether itbepositive ornegative). Then thefunctionywill receive acertain increment Ay. Thus, with thevalue oftheargu- ment xwe will have y=/(x), with the value ofthe argument x+Ax wewill have y-+Ay=/(x-+Ax). Let usfind the increment ofthefunction Ay: Ay=f(x+Ax)—Ff(2). Oy Forming theratio ofthe increment ofthefunction totheincrement oftheargument, weget by_fe+a—f)Ea (3) WethenfindthelimitofthisratioasAx—-0.Ifthislimitexists,itiscalled the derivative ofthegiven function f(x) and isdenoted I(x). Thus, bydefinition, (e) =lim4Y f(slim ae or Pea Wetan— le, “ Consequently, thederivative ofagiven function y=f(x) with respect totheargument xisthelimit oftheratio oftheincrement ofthe function Ay tothe increment oftheargument Ax, when thelatter approaches zero inarbitrary fashion. ; Itwill benoted that inthegeneral case, the derivative ’(x) has adefinite value for each value ofx,which means that the derivative isalso afunction ofx. The designation f'(x)isnottheonly one used foraderivative. Alternative symbols are 490 Te The specific value ofthederivative forx=a isdenoted f’(a) or ase operation offinding thederivative ofafunction f(x)is called diferentiation ofthefunction, n Derivative and Digerential Example 1.Given the function y=x*; find itsderivative y’: 1)atanarbitrary point x, 2)atx—3. Solution, i)Forthevalue oftheargument x,wehave y=at, When the value ofthe argument isx+Ax, wehave y-FAy=(-+Az)% Find the inerement ofthe tunction: y=(e-+Antat2at(a,Forming therato82,wehaveAy2A(ON oyayrs ‘Ar Betat. Passing tothelimit, wegetthederivative ofthegiven function: vstimMatin x442)26 arse bre Hence,thederivative ofthefunction y=x*atanarbitrary pointisy’=2x,2)Whenx=3wehave VVeoy=23=6. Example 2y=; findy. Solution. Reasoning asbefore, weget 1, 1,yapiytev= sgt [a ey ve AU=TERR ¥x@+as) FFE! aot. ax” eFax)" (=tim48im ft_}ot, PSseteaxarse|F0Pan| Note. Inthe preceding section itwas established that ifthe dependence upon time fofthedistance sofamoving point is expressed bytheformula s=1(0, thevelocity vattime ¢isexpressed bythe formula =lim$5 timLébad—1)ay ae Hence v=s=f' (f), or,the velocity isequal tothederivative*)ofthedistancewith respectto the time. “*)Whenwesay“thederivative withrespect.tox"or“thederivativethespe fo7wemeanthatincomputing thederivative weconsider the variable x(orthe time ¢,etc.) the argument (independent variable). Geometric Meaning oftheDerivative 3 SEC. 8.GEOMETRIC MEANING OF THE DERIVATIVE We approached the notion ofaderivative byregarding the velocity ofamoving body (point), that istosay, byproceedingfrommechanical concepts. Weshailnowgiveanolessimportant geometric interpretation ofthederivative. Todothiswemust first define aline tangent toacurve atagiven point, We take acurve with afixed point M,onit.Taking apointM,onthecurve wedraw thesecant M,M, (Fig.57).Ifthepoint M,approaches thepoint M,without linit, ‘thesecant M,M,will occupy various positions M,M,,M,M,, andsoon.If,inthelimitless approach ofthepoint M,(along thecurve) tothepoint M,from'either side, thesecant tends tooccupy the position ofadefinite straight line M,T, this line iscalled the tangent tothecurve atthepoint M, (the ‘concept “tends tooccupy” will beexplained later on). Let usconsider thefunction f(x) and thecorresponding curve y=F(x) inarectangular coordinate system (Fig. 58). Atacertain value ofxthefunction hasthevaluey=}(x).Corresponding tothesevaluesofxand yonthecurve wehave thepoint M,(x,y).Letusincrease a y Ly foley ae M cn Af||ae olxtae Fig. 57 Fig. 58. the argument xbyAx. Corresponding tothenew value ofthe argument, x-+4+Ax, wehave anincreased value ofthe function, y+Ay=f(x4-Aa}. Another corresponding point onthecurve will beM,(x-+Ax, y+Ay). Draw thesecant M,M, and denote by@the angle’ formed bythesecant and thepositive direction ofthex-axis. Formtheratio4%.FromFig.58itfollows immediately that AY tyestanQ a) ™ Derivative and Diferentiat Now ifAxapproaches zero, the point M, will move along the curve always approaching M,. The secant M,M, will turn about M,and theangle gwill change inAx. Ifas‘Ax—-0 theangle @ y approaches acertain limita,thestraightline passing through M,and forming an angle awith the positive direction of ger? theabscissa axiswillbethesought-forline tangent. Itiseasy tofind itsslope: mM tana=lim tang=lim4¢—=/' (@). 4 Hence, 7NO ig F@)=tan a, 2) Fig. 69. which means that the values ofthe derivative f'(x), for agiven value of theargument x,isequal tothetangent ofthe angle formed with the positive direction ofthex-axis bytheline tangent tothegraph ofthefunction f(x) atthecorresponding point M,(x, y). Example. Find thetangents oftheangles ofinclination oftheline tangent tothecurveyx?atthepointsM,(z:3}My(—1,1)Fig.59). Solution. Onthe basis ofExample },Sec. 4,wehave y’=2x; hence, tae|,antinaey| an?ee nent SEC. 4,DIFFERENTIABILITY OF FUNCTIONS Definition. Ifthe function y=f(x) a) hasaderivative atthepoint x=.x,, that is,ifthere exists imM4 fimLetad—lts) be ae @) wesaythatforthegivenvalue x=x,thefunction isdifferentiableor(which isthesame thing) hasaderivative. Iafunction isdifferentiable atevery point ofsome interval la,6]or(a,6),wesaythat itisdifferentiable over theinterval. ‘Theorem.’ Ifa function y=/(x) isdiferentiable atsome point x=x,, itiscontinuous atthis point, bigerentiabiity ofFunctions 8 Indeed, if im YopdiaSEP. ‘then ala d+y where yisaquantity that approaches zero asAx-+0. But then Ay=f(x,)dx-+yAx; whence itfollows thatAy0 asAx-+0; andthismeansthatthefunction f(x) iscontinuous atthepoint x,(see Sec. 9,Ch. Il). Inotherwords, afunction cannot haveaderivative atpoints ofdiscontinuity. The converse isnottrue; from thefact that at some point x=x, thefunction y=/(x) iscontinuous, itdoes notyetfollow thatitisdifferentiable atthispoint: thefunction f(x) may nothave aderivative atthepoint x,.Toconvince ourselves ofthis, letusexamine several cases. Example 1.Afunction f(x) isdefined inaninterval (0,2}asfollows (see Fig.OOF fax when0<¥<I, Ha2e—1 when xc? t= this function hasnoderivative although itis continuous atthis pint. indeed, when gx>0 we have timOFAN) jgBUFAN=N2A=N yyy20x. ae ae ate a arte ar when 4x<0 weget inOFAN) gyMAIN tmBE, wae any ge at tebe Thus, this limit depends onthesign ofax,and. this means that thefunction has toderivative") atthe point a=. Geometrically. this tein accord with thefhehatattepoint«Ye given“curve dowsnathave dete ne angen.Rowthecontinuity ofthefunction atthepointx1follows tromthe fact thatayeaxwhenar<0, ay=2ar when ar>0, and, therefore, inboth cases ay-+0 as4x0 *)Thedefinition ofaderivative requires thattheratioM2should (as ‘Ax-+0) approach one and thesame limit regardless ofthe ay inwhich ae Spproaches aero. 16 Derivative and Differential ye Example2.Afunctiony=j/x,theanhofwhichisshowninFig.61, isdefined and continuous for allvalues ofthe independent variable. eet us{ry tofind out whether this function has aderivative atx=0; to do‘this, wefind the values ofthetunction atx—0 and.atw=-0-Ax!at =O wehave y=0, atr=0+4Axwehavey-+Ay=j/(as)- y y ‘ ole jt, * * ‘Ol 7 x ~ Fig. 60. Fig. 61. Theretore, Ay=//H- Find the limit ofthe ratio ofthe increment ofthe function tothe incree ment oftheargument: yatim8imVOI im1pe.aenear” arse OE are aa ‘Thus, theratio ofthe increment ofthe function fotheincrement oftheargumentMihepoint+=-0approaches infinitya8Ax~0(hencethereisnolimit).Consequ:ently,tisfunctionisnotdifferentiable atthepoint#==0.Thelinetangenttothe cuveatthispointforms,withthes-axis, anangle<2,whichmeansthatit coincides with the y-axis. SEC. 5,FINDING THE DERIVATIVES OF ELEMENTARY FUNCTIONS,THEDERIVATIVE OFTHEFUNCTION y—x»,WHERE1ISPOSITIVE‘AND INTEGRAL Tofind the derivative ofagiven function y=/(x), itisneces- sary tocarry out the following operations (on thebasis ofthe general definition ofaderivative): 1)increase theargument xbyAx, calculate theincreased value ofthe function: ytdy=/(e+Ax) Finding the Derivatives ofElementary Functions n 2)find thecorresponding increment ofthefunction: Ay=f(e+Ax)—F(x); 3)form the ratio of the increment of the function to the increment ofthe argument: Ay_feta—f),ar an 4)find the limit ofthis ratio asAx—+0: te tim 4 tim Leta —Le) Here and inthe following sections, weshall apply this general method for evaluating the derivatives ofcertain elementary functions. Theorem. The derivative ofthe function y=x", where nisa positive integer, isequal tonx"-'; that is, ify=x",thenyf!=ne, 0) Proof. We have the function y=. 1)Ifxreceives anincrement Ax, then y+ Ay=(x+ Ax)". 2)Applying Newton's binomial formula, wefind Ay(e+Axx tA Era)+(eat or Ayanet"beSOE)errant... +(x)", 3)We find the ratio MenettpESDetvey+(Aa) 4)Then we find the limit ofthis ratio f=limS4— omfiae =lisat4MON)yaad oy 1 =Him,[ret 2OSDatae.(ay!) =ne, consequently, y’=nx"-!, and thus wehave proved thetheorem, 8 Derivative and Digerentiat Example 1yaat, y/=5e!-'=5x4, Example 2yx, y'—Is'=!, y'=1, The latter result has asimple geo- metric interpretation: the line tangent tothe straight Tine ye forany value GFcolncgey with,thilineang.consequent lorms‘withthe.postive direction ofthe x-axis anangle, the tangent ofwhich isl. Note that formula (1) also holds true when aisfractional or negative. (This will beproved inSec. 12). Example 3. y= Vz yar? then byformula (1), taking into consideration what wehave Just said, weget 1 iteyape or inYOoVE" 1 Example&y=. Represent yinihe form ofapower function: 4 gon Then eeaceei teta Cie ao SEC,6,DERIVATIVES OFTHEFUNCTIONS y=sin.xy=cos Theorem 1.The derivative ofsin.x iscosx, or ify=sinx, then y'=cosx. ap Proof. Increase theargument xbythe increment Ax; then 1)y+Ay=sin(x-+Ax);2)Ay=sin(x+Ax)—sin x=2sin?£42cosets.=?sinSt ar), =2sinSf-cos(2+4); Ax a2) gbaainfFeos(244%)_ang ay foo (if) an 4), 3)Me = aos (e+): ? Derivatives oftheFunctions y=sin x;y= cosx 79 sintt fmtimBemtim2.ti ar =hm,Semlin,dimcos(x+9).z but since assin in ang oh z wwe get (=limcos (x-+St)=cosx, =fimcos(«-+4) This latter equality isobtained onthegrounds that cosxisa continuous function. Theorem 2.The derivative ofcosx is—sinx, or ify=cosx, then y'=—sinx. uy Proof. Increase the argument xbythe increment Ax, then y+Ay=cos (x+Ax); ‘Ay=cos(x-+Ax)—cosx=—2sint42—* gintarts ginAE ar). =—2sinsin(244%); ay ar),Saaesin(4+): z =tim44—_5—tim—2 at) fi Ar). Y=finShoJimGesineGF)=—imsin(+92): z taking into account the fact that sinx isacontinuous function, wefinally get yf=—sing, # Derivative and Diferentiat SEC. 7,DERIVATIVES OF: ACONSTANT, THE PRODUCT OF ACONSTANT BY AFUNCTION. ASUM, APRODUCT, AND AQUOTIENT Theorem 1.The derivative ofaconstant isequal tozero; that is, ify=C,whereC=const,theny’=0. avy Proof.yOisafunction ofxsuchthatthevaluesofiareequaltoGforallx.Hence, forany value ofx y=/(x)=C. We increase the argument xbyanincrement Ax(Ax0). Since the function yretains the value Cforallvalues oftheargument, we have y+Ay=f(xt+Ax)=C. Therefore, the increment ofthe function is Ay=F(x-+Ax)—f(x)=0, the ratio ofthe increment ofthe function tothe increment ofthe argument auvrei) and, consequently, 7 Ay ooarm that is, y'=0. The latter result has asimple geometric interpretation. The graph ofthefunction y=C isastraight line parallel tothex-axis. Obviously, the line tangent tothegraph atany one ofitspoints coincides with this straight line and, therefore, forms with the x-axis anangle whose tangent y’iszero.Theorem. 2:Aconstant factormaybetakenoutside thederioa- tive sign, ie., ify=Cu(x) (C=const),theny’=Cu'(x). 1) Proof. Reasoning asintheproof ofthepreceding theorem, we have y=Cu(x); y+dy=Cule+Ax};Ay=Cu(x+ Ax)—Cu (x)=C[u(x +4x)—u (x), Derivatives of:AConstant, theProduct ofaConstant byaFunction 81 Ay culeban wncai v=limS4=limS2+89—40) |joy'Cu’(x), boAXtrae ar Example1.y=3va roof) na(,4)as(—1) Fe Sey—a(gp)-a(eF)-9(-4) Fag, or 3 ca ee Theorem 3.The derivative ofthesum ofafinitenumberofdiffe- rentiable functions isequal tothe corresponding sum ofthe derivatives ofthese functions. *) For the case ofthree terms, forexample, wehave yHue)to@) tela y'=u' +o’ w+0"@). (VI) Proof. For the values ofthe argument x youtotw {for thesake ofbrevity wedrop theargument xindenoting the function). For the value ofthe argument x+Ax wehave y+dy=(u+du)+(v+d0)+ (w+dw), where Ay, Au, Av, and Awareincrements ofthefunctions y,u, vand w,which correspond totheincrement Axintheargument x.Hence, - Ay_duae,bw dy=dutAv+dw, mot4S24Se, f= timS40 tim,844 tim82 aw YmBose its imaeimge or yf=u"(x)+0"(x)w(x). 1 Example 2:y=3xt—y 4,Ve *)Theexpression y=u(x)—v(x)isequivalent toy=u(x)+(—1)v(x) and yS[U) (1oaWFleGI=H"G@)0") By DertoativeandDiferential and so 14 yee tol.WF Theorem 4.The derivative ofaproduct oftwo differentiable functions isequal totheproduct ofthederivativeofthefirstfun- ction bythesecond function plus theproduct ofthefirst function bythederivative ofthesecond function; that is, ifyuo, then y!=u'v-+uv’, vty Proof. Reasoning asinthe proof ofthe preceding theorem, we get y=uo, y+Ay=(u+Au) (0-40), Ay=(u+ Au)(v+Av)—uv=Auv-+uAv-+ AuAv,Bybey4B24Ay2AteMypuht+ue,fetimMotimSy4timwAS4timdub?= CaataSaar Da ae =(im imS24timAwtim42 (dinSa)oalige fi,efi (since wand vareindependent ofAx). Let usconsider the last term onthe right-hand side: F im AP Since u(x) isa differentiable function, itiscontinuous. Consequently, lim Au=0. Also, in AdimgenFe Thus, the term under consideration iszero and wefinally get yisu'o+ue’. The theorem just proved readily gives ustherule fordifferentiating the product ofany number offunctions. Thus, ifwehave aproduct ofthree junctions ysuvw, Derivatives of:AConstant, theProduct ofaConstont byaFunction 89 then byrepresenting the right-hand side asthe product ofwand(oa),wegety’=u"(ow)+u(o1)'=u"ow-+u(0'w+vw’)=u'vw+u0'w+ wow.Inthis way wecan obtain asimilar formula for the derivative ofthe product ofany (finite) number offunctions. Namely, if Y=tty. yythen Yay oeally Et, oeagg FeeRU oegayle Example 3.Ifyoxtsins, then of=(29"sinxhsinx)’=Desinbcos, Example4.Ify=VFsincos,then ¥=(V2)sinxcosx+V%(sinx)’cosx-+Wxsinx(cosx)!= 1 -spy Vcosxcosx+Vxsinx(—sins)= u Z(costx—siatx)=4Vos apie caskVEcos—aaty=SOE4VFonde Theorem 5.The derivative ofafraction (that is,thequotient obtained bythedivision oftwofunctions) isequal’ foafraction whose denominator isthesquare ofthe denominator ofthegiven fraction, and thenumerator isthediference between theproduct of thedenominator bythederivative ofthenumerator, and. thepro- duct ofthenumerator bythederivative ofthe denominator; i.e., y=, thenyf=2S, wun Proof. IfAy, Au, andAv areincrements ofthefunctions y,wu, and v,corresponding totheincrement Axoftheargument x,then a+b y+dy=Ste, =itu uw_au~wavdu=oFa0 ow TAN * goumuae du,Av by ae ae"arw+ ~veFaH)* Re Bou elimanutim2 fmtimYetimBEBeanteaeal,ae o>Ioaem28,Cora~~vinwad rn Derivative andDigerentiat Whence, noting that Av—+ Oas Ar—+0, *)weget yates, Example 5Ity=2,then 1(eV cosx—a (cosx)_Setcosxt"sine ioncostx ‘cost 7 Note. Ifwe have afunction ofthe form y=", where thedenominator cisaconstant, then when differentiating this function wedonotneed touse formula (VIII); itisbetter tomake use offormula (V): ra(La\ el yet!y=(4u) etwas, Ofcourse, the same result isobtained ifformula (VIII) isapplied, Example 61fy=£282, then eos ___sing iis SEC. 8.THE DERIVATIVE OF ALOGARITHMIC FUNCTION Theorem. The derivative ofthefunction log,xis logse, thatis, ify=log,x,theny'=+log,e. (IX) Proof. IfAyisanincrement ofthe function y=log,x that corresponds tothe increment Axofthe argument x,then y+Ay=log, (x+x); y=log,(x+4x)—log,x=log,2*=log,(1+4); oy Ax’tmEtog,(144%). )TimAv=0sinceo(2)isadifferentiable and,consequently, continuousfunction. Derivative ofComposite Function 6 Multiply and divide byxtheexpression ontheright-hand side of the latter equality: dy_ tx ax) ot At)Sengmelee.(14%)=plow,(14). Wedenote thequantity 4¥interms ofa.Obviously, forthe given x,a—+0 asAx—+0. Consequently, syd re. fem x108. (1+) But, asweknow from Sec. 7,Ch. II, lim(I-+a)® =e, But ifthe expression under the sign ofthe logarithm approaches themumber_, thenthelogarithm ofthisexpression approaches loge (invirtue ofthecontinuity ofthelogarithmic function). ‘We therefore finally get y=limSYtim4tog,(1-+a)* =+log,e. dere dt got * Noting thatloge=->, wecanrewrite theformula asfollows: voidY=2a The following isanimportant particular case ofthis formula: ifa=e, then Ina=Ine=1; that is, ify=inx, thenyat, (xX) SEC, 9,THE DERIVATIVE OF ACOMPOSITE FUNCTION Givenacomposite function y=/(x), thatis,suchthatitmay berepresented inthe following. form: y=Flu), u=@(x) or_y=F{g(x)} (seeCh,I,Sec.8).Intheexpression y=F(u),uiscalled theintermediate argument. Let usestablish arule fordifferentiating composite functions. Theorem. /fafunction u=@(x) has, afsome point x,aderiva- tive u,=9(x),andthefunction y=F(u)has,atthecorresponding % Derivative and Diferentiat valueofu,thederivative y,=F’(u),thenthecomposite functiony=F[p(x)]atthegivenpointxalsohasaderivative, whichisequal to Ye= Fa(u) 9(x), where inplace ofuwemust substitute (he expression u=@(x). Briefly, =gia Inother words, thederivative ofacomposite function isequal to theproduct ofthederivative ofthegiven function with respect to theintermediate argument ubythe derivative ofthe intermediate argument with respect tox. Proof. For adefinite value ofxwe will have u=9(X),y=Fu). For the increased value ofthe argument x+Ax, utAu=p(x+Ax), y+Ay=F(u+ Au). Thus, tothe increment Ax there corresponds an increment Au, towhich corresponds an increment Ay, whereby Au—0 and Ay—0 asAr—0. Itisgiven that avy.im ju Ye From this relation (taking advantage ofthedefinition ofalimit) weget (for Au#0) vaata, ) where a—+0 asAu—+0. We rewrite (1)as Ay=yiAu+a Au. ) Equality (2)also holds true when Au=O for anarbitrary a, since itturns into anidentity, 0=0. For Au=O weshall assume a=0. Divide allterms of(2)'by Ax: a CC :May Brae. } Itisgiven that tim.=u, lima=0,aeeede get Derivative ofComposite Function 87 Passing tothelimit asAx—+0 in(3), weget Y=Yall () which istherequired proof. Example 1.Given afunction y=sin (x. Find y;.Represent thegiven function asafunction ofafunction as follows: yest, wast We find Y=Cosa, w=2x. Hence, byformula (4), Wemvuly=cosu-2e, Substituting, inplace ofa,itsexpression, wefinally get y=2.608(29) Example 2.Given thefunction y=(In 2).Find y. Represent this function asfollows: gow, using, We find amt ued. Hence, apa 1v= t=sins. Ia function y=f(a) issuch that itmay berepresented inthe form y=F(u), w=lr),C=Pir thederivative y,isfound byasuccessive application oftheforegoing theorem, Applying the’proved rule, wehave Feil Applying thesame theorem tofindui,wehave Sutstituting theexpression ofa,intothepreceding equality, weget tevuie, Cy We Fi)79) Wi Example 3.Given thefunction y=sin{(In.x)']. Find yj.Represent the function as follows: y=sinu, uso, v=ing, 8 Derivative and Digerential We then find gemcosu, uyndo%, oat. Inthis way, byformula @),weyet HeaWgiee=3 (osu)tL, orfinally, yj,60s[(Inx)*]-3(Inat Itistobenoted that: thefunction considered isdefined only forx>0. SEC, 10,DERIVATIVES OFTHE FUNCTIONS y=tan.z, yaeotx, y=Intxt Theorem 1.Thederivative ofthefunction tanxisgy orify=tanx, theny=abe (xD, Proof. Since sing I= cose bythe rule ofdifferentiation of@fraction [see formula (VIII), See. 7,Ch. III] weget +{sla2)’cosx—sin.x(cosx)__cosxcosx—sinx(—sin) oe See costetsintycosts cosFe* Theorem 2.The derivative ofthefunction cotxis 1 7 1ger oP y=cots, theny'=—sh-, (XID Proof.Sincey=<%,,wehave 1-_(608.2)sinx—cos x(sinx_—sinxsinx—cosxcosxy= ‘sinter = ‘sin? ~=en_sintesteastealate war Example 1.Ify=tan Vx, then \ sy tt j=) v=rn eee eae) AnImplicit Function and ItsDifferentiation 89 Example 2Ify=In cotx,then ie 1 1 2 Y= org (ot8)=Sore(-an)--ateat ee Theorem 3.The derivative ofthefunction In|x|(Fig.62)is+, orify=In|x|, theny'=1, (XH) y yeti ? if *N/** Fig. 62. Proof. a)Ifx>0, then |x|=x, In|x|=Inx, and therefore tat y=t. b)Letx<0, then|x|=—-x. But In|x|=In(—2). (Itwill benoted that ifx<0, then —x>0.) Let usrepresent thefunction y=1n(—x) asacomposite function byputting ySlnu; u=—x. Then Hemet (—N=t(—=t. And sofornegative values ofxwealso have theequation 1 waz: Hence, formula (XIII) has been proved for any value x40. (For x=0 thefunction In|x| isnotdefined.) SEC. 11,AN IMPLICIT FUNCTION AND ITS DIFFERENTIATION Letthevalues oftwo variables xand yberelated bysome equation, which wecan symbolise asfollows: F(x, y)=0. a) % Derivative and Diferential Ifthe function y=f(x), defined onsome interval (a,6),is such that equation (1)becomes anidentity inxwhen theexpres- y 9 fo ,, =a a, Fig. 63. Fig. 64. sionf(x)issubstituted intoitinplaceofy,thefunction y=f(x)isanimplicit function defined byequation (1). For example, the equation xt+yt—at=0 (2) defines implicitly the following elementary functions (Figs. 63 and 64): y=Vaae, @) y=—VEae. «) Indeed, substitution into equation (2)yields theidentity #4(a'—2')—a' =0. Expressions (3)and(4)wereobtained bysolving equation (2) fory.But notevery implicitly defined function may berepresente explicitly, that is,intheform y=/(x),*) where f(x) isanele- mentary function. Forinstance, functions defined bytheequations y—y—x'=0 or y—x—4siny=0 arenot expressible interms ofelementary functions; that is,these equations cannot besolved forybymeansofelementary functions. Note 1.Observe that the terms “explicit function” and “implicit function” donotcharacterise thenature ofthefunction butmerely theway itisdefined. Every explicit function y=f(x) may also berepresented asanimplicit function y—f(x)=0. *)Ifafunction isdefined byanequation oftheform y=/(x), one says that the Tunction isdefined explicitly orisexplicit, Dervoatives of«Power Function foranArbitrary Real Exponent 9k We shall now give the rule for finding the derivative ofan implicit function without transforming itinto anexplicit one, thatis,without representing itintheformy=f(x).‘Assume the function isdefined bythe equation xy'—at=0. Here, ifyisafunction ofxdefined bythis equality, then the equality isanidentity. Differentiating both sides ofthis identity with respect tox,and regarding yasafunction ofx,weget(via therule ofdifferentiat- ingacompositefunction)2x-+Quy’=0,whence yak. Observe that ifwewere todifferentiate thecorresponding explicit function y=Va—*, we would obtain cana which isthe same result. Let usconsider another case ofanimplicit function yofx: y—y—x*=0. Differentiate with respect tox: 6y'y’ —y—2x=0, whence otY=oT: Note2.Fromtheforegoingexamplesitfollowsthattofindthe value ofthederivative ofanimplicit function foragiven value oftheargument x,onealso hastoknow thevalue ofthefunction y foragiven value ofx. SEC. 12, DERIVATIVES OF APOWER FUNCTION FOR AN ARBITRARY REAL EXPONENT, OFAN EXPONENTIAL FUNCTION, [AND ACOMPOSITE EXPONENTIAL FUNCTION Theorem 1.The derivative ofthefunction x",where nisany real number, isequal tonx"-*; that is, ify=x", theny!=nx™™, a’) 2 Derivative and Diferentiat Proof. Let x>0. Taking logarithms ofthis function, “weget Iny=ninx, Differentiate, with respect tox,both sides oftheequality obtained, taking ytobeafunction ofx: Hants yaynt. Substituting into this equation thevalue y=x", wefinally get yan, Itiseasy toshow that this formula holds true also forx<0 provided x"ismeaningful. *) Theorem2.Thederivative ofthefunctiona*,wherea>0,is a*Ina; that is, ify=a*, then y’=a"Ina, (XIV) Proof. Taking logarithms oftheequality y=a*, weget Iny=xina, Differentiate the equality obtained regarding yasafunction ofx: ty=Ina; y=ylna or y'=a" ina, Ifthe base isa=e, then Ine=1 and we have the formula yae’, yer (xiv) Example 1.Given the function yae. Represent itasacomposite function byintroducing theintermediate argument u: ge, wast then Yume d=de and, therefore, Ypat deme de, *)ThisformulawasprovedinSec.5,Ch.IMl,forthecasewhenaIsapositive integer. Formula’) asnowbeen‘provedTorthegeneral case(or fny constant sumber n). Derivatives of@Power Function for anArbitrary Real Exponent 93 A.composite exponential function isafunction inwhich both thebase and theexponent arefunctions ofx,forinstance, (sinx)", xtinz, 2%,(Inx)*, and the like; generally, any function oftheform y=lu(x)su" isanexponential function (composite exponential function). *) Theorem 3. Ifyu,theny’=ou"-'u' +u°0!Inu (xv) Proof. Taking logarithms ofthe function y,wehave Iny=vinu. Differentiating the resultant equation with respect tox,we get ty=otw +o whence yay(o “40'Inu). Substituting into this equation the expression y=u, weobtain yf=vu?-'w’ +uo"Inu. Thus, thederivative ofanexponential function (composite expo- nential function) consists oftwo terms: the first term isobtained byassuming, when differentiating, that wisafunction ofxand v isaconstant (that istosay, ifweregard u”asapower function); thesecond term isobtained ontheassumption that visafunction ofx,and w=const (i.e., ifwe regard u®asan exponential function). Example 2.10y=a%, then y!=ae¥-1(e')-+28(e')In ory’=a*pe¥Inxme*(I-10) Example 3.Iy=(sinay", then yfx8(sinx)**~!(sinx)’+(sia.x)*(x4)Insine=A(sinx)60x+(sin2)"DeInsinx. The procedure applied inthis section for finding derivatives (first finding thederivative ofthelogarithm ofthegiven function) iswidely used indifferentiating functions. Very often theuseof this method greatly simplifies calculations. *)Inthe Russian mathematical literature this function isalso called an exponential-power function ofapower-ex ponential function Derivative and Diferential Example 4.Tofind thederivative ofthefunction (+0? V1 eee” Solution. Taking logarithms weget Ing=2in e++5IneI)—3IneAe Differentiate both sides ofthis equality: vo?) 13grit igh a Multiplying byyand substituting, inplace ofy,the expression HIV ET oatera" Mepoy VEs2|ot 9s [Aateeeye | Note.Theexpression'“—(Iny)', whichisthederivative, with respect tox,ofthe natural logarithm ofthe given function y=y(2), iscalled thelogarithmic derivative. SEC. 13. AN INVERSE FUNCTION AND ITS DIFFERENTIATION Take anincreasing ordecreasing function (Fig. 65) y=F() (a) defined insome interval (a,6)(a<6) (see Sec. 6,Ch. I).Let f(a)=c, f(6)=d. For definiteness weshall henceforward consider A aninereasing function.{ Letusconsider twodifferent valuesx,and x,inthe interval. (a,6). From the definition ofanincreasing function it follows that ifx,<x, and y,=f(x,), %=F(,), theny,<y,. Hence, totwo Y%%©%difierentvaluesx,andx,therecorrespond twodifferent valuesofthefunction, y, Fig.65. andy,.Theconverse isalsotrue: it Ww<% ¥,=F(x,), andy,—f(x,), then fromthedefinition ofanincreasing function‘itfollowsthatx,<x. Thus,aone-to-one correspondence isestablished between thevalues ofxand thecorresponding values ofy. Regarding these values ofyasvalues ofthe argument and the values ofxasvalues ofthe function, wegetxasafunctionofy: x=9). @ ‘An Inverse Function and ItsDiferentiation 9% This function iscalled the inverse function ofy=f(x). Itisobvi- ous toothat the function y=f(x) isthe inverse ofx=<(y). With similar reasoning itispossible toprove that adecreasing function also has an inverse. Note 1.Westate, without proof, that ifanincreasing (or de- creasing) function y=f(s)is continuous ontheinterval (a,6],whereI(a)=c, [|(b)=d,thentheinversefunction isdefined andiscontinuous onthe interval {c,d]. Example 1.Given thefunction y—x'. This function isincreasing onthe infinite interval —c»<x<; ithasaninverse function x=j/y(Fig.68). Ttwill benoted that the inverse function xg (y)isfound bysolving the equation y=/'(e) for x y yo3hg a eng 7 7 v —>x’ 7”yotng Daa ; 47 ¥ ¢ Fig. 66. Fig. 67. Example 2.Given thefunction y=e*. This function isincreasing onthe Infinite interval ao <x <i. Ithas an inverse x=iny. The domain of definition ofthe inverse function isO<y-<o (Fig 67). Note2.Ifthefunction y=f(x)isneitherincreasing nordecreas-ing onacertain interval, itcanhave several inverse functions.*) Example 3. The function y=s" isdefined on an infinite interval ecco. lis neither increasing nor decreasing and does nothave‘aninverseTunction.IfweconsidertheintervalO<x<'co,thenthefunctionhere,isincreasing andx=Vyisilsinverse. Butintheinterval —0<2<0 thefunctionisdecreasing anditsinverseiss=—Vy(Fig.68). Note 3.Ifthe functions y=f(x) and x=@(y) are reciprocal, their graphs arerepresented byasingle curve. But ifweagain “)Lettbenotedonceagainthatwhenspeaking ofyas9function ofxwe have inmind that yisasingle-valued function ofx. 96 Derivative and Diferentiat denote the argument ofthe inverse function byx,and thefunction byyand then construct them inasingle coordinate system, we willgettwodifferent graphs. gyrty Ttwill readily beseen thatthegraphs will besymmetric about thebisector of thefirst quadrantal angle. Example 4.Fig. 67 gives the graphs ofthefunctionye(ots—tny) and’ileinverse a) wer] yatingy which seconsidered” inExample2 9 *Letusnowprove atheorem thatper- mits finding thederivative ofafunction Fig.68. y=F(x) ifweknow the derivative of the inverse function, Theorem. Ifforthefunction y=10) ay there exists aninverse function x=) @) which atthepoint under consideration yhas anonzero derivative '(y),then atthecorresponding point xthefunction y=f(x) has aderivative f'(x)equaltorathatis,thefollowing formula istrue > 1 fO=Fm: (XVI) Thus, thederivative ofone oftwo reciprocal functions isequal to unity divided bythe derivative ofthesecond function forcorre- sponding values ofxand y.*) Proof. Differentiate, with respect tox,both sides ofequality (2), taking yasafunction ofx**): 1=9' W)ye *)When wewrite /"(3) ofyyweregard xastheIndependent variable whenevaluating thederivative; butwhenwewrite9!(y)orx),weassume that yisthe independent variable when evaluating the derivative. It,should benoted that after differentiating with respect to9,asIndicated ontheright sidofformula (RVD. f(a)mutbesubatated Tory. **) Actually, here we findthe derivative ofafunction ofxdefined implicitly bythe equation, xy) =O ‘AnInverse Function and ItsDifferentiation 7 whence ‘4 ata Noting that y,=f’(x)wegetformula (XVI), which may also be written as yyoo y=foo wae 8 The result obtained isclearly illustrated geometrically. Consider the graph ofthe 4 function y=f (x)(Fig. 69). This curve will also bethe graph ofthe function x=@(y), where xisnow regarded asthe function andyastheindependent variable. Take some 9/8 @ x point M(x, y)onthis curve. Draw atangent tothecurve atthis point. Denote byaandBFig.69. theangles formed bythe given tangent and the positive directions ofthex-and y-axes. On the basis ofthe results ofSec. 3concerning thegeometrical meaning ofaderivative we have I(x)=tana,\ . 8 9”W)=tanB. ® From Fig.69itfollows directly thatifa<-, then a Butifa>, then,asisreadily seen,B=9t—a, Hence, in anycase tanB=cota, whence tanatanB=tanacota=1, or ° 1ana arg. Substituting theexpressions fortanaand tanBfrom formula (3), weget yeahlO-eG: A-a9e0 * Dertoative and Diferenttat SEC. 14. INVERSE TRIGONOMETRIC FUNCTIONS AND THEIR DIFFERENTIATION 1)The function y=aresinx. Let usconsider the function x=sing ay and construct itsgraph bydirecting they-axis vertically upwards (Fig. 70). This function isdefined inthe infinite interval —co<y<+oo, Over the interval # —F<y<F, thefunction x=sinyisincreasing and itsvalues fillthe in- terval—1<x<1. Forthisreason, the jy-aresinx.-function x= siny has aninverse which is denoted by a. Jee y=are sinx.*) \ This function isdefined on the inter-F! val—1<r<l, anditsvalues fillthe xesiny interval —}<y<4.InFig.70,the Fig. 70. graph ofy=arcsinx isshown bythe heavy line.Theorem 1.Thederivative athefunction arcsinx isequalto i vrei ke, ify=aresin x,theny’=— +. (XV) 2 viz Proof. On the basis of(1)we have xj=cosy. Bytherule fordifferentiating aninverse function, ntYerscosy but cosy=Vi—sin'y=VI—x, ye benotedthatthefamiliar equationy=arcsinoftrigonomet: ‘8atotherayofwriting(I).Here(ara'given2)ydenotesthe(talityof Values ofangles whose sine 1Sequal to2. ° Inverse Trigonometrte Functions %9 therefore, yoo, =a the sign infront ofthe radical isplus because the function y=aresinx takesonvaluesintheinterval —<y<7,and,consequently, cosy>0. Example 1.y=aresine®, pag neve te“Timer vi-*" Example 2. yo(sesin)\, 1a iy 14 =2aresin +—1__ (1) 2~2aresint$ —1_. vracesols) wesleyae 2)Thefunctiony=arecosx.Asbefore, weconsider thefunction 7 x=cosy (2) andconstruct itsgraph withthey-axis extending G3upwards (Fig. 71). This function isdefined on the infinite interval—co<y<+oo. On the ae interval O<ye<a, the function x—cosy is decreasing andhasaninverse thatwedenote |}, y=arccosx. Mrecosy This function isdefined on the interval Fig. 72 —I<x<l. The values ofthe function fill the interval x>y>0. InFig. 71, thefunction y=arccosx is depicted bytheheavy line, _Theorem 2.Thederivative oftheJunction arecos.xis—rS ie, 1= {=m 1 ify =arccosx,theny’vw (XVUD), Proof. From (2)wehave xy=—siny, . 100 Derivative andDiderentiat Hence aa ee eyHe sing Vinee But cosy=x, and so p 1 BO Te Insiny=VT—cos¥y theradicalistakenwiththeplussign,since the function y=arccosxisdefinedontheintervalO<y<a and, consequently, siny>0. Example3.y=are.cos (tan2) ; i . 1 1 nn cre 3)The function y=arctanx, dey Weconsider thefunctionCe eed x=tany @) ayFtany and construct itsgraph (Fig. 72) This function isdefined for all ——S—— values ofyexcept y=(2k+ 1)8calire (h=0,ely2...) Onthe2 interval—3-<y< $thefunctiongp xmtanyisincreasing andhasan inverse: amyAetany y=arctans. =o |=~ Thisfunction isdefinedonthe-#) interval —oo<x<oo. Thevaluescara ofthefunction filltheintervalig. 72. x x " . —f<y< 4.InFig.72;the graph ofthefunction y=arctanx isshown asaheavyline, jTheorem 3.Thederivative ofthefunction arctanx is35 ify=arctanx, theny’=ta. (XIX) Proof. From (3)wehave sod 1a Inverse Trigonometrie Functions 101 Hence y=b=costy but ay u : 008!Y=eery=THTay since tan y=x, weget, finally, ao Y=Tee Example 4.y=(arctan.Hastanaaretanay=Ateretanat ha. 4)Thefunctiony=arecotx. yConsider thefunction Sn |)ee x=coty. (4)“ ° Peary Thisfunction isdefined forall 5 valuesofyexcepty=kx(k=0, 1,==0—-}t--= £2). The graph ofthis function isshown inFig.73.Ontheinterval NjcacoenO<y<x, the function x=coty is decreasing and hasaninverse: a y=arccot x. 3xncotyConsequently, thisfunction isde______-2|, ~~~fined ontheinfinite interval —oo< <x<oo, and its values fill the Fig.78. interval m>y>0. Theorem 4.The derivative oftheJunction arccotxistai ie. ifyarecotx,theny’=— >t. (XX) Proof. From (4)wehave 4aco] Hence a 2 1 1 Y=—SINY=—eG —Treaty * 102 Derivative andDiferentiat But coty=x. Therefore 0 1 we TE SEC. 15. TABLE OF BASIC DIFFERENTIATION FORMULAS Let usnow bring together into asingle table allthe basic for- mulas andrules ofdifferentiation derived inthepreceding sections. y=const,y’=0. Power function: yar, fsox"; particular instances: aa 9=V%,=e at, yahpag ame Trigonometrie functions: y=sinx, y’=cosx, y=coss, y'=—sinx, ot y=tans,=a, yrcotx, y'=—she. Inverse trigonometric functions: y=aresinx,‘7S yasreeess, Yah, y=arctanx, ve y=arccot x,yYe—the Exponential funetion: y=a", y'=a* ing; Parametric Representation ofaFunction 103, inparticular, yae, ymer. Logarithmic function: y=log,x, y=+log,¢; inparticular, y=Inx, vat. General rules for differentiation: y=Cu(x), y’=Cu' (x)(C=const), y=u+o—w, y’=u'+0'—w', yaw, y=u'v+uv’, y=4, yates, y=fu), \P Q w=) Hemfeu)(2), y=ur,youu! +uP! Inu. pifY=F@) xo), whereFandgarereciprocal funetions, then el _P= gigwherey=lOe). SEC. 18. PARAMETRIC REPRESENTATION OF AFUNCTION Given two equations: x=9(0, \ yao, 0 where ¢assumes values that lieintheinterval [T,, T,]. Toeach value of£there correspond values ofxand y(the ‘functions @ and 1pareassumed tobesingle-valued), Ifone regards the values ofxand yascoordinates ofapoint inacoordinate xy-plane, then toeach value of¢there will correspond adefinite point in theplane. And when ¢varies from 7,to7this point will de- scribe acertain curve. Equations (1)arecalled parametric equations ofthis curve, fis the parameter, and parametric istheway the curve isrepresented byequations (1). 104 Derivative andDiferential Let usfurther assume that thefunction x= (6)has aninverse, t=®(j). Then, obviously, yisafunction ofx; y=P[O(x)]. @) Thus, equations (1)define yasafunction ofx,anditissaid that ‘thefunction yofxisrepresented parametrically.__Theexplicit expression ofthedependence ofyonx,y=f(x),isobtained byeliminating theparameter ¢fromequations (1).Parametric representation ofcurvesiswidelyusedinmechanics. Ifinthexy-plane thereisacertain material pointinmotion and ifweknow thelaws ofmotion oftheprojections ofthispoint onthecoordinate axes, then z=0(0. | r <y=¥0) wy aN where theparameter ¢isthetime. Then equations(I’)areparametricequationsof wethetrajectory ofthemoving point. Eli- minating fromtheseequations thepara- ymeter ¢,weget the equation ofthe trajectory intheformy=f(x)or ———- F(x,y)=0. Bywayofillustration, let or¥€® ustake thefollowing problem. Fig. 74. Problem. Determine the trajectory and point“ofimpact ofaload‘dropped iroman aitplanemovinghorizontally ‘withvelocityopalanaltitudeYo(ait reslafance’isdstegar ded) Solution. Taking. scoordinate system asshown inFig. 74, we assume that the airplane drops the load atthe instant itcuts they-axis. Itis obvious that the horizontal translation ofthe load will beuniform and with constant velocity 24! eal Verticaldisplacement ofthefallingloadduetotheforceofgravitywillbe expressed Bytheforma . eay sat Hence the distance ofthe load from theground atany instant will be yan. The two equations xan, pu The Equations ofCertain Curves inParametric Form 105 will bethe parametric equations ofthe trajectory. Toeliminate theparame- letwendthevaluef= tromthefrstequation andsubstitute iinto thesecond equation. Thenwegettheequation ofthetrajectory intheform eel903 ‘This isthe equation ofaparabola with vertex atthe point M(O, ys), the y-axis serving asthe axis ofsymmetry ofthe parabola ‘We determine the length ofOC, denote the abscissa ofCbyX,and notethat‘theordinate ofthispoint’ isy=0. Putting these values intothe preceding formula, weget omy — Exart whence xu B.@ SEC. 17, THE EQUATIONS OF CERTAIN CURVES INPARAMETRIC FORM Circle. Given acircle with centre atthecoordinate origin and withradius1(Fig.75), Denote by{the angle formed bythe x-axis and theradius tosome point ‘M(x, 9)ofthe citcle. Then the coordinates ofany point on. the eircle will beexpressed interms ofthe parameter ¢asfollows: xereost,paraats Josectn Thesearetheparametric equation ofthecircle. Ifweeliminate, thepara. meter ffrom these equations, wewill have anequation ofthecircle contain- ing ‘only xandy. Squaring the parametric equations and adding, weget yer!(costsin) or steer, Ellipse. Given the equation ofthe ellipse Ethan o Set reacost, @ Putting this expression into equation (1),weget y=bsint. @ The equations xeacest,prbiag}ostst ® aretheparametric equations oftheellipse. 106 Derivative xd Diferentet etwtddouttegeomet meaning offheparameter Dustocd Siecle Ferg shat doe. sisi cntesathecoordinateorigiandwigadcandao, y y >“e aLRrr) EN TERS Fig. 75 Fi. 76 circle withthe same abscisa asM, Denote by¢theangleformedbythe isis Og" with theata From te Saute tis abrtiy tn cOP—acont this ieqution @ CQ=bsint. From(2)weconclude thatCQ; inotherwords, thestraight lineCM metaely tsSquations 2)Isanangleformed bythecadeOBand 1ea tne antl aaa eened SFeect nates a GN 5 F rer cyeteld. The eycold ts cure denecibed by»pet iiag enthe ccmferenteofacircleitthiscirclerollsuponastraight.linewithoutslidingtees cree Tutter lien Sen theDont Mofheang caneAR Peel cepa eer eatee eyeat 1 IE cee fe Meads ete coiling Sees woe eres reea?ae +=0P=08-~PB, but since the circle als without sliding, wehave 08=Bet, PB MKasiat.Hence,emat—asint=e (sla. The Equations ofCertain Curves inParametric Form 107 Further, Y=MP =KB=CB—CK=a—acos!=a(t—cos). The equations x=a(t—sint),poeta, }ost are the parametric equation ofthe eyeloid. As{varies between 0and 2x ihepia Aoi describe oneateofhecei.minting thepaameer firmtheTater equations, wegetxasanctiealofydirectly.IntheintervalO<¢<z, thefunctiony=a(1—cos t) Substituting theexpression forfinto thefirst ofequations (3),weget =aarecosS—4—a sin(arecosS=# xeaarecost toan(seconSt) or x=aarecos“4—Y2ay—H# when0<x<0. Examiningthegurewenoethtwhenna<x-<2a rataa—(aate co=!—VBy=F) . Xtwill benoted that the function s=a(t—sin hasan inves, butitisnotexprsile intems,ofelementary functions.Andsothefunction y==/(x) isnotexpressible intermsofelementary Note 1.The cyclotd clearly shows that incertain cases, tis more con- veoten to!une the parame equations forsdying Tetons and curves thane the lve reatonship‘of and (a8 alunelon ol2otFast function of9) Astrold. The astroid isa curve represented by the following. parametric sre Yoctcan © Raising the terms ofboth equations tothepower 2/8and adding, weget 108 Derivative and Diflerentiat q therelationship between xandy 8 ;2\\ aybeat Cy Va ateron(Sec.12, )itwil shown:fF tiaGivetthettmsownin ig.78.Itcanbeoblainedasthetrajector olscertain pointon.thecircumference of acircle ofradius a/4 rolling (without Hiding) upon another circle ofradius a (hemalicelealwaystemainyside Fig.78. Note2.Ttwillbe'noted thatequations(4)andequation (6)definemiorethanone functiony=/(x).Theydefine twocontinuous functions ontheintervaiwecr<ya.Onetakesonnonnegative values,theothernonpositive values. SEC. 18. THE DERIVATIVE OF AFUNCTION REPRESENTED PARAMETRICALLY Let afunction yofzberepresented bythe parametric equations =O(1)£28) cect, Let us assume that these functions have derivatives and that the function£=9(0)hasaninverse{=O(s), whichalsohasaderivative. “Thenthefunction yf3)definedbytheparametricequationsmay‘beregardedasa compositefunction: y=, (=O), {being the intermediate argument Bytherule fordifferentialing acomposite function weget HeH4f=VOOC. ® From thetheorem forthe differentiation otan inverse function, itfollows that j ©!(x)=. *ee) Putting this expression into (2), wehave ee a0) 0 =o ot : w=, (x1 The Derivative ofaFunction Represented Parametrically 109 The derived formula permits finding thederivative y;ofa function represented parametrically without having tofind the expression ofyasadirect function ofx. Example 1.The function yofxisgiven bytheparametric equations xmacost,pres r}O<tcem. Findthederivative 44:1)foranyvalueoff;2)fort=. Solution. + _(asing’ __acostDg=GR=Ss cotts ; x 2(Wi),ano f=1. Example 2.Find theslope of#tangent totheeycloid r=a(t—sing, y=a(1—cosf) a4anarbitrarypoint(Ot2m), Sotution. Theslope ofatangent ateach point isequal tothe value of thederivative yjatthatpoint; i.e.,itis Mt aoe ad But ; . sma(l—cost), ymasint. Consequently, 1cost sin£cos£ oeasin Tt ty(3-5 “aes yal 2 atlgz).sty Hence, the slope ofatangent tos cyclold atevery point isequal to tan($—), where¢1sthevalueoftheparameter corresponding tothispoint.Butthismeansthattheangleaoftheslopeofthetangenttothex-axisis taualto1 lorvalueoftyingbetween —nanda “Indeed, theslope1sequaltothetangent oftheangleofnclination<ntaefotheene,Andotanta(F-$)atonZ-$x forthosevaluesof¢forwhichTazliesbetween 0andx, 10 Derivative and Diferentiat SEC. 19, HYPERBOLIC FUNCTIONS Inmany applications ofmathematical analysis weencounter combinations ofexponential functions oftheform4(e*—e~*)and y("+e7%).Thesecombinations areregarded asnewfunctions and are designated asfollows: i enesinh x=Soe| cosh xate| The first ofthese functions iscalled the hyperbolic sine, the second, thehyperbolic cosine. These functions maybeused todefine twomorefunctions: tanhx=22%andcothx=Sh: tanhx=SS3—the hyperbolic tangent, —* 1)cothx=5—thehyperbolic cotangent. ” The functions sinh x,cosh x,tanh xare obviously defined for allvalues ofx.But the function coth xisdefined everywhere, except the point x=0. The graphs ofthe hyperbolic functions are given inFigs. 79, 80, 81. From the definitions ofthe functions sinh xand cosh x[formu- las (1)} there follow relationships similar tothose between the appropriate trigonometric functions: coshx—sinh*x=1, @cosh(a+6)=coshacoshb-+sinhasinh6, @sinh(a-+6)=sinhacoshb-+-coshasinhb. eB) Indeed, . tye (SHEE (fmensyt coshx—sinht 2m(SHE) (Et) PDEepe =bebe ee a Further, noting that orecosh(a+6)=e : Hyperbolic Functions m weget a2,> ab.2coshacosh6-+sinhasinhb=SETete”5oethet EO ent td pert eth getd gtob geredate pepeepete eee peneaH =cosh(a+6). y Theprove issimilar forrelation ‘A 8’). af Thename“hyperbolic functions”Sfx comesfromthefactthatthefunc-HStionssinhtandcoshtplaythe Lykdsameroleintheparametric \representation ofthehyperbola, MY eoy=1, 0 x yrcothx —-———L-— > Fig.79. lo x av a——5—— -----4--s7 7Taanhx ¥ —_——— Fig. 60. Fig. 1. asthetrigonometric functions sin¢and cos¢dointheparametric representation ofthecircle,at+yt=l 2 Derivative and Diflerentiat Indeed, eliminating theparameter ¢from theequations x=cost, y=sint, weget x-+y'=cos*f+ sin't or xt+y'=1 (the equation ofthe circle). Similarly, theequations x=cosht, y= sinht aretheparametric equations ofthehyperbola. Indeed, squaring these equations termwise and subtracting the second from the first, weget xt—y'=cosh"f—sinh? Since, onthebasis offormula (2), theexpression onthe right side isequal tounity, wehave eaytal, which isthe equation ofthe hyperbola. Letusconsider acircle with theequation x*+y*=1 (Fig. 82), Intheequations x=cost, y=sin, theparameter¢isnumerically equal tothecentral angle AOM ortothedoubled area SofthesectorAOM,sincet=25. A y G ft Isinht sint sae x Fig. 82. Fig. 23 Letitbenoted,without proof,thatintheparametric equations ofthehyperbola,x=cosht,y=sinhf, theparameter fisalso numerically equal tothe doubled area of the“hyperbolic sector” AOM (Fig. 83). The Diferentiat 13 The derivatives ofthehyperbolic ‘functions are defined bythe formulas (sinhx)’=coshx, (tanh) =e, (cosh.x)’ =sinh, (cothx)’= —ates (xxm which follow from thevery definition ofhyperbolic functions; for instance, forthefunction sinhx=“=!wehave (sinhay’=(25) =S$=coshx SEC. 20, THE DIFFERENTIAL Letthefunction y=/(x) bedifferentiable ontheinterval (a,6). The derivative ofthis function atsome point xof(a,6]is determined bytheequality in epdimgeal’). AsAx—+0, theratio4%approaches adefinite number /'(x)and, consequently, differs from thederivative J’(x)byan_infinitesimal: Mal (x)+a, where a—-0 asAx—0. Multiplying allterms ofthelatter equality byAx, weget Ay=/' (x)Ax+ adx. 0) Since inthegeneral case f’(x)0, foraconstant xand avariable Ax—0, the product f’(x)Ax isaninfinitesimal ofthe first or- derrelative toAx. But theproduct aAx isalways aninfinitesimal ofhigher order relative toAx, because im°9¥—limaceBnaeEnon Thus, theincrement, Ayofthefunction consists oftwo terms, of which thefirst is[when /’(x)40} theso-called principal pari of theincrement, and islinear relative toAx. The product /’(x)Ax iscalled thedifferential ofthefunction and isdenoted bydyor dj(x) (read, dyordfofx). m4 Derivative and Differential Andsoilafunctiony=/(x)hasaderivative['(x)atthepoint x,theproduct ofthederivative f’(x) bythe increment Axofthe argumentiscalledthediferential ofthefunction andisdenoted bythe symbol dy: dy=f" (x)Ax. (2) Find thedifferential ofthe function y=.x; here, y=(=1, and, consequently, dy=dx—=Ax ordx=Ax. Thus, thedifferential dx‘ofthe independent variable xcoincides with itsincrement Ax. The equality dr=Ax might beregarded likewise asadefinition ofthe differential ofanindependent variable, and then the fore- going example would indicate that this does notcontradict thede- finition ofthedifferential ofafunction. Inanycase, wecanwrite formula (2)as dy=f' (x)dx. But from this relationship itfollows that a) 4re=%. Hence, the derivative f’(x) may beregarded asthe ratio ofthe differential ofafunction tothe differential ofthe independent variable. Let usreturn toexpression (1), which, taking (2)into account, may berewritten thus: Ay=dy-+abx. @) Thus, the increment ofafunction differs from the differential of afunction by aninfinitesimal ofhigher order relative toAx. If F(x)#0, then aAx isaninfinitesimal ofhigher order relative to dy’and imYt lim 84% <1 4lim¢foediag =1+fim payee |+impay For this reason, inapproximate calculations one sometimes uses theapproximate equality Ay dy (4) or,inexpanded form, F(x+Ax)—F(x)mf(x)Ax, 6) thus reducing the volume ofcomputation. The Differential 5, Example 1.Find the differential dyand the increment ayofthefunction vee 1)for arbitrary values ofxand ax;2)for=20,x=0.1Solution. 1)Ay=(e-4Ax)t—24eAr+Ass,dy=(eh)x==2eae,2)Mtx=20,gx=04,theny=2-20-0.1 4(0.1)=4.01, dy=2-20-0.1 =4.0, Replacing ay by dy yields an error of0.01. Inmany cases, itmay beconsidered. smallcompared”toAy=4.01andthe- * teloredisregarded ° ax oti Bhai» clearpicture oftheabove 77ER777EA axrobles LInapproximate calculations, onealso Ymakes useofthefollowing equality, which V4 isobtained from (5): ES ferayeitr aa. © Y Example 2.Let /(x)=sinx, then f'(x)=cosx. Y nelsthtcasetheapproximate equality (6)takes < sin(++ax)slaxposxAs.0Pig88 Let uscalculate the approximate value of sin 48" pe ite, 46%45?$1 4.Substituti Putradia Z, gemita rh, A6tmase bite E45,.Substituting snto (7)wegetni FsoogBE sin4orsin(4) sinFtcos2S, or V3, Vix sin46=1?V20.7071+0.7071-0.017=0.7194. Example 3.Ifin(7)weput x=0, ax=a, weget the following approxi- mate equality: sinaa, Example &Mffe)—tans, thenby(@),wegstthefollowing approximateequality 1 tan(x-+Ax)=tana+SoOtforx=0,ax=a,weget tanaxe. Example 5.Iff(x)= Vx,then(6)yieldsoe 1Vitae Vata 16 Derivative and Diferential Putting x=1, Ax=a, wegettheapproximate equality ViFexi+ya The problem offinding thedifferential ofafunction isequiva- lent tothat offinding the derivative, since, bymultiplying the latter into thedifferential oftheargument wegetthe differential ofthefunction. Consequently, most theorems and formulas perta- ining toderivatives are also valid fordifferentials. Let usillustrate this. The differential ofthesum oftwo differentiable functions uand »isequal tothesum ofthedifferentials ofthese functions: d(u+v)=du+dv. The differential oftheproduct oftwodifferentiable functions u and oisdetermined bytheformula d(uv)=udv+0du. Byway ofillustration, letusprove thelatter formula. Ify=uo, ‘then dy=y’Ax=(wo’ +0u')Ax=uo'Ax+0u'Ax, but o'Arado, u’Ax=du, therefore dy=udv+vdu. Other formulas (for instance, the formula defining the differen- tial ofaquotient) are proved insimilar fashion: ity=4,thendy=2%=eee, Let ussolve some problems dealing with calculating thediffe- rential ofafunction. Example 6.ystants, dye?tanxhede. 1 1 Example7.y=VI+Inx,4mre Wefind theexpression forthedifferential ofacomposite function. Let y=f(4), w=9X), Ty=MlEHb The Geometric Significance oftheDifferential ur Then bythe rule forthe differentiation ofacomposite function, Ce BaFue Hence,. dy=fu(u)9(x)dx, but9’(x)dx=du, therefore dy=f" (u)du. Thus, thedifferential ofacomposite function has thesame formasitwouldhaveiftheintermediate argument weretheindependentvariable. Inother words, theform ofthedifferential does not de- pend onwhether theargument ofafunction isanindependent va- riable orafunction ofanother argument. This important property ofadifferential, called invariance oftheform ofthe differential, will bewidely used later on. Example 8Givenafunction y=sinVF.Finddy.Solution. Representing thegiven function asacomposite one: y=sinu, w=Vx, we find | ax; dymeosu rae, 1 but57demdnsowecanwrite dy=cosudu or dy=cos(Vx)d(V%). SEC. 21, THE GEOMETRIC SIGNIFICANCE OF THE DIFFERENTIAL Let usconsider the function y=) andthecurveitrepresents (Fig.85). Onthecurve y=f(x), take anarbitrary point M(x, y),draw a line tangent tothecurve atthis point anddenote byathe angle *) which theline tangent forms with the positive direction ofthe x-axis. Increase theindependent variable byAx;then thefunction *)Assuming that thefunction f(x) hasafinite derivative atthepoint x, wegetawt. 18 Derivative and Diferentiat willchange byAy=NM,. Tothevaluesx+Ax, y+Ayonthecurve y=}(x) there will ‘correspond thepointM,(x-+Ax,y-+Ay) From the triangle MNT we find NT=MN tana; since tana=/' (x), MN=Ax, wegetNT=f’(x)Ax; but bythedefinition ofadifferential /’(x)Ay=dy. Thus, NT=dy. The latter equality signifies that thediferential ofafunction f(x), which corresponds fothe given values xand Ax, isequal tothe y lay“Y ayoye ye dui “| ofLa|xwax x a xxt OX Fig. 85. Fig. a6. increment intheordinate ofthelinetangent tothecurve y=[(x)atthe given point x. From Fig. 85itfollows directly that M,T=by—dy. Bywhathasalreadybeenproved,“7—+0asAx—-0. ‘One should not think that the increment Ay isalways greater than dy.For instance, inFig. 86, Ay=M,N, dy=NT, and Ay<dy. SEC, 22, DERIVATIVES OF DIFFERENT ORDERS Letafunction y=f(x) bedifferentiable onsome interval (a,6}. Generally speaking, thevalues ofthederivative /’(x)depend onx, which istosaythat thederivative ’(x)isalso afunction ofx. Derivatives ofDifferent Orders 9 Differentiating this function, we obtain the so-called second derivative ofthe function f(x) The derivative ofafirst derivative iscalled aderivative ofthe second order orthe second derivative ofthe original function and isdenoted bythesymbol y*orF(2): ¥=UY =F. For example, ify=x, then yf=5x4; of=(6x4) =202 The derivative ofthesecond derivative iscalled aderivative of iethirorderorYeethirddeviating andIsdenoted by¢”orme Generally, aderivative ofthe nth order ofafunction f(x) iscalledthederivative (first-order) ofthederivative ofthe(n—1)storder and isdenoted bythesymbol y®or(x): y=") =7 (The order ofthederivative istaken inparentheses soastoavoid confusion with theexponent ofapower.) Derivatives ofthe fourth, fifth, and higher orders are also denoted byRoman numerals: y!¥, y¥,yl, ...Here, theorder ofthe derivative may bewritten without brackets. For instance, ify=." theny’ =5x', y=20x', "=60x", y!¥ay" =120r, yY=y= 120, yaya. =0.Example1.Givenafunctionye"(k—const). Findtheexpresionofits derivative ofany ofder 1Solution,y=ke,y=Ate, Yabo, Example 2.y=sinx, Find y. Solution. ysconxmsin (14Z), a—sinz=sn (424), y=—corsmain (x43), Masiaxmsin(2445), ymasin(enF). 120 Derivative andDiferentiat In similar fashion we can also derive the formulas for the derivatives ofany order ofcertain other elementary functions. The reader himself can find the formulas for derivatives of the nth order ofthe functions y=2", y=cosx, y=Inx. The rules given intheorems 2and 3,Sec. 7,are readily generalised tothecase ofderivatives ofany order. Inthis case we have obvious formulas: (utoym=uq 0,(Cuy=Cu™, Let usderive aformula (called the Leibniz rule, orformula) that will enable ustocalculate the nth derivative ofthe product oftwo functions u(x) 9(x). Toobtain this formula, letusfirst find several derivatives and then establish thegeneral rule forfinding the derivative ofany order: y=u0, ysuwotu’, yfsu'otu'y’ +u'o'uv"=u+2u'0"fue",yf"=u''0-+ uo!+2u'0'+2u'o'pu'v"+uo"= =u" +3u'o’ +3u'v" tun”, YWuly+duo"+6u'o"+duo”+u0!¥, The rule forforming derivatives holds forthederivative ofany order and obviously consists inthe following. Theexpression (u+v)" isexpanded bythe binomial theorem, andintheexpansion obtained theexponents ofthepowers ofiand varereplaced byindices that are theorder ofthederivati- ves, and thezero powers (u’=v'=1) intheend terms ofthe expansion are replaced bythefunctions themselves (that is, “derivatives ofzero order"): y=(uo)™=ue+nao! EESH yOmrgr 4.uo™, This isthe Leibniz rule. Arigorous proof ofthis formula may beperformed bytheme-thod‘ofcomplete mathematical induction (inotherwords, toprove that ifthis formula holds for the nth order itwill hold for the order n+ 1). Example 3,y=et*st. Find thederivative of Solutton. Waa, ofa2e, Differentials ofVarious Orders a weak, m2, at=atet*, om olV=,,,=0, y= +na! De+Aatte or y= e%(ax? 42na™~'x +n(n—1)a"~*), SEC. 23. DIFFERENTIALS OF VARIOUS ORDERS Suppose wehave afunction y=f(x), wherexistheindependent variable. The differential ofthis function dy=f"(x)dx issome function ofx,butonly thefirst factor, f’(x),can depend onx;thesecond factor, (dx), isanincrement oftheindependent variable xand isindependent ofthe value ofthis variable. Since dyisafunction ofxwehave the right tospeak ofthe differen- tial ofthis function, The differential of the differential ofafunction iscalled the second differential orthesecond-order differential ofthis function and isdenoted byd'y: d(dy) =d'y. Letusfind theexpression forthesecond differential. Byvirtue ofthe general definition ofadifferential wehave dy=f’(x)dx}dx. Since dxisindependent ofx,dxistaken outside thesign ofthe derivative upon differentiation, and weget d'y=f" (x)(dx)*. When writing thedegree ofadifferential it,iscommon todrop thebrackets; inplace of(dx) wewrite dx*tomean thesquare oftheexpression dx;inplace of(dx)' wewrite dx’, etc. The third differential otthe third-order differential ofafunction isthe differential ofitssecond differential: dy=d(dy) =[f (x)det de=f" (x)de’, Generally, thenthdifferential isthefirst differential ofadiffe- rential ofthe (n—1)st order: dty=d(d""y) =[f°(x)da"dx,dy=Faya". (1) 122 Derivative andDiferentiat Using differentials ofdifferent orders, thederivative ofany order may‘berepresented asaratioofdifferentials oftheappropriateorder:+Gail! re 1m(gya re=8 re... Mme=Z. @) Itshould, however, benoted that equalities (1)and(2)(forn>1) hold only forthecase when xisanindependent variable.*) SEC. 24, DIFFERENT-ORDER DERIVATIVES OF IMPLICIT FUNCTIONS AND OF FUNCTIONS REPRESENTED PARAMETRICALLY 1,An example will illustrate the finding ofderivatives of different orders ofimplicit functions. Let animplicit function yofxbedefined bytheequality eg£+8—1=0. a Differentiate, with respect tox,allterms oftheequation and re- member that yisafunction ofx: 24BtReo from this weget dy__ otg-—a (2) Again differentiate this equality with respect tox(having inview that yisafunction ofx): wy yw ae ee Substituting, inplaceofthederivative $2,itsexpression trom (2),weget oe ey Wayrs yo or, after simplifying,cae ty___oNatyt+bist)cr *Nevertheless, weshallaiowriteequality @)when. leagtanIndepen-xpression wua 1. dentvarlable;butinthlscase,theexpression £4,,,,,°4shouldb regarded assymbols ofderivatives. Different-Order Derivatives ofImplicit Functions 123 From equation (1)itfollows that aty! +bit =atb'; therefore thesecond derivative may berepresented as ty Bo ay Differentiating thelatterequation withrespect tox,wefind#4, ete. 2.Letusnowconsider theproblem offinding thederivatives ofhigher orders ofafunction represented parametrically. Let’ the function yofxberepresented byparametric equations x=9(t),\i<st<T; (3) y=, J ® thefunction x=@(f) has aninverse function t= (x)ontheinter- val[f,,T). InSec,18itwasproved thatinthiscasethederivative %is defined bytheequation ay ay_ a2-5: @) a Tofindthesecond derivative, £%,differentiate (4)withrespect tox,bearing inmind that ¢isafunction ofx: dyn dy, ay_a (a \_a (at\aea-4(8)-2(2)&- 6) ai a but dgae(49)ty4(42)aeddate di\ de )~ ax? ~ dx\* .ar (@) (zi) ded Sx. @ wo Dertoative and Diferentiat Substituting thelatter expressions into (5),weget deaty_ayateay_dda ax dx\* .(a) This formula may bewritten inmore compact form asfollows: dy_ gvOv Ov Og at ieOF . In similar fashion we can find the derivatives ay ty ae at : and soforth. Example. Afunction yofxisrepresented parametrically: seacost, y=Oslal, teat dydty Findthederivatives 2, 4, Solution. 45ast;Ymacm a *ae ‘ dy, day arco Shoosat dy_beostfyteeta ay_(asin) (—bsin)=(0cos9(—acos)_ obI ae(=asia0 SaSint SEC. 25. THE MECHANICAL SIGNIFICANCE OF THE SECOND DERIVATIVE. Letsbethepath covered byabody under translation asa function ofthetime; itisexpressed as s=f(0. mn ‘Aswealready know (see Sec. 1,Ch. III), thevelocity vofabody atany time isequal tothe first derivative ofthe path with respecttotime: on @a Atsome time ¢,letthevelocity ofthebody bev.Ifthemotion isnotuniform, then during aninterval oftime A¢thathaselapsed since¢thevelocity willchange bytheincrement Av. The Mechanical Significance oftheSecond Dertvative 128, The average acceleration during time Afisthe ratio ofthe increment invelocity Avtothe increment intime: ae a=45. Acceleration atagiven instant isthe limit ofthe ratio ofthe increment invelocity tothe increment intime asthe latter approaches zero: =lim4°; amfimae inother words, acceleration (at agiven instant) isequal tothe derivative ofthevelocity with respect totime: ae a=Z, butsincev=$$,consequently, 4 (ds) _d's a=4 (4)=, ortheacceleration oflinear motion isequal tothe second deriva. tive ofthepath covered with respect totime. Reverting toequation (1),weget a=F(). Example. Find thevelocity vandtheseceleration aofafreely falling body, ifthedependence ofdistances upon time ¢isgiven bythe formula sepattotts, Cc) where g=9.8 misect Istheacceleration ofgravity, and s4—se» 1sthevalue osat (6 Solution. Differentiating, wefind as vagnat ted C) from this formula Itfollows that v= (ites Differentiating again, wefind ante‘at an=8 Let Itbenoted that, conversely, ifthe acceleration ofsome motion is.con- stant, and equal tog, the velocity will beexpressed byequation (4), and thedistance byequation (3)provided that (pag and(shiesto 126 Derivative and Dierentiat SEC. 26, THE EQUATIONS OF ATANGENT AND OF ANORMAL, THE LENGTHS OF THE SUBTANGENT AND THE SUBNORMAL, Let usconsider acurve whose equation is y=F0), On this curve take apoint M(x, y,)(Fig. 87) and write the equation ofthe tangent line tothegiven curve atthe point M, assuming that this tangent isnotparallel totheaxis ofordinates. The equation ofastraight line with slope &passing through the pointMisoftheform y yah(e—x). TF00 Forthetangent line(seeSec.3) Meus k=F (x), and sotheequation ofthetangent A pf theformWP % yn=F) ex). Inaddition tothe tangent toa Fig. 87. curve atagiven point, one often has toconsider the normal. Definition.Thenormaltoacurveatagivenpointisastraight line passing through thegiven point perpendicular tothetangent atthis point. From thedefinition ofanormal itfollows that itsslope &,is connected with theslope k,ofthetangent bytheequation 1 baz “ 1 k=Pa: Hence, theequation ofanormal toacurve y=/(x) atapoint M(x, y,)isoftheform 1 yn pe Example 1.Write theequations ofatangent and anormal tothe curve gaze the point M(1,1)‘Solution. Since y’=3:, theslope ofthetangent IsW/)sa1=3. Therefore, theequation ofthetangent is y—1=3 1) oyar2. The Equations ofaTangent and ofaNormal wz Theequation ofthenormal is 0 1 y-t==3 1) or14 yee page ty 40) (sce Fig. 88). Thelength Tofthesegment QM Omni +7 (Fig. 87) ofthe tangent between the point oftangency and the x-axis iscalled thelength ofthe tangent. The projection ofthis ‘ segment onthe x-axis, that is, QP, iscalled thesubtangent; the Tenght ofthesubtangent isdeno- PAPtedbyS,.Thelength Nofthe tg88.segment MRiscalled thelength ofthenormal, while theprojection RPofthesegment RM onthe x-axis iscalled the subnormal; the length ofthe subnormal is denoted bySy. Letusfind thequantities 7,S,,N,Syforthe curve y=/(x) and thepoint M(x, y,).From Fig.87itwillbeseenthat nH OPay,cotam fem, therefore 7 w Pe lysence tofa+8-[; Ver¥i|. Itisfurther clear irom this same figure that PR=y, tana=yyi, and so Spe lyuils NaV b+Guay, Va|. These formulas arederived ontheassumption thaty,>0, yi>0. However, they hold iathegerieral case aswell. 128 Derivative and Differential Example 2.Find the equations ofthetangent andnormal, ‘the.lengths ofthetangent.and.thesubtangent, LEN the.Tengths ofthenormal "and a subnormstfortheellipse Ne,U reacost, y=bsint (1) atthepoint M(x, y,)forwhich Fig.89.vs =Gig.09) Solution. From equations (1)wefind de .ay, dy__ob ‘dy a+Ba-asins Havens Yatra (Hfance Wefind thecoordinates ofthepoint oftangency ofM: @ ° naw=e, neu) ata. aeVEOO VE The equation ofthe tangent is i-+(®-y3) yea TOV, orbxtay—ab VE=0. The equation ofthe normal is —ee 87a7s(*-72) or (ax—by) V2—at+0*=0. The lengths ofthesubtangent and subnormal are ® Vi|_« Sra| Ye) _2_. *iev2 sw-[te(~2) [este wl 7a(-2)are The lengths ofthetangent and thenormal are as _|VzVLay hee VaTTTE |(-2)41lpgVar aayaAwelpeV1+(-2)[rapeVR The Geometric Significance ofthe Derivative 129 SEC. 27, THE GEOMETRIC SIGNIFICANCE OF THE DERIVATIVE OF THE RADIUS VECTOR WITH RESPECT TO THE POLAR ANGLE Wehave thefollowing equation ofacurve inpolar coordinates: e=/(H). (yy Let uswrite the formulas for changing from polar coordinates torectangular Cartesian coordinates: y x=Qcost, y=esind Substituting, inplaceof@,itsexpression 7intermsof0fromequation (1),weget 4p" x=/(0) cost, . A y=F(0)sind, @ “A Equations (2)areparametric equations of9| * the giveri curve, the parameter being the Fig. 90. polar angle 6(Fig. 90). Ifwedenote by@the angle formed bythe tangent tothe curve atsome point M(g, 0)with the positive direction ofthe x-axis, wewill have ay _4y_ atang= t=O a or sind+0c0s0tang=j,———_- @)$8cos0—gsind a Denote byptheangle between thedirection oftheradius vector and thetangent. Itisobvious that w=o—6, tong=tae—tandTrtangtan0° Substituting, inplace oftang, itsexpression. (3)and making thenecessary changes, weget tampa'0/3180-400s.)cos0—(@"cos0—esind) sind_ an8=(&cos0—¢sinB)cosO-F(@’sinOF¢cos)sind—Q” or = ecotp. (4) 53008 130 Derlvative andDiferential Thus, thederivative oftheradius vector with respect tothe polar angle isequal tothelength oftheradius vector multiplied bythecotangent ofthe angle between theradius vector and. the tangent tothe curve atthegiven point. Example. Toshow that the tangent tothe logarithmic spiral eae intersects the radius vector ataconstant angle Solution. From the equation ofthespiral weget ete=aet*, From formula (4)wehave cotp=& ma;thatis,»=arccota=const. Exercises onChapter It Findthederivatives offunctions usingthedefinition ofderivativepyaat.Ans,Sh2yet.Ans,<4.3yeVF.Ans. : wy oeaie geUSVE. We 1 1 AyekyAns. Le.5.yasinte,Ans,sinecose,6.y=2at—x.aE} aye! ’ Determine thetangents oftheanglesofinclination oftangents to,the curves! 7.g=2"- a) When r=1. Ans. 3.b)When re—I. Ans. 3.Make a drawing. 8.y=4. a)Whenx=-p. Ans,—4.b) Whenxm.Ans,1. 1 3 wing.8.y=VEwhenx=Ans. Makeadrawing.9.y=VFwhenx r¢]Findthederivativesofthefunctions: 10,yoatSet,Ans.y's6e, TeyeGh—atAns,YelBtte. 2yee An.ym Sxt__ Oe eeoedl tne aSee ESE ansve. yetar EeAns.yma. 18.ymetparpoe,Ans.yf2k+ Ey 1 vs 11 10x* $2.16. y= Vets. Ans. Y= +ap-a- +lox*+! yaVRFT YEty.Ans.avaaya3B a ey ax Exercises onChapter II 131 val_m,2eOnt 3-0WF. padt AnsyfehBE 9,ymFIVE 46.COa 1 ot 6 Ve 5ot3ytmk.ww.yeMt Eas,yada dt Ve YVR VE VE got geet 1st+geee 2A,y=(LpAe8)(14204), Ans.yf4x(1443x4108), 22,y=x(2x1)Gx+2), Ans, y’=2(Qxt42—1). 28. y=2x1) 643). Ans. y=6xt—2c+12, “ah264 a—z Mhype.AasfAOESosyaEEaneva _t patO+th yatta 2O=TeaAns.=Teme ah1SEEans.FO)=aED6+4 ott ns,peeDODro ee es CO dr 2,wae Ans.yee 30.y=et~3)2, Ans.of=8x(2et—3). 31,ya(xtta%’. Ans.y=l0e(x* +a. 32y=x —_ a—ir =VEER Ans.¥=A. 98.yale)Vom.Ans.y=o, FaAns.f=pres BBymteteV eyfr THE ——— =2a) | HyeVE. ons.vy ae eee AEE a8VEEL tnfeBE ahpm ee ott y * —smuti Ans.¥(t4y) yonAVES. An.y= 1 1 1-—SS [+—ai('tr73)]- 3.yasiotx,Ans, yar 2VetVa\ 2V% yresin2e, 40. ye2sinx-+eose, Ans, y'—=2eose—Ssinde. 41. yas @ sins 1 stan(arb).Ans.Ymca MeyeSEES.Ans.ym. 43, yosin2e-cosSe. Ant. y’=2cos2xcos3e—3sinQesinde,44,y=cot*Sx,Ans.y'=—10cot52esc?5x.45.y=tsint-+cost. Ans,y’=teost.4.y=slotcost.Ans.y'=sint¢(3cost—sint1).47.yaaVED. Ans.f= tanScot= asin2x 12 -sasintZcos® grey =FER6meat$Ans,rpmscont, ym—2| Ps 132, Derivative andDifferential econeins(an5cotF) : An.f= a ee ws= u7 j fx. =Incosx. tesiot5cos.St.yeetants.Ans.yfemtanesectx 62yal Ans, of=—tanx.63.ymintane, Ans.yma?pe.Shyminsinh,Ans.y’= tans—1 , exes =2eotx.55.ed Ans.y=sinx+cosx.56.y=inVee. 1 Rye joc.58.yesineta Ans.fmaig.Shyerintan(FHF). dns,yey.68yosinetayXco8(r-ba).Ans.ymcos2(r-+a).” 69.J(x)—sin(ins). Ans.Pix)= =D) 6.payetan(ing.Ans.forSP).61Jay—sincos0. 1tant ar “ Ano,P(x)—sinx08C0501.62rmtanto—tangtg. AnsFmtantg 63. [(x)m(xcotx)tAns.f(a)2xcotxeotx—xesetx). 64y=In(ax+OF tat. yet = ere OC el a ear SeMex 2 jaBemcos ante ans.yg. 67ymlogyasin. Ans.oeeS Let 4 241 63.yanteAns.ymes68.ymin(xt).Ansvont ata? 10.yin20-45). Ans.Ymagers. Thymeins,Ans.y=lax tayetots, Ans,yfRE.1a,yoinkTER,Ans.mt ra! Vz _! TAymtndng). Ans.y=she.15.fay—inVTLS.Ans.Fd= Veyi—x 2 — 7 f(x) Ina, Ans. f(x)— Th.yaVOPR Ve=ite s View , atVaree esd Tee mainPEER ns,yVEEP,79,yinereVEROLEEE Vryat cose * oa Ans,yaVOFE ye SBEGSintanSd.Ans.veges sine Ltsintx 1eat , foyeh. Ans.ofaESE. giyeytanttincose,Ans. Stents, 8 yee Ans. ymae, BL ymete Ans, yfadele, BALgma™. Ans.2ca**Ina,88.yeTF. Ans,yf=2x+1)7*™In7,es an vi 18 T mel",Ans.ya—Duc"Ine,81ymae”A,Ans,ye—SamoF 86.y y e.yar yVa" Exercises onChapter 111 133 a’.Ans, ’ma'ina. 8% 10%png,atatIna_gina, areat,Ans,Patina. 8raal™*,Ans,Sal”RE gleaming, el .‘2e* 80yetaa,Ansmet2a9,91yehAns.PE. al 1 aurea ayang Ane verb. heb eT oF,aney= meeeT,ongenetAns.yecose05.yea,Ans,om =na"™"™™ sectnxina. 96.y=e*sinx. Ans.y’=e°*(cosx—sin® x). 97.y=e*Insinx. Ans. y’=e* (cotx-tinsinx). 98.y=x"e"™*, Ans. y= =x" (ntxcosx). 99.yax®. Ans.y’'=x¥(Inx+l1). 100.ya? anegee(ISIE),songes,Anemeetin tyme ns.fmettinn 10,y=(£)™AneyanZ)"(I4inZ). Bougeentes,Aneyfartes(MEpnecesx)8y(nAng =(sinx)¥ (Insinx-+-xcot x).106.y=(sinx)"*°*, Ans.y’=(sinx)'*°4 ae Z x(1+sectx Insinx).107,gatanoS.Ans.v=-7oe—eres cosVI-Re ane Ly:=ins.y’=—>2*In2.109.y=10%Ans.y= 108,yn YT=B.Ans.y=SEY ’ as.=ri0(torte). tuncfitd thederivatives ofthefunctions afterfirst’takinglogarithms ofthese anctions ees 1(FT (1Qe 2 ayeVEEP am VIER(Stith) - uit,gE og,ySEVERR(2048 Vu-3 Va—3 a4 Tea 2 +t ng, ye EENGHMe$5) saa): 88rmaggtegge aeveSS Vey =161x*+480x—271 MSye ns.y= r 5Ve Va OVeV a—9@ 14 Derivative and Diflerentiat 1ayeEEAnsyaR215, yaa(0-434)"(@—20)8 Ans.y!=aaa? x 1=5et(a$n%(a—2e)(at2ax—12),116,yoaresin.Ans.y=Seta+Ne2y0+ A.116.ya:An=S 117,y=Goresins).Ans.y= ©MB,yearecot(FHI).Ans.g'= y=(aesinx) yVYios y G+). y Ea ci ad cos(x! STEER TR:NMymarecot Ey.Ans.yaptay.120.ymarecos(e%. .= arccosx oeet VT=arecosx) dasyfmR. tmyeBEE,ans,y=SEE Rass AVangpet x 122,yoresin1Ans,yh 123,yasVm+ataresin® . ysare Vz “Visa y + ry Ansm2VER tye VORpaaresin Ans.ym ra arccat 22 du .ee 1 sV3 fares aearccotYEans,yeBEL tnyaraesing. Ans.of= Tae An vee 1hymearesin y x hlsesine+PEOB/la)matecos (in.Ans.PO)—es cone 1.[a)saresin Vang.Ans.f=Et, Ud 00Vama—ante , T=cosx 1 arecotx, i =srccotVattocx<m Ans 1h,yet Ans.y= ee iyyecatccot ee.Ans.yapPep.188.yometetine Tea1ymarccot SEAns,ymaaw.1089 : Ans,yfaxtresins (ESN «134, gare (sinx).Ans.y'= y (StEa): vescsto inn y eaf08EJFLinIstand4thquadrants. 359cop452Angyo TeeesT7{21te2dant30quadrants,BSA tgeoeagAnd 4 a ‘=a 2a!<sparey: Aymarccot +10VE Ans.yf=a. 17y= L4x\F_1 ot m=! -win(J#)"—artina, daegeri tan98S! TP 41 jet et Faretans,Am.fm 1.ymginES+pgarctan Exercises onChapter U1! 135 a! 1eeVBat V2pnypatV3 ns.yeah. a,yminEVEL 4oartan22ans,yt2, AnsfmegehyMO.yonBEYEESpaartanEU,ans,ym md Qn|x|"J.peacecosEhAns,28 MALy=are008Tey oT) Differentiation ofImplicit Functions 44, roan av_20 4ytect a a Ay aeergatytmatot,AnsLEE 145,gtByb20r Anshm—EatetpatyteatotAnsamE18,gtyb20rd. dy_a yer wiiVt iF AnsHae. M6ttytmetAns,He VE.a ere dyVz dy_iy *a?. Ans.= 2.M8,y*—2ey +0*=0. Ans.2a—4_, te ae * Pott ae pyt—tay=0, Ans,Yae=* cos (e-ty: Ans,Mo Me,tty —aaynd. AnsHE, 150,ymconte ty.Ans,___sineto) stays, Ans,YaLtvsinGy) T¥sin &+9)" tet.castp) aadx xsin(xy)* Find#offonctions repressted parametric: 182,xeacost;yabsint,Ans,abooth,158.ema(t—sind:y= ns.Lemcotte.164,xmacoetyyobs? ay aatimeonn, Ans Memeo £.154, xmacostt: ybaintt. Ans,Mm ° at Baty dy Ot .maton 185reMs yeBO. Ans,am 188,wm2incots: vatinscots. Showthat#ntan2s Find the tangents ofangles oftheslopes oftangent lines tocurves: : 1 3 157.x=cosy=sindatthepoints=—4,y=V3,a2drawingL a 3 Ans, L158, x=2008,y=sin¢atthepointx=1;y=—3,Ma wn¥ p gat. Mate0 1 x awing. Ans. 1 159, x—a(t—sinf), y=a(t—cos) when f=. Midrawing.Ans.75 (sin,ya(cost) F.Make4drawingAns.1,100,x=acos!f, y=asia’(whent=.Makeadrawing. Ans,—1.161.Abodythrownatanangleato,thehorizon (inairlessspace)described@curve,‘undertheforeeolgravity,whose’equationsare!=> 136 Derivative andDigerentiat meycotat, proper Blgn98apes).Knowingtheta6",==50msec,determine thedirectionofmotionwhen:1)m2sec;2)t=7sec.Makea.drawing.Ans.I)tang,0.88,9,43°30")2)tang=— 1.019,ei Find the differentials ofthe following functions: 162. y=(at—24, Ans. dy=—0x(at—2hds. 168,y=VIFFAns.dy= idxaa zing “jeara Ans,dymsecteds, 165.y=2IE4 neds Hina)Ans.dy=fede, Calculate the increments and differentials ofthe functions: 166. y=2x?—x when x=, Ax=0.01. Ans. Ay=0.0302, dy=0.03. 167. Gi- ven y=#42e. Find Ayand dywhen x=—lya#—0.02.Ans.Ay—0.098808,sins. Finddywhenx=, are. Ans.dy= dy=0.1, 168.Given y=sins. Find dywhen x=%,gx=. Ans.dy =£=0.00873.169.Knowingthatsin0°=30.066005;cos6=,find theapproximate valueofsinOU"andsin0°18- Compare therenulls withtabular data.Ans.sin60°3’=0.866461; sin60°18’=0.868643.170.Findthe approximate.valueoftan48°4'90"."Ans.1.00282,"171.Knowing. that jogy200230103findtheapproximate valueoflogy200.2.Ans.330146.Deilvatives of different’ orders, 172. “yoode—et-+Ox—i. Find Ans.18x—4.173.y=VR,Pindy"Ans.ABx6,174,yeast,Findy.ns115yn$.FindysAns,MOENE16,ymYARPid A 6 Ans,———". 171.y=2Vx.Findy,Ans.——7=. 178.y= waves 9 f ava 4 satpoxte. Findyf.Ans.0.179.f(x)=In(x1). Findf'Y(x).6 . ent sect =Insi Ans.—bape180.ytanFindy",Ans,sects—4see.18.y=nsins, Find. y*. Ans. 2cotxcsc!x,182.(x)=Vee.Find,f(a).Ans.(2)= SSR). 188.y=PEL.FindMe).Ans.GAL. 164.p= 24a arctan2,Find£2.Ans.At abet Het stetteyarctan. Find£4.Ans.tt. 185.yerHe), rind2.ans, 8hyeeosar, Findg%Ametco(aren ) 187. yak. Find. Ans, (Ina")a. 188, y=in(Its). Find y, —jy(= a! Bindy.Ans.2-0"2 Ans.(yt!De!16m,yetZE.Rind9.Ans.20—0"Gar 190, yaetx. Find y! Ans. ef(eta). 191, gaettInx, Find Exercises onChapter II! 137 Ans2M192,ymstatx.FindAns.—2"~"cos(2e-+-F). 108.y-—ssiox, Findy.Ans.xsin(x+Fn)—n cos(x4m).194.Iymetains, prove that24’+2y=0. 195.y*=dar. Find£4.Ans-<196,itatyt mete, FindTHand$4.Ans. are SER. on.att ytert.Find Gh.Ans.5.198.try=0.Find$4.Ans,0.190,@artan(9-+0). #0 ans,—25+80F-+ 30% secpeore=C. Find42 Find$9.An grtAe)00.secpeongac. FindFE. tanto—tan*@ so4ex4xind22.Ans,Lette 0) AnsRg Weteety.FindFe.Ans.OKT E) 2 = tans, —20, xa ((— 202.gttat—Sary=0. Find£4.Ans.—PH. 200.x=0(sind, y=a(l—con 9,Find$4.Ans... 204,x=acos2t, y=0sintt itose(F) :z ci xeacost, yasin, Find£4.Ans,—Soo8e Show that$4=0. 205. 4,yrasint, Find £4.Ans, oe, oe anes 206,Showthat25,(sinhx)=sinhx;Aor(sinhx)coshx. Equations ofaTangent and Normal. Lengths ofaSubtangent and aSubnormal 207.Writetheequations ofthetangentandnormaltothecurvey=x*—Aitx4.5atthepoint'm(32),Ans.Thetangent16,Ge—y-—22%0; the normal, x4-8y—19—~0, 208. Find the equations ofthe tangent and normalGF"theengioftheebtangentandsubmormaloftheicestgtonrtatthe pointM(x,,y,).Ans.Thetangentisxx,+yy,=r%; thenormalisx,y—yx=0;un 209,Show thatthesubtangent oftheparabola y*=4px atanypoint is divided into two bytheverlex, and thesubnormal 1sconstant and equal to 2p. Make'a drawing.210.FindtheequationofatangentatthepointMi(xy,yy): ana 2 a)Totheellipse +femt. Ans.SMa Fe Ans,MWe byTothehyperbola %3—Yr=t, Ans, tM a1, 188 DerivativeandDiferentiat 211,FindtheequationsofthetangentandnormaltotheWitchofAgnesyogitalthepointwhere220Ans.Thetangent24-2¢4athe normal ipeedt—3e. 212, Show thal the normal tothe curve Sy=Gr—Se* drawn tothe point a1(ik)posesteunesorte 2Stowtatetant totheere(£)"+ (Ea atepoi M(a,0)is244—2, 214. Find theequation ofthat tangent tothe parabola, yt=20e, which forma anangle of45" with thetanige Ans. yaa {at the point (S10) is.Pngineequations ofthosetangents oheciee 4fF,whieh areparalleltothestraightline2x-+3y==6. Ans.2x-+3y£260.Bor Find theequations ofthose tangents tothehyperbola 4:*—9y*m=96, platareperpendicular tothestraightline2y+5x=10. Ans.Therearenoue tangents‘217.Showthatthesegment(lyingbetweenthecoordinate axes)ofthetangent tothe hyperbola xy=m {divided into two bythe polnt oftangency Sis. Prove thet thesegment (between thecoordinate ake) ofa:Tangest totheauleroid 24pmaT inofconstant length 218.ALwhatangle@dothecurvesy=o*andyo=b*intersect?Ans.lange lta—inbTeIna-Inb” 220. Find thelengths ofthesubtangent, subnormal, tangent and normal ofthecycloid x=a(8—sin@), y=a(1—cos@) atthepointatwhich6-5. Ans. sp=4; sya; T=a V3 N=a V2. al. Find the quantities sp,sq,7and Wforthehypocycloid «—=4a‘os', pedasiats, Ans, sps—Aasiotf cost;ty=—tall;TeasettsW==asiatant, Miscellaneous Problems sor 1 Findthederivatives ofthefollowing functions: 222,y=piit— x ed aa aresin.Ans.y= wintn($—$) daeaachg.thpane!Anspte tose 2 meyecainginn, Am, feSEH. ms, gatax eera! ; xacetan(7/23tan5)(2>0,0>0)Antiymarpeaee. BEylel 5 — a Ans. yar. mat. yaatesin VTS, Ans, feSy rece YVR Exercises onChapter Itt 19 28 From theformulas forthe volume and surtace of«sphere, owed! andster Wfaoe hatae. Espa thagmat salar oftheotin asimilar rlatioratip between thearen of-ctle and the length oftheeltanton ! SH5°Th atriangle ABC, the sde as expresed interms ofthe othe two sider 2st he ae ten en Oytie Toma onVIREOTCA, For 6and ¢constant, side @isafunetion ofthe angle A.Show thatseoywerehythealaeofthelangecomesondag tthehae erprt this result geometrically. Bb,Uaeheerent cocet, dere thegn theapprox a Y=PPPS ats, fetbLatsy tery |08amunber small compared with 9 Tal.'tut jelod alcncllation'aa pendulum iscomputed bythe formula resVE. z Incaleuating the period 7,how will theeror beaffected byanero of1% 1pfeemensefement of 1)he leat ofthe pendca' Ee" sceleration Ogavlty gsAne aie: eSae Binfeant hahepaper itfranyoiloftthe semen ofthe tangent Tremaine cdntan” inength, Prove ths ohhe base lt 1)theequation ofthe tractrix inthe form ee co) raVI=P+E inVEE a>0} VERGO Vea Oe 2)theparametie equations ofthe curve xea(inten £40), ymasint 233. Prove that the function y=Ce*+C,e~™ satisfies the equation 145g 4b (hee OFand Gee cnaani) Bling pores eelcostprovebeequalities fme,=2. 238.Prove “thatthefunction y—sin marcia) satistes the‘equation (Byaymy 2e . 235,Provethat(aebapeFme,tanhan($b) CHAPTER IV SOME THEOREMS ON DIFFERENTIABLE FUNCTIONS SEC. 1.ATHEOREM ON THE ROOTS OFADERIVATIVE (ROLLE'S THEOREM) Rolle’sTheorem. /fafunctionf(x)iscontinuous onaninterval {a,6]and isdiferentiable atallinterior points ofthis interval, and vanishes [f(a)=f(b)=0] attheend points x=a and x=6) then inside (a,6)there exists atleast onepoint x=c,a<c<b, atwhich thederivative f'(x)vanishes, that is,f'(c)=0.*) Proof.Sincethefunction (x)iscontinuous ontheinterval (a,6], ithas amaximum Mand aminimum mon this interval. IfM=m the function f(x) isconstant, which means that for allvalues ofxithasaconstant value f(x)=m. But then atany point oftheinterval f'(x)=0, and thetheorem isproved, Suppose M%m.Thenatleastoneofthesenumbers isnotequal to zero. For the sake ofdefiniteness, let usassume that M>0 and that the function takes onitsmaximum value atx—c, sothat i()=M. Let itbenoted that, here, cisnot equal either toaorto6,sinceitisgiventhat{(a)=0, f(b)=0.Sincef(c)isthemaximum value ofthe function, f(c+Ax)—f(c)<0, both when Ax>0 and when Ax<0. Whence itfollows that Hetad—HO <9whenAx>0; a) Heban=HO 9whendx<0. Co) Since itisgiven inthe theorem that the derivative atx=c¢ exists, weget, upon passing tothelimit asAx—+0, limHetanle_7(<0whenAx>0; are timHekAN—LO pf(e)50whendx<0. But therelations f’(c)<0 and f’(c)>0 arecompatible only if '(c)=0. Consequenily, there isapoint cinside theinterval [a,6] atwhich thederivative /’(x) isequal tozero. +)Thenumber¢iscalledtherootofthefunction@(x)if@()=0. ATheorem ontheRoots ofaDerivative (Rolle's Theorem) 14 The theorem about theroots ofaderivative has asimple geo- mettic interpretation: ifacontinuous curve, which ateach point hasatangent, intersects the x-axis atpoints with abscissas aand 6,then onthis curve there will beatleast one point with abs-cissac,a<c<6,atwhichthetangent isparallel tothex-axis. y - ofay-foo erin fauo Oaco Gbx F al raed Fig. 91. Fig. 92. Note1.ThetheoremthatResetsbeenprovedalsoholdsfor adifferentiable function such that doesnotvanish attheend points oftheinterval (a,6],buttakes onequal values f(a)=f(b) (Fig. 91). The proof inthis case inexactly thesame asbefore. Note 2.Ifthefunction f(x) issuch that the derivative does not exist atallpoints within theinterval [a,6],theassertion ofthe theorem may prove erroneous (inthis case there might not bea point cin theinterval [a,6),atwhich thederivative f'(x) vanishes), For example, the function y=f()=1-VR | (Fig.92)iscontinuous ontheinterval (—1,yyandvanishes at theend points ofthe interval, yet the derivative 7 [= Vr within theinterval doesnot.vanish. Thisisbecause thereisapoint x=0 inside the interval atwhich the derivative does not exist (becomes infinite). The graph shown inFig. 93isanother instance of afunction whose derivative does notvanish in’theinterval [0,2] The conditions ofthe Rolle theorem are notfulfilled forthis function either, xbecause atthepoint x=1 thefunction has ' noderivative. Fig.93. we Some Theorems onDiferentiable Functions SEC. 2.ATHEOREM ON FINITE INCREMENTS (LAGRANGE’S THEOREM) Lagrange’s Theorem. Ifafunction f(x) iscontinuous onthe in- terval a,6)anddifferentiable atallinterior points ofthisinterval, there will be, within (a,6),atleast one point c,a<c<b, such that 1(6)—f(@)=F' ()(6—a). () Proof. Letusdenote by@thenumber L4—1), g=L=1o, 2) and letusconsider the auxiliary function F(x) defined bythe equation FQ)=/)—1(@)—@—a)Q ) What isthegeometric significance ofthe function F(x)? First write theequation ofthechord AB(Fig. 94), taking into account thatitsslopeisMQ andthatitpasses through the r point(a,f(a)): A yf(a)=Q(x—a); GgUhwhence lW) y=1(@)+Q—2). 70)ButFis)=f2)—[F(@) +Qe—a)}. 4Thus,foreachvalueofx,F(x)is A equal tothedifference oftheordinates KA \ ofthecurvey=f(x) andthechordoP a y=f(@)+Q(x—a) for points withoo a¢¥°5 *‘thesame abscissa. Fig.94. Itwill bereadily seen that F(x) iscontinuous ontheinterval [a,6), isdifferentiable within this interval, and vanishes at the end points oftheinterval; inother words, F(a)=0, F(b)=0. Hence, theRolle theorem isapplicable tothe function F(x). By this theorem, there exists within the interval apoint x=c such that F'()=0. But F=f )—Q And so F=f ©—Q=0, ATheorem ontheRatio oftheIncrements ofTwo Functions M43, whence .Q=F (), Substituting thevalue ofQin(2),weget {I-10_¢(9, ay whence follows formula (1)directly. The theorem isthus proved, See Fig. 94foranexplanation ofthe geometric significance of theLagrange theorem, From the figure itisimmediately clear that thequantity HH) isthetangent oftheangleofinclinationa ofthechord passing through the “points Aand Bofthegraph with abscissas aand 6, Ontheother hand, /'(c)isthetangent oftheangleofinclination ofthe tangent line tothecurve atthe point with abscissa c.Thus, thegeometric significance of(I')oritsequivalent (1)consists in the following: ifatallpoints ofthe arc AB there isatangent line, then there will be,onthis arc, apoint Cbetween Aand B atwhichthetangent isparallel tothechordconnecting pointsA and B, Now note the following. Since the value ofcsatisfies the condition a<c<6, itfollows that c—a <b—a, or c—a=0(b—a), where 8isacertain number between 0and 1,that is, o<e<i. But then c=a+0(b—a), and formula (1)may bewritten asfollows: £(6)—(a)=(b—a)f’ |a+8(b—a)], 0<0<1. ay SEC. 3, ATHEOREM ON THE RATIO OF THE INCREMENTS OF TWO FUNCTIONS (CAUCHY’S THEOREM) Cauchy's Theorem. //[(x)and(x)aretwofunctionscontinuous onthe interval (a,6)and differentiable within it,and 9"(x)does not vanish anywhere inside theinterval, there will befound, in la,6),some point x=c, a<c<b, such that LO)—Ha) _FeFO=90 “FO o 4 Some Theorems onDiferentiable Functions Proof. Let usdefine thenumber Qbytheequation aLb) 1a)e=F900)" @) Itwillbe,notedthat9(6)—g(a)+0,sinceotherwise(6)would beequal to(a), and then, bytheRolle theorem, the derivative g(x) would vanish intheinterval; butthis contradicts thestate- ment ofthe theorem. Let usconstruct anauxiliary function F(x) =F()—F(@)—Q [9(*)—9 (@)]- Itisobvious that F(a)=0 and F(b)=0 (this follows from the definition ofthe function F(x) and thedefinition ofthenumberQ). Noting that the function F(z) satisfies all the hypotheses ofthe Rolle theorem ontheinterval (a,6],weconclude thatthere exists between aand 6avalue x=c (a<c<b) such that F’()=0. But F’(x)= (x)--Qg" (x), hence F'(C)=F ()—Qg' (= 0, whence foQ=F6° Substituting the value ofQinto (2)weget(1). Note. The Cauchy theorem cannot beproved (asitmight appear atfirst glance) byapplying theLagrange theoremtothenumerator and denominator ofthe fraction 1(8) Ha) 9()—9(a)* Indeed, inthis case wewould (aiter cancelling out6—a) get the formula 16)—1 (a)_P(e) e(b)—e@) gle) Inwhich a<c,<b, a<c,<b. But since, generally, c,#c,, theresult obtained ‘obviously doesnotyetyield theCauchy’ theorem. SEC. 4.THE LIMIT OF ARATIO OF TWO INFINITESIMALS (EVALUATION OFINDETERMINATE FORMSOFTHETYPE.3) Letthefunctions f(x)andg(x), onacertain interval [a,6), satisfy theCauchy theorem and vanish atthepoint x=a of’this interval; {(a)=0 and g(a)=0. The Limit ofaRatio ofTwo Infinitely Large Quantities 148 Theratio£2tsnotdefined forx=a,buthasaverydefinite meaning forthevalues x=£a, Hence, wecan raise thequestion ofsearching torthelimit ofthis ratio asx—a.Evaluating limits ofthis type isusually known asevaluating indeterminate forms ofthetype* We have already encountered such problems, forinstance when considering thelimit lim“°*andwhenfinding derivatives ofele- mentary functions. Forx=0, theexpression “"*ismeaningless; thefunction F(x)="2* isnotdefined forx=0, butwehaveseen thatthelimitoftheexpression 2%asx—0existsandisequal tounity. L'Hospital’s Theorem (Rule), Let the functions f(x) and (x), insome interval, satisfy theCauchy theorem and vanish atsome pointx=a:f(a)=9(a)=0; then,iftheratiofehasalimitasx—+a, therealsoexistslimie,and in£2). =tmLDFn9c ay” Proof, Ontheinterval {a,B]take some point xa. Applying the Cauchy formula wehave ihe _Fe=0@~FO where&liesbetweenaandx.Butitisgiventhatf(a)=@(a)—0, and so to 1®wae" 0 UFx75a,thenEaalso,since&liesbetween xanda.And F a) im i iFTimSG=A,thenlimEGexistsandisequaltoA.Whence itisclear that imLO@tim£@—jimLO dinSeay=tiary=Limgray=A and, finally, jimHt £0Pei tmgray 146 Some Theorems onDifferentiable Functions Note 1.The theorem holds also for the case when the functions I(x) ofg(x) arenot defined forx=a, but limf(x)=0, lim@(x)=0. Inorder toreduce this case tothe earlier considered case, we redefine thefunctions f(x)and(x)attheeintx=asothat they become continuous atthe point a.To dothis, itissufficient toput F(a)=limF(x)=0;9(a)=lim@(x)=0, sinceitisobvious thatthelimitoftheratioLas x—adoes notdepend on.whether thefunctions f(x) and p(x) are defined atx=a. y Note2.Iff’(a)=@' (a)=0 andthederivatives f(x)andg’(x)satisfy theconditions that were imposed bythe theorem on the functions f(x) and g(x), then applying theL'Hospital rule tothe ratioFO),wearriveattheformula tim[4tim£4,and ee" xoaPOevaF&)" so forth, Example 1. smS05 ig(852 tySosSe_5. oe GP ee 88 Example 2. —_ timO42) StimPEE,(arsee a Example 3. timSee tyFER PimEHO timSHE2a 2D, poaee Tatts Taco 0, Sine ets eos 7 Here,wehadtoapply,the,Ltlspital rulethee,times,because, thecatiosalthetatsecond andthiderivatives at=Oyieldtheindeterminate form 2. Note 3.The L'Hospital rule isalso applicable if limf(x)=0 and lim@(x)=0. Indeed, putting rot, weseethatz—+0 asx—+0o and theretore lim1(4)=0 lim(4)=0. The Limit ofaRatio ofTwo Infinitely Large Quantities “7 (3) Applying theL*Hospital ruletotheratio 42 wefind*(7) 1 (4) (1tin12htinGE)oiPCE)(=H) (®) 1 wal T\ steNSE) ee(Z)(TE) 1 tly , =lim(r)_tim£&.,rig (Lytee which iswhat wewanted toprove. Example 4. wotreo(—4) . tinSEmtnNTE imbcosLa. SEC.6.THELIMITOFARATIOOFTWOINFINITELY LARGEQUANTITIES (EVALUATION OFINDETERMINATE FORMS OFTHETYPE=) Let usnow consider the question ofthelimit ofaratio oftwo functions f(x)and@(x)approaching infinity asx—- a(oras x—00), Theorem. Let the functions f(x) and @(x) becontinuous and differentiable forallxa intheneighbourhood ofthe point a: thederivative @'(x)does not vanish; further, let limf(x)=00, lim@(x)=00 and let there bealimit A imLe), Thenthereisalimittaea)and im£2)=timLOHiga=EOgrayA ® Proof. Inthe given neighbourhood ofthe point a,take two points aand xsuch that a<x<a (ora>x>a). By Cauchy's theorem we have Lw=Ha) _1 ®e®)-9@) Fe" 8 Some Theorems onDiferentiable Functions where a<e<x. We transform the left side of(3)asfollows: Ha) te—r@ _to!~Ter @@=9 @)~ 90) we)” 70) From relations (3)and (4)wehave fic) LO_Heo Fey Yo ~em [aera 70) Whence we find 12) te_to‘ee © ea VO (Te Tey From the condition (1) itfollows that foranarbitrarily small 250, @may. bechosen soclose toa that for allx=c where a<c<a, thefollowing inequality will befulfilled: rolra-4l<e or Ane<FOcate. ) Let usfurther consider the fraction 12) nom cl Fixing @insuch manner that theinequality (6)will befulfilled, weallow xtoapproach a.Since f(x)»ooand@(2)—r00 as x—a, we have 12@) ow _fine =!TG) and, consequently, forthe earlier chosen ©>0 (for xsufficiently close toa)wewill have 12) _ {TF1-—~Fal<e Te) The Limit ofaRatio ofTwo Infinitely Large Quantities 49 or 1a) 1mec—#) cite, 0)1H) Te) Multiplying together the appropriate terms ofinequalities (6)and (@, weget ae)fF(c) (x) 4-9-9<F9faSAFO +9)re) or, from (5), (4-919 <b<(a+eylte.Sinceeisanarbitrarily smallnumberforxsufficiently closetoa,itfollows from thelatter inequalities that im1.timian A or,by(1), ion£0)—tim£69— A60)LG iAn which iswhat had tobeproved. Note 1.Ifinpremise (I)A=co, that is, iyEe) then equality (2)holds inthis case aswell. Indeed, from the preceding expression itfollows that imVL)timao Then bythetheorem just proved im22) imFOfimFay=Him,Fray= whence im£2).: simgeo 150 SomeTheoremsonDiferentiable Functions Note 2.The theorem just proved isreadily extended tothe casewherex—oo.Iflimf(x)=00, lim@(x)=0o andtimte) exists, then im£2=tim£1)Oe ® Theproofisperformed byreplacing x=+, aswasdoneunder similar conditions inthecaseoftheindeterminate form$(see See, 4,Note 3). Example 1. ; timStim ©tim Fao. swetaneWYavet Note 3.Once again note that formulas (2)and (8)hold only if the limit’on the right (finite orinfinite) exists. Itmay happen that the limit on the left exists while there is.no limit onthe right. Toillustrate, letitberequired tofind timEES| This limit exists and isequal to1.Indeed, lim£488tim(1+%)=1 But the ratio ofderivatives (e-tsing)_beoss “w= 1 =1+cosx asx—+00 does not approach any limit, itoscillates between 0 and 2. Example 2. timSPtty2=& MN atad elec" Example 3. 1 tanx costx 1costae12-3cos3xsin3x Km,fanse, =3costae="3Deosesing 29Oetg ot rok cosSx sine Ssinde(—1)_g(—) (=). sisoselmsingSine The Limit ofaRatio ofTwo Infinitely Large Quantities 181 Example 4. Jeaain en? Generally, forany integral n>0, inStim tim BORDtoo, The other indeterminate forms reduce tothe foregoing cases. These forms may bewritten symbolically asfollows: a)0:00, b)0%, c)0%, d)1, @)co—co. They have the following meaning.a)Letlimj(2)-—0; lim@(x)00;itisrequiredtofind lim(f)9(a). This indeterminate form isofthetype 0-00. Iftherequired expression isrewritten asfollows: Jimf(x)9(2)]=tim_LO) ee) orinthe form Tim()9(x)=1im22, Tw thenasx—-aweobtain theindeterminate form2or2, Example 6. as img"Inxt,=tim2mim0, ear ar b)Let limf (x)=0, limg (x)=0; itisrequiredtofindlim[Fy] or,aswesay, toevaluate the indeterminate form 0°, Putting y=Fore, 162 SomeTheoremsonDiferentiable Functions take logarithms ofboth sides oftheequality: Iny=@(x)[Inf(x)]. Asx—+a weobtain (on the right) the indeterminate form 0-00. Finding limny, itiseasy togetlimy. Indeed, byvirtue ofthe continuity ‘ofthelogarithmic function, limIny=Inlimy andif Inlimy= 6,itisobviousthatlimy=e>.If,inparticular, b=+00 or—co, thenwewillhavelimy= +00or0,respectively. Example6.Itisrequiredtofindlimx*.Puttingy=2*wefindInlimy==limIny=limin(e*)=1imxin.xy;7" . as times) 2timFine, a og consequently, Inlimy=0, whence limy=e"=1, of lim x*= 1. ‘The technique issimilar forAiding limits inother cases. SEC. 6,TAYLOR'S FORMULA Let usassume that the function y=f(x) has allthe derivatives uptothe (n+ I)th order, inclusive, insome interval containing thepoint x=a. Let usfind apolynomial y=P,(x)ofdegreenot above n,the value ofwhich atx—a isequal tothevalue ofthe function’ f(x) atthis point, and the values ofitsderivatives up tothe nth order at-x—a are equal tothe values ofthe corresponding derivatives ofthefunction f(x), atthis point: P,(a)=f(a), P,(a)=f (a),P,(a)=F(a),«--5PY(a)=f(a).(1) Itisnatural to’expect that, inacertain sense, such apolynomial is“close” tothe function f(x). Letuslook forthis polynomial intheform ofapolynomial in degrees of(x—a) with undetermined coefficients: Py(4)=C,+C,(xa)+C,(x—a)*+C,(xa)"+eee $C, (x—a)", Q) Wedifine the undetermined coefficients Cy,Cy, ...) C,sothat they will satisfy conditions (1). Taylor's Formata 169 Let usfirst find the derivatives ofP,(x): P,(2)=C,+2C, («—a)+3C,(x—a)"+...nC,(x—a)"*,Pi,(x)=2C,+3-2C,(x—a) +...+n(n—1)C, (x—ay""*, @) fiejpgeas descenbebaaphaeSsaadyat Substituting, into theleft and right sides of(2)and (3), the value of.a inplace ofxand replacing, byequalities (1), P,(a) byf(a), Pn(a)=f' (a),etc., weget fa=C,, r@=c, Pa@)=2:16,, f"(a)=3-2-1C,, f(a)=n(n—1)(n—2) ...2-1C,, whence we find C=f@, C=f@. C=ah@, Carga @,ooGap. 4) Substituting into (2)the values ofC,, C,, C,that have been found, we.gettherequired polynomial: Pyley=fla)+72F(a)+=pa)+2SUpray... ES. ) Designate byR,(x) the difference ofthe values ofthegivenfunction f(x)andoftheconstructed polynomial P,(x)(Fig.95): Ry)=F(e)—P (2), whence Fe) =P)+Ry(2) or,inexpanded form, . Heo=Hat Fr@+2S"rat... M+R. © 14 Some Theorems onDiferentiable Functions R,(x)iscalledtheremainder. For av orto thosevaluesofx,forwhichtheix) remainder R,(x) is.small, the= polynomial P.(x) yields anapproxi-Wal) Taterepresentation ofthefunction . F(x). Thus,formula(6)enablesone Fax)toreplace the function y=/(x) by the polynomial y=P,(x) to an appropriate degree of accuracy equal tothevalue ofthe remainder a x x R,(*). ‘Ournextproblem istoevaluate Fig.96. the quantity R,(x) for various values ofx. Let uswrite the remainder inthe form wear! Ry(x)=HEUH), a) where Q(x) isacertain function tobedefined, and accordingly rewrite (6): fe=1@+22@+45"r@+...=a"pm(gy4=I" ° et SM O+aay ee) 6) For fixed xand a,thefunction Q(x) hasadefinite value; denote itbyQ.Let‘usfurtherexamine theauxiliary functionof¢(flyingbetween aand x): FOF OSE ro...(=1"pm(gy= ooASE pmgH, where Qhas thevalue defined bytherelationship (6’); here we consider aand xtobedefinite numbers. Wefindthederivative F’(f):Hiyaty)abr) 4EE! Fo=—rotl oFror ero- ESOItooSOEpmyO pm2=0" pnenyy4ENE=O" aT Ot Gee Taylor's Formula 158 or,oncancelling, . . F(= SSO pny EM Q, ® Thus, the function F(f) has aderivative atallpoints #lying near the point with abscissa a. Ttwill further benoted that, onthebasis of(6'), F(x=0, F(a)=0. Therefore, the Rolle theorem isapplicable tothe function F(/), and, consequently, there exists avalue ¢=§ lying between aand xsuch that F'(E)=0. Whence, onthe basis ofrelation (8), weeet ; =ESB" pen 4SEMQ=0, andfromthis a=") Substituting this expression into (7),weget aEO ney Ra(1)=FSO FOE), This istheso-called Lagrange form oftheremainder. Since £lies between xand a,itmay berepresented inthe form)) §=040(e—a) where 6iganumber lying between 0and 1,that is,O<b<1; then the formula ofthe remainder takes the form Hat nen —Ral)="GeyMt”a+Oa). The following formula He=f)+27@+2S"rat... 2FEET +SIpreya0ay)O iscalled Taylor's formula ofthefunction f(x). IfintheTaylor formula weputa=0 wewill have 1)=O+4F O+EF Ot. my 4 eyweFGFO+qpml (8x)(10) where @lies between 0and 1.This special case olthe Taylor formula issometimes called Maclaurin's formula, +)SeeendofSec.2ofthischapter. 156 SomeTheorems onDiferentiable Functions SEC, 7.EXPANSION OF THE FUNCTIONS ex, SIN x,AND COSx IN ATAYLOR SERIES 1.Expansion ofthefunction f(x)=e*. Finding thesuccessive derivatives off(x), wehave a)=e [O=1, Fae, FO=1 Pr(yme*, PPO =L. Substituting theexpressions obtained into formula (10), Sec. 6, weget soe mynal+tteGtat ++ ape 0<e<i. If|z|<1, then, taking n=8, weobtain anevaluation oftheremainder: 1 Ri< 3: Forx=1wegetaformulathatpermitsapproximating thenumbere: emltligtgyt tai evaluating tothefifth decimal place, wehave e= 2.71827, Here there arefour significant digits, since theerror does notexceed #,oF0.00001. Observe that nomatter what xis,the remainder Remet 0asneo. Indeed, since @<1, theguantity eforfixed xisbounded (it Isless than e*forx20, and less than 1forx<0). Weshall prove that, riomatter what thefixed number 2, wen? asn—+co, Indeed, ae ae eeeleemle|4-2- Fal Ifxisafixed number, there will beapositive integer Nsuch that Ix]<M Expansion ofFunctions e, sin, and cosx inaTaylor Series 187 Weintroduce thenotation l=9;then,notingthat0<q<1, wecan write (for n=N+1, N+2, N+3, ete.): bGtle|f -$-4 ck emla|t-3°3 tl xee x NT newetSTDS a WHT for the reason that lil-a lwnl<o oodlitl<e Butoonisaconstantquantity; thatistosay,itisindepend-entofn,whileg"-“** approaches zeroasn—oo.Andso dimem=© 0) Consequently, R,(x)=e"<2775 alsoapproaches zeroasapproaches infinity. From theforegoing itfollows that forany x(ilasufficient num. berofterms istaken) wecanevaluate e*toanydegree ofaccuracy. 2.Expansion ofthe function f(x)=sin x. Wefindthesuccessive derivatives of/(x)=sinx: Te) =sinx, 1(0)=0, re=cosx=sin(x+F), fO=1, rw==sinxmsin(x+24), fF0)=0, I"(x)=—cosx=sin(x4+34), r@=-1 P*(9)=sinx=sin(x+44), 7(0)=0, pr(y=sin (x05), = sinnS, raya sin(xee4D FZ).eere=sin[E+a+n 4S]. 158 SomeTheoremsonDifferentiable Functions Substituting the values obtained into (10), Sec. 6,wegetan expansion ofthefunction f(x)=sinx bytheTaylor formula: eo sine=x—Ftyp— ee stersinns+Aein[E+@4+ 5]. Since|sin[e+e+n9] [<1wehavelimR,(x)=0 forall values ofx. Letusapplytheformula obtained foranapproximate evaluation ofsin20°.Putn=3, thusrestricting ourselves tothefirsttwoterms oftheexpansion: Pasin tae tot(2)o sin20°=sinFe F—3;(F)0.343. Evaluate theerror, which isequal tothe remainder: 1Ral=|(4)* grsing+2m|<(J)zp—=0.0006<0.001. \ij ' 7 Hi \ an ee,\ gon7 Seoll ‘ 7 I \Zz i \ a2 +t a NE 7 iy, \sine \ Va sage 7 \ , u Fig. 96. Hence, theerror isless than 0.001, and sosin20*=0.343 tothree places’ ofdecimals. Fig. 96shows thegraphs ofthefunction [(x)=sinx and the Exercises onChapter 1V 159 firstthreeapproximations: S,(x)=2; S,(x)=2—; S,()=x— a 3.Expansion ofthefunction f(x) =cos x. Finding thevalues ofthesuccessive derivatives forx=0 ofthe function [(x)=cosx and substituting them into the Maclaurin formula, weget the expansion: cose=IPE... +Hcos(nF)+ +foes[s+e4+n5] . 1El<lel Hereagain,limR,,(x)=0 forallvaluesofx. Exerelses onChapter 1V Verify thetruth ofRolle's theorem forthefunctions: 1.y= x4—3x-+2 on the interval (I,2]. 2.y=a*-+5s!—6x onthe interval (0,I],3.y=(x—1) G-2)(5—9) othe interval I)3)4.yrsiats ontheinterval [0, %.The function f(x)==44°-4'2—ax-—1 has rootsIand—1.Findtheroot ofthe derivative /’(s) mentioned inRolle’s theorem. 6.Verify thatbetween therootsofthefunction y=j/s"—Sx-F6 liesthe root ofits derivative. 7.Verify the truth ofRolle's theorem for the function y=cos*x onthe ne interval[-4.+4]. _ 8Thefunction y=1—j/x* becomes zeroattheendpointsoftheinter val (I, 1},Make itcleat that the derivative ofthis function does not ennsaywhere totheinterval (1,1) Explain whyRotles theorem istot applicable hereaForm Lagrange’s formula forthefunction y=sinx ontheinterval liu tal, Ans. singing, =U—2,)cone,x,<e<Hy 10"Verily thetruth ofthe ‘Lagrange formula forthe function y=2x—x* onthe interval [0,1] M1:Atwhat point Isthe tangent,to thecurve y==2" parallel tothechord from’ point M,(0, 0)toM,(a, a")? Ans. At the point with abscissa wawhatpointithetangenttothecurveyin.paralleltothechordlinkingthepoiksMy(l,0)andAlte,PAs.“Atthe:pointwithabciae cue‘Applying.theLagrangetheorem,provetheinequalities: 13.ef>1-5, rinses) <aSO)1.OAMnb(O—2)forb>.16.are lane <n 10 Some Theorems onDigerentiadle Functions 17,WritetheCauchy formula forthe functions), g(0)=4¥ onthe interval {1,2)andfindc.Ans,cmt Erauate thefollowing Hits:18,timS—" Ane, 9,gmPA Ans.2.20. im8922, Ans, 2.tim£91 Ans, 2. Shean PS coge=T 22,tinSBE. Ans.There1snolimit(VEasx++0,VEInsiax 1 nut a =), 28, tim . Hh. imFH, « 8immaa AMyeMHI Aneng 25, uimz—aesine ang, 1 2gtimSME—SING Ane, cosa. sty inte 3 Baa im28091 Ane9.96,tim8X 1 3e—1 7inae A Spa AgM Ans.790Ngee(wmeren>0).Ans.0.31.UOtagAM CeENEayPOI ORES 82.lim. x).Ans.=.lim2. Ans.fora>0;ofora< canesnie imEEX Ans.1,35,tim2803 Ang1,96,timIntanTx MeUnmere Am ain AM88ITetandns1.37tinBEDE, ans0,98,tmmayanSE.Ane,2 oaFh ® ar 24 i Lis wean[sey-z]- A2:Oe[nace]: Atot 24 1 fitmoepmtngh donmeia[poh]. amb. 43,limxcot2x. Ans. 1.44. umxte®. Ans, @.45.limx¥, Ans. Lea ras a 7 teaVBamet attn(LYM atat(142), Ans.e%49,lim(cotx)™%,— Ans.$.0.im(cos)? Ans.1. Exercises onChapter 1V 161 St.limsee)e. Ans.Lo.52.tim(tant) Ans.4, 58, Expand, inpowers ofx—2, the polynomial xt—5e*+5xt42 42, Ans, 27 (x—2)—(x—2)843 (x—2)"+(2—2)*. ‘54.Expand, inpowers ofx-+1, thepolynomial x*-2x'—x?4e41, Ans. (eee Se HIE DEED55.WriteTaylor'sformulaforthefunctiony=VFwhena=1,n=3. coy gkol L_G=bF 1=n8Gt5| anVret42. gat tae eoaT Ig8x x(e-ID) *,0<0<1. 56.Write theMaclaurin foimula forthefunction y=VIE when n—2. Ans.ViFial+ pegt¢—= 00<1, 16(1+-6x)* 57.Usingthesults ofthepreceding. exercise, evaluate theerroroftheapproximateequalityVTFESL-+4 etewhen#02AnsLesthan1 rie Determine the origin oftheapproximate equalities forsmall values ofandevaluntetheerooftheeequalities:68,Incase2 ed cas Ea 8.tines SE. covwesineeet Sot.actanes x, Seer tt — oaTHE EE63,netVimar. Using Taylor's formula, compute thelimits ofthefollowing expressions: emASH ans.1,65,tigEOED—SEAng, eeeae z imtags Hans,67,tm[eatin(14L4 66.im, = ©Ans.1,67.im n(14t)]. Ans,im(1802) Ang, ae Ae tin(44-824). ans00um(Lente)ans2 6-a308 CHAPTER V INVESTIGATING THE BEHAVIOUR OF FUNCTIONS SEC. 1. STATEMENT OF THE PROBLEM Astudy ofthe quantitative aspect ofnatural phenomena leads tothe establishment and study offunctional relations between the variables involved. Ifsuch afunctional relationship canbeexpres- sedanalytically, that is,intheform ofone ormore formulas, we arethen inaposition toinvestigate itwith thetools ofmathema- tical analysis. For instance, astudyofthefightofashellinempty space yields aformula that gives thedependence oftherange R upon theangle ofelevation aand theinitial velocity v,: vfsin2a,R= (gistheacceleration ofgravity). With this formula wecan determine atwhat angle atherange Rwill begreatest, orleast, and what the conditions must befor therange toincrease astheangle aisincreased, etc. Let usconsider another instance. Studies ofoscillations ofaload ‘onaspring (ofatank orautomobile) yielded aformula showing howthedeviation yoftheloadfromaposition ofequilibriumdepends onthe time ¢: y=e~™ (Acoswf+Bsinaf). The quantities &,A,B,@that enter into this formula have avery definite significance’ for agiven oscillatory system (they depend upon theelasticity ofthespring, the load, etc., butdonotchange with time 1)and for this reason are considered constant. On the basis ofthis formula we can find out atwhat values of tthedeviation ywill increase with increasing f,how themaximum deviation varies asafunction oftime, forwhat values oftwe observe these maximum deviations, forwhat values of¢weobtain maximum velocities ofmotion ofthe load, and anumber ofother things. Allthese questions areembraced bytheconcept “investigating thebehaviour ofafunction”. Itisobviously very difficult tode- termine allthese questions bycalculating thevalues ofafunction atspecific points (like wedidinChapter II).The purpose ofthis chapter istoestablish more general techniques forinvestigating the behaviour offunctions. Increase and Decrease ofaFunction 163 SEC. 2 INCREASE AND DECREASE OF AFUNCTION InSec. 6ofCh. Iwe gave adifinition ofanincreasing and a decreasing function. We will now apply the concept ofthe derivative toinvestigate the increase and decrease ofafunction. Theorem. /fafunction f(x), which has aderivative onthein- terval {a,6}, increases onthis interval, then itsderivative on la,6}isnot’negative, that is,f'(x)=0. 2)Ifthefunction f(x) iscontinuous ontheinterval (a,b|and isdifferentiable on(a,6), where f'(x)>0 for a<x<b, then this function increases onthe interval (a,6}. Proof. Letusfirst prove thefirst part ofthe theorem. Let/(x) increase onthe interval (a,6).Increase the argument xbyAx and consider the relation Pet ax)fe)is : 0) Since f(x) isanincreasing function, F(e+Ax)>F(x) for Ax>0 and fetAx)<f(x)forAx<0. In both cases Hepanalo so, ® and consequently fim[etAI—10) 9 aneOF which means /’(x)0, which iswhat wesetouttoprove. {Ifwehadf'(x)<0,thenforsufficiently smallvaluesofAx,relation (1)would benegative, but this would contradict relationship (2).] Letusnow prove thesecond part ofthetheorem. Let[’(x)>0 forallvalues ofxontheinterval (a,6). Let usconsider any two values’ x,and x,x,<4,, onthe interval (a,6}. Bythe’ Lagrange theorem onfinite increments wehave H)—f&)=l Oy —H) 4<B<Hy Itisgiven that /’(&)>0,hence/(x,)—/(x,) >0,andthismeans that f(x) isanincreasing function. There isasimilar theorem foradecreasing (differentiable) function aswell, namely: Iff(x) decreases onaninterval a,bj,then |’(x)<0onthis interval. Iff'(x)<0 on(a,6),then f(x) decreases on(a,6}.[Of ° 1 tnveligatig he Behan ofFunctions course, we again assume that the function iscontinuous atall points of(a,6]and isdifferentiable everywhere on(a,6).) Note. The foregoing theorem expresses the following geometric fact. Ifonaninterval {a,6)afunction f(x) increases, then the tangent tothecurve y=/(x) ateach point onthis interval forms y y pi 7 (@ (o) fie. anacute angle @with thex-axis or(atcertain points) ishorizon- tal; the tangent ofthis angle isnot negative: f'(x)=tan @>0 (Fig. 97,a).Ifthefunction f(x) decreases ontheinterval (a,6), then the angle ofinclination ofthetangent forms anobluse angle (or, atsome points, the tangent ishorizontal); the tangent of this angle isnotpositive (Fig. 97,6).Wecanillustrate thesecond part ofthe theorem insimilar fashion. This theorem permits judging theincrease ordecrease ofafunction bythesign ofits derivative. Example. Gelermine the domain ofincrewe and decrease ofte function faa Solution, The derivative iequal to ya4e; for25-0 wehave y'> 0and the function tneeaes: HeeS28 ogRave 028 ae nation nets ig my SEC. 3. MAXIMA AND MINIMA OF FUNCTIONS Definition ofamaximum. Afunction /(x) has amaximum at the‘point x,ifthevalue ofthefunction f(x) atthepoint x,is greater than itsvalues atallpoints ofacertain interval contain- ing the point x,Inother words, thefunction f(x) hasamaxi- Maxima and Minima ofFunctions 165 mum when x=x, iff(x,+Ax)<f(x,) forany Ax(positive and negative) that aresufficiently small inabsolute value.*) For example, the function y=f(x), whose graph isgiven in Fig. 99, hasamaximum atx=x,. Definition ofaminimum. Afunction f(x) has aminimum at x=r, if F(x,+Ax)>F(x) forany Ax(positive and negative) that aresufficiently small in absolute value (Fig. 99). For instance, the function y=a* considered atthe end ofthe preceding section (see Fig. 98) has aminimum forx=0, since y=0 when x=0 and y>9 forallother values ofx. y y ext g q yx ay ee ¥ Fig. 98. Fig. 99 Inconnection with the definitions ofmaximum and minimum, note thefollowing. 1.Afunction defined on an interval can reach maximum and minimum values only for values ofxthat liewithin the given interval. 2.One should not think that the maximum and minimum ofa function areitsrespective largest and smallest values over agiven interval: atapoint ofmaximum, afunction hasthelargest value only incomparison with those Values that ithas atallpoints sufficiently close tothepoint ofmaximum, and thesmallest value *)Thisdefinition issometimes formulated as-follows: thefunction ji) has ‘amaximum atx, ititispossible tofind. aneighbourhood (a, )ol 2G<4)<8)Suchthtforallpointsofthisneighbourhood diferenttrom Hitheinequality (4)</(x)isfulfilled. 166 Investigating theBehaviour ofFunctions only incomparison with those that ithas atallpoints sufficiently close tothe minimum point. Toillustrate, take Fig. 100, which shows afunction defined on the interval [a,6),which atx=x, and x=x, has @maximum; atx=x, and x=x, has aminimum, but the minimum ofthe function atx=x, isgreater than the maximum ofthe function atx=x,. Atx==6, the value ofthe function isgreater thananymaxi- y mum of the function on the interval under consideration The generic terms for maxima and minima of afunction are extremum (pl. extrema) orextreme values ofthe function, Tosome extent, the extrema of afunction and their positions on the interval (a,6)characterise the oe %variation ofthe function versusCTT changes intheargument. Fig100. Below we give amethod for finding extrema, Theorem 1.(Anecessary condition for the existence ofanextremum). ifafthepointx=x,adifferentiable function y=|(x) eslear tac derivative vanishes althispoint: "(x)= 0. Proof. For definiteness, letusassume that atthepoint x=x, the lunction has amaximum. Then, for sufficiently small (in absolute value) increments Ax(Ax%0)' wehave Pj +Ax)<f(x), that is, fo,+49-1(4)<0. But inthis case thesign oftheratio Ja, ban—lex) ‘ar isdetermined bythesign ofAx, namely: Martan=H8) 0whenAx<0 tetant) <9whenAx>0. Maxima and Minima ofFunctions 167 Bythe definition ofaderivative wehave Peaystio,Heta0) | Iff(,) hasaderivative atx—x,, the limit onthe right is independent ofhow Ax approaches’ zero (remaining positive or negative). But ifAx—+0 and remains negative, then f(x) <0. But ifAx—+0 and remains positive, then P(e)=O. Since f’(x,) isadefinite number that isindependent oftheway inwhich Axapproaches zero, the latter two inequalities are compatible onlyif Fe)=0. The proof issimilar for the case ofaminimum ofafunction. Corresponding tothis theorem isthefollowing obvious geometric fact: ifatpoints ofmaximum and minimum, afunction f(x) has aderivative, the tangent: line tothe curve y=f(x) atthese points isparallel tothex-axis. y Indeed, from the fact that f’(x,)=tang=0,where@istheanglebetween thetangent line yeand thex-axis, itfollows that p—0 (Fig. 99). From Theorem |itfollows straightway that ifforallconsidered values oftheargument x thefunction f(x) has aderivative, then itcan A have anextremum (maximum orminimum) only 7 ¥ datthose values forwhich thederivative vanishes. The converse does not hold: ifcannot be said that there definitely exists amaximum ormini- mum forevery value atwhich the derivative vanishes. For instance, inFig. 99 wehave a function for which the derivative atx=x, vanishes (thetangent lineishorizontal), yetthe piayfunction atthis point is.neither amaximum eee nor aminimum. Inexactly thesame way, thefunction y=2" (Fig. 101) atx=0 has aderivative equal tozero: Y)eno =(3%*)cn0 =0, but atthis point the function has neither amaximum nor a minimum, Indeed, nomatter how close the point xistoO,we 163, Investigating theBehaviour ofFunctions will always have x*<0 when x<0 and x'>0 when 1>0. We have investigated the case when afunction has aderivative atallpoints onsome closed interval. Now what about those points atwhich there isnoderivative? The following examples will showthatatthesepoints therecanonly ybeamaximum oraminimum, but there yl maynotbeeither oneortheother. Example 1.The function y=[x| has no derivative at“thepoint20at.ths“point the curve does nol have adefinite tangent 3 %F line), butthefunction hasaminimum atthis point.y==0when20,whereas foranyother Fig. 102 Point xdifferent trom’ zero, we have y>0 (Fig, 102) Lue Example2.Thefunction y=(I—x*}""has noderivative atx=0, since v=—(1—x"9*"* becomes infinite atx=0, butthefunction hasa‘maximum atthispoint:((0)=1, f(x)<Iatxdifferent fromzero(Fig.103).Example 3.Thefunction y=}/xhasnoderivative atx=0 (y+@ 2sx—.0). Atthis point the function does not have either amaximum oraimintonim: /(0)=0:7(2)<0forx-<0;[(2)>0forx>0(Fig.104). y , y p=(1-xByeyn oa ¥ 7 a ¥ Fig. 103 Fig. 104. Thus, afunction can have anextremum only intwo cases: either atpoints where the derivative exists and iszero; orat points where thederivative does notexist. a Itmust benoted that ifthe derivative does not exist atsome point (but exists atclose-lying points), then atthis point the derivative isdiscontinuous. The values oftheargument forwhich the derivative vanishes ‘orisdiscontinuous are called critical points orcritical values. Maxima and Mintma ofFunctions 169 From what has been said itfollows that not forevery critical value does afunction have amaximum oraminimum. However, ifatsome point the function attains amaximum oraminimum, this point isdefinitely critical. And sotofind theextrema ofa function doasfollows: find allthe critical points, and then, investigating separately each critical point, find out ‘whether the function will have amaximum oraminimum atthis point, or whether there will be neither maximum nor minimum. Investigations offunctions atcritical points isbased onthe following theorem. Theorem 2.(Sufficient conditions forthe existence ofanextre- mum). Let there beafunction [(x) continuous onsome interval containing acritical point x,and diferentiable afallpoints of this interval (with theexception, possibly, ofthepoint x,itself). Ifinmoving from left toright through this point thederivative changes sign from plus tominus, (hen atx=x, the function has amaximum. But ifinmoving through the point. x,from leftto right thederivative changes sign from minus foplus, thefunction has aminimum atthis point. ‘And so tay {1>0when<x,a))pO)<0whenx>x, then atx,the function has amaximum; yy {FOS whencay,1D))p(y>0whenx>x,, thenatx,thefunction hasaminimum. Noteherethatthecon- ditions a)orb)must befulfilled forallvalues ofxthat are sufficiently close tox,,that is,atallpoints ofsome sufficiently small neighbourhood ofthe critical point x,. Proof. Let usfirst assume that thedetivative changes sign from plus tominus, inother words, that forallxsufficiently close to x,wehave Ff(x)>0 when x<x,, F<0 when x>x,. Applying. theLagrange theorem tothedifference /(x)—f(x,) we have 11a) ="Oe—«) where §isapoint lying between xand x,. 170 Investigating theBehaviour ofFunctions 1)Letx<.x,; then E<x, /®>0 /OE—x)<0 and, consequently, F@)—T(%)<9, or 1) <1). O) 2)Letx>x,; then b>ay, FE)<0, FE(e—x) <0 and, consequently, fe)—f(e)<0 or Fa)<fix,). (2 The relations (1)and (2)show that forallvalues ofxsuffici- ently close tox,thevalues ofthe function are less than those atx,.Hence, thefunction f(x) hasamaximum atthe point x,. The second part ofthetheorem onthe sufficient condition for @minimum isproved insimilar fashion. Fig.105illustrates themeaning 7ofTheorem 2. Atx=x,, letthere bef(x) =0 and let the following inequalities befulfilled forallxsufficiently close tox,: F()>0 when x<x, a a7 NBye F(x)<0 when x>x,. Fig.105. Then when x<x, thetangent to the curve forms with the x-axis anacute angle, and thefunction increases, butwhen x>x, ‘the tangent forms with the x-axis anobtuse angle, and the func: tion decreases; atx=x, the function passes from increasing to decreasing, which means ithas amaximum. Ifatx,wehave }'(x,)=0 andforallvalues ofxsufficiently close tox,thefollowing inequalities arefulfilled: 1(@)<0 when x<x,, FX) >0 when >44, Testing aDierentiable Function forMaximum and Minimum 171 then atx<x, the tangent tothe curve forms with the x-axis an obtuse angle, the function decreases, and atx>-x, the tangent tothe curve forms anacute angle, and the functfon increases. Atx=x, thefunction passes from decreasing toincreasing, which means it’has aminimum. Ifatx=x, wehave f'(x,)=0 and forallvalues ofxsufficiently close tox,the following inequalities are fulfilled: F(x) >0 when x<x,, F(#)>0 when x>x,, then thefunction increases both forx<x, and forx>x,, There- fore, atx=x, the function has neither 4maximum noramini- mum. Such is'the case with the function y=x* atx=0. Indeed, thederivative y’=3x*, hence, Y')ene=0, U'xco>0, Were >0, and this means that atx=0 the function has neither amaximum nor aminimum (see above, Fig. 191). SEC. 4.TESTING ADIFFERENTIABLE FUNCTION FOR MAXIMUM AND MINIMUM WITH AFIRST DERIVATIVE The preceding section’ permits ustoformulate arule fortesting adifferentiable function, y=f(x), for maximum and minimum: 1.Find thefirst derivative ofthe function, i.e. f'(x). 2.Find the critical values ofthe argument’ x;todothis: a)equate thefirst derivative tozero and find thereal roots of theequation f*(x)=0 obtained; b)find thevalues ofxatwhich the derivative f’(x)becomes discontinuous. 3.Investigate thesign ofthe derivative onthe left and right of‘the critical point. Since the sign ofthe derivative remains constant ontheinterval between twocritical points, itissufficient, forinvestigating thesign ofthederivative ontheleftandright of,say,theerilical pointx,(Fig.105),todetermine thesignof the derivative atthe points aand B(x,<u<xy t,<B<ty where x,and x,aretheclosest critical points). 4.Evaluate thefunction /(x)for every critical value ofthe argument. This gives usthefollowing diagram ofpossible cases: ' eer |+ f=0 —|Maximumpoint |- {ado +|Minimumpoint + F(4) =0 + Neither maximum nor = Payee ease)orisions |—|REHE?eominmnr ‘minimum (function de- creases) oe ea Yar —tet 2)Find the real roots ofthe derivative Bated 80. Consequently nek nea The derivative Iseverywhere continuous and sothere arenoother itis role Investigate thefirstcriticalpointx,—=1.Sincey’=(x—1) (x—3), forx<1 wehave y’=(—)(—)>9, fora>1wehave f= (4)4=)<0 Thus; when pssing rom tet toight) through the value sat the deivativechanges SignomplusYominus, ‘Hence,‘at x1"thefunction Basa 2 Om=F" when x<3 wehave y’=(+)(—) <0, when x>3 wehave y’=(+)4-+)>0 Testing aDifferentiable Function forMaximuri and Minimum 173 Thus,whenpassing through thevaluex—5thederivative changes sign from minus toplus. Therefore, at#9 the function tas a:minimum, Namely: Wxea=1. Thisinvestigation yieldsthegraphofthefuntion (Fig.108). Example 2. Testfotmaximum and minimum thetunction y=6—0 YF. y Solution. 1)Find thefirst derivatives Qa—1)_ 5e—2 v=vee - 2Ve 37e y-B-zatesuet y yn VE 9 ¥ et loaT ase Fig. 106, Fig. 107. Find the critical values ofthe argument: )find the points atwhich a2 2, art 7 re )find the points atwhich the derivative becomes discontinuous (in this instance, itBecomes infinite). Obviously, that point is =0. (Utwill benoted that forx2—0 thefunction isdefined and continuous.) "There arenoother critical points 8)'Tavestigate’ the character ofthe critical points obtained, Investigate thepointsy. Notingthat VW) 4<% W)_ 2>% eck eo 14 Investigating the Behaviour ofFunctions 2 i weconclude thatatx=Z thefunction hasaminimum. Thevalue ofthe function attheminimum point is 2_,\3/t__2a/a.o,1=(§-1) Va--iVE Investigate the second critical point x=0. Noting that Wree>O Weae <0 weconclude that atx=0 the function has amaximum, and (y)ses=0. The graph ofthe investigated funetion isshown inFig. 107. SEC. 5.TESTING AFUNCTION FOR MAXIMUM AND MINIMUM WITH ASECOND DERIVATIVE Letthederivative ofthefunction y=f(x) vanish atx=.x,; we have f’(x,)=0. Also, letthe second derivative /*(x) exist and be continuous insome neighbourhood ofthe point x,.Then the fol- lowing theorem holds. Theorem. Let f'(x,)=0; then atx=x, thefunction hasa maximum iff*(x,)<0, and aminimum iff"(x,)>0.Proof. Letusfirstprove thefirstpartofthetheorem. Let F(x,)=0 and ft(x,) <0. Since itisgiven that /"(x) iscontinuous insome small interval about the point x=:x,, there will obviously besome small closed interval about the point x=x,, atallpoints ofwhich the second derivative f'(x) will benegative.Sincef"(x)isthefirstderivative ofthefirstderivative, [*(x)==(f'(2))',itfollowsfromthecondition (f’(x))’<0that,/’(x)decreases ‘ontheclosed interval containing x=x,(Sec. 2,Ch.V). But /'(x,)=0, and soonthis interval wehave /'(x)>0 when x<x, and when x><x, wehave |'(x)<0; inother words, the derivative /'(x)changes’ sign from plus tominus when passing through thepoint x=x,, and this means that atthepoint x,the function f(x) has amaximum. The first part ofthe theorem is proved. Thesecond partofthetheorem isproved insimilar fashion: iff(x,)>0 then /’(x)>0 atallpoints ofsome closed interval about the point x,,but then onthis interval ["(x)=(f'(x)>0 and, hence, f’(x) increases. Since f’(x,)=0 thederivative f’(x) changes sign from minus toplus when passing through the point %,ive, the function f(x) has aminimum atx=x,. Ifatthecritical point f’(x,)=0, then atthis point there may beeither amaximum oraminimum orneither maximum nor_ Testing @Function forMaximum and Minimum 175, minimum. Inthiscase,investigate bythefirstmethod (seeSec.4,Ch. V). The scheme forinvestigating extrema with asecond derivative isshown inthe following table. 0|=|mexiumpein ° + Minimum point3$|Unknown Example 1.Examine thefollowing function formaximum and minimum y=2sinx-+c0s2 Solution. Sincethefunction tspecodie withapeiod of2x,iisulle ¢ cient toinvestigate the function inthe interval (0, Sx 1)Find the derivative yf=208x—2sin2v=2(cosx—2sinxcosx)=2cosx(1—2sinx). 2)Find the critical values ofthe argument: 2e08 x(1—2sinx)=0, a4, get: 4a, aae Ei ae Zi as ae Z. 3)Find the second derivative: yf=—2sinx—4cos2. 4)Investigate thecharacter ofeach critical point: Opeth bens co. Hence, atthepoint x,=% wehave amaximum tis Oa atzne: Further, tm 21pbla2>0 ‘Andsoatthepointxy—3 wehaveaminimum: W ,=2l—leb AinSEwehave La Oe =F Ga3.K0. W) meg tylidnr a WY) w=—2(—N—-4(— 1)=6>0. W) y= 2(—N-1=- 3. af’ yr2sinx+cas2x AYN (“\ oe oe 7 The following examples will show that ifatacertain point k=x, wehave f’(x,)=0 andF'(x,)=0, then atthis point thefunction f(x) can have either amaximum oraminimum orneither. Y= 1F, yee 0. Testing aFunction forMaximum and Minimum \7 11isthusimpossible heretodetermine thecharacter ofthecritical point by means of the sign of the second derivative3)Tavestigate thecharacter ofthecriticalpointbythefirstmethod(see see.annch.W):Weeo>%Wes4<O Consequently, atx-=0 the function has @maximum, namely Wano=! The graph ofthis function isgiven inFig. 109. y Hy a ¥ yoxt yot-xt a ¥ Fig. 109. Fig. 110. Example 3.Test for maximum and minimum the function pax Solution. By the second method we find 1)y=6H, y=6:=0, x=0; 2)y=80r, WIyne=0. Thus, the second method does not yield anything. Resorting tothefrst method weget Wreee <% Wes e> siTMeefoe ax=0thefunction hasminimum “ «Fi ee‘Exampie 4.Testformaximum andminimum Ox the Tunetion yal. Solution. Second method: ¥=3(x—1)? 3e—1P=0, x=1; a a P=61), Wei=0. Thus, the second method does not yield ananswer. Bythe first method weget Weer>O Wey 10 Consequently, atx=1 the function does not « have eller @maximum oraminimum (Fig: 111). Fig. 111, 178 Investigating theBehaviour ofFunctions SEC. 6.MAXIMA AND MINIMA OF AFUNCTION ON AN INTERVAL Let the function y=/(t) becontinuous onthe interval {a,6). Then the function onthis interval will have amaximum (see Sec. 10, Ch. II). We will assume that onthe given interval the function f(), has afinite number ofcritical points. Ifthe maximum. is reached within the interval [a,6), itisobvious that this value will beone ofthe maxima ofthefunction (ifthere areseveral maxima), namely, thegreatest maximum. But itmay happen that the max- imum value isreached atone ofthe end points ofthe interval Tosummarise, then, onthe interval (a,6]the function reaches itsgreatest value either atone ofthe end points ofthe interval, oratsuch aninterior point asisthe maximum point. The same may besaid about theminimum value ofthefunction: itisattained either atone oftheend points oftheinterval oratyortaeed aninteriorpointsuchthatthelatteristheq minimum point.Fromtheforegoing wegetthefollowing 5 rule: ifitisrequired tofind the maximum of continuous function onaninterval [a,6),do the following: 1)Find allmaxima ofthe function onthe interval 2)Determine the values ofthe function at the end points ofthe interval; that is,eval- % 1X wate f(a) andf(b). 3)Ofallthevalues o!thefunction obtained choose the greatest; itwill be the maxi- mum value ofthe function on the interval. The minimum value ofafunction on an interval isfound insimilar fashion. Example. Determine themaximum andminimum ofthe: function y=x?’—Se-+-3" on the interval Solution. 1)Findthemaxima andminima ofthefunctionontheinterval[-33: YaWW3, &IH3=0, el, K=— 9=68 Wear=8>0. Thus, atx=} there isaminimum: enh 5 Further, Fig. U2, ewe 6<0, Applying theTheory ofMaxima and Minima 179 And so atx=—1 we have maximum: Wen-1=8. 2)Determine the value ofthe function atthe end points ofthe intervals Cer kee ‘ i 3 ‘Thus,thegreatestvalueofthisfunction ontheinterval [—3,5]tsNea=5, andthesmallest valueis u Wxa-1=— 15. ‘The graph ofthe function isshown inFig. 112. SEC. 7.APPLYING THE THEORY OF MAXIMA AND MINIMA OF FUNCTIONS TO THE SOLUTION OF PROBLEMS The theory ofmaxima and minima isapplied inthesolution of many problems ofgeometry, mechanics, and soforth. Let us examine afew. Problem 1.The range R=OA (Fig. 113) ofashell (inempty space) fired with aninitial velocity v,from agun inclined tothe horizon atanangle g,isdetermined bytheformula — ofsin 29 Z (gistheacceleration ofgravity). | ¥ ®Determine the angle @atwhich the range Rwill beamaximum fora Fig.113. given initial velocity v,. Solution. The quantity Risafunction ofthevariable angle @, Testthisfunction foramaximum ontheinterval O<@<J: dR_2vbcos2p vtcos29 a onGetgg OSetiticalvaluep=35 eR __‘ehsinty are 40Wwe (i), =ese Hence, forthevalue p=% thefunction Rhasamaximum ®2-F \ 180 Investigating theBehaviour ofFunctions The values ofthefunction Rattheend points ofthe interval [o.#]are Rgeo=0 (R),_n=0 ‘Thus, themaximum obtained isthesought-for greatest value ofR. Problem 2.What should thedimensions beofacylinder sothat foragiven volume 9itstotal surface Sisaminimum? Solution. Denoting byrthe radius ofthe base ofthe cylinder and byfhthealtitude, wehave S=2nr* +2arh. Since the volume ofthe cylinder isgiven, foragiven rthe quantity isdetermined bytheformula v=ar'h, whence ° had. Substituting thisexpression offintotheformulaforS,wehave S=2nr*+ 2ar or s=2(ar+2). Here, visgiven, so_we have represented Sasafunction ofa single independent’ variable r."Findtheminimum valueofthisfunction ontheinterval0<r<oo: as °Bao (22-4) : or—t=0, n=VE, as w(2), =?(28+8),_., 2% Thus, atthe point r—r, the function Shas aminimum. Notic- ingthat limS=0o and limS—0o; that is,that asrapproaches zerootinfinity thesurfaceSincreaseswithoutbound,wearrive attheconclusion that atrr, the function Shas aminimum. Testing @Function forMaximum and Minimum 181 Buttfr=gzthen hada2 Ve=n. Therefore, forthetotal surface Sofacylinder tobeaminimum foragiven volume v,the altitude ofthe cylinder must beequal to its diameter. SEC. 8. TESTING AFUNCTION FOR MAXIMUM AND MINIMUM BY MEANS OF TAYLOR'S FORMULA InSec,5,Ch.V,itwasnoted thatifatacertain point x=awehave f’(a)=0 and f'(a)=0, then atthis point there may be either amaximum oraminimum orneither. And itwas noted that inthis instance the problem issolved byinvestigating bythefirst method; inother words, bytesting the sign the first derivative on the left’ and onthe right ofthe point x=a. Now wewill show that itispossible inthis case toinvestigate bymeans ofTaylor's forinula, which was derived inSec. 6,Ch. IV. For greater generality, weassume that notonly (x), butalso allderivatives uptothe nth order inclusive ofthe functions /(x) vanishatx=a:P@=P@=...=/(a=0 oy and pr (a)40. Further, assume that /(x) has continuous derivatives uptothe (n 1)st’order inclusive inthe neighbourhood ofthe point x=a. Write theTaylor formula forf(x), taking account ofequality (1): (=a)"** je Hay=1(a)+SHA pow, @ where &isanumber that lies between aand x. Since /"*(x) iscontinuous inthe neighbourhood ofthe point aand [**(a)40, there will beasmall. positive number Asuch that forany xthat satisfies theinequality |x—a|<A, there will bef**”(x)#0.Andifporn(a)>0,thenatailpointsoftheinterval(a—A,ath)wewillave}"*%(x)>-0;iff(a)<0,thenatallpointsofthisinterval wewillhavef+(x)<0,Rewrite formula (2)inthe: form (ea sng, , : 1)—Ka)= “we Mt") @y and consider various special cases, ise Investigating theBehaviour ofFunctions Case 1.nisodd, a)Let f"*" (a)<0. Then there will beaninterval (a—h, a+h) atallpoints ofwhich the(n+1)st derivative isnegative. Ifxisapointofthisintervalthen&likewiseliesbetweena—Aanda-+h and, consequently, f"*"(&)<0. Since n+1 isaneven number, (x—a)"*">0 forxa, and therefore the right side offormula (2) isnegative Thus, forxa atallpointsoftheinterval(a—h,a-+A)wehave F(x)—F(@ <0, and this means that at_x—a the function has amaximum. b)Letf"*"(a)>0.Thenwehavef"*"(E)>0forasufficiently small value ofAatallpointsxoftheinterval(a—h,a++h).Hence, theright side offormula (2') will bepositive; inother words, for x#a wewill have the following atallpointsinthegiveninterval: F(x)F(a)>0. and this means that atx=a the function has aminimum. Case 2.niseven. Then n-+1 isodd and the quantity (c—a)"*" hasdifferent signs for x<a and x>a. Ifhis sufficiently small inabsolute value, then the (n-++1)st derivative retains the same sign asatthe point aatallpointsof theinterval (a—A, aA). Thus, f(x)—f(a) hasdifferent signs for x<a and x>a. But this means that there isneither maximum nor minimum at x=a Itwill benoted that iff"*(a)>0 when niseven, then F(x) <f(a) for«<a and f(x)>f(a) forx>a. But if**" (a)<0 when niseven, then f(x)>f(a) forx<a andF(x)<f(a)lorx>a.The resuits obtained may beformulated asfollows. Ifatx=a we have F@=F@=...=/"(a=0 andthefirst nonvanishing derivative /"* (a)isaderivative ofeven order, then atthe point'a (2)hasamaximum iff"*"(a)<0,F(x) hasaminimum iff** (a)>0. But ifthefirst nonvanishing derivative /°*" (a)isaderivative ofodd order, then the function has neither maximum nor minimum. atthe point’a. Here, F(x) increases iff**" (a)>0, T(x)decreases iff*" (a)<0. Convexity and Concavity ofaCurve 183, Example. Test the following function formaximum and minimum: I()axt—4et68art1. Solution, Let us find the critical values of the function F(x)4x—12412x—4=4(x83+3e—1), Fromequation Ate p3x—1)=0 weobtain the only critical point ral inge this equation has only one real_r00! (ofaveatigate thecharacteroftheicalpointx=1: [iG)a12et—24412—0fors—t, fr (y= dae 24=0 fors=1, 1Y(2)=24>0 forany &. Consequently, forx=1 the function /(x) has aminimum. SEC. 9,CONVEXITY AND CONCAVITY OF ACURVE. POINTS OF INFLECTION Letusconsider, inaplane, the curve y=/(x), which isthe graph ofasingle-valued differentiable function f(x). Definition 1.We say that.a curve isconvex upwards onthe interval (a,6)ifallpoints ofthe curve liebelow any: tangent toiton this interval. Wesaythat thecurve isconvex y downwards onthe interval (b,c) ifallpoints ofthe curve lieabove * Aanytangenttoitonthisinterval. 7ZL Weshall callacurve convex up, ; aconvex curve, and acurve convex Fig. 114 shows acurve convex ‘ ontheinterval (a,6)and concave 74 °° * ontheinterval (6,c). Fig.114.‘Animportant ‘characteristic of ie. UN. the shape ofacurve isitscon- vexityorfore Thissection willbedevoted toestablishingthecharacteristics bywhich,wheninvestigating afunction y=f(x),one can judge ofthe convexity orconcavity (direction of,bulge) on various intervals. We ‘shall prove the following theorem.Theorem 1.Ifatallpointsofaninterval(a;6)thesecondderiv-ativeofthefunction f(x)isnegative, i.e.,#(x)<0,thecurvey=F(x)onthisinterval isconvexupwards (thecurveisconvex). 184 Investigating theBehaviour ofFunctions Proof. Inthe interval (a,6)take anarbitrary point x=x, (Fig. 114) and draw atangent tothe curve atthe point with abscissa x=x,.Thetheorem willbeproved provided weestablishthat allthe points ofthe curve onthe interval (a,6)liebelow this tangent; that is,that theordinate ofany point ofthecurve y= (x)isless than theordinate yofthetangent line foroneand the same value ofx. The equation ofthecurve isoftheform y=1 0. wy But the equation ofthe tangent tothe curve atthis point x=x, isofthe form 9-f)=!&)4) or - G=fG)+F HOH). @ From equations (1)and (2)itfollows that the difference ofthe ordinates ofthe curve and the tangent forthesame value ofxis y—9=1@)—1e) Pf6)e—). Applying theLagrange theorem tothe’difference f(x)—f(x,), weget _. 9-9=P ©)—4)—F HE) (where ¢liesbetween x,and x)or y—9= OF &)E—*)- ‘Weagain apply theLagrange theorem tothe expression inthe square brackets; then y-9=P (4)(C—*)&—2,) @) (where c,liesbetween x,and c). Letusfirst examine thecase when x>x,. Inthis case, x,< <c<x; since x—x,>0, c—x,>0 and since, inaddition, itisgiven that . Fe)<d, itfollows from equality (3)thaty—7<0. Now letusconsider thecase when x<x,. Inthis case x<co< <c,<x, and x—x,<0, c—x,<0, and ‘since itisgiven that Convexity and Concavity ofaCurve 185 F'(c,)<0, then itfollows from (3)that y—9<0. We have thus proved that every point ofthe curve lies below thetangent tothecurve, nomatter what values xand x,have on the interval (a,6). And this signifies that the curve isconvex. The theorem isproved. The following theorem isproved insimilar fashion. Theorem 1’.Ifatallpoints ofthe interval (6,c), thesecond derivative ofthefunction f(x) ispositive, that is,f'(x)>0, then thecurve y=[(x) onthis interval isconvex downwards (the curve isconcave). Note. The content ofTheorems 1and 1’may beillustrated geometrically. Consider thecurve y=/(x), convex upwards onthe interval (a,6)(Fig. 115). The derivative /’(x) isequal tothe iY y ¢aA ~ofaa 1bx atlKeey % CG Oe Fig. 115. Fig. 116. tangent ofthe angle ofinclination aofthe tangent line atthepointwith-abscissa x,orf'(x)=tana. Forthisreason, .(x)==[tana];.1f/"(x)<0forallxontheinterval(a,6),thismeansthat tana decreases with increasing x.Itisgeometrically obvious that iftanadecreases withincreasing x,thenthecorresponding curveisconvex. Theorem 1isananalytic proof ofthis fact. Theorem 1’isillustrated geometrically insimilar fashion (Fig. 116). Example 1.Establish the intervals ofconvexity and concavity ofacurve represented bythe equation pata Solution. The second derivative yo2<0 forallvalues ofx.Hence, thecurve iseverywhere convex upwards (Fig. 117). 186 Investigating theBehaviour ofFunctions Example 2.The curve isgiven bythe equation pae. Since yae>o forallvalues ofx,the curve istherefore everywhere concave (bulges, oris convex, downwards) (Fig.18). Example 3.Acurve 1sdefined bythe equation y=, Since yor, y¥'<0for«<0andbieforx>0.Hence,for«<0thecurveisconvex ipwards, and forx>0, convex down (Fig. 119}. y d iy ye afi Vex Yyet 7 ¥ great) * Fig. 117. Fig. 118. Fig. 119. Definition 2.The point that separates the convex part ofa continuous curve from the concave part iscalled thepoint of inflection ofthe curve. ‘InFigs. 119 and 120 thepoints ©and Barepoints ofinflection. Itisobvious that atthe point ofinflection thetangent cuts the curve, because onone side the curve lies under the tangent and onthe other side, above it. Let usnow establish the sufficient conditions foragiven point ‘ofacurve tobeapoint ofinflection. Theorem 2.Letacurve bedefined bytheequation y=f(x). Iff'(a)=0 orf"(a)doesnotexistandifthederivative f”(x)changessign when passing through x=a, then thepoint ofthecurve with abscissa x=a isthepoint ofinflection. Proof. 1)Let f"(x)<0 forx<a and ‘f"(x)>0 forx>a. Then forx<a thecurve isconvex upand forx><, itisconvex down. Hence, the point Aofthecurve with abscissa x=a isthe point ofinflection (Fig. 120). Convexity and Concavity of@Curve 187 2)Mf(x)>0 forx<b and f"(x)<0 forx>6, then forx<6 the curve isconvex down, and forx>6, itisconvex up. Hence, the point Bofthecurve with abscissa x=6 isthepoint ofinflection (see Fig. 121). y y A8B aa al ¥ Fig. 120. Fig. 121. Example 4.Find the points ofinflection and determine the intervals of convexity and concavity oftheeurve yaer*" (Gaussian curve), Solution. 1)Find the first and second derivatives: yf=—2ee-*", ¥a2e-* Qxt1). y2)Thesecondderivativeexistseverywhere. Findthevaluesofxforwhich 2e-** (2e*—1) =0, neck, nela nc 3)Investigate the values obtained: 1 for 2<— atwe hi .<Vr lavei>0, 1 fors>——1=wehave7<0; TF v<0 thesecond derivative changes signwhenpassing through thepoints. Hencelorx———he ,thereisapointofinflection onthecurve;itscoordi brn=75 Pe jectiononthecurve;itscoordinates (2,07 1 For * °,<yr v< 1 for s>te youDye? 188 Investigating the Behaviour ofFunctions 1 Thus,thereisalsoapointofinflectiononthecurveforx=—=j Itscom r VE ordinatesare(ys:o’).Incidentally,theexistenceofthesecondpointotinfection follows directly tromthesymmetry ofthecurvesbout the ganis 4)From the foregoing itfollows that 1 for —eo<x<—te the curve is concave:<< 75 Iscone 1 H for he <x< te the curve Isconvex; 1v2 vz for=<x<othecurveisconcave, vi< 5)From the expression ofthe first derivativefnDee-2* 1follows that , for<0 y'>0, the function Increases; forx50 9’<0, the function decreases: fors=0 20, Atthis point the function hasamaximum, namely,yt.Theforegoin analysis Makes iteasytoconstruct agraphofthecurve(Fig.12). y yer +4 Ts * Fig. 122, Example 5,Test thecurve y=x* forpoints ofinflection, Solution. i)Find thesecond derivative: yale, 2)Determine the points atwhich y=0:1240;x=, 43)Investigate the ‘value x0 oblained: forx<0 y'>0, the curve isconcave; for#50 950, the curve is.concave, Thug,thecurvehas,nopointsafnection (Fg,123). ‘Example 6.Investigate thefollowing curve forpolnts ofinfléctlon y=u—y Asymptotes 189 Solution. 1)Find the first and second derivatives: Fee eeeyayoo hsva Zoy 2)The second derivative does not vanish auywhere, but atx=1 itdoes notexist(y"=40). y yoxt YYyate-y a 7 a * Fig. 123. Fig. 124. 3)Investigate thevalue x=1: for x<1 >, the curve Isconcave: forx>1 YO, the curve isconvex Consequently, atx1 there isapoint ofinflection (1,0)Iewillbenotedthatforx=1yar; thecurveatthispointhasaver tical tangent (Fig. 124). SEC. 10, ASYMPTOTES Veryfrequently onehastoinvestigate theshapeofacurve y=F(x) and,consequently, thetype ofvariation ofthecorrespond- ing function inthe case ofanunlimited increase (inabsolute value) oftheabscissa orordinate -ofavariable point ofthecurve, ‘oroftheabscissa and ordinate simultaneously. Here, animportant special case iswhen the curve under study ‘approaches agiven line without bound asthe variable point ofthe curve recedes to infinity. * Definition. The straight line Aiscalled anasymptote toacurve, ifthedistance 8from the variable point Mofthecurve tothis straight line approaches zero asthe point Mrecedes toinfinity (Figs. 125 and 126) ~+)WesaythevariablepointMmovesalongacurvetoinfinityifthe distance ofthepointIromtheoriginincreases without bound, 190 Investigating theBehaviour ofFunctions Infutureweshalldifferentiate between vertical asymptotes (paral- let totheaxis ofordinates) and inclined asymptotes (not parallel totheaxis ofordinates). cw y 2) |"6s¢ al ig 7a 7 Fig. 125. Fig. 126. 1.Vertical asymptotes. From thedefinition ofanasymptote itfollows that iflimf(x)=00 orlim(x)=00 orlimf(x)=00, esaed rand me then thestraight line x=a isanasymptote tothecurve y=f(x); y and, conversely, ifthe.straightline’ x=a isanasymptote, then 2 oneoftheforegoing equalities isee fulfilled.es Consequently, tofindvertical asymptotes one’hastofindvalues olofx=a such that when they are%approached by—thefunctiony=l(x) the‘latter—approachesinfinity, Then the straight line x=a will beavertical asymptote. Example 1.Thecurveyzhasa Fig, 127. vertical asymptote x=5, since yosr5(Pig12/). usenzAmPle 2Thecurveymtan «has Tani numberof vertical asymp- x Bn, Sneeahs ens eaBs... This follows from the fact that tanz—-c asxapproaches the values Be ee Be eg100, Asymptotes 191 Example 3.Thecurve y=e* hasaverticalasymptotex=0,sincelime= =e (Fig. 129) y ] grtanxi Dx10x(ox([ Fig, 128. IL.Inclined asymptotes. Let the curve y=/(x) have an inclined asymptote whose equation is yoko. (yy y : Mewip yeeAilA H sae % a ¥ Fig. 129. Fig. 130. Determine thenumbers &andb(Fig. 130). LetM(x, y)beapoint lying onthecurve andN(x, 4),apoint lying ontheasymptote, The length ofMP isequal’ tothedistance from thepoint Mto 192 Investigating theBehavtour ofFunctions the asymptote. Itisgiven that limMP=0. c) Designating theangle ofinclination oftheasymptote tothex-axis byg,wefind from ANMP that Mp wo=Ae, Since@isaconstant angle(notequalto)byvirtueofthe foregoing equation limNM =0 @’) and, conversely, from (2') weget (2). But NM=|QM—QN|=ly—9|=|f()—(ke+6)|, and (2’) takes the form lim(f(@)—kx—6]=0. @) Tosummarise: ifthe straight line (1)isanasymptote, then (3) isfulfilled; and conversely, if,given constants &and 6,equation (3)isfulfilled, then thestraight line y=Ax-+ isanasymptote, Let usnow define &and 6.Taking xoutside the brackets in (3), weget timx[42—*—2] =o. Sincex—++00,thefollowing equation mustbefulfilled: lim[2-2-2] =0. For6constant, lim=0.Hence, lim(i4]=0, or featim1), @ Knowing &,wefind 6from (3). b=lim(/(kx). ® Asymptotes 199 Thus, ifthe straight line y=kx-+6 isanasymptote, then &and 6may befound from (4)and (5).Conversely, ifthelimits (4)and(5) exist, then (3) isfulfilled and the straight line y=kx+6 isan asymptote. Ifeven one ofthe limits (4)or(5)does not exist, then thecurve does nothave an yasymptote. Itshould be noted that we carried out our investigation as applied toFig. 130, asx—+-++co, butallthearguments holdalso au forthecasex—+—oo. ix24Oe-4 vtExample4.Findtheasymptotes {7* yates y Solution. 1) Look for vertical asymptote: when =0yet: whenseo jote ay = Therefore, the straight line x=0 isavertleat asymptote2Lookforinclined asymptotes: emtimLoetiBEENEbola Fam tim[i424] = din[ga] o} Fa13h that is, kel, Olinto etmtin[EMI] ig[atte pe re bers pS ee =tin[2-2]a2or,finally, vee baa, Therefore, the straight line garg2 isaninclined asymptote tothegiven curve “Toinvestigate themulual postions cf curve. and anasymptote, letus consider thediference oftheordinates ofthecurve andthesaymptote tor Set eget Thisdiference inegative for+>0,andpositive forx<0;andsofor1>0thecurve liesbelow theasymptote, andforx<0, Itlesabove theasymptote (Pigs 13). Toa 194 Investigating theBehaviour ofFunctions Example 5.Find the asymptotes ofthe curve yae*sinxbe, Solution.1)Itisobvious thattherearenovertical asymptotes. 2)Look for inclined asymptotes: femtimtotimsAserbe lim.(ean b=lim[e-*sinx-+-x—z]= lime-*sinz=0. Hence, thestraight line oe 1saninclinedasymptote asx-+4:0Theglvencurvehasno-asymptote asx-+—oe, Indeed, thelimit tim2 doesnotexist,sinceLaysinx41. (Here,thefrsttermincreaseswithout bound asx-+—o and, therefore ithas nolimit.) SEC. 11, GENERAL PLAN FOR INVESTIGATING FUNCTIONS ‘AND CONSTRUCTING GRAPHS The term “investigation ofafunction” usually implies the finding of: 1)thenatural domain ofthefunction; 2)the discontinuities ofthe function; 3)the intervals ofincrease and decrease ofthe function; 4)themaximum point and the minimum point, and also the maximal and minimal values ofthe functions; 5)the regions ofconvexity and concavity ofthegraph, and points ofinflection; 6)theasymptotes ofthegraph ofthefunction. The graph ofthe function isconstructed onthe basis ofsuch aninvestigation (itissometimes wise toplot elements ofthe graph inthe very process ofinvestigation). Note 1.Ifthe function under investigation y=f(x) iseven, that is,such that upon change ofsign ofthe argument thevalue ofthe function does not change, i.e.,if H—2)=10), then itissufficient toinvestigate the function and construct its graph for positive values ofthe argument that liewithin. the domain ofdefinition ofthefunction. For negative values oftheargument, thegraphofthefunction isconstructed onthe.groundsthat the graph ofan even function issymmetric about the ordinate axis. Generat Pian forInvestigating Functions and Constructing Graphs 195 Example 1.The function yx" iseven, since (—x)*=x" (see Fig. 5). Example 2.Thefunction y==cos is’even,sincecos(—x)=608% (see Fig. Note 2.Ifthe function y=/(x) isodd, that is,such that for any change inthe argument the function changes sign, i.e.,if N—)=—f@), then itissufficient toinvestigate this function inthe case of positive values oftheargument. The graph ofanodd function is symmetric about the origin. Example 3.The function y=" isodd, since (—2)*=—1" (see Fig. 7)Example&:Theluncton’gesaitsisodd,scetin(—a)e—oe ace Fig. 10) Note 3.Since aknowledge ofcertain properties ofafunction allows ustojudge ofthe other properties, itissometimes advi- sable tochoose the order ofinvestigation onthe basis ofthe specific peculiarities ofthe given function. For example, ifwe have found out that thegiven function iscontinuous and differen- tiable and ifwehave found the maximum point and themini- mum point ofthis function, wehave thus already determined also therange ofincrease and decrease ofthefunction. Example 5.Investigate the function oT and construct itsgraph. Solution. 1)The domain ofthe function isthe interval —co<x<o», Itwill straightway benoted that forx<0 wehave y<0, and forx>0 wehave y>0.2)Thefunction iseverywhere continuous. 43)Test the function formaximum and minimum, from theequation a=t=ial =ae Find theeritical points: nach aeh Investigate thecharacter ofthecritical points: for%<—I wehave y’<0; forx>—Iwehavey’>0. Hence, at2=—1 the function has aminimum: Srin=(Wen—1 And forx<1 wehavey’>0; forx>1 wehave y’<0. ” 196 Investigating theBehaviourofFunctions Hence, atr=1 the function has amaximum: Vmax (W)enr=1 4)Determine the domain ofincrease and decrease ofthe funetion: for—co<x<—I wehavey’<0,thefunctiondecreases; for —1<x-<1 we have y’50, the function increases: for 1<x<e wehave y<0, the function decreases, 5)Determine thedomains ofconvexity andconcavity ofthecurveand the points ofinflection: from the equality 7Bet 3)_v="Tay =? weget ne V3 nO, V3 Investigating y*asafunction ofxwefind that tor—«@<x<—V@ ¥<0, thecurve isconvex; for —V3<x<0 —g>0, thecurve isconcave; for O<x<V3_ <0,thecurve isconvex; forY3<r<ew —y'>0, thecurve isconcave. Thus,thepointwithcoordinates x=—V3,y=—03is&pointof infection: inexactlythesameway,thepoints©,0)and(3,43)are points ofinflection 16)Determine the asymplotes ofthe curve: for r++0 9+0, for s+—0 y 0. Consequently, the straight line y=0 isthe only inclined asymptote. Thecurve hatno.vertical asyinptotes. because thefunction doesaotapproach fnfinity forasingle finite value of=, x 7YFexE v5 -4 0 1 Ls “15 Fig. 182 ‘The graph ofthecurve under study isgiven inFig. 132. Example 6.Investigate the function y=tate and construct itsgraph, Generat Plan forInvestigating Functions and Constructing Graphs 197 Solution. 1)The function isdefined forallvalues ofx. 2)‘The tunetion iseverywhere continuous. 3)Test the function for maximum and minimum: yogic tae moten ee _ 3/Gar—eP 3{/xQa—aF There isaderivative everywhere except forthe points s=0 and x,=22, Snvestigate thelimiting values ofthederivative ax-+—0 and rete . ,fa— so—e tim Ew, tin =+0; Wives A.O75Vaare O37 Year for¢<0 y/<0, and fors>0 v>0. Hence,” atx0 the function figs aminimum, The value ofthe function atthis point iszero, Nowinvestigate thefunction attheothercritical pointxy=2a. Asx22 the derivative Also approaches infinity. However, in(his cave, forallvalues ofcloveto2a(athontherightandlelof20),thedenvatve isnega- fiver Therefore, atthis point the function ‘hss neither amaximum nota Ininimum. Atand about the point x4—28 thefunction decreases; thetangent {othe curve atthis point isvertical Atrafthederivative vanishes. Letusinvestigate thecharacter of this eritieal point. Examining theexpression ofthefirst derivative, wenote that forr<42y>0,andfors>By<0. than, ce42tefuetionBe«masa 2oaq tain da77. 4)On the basis ofthis study we get the domains ofincrease and decrease ofthe funetion! for ew <x-<0 the function decreases tor0.x<Mthefuetion ieee for$2<x<cothefunctiondecreases. 5)Determine thedomains ofconvexity andconcavity ofthecurveand the poinis ofinilection: the second derivative fe 92@a—x)* 196 Investigating theBehaviour ofFunctions does not vanish atasingle point. Yet there are two points atwhich the Second derivative isdiscontinuous 4-0 and. 2a Untinvstgele thesignofthesecondevvative neareashofthee forx<0 wehave y<0 and thecurve isconvex up; for>0 wehave y<0 and the curve isconver up. Hence, the point with abscissa £0 Isnot apoint ofinflection. Forx<2a wehavey’<0andthecurve isconvex upwards;forx20 wehave gf50 and the curve isconver: down, \y Fa,“iSse|gcbeara 7 ae Fig. 198. Hence, the point (2a, 0)onthe eurve isapoint ofinflection. 6)Determine the fsymptotes ofthecurve: Peed — feotinLatigVEE igYB =, eT. ae I VE = lim [3/%ar—F +x]= batn[Ymae=F +] '1 2axt—at4x8 _ = timptih. Via Via 8 Thus, thesteaight tine 5 22 gooey 1sanInclined asymptote tothecurve y=}/Zae7=aF Thegraph ofthis function isshown inFig. 133. Investigating Curves Represented Parametrically 199 SEC, 12, INVESTIGATING CURVES REPRESENTED PARAMETRICALLY Let acurve begiven bythe parametric equations x=9(t)\ 1 y=¥0. 0 Inthis case the investigation and construction ofthecurve is carried out just asforthe curve given bytheequation y=F). Evaluate the derivatives ay ane wv _y CO)Yaw. For those points ofthe curve near which itisthegraph ofa certain function y=f(x), evaluate thederivative dy wi)creataod ®) Wefind thevalues oftheparameter (=f,,f,,...»4forwhich at least one ofthe derivatives g’(¢) orw’(t) vanishes orbecomes discontinuous. (We shall call these values of¢critical values.) Byformula (3),ineach oftheintervals (fy,f,);(trfai«++i(Ceoas fa) andhence, ineachoftheintervals (x,,ay(eqHs2peea), (wherex,=(t;)),wedetermine thesignof$2,inthiswaydetermin- ingthedomain ofincrease and decrease. This likewise enables us todetermine thecharacter ofpoints that correspond tothevalues ofthe parameter fy,fy,«+41 tyNext, evaluate dy_¥OeOFOVD eya ieor . From this formula, determine the direction ofconvexity ofthe curve ateach point. To find the asymptotes determine those values of/,upon approach towhich either xoryapproaches infinity, and those values of¢upon approach towhich both xand yapproach in- finity. Then carry out the investigation inthe usual way. The following examples will serve toillustrate some ofthe peculiarities that appear when investigating curves represented parametrically. 200 Investigating theBehaviourofFunctions Example 1.Investigate the curve given bythe equationsmacos"t, . . pease f “ Solution. Thequantities xandyaredefined forallvalues of¢.Butsince the functions of‘cos! and sint¢ are periodic, with aperiod 2x, ittssul cient {0consider the variation ofthe parameter 1inthe range trom 0'to 2x: here. the interval {-—a, a}isthe range of+and the interval [—a, a]fsthe range oly.Consequently, this curve has noasymptotes. Next, wefind 4sacosttsint, dy : 2)4450sinttose Thesederivatives vanishatt=0,3,x,9%,On,Evaluate dy_3asin®cost 'de~—3acostsin?~~*"# oe Onthe basis of(2) and (3') wecompile the following table: Sip] type clvariation Rangeot|Corresponding |Corrspondin are, rcoments|cragengre|oB[Peegata octcd |a>xr0 |o<y<a |—|decreases $<tcn |0>2>-0 a>y>0 |+|Increasesetc |-ocx<o |osys—a |—|Decreases Berean |ocx<a |—a<y<o |+|increases Fromthetableitfollowsthatequations(I')definetwocontinuous functionsatthe type,y=1a)forOatam #0(iefrattwoHinesathetale)foe<teonVO(seetwolastHinesofthetable):From(3’)itfollows 2 ay.tim {4-0 oe and 7im4oofee [Atthese points the tangent tothecurve Isvertical. We now find ay) no, 4] mo, | =Hemom HE[sm0™? H|sm20 = Investigating Caroes Represented Paramerealy 1 Athese points the tangent tothe curve is .horizontal, Wethenfind y ay dat™3acos*?sint* Whence itfollows that a foro<t<a F4>0 thecurveisconcave, torn<t-<2n#6.<0ibecurveisconve (Onthe basis ofthi investigation. wecan contacts cute(ig.18),shioeed Fig106, Example 2Construct acurvegivenbythefollowing equations (ll ofDescartes): tot att *"T4e) Y=TER ty) Solution. Both functions aredefined forallvalues of#except f=-—1, andsat wat ee eeee ere ahaRa ms, im ya te.al tat Further note that when {=0 rm0, yd, when f= +o x+0, yd, when t=—@ x0, yO. aeatt, Find aanda Liaae_&(3-" dy_satQ—0) eoa Uh oa For ¢weget the following crtieal values = = =72 WeckheO hey. ueVE Then we And dy dydt_12-0)#7a3(Toy ” a *(2-*) 202 Investigating theBehaviourofFunctions Onthe basis offormulas (1"), (2'), and (3°) wecompile thefollowing table: Sig|typeotvara angeot|Carraponcing |Corraponting |SU,[Jy2e0!saraton —w<i<-1|0<r<te |0>y>—« |—|Decreases=r<r<o |—e<x<o |tesy>0 |—|DecreasesoKt<oeo<x<af/G jo<y<a}/3| 4|Increases yes? af/F>x>ay/ Waf/B<y<aj/a] —|Decreases VY2<t<@ |aj/2>x>0| aji/i>y>o| +|Increases From (3°) wefind . dy=0(¥) =o. can 4) saco Ge) Ga) Thus,thecarvecultheoriginteewiththe tangent parallel tothesax and with the tangent parallel tothey-axis. Further ay(#3. 7%Va (ena /G Atthis point thetangent tothecurve isvertical. dy(B).- enti2 [Atthis point the tangent tothecurve ishorizontal. Letusinvestigate the Question Bfthe existence ofanasymptote: Saft(1+e) be tim Low |ne iMSarr) m [salt dat bmtny—tedmtin,[PEA(—0en]= jim F841) tim824g,=n, PGP] =ree Exercises onChapter V 203 Hence, the straight line:-y=—s—a isanasympiote toabranch ofthecurve as x4 +e. ‘imifarly wefind bemlim£1, b= tim y—ky=—a. ¥ Thus, the straight line isalso anasymp-totetoabranch ofthecurve asx—>—er, (On the basis ofthis. investigation we conte civ(Pg.18) Some. problems involving investigation of curves-wil again: be. disetsed inChaplet Vill *singuiat Points ofaCurve™ Fig. 195. Exercises onChapter V Find theextremes ofthe functions: 1.yaat—2c+3. Ans. gaia? at rel2yee DSH AasmeeateldyeSeie Ans.Yax=10 atx=1,Yuin—22atx=5.4.y=—x4428.Ans.Yae=t atx2, Yuin0at20.5yax'—Bet4 2.Ans.Yoar=2a0, tmnt AMES he.pode—tobepalGON.Anenieamdand x=3, minatx=—3 andx—=4. 7%y=2—(—1)*. Ans.Ymax==2 atr=1. 8.y=3—2 (041). Ans.Nelth rin SEE. Anemin at=V3,maxatr=2—VE. 10,yaF=AG—A) Ans,maxatxl? Meym2eper®, Ans.minat=—!32. 12,yapX. Ans.Yminme at roe1premtsing(—Fcrc J).AnspauVEatet, aex x 4yosinde—e(—FcrcF), Ans.maxated,minatx=a—Z, 18,yoxstans. Ans, There is)neither max nor min, 16. y=eFsins, Ansmigatvein, maxalmDhnd2,7.poet!2242,Ans max when 2=0;twomin whenf=—1andwhenf=1.18y=(e—2)(e+1,Ans.Ymig=—8.A whenx=q. 18,yoatt. Ans.minwhenx1;max ‘ é = ygOw wienxem1MhyaatO—H!Ansdaghen=FYa=Ohen 0andwhensma.ay=ec.Anemaxwhenrms:min whenrms.28,yesVISEAnsgmcwhenx15Yuin—t 204 Investigating theBehaviour ofFunctions — 25/7 2 when x=1.2yeeVIFCD.Ans.gamV/Ewenod, 2pepe: Ansminwhengm—I;maxwhenxe.28youxins, Ans minwhenx=.28.yaexintx. Ans.maxwhenx=e?;minwhenx= 27,y=Inz—ate tan x.Ans. The function increases. 28 y=sin3x—3 sinx.Ans,minwhenx=;maxwhenx=3%.29,ya2e-tare tanx.Ans.No extrema.$0,yersinxcosts.Ans.minwhenx=;twomax:when rearecos V2andwhensanceos(— V2) Styearein(sin). Ans,maxwheneS408;minwhenxmAS) Find the maximum and minimum values ofthe function onthe indicated intervals: $2y=BFE (Pee2).Ans.Maximumy=?at x=,minimum y=—2atr=42.38.ya—2et43x41(—1Se<8). AnsMaximumvaluey=%atx=,minimumvaluey=—!2atx=—t yeEZ] O<e<4),Ans.Maximumvaluey=atx=4,minimum x x valuey==Uat60.88yosinde—e(—Fecec). Ans.Maximum x x x x vateyo%atcm,minimumvaluey=—%at= 36.Using square tinsheet with aside a,make atopless box ofmaximum volume bycutting equal squares atthecomers and removing them and{hen Bending the tinsoa8toform thesides ofthe box. What will the length. of 4sideofthesquaresbe?Ans.© 37.Provethatofallrectangles thatmaybe.inscribed inagivenciscle, thesquare hasthegreatat area.Alsoshowthat“heaquare willhavethe maximum perimeter aswell.38.ShowthatofallIsoscelesrangesinscribedinagivenercle,anequ lateral triangle has the largest. perimeter:38.Find righttriangle ofmaximum areswithahypotenuse A.Ans. it Length ofeach side, 40,Find theheight ofaright cylinder with greatest volume that can be inseribedinasphereofradiusR.Ans.Height,we 41,Find theheight ofaright cylinder with greatest lateral surface that imaybeinscribed inagivenspreofadiRUAns.Height, V2 42.Find the height ofaright cone with least volume circumscribed about agiven sphere ofradius R.Ans. 4R(the Volume ofthe cone isequal totwo ‘olumes ofthe sphere). 443A reservoir with asquare bottom and open top istobelined inside with lead. What are the dimensions ofthe reservoir (fohold 32litres) that Exercises onChapter V 205 willrequirethesmallestamountoflead?Ans.Height,02metre,sideofBase, 04 mette (the side ofthe Base must betwice the height). “AA rooter wants {0make anopen channel” ofmaximam capacity with bottom ana sides 10em in-widity and. withthe ‘sides inclined atihe some Engle tothe’ bottom. What isthe width ofthe channel atthe top? Ars aoe » 7Bprove that aconie tent ofgiven storage capacity requires theleat material when itsheight tsV2times theradius ofthebase fot is.required (omake. acylinder, open atthe top, the walls and bottom ofwhieh have a'given thiekaess, What should ‘the Ghnensions ofthe cylinder besothat foragiven storage capacity itwill require the least Tater? "Ane HR is(hiner tadier ofthebase, thetower volume of thecylinder, thenR=VE. 47. ItIsrequired tobuild aboiler out ofaeylinder topped bytwo henge and’with,ale ctl hchnts Satire, lume ori hould kove minimum outer surtate. Ane: Itshould have’ theshape ot 4spherewithInnerradiusR=J/ 48,Construct anIsosceles trapezoid, which foragiven area Shas aminirumperimeterstheangleatthe?baseofthe(rapessidsequaltoa.AasThelengthofoneofthenonparallel sidesisViz 40,Inseribe InagivensphereofradiusRaregular teangular prisofrmarimum volume, Ans, The altitude ofthe prism tsmen Va 50. Itisrequired tocircumscribe about ahemisphere ofradius Racone offlainain "lug: hepaneofthe baseofteCoecolnldes withhat ofthe hemisphere; find the altitude ofthe cone. Ans. The altitude ofthe coneisRV3. St. About 2given cylinder ofradius ¢citcumscribe aright cone ofmini rum volume: wesssunic the planes and ‘ceres ofthe eitetar bases ofthe Spina adie coneCond!" Ans.Theradsofthe baseofthecones equal toSr $2,Outofsheemeal, having theshapeof9celofradius cut aseclor such that itmay beDent info‘ funtel ofmaximum storage capacity. Ans,ThecentralangleofthesectortsanY/2, 53. Ofallcircular eylinders inscribed inagiven cube with side asothat thelr anes coincide wit the dlagonal of{he cube and. the circumferences of the bases touch itsplanes, find the cylinder wilh maximum volumes Ans The atteofthecylinderIsequalto23,theradiusofthebaseis52.54.Given, ina rectangular coordinate system, apoint (x.49)Iyinginthe fedganda Drawaight fine.houghthspointSo"thalitfoams atriangle ofleast area with the positive directions ofthe axes. Ans. The Straight tine inercepts ontheaxes thesegments 2yand 2ye:thuss ihasthe Ear equation4gow. 206 Investigating theBehaviour ofFunctions 58.Given apoint onthe axis ofthe parabola y*=2px_at adistancea otheeen tnd“he!shai “ofthepointofthe Curecose fi 50.Assuming thatthestrength ofabeamofrectangular cross-section isdirectiy" proportional tothewidthandtothecubeofthealtitude, findthe Sid of2beam ofmaximum strength that may becutout ofa logofdiameter Te'em. Ans. The width ts8cm. Sr.Atorpedo boatisstanding atanchor 9kmfromtheclosest pointof thetote: aiesenger, hastobesent'tg4camp1o-km (along theshore) {tom the point oftheshore closest iothe boat. Where should, tne messenge? land 5028 fogsttothecamp intheshortest possible timer llhe.does'8 kite Walking and 4‘krjtr towing” Avs. Ala point 3kimom thecacy Se point moves over'a plane inamedium situated outside fheline Ma with velocity. Oy and along. the line "MN with. velocity oy.What’ pathbetween AandB,situated onMN,willitcoverinthe:shortest time?TheGitesAnt“ the"estancete‘potionwotremtheWi TmmA AGAhaa heeltaeomoo 24 forSoS andacme tor9<th.59.AlodwishoistedbyaleverforceFisappliedtooneend,the intofsupportisattheatherendoftheleverIftheloadfssuspended Fim polit’ centimetres (romthe.fulerum, and.theleverrod,weghs © grams percentimetre oflength, what" should thelength oftherodbeforthe force(required toraisetheload)tobeaminimum? Ans.x=V/cm, 60. For nmeasurements ofanunknown quantity «the following’ readings have been blained: Xjfy,vey fqShow that the’sum ofthe squares ofthe tirors GayoesteceeGeaywillbeTeastifor#'wefakethe number Stat.ty 61, Toreduce the friction ofaliquid against thewalls ofachannel, the ares inContact with the guid crust beaiaimumn, Show that the beat shape ata" open rectangular channel with: given crostsetional aren ithat for hich the width ofthe channel istwice ite altitude Betermine thepoints ofinflection and theIntervals ofconvexity andcon- cavity ofthe curves ed.yest. Ans. Forx<0 thecurve isconvex; forx>0 thecurve iscon-cavesalZaohereloapointof,infection. G2.yori Ans.The‘carve fseverywhere convex. Gh.y=a!—Sx'—9x+9, Ans.Pointofinflection atPaIVSG OeAns.Beatotnestion arm.tsyadns.The cave reg emer. pay An,Plt olSteen xm t——. 68. getanx. Ans. Point of inflection atx=nx. 69. y=xe~*.£Vach . y Ans. Point ofinflection atr=2 70.y=a—/x—6. Ans. Point ofinflec- tionatr=b.71.y=a— j/(e—b). Ans.Thecurve hasnopoint ofinflection. Tiedtheesate tefllorng cares port. Anezm 1 om0.1.yey. Ans.x2yO.yetoes. Ans.md, 9-0.2.yet A2,90. yetay.Ans.« Exercises onChapter V 20 yeTSyeeF1Ans0,yO.16,ymin,Ansrnd TePesta Ansyard? 78ylaataw Ans.ytenn, 79.gage Ans.x=2a,80,y*(x—2a)=x4—a". Ans.x=2a,y=4(x40). Investigate the following functions and construct their graphs: Bye teHI yeas. 8yee. ye, str __* et? # we 85.yAEE06,yeah. otymEE?a8,ye.Operon, 90.yg yeVHD tO.yeeVIET. yeVE. M, yase-*, 95. yale. 96, yor—In(etl). 97. yan (et+tBepatina. bo.ySegemm, foo!yeeaae, toi.Symesad et, es toe,y=insing, 108.y=! 104. 1,ts.{70 Fa gate yexea(—sinh, xmat!cost, ‘0s.{jZeumene ae{poeta Additional Exercises Findtheasymptotes ofthefollowing lines:108.y=2"!Ans.x=—t: Yee. 10, yaeten®. Ans. yor. M0, 2y(efIPax, Ans x=; gaye. AM,y'sat—at. Ans.x+y=0.2.ye-™sine.Ans.y=O, M3, yeerFsndebe,Ansyas1M.gmcin(ept). Ansc=—t; perth. us.gese™, Ans.xm0;yor.116.ce2,ye,eee iam "Tor Ans,gab px—y* Investigate andgraph thefollowing funetions: 117.y=Lx|. 118.y=n|x. M9,gtoat—x. 120,ym(eI)(x—2).2.yetlx].122,ysVxt—x.— at ~Fin 4 3,yaVFFT. 14,y=Eine, 125.ye Bing128,yal. tareyeeptahpoebE,tap,yeringBDymetmeBLymwelsingel.182g=t2Z, 18,georaretans, 14,yoe—2aetane, 135, ymen™sinSe.196,y=|sin|+x.197,ymsinat,138.y=cos*x+sin*x. CHAPTER VI THE CURVATURE OF ACURVE SEC. 1,THE LENGTH OF AN ARC AND ITS DERIVATIVE Let the arcofacurve MyM (Fig. 136) bethe graph ofthe function y=f(x) defined ontheinterval (a,6).Let usdetermine the are length ofthecurve. On the curve M,M take thepoints MyM, Myy eee) Mics MyoyMay, M.Connecting the pointswegetabroken lineM,M,M,...M,_,M;...M,_,M inscribed in ~ theareM,M: Denote’the length ofthis Mp Ms broken linebyP,. The length ofthearcM,M isthelimit (we denote itbys)approached bythe length ofthe broken line asthe largest Mot) ofthelengths ofthesegments ofthebro- kenlineM,_,M, approaches zero,ifthis M®Jimitexistsandisindependent ‘ofany Fig 196, choice ofpoints of the broken line M,M,M,...M,.,M,..-M,—.M. Itwill benoted that this definition of‘the arelength ofan arbitrary curve issimilar tothe definition ofthe: length ofa circumference. InCh. XII itwill beproved that ifafunction f(x) and its derivative /’(x)arecontinuous onaninterval [a,6],then the arc ofthe curve y=f(x) lying between the points ‘fa,f(a)] and 1b,F(O)] has adefinite length; amethod will beshown forcom- puting this length. There also, itwill beestablished (asacorollary) that under thegiven conditions theratio ofthelength ofany arc ofthis curve tothelength ofitschord approaches unity when the length ofthechord approaches zero, that is, limnatnta_ MM»length MyM . This theorem may beteadily proved forthe circumference *)of *)Consider the arc AB, thecentral angle ofwhich Is2a. (Fig. 137). The length ofthisareis2Ra(R istheradiusofthecircle),andthelengthofitschordis2Rsina.Therefore, lim108thAB.jim2Ra__y_ a? length AB a+IRsing The Length ofanArc and ItsDerivative 209 acircle; however, inthe general case weshall accept itwithout proof (Fig. 137). Let usconsider the following question. Onaplane wehave acurve given bytheequation y=f(a). LetM,(x,, y,)besome fixed point ofthecurve and M(x, y), some variable point ofthe curve. Denote bysthe arclength MM (Fig. 138). y % wte yfB ee 5 <jm 8 , " . yA Ol%xxd* Fig. 197. Fig. 138. The arc length swill vary with changes inthe abscissa xof thepoint M;inother words, sisafunction ofx.Find thederi- vative ofswith respect tox. Increase “xbyAx. Then the arcswill change byAs=the length ofMM,. LetMM, bethechord subtending this arc. In ordertofindlim&doasfollows: fromAMM,Q find MMi=(Ax)?+(Ay)*. Multiply and divide theleft-hand side byAs*: MM,\* 2 2 2(BBY ast=(an'+ant, Divide allterms ofthe equation byAx*: ‘HM,*(as)*_‘ay)* (BE) (a)=1+()- FindthelimitoftheleftandrightsidesasAr—-0. Taking intoaccount thatlim“#4—1andthattim44=4 wegetin. a araOtde ds) ay(@)'=1+(2) 210 TheCurvature ofaCurve or as taysa/ 1+(#)- a) For thedifferential ofthearcwegetthe following expression: at as=14(%)‘ae oy or*) ds=Vax' $y. (2) We have obtained anexpression for the differential ofarc length for the case when the curve isgiven bytheequation y=f(x). However, (2')holds also forthecase when thecurve is Tepresented byparametric equations. Ifthecurve isrepresented parametrically, x=9(t), =v), then dx=q' (dt, dy=y' (tat, and expression (2')takes theform : ds=Vig OFF¥Ordt. SEC. 2,CURVATURE One oftheelements that characterise theshape ofacurve is, thedegree ofitsbentness, orcurvature. Let there be acurve that does not intersect itself and has adefinite tangent ateach point. Draw tangents tothe curve at any two points Aand Band denote theangle formed bythese tangents byafor, more precisely, the angle through which the tangent turns from AtoB(Fig. 139)]. This angle iscalled the ‘angle ofcontingence ofthearcAB. Oftwo arcs ofthesame length, that arcismore curved which has agreater angle of contingence (Figs. 139 and 140). Ontheother hand, when considering arcs ofdiferent length we cannot evaluate thedegree oftheir curvature solely bytheappro- *)Strictly speaking, (2°) holds only for the case when dx>0. But if dx<0, then ds——Vdst+dyt. Forthis reason, inthegeneral case this formula ismore correctly written as[ds|=Vaxtdy*, Curvature an priate angles ofcontingence. Whence itfollows that acomplete description ofthe curvature ofacurve isgiven bythe ratio of theangle ofcontingence tothe length ofthe corresponding arc. be a @ 4 A Fig. 139. Fig. 140, Definition 1.Theaverage curvature K,,ofanarcABisthe ratio ofthe corresponding angle ofcontingence «tothe length of the are: Kiesao For one and the same curve, the average curvature ofitsdiffe- rent_parts (arcs) may bedifferent; forexample, forthecurve shown inFig. 141, the average curvature of theareABisnotequalto_theaveragecurvature ofthearcA‘B,, although o thelengths oftheir arcs arethesame. 8) What ismore, atdifferent points the curvature ofthe curve differs. To cha- racterise the degree ofcurvature ofa given line inthe immediate neighbour- hood ofagiven point A,weintroduce Papptheconcept ofcurvature’ ofacurve at ba agiven point. Definition 2.The curvature K,ofaline atagiven point Ais thelimit oftheaverage curvature oftheareABwhen thelength ofthis areapproaches’ zero (that is,when the point Bapproa- ches the point A): K,= limKyy= lim2%), pea abe AB *)Weassume that themagnitude ofthelimit does not depend onwhich side ofthe point A'we take the variable point Bonthecurve- 212 The Curvature o}aCurve Example. Foracircle ofradius r:1)determine theaverage curvature of the are ABsubtending the central angle a(Fig. 142); 2)determine the curvature atthe point’ A. * Solution. 1)Obviously theangleofcontingence oftheareABisa,the length ofthe are isar. Hence, Kant Qeaear Zm~8ot y 1 Kea. | 2)The curvature atthe point Ais Kea tim1 anear Fig. 142. Thus, the average curvature ofthe arcofacircle ofradius ris independent ofthe lengih and po- 1 sition oftheare, and. forallares itis equal tol. Likewise, thecurvature ofacircle atanypoint isindependent ofthe choiceofthispointandisequalto+ Note. Itshould benoted that, generally speaking, forany curve the curvature atitsvarious points differs (this will beseen later). SEC. 3.CALCULATION OF CURVATURE Let usdevelop aformula forfinding thecurvature ofany line atany point M(x, y).Weshall assume that thecurve isrepresen- ted inthe Cartesian coordinate y system byanequation oftheform y=f) 0) andthat thefunction f(x) hasa | continuous second derivative Draw tangents tothecurve atthe points Mand M, with abscissas x and-x+Ax and’ denote by@and . @+Ag the angles ofinclination of +d these tangents (Fig. 143). a ¥ Wereckon thelength ofthe fig.18areM,Mfromsome fixed point M, ne anddenote itbys;then As=M,M,—M,M, and |As|=MM,. ‘Aswill beseen from Fig. 143, theangle ofcontingence corres- Cateutation ofCurvature 213 ponding tothearcMM, isequal totheabsolute value*) ofthe difierence ofthe angles @and @-+Ag, which means itisequal to|Ag|. According tothe definition ofaverage curvature ofacurve, on the segment MM, wehave =|4e1_|ae,Keo=Tas]“|iI. Toobtain the curvature atthe point M,itisnecessary tofind the limit ofthe expression obtained on the condition that the arelength MM, approaches zero: ae Kesi. Since thequantities @and sboth depend onx(are functions ofx),@may thus beconsidered asafunction ofs.Wemay con- sider that this function isrepresented parametrically bymeans ofthe parameter x.Then ‘im 88.49was a and, consequently, ak=|s|- @ Tocalculate $2,wemake useoftheformula fordif- ferentiating afunction represented parametrically: 4gdo_aeana Ge Toexpressthederivative $2intermsofthefunction y=/(x), we notethattang=%! and,therefore, g=are tan. Differentiating thelatter equality with respect tox,weget ay do __ae aug *)Itisobvious thatforthecurvegiveninFig.143,|Ap|=Aq since ae >o. 20 The Curvature of@Curve Asregards thederivative $£,wefoundinSec.1,Ch.VI,that ds Taya7Vi+%) . Therefore, ay aagay aywee) a ae att ayye&Vi+() [+(%)] or,sinceK=|32],wefinallyget (zaxe ®b+(@)] Itisthus possible tofind the curvature atany point ofa curvewherethereexistsasecond derivative £4andwhereitis continuous. Calculations are done with formula (3). Itshould be noted that when calculating the curvature ofacurve only the arithmetical (positive) value oftheroot inthedenominator should betaken, since thecurvature ofaline cannot (by definition) be negative. Example 1.Determine the curvature ofthe parabola y*=2px: B)atthepoint¥,(0,0); 6)atthepointmy($.0) Solution. Find thefirst andsecond derivatives ofthefunction y=V2pe: dy__p_ dy GeV 2px" Ge pay” Substituting theexpressions obtained into (3), weget a » k=, JBeet 1 b)Krewe : 1 0Ket aie Calculation oftheCurvature of@Line Represented Parametrically 215 Example 2.Determine thecurvature ofthestraight line y=ar+0 atan arbitrary point (x,y)- Solution. eyma, 0. referring to(9) we get"e2« K=O. Thus, astraight line isa“line ofzero curvature”. This very same result ts readily obtainable directly from the definition ofcurvature, SEC. 4,CALCULATION OFTHE CURVATURE OFALINE REPRESENTED PARAMETRICALLY Let acurve berepresented parametrically: x=91, Y=Vl). Then (see Sec. 24,Ch. III). tyV0dy_venweax Fl) oe oro Substituting theexpressions obtained into formula (3)ofthe preceding section, weget lve—w¢ K=eh * . 1eter 0 Example, Determine thecurvature oftheeycloid x=a(t—sin!), y=a(l—cos ft) atanarbitrary point (e,9) Solution. ae ate 4 eyFaat—cosy, Fmasint, Ymasint, fmacost. Substituting theexpressions obtained into (3), weget ala(l—cosi)acost—asinteasint| eost—t)=ell—eosi)acost—asintasin tlleast Ta(L—cos OFatsiat(7Pha—cosHi 1 1 “Thales? alain |”copuF] SEC. 5.CALCULATION OF THE CURVATURE OF ALINE GIVEN BYAN EQUATION INPOLAR COORDINATES Given acurve represented byanequation oftheform e=F 0). qt) 216 The Curvature of@Curve Write the transformation formulas from polar coordinates to Cartesian coordinates: x= Qc0s6,a ® y=esinb. Ifinthese formulas wereplace gbyitsexpression interms of6,i.e, £(6), weget x=(6)cos0, 3 y=f() sind. ® The latter equations may beregarded asparametric equations ‘ofcurve (1), the parameter being 0. Then =Weost—osins, “=48sind+ecos6, . $8=FBcos28sinb—gcos), Fu£8sind+258cosd—gsind. Substituting the latter expressions into (1) ofthe preceding section, wegetaformula forcalculating thecurvature ofacurve inpolarcoordinates: eedK=Let2er eelrey © Example, Determine thecurvature ofthespiral ofArchimedes @=a8(a>0) atanarbitrary point (Fig. 144). Solution. 8 4g dgOe Bao, ao. e Hence Pamolas (Or aye OI Itwill be noted that for -large values ofOwe have theapproximate equalities ot?O411SETA;therefore,rep- Jacing 0°42 byO*and O41Fig.144, byOFintheforegoingformals, The Radius and Circle ofCurvature, Evolute and Involute 217 wweget anapproximate formula (lor large values of®) LeoKogyn ab Thus,forlargevalues of@thespiral ofArchimedes has,approximately, the same curvature asacircle ofradius a8. SEC. 6,THE RADIUS AND CIRCLE OF CURVATURE. CENTRE OF CURVATURE. EVOLUTE AND INVOLUTE Definition. The quantity R,which isthe reciprocal ofthecur- vature Kofaline atagiven point M, iscalled theradius of curvature ofthe line atthe point inquestion: : 1 R=x Oy or ay)*) la]ae Draw anormal, atthe point M,toacurve inthedirection of theconcavity ofthe curve, and ‘layoffasegment MC equal to theradius Rofthe curvature ofthe curve atthe point M.The iY yBi C(@,f) Min) a ¥ Fig.145) Fig.146. point Ciscalled the centre ofcurvature ofthegiven curve atM; thecircle, ofradius R,with centre atC(passing through M)is called thecircle ofcurvature ofthegiven curve atthepoint M (Fig. 145). From the definition ofcircle ofcurvature itfollows that ata given point thecurvature ofacurve and thecurvature ofacircle ofcurvature are the sate. ae The Curvature ofaCurve Let usderive formulas defining thecoordinates ofthecentre of curvature. Let acurve begiven bythe equation y=f(x). (3) Take apoint M(x, y)onthis curve and determine the coordi- nates aand Bofthe centre ofcurvature corresponding tothis point (Fig. 146). Todothis, write theequation ofthenormal to the curve atM: Y¥—y=—4(x—2). 4 (Here, Xand Yare the moving coordinates ofthe point ofthe normal.) Since the point C(a, B)lies onthe normal, itscoordinates mustsatisfy equation (4): j B—y=—} (a—2). ) Further, the point C(a, B)isseparated from M(x,y) bya distance equal totheradius ofcurvature R: (aa) +(G—y)"=RE CC) Solving equations (5)and (6)simultaneously, wefind aand B: =x+ja@—9=R,a ype (oatsea Whence fl wrt LR, Beye amakyi BavFrae q agei andsinceRTT , bayomg, parte. Inorder todecide which signs (top orbottom) totake inthe latter formulas, we must examine the case y’>0 and the casey'<0.Ify7>0, thenatthispointthecurveisconcave, and,hence, B>y (Fig. 146), and for this reason wetake thebottom signs. Taking into account that inthis case |y"|=y’, theformulas ofthe coordinates ofthe centre ofcurvature will be ayty)oe | )peyttte. f The Radius and Circle ofCurvature. Evolule ond Involute 219 Similarly, itmay beshown that formulas (7)will hold forthecasey’<0aswellIfthecurve isrepresented bythe parametric equations x=9(0, Y=vl), then thecoordinates ofthecentre ofcurvature arereadily obtain- able from (7) by substituting, inplace ofy’and y’, their expressions interms oftheparameter 1h xvifea vot. x * Then ane OE)eyTF0,* .saystaehy (7)Sut ara ee* Example 1.Todetermine thecoordinates ofthecentreofcurvatute oftheparabola P=2px: 2)atanarbitrary point M(x, y);b)atthe point M,0,0);c)atthe pointa,($9) - Solution. Substituting thevalues£2and{Yinto(7)weget(Fig.147: exh y a=3r+p, B=; ph 9emdete, PAS y b)atx=0 wefind a=p, B=0; D 5p o)atrad wehavea=, p=—p ” IfatM,(x, y)ofagiven linethecur- c *vature differs fromzero,thenaveryde- ap)finite centre ofcurvature C,(a,B)corres- ponds. tothis point. The totality ofall centres ofcurvature ofthegiven line forms acertain new line, called theevolute, with Fig.197. respect tothe first. Thus, thelocus ofcentres ofcurvature ofagiven line iscalled the evolute. Asrelated toitsevolute, thegiven line iscalled the evolvent orinvolute. Ifagiven curve isdefined bytheequation y=f(x), then equa- tions (7)may beregarded astheparametric equations oftheevo- 220 TheCurvature ofaCurve lute with parameter x.Eliminating from these equations thepara- meter x(ifthis ispossible), we get animmediate relationship between thecoordinates oftheevolute aand B.But ifthecurve isgiven byparametric equations x=(), y—(t), then equa- tions (7’) yield the parametric equations ofthe evolute (since the quantities x,y,x’,y',x’,y°are functions off). Example 2.Find the equation ofthe evolute ofthe parabola y=2px. Solution, Onthe basis ofExample 1we. have, for any point (&, y) ofthe parabola, a=3e+p, Ay sa i, p=, IP Ve Eliminating the parameter xfrom these ( equations,“weget8ip x < Thisisthe equation ofasemicubica OXY parabola(Fig.148). Example3.Findtheequationofthe evolute alanellipse represented bythe parametric equations maces, y=bsint. Solution. Evaluate the derivatives of« and yWith respect to , m—asing, 4=0cost;ote a—acost,’ y=—bsint, Substituting theexpressions ofthederivatives into (7'), weget Beanflatsint +btcost) 008Tesinkpadcost scant—acos¢snt¢—5conttm(a—22)cost Thus, .a=(0-2) costs Similarly weget p=(8) aut, The Properties ofan Evotue Pa Eliminating the paramete t,weget theequation ofthe evolute ofthe elies mee ay, (BY (aman(5)"+(2)"-(3") Here, @and Bare thecoordinates oftheevolute (Fig. 149). Example 4.Find the parametsic equations oftheevlue ofthe eycoid rceit—aa) y=a(1—cosf). y Solution Fma(\—cos; yma satsPoarofee, IAN Substituting the expressions obtainedinto(7),weget AW a=a(t+siné), AX porate SS):Rearrangethevariables,putting \\ ares \ Sraface V then the equations oftheevolue will Bez (Re fen taa(e—s 9, Fe. 10."yea(icon: « they deing, Incoordinates &,9.acyclo withthe same generating circle oftadlusa,‘Thus,theevolute of8cyalold isthalsamecyclo displaced along Taeadis Byte Sed tocg theaed yates vei n y wove = 7 Fig. 180 SEC. 7.THE PROPERTIES OF AN EVOLUTE Theorem 1.The normal toagiven curve isatangenttoitsevolute. Proof.Theslopeofthelinetangent toanevolute defined by the parametric equations (7) ofthe preceding section is 22 The Curvature ofaCurve equal to4 48 ap_a& aa aa & Noting that [byvirtue ofthesame equations (7’)] da Sy"yyyy", Butyyyy"Be eee syvy"vy" cr (2) wegettherelationship a day" But y’istheslope ofthe line tangent tothe curve atthecorre. sponding point; ittherefore follows from therelationship obtained that thetangent tothe curve and the tangent toitsevolute at thecorresponding point are mutually perpendicular; that is,the normal toacurve isthetangent totheevolute. Theorem 2./f,over acertain segment M,M, ofacurve, the radius ofcurvature varies monotonically (i.e., either only increases oronly decreases), then theincrement inthearclength oftheevo- lute onthis segment ofthecurve isequal (inabsolute value) to thecorresponding increment intheradius ofcurvature ofthegiven curve. Proof. From formula (2'), Sec. 1,Ch. VI, wehave dst=dat+-dB* where dsisthe differential ofthe arc length ofthe evolute;whence Pye(a)-(@)+(2y- Substituting, here, theexpressions (1)and (2), weget as)" wy(Stay muy"($)'-aty y(ear. @) Thenfind($#)".Since , ;at pieMeyRate, Ra. Differentiating both sides ofthis equation with respect tox,we getthefollowing (after appropriate manipulations): aR_2+y")*Gy'y*@—y'" —yy") Ra WF . The Properties ofanEvolute 223 Dividing bothsidesoftheequation byaRa2 wehave AR_(yyy —y"v"') aera p Squaring, weget (8)a+(eer. ) Comparing (3)and (4), wefind aR\* ‘ds\*(ey) whence aR_ ods ata Itisgiven that$%doesnotchange sign(Ronlyincreases or onlydecreases); hence, “4doesnotchange signeither. Forthe sakeofdefiniteness, let$2.<0, $450 (which corresponds to Fig.151).Hence, 2=—4, Let_the point M,have abscissa x,andM,have abscissa z,.Apply the Cauchy theorem tothe functions s(x) and R(x) onthe interval [x,,x]: ds RGI—RO) ® , ae eat where &isanumber lying between x,and x,(x,<E<x,). ‘We introduce the designations (Fig: 151) s(x)=5, s(x,)=s, R(x)=R, R(x,)=R,. Then#==—1, ors,—s,=—(R,—R,). Butthismeans that 15-5 1=1R,—Ry |. Thisequality isproved inexactly thesamemanner iftheradius of curvature increases. We have proved Theorems 1and 2forthe case when thecurvaisgivenbyanexplicitequation, y=f(2). 24 The Curvature ofaCurve Ifthe curve isrepresented by parametric equations, these theorems also hold, and their proof isexactly the same, Note. The following isasimple mechanical method forconstructing acurve (involute) from itsevolute. si 3]3 4s, | eSwe Me 7 AG Lov a % @ Fig. 151. Fig. 152. Let aflexible ruler bebent into the shape ofanevolute C,C,(Fig. 152).Suppose oneendofanunstretchable string isattached totitepoint C,and bends round theruler. Ifwehold thestring taut and unwind it,theend ofthestring will describe acurve M,M,, a which istheinvolute (orevolvent, the name coming from this process of“evolving”). Proof c that this curve isindeed an ; involute may becarried outbyrs meansoftheabove-establishedMmpropertiesoftheevolute. [|\A Itshouldbenotedthattoa Kr} single evolute there correspondoPXaninfinitude of—variousinvolutes (Fig. 152). Example. Let there beacircle of radius a(Fig. 153). Take theinvolute Ofthiscirclethat’passesUhrough the Fig. 153. point M,(a,0). Approximating tieReal Roots ofanEquation 23 TakingintoaccountthatCM=CM,=at, itiseasytoobtaintheequations ofthe involute ofthe circle: OP=x=a(cost+tsin‘), PM =y=a(sint—tcos. Itwill benoted that the profile ofatooth ofagear wheel ismost often inthe shape ofthe involute ofacircle. SEC. 8.APPROXIMATING THE REAL ROOTS OF AN EQUATION Methods ofinvestigating thebehaviour offunctions enable usto approximate the roots ofanequation: F(x) =0. Ifthe equation isanalgebraic equation*)ofthefirst,second, third,orfourthdegree, thereareformulas whichpermitexpressing theroots oftheequation interms ofitscoefficients bymeans of afinite number ofoperations ofaddition, subtraction, multiplica- tion, division and evolution. Generally speaking, there arenosuch formulas forequations above the fourth degree. Ifthecoefficients ofany equation algebraic ornonalgebraic (transcendental) arenot literal but numerical, then theroots oftheequation may becal- culated approximately toany degree ofaccuracy. Itshould benoted that even when theroots ofanalgebraic equation areexpressed interms ofradicals, itissometimes better, practically speaking, toapply anapproximation method ofsolving theequation. Below wegive some methods ofapproximating the roots ofanequation. T,Method ofchords. Let there beanequation F(x)=0 (wy where f(x) isacontinuous, doubly differentiable function ontheinterval[a,b].Supposethatbyinvestigating thefunctiony=f(x) within theinterval (a,b]weisolateasubinterval [2,4]suchthat within this subinterval thefunction ismonotonic (either increas- ingordecreasing), and attheend points the values ofthefunc- tion f(x,) and /(x,) areofdifferent signs. Fordefiniteness, wesaythatf(x,)<0, f(x,)>0 (Fig.154).Sincethefunction y=f(x)iscontinuous ontheinterval {r,,%) itsgraph‘willcut.the¥-axis insome one point between x,and x,. Draw achord AB connecting the end points ofthe curve y=F(x), which correspond toabscissas x,and x, Then the *)The equation /(s)=0 is called algebraic i f(x) Isa polynomial (seeeeTSeatenLe) le £(2)is@polynomial ( ©~a208 226 TheCurvature ofaCurve abscissa a,ofthepoint ofintersection ofthis chord with thex-axis will betheapproximate value oftheroot (Fig. 155). Inorder to find this approximate value letuswrite theequation ofthestraight line AB that passes through two given points A(x, f(x,)] andBly,Fle): 4 ; =H) _= TF) ae /, 9 fi) |10%) %/| 97 a | 7] reed Yer *Aria) Fig. 154,. Fig. 155. Since y=0 atx=a,, itfollows that =H) ama TRH) HH" whence (ama) fx) = TT)” @ Toobtain amore exact value ofthe root, we determine /(a,). Iff(@,)<0, then repeat the same procedure applying formula (2) totheinterval [a,,x].Iff(a,)>0, then apply thisformula totheinterval[x,,a,].Byrepeating thisprocedure severaltimeswewill obviously obtain more and more precise values ofthe root a,, etc. Example 1.Approximate the roots ofthe equation Ha)=2"— 64+2=0, Solution. First find thesegments where the function’ f(x) Ismonotonic.Evaluating’ thederivative ffG@)=3e—6, wefind:that{tispositiveforx<—V3, negative for—V2<x<4Y2andagainpositiveforx>VE (Fig,189).‘Thusthefunction hastheeesegments ofmonatonlcty, oneachof Tomakethecalculaiions moreconvenient, letusnarrow thesesegments ofmonotonicity (but insuch manner that there should beacorresponding Approximating the Real Roots ofanEquation 27 root oneach segment). Todothis, substitute into expression /(x), atrandom, Some values of£,then isolate (within each segment ofmonotonicity) such Shorter intervals) that the functions at the end points will have different signs: \y x=, FO)=2, belBot pipzts } — 4=-3, f(—3)=—-7, @ R=? (—=6, 7 =? 1Q)=—2, } 74=3, F@=U, 4 Thus,theroots liewithin theintervals 7 @,1), (-3,-2, @,3). 4 Find the approximate value ofthe root inthe 3 Interval (0,1);from formula (2)wehave 2 ano2502294, ¥|SSI 3a o% Sinceia 1(0.4) =0.4°—6-0.44+-2——0.335,f(0)=2, IfollowsthattherootliesBetween 9and04Again 2 applying(2)tothisinterval,wegetthefollowinapproximation: ® . ied=0—CASO PS,0.542,ete. is Similarly we approximate the roots inthe other intervals Fig. 156. 2.Method oftangents (Newton's method). Again, letf(x,)<0, {)>0; andontheinterval [x] thefistderivative déesnot change sign. Then there isoneroot oftheequation f(x) =0inthe interval (x,, x,). Let usassume that thesecond derivative does not change sigiftheinterval [xx4]ether: thiscanbeachieved by reducing the length ofthe interval within which the root lies Retention ofthe sign ofthe second derivative onthe interval [x,, *] means that the curve iseither only convex oronly concave on[x,,x. Draw atangent tothe curve atthe point B(Fig. 157). The abscissa a,ofthe point ofintersection ofthe tangent with thex-axis willbeanapproximate value oftheroot. Tofindthis abscissa write theequation oftheline tangent atthepoint B: yh) =F(%)(#4). Noting that x=a, aty=0, wehave a=1,7, ®) o 298 TheCurvature ofaCurve Then, drawing the line tangent atthe point B,, weanalogouslyfind@more exact value oftheroota,.Byrepeating thisprocedure y Ay 8 7| 7% of ; GN 444 e 7 — O} / * A A Fig: 157. Fig, 158 wecan calculate theapproximate value oftheroot toanydesired degree ofaccuracy. Note the following. Ifwedrew the line tangent tothecurve not atthepoint Bbut atA,itmight appear that thepoint of intersection ofthetangent with thex-axis A liesoutside theinterval (x,,x). From Figs. 157 and 158" it‘follows that the tangent should be:drawn attheend oftheareatwhich thesigns ofthefunc- | tion and itssecond derivative coincide. hg Since itisgiven that ontheinterval [x,, x,]thesecond derivative retains itssign, athe signs ofthe function and the second Pa derivative must coincide atoneoftheendait points. This rule also holds forthecase whenf'(x)<0. Ifthelinetangent is drawn atthe left end point ofthe interval, then informula (3) wemust putx,inplace ofx,: a=x,—fed, @) When there isapoint ofinflection Cintheinterval (x,, x,), themethod oftangents can yield anapproximate value ofthe root lying without theinterval (x,,x,)(Fig. 159). Example 2.Apply formula (8)tofinding the root ofthe equation He) =x! 6r2=0 within the interval (0,1).We have 10=2, /O=@—6|,..=—6, Exercises onChapter VI 229 and sofrom (3)weget 201 a=0-3.=4=0.333. 3.Combined method (Fig. 160). Applying atthesame time on theinterval (x,,x,]themethod ofchords and themethod oftan- gents, wegettwopoints a,and ya,lying oneither sideofthe 8 desired root a,since f(a,) and H(@,) have different signs.’ Then,on‘theinterval [aq,]again fie,apply themethod ofchords and the method oftangents. Thisyields twonumbers: a,anda,, iy, which are still closer to the value oftheroot. Wecontinue b xinthismanneruntilthedifference (by) iSbetween theapproximate values 1G)found isless than therequired degree ofaccuracy. Ttwill be noted that inthe combined method weapproach Fig.160. the sought-for root from two . sides simultaneously (i.¢., atthe same time weapproximate the root with anexcess and with adeficit). To illustrate, inthe case we have examined itwill beclear that by substitution we have F(0.333) >0, (0.342) <0. Hence, theroot isbetween theapproximate values obtained: 0.333 <x <0.342 Exereises onChapter VI Find the curvature ofthe curves atthe indicated points: 1.Betbattatbtatthepoints(0,6)and(a,0.Ans. at©,6); Ato. % 2xy=I2 atthepoint (3,4).Ans. Zh. 6x, 8yestattheolntteeweAnsoe 4.W6ytetst—a¥ abthepolnt@,0}Ans.2, 230 TheCurvatureofaCurve Fyyteat 1 5.xty?aa"atanarbitearypoint,Ans.—1 3(axy)* Find the radius ofcuryature, ofthe following curves at,the indicated points: draw each curve and construct the appropriate circle ofcurvature . V0 6.ytas! afthepoint (4,8).Ans, Ra VD 7,stesfay atthe point (0,0).Ans. Rew2a. 8.bit —atyt=ath"atthepoint(xy,y,).Ans.panto, 9.y=Inx atthepoint (1,0).Ans. R=2 V3. Tonyaathepoint(3.1).AneRa 1,STO}forttyAns.R=Bastaost Find the radius ofcurvature ofthe indicated curves: rast1aSOM VortakAntsRa. 18.Cirle gmasin®. Ans.Rad. ;tp atM4.SptofArchimedes quad,Ans,R=te, 18Cardotd gma(t—cos®). Ans.R=2VaR. 16,Lemniscate tateos28.Ans.R=S™.‘ o :eec 17,Parabola gmasee". Ans. Retasect 2. 18.qmasint$ Ans,Radasint2, Findthepinsofcursatwhiehtheausofcurvature Is2minimum:Zt v9.pains.ans.(42,—5n2) Ling, V2 wogmet,Ans,(Fin, 42). a.VE4VG=VE. Ans.(4.4). aayean(1Z). AnsAttepint0,0)Rae. Findthecoordinates ofthecentreofcurvature (a,B)andtheequation of the evolute Toreach ofthefollowing curves: Paeae rrnsaee Mattyhaa?. Ans. a=x43x 9%;Paytar"ys Everts onChapter VI 2a raat attI5y',payor Bytaete AneoH, 5S rat, on pose? t AA sabincol $—keost, ict a {SRI Bag,gk (FEF) crate, OVyetsint sialon +n, : am{FTEs ayygaacont: Pest x=acos't, 7 Pts oo.{Faas ‘ans.aacon(908a masite20costint 3FtdtheToletthe equation 244-420 totveedecimal places. Ane cates esate eae Su"Yor theeaston f()'se—-s02=0, approximate the root inthe interval Cay ane L088 SEeels herolfieequation at4.2—6r+ 20totwodeco placesAns098rySOSHNe,© SSsalvethe,gaiationit—350"appronimately, Ans,«51.71, =laiV3 noite S 24,Approximate therootoftheequation*—tanx=0lyingbetween0 and38Ans,4.4905 35:Evaluate the soot oftheequation sins=1—x tothree plas ofdec mals. Hint. Reduce the equation totheform f(x)=0.' Ans. 0.5110<x< east ‘Miscellaneous Problems . 86.Show that ateach pont ofthe Jemriscate ofa"cosBpthecurvature isproportional Yothe waste actor ofthe ona OPPPind he” genes value ot the® dis of curvature ofthe curve emaset Ans,R=. 28,Find thecoordinate ofthecentre ofcurvature ofthe curve g=zins atfhepointahayt nseh 5: Bove tha olpois ofinspiel ofArchimedes ¢=a9 as9 othe mmagetade otthedefence between ie radius weet a8a (he fads ofcute kre approaches er,WORK parasia yart4-be-, which hascommon tangent and cur atorwiththesiecurvey=snxatthepoint(1). Makeadeawing ate a Ans.y=— Ft41-F. 41. The function y= (x) isdefined asfollows: [()=# inthe inter—o<2ch, [e)merttorteintheiteraIecto. Whatpuso and¢beforheLinegos) toRove_contnuos curate Netreberhaie's drawing. ne 223, 6 3 at The Carvatare of@Carve (2,Show thatthe eau of«curvature ofacylold atany oe ofits point ese the length ofhe somal sttat plas {3Wate‘he equnon ote cel fuser ofthe parabola ytathepointLb,An.8+(gE) =, 4,Wate theequation ofthe clef cxrvte ofthe carve yestans at thepoint(1).Ans.(2-220), (2)"128 epoint(2,1).ans.(x2512)"(y—2)'a15.Find thelahotTheenti.clueof4lpnwheesears 46.Find the approninate value ofthe roots ofthe equation aefa2 toRa oeCL UTPindfehopeatdate valuetthe volsof'tequation stnz=08 towin" Got” ARE*Fbeeuaton Ns, ony ‘one fal tott stk Tava ihe Spprontnal aoe ofheYO othe cunt tan2=1tominisO48).“M0"FResguntin asay"oneealon2Fos CHAPTER VII COMPLEX NUMBERS. POLYNOMIALS SEC. 1,COMPLEX NUMBERS. BASIC DEFINITIONS Acomplex number isthe expression a+bi (1) where aand 6arerea! numbers, iistheso-called imaginary unit, which isdefined bytheequalities i=Vo1or*=—1; (2) aiscalled thereal part, andbi,theimaginary part ofthecomplex number. Two complex numbers a+6i and a—bi that differ onlyinthesignoftheimaginary partarecalledconjugate. Tia=0, the number 0+bi=bi iscalled apure imaginary; if b=0, weget areal number: a+0-i—a, Weagree upon thetwo following basic statements: 1)two complex numbers a,+6,i and a,+6,i areequal if a,=4,, b,=6,, that is,iftheir real parts areequal and their imaginary parts are equal; 2)acomplex number isequal tozero: a+bi=0 ifand only ifa=0, 6=0. 1.Geometric representation ofcomplex numbers. Any complex number a-+6i may berepresented inanxy-plane asapointA(a,6) with coordinates aand 6(Fig. 161); and. conversely, any point Ma, 6)inanxy-plane may beregarded asthe geometric image ofacomplex number a+6i. But iftoeach point A(a, 6)there corresponds acomplex number abi, then, totake 2specific case, topoints lying on the x-axis there correspond real numbers (b=0). But if@point lies onthe y-axis, itrepresents apure imaginary number, since a=0. For this reason, when complex numbers are represented in the plane, the y-axis iscalled the imaginary axis oraxis of imaginaries, and thex-axis, thereal axis (axis ofreals). Joining thepoint A(a,6)with theorigin, wegetavector OA. Incertaini nstances, itisconvenient toconsider the vector OAas thegeometric representation ofthe complex number a+-bi. 204 Complex Numbers. Polynomials 2.Trigonometric form ofacomplex number. Denote by@and r(r>0) the polar coordinates ofthe point A(a, 6)and consider theorigin asthe pole and thepositive direction ofthex-axis, the polar axis. Then (Fig. 161) wehave theqe)'amiliarrelationships:7 a=rcosg, b=rsing, o and, hence, thecomplex number may be z ygiven intheform a+bi=r(cosg+i sing). @) Fig.161. Theexpression ontheright-hand side is called thetrigonometric formofacomplex numbera+6i. Thequantities rand@areexpressed intermsofa and 6,bythe formulas r=VEFO, g=arctant and arecalled: r,the modulus, @,the argument (amplitude or phase) ofthecomplex number a+-bi. The amplitude ofacomplex number, theangle ,isconsidered positive ifitisreckoned from thepositive x-axis counterclockwise,andnegative,intheoppositesense,Theamplitude@is obviously not determined uniquely but towithin the accuracy ofthe term Qnk, where &isany integer. The modulus rofthe complex number a-+bi is,sometimes denoted bythesymbol |a+6i|: r=|a+bi| Itwill be noted that the real number Acan also be written in the form (3), namely: A=|A|(cos0-+isin0) forA>0, A=|A|(cos-+isina)forA<0. The modulus ofthe complex number 0iszero: |0|=0. Any angle @may betaken foramplitude zero. Indeed, forany angle 9wehave theequality0=0-(cosp+é sing). SEC. 2.BASIC OPERATIONS ON COMPLEX NUMBERS 1,Addition ofcomplex numbers. The sum oftwo complex numbers a,+6,i and a,+6,/ isacomplex number defined by theequality2,46,)+,+6,=(@,+4)+(,+6)i O) Basie Operations onComplex Numbers 235 From(1)itfollows thattheaddition ofcomplex numbers given invectors isperformed bytherule ofthe addition ofvectors. 2,Subtraction ofcomplex numbers. The difference oftwo com- plex numbers a,-+6,i and a,+6,i isacomplex number such that when itisadded toa,+6, ityieldsa,+bi iy _ Itiseasy toseethat owas (a,+6,))—(@, +6,1)=(a,—a,)+(6,—)i- ‘ ae’,(2) G Itwill benoted thatthemodulus ofthe aa difference of two complex numbers a ¥ V@—ay+,—6,) isequal tothe i ib,distance between thepointsrepresenting (yt) -Cay+iby) these numbers intheplane ofthecom- Fig.162. plex variable (Fig. 162). 3.Multiplication ofcomplex numbers. The product oftwo com- plex numbers a,-+6,i anda,+6,i isacomplex number obtainedwhen these twonumbers are’multiplied asbinomials bytherules ofalgebra, provided that Pel @=(-lis—h = (—)Ms—PH1h Pai ate, and, generally, forintegral &, ied; MG eH; From this rule weget (a,+6,i)(a,+6,i)=a,,+6,4,+4,b,i+4,6,7, or (4,+6)(+6) =(,4,—0,6,) +(a,+4) 3) Ifthecomplex numbers arewritten intrigonometric form, we have 1,(cos,+4sin@,)r,(cos@,+ising,)==r,[cos9,c0s¢,+ising,cos@,+icos9,sin@,+i"sing,sing,]= =r,[(cosg,cosy,—sing,sinp,)+i(sincosp,+cos@,sing,)]= =r,[cos9,+9.)+isin(@,+¢))]- Thus, 1,(cos@,+4sin9,)r,(cosp,+4sin@,)==r,[008(9,+)+/sin(9,+9) eB) the product oftwo complex numbers isacomplex number, the modulus ofwhich is.equal tothe product ofthe moduli ofthe 236 Complex Numbers. Polynomials factors, and theamplitude isequal tothe sum oftheamplitudes ofthefactors. Note 1.By virtue of(3), the conjugate numbers a-+6i and a—bi satisfy theequality (a+ ib)(a—ib)=a+6% the product ofconjugate complex numbers isequal tothesum of thesquares ofthe moduli ofeach ofthem. 4.Division ofcomplex numbers. The division ofcomplex numbers isdefined astheinverse operation ofmultiplication: if ath ty (where Va 63%0), then xandymust besuch astofulfil the equality a,+bi=(a,+6,1)(x+yi) or 4,+b=(a,x—by)+(ay+b,2)i. Consequently, a=ax—by, b=bxtay, whence we find aMidatbids yaad—aybyeave ape and finally weget autbud_aaybaby4aybyaby,arbi atte tor” ” Actually, complex numbers are divided asfollows: todivide a,+ib, bya,+ib,, multiply thedividend and divisor byanum- berconjugate tothedivisor (that is,bya—i6,). Then thedivisor will beareal number; dividing the’ real and imaginary parts of thedividend byit,wegetthequotient atbd(0+bul)(Gy—bal)_(0104+6464)+(0b,0469)iFO y+ i)(byl) — ano dy biby ay,—aby a“aee taper For thetrigonometric form ofacomplex number wehave Fy(cos @+ising)_ty ieosgebTsin gy)=7(OS(P=Fs)+4Sin(9,—9,)]- Basic Operations onComplex Numbers 21 Toverify this equality, multiply thedivisor bythequotient: 1,(cos@,+isin5)[008(9,9) +isin(9,—9,)] = =HF[05(9,+9,—4)+4in(9,+%—BN=P,(C08,Fésing).Thus, themodulus ofthe quotient oftwo complex numbers is equal tothequotient ofthemoduli ofthedividend and thedivisor; the amplitude ofthe quotient isequal tothedifference between theamplitudes ofthe dividend and divisor. Note 2.From therules ofoperations involving complex num- bers itfollows that theoperations ofaddition, subtraction, multi- plication anddivision ofcomplex numbers yield acomplex number. Itherulesofoperations oncomplex mumbers areapplied to real numbers, regarding the latter asaspecial case ofcomplex numbers, these rules will coincide with the ordinary rules of arithmetic. Note 3.Returning tothe definitions ofasum, difference, pro- duct and quotient ofcomplex numbers, itiseasy toshow that if each complex number inthese expressions isreplaced byitscon- jugate, then theresults oftheaforementioned operations will yield conjugate numbers, Whence follows (asaparticular instance) the following theorem. Theorem. /finapolynomial with reat coefficients ASMEARE A, weput the number a+-bi inplace ofx,and then theconjugate number a—bi inplace ofx,theresults ofthese substitutions will bemutually conjugate. SEC. 3,POWERS AND ROOTS OF COMPLEX NUMBERS 1,Powers. From formula (3)ofthepreceding section itfollows that ifnisapositive integer, then [r(cos+isin@)]"=r"(cosn@-+i sinng). Ww This formula iscalled DeMoiure's formula. Itshows that when acomplexnumberisraisedtoa_positive integralpowerthe modulus israised tothis power, while theamplitude ismultiplied bytheexponent. Now consider another application ofDeMoivre’s formula. Setting r=1 inthis formula, weget (cos@+ising)"=cosn@+isinng. Expanding the left-hand side inabinomial expansion and equating thereal and imaginary parts, wecan express sinnpand 238 ComplexNumbers.Polynomials cosmpinterms ofthe powers ofsin@and cos. For instance, if n=3 we have cos*p-+ i3cos*@sinp—3cos@sin?p—isin’p=cos3p-+isin3g;making useofthecondition ofequality oftwo complex numbers, weget: cos39=cos’p—3cossin’, sin3p=—sin’+3.cos*@ sing. 2.Roots. The nth root ofacomplex number isanother complex number whose nth power isequal totheradicand, or V7COS@+iSING)=e(cosp+isin yp), il o"(cosnp-+isinnw)=r(cos@+ising). Sincethemoduliofequalcomplexnumbersmustbeequal, while their amplitudes may differ byanumber that isamultiple of2n, we have e=r, np=o+ 2x. Whence we find e-V7,yatee, where kisany integer, j/7 isthe arithmetic (real positive) value oftheroot ofthe positive number r.Therefore, Vr(eosGFising)=f/7(cosSHAE4.5sin2424)-2 Giving &the values 0,1,2,..., n—1, we get ndifferent values oftheroot. For theother values of&,theamplitudes will difier from those obtained byanumber which isamultiple of2x, and, for this reason, root values will beobtained that coincide with those considered. Thus, thenthroot ofacomplex number hasndifferent values. The nth root ofareal nonzero number Aalso has nvalues, since areal number isaspecial case ofacomplex number and may berepresented intrigonometric form: ifA>0, then A=|A| (cos0-+/sin0); ifA<O, then A=|Aj(cosn-+isinx), Example 1,Find allthevalues ofthecube root ofunity. Solution. Werepresent unity intrigonometric Torm; 1=c0s0+sind. Powers and Roots ofComplex Numbers x9 Byformula (2)wehave " YDi/OFTaDcosEINE4igFPAiN Settingagate0,1,wendtheewaar(Ne\ afthe’root 4 y ¥ 2H san? 8san . symeonisin rig16 Noting that Qe oeVB,oggHL. 4x_V3canada aI: cost: ant I3, weget 1 13notin bi, yet, ItEe163thepoints A,B,Caegeometrie representations oftherots obtained! 3.Solution ofabinomial equation. An equation ofthetorm aA iscalled abinomial equation, Let usfind itsroots. IfAisa real positive number, then =VA(cos+isin228) (e=0, 1,2 2, aD. The expression inthe brackets gives allthe values ofthe nth root of I. IfAis areal negative number, then sm1(cos#2!4sinEEE) ‘The expression inthe brackets gives allthe values ofthe nth root of —1. IAisacomplex number, then the values ofxarefound trom formula (2). Example 2,Solve theequation eo. 20 Complex Numbers. Polynomials Solution. saf/STRATTainTe=cosAE4ain7A, Setting &equal 100,1,2,8,weget sy c0804/sin0=1, 2m 2ksyneossinBai, 48nt syncs pein a1, 6ig8 nmcos a Bm i, SEC. 4.EXPONENTIAL FUNCTION WITH COMPLEX EXPONENT AND ITS PROPERTIES Let z=x+iy. Il_x and yarereal variables, then ziscalled a complex variable. Toeach value ofthecomplex variable zinthe xy-plane (the complex plane) there corresponds adefinite point (see Fig. 161). Definition. Iftoevery value ofthe complex variable z,out of acertain range ofcomplex values, there corresponds adefinite value ofanother complex quantity w,then wisafunction ofthe complex variable z.The functions ofacomplex argument are denoted byw=/(z) orw=w(z).We introduce theconcepts ofthe limit ofafunction ofacom- plex variable, ofthederivative, oftheintegral, and soforth. Here, we consider one function ofacomplex variable, the exponential function: w=e or w=ert, The complex values ofthefunction waredefined asfollows:*)eft)=e(cosy+isiny), (ly that is w(z)=e* (cosy+isiny). (2) Examples: amt OFnelcosSttsin)me(4242), «Tie advgsility ofthis deinition ofthe exponenia function of9 compler vatlable will also beshown later on, Sec. ZieCh. XIll and Sec. 16, ari Exponential Function with Complex Exponent a 2em0thn oFae cos%tisina, Boselti, ettmet(eos|+isin1)=0.544(0.83, 4.24 isareal number.+PZe(cos0-1sin0)me* Isanordinaryexponential function Properties ofanexponential function. 1.Ifz,and z,aretwo complex numbers, then enteret, @) Proof. Let aA HANTS then eamCEHIUHII)mmCEEEIH!ittserver(cosy,+4.)+isiny,+y,)]- (ay Ontheother hand, bythetheorem oftheproduct oftwo complex numbers intrigonometric form wewill have eet =etrtinentin =eX(cosy, +isin y,)e%(cos y,+isin y,)= =erets [cos(y,+y,) +isin y,+4,)]- (5) In(4)and (6)theright sides are equal, hence theleftsides are equal too: ett=eter, etc, 2.The following formula issimilarly proved: - aeraeta © 3.Ifmisaninteger, then (ey en, a) For m>0, this formula isreadily obtained from (3); ifm<0, then ifisobtained from formulas (3)and (6). 4.The identity entat (8) holds. Indeed, from (3)and (1)weget ett=eet=¢(cos2n-+isinQn)=e. From identity (8)itfollows that the exponential function eisa periodic function with aperiod of2ni. 5.Letusnow consider thecomplex quantity w= u(x)+iv(x), 242 ComplexNumbers.Polynomiats where w(x) and (x) arereal functions ofthereal variable x. This isthecomplex function ofareal variable. a)Let there exist the limits limu(x)=u(x,),limv(x)=0(x,). Thenu(x,)+i0(x,)=w,iscalledthelimitofthecomplexvariablew. b)Ifthe derivatives 'u’(x)and v’(x) exist, then weshall call the expression wi=u(x)+10"(x) O) the derivative ofthe complex function ofareal variable with respect toareal argument. Let usnow consider the following exponential function: wwc0rtiBeaglaBhs, where aand Bareconstant real numbers, andxisareal variable. This isacomplex function ofareal variable, which function may berewritten, according to(1), asfollows: wae[cosBx-+isinBx] or w=e cosBx+ie*sinBe. Letusfind thederivative w,,From (9)wehave w= (e%*cosBx)’+-i(e"* sinBx)’= =e*(acospx—BsitiBx)+ie"(asinBx-+BcosBx)— =a[e"*(cosBx-+ésinBx)]+iBe*(cosBx-+isinBx)) ==(+B)[e"*(cosBx-+i sinBx)]=(a+i)e"*” *, Tosummarise then, ifw=e'** then w'=(a-+iB) e*** or [ert #1=(a+ip)err? (10) Thus, if&isacomplex number (or, inthespecial case, areal number) and xisareal number, then (ey =e, @) Wehave thus obtained theordinary formula fordifferentiation of anexponential function. Further, (ey=[letY=(ey=hte andforarbitrary (ey=ates, We shall need these formulas later on. Euler's Formula. The Exponential Form ofaComplex Number 243 SEC, 6,EULER'S FORMULA. THE EXPONENTIAL FORM OF ACOMPLEX NUMBER Ifweputx=0 informula (1)ofthepreceding section, weget e¥=cosy +isiny. () This isEuler's formula, which expresses anexponential function with animaginary exponent interms oftrigonometric functions. Replacing yby—y in(I)weget e-=cosy—isiny. @) From (1)and (2)wefind cosy and siny: Oper cosyatte?|nytt” | ® siny=25| These formulas areused, among other things, toexpress thepow- ers ofcos@ and. sing’ and their products’ interms ofthesine and cosine ofmultiple arcs. 40-0) examplescatyo(2EEYV ewrgrgerayen =F[00829-4isha24)+2-+(cos2y—Fsin2y))= a1cos2=H(1+e0524)- epet)sfeet 2.cosgstg=(HE (5) = (meine egya Lect et The exponential form ofacomplex number. LetusYepresent a complex number intrigonometric form: z=r(cosp-+i sing), where risthemodulus ofthe complex number and @istheam- plitude ofthecomplex number. ByEuler's formula, cos@+ising=e", Thus, any complex number may berepresented intheso-called ‘exponential form: rere, Examples. Represent thenumbers 1,1,—2,—iintheexponential form. au Comptes Numbers, Polynomials Solution. 1=cos2kat+/sin2h=et", tescoEisinmet, —2=2(cosm+isinx)=2e%, x ntl. incon <isnae SEC. 6.FACTORING APOLYNOMIAL The function Fe)=ARMAWEobAy where nisaninteger, isknown asapolynomial orarational integral function ofx,the number niscalled thedegree ofthe polynomial. Here, the coefficients A,, A,, ..., A,arereal or complex numbers; the independent variable xcan also take on both real and complex values. The root ofapolynomial isthat value ofthevariable xatwhich the polynomial becomes zero. Theorem 1(Remainder Theorem). Division ofapolynomial f(x) byx—a yields aremainder equal tof(a). Proof. The quotient obtained bythedivision off(x) byx—a will beapolynomial f,(x) ofdegree one less than that off(x), and the remainder will beaconstant R. We can thus write FO)=(e—a)f, (+R. a) This equality holds forallvalues ofxdifferent from a(division byx—a when x=a ismeaningless). Now letxapproach a.Then thelimit oftheleftside of(1) will equal f(a), while thelimit oftheright side will equal R. Since the functions f(x) and (x—a)f,(x)+R are equal forall x#a, their limits arelikewise equal asx—+a, that is,f(@)=R. Corollary. Ifaisaroot ofthepolynomial, that is,iff(a)=0 then x—a divides f(x) without remainder and, hence, f(x) is represented intheform ofaproduct F(x)=(ea)f,(@) where f,(x) isapolynomial. Example1.Thepolynomialf(s)<i—6e441146becomeszeroforx=1; snus TUS=0, and sl divides this polynomial without remainder #60+IL-6=(x—1) (#7—5x+6). Letusnow consider equations inone unknown, x. Any number (real orcomplex) which, when substituted into the equation inplace ofx,converts theequation into anidentity is called aroot oftheequation, Euler's Formula, The Exponential Form ofaComplex Number 245, my Sa) on Example2,Thenumbersym;xm5Z;xa%H,...,aretherootsof the equation cos z=sin x Ifthe equation isofthe form P(x)=0, where P(x) isapoly- nomial ofdegree n,itiscalled analgebraic equation ofdegreen. From thedefinition itfollows that theroots ofanalgebraic equa-tionP(x)=0 arethesameasaretherootsofthepolynomial P(x).Quite naturally the question arises: Does every equation have roots? Inthe case ofnonalgebraic equations, the answer isno:there are nonalgebraic equations which donot have asingle root, either real orcomplex; forexample, the equation e*=0. *) But inthecase ofan’algebraic equation theanswer isyes. This isgiven bythe fundamental theorem ofalgebra. Theorem 2(Fundamental Theorem ofAlgebra). Every rational integral function f(s)hasatleastoneroo,realorcompiles. The proof ofthis theorem isgiven inhigher algebra. Here we give itwithout proof. With theaidofthefundamental theorem ofalgebra itiseasy toprove thefollowing theorem. Theorem 3.Every polynomial ofdegree nmay befactored intonlinear factors oftheformx—aandafactorequaltothecoefficient ofx". Proof. Let/(x) beapolynomial ofdegree n: FRAP EAE. +A, By virtue ofthe fundamental theorem, this polynomial has at least one root; wedenote itbya,.Then, bythecorollary ofthe remainder theorem, wecan write F(x) =(@—a,) F,(*) where f,(x) isapolynomial ofdegree n—1; f,(x) also hasaroot, Wedesignate itbya,.Then h@O=&—-a) i) where f,(x) isapolynomial ofdegree n—2. Similarly, A)=(*—a,)f, (2). *)Indeed, ifthe number, xy—a++bi_ were theroot ofthis equation, wewould have theidentity e*+—0 or(byEuler's formula)e*(cosb+isin6)=0. But e*cannot equal zeto forany real value ofa;neither iscos-Fisinb equal tozero(because themodulus ofthisnumber isV'e0s"b-sin"—1 forany »).Hence, the product e#(cos6+isinb)#0,i.e.,e**%40;butthismeans hat theequation e*—0 has noTools. 26 Complex Numbers. Polynomials Continuing thisprocessoffactoringoutlinearfactors,wearrive at the relation Pros (= (Oy) fa where f,isapolynomial ofdegree zero, i.e.,some fixed number. This number isobviously equal tothe coefficient ofx"; that is, La Ay. Onthe basis oftheequalities obtained wecan write (x)=A,(e—a,) (x—a,) ...(e—a,). 2) From theexpansion (2)itfollows that thenumbersa,,a,,...,a, areroots ofthepolynomial /(x), since upon thesubstitution x=a,, x=4,, ...,x=4, theright side, andhence, theleft, becomes zero. Example3.Thepolynomial f(x)=x'—bet+I1x—6becomeszerowhen zal, 222, x53, ‘Therefore,PGi$e6=1)(2-2)(#9), Novalue x=a that isdifferent from a,,a,..., a,canbearootofthepolynomial f(x),since nofactor ontherightsideof (2)vanishes when x=a. Whence thefollowing proposition.‘Apolynomial ofdegreencannothavemorethanndistinctroots.But then the following theorem obtains. Theorem 4./fthevalues oftwo polynomials ofdegree n,9,(x) and @,(x), coincide forn+1 distinct values a,a,Gy,... @,Of theargument x,then these polynomials are identical. Proof. Denote thedifference ofthepolynomials byf(x): FO) =, (&)—@, (*)- Itisgiven that f(x) isapolynomial ofdegree nothigher than rnthat becomes zero atthepoints a,,..., a,Itcan therefore be represented inthe form [(2)=A,(«—a,) (4—a,) -..(@—a,). But itisgiven that f(x) also vanishes atthepointa,.Thenf(a,)=0 and not asingle one ofthelinear factors equals zero. Forthis reason, A,=0 and then from (2)itfollows that the polynomial F(x) isidentically equal tozero. Consequently, 9,(x)—9,(2)=0 or@,(x)= (2). Theorem 5./f@polynomial P(x) ARH ALE FA TAR isidentically equal tozero, allitscoefficients equal zero. The Multiple Roots ofaPolynomial MT Proof. Let uswrite itsfactorisation using formula (2): P(X)=A FAL.FA,e+A=A,(4—a,)-.(XG).(1) Ifthis polynomial isidentically equal to’zero, ‘itisalso equalto zeroforsomevalueofxdifferent froma,,...,d,.Butthennone ofthebracketed values x—a,, ...,x—a, isequal tozero, and,hence,A,=0.Similarly itisproved that A,=0, A,=0, and soforth.Theorem 6.Iftwopolynomials areidentically equal, thecoeffi- cients ofone polynomial areequal tothecorresponding’ coefficients ofthe other. This follows from the fact that the difference between the polynomials isapolynomial identically equal tozero. Therefore, from thepreceding theorem allitscoefficients are zeros. Example 4.Ifthe polynomial az-+bst-bex-fd Isidentically equal tothepolyaomin ba theneebat,eee anddoy 8 SEC, 7.THE MULTIPLE ROOTS OF APOLYNOMIAL If,inthefactorisation ofapolynomial ofdegree ninto linear factors P(e)=A,(e—a,)(x—a,)...(x—a,) ay certain linear factors turn out thesame, they may becombined, and then factorisation ofthe polynomial will yield FQ)=A,(x—a,)"(ea, (xa,)hm, ay And bythe then Inthis case, theroot a,iscalled aroot ofmultiplicity &,,ora k,-tuple root, a,,aroot ofmultiplicity &,,ete senExample, The,pelynomial (2)=st—5:48x—4 maybeTacored intothe following linear factors: F(x) =(2—2)(x—2)(x—1). This factorisation may bewritten asfollows: 1)=2=D). The root a,=2 isadouble root, a=1 isasimple root. Ifapolynomial has aroot aofmultiplicity #,then wewill consider that thepolynomial has&coincident roots. Then from the theorem offactorisation ofapolynomial into linear factors weget thefollowing theorem. Every polynomial ofdegree nhas exactly nroots (real or complex). 248 ComplexNumbers.Polyromiats Note. All that has been said oftheroots ofthepolynomial P= AHA +FAR may obviously beformulated interms ofthe roots ofthe algebraic equation ARTE ARE AgOe Letusfurther prove thefollowing theorem. Theorem. If,forthepolynomial f(x), a,isaroot ofmultiplicity h,>l, then forthe derivative f(x) this number isaroot of multiplicity k,— 1. Proof. Ifa,isaroot ofmultiplicity &,>1, then itfollows from formula (1') that Fix)=(x—a,hq (x) where p(x) =(x—a,)* ...(x—a,)"m doesnotbecome zeroatx=a,; that is,@(a,) #0. Differentiating, weget F(x)=, (2a)? (x)+(xa) (x)==(x—a,)h"" [kp(x)+(e—a,)9(*)]- Put ; V=h9(2)+(X—a,) 9"). Then P= a" y(n) and here , 2.)=k,(@,)+(4,—a,)9"(a,)=2,9(a,)#0. Inother words, x=a, isaroot ofmultiplicity &,—1 ofthe polynomial j’(x).From theforegoing proof itfollows that if, —1,thena,isnotarootofthederivative /'(x).From theproved theorem itfollows that a,isaroot ofmulti-licity&—2forthederivative /"(x),arootofmultiplicity &,—3forthederivative f(2)...andarootofmultiplicity one(sifaple root) forthe derivative j~" (x)and isnot aroot forthe deri- vative f(x), or F@,)=0, F(@,)=0. F(@)=0, ..-,f—@,)=0, but F(a.) #0. SEC. 8.FACTORISATION OF APOLYNOMIAL IN THE CASE OF COMPLEX ROOTS Informula (1), Sec. 7,Chapter VII, theroots a,,a,,...,a,may beeither real orcomplex. Wehave thefollowing theoremTheorem. Ifapolynomial f(x)withrealcoefficients hasacomplexroota+bi,italsohasaconjugate roota—bi. Factorisation of@Polynomial U9 Proof. Substitute, inthepolynomial /(x), a4-bi inplace ofx, raise to’a power and collect separately terms containing ¢and those not containing é;wethen get Fat bi)=M+Ni, where Mand Nare expressions that donot contain é. Since a+6i isaroot ofthe polynomial, wehave F(a+b)=M+Ni=0 whence M=0, N=0. Now substitute the expression a—6i forxinthe polynomial.Then(onthebasisofNote3attheendofSec.2ofthischap.ter)wegetanumberthatisaconjugate ofthenumber M+Wi,or Ka—bi)=M—Ni. Since M=0 and N=0, wehave /(a—bi)=0; a—bi isaroot of the polynomial. Thus, inthe factorisation F(x)=A,(xa)(x—a,)«..(xa) ‘the complex roots enter asconjugate pairs. Multiplying together the linear factors that correspond toa pair ofcomplex conjugate roots, wegetatrinomial ofdegree two with real coefficients: [x—(a+61)][x—(a— 6)]= =[(x—a)— bi][(x—a) +bf]==(ea)potaaDartapbtext prdg, where p= —2a, q=a"+b*arerealnumbers. Ifthe number a+6iisaroot ofmultiplicity &,the conjugate number a—bimust bearoot ofthesame multiplicity &,sothat factorisation ofthe polynomial will yield the same number of linear factors x—(a+-6i) asthose ofthe form x—(a—bi). Thus, @polynomial with real coefficients may befactored into real factors ofthe first and second degree ofcorresponding multiplicity; that is, (x)=A,(ea, (xa)... see(Ea) (x8pet gh...(+petq,)s where Rp hyp ee php FUb e$Qa 250 ComplexNumbers.Polynomials SEC. 9.INTERPOLATION. LAGRANGE’S INTERPOLATION FORMULA Let itbeestablished, inthestudy ofsome phenomenon, that there isafunctional relationship between thequantities yand x whichdescribes thequantitative yaspect ofthephenomenon; the :y-p0of function y=@(x) isunknown, butexperiment hasestablished aDthevalues ofthis function y,,SEy=PUK)ysYaroooYqforcertainvalues DT J of’iheargument x,,2,x5,.-5 7Xq._intheinterval[a,6]. | in ‘Theproblemistofindafunc-a ittion(assimpleaspossiblefromoa % 1®®thecomputational standpoint; for navies example, apolynomial) whichwould represent the unknown function y= (x)ontheinterval {a,6]either “exactly orapproximately. Inmore abstract fashion theproblem may beformulated asfollows: given ontheinterval la,6)the values ofanunknown function y=q(x) atn-1 distinct points xy,yy05Xq! Y= PK)y He=P) oerYn=P(i itisrequired tofind apolynomial P(x) ofdegree <n that ap- proximately expresses thefunction @(x). For such apolynomial, itisnatural totake apolynomial whose values atthe points x,,x,,%,..., x,coincide with the corre- sponding values yy,yy.Yyr ---+YqOfthefunction @(x) (Fig. 164). Then the problem, which iscalled the “problem ofinterpolating ‘afunction”, isformulated thus: for agiven function @(x) find a polynomial’ P(x) ofdegree <n, which, forthegiven values of XySyoss pyWill take onthe‘values Y=P(E)Y=P(E)viesa=Ende For the desired polynomial, take apolynomial ofdegree nof ‘the form P(x)=C,(x—X,) (2%) ---(Hq)+ $C,(x—x,)(X=). R—2,)+ $C,(tx,(84)(28)ooRa)EoseFC (Ee) (HH) 6GE) 0) and define thecoefficients C,,€,,..., C,sothat thefollowing Interpolation, Lagrange’s Interpolation Formula 1 conditions are fulfilled: PU) =I PO) Ate vee Phe) Yee @ 1.(1)putx24; then, taking into account equality (2), weget Yo=Cy(y=%,)(HyH)«+(Ha) whence aORR aa ae Then, setting x=.x,, weget My=Cy(4,¥4)(4,4)0(Ha) whence - ”OBA Inthesame way wefind =OBR a DF =CnGRR Ge” Substituting these values ofthecoefficients into (I), weget ery ern PO)Fe eaa eet (x=) (X= 2) (8=n)tae) aa et (arg er) ee) (es) FaaadmagYet (4x9) (445) -(n=¥n=) aaac) [ree gin @) This formula iscalled theLagrange interpolation formula. Let itbenoted, without proof, that ifg(x) has aderivative of the(n-+I)st order ontheinterval (a,6),theerror resulting from replacing ‘the function. (2) bythe polynomial P(x), i.e, the quantity. (x)=q(x)—P(2), satisfies the inequality 1 mse TRO)<1Gx)(ea) OMaapmaxOMOL Note, From Theorem 4,Sec. 6,Ch, VII, itfollows that the polynomial P(x) which wefound’ isthe only one that satisfies the given conditions. 252 ComplexNumbers.Polynomials Example. From experiment weget the values ofthe function y= (x:schorimeleponetorndyaforom Ttisrequired to.represent the function “—q(s) approximately bya potyooialofdegre two! ution. From (3)wehave (for n=2): ayEPDM 9IN) (x=1)@—2) P=(ayaa ste=nara!— 9+Caan—a—y! 39, 123, 252PyB18BE SEC. 10, ON THE BEST APPROXIMATION OF FUNCTIONS BY POLYNOMIALS. CHEBYSHEV'S THEORY Anatural question follows from what has been discussed inthe previous section: Ifacontinuous function @(x) isgiven onthe closed interval a,6],can this function berepresented approxi- mately inthe form of@polynomial P(x) toany preassigned de- gree ofaccuracy? Inother words, isitpossible tochoose apoly- nomial P(x) such that the absolute difference between (x) and P(x) atallpoints oftheinterval {a,6]should beless than any preassigned positive number e?The following theorem, which we give without proof, answers this question inthe affirmative. *) Weierstrass’ Approximation Theorem. /fafunction @(x) iscon- tinuous onaclosed interval a, 6}, then for every 2>0 there exists apolynomial P(x)such that |f(x)—P (x)|<e, forevery x inthe interval. The outstanding Soviet mathematician Academician S.N.Bern- stein gave the following method ofdirect construction ofsuch polmomials thatareapproximately equaltothecontinuous func: jion@(x) onthegiven interval. Let @(x) becontinuous onthe interval (0,1].We write theexpression . 8,)=9(%)cae"(lays, Here,CZarebinomial coefficients, (1) isthevalueofthe given function atthepoint x=". Theexpression B,(x)isannth degree polynomial called the Bernstein polynomial. +)ItwillbenotedthattheLagrangeinterpolation formula{see(3)Sec.9]cannot yetanswer this question. Itsvalues areequal tothose ofthefunction St'the points sy:tyegets sso but they” may bevery far{rom the values ithe Function" other points‘of the interval [o,8) Exercises onChapter Vit 253 Ifanarbitrary e>0 isgiven, one can choose aBernstein poly- nomial (that is,select itsdegree n)such that forallvalues ofx onthe interval [0,1],the following inequality will befulfilled: 1B,)—9(2)|<e Itshould benoted that consideration ofthe interval [0,1],and not anarbitrary interval [a,6},isnot anessential limitation of generality, since bychanging the variable x=a+/(b—a) itis possible toconvert any interval [a,6]into (0,1}.Inthis case, the nth degree polynomial will betransformed into apolynomial ofthe same degree. The creator ofthetheory ofbest approximation offunctions by polynomials isthebrilliant Russian mathematician P.L.Cheby- shev (1821-1894). Inthis field, heobtained themost profound results, which exerted agreat influence onthework oflater mathe- maticians. Studies involving thetheory ofarticulated mechanisms, which arewidely used inmachines, served asthestarting point ofChebyshev’s theory. While studying these mechanisms hearrived atthe problem offinding, among all polynomials ofagiven degree with the leading coefficient equal tounity, apolynomial ofleast deviation from zero onthegiven interval. Hefound these polynomials, which subsequently became known astheChebyshev polynomials. They possess many remarkable properties, and at present areapowerful tool ofinvestigation inmany problems of mathematics and engineering. Exercises onChapter Vit 1.Find@rsnd—9 avt 2.Find(64119 (7+31.Ans.9495i. 3.FindPoh. Ans.G—GPh AFind 7)" Ans. 524471 5.FindVT.Ans2H 6.FindYIS—TM. Ans.+Q—39.7.Re duce the following ‘expressions to trigonometric form: a) 1+ Ans.WE(coog-+isin);by1—t.dns.VE(coeZt+isinZ).8.Find3/7.Ansay,isYE0,expressthefollowingexpres:sionsintermsofpowersofsinxandcosx:sin2x,cos2x,sin4x,cos4x,finx,cosSx.10.,Expressthefollowingintermsofthesineandcosineofmultiple arcs: Costx,Cos!x,cost,costarslat,sin?slatx,six.1,Divide f@)=t—4t48x—1byx44.Ans.(x)=(x+4)(x?—8x+40)—161, that tathequotient lsequal to.s*—8i--40; and the remainder is"/(~#) =— 16112.Divide |f(x)—=at+ 12x7454x*4108x481 byx43.Ans.f(r)= TerDreedszeg2,18,DividePaymentbyx1dns.Te) H=6—-DGtbepepe ete tl). 254 Complex Numbers. Polynomials Facto thefollowing, plynomias: 14,Iie1.dns,[01 ROH t6Tjeea.Sas.FG ED iBFey an.heeetGe) - 1H "Experiment yielded the following values ofyasafunction ofx: n= 4for x=0, ne 6for mal, W=10 for x=2. Represent (approximately) the function byasecond-degree polynomial. Ans, Pope41a.Find 2polynomial ofdegree four that takes onthe values 2,1,—1, 5, Oferet,284,5, epectvely. Ans.at—tte 2te 4. 19,Find polynomial ofthe lowest possible degree that takes onthe values 3,7, 9,19"for x=2, 4,5,10,respectively. Ans. 2¢—1. 30,FindtheBernsein ‘polynomials otdegre 1,33 andforthefunc. tion y=sinsix on the interval [O, I}. Ans. By(2)=0;, By(s)—=2e(1—2), ayy2PS(1a;By(2)=2«(1—2) (2VE—3)¢@—2VF—3)x4VE. CHAPTER VIII FUNCTIONS OF SEVERAL VARIABLES SEC. 1.DEFINITION OF AFUNCTION OF SEVERAL VARIABLES When considering afunction ofone variable wepointed out thatinthestudyofmanyphenomena oneencounters functions oftwo and more independent variables. Some examples follow. Example 1.The area Sofarectangle with sides oflength xand yis expressed bytheformula Sexy. Toeach pairofvalues ofxand ythere corresponds adefinite value ofthearea=fipafnetionofiovatabes imsesipedwit' srample2.ThevolumeVofarectangular parallelepiped withedgeso lengthsy.#isexpressed bytheformula mp ‘ass Vemsye. Here, Visafunction ofthree variables, x,y,2. Example 3.The range Rofashell fired with initial velocity vfrom a gun, whose Barrel isinclined tothe horizon ‘atanangle @,isexpressed by the' formula pewisn? itsisdirded).Heisterationofrit}air resistance is disregarded). Here, gis theacceleration of gravity. (ai poeeverypaleefealucs ofo¢sodghisformota. yields dedsite value ofRyinolher words, Ris afunction oftwo variables, v,and q. Example 4. gottebete< Vite" Here, uwisafunction offour variables x,y.2,f. Definition 1.Iftoeach pair (x,y)ofvalues oftwo independent variable quantities xand y(from some range D)there corresponds adefinite value ofthe quantity z,wesay that zisafunction of thetwo independent variables xand ydefined inD. ‘Afunction oftwo variables issymbolically given as z=/(x, y),2=F(x,y)andsoforth. Afunction oftwo variables may berepresented, forexample, bymeans ofatable oranalytically (by aformula) asinthe four examples given above. The formula may beused toconstruct 256 Functions ofSeveral Variables atable ofvalues ofthe function for certain number pairs ofthe independent variables. From Example 1we can build the following table: Saxy ne \RPE |EL:2 of 2] 3 4 6 3 of] 3] 4s 6 8 4 ola] 6 8 2 ° Inthis table, the intersections ofthe lines and columns, which correspond todefinite values ofxand y,yield the corresponding values ofthe function S. Ifthe functional relation 2=f(x, y)isobtained asaresult of changes inthequantity zinsome experimental study ofaphe- nomenon, westraightway get atable defining zasafunction of two variables. Inthis case, the function isspecified bythetable alone. Asinthecase ofasingle independent variable, afunction of two variables does not, generally speaking, exist forallvalues of xand y.Definition 2.Thecollection ofpairs(x,y)ofvalues ofxand y,forwhich the function z=f(x, 9) isdefined, iscalled thedomain ofdefinition ofthis function. Thedomain ofafunction isapparent whenillustrated geomet.tically. Ifeachnumber pair«andyisgivenasapointM(xy) inthexy-plane, then the domain ofdefinition ofthe function will beacertain collection ofpoints inthe plane, We shall also call this collection ofpoints thedomain ofdefinition ofthefunc- tion. Inparticular, theentire plane may bethedomain. Infuturé weshall mainly have todowith such domains asare parts of theplane bounded bylines. The line bounding thegiven domain weshall call theboundary ofthedomain. The points ofthedo- mainnot‘lyingontheboundary weshallcallinterior pointsofthedomain, Adomain consisting solelyofinteriorpointsiscalled anopen domain; that which includes the points ofthe boundary iscalled aclosed domain. Definition ofaFunction ofSeveral Variables 257 Example 5.Determine the natural domain ofdefinition ofthe function rade—y. The analitic expression 22—y ismeaningful for allvalues of and y. Therefore, the entire sy-plane isthe nalural domain ofthe funetion. Example 6.z=Vi-s?—g. For z'to have areal value itisnecessary that the radicand beanonne- gative number; inother words, xand ymust salisfy the inequality I-80, ofattyt. AllthepointsM(x,y)whosecoordinatessatisfythegiveninequalitylie igavace‘ofradius|SianMceeueeatthe:origi’andGo"theBoundaryof this circle. Bramble cay y sain ty). ysineloganitnsaredenedonlyorpostiveYY numbers, the following inequality must. be 'fatisted! .xty>0 ofy>—s. © This means that the natural domain of ‘definition ofthe function zisthe half-plane A above thestraight line y=—s, the line itsell ——g A notcluded (ig.68) Example 8. Thearea ofthe triangle Sisafunctionofthebasexandthealtitulege »psat ~ 2 Fig.165. The domain of this function isx>0, y>0Gincethebase.ofatriangle andiisalfitade cannot benegative orzero). We notice that the domain ofthis function does not “coincide with the natural domain. ofdefinition ‘ofthe analytic expression used. to”define. the function, because the natural domain of theexpression *Y1sobvlouly theentire2y-plane Itiseasy togeneralise the definition ofafunction oftwo variables tothe case ofthree ormore variables. Definition 3.Iftoevery collection ofvalues ofthe variables X,Ys2s+ tht there corresponds adefinite value ofthe vari- able'w, weshail then call wthefunction ofthe independent vari- ables x,y,z,..., u,¢and write w=F(x, y,2... u,f)oF w=f(x, 9,2,u,2),and soon. Just ‘as’ inthe case ofafunction oftwo variables, we can speak ofthe domain ofdefinition ofafunction ofthree, four and more variables. Totake anexample, forafunction ofthree variables, the do- main ofdefinition isacertain collection ofnumber triples (x,y,2). Letitbenoted that each number triple isassociated with some point M(x, y,2)inayz-space. Consequently, the domain of 93368 258 FunctionsofSeveral:Variables definition of afunction ofthree variables issome collection of points inspace. Similarly, one can speak ofthe domain ofdefinition ofafunc- tion offour variables u=f(x, y,2,t)asofacertain collection ofnumber quadruples (x, y,2,f).However, the domain of definition ofafunction offour oralarger number ofvariables no longerpermits ofasimplegeometric, interpretation. Example 2gives afunction ofthree variables defined forall values ofx,y,2 InExample’ 4wehave afunction offour variables. Example 9 Herewisafunction ofthefourvariables x,y,2,udefined fotvalues of the variables that satisly therelationship Iaxtayttuto, SEC. 2,GEOMETRIC REPRESENTATION OF AFUNCTION OF TWO VARIABLES We consider the function z=F(x, Wy (1) defined inthe domain Ginthe xy-plane (asaparticular case, this domain may betheentire. plane), and asystem ofrec: z id P attdt \&s/ x 9 AUate ‘ 7 ‘4 Fig. 16 Fig. 167. tangular Cartesian coordinates Oxy2 (Fig. 166). Ateach point (x,y) erect aperpendicular tothexy-plane and onitlayoffasegment equal tof(x, y). This gives’ usapoint Pinspace with coordinates XI Z=F (x,Ye The locus ofpoints Pwhose coordinates satisfy equation (1) isthegraph of function oftwo variables. From thecourse of Partial and Total Increment ofaFunction 259 analytic geometry weknow that equation (I)defines asurfaceinspace. Thus, thegraph ofafunction oftwovariables’ isa surface projected onto the xy-plane inthe domain ofdefinitionoftheFunction, Eachperpendicular tothexy-plane intersects thesurface z=/(x, y)atnot more than one point. Example. Asweknowfromanalytic geometry, thegraphofthe functionsoattgh isaparaboloid ofrevolution (Fig. 167) Note. Itisimprossible todepict afunction ofthree ormore variables bymeans ofagraph inspace. SEC. 8.PARTIAL AND TOTAL INCREMENT OF AFUNCTION Consider the line ofintersection PS ofthe surface z=/(x, y) with theplane y-const parallel tothexz-plane (Fig. 168). Since inthis plane yremains constant, 2will vary along the curve PS depending oniy onthechanges inx.Increase the inde- pendent variable xbyAx; then 2will beincreased; this increase iscalled the partial increment ofzwith respect toxand itis denoted byA,2 (the segment SS’ inthefigure), sothat L gyAzefetax W—Fe 9). EX Similarly, ifxisheldconstant a and yisincreased byAy, then z HE isincreased, and this increase is icalledthepartialincrement ofzShLotctag withrespecttoy(symbolised bySs)~L---gstA,z,thesegmentTT’inthe4NXcas figure):Lay A,z=F(x, y+Ay)—F(x, 9). (2) a Thefunction receives thein- Fig.168crement A,z“along theline” of ; intersection’ ofthe surface 2=f(x, y)with the plane x-const parallel totheyz-plane. Finally, increasing theargument xbyAx,andtheargument y bythe increment Ay, wegetfor2anew increment Az,which is called thetotal increment ofthefunction zand isdefined bytheformula Az=f(x+Ax, y+Ay)—f(x, y). (3) InFig. 168Azisshown asthesegment QQ’. a 260 Functions ofSeveralVariables Itmust benoted that, generally speaking, the total increment isnotequal tothesumofthepartial increments, Az#A,2+<,2. Example. 2=ay ate eax) yyy bx,Azexutay— =Fby,Bes(et09Wtby)—ay=y OxbytAxAy Forx=1, y=2, Ar=0.2, Ay=0.3 wehave A,2=0.4, Ay2—=0.3, 42=0.76, Similarly wedefinethepartialandtotalincrements ofafunction ofany number ofvariables. Thus, forafunction ofthree variables u=f(x, y,6)wehave Au=f(et+Ax,y,O—f(x4,0), Ayu=f(x,y+ dy, f(x, yO), Ayu=](x,y, t+At)—F(RY,ty Au=f(x+dx, ytdy, t+O)—1(x,9,). SEC. 4,CONTINUITY OF AFUNCTION OF SEVERAL VARIABLES We introduce animportant auxiliary concept, that oftheneigh- bourhood ofagiven point. The neighbourhood, ofradius r,ofapoint M,(x,, y,)isthe collection ofall points (x,y)' that satisfy the” inequality V@—xFFU—y)<r:thatis,the setofallpoints that lieinside a‘circle ofradius rwith centre inthe point My(Xo:Ya) ob0 ifWwe'say that afunction f(x,y) possesses someproperty “nearthepoint (¢ y (x,4)”oF“intheneighbourhood ofthepoint’ (x,,y,)” we mean that there isa circle with’ centre at(x,, y,), atall points ofwhich circlethegivenfunction a¥—possessesthegivenproperty. Fig. 169. Before considering the concept of continuity ofafunction ofseveral variables, let usexamine the notion ofthe limit ofafunction ofseveral variables.*) Let there beafunction z=f (x,y) defined insome domain Gofanxy-plane. Letusconsider some definite point M,(x,, 4.)inGoronits boundary (Fig. 169). *)We shall mainly consider functions oftwo variables, since three and more variables donot introduce any fundamental changes, but “dointroduce ‘additional {echnical difficulties Continuity ‘ofaFunction ofSeveral Variables 261 Definition 1.ThenumberAiscalledthelimitofthefunctionF(x, y)asM(x, y)approaches M,(xy,y,)ifforevery e>0 there isanr>0 such that forall points M(x,y) forwhich the inequality MM,<risfulfilled wehavetheinequality IF,yA]<e.ItAisthelimitoff(x,y)asM(x,y)—+M,(x,, y,),thenwewrite limf(x, y)= A. Definition 2.Let thepoint M,(x, y,)belong tothedomain ofdefinition ofthefunction f(x,y).Thefunction 2—f(x,y) iscalledcontinuous atthepointM,(cy,y,)ifwehave timFe,)=F(0Yds a)nh and M(x, y)approaches M,(x,,y,) inarbitrary fashion allthe while remaining inthedomain’ of‘thefunction. Designate x=x,-+Ax, y=y,-+Ay, then (1)may berewritten as, follows: timp +e,Yo+Ay)=F(XeYo) a’yFree] or simf+4x,YetAy)—F(Xo¥o)]=0- a’Fv] : ‘WesetAg= V(Ax)*+(Ay):(seeFig.168).AsAx—+0andAy—0, Ag—+0; and conversely, ifAg—+0, then Ax—-0 and Ay—0. Noting further that the expression inthesquare brackets in(1") isthetotal increment ofthe function Az, (I*) may berewritten inthe form lim Az=0. ay Fray Afunction continuous ateach point ofsome domain iscontinuous inthe domain. IfatsomepointN(x,,y,)condition (1)isnotfulfilled, thenthepointNV(x,,y,)iscalledapointofdiscontinuity ofthefunction2=/ (x,y).For example, condition (I') may notbefulfilled inthe following cases:1)z=](x,y)isdefinedatallpointsofacertainneighbourhood ofthepoint N(x, y,)with theexception ofthepoint N(x,, ¥,) itself; 262 Functions ofSeveral Variables 2)thefunction z=f(x,y) isdefined atallpoints ofaneigh-bourhood ofthepointNV(x,,y,)butthereisno.limitlimf(x,y); 3)thefunction isdefined atallpointsoftheneighbourhood of, N(x.y,)andthelimitexists:limFey),but limje,efCraWreed Example 1.Thefunction rartty {scontinuous for allvalues ofxandy that is,ittscontinuous atevery point inthe xy-plane Indeed, nomatter what thenumbers xand y,Axand Ay,wehave AEFAMEOEAMIE 2sAeb2yAU+OeON. Consequently, Jim 42=0. Prk at The following isanexample ofadiscontinuous function. Example &.The function 7me {sdefined everywhere except atthepoint x=0, y=0 (Figs. 170, 171. z y L—— y?id iCo-Bo\ a % v Fig. 170. Fig. 171. Letusexaminethevaluesofzalongthestraightliney=Ax(k=const).‘Obviously, alongthisline oJ .’= =FE=const, arr ea Partial Derivatives ofaFunction ofSeveral Variables 263 This means that afunction 2along any straight line passing through theGriginretains aconstant. valuethatdepends upontheslope’ oftheline Ths, approaching theorigin along. different paths we. will obtain different limiting values, and this means that thefunction. /(x, y)has. nolimit when thepola(zsithesyplane approaches theorigin. Thus,thefunction is discontinuous atthis point. Itisimpossible toredefine this function atthe coordinate origin sothat itshould become continuous. On the other hand, it isreadily seen that the function iscontinuous atallother points. SEC. 5.PARTIAL DERIVATIVES OF AFUNCTION OF SEVERAL, VARIABLES Definition. The partial derivative, with respect tox,ofa{unction z=f(x,y) isthelimit oftheratio ofthe partial increment A,2, with respect tox,tothe increment Ax asAxapproaches zero. The partial derivative, with respect tox,ofthe function 2=[(x,y) isdenoted by‘one ofthesymbols ; ar. afZsfew Fiz. Thus, bydefinition, 22jigAtmjimLedeIeae(imge ar : Similarly, thepartial derivative, with respect toy,ofafunctionz=}(x,y)isdefinedasthelimit‘oftheratioofthepartialincre-ment ofthefunction A,zwith respect toytotheincrement ofAy asAyapproaches zero.” The partial derivative with respect toy isdenoted byone ofthefollowing symbols: gf: %; & iNGESSe Thus, a i i =fim42=fimLeytan—/e. 0 BygyineAF—aye ay : Noting thatA,ziscalculated withyheld constant, andA,2with xheld constant, wecan formulate thedefinitions ofpartial deri- vatives asfollows: thepartial derivative ofthefunction2=/(x,y) with respect toxisthederivative with respect toxcalculated on theassumption that yisconstant. The partial derivative ofthe function z=F(x, y)with respect toyisthederivative with respect toycalculated ontheassumption that xisconstant. Itisclear from this definition that the rules forcomputing partial derivatives coincide with the rules given forfunctions of one variable, and theonly thing toremember iswith respect to which variable thederivative issought, f 264 Functions ofSeveralVariables Example 1.Given the function 2—stsiny; find the partial derivatives Oe2oySolution. oeGeaeesiay,mstcory. Example 2.2=2. Here Saye, Faeins. The partial derivatives ofafunction ofany number ofvariables aredetermined similarly. Thus, ifwehave afunction woffour variables x,y,2,f:uaF(ty26) then 96timHetane OMe HeeT gewe ar 2timHeeb awNM 29,andsoforth,Maeno ay Example3. wettyttate, eu O45, Hays Ot,GeneteatsFmt, SmSelet,|mast, SEC. 6,THE GEOMETRIC INTERPRETATION OF THE PARTIAL DERIVATIVES OF AFUNCTION OF TWO VARIABLES Let theequation z=f(xy) betheequation ofasurface shown inFig. 172. Draw the plane x=const. The intersection ofthis plane with thesurface yields theline PT. For a_given x,letusconsider a certain point M(x, y)inthexy-plane. Tothe point Mthere cor- responds apoint P(x, y,z) onthe surface z=f(x, y).Holding x constant, letusincrease thevariable ybyAy=MN=PT’. Then the function 2will beincreased by Ayz=TT’ [tothe point N(x,y+Ay)therecorresponds apointT(x,y+A,,2+4,z) on thesurface 2=f(x, y)]. Total Increment and Total Diflerentiat 265; io4 Z Theratio“%isequaltothetan- A. gentoftheangleformed bythe DBE secantlinePTwiththepositive LI7,y-direction: Y Ay opFate Ter. 4 Consequently, thelimit 8 v tim272! Gapne a9 isequal tothetangent oftheangle: * Bformed by the tangent line PB fothecurve PT atthepoint P afwith the positive y-direction: A tan B. Fig.172. Thus,thepartial derivative $isnumerically equaltothetan- gent oftheangle ofinclination ofthe tangent line tothecurve resulting from the surface z—f(x,y) being cut bythe plane x=const. Similarly, thepartial derivative %isnumerically equaltothe tangent oftheangleofinclination aofthetangent Tinetothe surface z=f(x, y)cutbytheplane y=const, SEC, 7,TOTAL INCREMENT AND TOTAL DIFFERENTIAL By the definition ofthe total increment ofthe function z=/ (x,y)wehave (see Sec. 3,Ch. VIII) Az=f(x+Ax, y+dy)—f (x,y). a Letussuppose that f(x, y)hascontinuous partial derivatives at thepoint (x,y)under consideration. Express A?interms ofpartial derivatives. Todothis, add to and subtract from theright side of(1)F(x,y-+Ay): Az=[f(x+ Ax,y+dy)—F(x, y+Ay)]+[F(xy+Ay)—F(x,9].(2) The expression Fe, ytdyy—Flx,¥) inthesecond square brackets may beregarded as‘thedifference between {wo values ofthe function ofthe variable yalone (the 266 FunctionsofSeveratVariables value ofxremaining constant). Applying tothis difference the Lagrange theorem, weget - He9+AT, =AyLED, a) where yliesbetween yandy+Ay. Inexactly the same way the expression inthe first square brackets of(2)may beregarded asthe difference between twovaluesofthefunctionofthevariablexalone(thesecondargumentretains thesamevaluey+Ay).Applying theLagrange theorem tothis difference, we have FleAx,y+Ay)—I(x,y+dy)=dxAELOM, (4) where %lies between xand x-+Ax. Introducing expressions (3)and (4)into (2)weget ae. atteD demalata 4led | 6 Since itisassumed that thepartial derivatives are continuous, tim2E.y+a9_Fe.9 Aroor oe ast- © tim20D _af#9) ae oy (because xandyrespectively liebetween xandx+Ax, andy and y+y, xand yapproach xand y,respectively, asAx—0 and Ay—+0). Equalities (6)may berewritten intheTorm ae. ae,1GgaM4y,, °AGD_Ae tyTOaEy, where the quantities y,and y,approach zero asAxand Ayapproach zero(thatis,asA=VAx?+ dy'—0).Byvirtue of(6'), relation (5)becomes zmLD ne LE ay4yBetysby 6) ‘Ox‘ay ifbad The sum ofthetwo latter terms ofthe right side isaninfini. tesimal ofhigher order relative toAg=VAx'+ Ay*.Indeed, the ratioBAF0asAg—-0, sincey,isaninfinitesimal andg Total Increment and Total Diferentiat 267 isbounded (|42|<1). Insimilar fashion itisverified that yay. nado,thesumofthefirsttwotermsisalinearexpression inAx and Ay.Forf(x, y)#0 andf,(x, y)%0, thisexpression isthe principal part’ ofthe increment, differing from Azbyan infinitesimal ofhigher order relative too=V Ary ay. Definition. The function z=f(x, y)[the total increment (Az) of which atthegiven point (x,y) may berepresented asasum of two terms: alinear expression inAxand Ay, and aninfinitesimal ofhigher order relative toAg]iscalled differentiable atthegiven point, while the linear part ofthe increment isknown asthe total differential and isdenoted bydzordf. From (5') itfollows that ifthe function f(x, y)has continuous partial derivatives ata.given point, itisdiferentiable atthis,Pointandhasatotaldifferential: dz=F,(x,y)Ax+8,(x,y)dy. Equality (6°) may berewritten inthe form Azed2+y,Ax+y,dy, and, towithin infinitesimals ofhigher order relative toAg,we may write the following approximate equality: Az=dz. ‘We shall call the increments ofthe independent variables Ax and Aydifferentials ofthe independent variables xand yand we shall denote them bydxand dyrespectively. Then theexpression ofthe total differential will assume the form of of demddx+Ldy. Thus, ifthefunction z=f(x, y)hascontinuous partial derivatives, itisdifferentiable atthepoint (x,y),and itstotal differential is equal tothesum ofthe products ofthe partial derivatives bythe differentials ofthecorresponding independent variables. Example 1.Find the total differential and the total increment ofthe function 2=ay atthe point (2,3)forax=0.1, ay=02. ‘Solution. Bz=(e+48)(y+Ay)xy=yAetxAUTOtOY tate dyaydetadymuartsay. 268 Functions ofSeveralVariables Consequently, 42=3-0.142-0.240.1.0.2=0.72; @2=3.01$2.02=0.7 Fig. 173 isanillustration ofthis example. The foregoing reasoning and definitions are appropriately generalised fofunctions ofany number ofarguments. p Ifwehave afunction ofany number br by Viables xy NN Wl(X,Y,2,UyveerA N andallpartial derivatives%,awn iy NY4x. arecontinuous atthepoint (x,y,2,4, N seu8),theexpression N Nteay42 a Ndw=HderEdy+Bde...Hat *isthe principal part ofthetotal increment Fig. 173. ofthe function and iscalled the total differential. Proof ofthe fact that the difference Aw—dw isaninfinitesimal ofhigher order than V(Ox)+(49)+-..+(BA)?isconducted inexactly thesamewayasfor afunction oftwo variables. Example 2.Find thetotal differential ofthefunction u=e"#*sintz of three variables x,4,2 Solution. Noting’ that the partial derivatives Maer esint2, Haetoray sint2, $emet479sn2cos2—0847"ln22 are continuous for allvalues ofx,y,2,wefind that a=e+ay+deme"(esateda2yat2dy+s22de). SEC. 8,APPROXIMATION BY TOTAL DIFFERENTIALS Let the function 2=f(x, y)bedifferentiable atthepoint (x,y). Find the total increment ofthis function: Az=i(x+Ax,y+dy)—f (x,y) Approximation byTotal Differential s 269 whence Het Ax, y+Ay)=f(x, 9)+42, a We had theapproximate formula Arde, @ where Cc) of dem$xtLay. @) Substituting, into formula (1), theexpanded expression fordzin place ofAz, wegettheapproximate formula Hetas,y+dyf(e, +29 neoTEMay,(4) towithin infinitesimals ofhigher order relative toAxand Ay. We shall now show how formulas (2) and (4) are used for approximate calcylations. Problem, Calculate the volume ofmaterial needed tomake acylindcical lass ofthe following dimensions (Pig. 174): radius ofinterior cylinder R, altitude ofinterior cylinder i, thickness ofwalls and bottom ofglass &. Solution. We give two solutions ofthis problem: exact and approximate. 2)Exact solution. The desired volume visequal tothedifference between the” volumes ofthe exterior cylinder and. interior f cylinder. Since theradius oftheexterior cylinder is iequal toRA, and the altitude isH-fAy 'TH ven(R+RH +R)—ARH Ro a 14v=QRHR+RR-+ HAE+ORKWY) ©.Hl bd)Approximate solution. Let usdenote by fthexoltneOEThTinterinylderthenfestThi j Whanctionoftwo variables Rand Hl. Itwe increase andHby&,then thefunction fwili increase byAf; Fig. 174. but this will bethesought-for volume v,v=Af. ‘On. the basis “oftelalion (I) we have the approximate equality ond? or wtapsotOR AR+57 OH. But since oF oF 2Ghaoarn, Honk, sk=aH=, 270 Functions ofSeveral Variables weget om (2RHR+ Ri). ( Comparing the results of(5)and (6), wesee that they differ by the quan- {ityx(HRRRHD,whichconsistsoftermsofsecondandthirdorderof Let us apply these formulas to. numerical examples.LeRedgy,220cm,keOcm . Applying (), weee, exactly, Ua(2-4-20-0.1-4-42-0.1 +.20-0,18-4-2-4-0.124 0.14)=17.8812, Applying formula (6), wehave, approximately,em(4.20-0.144-0.1)= 17.65, Hence, the approximate formula (@)gives ananswer with anerror less than O.8n,whichis100+7PS%— 9,whichislessthan2%ofthemeasured quantity. SEC. 8.ERROR APPROXIMATION BY DIFFERENTIALS Letsome quantity ubeafunction ofthequantities x,y,z, ...,f uP (XY 2ot) and letthere beerrors Ax, Ay, ..., Afmade indetermining the values ofthe quantities x,y,2...., Then the value ofw computed from the inexact values ‘of’the arguments will be obtained with anerror AueAX,YAY,ooo)2+B2,14+MDFOW,2De Below weshall investigate theevaluation oftheerror Au,provided the errors Ax, Ay, ..., Atare known. For sufficiently small absolute values ofthe quantities Ax, Ay, ..., Afwecan replace, approximately, the total increment bytheotal differential:afapA a bundax+Syt.+FAL. Here, thevalues ofthe partial derivatives and the errors ofthe arguments may beeither positive ornegative, Replacing them by the absolute values, wegetthe inequality B a a ou}<[$e]alt|Z]iaul+-..+]5|1A¢1- a) Ifinterms of|Atz|, [Aty|, ..., |A*u| wedenote themaximum absolute errors ofthecorresponding quantities (the boundaries for the absolute values ofthe errors), itisobviously possible totake sul=|}ace|hae Balen 3 ral=|5¢|rare]geiatul-+«+[sellace.® Exror Approximation byDifferentials mn Examples,Tekebamety-te,thenatal=1 Atx|-+1 Aty|+1 A*l- 2.Letusa—y, thenJatul=1a*%1+]Atal 3.Let u=ay, then Vatul=tell styltlylated. 4.Letw=, then y supaltater]Slyeypaeteedeae latai=|flaret+|S]iarvt =! 5,Thehypotenuse candthelegoof2righttriangle ABC,determined with maximum absolute errors |A*e|=02, [A*al=0.1, are, respectively, -c=75, am22, Determine theangleAtromtheformula slaA=; anddetermine the maximum absolute error |ZA| when calculating theangle A.Solution,sinA=,Amaresin®, hence, eae oaVarma 8 CVema From formula (2)weget VERmeet Ostpe +0.20.0007S radoner Vaya| 5VCE . ° Thus, Anaresin2&938", 6._In the right triangle ABC, let the leg b—=121.56 and the angle A=25°21'40", and the maximum absolute error indetermining the leg 6is {'a*b|=0.05 ‘metre, themaximum absolute error in‘determining theangle A ata [=i Determine the maximum absolute error incalculating the legafrom the formula a= tan A. Solution. From formula (2)wefind safetten Ayat|+2eLy ae Jata|—[tan A]A%b1+2501aA| Substituting theappropriate values (and remembering that |A*A| must be ‘expressed inradians), weget Ya|=tan25°21" eetee re [ata]tan25°21°40'-0.05-+rosea agETES =0.0237+0.0087=0.0324metre. 272, Functions ofSeveral Variables The ratio oftheerror Axofsome quantity tothe approximate value ofxofthis quantity iscalled the relative error ofthe quantity. Let usdesignate itdx, oxen, The maximum relative error ofaquantity xisthe ratio ofthe maximum absolute error tothe absolute value ofxand isdenoted by[2*21, Jote|= lara, @ Toevaluate themaximum relative error ofafunction u,divide allnumbers of(2)byJuJ=[f(x, 4,2,--O| afaf ar tet[arcs hiarote+lar, (4) but a ar a eo a %_ aFa$Finit =3nifveFeFinifl. Forthis reason, (3)may berewritten asfollows: . a . a * a +Voru)—|z in|s|[la%xl +|5Inif||y+...+|ZInf[Ae6) orbriefly, [d*u|=[A*In| fl. ) From both (3)and (5)itfollows that themaximum relative error ofthe function isequal tothe maximum absolute error ofthe logarithm ofthis function, From (6)follow therules used inapproximate calculations. 1.Let u=xy. Using theresults ofExample 3,weget oyfeELAS] yLytAtyt_ Lately LAY yee. Fy|eellen 7 7 ld that is,the maximum relative error ofaproduct isequal tothe sum ofthe maximum relative errors ofthe factors. 21und,then,usingtheresultsofExample 4,wehave dtu |=16x] +15%. The Derivative ofaComposite Function m3 Note, From Example 2itfollows that ifu=x—y, then a 11xandyareclose,itmayhappen that|Q*u|willbeverygreat compared with the quantity x—y being determined. This should betaken into account when performing thecalculations. Example 7.The oscillation period of@pendulum is Vtran/Z, where1isthelengthofthependulumandgistheacceleration ofgravity.‘WatrelativeerrrwilBe'madeindeermining 7whenusingthisTor mulaitwetakex=S:14 (accurate,to 0.005),{=fm(accurate {o0.01t), g=9.8 msect (accurate to0.02 m/sec) Solution. From (6)the maximum relative error is [8711 atl |. But eratozsinne tinting. Calculate |A*In |. Taking into account that x~3.14, ata—0.005,1ai'm,A100 m,g=98misect,Ang—0.02msec,wegel Thus, the maximum relative error is (8°T=0.0076=0.76%. SEC. 10. THE DERIVATIVE OF ACOMPOSITE FUNCTION. THE TOTAL DERIVATIVE Let usassume that intheequation z=F(u, 0) (0) uand oarefunctions oftheindependent variables xandy: u=@(x, ysU=P(x, y) (2) Inthis case, 2isacomposite function ofthearguments xand y. Ofcourse, 2canbeexpressed directly interms ofx,y;namely, 2=F le,ys0 Yh ® Example 1.Let peut’ pudh waxttys vee 4; then . FEED EVEL a Functions ofSeveral Variablet Now suppose that the functions F(u, 0),p(x, y),(x,y) have continuous partial derivatives with respect toalltheir arguments, andweposetheproblem: evaluate 5and onthebasis of equations (1)and (2)without having recourse toequation (3). Increase the argument xbyAz, holding thevalue ofycons- tant, Then, byvirtue ofequation (2), wand vwill increase by A,tt and Ayo. But ifwand vreceive increments A,u and A,o, then thefune-tionz=F(u,v)willreceive anincrement Azdefined byformula(6), Sec. 7,Ch. VIII: =F Aut EAo+yA yA. Divide allterms ofthis equality byAx: A?OFAgu)OFAgd|Ag Ag .BraeeetooaeReTe ItAx—0, then A,u—+0 and Azo—0 (by virtue ofthe conti- nuity ofthefunctions wand v).Butthen y,and y,also approach zero. Passing tothe limit asAx—+0, weget iim822«ipMa2»ignAa?22meOeltaeGe geOe limy,=0; limy,=nwo: fimyno and,consequently, dz _OFou,OFov 7 deGudet50Ge" 4 If_we increased thevariable ybyAyand held xconstant, then bysimilar reasoning wewould find thata:_OFdu,oFav 5nate “) Example 2. .FeInuto} ume, omstty: a2 mw aeau"eetGoa Ouaxayt, OH oes 20 og,Sener, Manners Bare Hat, Using formulas (4)and (4) wefind az pea yt eerBeat aren DH Ot Wo eeayty 1 toy rpraytSnape amares CetD. The Derivative ofaComposite Function 5 Formulas (4)and (4") are.readily generalised tothe, case ofa larger number ofvariables. For example, ifw=F(z, u,v,5)isafunction offour argu ments 2,u,9,s,and each ofthem depends onxand y,then formulas (4)and (4’) assume the. form du_dwdz,dwdu,dwdv,dwdsGe~a2detaudetBoOeasae" =du _dwd2-,dwdudwdv,dwds © ay—deBy+BuByBoay*Bsdy” Ifafunction isgiven z=F(x, yu, 0),where y,u,9inturn depend onasingle independent ‘variable (argument) x: y=F(x);u=@(x);0=P(X), then zisactually afunction only oftheone variable x,and we mayposethequestion offinding thederivative 42. This derivative iscalculated from the first ofthe formulas (6): dz_d20x,d2dy,d2du,dzd0d=GedeGye*Gude+a0ae But since y,u,0arefunctions ofxalone, the partial derivatives become ordinary derivatives; inaddition $=1.Forthisreason, dz_a,dedydedu,deo Geet IydetGudetdodx" 6) This formula isknown asthe formula forcalculating thetotal derivative $2(incontrasttothepartialderivative 2), Example 3. zeet4VG yess, Fog, NW cos, Be ATVG|de Formula (6), here, yields the following result 084dyogyIcopedx+ cos, Bt aeTVG arsarse 276 FunctionsofSeveralVariables SEC. 11, THE DERIVATIVE OF AFUNCTION DEFINED IMPLICITLY Let usbegin this discussion with theimplicit function ofone variable.*) Let some function yofxbedefined bythe equation F(x, y)=0. We shall prove the following theorem. Theorem. Let acontinuous function yofxbedefined impli- citly bythe equation F(x, y)=0 where F(x, y), Fy(x, 4),Fy(%, y)arecontinuous functions in some domain Dcontaining thepoint (x,y)whose coordinates satisfy equation (1); also, atthis point F(x, y)#0. Then the function yofxhas the derivative . ye Ee* Fy)" Proof. Letthevalue ofthefunction ycorrespond tosome value ofx.Here, F(x, y)=0. Increase the independent variable xbyAx. Then thefunction y will receive anincrement Ay; that is,tothevalue oftheargu- ment x-+Ax there corresponds thevalue ofthe. function y-+Ay. Byvirlue ofequation F(x,y)=0 weshall have Flet+ Ax, y+ Ay)=0. Hence F(e+ Ax, y+Ay)—F(x,)=0. The left member ofthe latter equality, which isthe, total incre- ment ofthefunction oftwo variables byformula (5'), Sec. 7,may berewritten asfollows: Flet An, ytdy—F(x,WaFEOx+Fdy+y,dety.dy, where y,and y,approach zero asAxand Ayapproach zero. Since the left’ side ofthe latter expression isequal tozero, wecan *)InSec. 11,Ch, Ill,wesolved theproblem ofthedifferentiation of anImplicit function ‘of‘one variable. Weconsidered individual cases. and thaotfindgeneral foe thatwouldicithederivative ofmpi Eitfunction; likewise wefailed toclarity the conditions oftheexistence’ of this derivative, The Derivative ofaFunction Defined Implicitly an write Farr Fdy-+yde+ydy=0. Divide thelatterequality byAxandcalculate 44: oF ay_at ax” (OF . att Let Axapproach zero. Then, taking into account that y,and y, alsoapproach zeroandthat$°-40, wehave,inthelimit, Ld ._ und. 0) Oy Wehave proved theexistence ofthederivative y;ofafunction defined implicitly, and we have found the formula for calcu- lating it. Example 1.The equation s4y'—1=0 defines yasanimplicit function ofx.Here, oF_»,.oF Fnpastey—l, Fate, Fany, Consequently, from (1), dye ay 7" Itwill benoted that the given equation defines two different functions [since “toevery value ofxintheInterval (—l, 1)there correspond two values ofy};however, thevalue that wefound ofy,holds forboth functions. Example 2Anequation isgiven that connects xand y: Ome try =0, Here,F(x,y)=e?—e*+xy, oF Gane nFanos, Consequently, from formula (1)weget dy_ertyey dx Fx OF" 2 Functions ofSeveral Variables Let usnow consider anequation oftheform F(x, y,2)=0. (2) Iftoeach number pair xand yinsome domain there correspond one orseveral values ofzthat satisfy equation (2), then this equation implicitly defines one orseveral single-valued functions 2ofxand y. For instance, theequation stytt2t—Rt=0 implicitly defines two continuous functions 2ofxand y,which functions may beexpressed explicitly bysolving theequation for z;inthis case wehave 2=VRoe—¥ and =—YVR—¥ Hh, Letusfind.thepartial derivatives 3and3oftheimplicit function 2ofxand ydefined byequation (2). Whenweseek%,weconsider yfixed,Andsoformula (1)is applicable, provided xisconsidered the independent variable and 2the function, Thus, oF ina=. Ea Inthe same way’ wefind aa) =H 7 Similarly, wedetermine theimplicit functions ofany number ofvariables and find their partial derivatives. Example 3.Styte—R=0, Bes ee lee lee Differentiating this function asan_explicit function (alter solving the equation for2),wewould obtain theVery same result. Example 4. Ce eyte45—0, Partial Derivatives ofDiferent Orders 279 Here, F(x, y,2)=e? +2ty +245, OF_9,FeFpea,Fane:Feat:Emettt: a2 dy detBeFHTayRT SEC. 12, PARTIAL DERIVATIVES OF DIFFERENT ORDERS Let there begiven afunction oftwo variables: z=f(x, y). Thepartialderivatives $=f,(x,y)andGh y)are,gene- rally speaking, functions ofthe variables xand y.And sofrom them we can again find partial derivatives. Thus, there arefour partial derivatives ofthe second order ofafunction oftwo vari- ables,sinceeachofthefunctions$¢andHamaybedifferentiatedboth with respect toxand with respect toy. The second partial derivatives aredenoted asfollows: Smfale,y);herefisdifferentiated twicesuccessively with ; respect tox;aynhey);herefisfirstdifferentiated withrespecttoxandthen the result isdifferentiated with respect to % aanbie(x,y)jherefisdifferentiated firstwith respect toyand thentheresultisdifferentiated withrespecttox; Soefiny);herethefunctionfisdifferentiated twicesucces sivelywithrespect toy. Derivatives ofthe second order may again bedifferentiated both with respect toxand y.Wethen getpartial derivatives of thethird order. Obviously, there will beeight ofthem: aDat|Gxtdy*Tadydx *Txdy**DyOx**Tydxdy*Iy*x*By?* Generally speaking, apartial derivative ofthenthorder isthe first derivative ofthe derivative ofthe (n—1) storder. For exam- ple,mn isaderivative ofthenthorder;herethefunction 280 FunctionsofSeveralVariables 2was first differentiated ptimes with respect tox,and then n—p times with respect toy. For afunction ofany number ofvariables, the higher-order Partial derivatives aredetermined insimilar fashion. Example 1.Compute the second-order partial derivatives ofthe function He,N=atytet. Solution. We find successively oF oF 2;Hae Fatt of oy,a}__9(2xy)_». oF_O(et+3y)_5. OFGe Seay ay GyoeH a a ae Example2.Compute 255andgoairaybatt, Solution. We successively find a a Oey ope a .Favepiy Bayete ays aay2H+8, oe2OFoie oe . Eien AByox=8+Gxy*, Byatt Gud a Example 3.Compute 2p umters7", Solution. Ou sagcay, OM areas 8!agacaMg peat Seater Tae’, gag te,agg ee The natural question that arises iswhether the result ofdiffer- entiating afunction ofseveral variables depends ontheorder of differentiation with respect tothe different variables; inother words, will, forinstance, thefollowing derivatives beidentically equal: Cia Caaay aE or Ml 9ang Mhery. O‘OxdyOt OtOxdy* and soforth. Itturns out that the following theorem istrue. Theorem. Ifthe function z=[(x, y)and itspartial derivatives FoPyFyandf,aredefined and‘continuous atapoint M(x, y) and insome neighbourhood ofit,then atthis point oy at = dragayazw=Ful Partial Derivatives ofDifferent Orders 81 Proof. Consider theexpression Ax[fetdx, y+dy)—fet dx,Mle, y+d)—Ks, De Ifwe introduce an auxiliary function (x) defined bythe equality P)=F, ytay—T(e, Ys then Amay bewritten inthe form A=9(x+Ax)—9 (x). Since itisassumed that f,isdefined in.theneighbourhood of the point (x,y), itfollows that @(x) isdifferentiable ontheinterval [x,x4Ax];butthen,applying theLagrange theorem,weget A=Axg’ (x) where xliesbetween xandx-+Ax. But *M=f& y+AN—f(& 9). Since f;,isdefined inthe neighbourhood ofthepoint (x,4),f,isdifferentiable ontheinterval[y,y+Ay];andsobyapplyingonce again theLagrange theorem (with respect tothevariable y) tothe difference obtained we have fe, y+dy)—fe&9)=Aufey%Ds where 9liesbetween yand y+Ay. Consequently, theoriginal expression ofAis A=Axhyfy (,9). Oy Changing theplaces ofthemiddle terms weget A=([f(e+Ax, y+Ay)—F(x, ytdyJ—[Fe+Ax, y—flx y)} Introducing the auxiliary function : VW=K(xtAx,YT, Yr we have A=~y+dy)—¥l). Again applying the Lagrange theorem weget - A=Ayy), where yliesbetween yandy+Ay. 282 FunctionsofSeveralVariables But - = = VG=hletdx, Fhe D. Again applying theLagrange theorem weget filet dx,D—fle Y=dhe &Ds whereXliesbetween xandx-+Ax.Thus, the original expression ofAmay bewritten inthe form AaAy Arh, &,9). 2 The left members ‘of(1)and (2)are’ equal toA,therefore the right ones are equal too; that is, AxAufey(%Y=AyAthyCY), whencenesfey&0)=fe&¥)- Passing tothe limit inthis equality asAr—+0 and Ay—0, weget oe _ Jimfey=limfoe¥)- arse ane astgst Since thederivatives f,,and f;,atecontinuous atthepoint (x,y),wehaveim,fey&=fey(ey) ast a and lim fie, yy=foe(x, y).And finally weget fer =fixlt, Ds asrequired. Acorollaryofthistheoremisthatifthepartialderivatives —Zt and—2! arecontinuous, then Oeaye Oy—FOeot_attBay aha Asimilar theorem holdsalsofora’function ofanynumber ofvariables. Example4.Find245and4itumesinz.Solution. AT yee2yet)sing,PHmesinspayer?sine(Lp2y)singan *Oxdy ud.* Level Surfaces 293 sfgatTeartaptetae(ement Hence, Ou Tedy02SybzOx (also seeExamples 1and 2ofthis section). SEC. 18. LEVEL SURFACES Inaspace (x,y,2)let there bearegion Dinwhich the function u=u(x, y,2) rt) isdefined. Inthis case we say that ascalar field isdefined in the region D.If,forexample, u(x, y,2)denotes the. temperature atthe point M(x, y,z),then wesay that ascalar field oftem- peratures isdefined; ifDisfilled with aliquid orgas and uC, y,2)denotes pressure, we have ascalar field ofpressures, ete. Consider the points ofaregion Dinwhich the function u(t, y,2)has afixed value c: u(x, 2)=e (2) The totality ofthese points forms acertain surface. Ifadifferent value ofcistaken, weobtain adifferent surface. These surfaces are called level surfaces. Example 1.Let there begiven ascalar eld wenateSae. Here, the level surfaces are a eeatten orelipsolds with semi-axes 27, 3VE, 4VEIfthefunction wisafunction oftwovariables xandy, umule, then the level “surfaces” arelines onthe xy-plane: 4, =o e which are called level lines. 284 Functions ofSeveral Variables Ifweplot values ofuonthe z-axis: zaulk, os the level lines inthe xy-plane will beprojections oflines obtained atthe intersection ofthe surface z=u(x, y)with the planes z=c(Fig.175).Knowing the Zz& level lines, itiseasy tostudy Hf thecharacter ofthesurface age. ofthe FutfiectEN F [ee W/ LEEDS ‘Tt iPies? =ZoiletBA v r 4 Fig. 178 Fig. 176. Example 2.Determine thelevel lines ofthefunction z=1—x*—y*. They arelTines with equations 1rat<ptare whlch are(hig. 178)ctcles with radios Vi=c. inparticular, when c=0 wegetthecircle x#-+y?=1. SEC. 14.DIRECTIONAL DERIVATIVE Inaregion D,consider thefunction u=u(x, y,z)and the pointM(x,y,z).DrawfromMavector$whose direction cosinesarecosa, cosB, cosy (Fig. 177). Onthe vector §,atadistance 2 Lsy/| 5ery ay a 7 Fig. 7. Directional Derivative 235 Asfrom itsorigin, letusconsiderapointM,(x+Ax,y+Ay,2+Az). Thus, As=VAx+dy?+Ae, We shall assume that the function u(x, y,2)iscontinuous and hascontinuous derivatives with respect totheir arguments inthe region D. ‘Asin Sec. 7,we will represent the total increment ofthe function asfollows: IuayOH bumFEAx-+FtAy+3Ae+eA2+2,dy+e,A2, ) where ¢,,©and e,approach zero asAs—+0. Divide allterms of (1)byAs: Auduax,day, WAZ), Ars,AY,Az eet teateetek. @) Itisobvious that a Ay ae Mercosa, Sf—cosp, Secosy. Consequently, equation (2)may berewritten as acosa+ cosB-+% cosy-+e,cosa+e,cosB+e,cosy. (3) ThelimitoftheratioS¢asAs—+0 iscalled thederivative of thefunction w=u(x, y,2)atthepoint (x,y,2)along thedirection ofthevector$andisdenoted by%;thus im 84-2 So, passing tothe limit. in(3), weget 3cosa+3cosB-+-$fcosy. ©) From formula (6)itfollows that ifweknow thepartial derivatives itiseasy tofind the derivative along any direction §.The partial derivatives themselves areaparticular case ofadirectional derivative. Forinstance, when a=0, B=, y=%, weget FE=Ff0080+$fcos5+$4cos%=. 286 Functions ofSeveral Variables Example. Given afunction wattytetFindthederivative2atthepointML.1,tf)slongthedestionofthevector$,—21-4/-+ 3A;b)alongthe 2 Ss GirectionoftievetborSEPSolution. )""Find "the “direction cosines ofthevector 2 ywan .coepeAi s VizieosVi" (a 1 3ea Fe CosBapCOSY: a Que24d1,3 Fig.178. a,RV Va eV Thepartial derivatives atthepoint M(1.1.1)are au ou duo,Sar, Hay, Satedu du au (eure (yee (%)a2 Thus, Ou, 2 1 3 a Laaoe BO TR VR a Via 1)Find the direction cosines ofthe vector $2 enemies testscorres. dual 1 16a aoaoeea ee ae ae Wenotehere(anditwillbeneededlateron)that2VS>7 ig. 178) SEC, 15, GRADIENT At every point ofthe region D, inwhich the function u=u(x, y,2)s-given, wedetermine ‘thevector whose projections onthe coordinate axes arethevalues ofthepartial derivatives Gradtent 281 2.z.%ofthisfunction attheappropriate point:auyOu7Ou grada=$14Se43 (y This vector iscalled thegradient ofthe function u(x, y,2).We saythat avector field ofgradients isdefined inD.Letusnow prove the following theorem which establishes arelationship between thegradient and the directional derivative. Theorem. Given ascalar \field u=u(x, y,2);inthis field, let there bedefined afield ofgradients graduHijeSe. Thederivative %alongthedirection ofsomevector$isequalto theprojection ofthe vector gradu onthevector S. Proof. Consider the unit vector S*, which corresponds tothe vector S$: 4 S*=icosa+-jcos B+kcosy. Findthescalar product ofthevectors graduandS*: gradu-S*= 5cosa+5cosB+3cosy. @ The expression onthe right isaderivative ofthe function u(x, y,2)along the vector S.Hence, wecan write gradu-S*= 3. Ifwedesignate the angle between the vectors grad uand S*by @(Fig. 179), wecan write Igradu|cosp= @) or projection S*gradu=% (4) and the theorem isproved. This theorem gives usaclear picture ofthe relationship between thegradient and the derivative, atagiven point, along any direction. Referring toFig. 180, construct the vector gradu atsome point M(x, y,2). Construct asphere forwhich gradu isthediameter. Draw ‘the vector $from M.Denote byPthe point ofintersection ofSwith thesurface ofthesphere. Itis 288 FunctionsofSeveralVariables then obvious that MP=|gradu|cos@, if@isthe angle between thedirections ofthegradient andthesegment MP(here, p<) , orMP=%. Obviously, when‘thedirection ofthevector Sis 'e Fig. 179. Fig. 160. reversed the derivative changes sign, while its absolute value remains unchanged. Letusestablish certain properties ofagradient. 1)The derivative atagiven point along the direction ofthe vector S$has amaximum ifthedirection ofS$coincides with that afthegradient; thismaximal valueofthederivative isequaltojeradal.Thetruthofthisassertion follows directly from(3):$¢willbe amaximum when @=0, and inthis case #=|gradu}. 2)The derivative along avector that istangent toalevel surface iszero. This assertion follows from formula (3). Indeed, inthis case, g=F,cosp=0 and$=|gradulcos=0. Example 1,Given the function astty+e. 4) Determine the gradientatthepointM(t,1,1).Theexpressionofthe gradtentofthistuneflonatanarbitrarypointwillbepression hegradu=2ct-+-24f-+-2ch, Gerada)=2+2/42,[graduly?VS 'b)Determine thederivative ofthefunction uatthepointM(I,1,1) slong the direction ofthe gradient. The direction cosines ofthegradient Gradient 280 wall be 08oN cospak, cosy= es ee vs" vo And so u_yp typ tio 1 adng 2 yo tno,ahi cai ne or aBradt Note. Ifthefunction u=u(x, y)isa function oftwo variables, thenthevector y gradu=314387 raduno lies inthexy-plane. We shall prove that grad uisperpendicular tothelevel line wy u(, y=e lying inthe ay-plane and passing through the corresponding point. Indeed, theslope &,ofthetangent tothe — % levellineu(x,y)=¢ willequal k,=—“F Fig.181. ri‘ (seeSec.11).Theslope&,ofthegradientisk,=“!. Obviously, bty=—1. Thisproves ourassertion (Fig.181).A’similar prop-erty ofthe gradient ofafunction ofthree variables will be established inSec. 6ofChapter IX. a —VrSetey cls ; 0 NUD oeH LiesLS TApp — M24) gradu Fig. 182. Fig. 188. Example 2.Determine thegradient ofthefunction w=’+4(Fig.182) atthe point M(2, 4). 10-2360 290, Functions ofSeveral Variables Solution. Here Sael aa,Ha 2y|=f el ATT" gradunary Sy, Theequation ofthelevelne(Fig.18)passing through thegivenpoint\ #yg2 £4022. SEC. 16, TAYLOR'S FORMULA FOR AFUNCTION OF TWO VARIABLES. Let there beafunction oftwo variables z=f(x, y) which iscontinuous, together with all itspartial derivatives up tothe (n+)st order inclusive, insome neighbourhood ofthe point M(a, 6). Then, like the case ofone variable (see Sec. 6, Ch. IV), represent thefunction oftwo variables intheform ofa sum ofannth degree polynomial inpowers of(x—a) and (y—6) and some remainder. Itwill beshown below that for the case ofn=2 this formula has the form 1, y= Ay+D(x—a)+EY)+ +HIA(—a)*+2B(ea)(y—6)+CY—H'J4R, (1) where the coefficients A,,D,E,A,B,Careindependent ofx andy, while R,istheremainder, thestructure ofwhich issimi- larfothe structure ofthe remainder inthe Taylor formula fora function ofone variable. Let usapply the Taylor formula forafunction f(x, y)ofone variable yconsidering xconstant (we shall confine ourselves to second-order terms): 1=O, b+2F*FOb+ a =+P heOFS len), @) where n,=6+8,(y—5), 0<0,<1. Expand thefunctions f(x, 6), f,(&, 6),Fy(*,6)inaTaylor's series inpowers of(x—a), con- fining yourself tomixed derivatives uptothethird order inclu- sive: Taylor's Formata for@Function ofTwo Variables 221 f(x, 6)= =Hla,+ATle,+EEMKa,)+FSSRal @) where E,=2+0,(x—a), 0<0,<1; Kyl, 0)=F,(a,6)+279fog(as6+ETSizelas),(4) where B=x40,(x—a), 0<0,<1; Fake,0)=Fv(@,6)+27"faux(Bx8), © where Be=x4+0,(x—a), 0<6,<1. Substituting expressions (3), (4)and (5)into formula (2),weget Fe,=F(a,6)+272fala,6)+2Sf(a,6)+ $FBEcb+E[ya+25feel,6)+ SPfrean00]+2EP™[fev.0-422Fiebas8]+ +S heiw(es1). Arranging thenumbers asindicated informula (1), weget Fe, )=1 (@,6)+(xa)fe(a,6)+(Y—)fy(a,6)++ala) fer(a,6)+2(x0(x—6)fay(@,6)+ $Y)" fy(@,6+ea)"Foc(Bas8) $3(e—a)*(t=)feayBay6)+3(xa)(Y—)*Fooy (B48)+ +6) fw, Ie © This isTaylor’s formula forn=2. The expression R=HeU0)"FeesBye0)+3(x—0)*YB)FoeyBar8) +3(4-4)Y—6)"faayEas0)+U5)Faw(@I to 2 Functions ofSeveral Variables iscalled the remainder. Further,letusdenotex—a=Ax,y—b=Ay, be=V(xy+(Ay).Transform R,: a [at yr, astay 'Rmae[BsfoeByeB+3SEMfig(En8)4 $3SAEfetyEar0)-+2onusWD)Be. Since |Ax|< Ag,|Ay|< Agand thethird derivatives arebounded {this isgiven), the coefficient ofAg*isbounded inthedomain under consideration; letusdenote itbya,. Then we can write R,=a,Ao'. Inthis notation, Taylor's formula (6)will then, forthecasen=2, take the form T(x,=1(@, 6)+Axfe(a,6)-+Ayfy(a,6)+ +Ax"fx(a,6)+2AxAyfy(a,6)+Ay*foy(a,6)+a,00". (6") Taylor's formula isofasimilar form forarbitrary n. SEC. 17. MAXIMUM AND MINIMUM OF AFUNCTION OF SEVERAL VARIABLES ~ Definition 1.We say that thefunction z=f(x,y) hasamaxi- mum atthepoint M,(x,,y,) (that is,when x=x, andy=y,) if Fey) >fey) forallpoints (x,y) sufficiently close tothe point (x,,y,) and different from it. Definition 2:Quiteanalogously wesaythatafunction z=f(x,y)has aminimum atthepoint M,(x,,y,) if Fe W)<He9) forallpoints (x,y)sufficiently close tothepoint (x,,y,) and different from it. The maximum and minimum of afunction are called extrema ofthe function; we say that afunction has anextremum ata given point ifthis function has amaximum orminimum atthe given point. Example 1.The function =D Y—2P—1 attains aminimum atx=1, y=2; i.e., atthepoint (1,2).Indeed,/(1,2)=—1, Maximum and Minimum ofaFunction ofSeveral Varlabes 295 and since (¢—1)¥ and (y—2)* are always positive forx#1, y#2, 50 IE y—2)-1>-1, that is, Fey >10.2 The geometric analogy ofthis case isshown inFig. 184. zal yt \f b-sinrtey) 9 fon----4-- Sh hy LF § a [s Fig. 184. Fig. 185. Example 2The function rodmaetten forx0, y=0 (coordinate origin) attains amaximum (Fig. 185). Indeed, 10,=>. Insidetheciceat-gt—=%lelustakethepoints,9)dierentfom thepoint(0,0).ThenforO<attyt<t., sin (4+ y?)>0andtherefore arn Ne,p=sin ehV<E or 1, <F(0, 0). The definition, given above, ofthe maximum and minimum of afunction may berephrased ‘asfollows.Letx=x,+Ax;y=y,-+Ay;then FO DFesYo)=Fe+AX,YetAY)—F(HesYo)=AF. 1)If4f<0 forallsufficiently small increments intheindepend- ent variables, then the function [(x,y) reaches amaximum at the point M(x,.y,)- 294 Functions ofSeverai Variables 2)IfAf>0 forallsufficiently small increments intheindepend- ent variables, then the function f(t, y)reaches aminimum at the point M(x,,y,). These formulations may beextended, without any change, to functions ofany number ofvariables. Theorem 1.(Necessary conditions ofanextremum). Ifafunction z=f(x,y) alfains an extremum atx=x,, y=y, then each first- order partial derivative with respect to2either vanishes forthese values ofthearguments ordoes notexist. Indeed, give the variable yadefinite value y=y,. Then the function ‘f(x, y,)will beafunction ofone variable, x.Since at x=x, ithas anextremum (maximum orminimum), itfollows that(3),_,,iseitherequaltozeroordoesnotexist.Inexactly thesamefashion itispossible toprovethat(3)caniseither equal tozero ordoes not exist.“ This theorem isnot sufficient for investigating the extremal values ofafunction, but permits finding these values forcases inwhich weare sure ofthe existence ofamaximum orminimum. Otherwise, more investigation isrequired. Fortans thefnetin 2yt—at asdrvtves Em; B=429, which vanish atx=0 and y=0. But forthe given values, this funetion has ‘neither maximum nor minimum. Indeed,” this 2y function isequaltozeroattheoriginand NN{kes bothpositiveandnegativevaluesat [Ss 4 pointsarbitrarily closeto:the’origin. Hence,fie value zero. ts neither amaximum nor a O %minimum (Fig. 186). Points atwhich2£=0 (ordoesnot exist)and360(ordoesnotexist)are Fig.186. called critical points ofthe function z=/(x,y). Ifafunction reaches an extremum atsome point, then (byvirtue ofTheorem 1)this can occur only atacritical point. For investigating afunction atcritical points, letusestablish sufficient conditions for the extremum ofafunction oftwo vari- ables: Theorem 2.Let afunction f(x, y)have continuous partial deri- vatives upfoorder three inclusive inacertain domain containing the point M,(x,.¥,); inaddition, letthepoint M,(x,,y,) bea Maximum and Minimum of@FunctionofSeveralVariables 295 critical point ofthefunction f(x,yj;that is, (tas Yo)9,AftesYo)— Testo, Tesh)<9, Then forx=x, y=: 1)F(x, y)has@maximum if OF vo), 4Gees (2 Une. ¥e))* 2126. 0) <0,PEt Teed—(Ate) 0andMGHo, 2)F(x, y)has @minimum if 21) Os vo)_ (2te ¥0))* 16. Ws 9,gets 1Goe)(Mae"0andMaw>0, 3)[(%,y) has neither maximum nor minimum if Esa.Ho)OMa.ve)_(2Lee.W)*—0, Sate (PBS) <0: 4)ipMesto Ge,go)(21ste)", thentheremayor maiy notbeanextremum (inthis case, anadditional investigation isrequired). Proof. Let uswrite the second-order Taylor formula forthe function f(x, y)[Formula (6), Sec. 16]. Assuming A=kyD=Yy X=H+ Ax, Y=Yt Ay we will have Fite, WtBUHeyypLEEdgMeeay 1[°F(xs,yo) PF(Xa.Ye) Pte. Wo)Aye 2 +[Pe nat42a aay4Asay")0,(A0)", where Ag=VArFAganda,approaches zeroasAg—-0. Itisgiven that I ysWd) 2, eaetd)Meet)9,Meewo, Hence Atm fle+Ax,Yt AN)—FliuY= Uea+af Mi* \*on[gaa+2909,Ardy+55Au]+0,(Bo. a Letusnowdenote thevaluesofthesecond partial derivatives, atthepoint M,(x,, y,)interms ofA,B,C: ‘ OF) 4. (a1) _p (at(54)4(eha8(Sue 296 Functions ofSeveratVariabtes Denote by@theangle between thedirection ofthesegment M.M, where Misthepoint M(x,+Ax, y,+Ay), and thex-axis; then Ax=Agcosg; Ay=Agsing. Substituting these expressions into the formula forAf,wefind Af-=}(Ao)'[A cos*+2Bcos@sing-+C sin'g-+2a,Ael. (2) Suppose that A%0. Dividing and multiplying by Athe expression inthe paren- theses, we have Af=4(do![4tetBainoyHACKOYHO+95,A0]. (3) Let usnow consider four possible cases, 1)Let AC—B*>0, A<0. Then inthenumeratorofthefraction wehave asum oftwo nonnegative quantities. They donotvanish i 4 simultaneously because thefirstterm vanishes fortang=—4,whilethesecondvanishes forsinp=0.IfA<0, then thefraction isanegative quantity that does not vanish. Denote itby—m'*; then Af=(Ae)*[—m"+2a,A0), where misindependent ofAg, a,Ae—r0 asAg—+0. Hence, for sufficiently smallAgwehave‘Af<o or F(x.+Ox,yyt+BY)—F(tyYy)<0. Butthenforallpoints (x,+Ax, y,+Ay)sufficiently closetothepoint (x,,y,) wehave the’inequality Fx,+Ox,y+49<F(XsYs which means that atthe point (x, y,)the function attains a maximum. 2)Let AC—B*>0, A>0. Then, reasoning inthesame way, weget ardAf=z(A0)*lm?+-2a,Ae] or Fy+Ax,y+Ay)>T(tyYo) thatis,f(x,y) hasaminimum atthepoint (x,,y,). Maximum and Minimum ofaFunction ofSeveral Variables 297 3")Let AC—B*<0, A>0. Inthis case thefunction hasneither amaximum nor aminimum. The function increases when we move from the point (x,, y,)incertain directions and decreases when wemove inother directions. Indeed, when moving along theray =0, wehave f=(Ao)[4+2,Ael>0; when moving along this ray the function increases. But ifwe movealongarayp=, suchthattang,=—A, thenforA>0 we have +[ACHBYins Af=7(oy[45inte,+2040] <0; whenmoving alongthisraythefunction decreases, 3")Let AC—B*<0, A<0, Here thefunction again hasneither a maximum nor aminimum. The investigation isconducted inthe same way asfor3’. 3”)LetAC—-B*<0, A=0. ThenB%0,andequality (2)may berewritten asfollows: Af=(Ao)[sin9(2Bcosp+Csift)+2a,A}. For sufficiently small values of@theexpression inthe parenthe- sesretains itssign, since itisclose to2B, while thefactor sing changes sign depending onwhether @isgreater orless than zero (alter the choice ofg>0 and @<0 wecantake @sosmall that 2a,will not change thesign ofthe whole square bracket). Conse- quently, inthis case, too, Afchanges sign fordifferent @,that is,fordifferent Axand Ay; hence, inthis case toothere isneither amaximum nor aminimum. Thus, nomatter what thesign ofAwealways have thefollow- ingsituation: ifAC—B*<0 atthepoint (x,,y,), then thefunction hasnei- ther amaximum nor aminimum atthis point. Inthis case, thesurface, which servesasagraphofthefunction, can,neatthis point, have, say, theshape ofasaddle (see Fig.” 186). The func- tion atthis point issaid tohave aminimax. 4)Let AC—B'=0. Inthis case, byformulas (2)and (3), itis impossible todecide about the sign ofAf.For instance,’ when A#0 wewill have 1 cqoys f(Asos@+8sing)# Af(Ao)'[(A2¢Ps2)" +20,a0], 298 Functions ofSeveral Variables when@=are tan(-4). thesignofAfisdetermined bythe sign of2a,; here, aspecial additional investigation isrequired (for example, with theaid ofahigher-order Taylor formula orin some other way). Thus, Theorem 2isfully proved. Example 3,Test the following function formaximum and minimum: zext—aytyt3e—y tl. Solution. 1)Find the critical points a aFoes Fart Solving the system ofequations By t3=0,rye} weget xe—4; yet=—fiv-z: ; 44 2Findthesecond-order derivatives atthecriticalpoint(—4,5) and determine the character ofthe critical point: 5 pt. on,Anant aaPoa CHaa: AC—B*=2-2—(—1)*=3>0. Thus,atthepoint(—.,)thegivenfunctionhas@minimum, namely an=—> Example4.Testforamaximum andminimum thefunction z=2*+y*—3xy. Solution. 1)Find thecritical” points using the necessary conditions ofan extremum: a a3 =0, a asyt—armo Whence weget{wo critical points: al, Wel and 4=0, 40. 2)Find thesecond-order derivatives: a a a Fins, tna, Fimo. Maximum and Minimum ofaFunction ofSeveral Variables 259 2)Investigae the character ofthe fist critical point:ate ae ate AC—BYm369270; ADO. Hence, atthe point (1,1) thegiven function has aminimum, namely: Fain—le 4)Investigate the character ofthesecond critical point My(0,0): Aw0; Ba 3; C=O; AC—B'=—9.<0. Hence, atthesecond critical point the function has neither amaximum nor 2minimum (minimax). ‘Example 6Decompose «givenpositivenumberaintothreepostiveterms so that their product isa.maximumSolution, Benote thefirsttermby2,thesecond byy;thenthethirdwill bea—s—g. The product ofthese terms is wary (2-9. Usgiventhat10,9>0,o—x—y>0, thatissta, 40,Hence andy Can avsume values inthe domain Bounded bythe straight lines x=0, y=0, Shysa, tnd ihe partial derivatives ofthefunction u: Hay(o—2e—-y), au $are—ty—0. Equating thederivatives tozero, we.get«system ofequations: (a2) =0;«(a—2y—x)=0. Solving this system weget the critical points: HHO, y=0, My, 0} n=O, yea M40, a);neayO,My(a,0% a a aenad. ung, m(S. 4). The first three points tieontheboundary ofthe region, the last one, inside. Ontheboundary oftheregion, the function uisequal tozero, while inside i'spstve meq attepont(2 thence ets¢ms imum (since itisthe only extremal point inside the triangle). The maximumvalueottheproduct "7 ms "aa(, 22) @Yan$5(9-$-$)"F ao FunctionsofSeveratVariables Investigate the character ofthe critical points using the sufficiency condi- tions. Find the second-order partial derivatives ofthe function 4: a a aBeenay,POatx0y;Meoe Oup_ou Ou AtthepointM,(0,0)wehaveAmSHO;Bmgetma,CmStno, AC—Bt=—at<0, Hence atthe pointMytheres neither amaximtin nor 8minimum, AtthepointMO.) weMveAmSt=—2e; Begrimme: ca24ao,ay AC—Bt= ato. Which means that atthe point M, there isneither amaximum nor amini- mum, Atthe point M,(a, 0)wehave A=0, B=—a, C=—2a: AC—Bt= —a'<0. AtMytoo,thereisneither, maximum notaminimum, Atthepointfa ee oam($$)wehaveA=—38;p=—4; C=; faa So,Ac—ata4@!_ 50;Aco. Hence, atM,wehave amaximum SEC. 18. MAXIMUM AND MINIMUM OF AFUNCTION OFSEVERAL VARIABLES RELATED BYGIVEN EQUATIONS (CONDITIONAL MAXIMA AND MINIMA) Inmany maximum and minimum problems, one hastofind the extrema ofafunction ofseveral variables that arenot indepen- dent, but are related toone another. byside conditions (for example, they must satisfy given equations). Byway ofillustration letusconsider the following problem. Usingapleseoftin2ainareaitisrequired tobuild@closed box inthe form ofaparallelepiped ofmaximum volume. Denote the length, width and height ofthebox byx,y,and z. The problem reduces’ tofinding the maximum ofthe function vm aye provided that 2xy-+2x2 42ue—2a. The problem here deals with aconditional extremum: thevariables x,y,zarerestricted bythe condition that 2xy+2xz4+2yz=2a. Inthis section weshall con- sider methods ofsolving such problems. Let usfirst consider the question ofthe conditional extremum ofafunction oftwo variables ifthese variables are restricted by asingle condition. Maximum and Minimum ofaFunction ofSeveral Variables 301 Let itberequired tofind themaxima andminima ofthefunc tion u=i(x,y) ay with theproviso that xand yareconnected bytheequation 9(x, y)=0. Q) Given condition (2), ofthe two variables xand ythere will be only one which isindependent (for instance, x)since yisdeter- mined from (2)asafunction ofx.Ifwesolved equation (2)for yand put into (1)theexpression found inplace ofy,wewould ‘obtain afunction ofone variable, x,and would reduce the prob- lem toone that would involve finding themaximum and minimum ofafunction ofone independent variable, x. But theproblem may besolved without solving equation(2)for xory.For those values ofxatwhich the function ucan have amaximum orminimum, the derivative ofuwith respect tox should vanish. From(1)wefind$4,remembering thatyisafunction ofx: du_of,ofdsanataya Hence, atthepoints oftheextremum at, ataHriit=o. (3) From equation (2)wefind a P+Remo. Oy This equality issatisfied forallxand ythat satisfy equation (2) (see Sec. 11,Ch. VIII). Multiplying theterms of(4)byan(asyet) undetermined coef- ficient 4and adding them tothecorresponding terms of(3), we haveTeat dy 29,9dy)_(s+55Se)+4(S57) = or ty48)1(at44aR)dy_(SE-+432)+(55-4457) a0. ® Thelatterequality isfulfilled atallextremum points. Choose 4 suchthat forthevalues ofxandywhich correspond totheextre- 02 Functions ofSeveral Variables mum ofthefunction w,thesecond parentheses in(6)should vanish: *) Ff40Lan Bao. But then, forthese values ofxand y,from (5)wehave FeataFtd0. Itthus turns out that atthe extremum points three equations (with three unknowns 2,y,4)aresatisfied: 14128=0, at442 6 athay=o 6) (x, y)=0. From these equations determine x,y,and 4;thelatter only played anauxiliary roleandwillnotbeneeded anymore,From this conclusion itfollows that equations (6)arenecessary conditions ofaconditional extremum; orequations (6)aresatisfied atthe extremum points. But there will not beaconditional extre- mum forevery xand y(and 4)that satisfy equations (6).Asup- plementary investigation ofthenature ofthe critical point isre- quired. Inthesolution ofconcrete problems itissometimes pos-Sibleioestablish thecharacter ofthecritical pointfromthe statement oftheproblem. Itwill benoted that theleft-hand sides ‘ofequations (6)arepartial derivatives ofthefunction Fay M=le W+ho(e, 9) @ with respect tothe variables x,yand 2. Thus, inorder tofind the values ofxand ywhich satisfy con- dition (2), forwhich the function u=f(x, y)can have acondi- tional maximum oraconditional minimum, one has toconstruct anauxiliary function (7), equate tozero itsderivatives with re- spect tox,y,and 2,and from the three equations (6)thus obtained determine thesought-for x,y(and theauxiliary factor 2). The foregoing method canbeextended toastudy ofthecondition: alextremum ofafunction ofany number ofvariables. Letitberequired tofindthemaxima andminima ofafunction ofnmvariables, u=f (x, %-., ,) provided that the variables *)For the sake ofdefiniteness, weshall assume that atthe critical points ogGr. ‘Maximum and Minimum ofaFunction ofSeveral Varlables 303 Ky,Nyyveey Xqareconnected bym(m<n) equations: (yrKyveeyq)=O, 2(ys KayveeyXn)=O,elieBeme ®Pa(XysKyvey %)=O. Inorder tofind thevalues ofx,x4,.++ Xmforwhich there may beconditional maxima and minima, one’ has toform the function FexyyXqyeenSanByereyhag)Pye reyEa)EADlsoeeeXa) SC ee dC equate tozero itspartial derivatives with respect tox,,Xy+++)Xq! afon, -tae. +n,Beno, af 29,tne. +2,Geno, 5) ee oo) and from them+n equations (8)and (9)determine x,,£4,«+5 andtheauxiliary unknowns 4,,..., AgeJust asinthecase ofa function oftwo variables, we’ shall, ‘Tnthe general case, leave undecided the question ofwhether the function, for the values found, will have amaximum or minimum orwill have neither. We will decide this matter onthe basis ofadditional reasoning. Example 1.Let usreturn tothe problem formulated atthe beginning of this tecllon: fo find the maximum of the function vane provided that xytxzt+y2—a=0 (x>0, y>0, 2>0). (10) We form the auxiliary function Fe, y,Nmxye+hleybazy2—0). Find ilspartial derivatives and equate them tozero: wth +2)=0,az(x+2)=0, \ ayxyth(x+y)=0. Theproblem reduces tosolving asystem offourequations (10)and(11) 1mfour unknowns (xy, 2and 2).Tosolve this system, multiply thefrst of aot FunctionsofSeveratVariables equations (U1)bythesecond byy,thethdby2,andaddstaking (10)intoaccountwefindthathaz—S22,Puttingthisvalueof}intoequations (11) weget we[gu +s]=0, ay 7 x[I-Bera] =o, eS5[-Ee+n] =0. Since itIsevident from the slatement ofthe problem that ,y,zarediffe fent trom zero, weget from the latter equationsar wy Buaye Futon Yotoen Zeta. Fromthefrsttwoequationswefindx=y,fromthesecondandthirdequations,yazButthentromequation(10)vegetx=yaz= V/©.Thisisthe onlysystem ofvalues ofs,y, ands, for which there can beamaximum or ininimim: Ilcan beproved that thesolution obtained yields amaximum. Incidentally, this isalso evident from geometrical reasoning (Ihe statement ofthe problem Indicates: that the volume ofthe. box cannot. bebig. without bound; itis therefore nalral toexpect thatorsmeete vlusoftheneste‘Thus, forthevolume oftheboxtobeamaximum, theboxmustbea cube,anedgeofwhichtsequalto/%. Example 2.Determine themaximum valueoftheathrootofaproduct ofnumbers, rowdedthaltheiouigalfogianm ber'a. Thus, the‘probiem”is stated asfollows: itisrequired tofind the max- imum ofthefunction u=f/x, <-%,onthecondition that HAH.bea=0 2 (>0,42>0,ot>0) a Form anauaillary function FbyseventasMmP/FBetMtshighveetg. Find itspartial derivatives: Fad Atiepe 20ofuenk, (vee a)™ Fat ttn=o ofw=nh, Fy=bEtano ofunt. Singular Points ofaCurve 208 From the foregoing equations wefind eqn ate and from equation (12) wehave need. By the meaning ofthe problem these values yield amaximum ofthe funetion $/%--Fyequalto Thus,foranypositive numbers xy,#5,...1%_connected bytherelation ship Fayboe tga, the Inequality Va mse (3) {sfulfilled (sinceithasalready beenproved that=isthemaximum ofthis function). Now substituting into (13) thevalue ofaobtained from (12), weget This inequality holds forallpositive numbers xy)sy,.... 49.The expression ‘onthe left-hand side of(14) iscalled the geometric mean ofthese numbers. ‘Thus, thegeometsic mean ofseveral. positive numbers. is‘not greater” than thelt arithmetic mean. SEC, 19, SINGULAR POINTS OF ACURVE The concept ofapartial derivative isused ininvestigating curves. Let acurve begiven bythe equation F(x, y)=0. The slope ofthetangent tothe curve isdetermined from the formula OF ty a> oF %y (see Sec. 11,Ch. VIII), Ifatagiven point M(x, y)ofthecurve under consideration, atleastoneofthepartial derivatives $°and$Fdoesnotvanish, thenatthispointeither$¢or$iscompletely determined. The curve F(x, y)=0 hasavery definite line tangent atthis point. Inthis case, thepoint M(x, y)iscalled anordinary point. 306 Functions ofSeverat Variables But ifatsome point M,(x,, y,)wehave oF oF(Beng and(Bean then theslope ofthe tangent becomes indeterminate, Definition. Ifatthepoint M,(x,, y,)ofthecurve F(x, y)=0, bothpartialderivativesSe andSFvanish,thensuchapointiscalled asingular point ofthe curve. Thus, asingular point ofacurve isdetined bythesystem ofequations .OF9.OFF=0, Fao, Fao. Naturally, notevery curve has singular points. For example, fortheellipse “og4-$-1-0, obviously, eg .OF_%&,OF_%y, Fan-Sth—n a8, Tak: thederivatives aeand2%vanishonlywhenx=0,f=0,butth derivatives 3and$Fvanishonly =0,y=0,butthese values ofxand ydonot satisfy the equation oftheellipse. Consequently, theellipse does nothave any singular points. Without undertaking adetailed investigation ofthebehaviour ofacurve near asingular point, letusexamine some examples ofcurves that have singular points. Example 1,Investigate thesingular points ofthecurve y'—x(x—a)*=0 (a>0). Solution. Here, F(x, y)=yt—x(e—a)* and therefore oF oFFava ean Fmry Solving the:three equations simultaneously, oF oF Funnn0, Fao, $0, ‘wefind theonly system ofvalues ofxand ythat satisfy them: =a, 440. Consequently, thepointMs(a,0)issingularpointofthecurve. tetasWreeigaethe,bekavlouyofUscurveneatsingularpointand then construct the curve. . Singular Points ofaCurve or Rewrite the equation inthe form pet VE. Fromthisformula itfollows thatthecurve:1)isdefined onlyforx>=0: 2)issymmetrical about thex-axis; 8)cuts thex-axis atthepoints (0,0)and (@,0)The latter point issingular, aswe have pointed out Lei usfirst examine that part ofthe curve which corresponds totheplus sign: yaa) VE Find thefirst and second derivatives ofywith respect tox: yaSS peed Soe ane For x=0 wehave y=os.Thus, the curve touches they-axis attheorigin, A A a Forx= wehave y'=0, °>0, which means thatforx= thefunc- ton yhas @minimum: a2V5. v-3V 5" Ontheinterval O<r<a wehavey<O;fors>S y>O; axe yom Forx=a wehave y=VG, which means thatatthesingular point M,(a, 0)thebranchofthecurvey==-+(e—a) Vxhasatangent y=VG (ea). Since thesecond branch ofthecurve y=—(2—a) Vis symmetrical with the first about the x-axis, the Curve has also asecond tangent (lothe Second branch) atthesingular’ point y=—Va(e—a). iy The curve passes through the singular. pointtwiceSuch'a"polet icalleda'nodat port” poxte-at The foregoing curve isshown inFig, 187, Example. Tetforsingular" points thecurve (semicubleal parabola) yoo. Solutton. The coordinates, of the singular pointe aredetermined fomthefolowing setof %as"ytax'=0; axt=0;Y=0.Consequently, Mg(0,0)isasingularpoint. / aturreviteShegivenequation & yaa VE Toconstruct thecurveletusfestinvestigate thebranch fowhich theplus sign intheequation Fig. 187. 308 FunctionsofSeveratVariables corresponds, since the branch ofthe curve corresponding tothe minus sign isSymmetric withthefirstaboutthexaxie, " “thefunctionyisdefinedonlyforx20,itisnonnegative andincreases as xIncreases, Letusfindthefirstandsecond derivatives ofthefunction y=V7 p=3V% yasyaaVE vayVe Forx=0 wehave y=0, y/=0. And sothegiven branch ofthecurve has a tangent y=0 attheorigin. Thesecond branch ofthecurve y=—VP also : passes through theorigin andhasthesametangent =0 y Tihas, two different branches ofthe curve meet atthe ori- gin, ‘have the same tangent, and aresituated ondifferent Sides ofthetangent, This. kind ofsingular point called acusp ofthe first kind (Fig. 188). Note.Thecurvey*—x'=0 mayberegarded asalimit ‘ingcase ofthecurve y*=x (x—a)*=0 (considered inExam-ying Ble,1)asa0; thatis,when theloopofthecurve is'y?-*8=0 contracted intoapoint. 7 Example 3.Investigate thecurve y—sh—#4=0. Solution. The coordinates ofthesingular points are de- fined bythefollowing setofequations: Str y—x)—5r4=0; 2-0, Fig.198. which hasonly onesolution: x=0, y=0. Hence, the origin 4singular pot. Rewrite thegiven equation in theform gate VE From this equation itfollows that xcan take onvalues from 0to+o.Tatusdelermine thefratandsecond derivatives: 5 yas SVRvaraBVe Investiat, separately, thebranches ofthecurvecoresponing fo,pusand minus, In‘both ‘cases,whenx=0wehavey=0,’=0, which means thatTor Both branches the x-axis isatangent. Let usfirst consider the branch gat tV Asx increases from 0too»,yinereases from 0toeo. The second branch yan VF cuts the x-axis atthe points (0,0)and (1,0). Forx=jgthefunction y=at—V# hasamaximum. Ifx+-2, then yo-e. Singular Points ofaCurve 309 Thus, inthis case the two branches ofthe curve meet atthe origin; both branches have thesame tangent and_are situated onthe same side ofthe tangent nearthe’pointoftangency- This,Kind.ofsingular pointiscalled usp ofthesecond kind. The graph ofthis ionetion isshowninFig.189: y|Example4.Investigate thecurve parte yotteted. ¢ Solution. Theoriginisasingular point. Toinvestigate thecurve near this point. re- write the equation ofthe curve inthe form x yaa VISE q Siac,theequationofthecurve,contains onlyevenpowersofthevariables, thecurve tssymmetric aboutthessordinats axes,and, ye consequently, itissufficient toinvestigatethatpartof‘thecurvewhichcorresponds to Fig.189. the positive values ofxandy. From the latter equation itfollows that %can vary over theinterval {rom 0to1,that i,O<eal. Let'us evaluate thefirst derivative forthat branch ofthe curve which is agraph ofthefunction y=+2"Via ytae)vi-#* Fortm0wehayey=0,y/m0.Thus,thecurvetouchesthex-axisattheorigin.Forx—1wehavey=0,y’=o;consequently, atthepoint(1,0)thetangentisparalleltothey-axis.Forx=Y/%thefunctionhasamaximum Fig, 190). ; , ‘Attheorigin (atthe singular point) the two branches ofthe curve corre- sponding toplus andminus infront oftheradical signaremutually tangent, ‘Asingular"pointofthis.kind.is yjcalled: apoint of-osculation "(alsoknownastacnodeordoublecusp). yextextoo‘Example 5.Investigate thecurve v8 e—1)=0, xX|.Solution. Letuswritethesys- fg I temofequations defining thesia- gular points: B>RE—N=0 Fig. 190. —Se}2850, 2y=0. This system has the solution x=0,yo=0.Therefore, the.point (0,0)Issingular pointofthecurve, Letus.rewrite thegivenequation in yoasVErt. 10 Functions ofSewrat Variables all,ghvlon thatxcanvarytom1toandasotkethevaluO(n whichcasey=0),Letusfhvestigate thebranch ofthecurvecorresponding totheplussign infront ‘otthe radical. Asxincreases fom Ito cosy inceases from 0to& 4‘The derivative ya poten ae When xm we have y—oo; hence, atthe point (1,0)the tangent isparallel tothe y-axis. ‘The. second branch ofthe curve corresponding totheminussigns symietic wihthefirsou Thepoint (0,0)hascoordinates thatsatisly the qq *equationand,consequently, belongstothecurve, But near itthere are noolfer points ofthe curve (Fig 191), This kind ofsingular potnt tscalledan folated.singular'potnt. Exercises onChapter VIII Fig.191 anf hepatil derivatives ofthefllowing =sin? oz ty,Baw a2", a Lereetsoty, Ans,Botesiatyy,Batsn2y 22"Ans.FE a au gersSeeing8wAns.Siaee‘ Banayetores, 2neestortent, 4uaVEte. ou x oz y a x AnsmE. &rearetan(iy).Ant.anHea:ee. ae"Vag ©PTMetny). Ans.omSaag ay“Tah trea tan2.Ans,Feet; Ma. 7,eminYEEPHE x OER Vetere a 2 FF anMONG: Ans,2 =—: f=. Bus oY.Ans.Sammeft aVee Vee oF ay wo G2, MG mare sia ieBRR Bayweemaresin(ety). Ans,Sem 1a Foe ayv oS =2%,10.z=aretony Ans,%=_, “Virere re reers en a" aoe Find thetotal derentials ofthe'fellowing functions: 11,2=ea*-pay*-tsiny.Ansdem(QeyPideeybonny)dy.12enin(y).Ansdell Exercises onChapter VIII atl 13,zee*4", Ans, dzm2e*+*(xdx+ydy). 14.u=tan(3x—y)+6**,sae 1 ns.dum _5(1 _perer Fin6de.15.w= Ans,dumart +(—agrgemq tO"In6)dy+074"Ino aresin ,Ans.dw=2Ge—2ay | y iviVF—# 16.Evaluate f,(2, 3)andf,(2,3)iff(x.y=atty® Ans. (52,3)=4, Fy 9=27. 17,Evaluatedfx.)fors=1.y=Cidem 5dymyf(y)=VER. 1Ans.4. 18. Form aformula which, forsmatl absolute values ofthequantities x, ; Tt and2,yieldsanapproximateextVae 1and2,yieldsanapproximate expresionforV/EEE. ns.1 1 +792). TF 1 18,DothesametorY/EE.Ans.14heyaasag2 20.Findandafozeuto% uaxt+siny, veln(x+y). a 1a 1 .Hareg ao; Bo, en Ans,mtry20=;Bmcosy+20 a 8 TEE cos: a 1a=—l.; %a0. wz)an ae : z2.Find Band itamet, wasn, omitytAns meC08x60),met02-29)meAye 23, Find the total derivatives ofthegiven functions: 2—are sin(u-+-0}; a x 2 wesinxcosa;omcosising, Ans,Satiftka—F<xtactnt%,ae x XoyaW=), Shan)itesF<xtac@etyath. tu 9;yoasing zecose,Ans.Moetsins,25.2aln(I—ati x=YHOO;=—2tan0, Find thederivatives ofimplicit functions ofxgiven bythefollowing equa-ions96.2242 usa Be 4 tions: 26.B+—1=0. ans, Ha SA on,BLat. ans, Hb ate Faw. Ans.Ya —FIng =o mph.8favAnsCoE a8,sin(ay)Pay0. 312 Functions ofSeveral Variables ns,Bambee yg,Beate; find anddeGedecty 0:ting2%angpngOt Ans.Seate!By—Bs‘31.u—vtanaw=0;find5and55Ans.a7 _etaw do sindaw on2 yore sd, 1021See, eo2aantLmPHshowtatHELTOL, ws£nP(4); sowtatty, nomervitteiret funtion F. Compute the second-order partial derivatives: Meexttty$5y4.Ans,a6r—ay;, 2=ax;210.- BROS ae88a=" . te sayoe_e*,cosy 38.reefIngtsingnx.Ans,ZEmeting HL, Seayey, .otsing oe9agagsygte FieG—sinyin« 1ououoe 36. Prove that ifw=!then2424Hg, Virege antaptee3,Provethatit2—2¥%thenxBeypepema ; , oe oe_ 98,Provethatifzetn(ety,then24.2%20,O_O 39.Provethatifz=@(y+ar)+(y—ax), thena’igo? forany doubly diferentible @and ¥. taindedeeoPtetnctionstay ttthepita2 Inhedetonate anandof6Oathewarae.642025 At, Find thederivative ofthefunction 2=5:'—3r—y—I atthepoint 12.1 heeto ramtspoiotepatNG.8, Ansone, 42, Find the derivative ofthe function f(s, y)inthe direction of: 1)the 1apa virheanal aneOana (42) 2)thsete vente Ontn.f. oofenvastte-b sete, Showthatatepit(2,4) te derivative inany direction isequal tozero (the “function isstationary") ‘4. Of all_teangles with the same perimeter 2p, determine the triangle with greatest ares. Ans, Equilateral tangle. 45.Find arectangular parallelepiped ofgreatest volume foragiven total surfaceS.Ans.Acubewithedge/S. Exercises onChapter VIII 313 48,Find the distance between two straight lines inspace whose équations BSNL YE BLY LE gsVE weeer Tara Am oe Test for maximum and minimum the functions: a1.rmety(a—s—y), Ans.Maximum 2atr=Ziy= tit 1 as.aetpaytytt by.Ans.Minimum2atxy—she. ay V3 ©.rasinxtsing tsinety)Ose8;0<y<4),Ans.Maximum patrayad. 50. z=sinx sing sin(x+y)(0<x<m;Oye). Ans.Maximum 2at ray =%. Findthesingular points ofthefollowing curves investigate theircharacter and form equations ofthe tangents atthese points: Bi;ttt Sar=0.Ans.My(0,0) isanode;x=0, y=Oarethe equationsofthetangentsG2yPeeat(at—e). Ans.Adoublecuspattheorigin;thedoubletangent yoo. : 53.yteg—z- Ans.M,(0,0) isacuspofthefirstkind;y*=0isa tangent netgiast—x. ‘Ans.Mg(0,0) isanode;y=43xaretheequations of etangents 85.x*—2ax'y—ary*+at*=0, Ans. M,(0,0) is cusp ofthesecond kind; y'=0 isadouble tangent. 56.y*(a-+x')=x"(at'—x4),Ans.My(0,0)isanode;y=+xarethe equations ofthe tangents. 87.bie?+aYytmxty?, Ans. Mg(0,0) isanisolated point. 58.Show that the curve. y=txinz hasan end point’ atthe coordinate origin and atangent which isthey-axis. 59.Showthatthecurvey=—!— has nodal point attheotigin and Leer that thetangents atthis point are: ontheright y=0, ontheleltyx. CHAPTER IX APPLICATIONS OF DIFFERENTIAL CALCULUS TO SOLID GEOMETRY SEC. 1,THE EQUATIONS OF ACURVE INSPACE _,bet usconsider thevector0A=rwhoseoriginiscoincident withthecoordinate originandwhoseterminus isacertain point A(x, y,2)(Fig. 192). Avector ofthis kind iscalled aradius vector. Letusexpress this vector interms oftheprojections onthe coordinateaxes: raxityj+ek. )¥_acy2)Lettheprojections ofthevectorrbe functions ofsome parameter: f: di x=(1),7 y=9(t), 2) % 2=4(0- ig. 1,Then formula (1)may berewritten asfollows: " r= ite i+xOk ay cr, inabbreviated form, rer(t). ay As¢varies, x,y,and zvary; and the point A(the terminus of thevectorr)willtraceout'aTineinspacethatiscalledthe hodograph ofthevector r==r(f). Equation (1°)or(1*) iscalled thevector equation oftheline inspace. Equations (2)areknown astheparametric equations oftheline inspace. With theaidof these equations, thecoordinates x,y,zofthecorresponding point ofthe curve are determined for each value of¢. Note. Acurve inspace can also bedefined asthe locus of points ofthe intersection oftwo surfaces. Itcan therefore be given bytwoequations oftwo surfaces: ®,(x,y,2)=0,(8, 4,2)=0. t Thus, forexample, the equations Boy pte, z=1 are the equations ofacircle obtained atthe intersection ofa sphere and aplane (Fig. 193). Thus, acurve inspace may berepresented either byparamet- ricequations (2)orbytwo equations ofsurfaces (3). Ifweeliminate theparameter ¢from equations (2)andgettwo equations connecting x,y,2,we will thus make the transition from the parametric method ofrepresenting aline tothesurface F Er Fig. 19. method. Andconversely, ifweiex=9(t), where@(f)isanar- bitrary function, and find yand zasfunctions of¢fromequations ®,[9(4),4.2]=0,®,[9(0,4,2]=0, wewill then make thetransition from representation ofalineby means ofsurfaces toitsparametric representation. axis coincides with thez-axis (Fig. 194). Onto this cylinder wewind aright tihng eteSiac ahte eat ee 316 Applications ofDiferential Catculus foSolid Geometry Let uswrite theequation ofthe helix, denoting byx,y,and 2the coor- inates ofitsvariable point Mand byfthe angle AOP (see Fig. 198). Then xacost, y=asint, 2=PM=AP tan0, where0denotes theacuteangleofthetriangleC,AC.NotingthatAP=at,sinceAPisanarcofthecircleofradiusacorresponding tothecentralangle and. designating tan 8inerms ofm,weget the parametric equations of the'helix in the form x=acost, y=asin!, 2amt 2 ¢ E ar3S 7G? 9 Fig. 194. (here tisthe parameter), orinthe vector form: retacost-+Jasiat-+kamt, {tisnot difficult toeliminate the parameter ¢from the parametric equa- tions ofthe helix: square the first two equations and add, Wefind 2"4-y*—a", This isthe equation ofthe cylinder onwhich the helix lies. Then, dividing ermwize thesecond equation bythefirst and substituting into theobtained Equation thevalueofffound,tromthethirdequation, wefindtheequation of another surface cn which the helix lies: Leung, THis isthe so-called helicoid. Itisgenerated asthe trace ofahalf-tine pataiel tothesy-plane itheendpoint thlshalting isontheanand {the half-tine self rotates about the z-axis at-a constant angular velocity,andriseswithconstant velocity s0thatitsextremity istranslated. alonghez-axis.Thehelixisthelineofintersection ofthesetwosurfaces, and30can ierepresented Bytwo equations: styled, Latnd. The Limit and Derivative ofthe Vector Function a7 SEC. 2.THE LIMIT AND DERIVATIVE OF THE VECTOR FUNCTION OF ASCALAR ARGUMENT, THE EQUATION OF ATANGENT TO ACURVE, THE EQUATION OF ANORMAL PLANE Reverting totheformulas (1’)and (1”)ofthepreceding section, we have r=e (Hit vOsrxOk or r=r(t). Whentvaries,thevectorrvariesinthegeneralcasebothinmagnitude and’ direction. We say that risavector function of the scalar argument ¢,Let ussuppose that lim@()=9., ra foe timy=. Pe lim’ (= 45.lin'x =ty Then wesaythatthevector r,=9,i-+ba/-+ 0 y +xgkisthe limit ofthe vector r=r(t) ‘and 5 we'write (Fig. 195) Fig.198. limr@®=r,. fol From the latter equation follow the obvious equations lim|(—r,|=hin VIP@=eT FOOT FeOHP=O and , limIrO1=Ih Let usnow take upthe question ofthe derivative ofthe vector function ofascalar argument, FO=EOI+VOI+~OR, a) assuming that theorigin ofthevector r(t) lies atthecoordinate origin. Weknow that thelatter equation istheequation ofsome space curve. ; Letustakesomefixedvalue£corresponding toadefinite point Monthe curve, and letuschange bytheincrement af;we then get the vector Ft MD=GUT ADIFVE+ ANI+4+A)h, 318 Applications ofDifferential Calculus toSolid Geometry which defines acertain point M, onthe curve (Fig. 196). Letus find the increment ofthe vector brar(t+A)—r()= =[et+A)—9 (i+ g ++ 4—~ (i+ a +ixQ+AN—x (Ok.| wyInFig. 196,whereOM=r(t),OM,= =r(t+At), this increment isshown’ by 5 \y, thevector MM,=Ar(1). Gi) Letusconsider theratio45oftheincrement of avector function to the Fig.196. increment ofascalar argument; this is obviously avector collinear with the vector Ar(t), since itisobtained from thelatter bymultiplication withthescalarfactor1.Wecanwritethisvector asfollows: Ar)_et+at)—o(t) bt+aH—Vvit) netA)—x()ia ry rehr Ifthefunctions p(t),p(t),x(t)havederivatives forthechosenvalue off,thefactors ‘ofi,, will inthelimit become thede- rivatives (0, '(1),x'(f) asAf—-0. Therefore, inthis case the limitof4FasA¢—-Oexists andisequaltothevectorg’(4+(O/-+ +X (OR Him=o (ity Oi+H Ob. The vector defined bythelatter equation iscalled thederiva- tive ofthevector r(t) with respect tothescalar argument ¢.The derivative isdenoted bythesymbol $forr’, Thus, Har =oOle Os+x (OR @) or ar dey dy de .aralitgitge @) Letusdetermine thedirection ofthevector4°. Since asAt—+0 thepoint M,approaches M,thedirection of thesecant MM, yields, inthelimit, thedirection ofthetangent. The Limit and Dertoative ofthe Vector Function 319 Hence, thevector ofthederivative $Fliesalongthetangent to thecurveatM.Thelength ofthevector 4isdefined bythe formula *) |F|-Vie Ort Orth OF. @) From theresults obtained itiseasy towrite theequation ofthe tangent tothe curve raxityjt+ ck atthe point M(x, y,2),bearing inmind that inthe equation of the curve x=@(f), y=p(t), 2=x(0). The equation ofthe siraight line passing through thepoint M(x, y,2)isofthe form Xax_Yoy_ 2-2 m a e¢ where X,Y,Zarethecoordinates ofthevariable point ofthe straight line, while m,n, and pare quantities proportionaltothe directioncosines ofthisstraight line(thatistosay,tothepro- jectionsofthedirectional vectorofthestraight line). On the other hand, we have established that the vector dr_dey,dy,de aS i+ ee isdirected along thetangent. For this reason, theprojections of this vector are numbers that are proportional tothedirection co- sines ofthe tangent, hence also tothenumbers m,n,p.Thus, theequation ofthetangent will beofthe form Xax_Y—y 2-2 “ay a “ aa & Example 1.Write the equation ofatangent fothe helix xmacos!, y=asinl, 2=ant foranarbitrary valueof¢andfor(=2, Solution. ae ay a=—asint,MYacest, f=om. “WEalaumetinahepintsunderconsideration [26]0, 320 Applications ofDifferential CalculustoSolidGeometry From formula (8)wehave X=acost_Y—asint _Z—amt—asiné~ acost am Inparticular, for1=%weget oV2 )_eV3 ARare Yoene ava ava om ornra Just asinthecase ofaplane curve, astraight line perpendi-culartoatangent andpassing ‘through thepoint»oftangency iscalled anormal tothespace curve atthegiven point. Obviously, one can draw aninfinitude ofnormals toagiven space curve at'agiven point. They alllieintheplane perpendicular tothe tangent line. This plane isthe normal plane. From the condition ofperpendicularity ofanormal plane toa tangent (4), wegettheequation ofthenormal plane: ae 4 aeHX) +5Y9)+(Z—2)=0. ©) Example 2.Write the equation ofanormal plane toahelix atapoint forwhieht=, Solution. From Example |and formula (5)weget V3/y_aV2), V3 aV2 xTE(x2) +(v9) +m(z-amFZ)=0. Let usnow derive theequation ofatangent line and the nor- mal plane ofaspace curve forthe case when this curve isgiven bythe equations O,(x,y,2)=0, O,(x, y,2)=0. (6) Let usexpress the coordinates x,y,2ofthis curve asfunctions ofsome parameter t: x=9(t), v=), Z=244). @ Weshallassume that(0,#(0,x(0)arediferentiable functions of t. Substituting into equations (6), inplace ofx,y,2,their values forthepoints ofthecurve expressed interms of#,weget two identities int: ®,[p(4), vit), x) =0, (8a) 19, VO), x) =0. (8b) Differentiating the identities (8a) and (8b) with respect tot,we The Limit and Derivative ofthe Vector Function a2 tUi 90,dx|90,dy,00,de_9 oedetdyat*Oeap ® Byde5904dy|90442 Ge dt toyatt oeai From these equations itfollows thatdx90,00,2,90,dy20,90,90,00,Wa wy.it_“oeoeieoe (10) dz30,90;_90,00,dz30,90,90, 90,° aieOyOyOe “OeOyay Here,wenaturallyassumethattheexpression 922222-20:9s4. 30; however, itmay’ beproved that the final formulas (I1) and (12)' (see below) hold also forthe case when this expression is equal tozero, provided that atleast one ofthe determinants in the final formulas differs from zero. From equations (10) wehave dx dy dz a a 3,I,Wb,90,~30,9D,_7D,VO,~VW,TH,FO,9,” dyorOeOy“OeOeOx2OEOyOyOF. Consequently, from formula (4)theequation ofthetangent line wili-have the form X—« Voy Zz30,5D,—90,3D,~3,WD,_0,5B,—F.5D,1,5," tyGea ay“GadeaeOeyy or,using determinants, Kar Yay ap9,3D=730,98]79H,I,* ayde aOe oroy 20,20,| —|a0,00,] a,20,fayae|oroe||Beoy The normal plane isrepresented bytheequation 2,204) 20,00, 20,20,Wy aOF aynala+-olepen|lanes=o(12) Gy ars oeoy These formulas are meaningful only when atleast one ofthe determinants involved isdifferent from zero, But ifatsome point Naa90e 32 Applications ofDifferential Calculus toSolid Geometro o}the curve all three determinants 2B,20,)120,00,)20,a0,dy||aa||oxoya,2,|"|a0,00,|"|a0,a0,Gyae|[etoe|Getaye vanish, this point iscalled asingular point ofthe space curve. Atthis:(oethecurvemaynothaveatangent atall,aswasthecasewithsingular points inplanecurves (seeSec.19,Ch,VIL).Example 8.Findtheequations ofatangent line.andanormal’ plane{o the line’ ofintersection ofthe sphere x'y2*=d4r* and the cylinder : Sbylooty atthepoint Mint 7 V2)Fig.19). -——Solution. : ®, (x,y,eatytpeta, >Ole, 9 =e bo—2ry, t a20, 30, 9, fo |.Fate Gate Sine, K——— Frmy—a, Brno, i Thevalues ofthederivatives atthegiven H point Mwill be an 2, 9, 2%i, Bina, Baw, BarvE 2», 2, Fig.197. Prar, Pino, rao, For this reason theequation ofthe tangent tine has the form Kar Yor Zor VEaik7ial The equation ofthenormal plane is V2W—n-@—r Vj=0. SEC. 3,RULES FOR DIFFERENTIATING VECTORS (VECTOR FUNCTIONS) ‘Aswehave seen, thederivative ofavector r=91+ VOI+4Ok, ) is,bydefinition, equal to PO=9 OLY OJt1 OR @) Rules forDiferentiating Vectors (Vector Functions) 323, Whence itstraightway follows that the basic rules for differen- tiating functions hold forvectors aswell. Here, we shall derive theformulas for differentiating asum and ascalar product of vectors; the other formulas we shall write down and leave their derivation for the student. I.The derivative ofasum ofvectors isequal tothesum ofthe Gerivatives ofthe vectors. Indeed, let there betwo vectors: 7,(0=9,O1+¥,OJ+%,OR\ my (FO =H OEE (OI Ma(OR: their sum is HOF O=(, OFF OUT O+HOt buO+% OM Bythe definition ofaderivative ofavariable vector, wehave AAOFALig,(+9,OFEEN,OFMOFF OFOE or AOPROLig)1)+QUE (HDFOOSE KObasOk= HHO OIG ORFROLEROSH ORSK+e Hence, din+n()_drysdry a tata o II,Thederivative ofascalar product ofvectors isexpressed by the formula SPatntne an Indeed, if7,(4),7,(0)aredefined byformulas (3), then, aswe know, thescalar product ofthese vectors isequal to FOO =P+WRF Kile Forthisreason UA Stak? a ee hh cree CO. + Ft) +OEE WV = =OEE VERT TETREOLEIELANES ER)=any, dreantag: The theorem isproved. From formula (II)wehave thefollowing important corollary, Corollary. /fthevectoreisaunitvector, thatis,|e|=1, then ilsderivative isavector perpendicular toit. 1" . 2 Applications ofDiferentiat Calculus toSolid Geometry Proof. It@isaunit vector, then ee=l. Letustake thederivative, with respect to¢,ofboth sides of the latter equation: de,de et+He=0, or 4e2e4=0, that is,the scalar product deet=0, andthismeans thatthevector{¢isperpendicular tothevectore. III. The constant numerical factor may betaken outside the sign ofthederivative: dart) 4) apt10a4mar’(H. a IV.The derivative ofavector product ofvectors r,and r,is determined bytheformula dinxnd dn dryXtal Sr trXE av) SEC, 4,THE FIRST AND SECOND DERIVATIVES OF AVECTOR WITH RESPECT TO THE ARC LENGTH. THE CURVATURE OF ACURVE. THE PRINCIPAL NORMAL Thearclength*)ofaspacecurveM,A=s Fig.198)isdeter- mined just asinthecase ofcurves in&plane, When avariable point A(x, y,z)moves along acurve, the arc length svaries; conversely, when svaries, thecoordinates x,y,2ofavariable point Alying onthecurve also vary. Therefore, the coordinates x,y,2ofavariable point Aofthe curve may beregarded as functions ofthearelength s: x=9(5), y=V(s), z=4S). *)The are length ofaspace curve isdefined inexactly the same way as ‘he arc length ofaplane curve (see Sec. 1,Ch. VIand Sec. 3,Ch. XII). First and Second Derivatives ofVector with Respect toArc Length 325 In-these parametric equations ofthecurve, theare length sis theparameter. The vector OA=r is,accordingly, expressed as P=9()I+V)ITK(S) or r=r(s). a) Thus thevector risafunction ofthearclength s. 2{ 4g, 8 a ava ycx). “i 9 y 7 hs o Fig. 198. Fig. 199. Let usfind out thegeometrical meaning ofthe derivative Asisevident from Fig. 198, wehave thefollowing equations: MA=s, AB=As, MB=s+As, OA=r(s) OB=r(s+As), AB=Ar=r(s+As)—r(s), ar_4Bana Wehavealready seeninSec.2thatthevector{f=limSis Base inthedirection ofthetangent tothecurve atthepoint 4towards increasings.Ontheotherhand,wehavetheequalitytin||=1 3B [the limit oftheratio ofthechord length tothearclength)]. *)inSec.1,Ch.V1,wementioned thisrelationforaplanecurve.It alsoholdsTor.space curve: r(d)=o (LEP ()/+1(0R ifthe functions (0,(0andX(0)havecontinuous derivatives thatdonotvanish simultayfeously. 326 Applications ofDifferential Calculus’ toSolidGeometry Hence, £isaunitvector inthedirection ofthetangent; letus denoteitby0: Hao, CO) Iithe vector risrepresented bythe projections raxityjtzk, then onFit ite, @ and dx\*)(dy\* ayV® +(#)+(Z) <1 Letusnowexamine thesecond derivative ofthevector func- tion4%,thatis,thederivative withrespect to#,anddeter mine itsgeometric significance. From formula (2)itfollows that a_4[ar]_ao e-sa)—a Consequently, wehavetofindlim4. anne ds From Fig,199wehave AB=As, AL=o, BK=0+Ao. Draw from thepoint Bthe vector BL,=o. From the triangle BKL, we find BR=BL,+0R or “ o+do=o4L,R. Thus, L,K= Ao.Since, bywhat hasbeen proved, the length of thevector 6does notchange, |o|=|0-+Ae|; hence, the triangle BKL, isanisosceles triangle. The angle Ag atthevertex ofthetriangle istheangle through which the tangent tothe curve turns from the point Atothe point B;inother words, itcorresponds tothe increment inthe arelength As.From thetriangle BKL, wefind L,K=|40|=2/0|| sin42|=2|sin-42| (since |oj=1). First andSecond Derivatives ofVector with Respect toAre’Length 327 Divide both sides ofthelatter equation byAs: Ae)_9/2|_|2||a9) les|=2)l=|aer(lat r Letusnow pass tothelimit onboth sides ofthelatter equation asAs—+0. On the left side we have ae)_|da de[3e|=[22|- Then sn88tim |?) =1 +0) AP " wel sinceinthiscaseweconsider curvessuchthatthereexistsalimitlim#2and,consequently, Ap—+0asAs—+0.Thus,afterpassingto‘thelimit wehave 40)—tim|4® lel-amlal: © The ratio ofthe angle ofturn Agofthe tangent, when thepoint Agoes tothe point B,tothe length Asofthe arcAB (in abso- lute value) iscalled (just asitisinthe case ofaplane curve) theaverage curvature ofthegiven line onthesegment AB: =|s¢ averagecurvature =|48|. The limit ofthe average curvature asAs—+0 iscalled thecurva- ture ofthe line atthe point Aand isdenoted byK: K=jim|42|. Butthenfrom(4)itfollows that$2—=K; which means thatthe length ofthederivative ofaunit vector*) ofatangent with res- pect tothearclength isequal tothe curvature oftheline atthe given point. Since thevector oisaunit vector, itsderivative &&isperpendicular toit(seeSec.3,Ch.1X,Corollary). It should beremembered that the derivative ofavector isavector and forthis reason wecan speak ofthe length ofthe derivative, 328 Applications ofDifferential Calculus toSolidGeometry Thus,thevector 4%isequal, inlength, tothecurvature ofthe curve, and, indirection, isperpendicular tothe vector ofthe tangent. Definition. The straight line that has the same direction asthe vector“€andpasses through thecorresponding pointofthecurve iscalled theprincipal normal ofthe curve atthe given point. We denote bymthe unit vector ofthis direction. Sincethelength ofthevector 4%isequaltoK,which Isthe curvature ofthe curve, we have BaKn. The reciprocal ofthecurvature iscalled the radius ofcurva- tureofthelineatthegivenpointandisdenoted byR,x=R. So we can write eaEne i) From this formula itfollows that 1_(ar)mn(%)- C) But dr_d’e dy dizBB BitGe Hence, 1 Ty (Ty ,a-V (@)+(@)+@)- 6) This formula enables ustocompute thecurvature ofaline at any point provided that this line isrepresented byparametric equations inwhich theparameter isthe arc length s(inother words, iftheradius vector ofthevariable point ofthegiven line isexpressed asafunction ofthearclength). Let usconsider thecase when the radius vector risexpressed asafunction ofanarbitrary parameter ¢: r=r(). Inthis case thearelength swill beregarded asafunction of theparameter ¢.Then thecurvature iscomputed asfollows: dr_drdsaaat* o First and Second Derivatives ofVector with Respect toArc Length 329 Since - laj=t9 we have ‘dr\* ‘ds\*(#)'-G)- - ® Differentiating theright and leftsides of(8)and reducing by two, weget arate _dsd'saeaae ® Further, from formula (7)itfollows that arid asdi ds" a Differentiate, with respect tos,both sides ofthis equation: a's adr 1araa7 Tas ai [ayGy “Gy Substituting intoformula(6)theexpression obtainedfor$5weget dts 9% 1as1ardeRY|Gt7ds\*—di[ds= Gy *G) a'r)* (ds)\*_pd°rdrdsd's,(dr)*(4°s)* (2)@)aaa e+(%) (@) = (ay ; ii) Thi de ar ~)Thisequation followstromthefactthat[$¢]—im|e].Butar 1schord subtending anareoflengthas:Therefore ©approaches 1as as, 00 Applications ofDiferential CalculustoSolidGeometry Fi 48ana Expressing $$and£%byformulas (8)and(9)interms ofthe derivatives ofr(f), weget*) ar\*(ar\*_( dtrar\*-[#) (a)-(S47) (10)(zt Formula (10) may berewritten asfollows: **) dr aryoget] ay{@)} We have obtained aformula that enables us to calculate the curvature ofagiven line atany point foranarbitrary paramet- rierepresentation ofthis curve. Ifinaparticular case thecurve isaplane curve and lies in thexy-plane, then itsparametric equations have the form x=0(), 9=9(0, 2=0. Putting these expressions ofx,y,zinto formula (11), wegetthe earlier derived (inCh. V1) formula that yields thecurvature ofa plane curve represented parametrically: Kale OY=v OF!(oO ONY * Example, Compute thecurvature ofthehelix rata cost-+Jasint+-kamt atanarbitrary point, *)We:tamormthedenominator asttlows($f)'—={(#4)"I= ={() Yeewecannotwrite("By ($fwemeanthesee squareofthevector$F:by{(Zf)'}' thethirdpowerof($F)".Theexe **)Weutilised theidentity a**—(ab)*=(axb)*whosevalidityisreadily recognisable lone rewrites theIdentityssfollows:a*b*—(abcosg)*=(ab sing) Osculating Plane. Binormal. Torsion a Solution. ar Gam Hasin +Jacost+kam, aegaa—$acost—Jasint, dr. dtr‘ ee 4 i 1 . 2,afGF_|—asint costam|4atmstat—fatmcoshat,acest —asint 0 dr de)(SfxGh)matonen, (EE)motantatctatmtmatttm. Consequently, ane TL atime) ,RamaPam R=a(1-+m*)=const. Thus, the helix has aconstant radius ofcurvature. Note. If2curve lies inaplane, then without violating genera- lity, wecan assume that itlies inthexy-plane (this can always beachieved bytransforming thecoordinates). Now ifthecurveliesinthexy-plane, thenz=0;butthen$4=0 alsoand,conse- quently, thevector mlikewise lies inthexy-plane. Wethus con- clude that ifacurve lies inaplane then itsprincipal normal lies inthe same plane. SEC. 5.OSCULATING PLANE, BINORMAL. TORSION Definition 1,The plane passing through thetangent line and the principal normal toa given curve atthe point Aiscalled an osculating plane atthe point A. Fortheplane ofacurve, the osculating plane coincides with the plane ofthe curve. But ifthe curve isnot aplane curve, and ifwetake two points onit,Pand P,, wegettwo different osculating planes that form adihedral angle p.The bigger the angle wt,themore thecurve differs inshape from aplane curve. Tomake this more precise, letusintroduce another definition. Definition 2,The normal (toacurve) perpendicular toanoscu- lating plane iscalled abinormal. On the binormal let ustake aunit vector 6and make its direction such that the vectors 6,m,&form atriple with the 382__Applications ofDiflerentiat Calculus toSolid Geometry same orientation astheunit vectors é,J, lying onthecoordi- nate axes (Figs. 200, 201). ke o ‘& 4 Fig. 200. Fig. 201. Byvirtue ofthedefinition ofavector and scalar product of vectors we have b=oxn; bb=1. a) Wefindthederivative of#2.Byformula (IV),Sec.3, do _d(oxn)_do dn10x) xnsoxt. @ do_n But42=(seeSec.4),therefore Sxn=znxn=0, and formula (2)takes theform db yd faoxe. (3) From this itfollows (by the definition ofavector product) that4isavector perpendicular tothevector ofthetangent o. Ontheotherhand,sinceisaunitvector,2°isperpendiculartob(seeSec.3,Corollary).Thismeans thatthevector %isperpendicular bothto¢and to6;that is,itiscollinear with the vector 1 Letusdenote thelength ofthevector $2by7;weput ialalr then . db_o1 ain 4) Oscutating Plane, Binormal. Torsion a3 Thequantity 7istheforsion ofthegivencurve. The dihedral angle 1between the osculating planes that corre- spond totwo points ofthe curve isequal totheangle between the binormals, Byanalogy with formula (4), Sec. 4,Ch. IX, one can write 40) tim|ae|=dim,résr- Tosummarise, then, the torsion ofacurve atapoint.Ais equal, inabsolute value, tothe limit which isapproached (asAs—+0), bytheratio oftheangle between theosculating planes atthepoint Aand theneighbouring point Btothelength [As] oftheareAB. Ifthecurve isplane then theosculating plane does notchange itsdirection and,consequently, thetorsion isequaltozero. From the definition oftorsion itisclear that itisameasure ofthe deviation ofaspace curve from aplane curve. The quantity Tiscalled theradius oftorsion ofthecurve. Let usfind aformula forcomputing torsion. From (3)and (4) itfollows that pamoxst. Multiplying scalarly both sides by,weget pan=alox] Ontheright side ofthis equation wehave theso-called mixed (ortriple) product ofthreevectors m,0and42.Inaproduct ofthis kind thefactors, asweknow, may becircularly permuted. Inaddition, taking into consideration that aa=1, werewrite the latter equation inthefollowing form: 1_ faern9[xn or poao[ax Z). ) Butsincea=R4, wehave dn_p dit yan a GRetae 394 Applications ofDifferential Calculus toSolidGeometry ande anOr|pdiraRdr [ax]=-RSax{R e+Gale afar. ar). pak ar air=R’(xF]+Ro[sexs]: But since thevector product ofavector into itself isequal to 2010, ar ar[ax] =O, Thus, 4 niein’ 2fderer[axe]=a"[SexSr]. Noting thato=% andreverting to(5),weget 1 adel dr dtrpo eax (6) Ifthefactor risexpressed asafunction ofanarbitrary param- eter f,itmay beshown, *)much like was done inthepreceding *)Indeed, dr_drdsanaear Differentiating this equality once again with respect tof,weget Sead(2)te,erdteee(ataedeamas \s)aataar— ar \a) to ae Differentiate itonce more with respect tof: a'r_d (a)ds(ds\", dirdsd's,d(dr)dsd's,drd's aas(4) (4)tenats (8)aaetasae@r(ds\",4drdsd's,drd's<o(2)orga eae Let usnow form atriple product: (St x8t)maarae)= ards{[dtr(ds),drd's),[dtr(ds\*,5dirdsd's,drair~Sii{[&(8)+2]«[(B) +355anesae]}- Opening the brackets ofthis product bythe rule ofmultiplying polyno-mials,anddisregarding thosetermsthatcontaineventwoidenli¢alvectorHelos (ncehetripleproduct ofthrefactors whereatleastwpareequalPON tee ade(eae)(1) aoexar)ds\ast“as)ai)* Oseutaiing Plane. Binormal. Torston 205 section, that arfdtr dirdr[dr dtr) _atlar ae[axr]=ayy(@) Putting this expression into formula (6)and replacing R®by itsexpression from formula (11), Sec. 4,wefinally get arf dtr ae 1__alexa]Ta a a[ea This formula makes itpossible tocompute the torsion ofthe curve atany point ifthe curve isrepresented byparametric equations with anarbitrary parameter ¢. Concluding this section, we note that the formulas which express the derivatives ofthe vectors 6,6,mare called Serret- Frenet formulas: do_n 4on dno 6 BR) GAT! ERT The last one ofthem isobtained asfollows: n=bxo, dn_d(bxa)_ db do_n nBas aXOFOXE =FXOFOxe= 1 L sem: =PAxo+POxn; but nxo=—6; bxn=—o, therefore Sr as TR Finally, noting that ds\*_(dr\*(a)'-@)’ ds)*_ (dr)(a)~{(a7)F- weoblain therequired equality 226 Applications ofDifferential Calculus foSolid Geometry Example. Compute the torsion ofthehelix retacost-+Jasint +hamt, Solution, —asiné —acostam ar fair dtrae He)loacestasin’0[matn, [xa] ‘=a!(1-4+-m4)(seeExample,Sec.4). Consequently, raSt alt) | SEC. 6.ATANGENT PLANE AND NORMAL TO ASURFACE Let there beasurface given byanequation oftheform F(x, y,2)=0. O) We introduce the following definition. Definition 1.Astraight line isafangent toasurface atsome _point P(x, y,2)ifitisatangent tosome curve lying onthe surface and passing through P. Since aninfinitude ofdifferent curves lying onthesurface pass through the point P,then, generally speaking, there will also be ‘aninfinitude oftangents’ tothe surface passing through this point. Weintroduce the concept ofsingular and ordinary points of asurlace F(t, y,2)=0. IfatthepointM(x,y,2)allthreederivatives ,i$are equal tozero oratleast one ofthese derivatives does notexist, thenMiscalledasingular pointofthesurface. IfatMx,y,2)allthreederivatives 3°,$°,3Fexistandarecontinuous, andat least one ofthem differs from zero, then Misanordinary point ofthe surface. We can now formulate the following theorem. Theorem. Alltangent lines toagiven surface (1)atanordinary point ofitPlieinone plane. Proof. Let usconsider, onasurface, acertain line L,(Fig. 202) passing through agiven ‘point Pofthesurface. Letthis curve be Tepresented byparametric equations: F=9) Y=VO, 2=*(D. @ ATangent Plane and Normal toaSurface 37 Atangent tothecurve will beatangent tothe surface. The equations ofthis tangent have theform Xax_Y¥—-y_ 2-2 “EG a a ou @ Ifweputexpressions (2)into equation (1), thelatter will be- come anidentity inf,since the curve (2)lies onthesurface (1). Differentiating itwith’ respect to¢,weget*) :OFde,OFdy,OFdeRataattoaa”®PS#. mausaneaethevectorsNand$F740, thatpass throug 3 in atate Be Fi FHE @VB /e Theprojections ofthisvector$F,2,2" yy depend onx,y,2,which are thecoordinates of;itwillbenotedthatsincePisanordinary point,these projections atthe point Pdonot simultaneously vanish and therefore TaN CLGWWGland ini=V(3)+(37)+(e)#0 The vectorar_dx),dy;de ae ee 6 istangent tothecurve passing through thepoint Pand lying on thesurface. The projections ofthis vector are computed from equations (2)with thevalue oftheparameter ¢corresponding to thepoint P.Let uscompute thescalar product ofthevectors N and4,whichproduct isequal tothesumoftheproducts of like projections: oFde,OFdy,OFdzNaraatwatea: *)Here weapply therule fordiferentiating acomposite function ofthree vais ThisueIsaplcble eresinelltepartialdvvatives 2 SEBEae,asstated,continuous. 08 Applications ofDigerential CalculustoSolidGeometry Onthebasis of(3), the expression onthe right isequal to zero; hence ar ni=o, From the latter equality itfollows that the vector Nand the tangent vector “ftothecurve(2)atthepointPareperpendicu- lar. The foregoing reasoning holds for any curve (2) passing through thepoint Pand lying onthesurface. Therefore, every tangent tothe surface atthe point Pisperpendicular tooneand % thesame vector Nandforthisreason3% allthese tangents lieinasingle planeRythat_isperpendicular tothevector ZY A N.Thetheorem isproved,< as Definition 2.TheplaneinwhichFSP earslieallthetangentlinestothelines[‘» onthesurfacepassingthroughthegiven point Piscalled thetangent plane tothesurface atthepoint P (Fig. 208). Par Ifshould benoted that theremay not exist atangent plane at thesingular points ofthesurface. Atsuch points, thetangent lines tothesurface may not lieinone plane. For instance, the vertex ofaconical surface isasingular point. The tangents to theconical surface atthis point donot lieinone plane (they themselves form aconical surface). Let uswrite the equation ofatangent plane toasurface (1) atanordinary point. Since this plane isperpendicular tothe vector (4), itsequation has the form XN +EVtFEZ—2)=0. © Iftheequation of@surface isgiven inthe form 2=f (x,y), or 2—f(x, y)=0, then ora oF__ at oF _yioe ayy aah and theequation ofthe tangent plane isthen ofthe form a af , 2-2(xa rhy—y. @) ATangent Plane and Normal toaSurface 339 Note. Ifinformula (6') we put X—x=Ax; Y—y=Ay, then this formula will take the form aApa ay:Zam Fart Fby; itsright side isthetotal differential ofthe function 2=/(x, y). Therefore, Z—z=dz. Thus, the total differential ofafunction of two variables atthe point M(x, y), which corresponds tothe increments Axand Ayoftheindependent variables xand y,is equal tothecorresponding increment onthe2-axis ofthetangent planetothesutface whichisagraphofthegivenfunctionDefinition 3.Thestraight linedrawnthrough thepointP(x,y,2) ofsurface (1)perpendicular tothe tangent plane iscalled thenor mal tothe surface (Fig. 203). Let uswrite theequations ofthe normal. Since itsdirection coincides with that ofthe vector N, itsequations will have the form Xax_Yay2-2 FF a ey Iftheequation ofthesurface isgiven intheform 2=/(x, y),or 2-1 (x,y)=0, then theequations ofthe normal have the form Xax_Yoy_2-2 rr oe oy Note. Let the surface F(x, y,z)=0 bethe level surface for some function ofthree variables u=u(x, y,2);that is, Fey, 2)=ule y,2)—C=0. Obviously, the vector Ndefined byformula (4)and inthe direction ofthenormal tothelevel surface F=u(x, y,2)—C=0, will be oupu5,OuNeagltgitae that is, N=gradu. Wehave thus proved that thegradient ofthefunction u(x, y,2) isinthedirection ofthenormal tothelevel surface passing through thegiven point. M0 Applications ofDierenti CalculustoSolidGeometry Example,Writetheequationofthetangentplaneandtheequationsof Nhesarial tothesriace ofthespherex+-ryetate 4atthepointPU,%3 seyteet—and; FaveFoy,Foor Pap amateyttied; Eons Fang: Fate; forx=1, y=2,2=3wehave OF_»OF_, OFam am*a8 Therefore, theequation ofthe tangent plane will be 2(x—1) $4(y—2)46(2—3) 0orx2y+32—14=0. The equations ofthe normal ate x=1_y—2_2-3 rr xol_y—2_ 2-3Sears Exercises onChapter 1X Findthederivatives ofthevectors: 1.resdeot¢4+Jarctant.Ans.Pensitetahberelebjtmin.Ans,rie yak rela hk,ansPate2 4.Find the vector ofatangent, the equations ofthe tangent and_ the equations ofthenormal plane tothecurve r=ti+¢%/+itk atthepoint G9, Ans,Pat+6/+IR; tangent: PatTPAISZ;normal plane: x-+6y-4272—78. 5:Find the vector ofatangent,theequationsofthetangentandtheequation ofthenormalplanetothecurverafcosS44Jsint+hsia5. 111eosbs Ans.Pamaaisint+7Jcost+yhcos55theequation ofthetangent X-cot Y—psint Zs EE RFE theequationoftheoralplane: paint—Ycos2008Lemaneycontbaconwherea,gszarethe coordinates ofthat point ofthe curve atwhich thenormal plane Ysdrawo od A (iat, enc, phate, sma). &.Find theequations ofthetangent tothe curve rent—sint, y=l—cos zeta 4;andthecosinesoftheanglesthatitmakeswiththecoordinate Exercises onChapter IX oH ares,Ans,AEXV=Ve2%copaersint2;cospocbsinty sin2cos2cot U p z 2 ? cosy=cos4.7.Findtheequation ofthenormal planetothecurvez_xt—y3, yx attheorigin. Hint. Write theequations ofthecurveinparametric’ form. Ans. x+y=0.8Find©,mbatthepoint=forthecurverad(cossta)+ 1=oay—k sin(1cos)—Reost. Ans.=(ttyeaa : +Jsin( ) Pattie: aHSBonus it 9.Find theequations ofthe principal normal ‘and the binormal tothe e e eo IXwurve rele; gett; cmt at the point Ui yn2%). Ans,2mcurt Tiga ritthepoint(xye)AnsrT a ee 1=4 —aR" a 10.Find the equation ofthe osculating plane tothecurve =x; xt=z a4thepointAiles Th,AnsOxmndystde 11,Findtheradiusofcurvature foracurverepresented bytheequationssHyt—4e0, etyend.Ans.R=2. a12,Findtheradiusoftorsionofthecurve:rafcos!-+/sin+k—"— en!Ansroe. 13Find theradius ofcurvature and thetorsion forthe curve r=74-214, Ans.Rat+915!,P=14,Provethatthecurve,ra(aj"-4byl+e)fb(al!-+Oyba)+ +(ayi+bytbey)hisplane.Ans.7”=20;thereforethetorsionisequaltozero, 15.Find thecurvature andtorsion ofthecurve x=e!, y=e-!, z= V2. 16,Find thecurvature and torsion ofthecurve x=e~'sint, yee! costs zee.Ans.Thecurvature isYathetorsionIspe! 17,Findtieequation ofthetangent planetothebyperbolotd 2y—#P -e1 atthepoint(xyyyy4).Ans.SM2oy, 8Findtheequation ofthenormal tothesurface x*—4yp22=6 at the point (2,2,3).Ans. y+4x—10; 3x—2—3. 19Findtheequation ofthetangent planetothesurface 2—=2:"-+4y* at the point M(2,1,12). Ans. &x-+8y—z2=12. 20. Draw {o'the surlace x*-+2/*++28=1 atangent plane parallel tothe planex—y-tiem0. AnexytiemVE, CHAPTER X INDEFINITE INTEGRALS SEC, 1,ANTIDERIVATIVE AND THE INDEFINITE INTEGRAL InChapter IIIweconsidered aproblem like thefollowing: Given afunction F(x), find its derivative, that is,the function F(x)=F"(a).Inthis chapter weshall consider thereverse problem: GiventhefunctionTe,itisrequiredtofindafunctionF(x)suchthatitsderivative isequal tof(x), that is, Fe)=!0). Definition 1.The function F(x) iscalled the antiderivative of thefunction /(x) onthe interval [a,b]ifatall points ofthis interval theequality F’(x)=/(x) isfulfilled. Example. Findtheantidrivative ofthefunstion f()=,Fromthe delnition ofanantiderivaive follows thatthefunctionF(a) tsanantiderivtive, since(%)aX Itiseasy toseethat ifforthegiven function f(x) there exists anantiderivative, then this antiderivative isnot theonly one. Intheforegoing example, we,could takethefollowing functions asantiderivatives: F(x)=3+1; F(x)=5—7 or,generally, F(x)=-4€ (where Cisanarbitrary constant), since ($40) =x. Ontheother hand, itmay beproved that functions oftheform %+Cexhaust allantiderivatives ofthefunction x*,Thisfollows from thefollowing theorem. Theorem, IfF,(x) and F,(x) are two antiderivatives ofthe Junction f(x) on‘the interval (a,6], then the difference between them isaconstant. Proof. Byvirtue ofthedefinition ofanantiderivative wehave F;watery " F@)=f(e) a foranyvalue ofxontheinterval (a,6). Antiderivative and theIndefinite Integral 343 Letusput F,)—F,(2)=9(2). @)Then by(1)wehave Fi) Fils) =F(x) f(a) =0 or #()=[F, ()—F,(x)=0 foranyvalueofxontheinterval (a,6].Butfrom9’(x)=0itfollows that p(x) isaconstant. Indeed, letusapply theLagrange theorem (see Sec. 2,Ch. IV) tothefunction @(x), which, obviously, iscontinuous anddifleren- tiable onthe interval [a,6}. Nomatterwhatthepointxontheinterval [a,6],wehave, byvirtue ofthe Lagrange theorem, 9(x)—@(a)=(x—a)9(E), where a<&<x. Since 9(&)=0, ()—9(a)=0 or 9)=9(a). @) Thus, the function (x) atany point xoftheinterval (a,6] retains thevalue (a), and this means that thefunction p(x) is constant on[a,6].Denoting theconstant @(a) byC,weget, from (2)and (3), FLQ)—FW)=C. From theproved theorem itfollows that ifforagivenfunction F(x) some one antiderivative F(x) isfound, then any other anti- derivative off(x) has theform F(x)+C, where C=const. Definition 2.Ifthefunction F(x) isanantiderivative off(x), then the expression F(x)-+C isthe indefinite integral ofthe function f(x)andisdenoted bythesymbol |f«x)dx.Thus, by definition Jie)dx=F(2)+6, if F’(x)=f(x). Here, thefunction f(x) iscalled theintegrand, f(x)dxistheelement ofintegration (theexpression under theintegral sign), and{is theintegral sign. Thus, anindefinite integral isafamily offunctions y=F(x)+C. ud Indefinite Integrals From thegeometrical point ofview, anindefinite integral isan assemblage (family) ofcurves, each ofwhich isobtained bytrans- lating one ofthecurves parallel toitself upwards ordownwards (that is,along the y-axis). Anaiural question arises: doantiderivatives (and, hence, an indefinite integral) exist forevery function f(x)? The answer isno. Lel usnote, however, without proof, that ifafunction f(x) is continuous onthe interval (a,5], then’ there isan antiderivative ofthis function (and, hence, there isalso anindefinite integral). This chapter isdevoted toworking out methods bymeans of which wecan find antiderivatives (and indefinite integrals) of certain classes ofelementary functions. The finding ofanantiderivative ofagiven function /(x) is called integration ofthefunction f(x). Note the following: itthe derivative ofanelementary function isalways anelementary function, then the antiderivative oftheelementary function maynotprovetoberepresentable byafinitenumber ofelementary functions. We shall return tothis question attheend ofthechapter. From Definition 2itfollows that: 1.The derivative ofanindefinite integral isequal tothein- tegrand, that is,ifF’(x)=f(x), then also (Sfepdr)’ =F+0y=F ) This equation should beunderstood inthesense that thederiva- tive ofany antiderivative isequal totheintegrand. 2.The differential ofanindefinite integral isequal tothe expression under theintegral sign: (Jfxrdr) =/(xdde. 6) This results from formula (4), 3.The indefinite integral ofthedifferential ofsome function is equal tothis function plus anarbitrary constant: SF(x)=FQ)+C. The truth ofthis equation may easily bechecked bydifferentia- tion [the differentials ofboth sides areequal todF(x). SEC. 2,TABLE OF INTEGRALS Before starting onmethods ofintegration, wegive thefollowing table ofintegrals ofthesimplest functions.Thetableofintegrals followsdirectlyfromDefinition 2,Sec.1.Ch, X,and from” the table ofderivatives (See. 15,Ch. Ill). Table ofIntegrals as (The truth oftheequations can easily bechecked bydifferentia- tion: toestablish that the derivative oftheright side isequal to the integrand). LJvde=2+C(a%—2). (Hereandintheformulasthatfollow, Cstands foranarbitrary constant.) 2.JF=injx|+c. 3.{sinxdx= —cosx+C. 4.Scosxde=sinx+C. 5.fAostanx tc. 6.|Siem cote$C. 7.[tanxde=In|c0sx|+C. 8.[cotxdr—In| sinx|+C. 9.Jetdxmet+c. 10.fatdx=So+. 11.[-Hpmaretans+C. de oh 71,(stat aretane+c.de_1plate 12.(Sag in|te]+c. ae 1 13Jpera sine$C. . 13"Jpegmarcsin Ete. de a 5page[x+V¥Ee]+0. Note. The table ofderivatives (Sec. 15,Ch. III) does not have formulas corresponding toformulas 7,8,11’, 12, 13°and 14. However, differentiation will readily prove the’truth ofthese as well. Inthe case offormula 7we have (—In|cosx|)’=—SS*mtane, consequently, {tanxdx—=—In|cosx|+C. 6 Indefinite Integrats In the case offormula 8 (in|sinx])’=SS"=cotx, consequently, {cotx=Injsinx|+C. Inthecase offormula 12, (dm|223|)-dnte+x1—inla—xiy = ipa 1a7Ftd) ao therefore, de lateSate =|] +0. Itshould be noted that the latter formula will also follow from thegeneral results ofSec. 9,Ch. X. Inthe case offormula 14, iVe|)-—I— (1+>), (aleVaree a+Vergo(i7H) Vesahence, .Speen nietVeI+¢. This formula likewise will follow from thegeneral resultsofSec.11. Formulas 11’and 13’may beverified insimilar fashion. These formulas will later bederived from formulas 11and 13(see Sec.4, Examples 3and 4). SEC. 3,SOME PROPERTIES OF AN INDEFINITE INTEGRAL Theorem 1.The indefinite integral ofanalgebraic sumoftwo orseveral functions isequal tothesum oftheir integrals SU) +fWlde=fF,Code+Ff,Code. a For proof, letusfind thederivatives oftheleft and right sides ‘ofthis equation. Onthe basis of(4)ofthe preceding section we have (SA@+h 0]dr)=+h (Sh)detGf,(2)dx)’= =(Sh) de)’+{hede)’=f+h0). Some Properties ofanIndefinite Integral “7 Thus, thederivatives ofthe left and right sides of(1)are equal; inother words, the derivative ofany antiderivative ontheleft- hand side isequal tothederivative ofany function ontheright- hand side ofthe equation. Therefore, bythetheorem ofSec. 1, Ch. X,any function onthe left of(1) differs from any function onthe right of(1) byaconstant term. That ishow weshould understand (1). Theorem 2.The constant factor may betaken outside theintegral sign; that is,ifa=const, then Saf(x)de=aff(2)de. ) Toprove (2), letusfind the derivatives oftheleft and right sides: (Saltode)’=af(x), (affdr)’ =a({F()dr)’=af(x). The derivatives oftheright and left sides areequal, therefore, asin(1), the difference ofany two functions on the left and right isaconstant. That is-how weshould understand equation (2). ‘When evaluating indefinite integrals itisuseful tobear inmind thefollowing rules. Lif ffide= Fix)+C, then Ji(ax)demLF(ax)+0. ® Indeed, differentiating the left and right sides of(3), weget (§/(ax)de)’=F(ax), (4F(ex)=4Flan=fF(axya=F’ (ax)=fax). The derivatives ofthe right and left sides areequal, which is what wesetout toprove. I if [feidr=F@)+C, then JPle+0)demF(x+6)+C. “ ea Indefinite Integrate Mh. tf fiede=F ()+C, ‘then Sflax+b)demZFax+b)+C. ) Equations (4)and (5)areproved bydifferentiation oftheright and left sides. Example 1. §eeosine+5VFarm2xtde—(ssineast 5VFar= n2fde—3fsincets[xarm =PE8(eos5—$mettseosatPxVELC.gt! Example 2. ; u u Vi)dem Ae 7Bt a=S(petapate Vieensfe“etyfeactfaae aaa ee eeYrsVasko vise, —gtl "gti Gal3 at gq Example3. JHeamietsi te. Example 4 Jcosedxsin40. Example 6. Jarod —beoteor46. SEC, 4,INTEGRATION BYSUBSTITUTION (CHANGE OFVARIABLE) Letitberequired tofind theintegral SF)de wecannot directly select the antiderivative of/(x) butweknow that itexists, Integration bySubstitution M9 Let uschange the variable intheexpression under the integral sign, putting x=), i) where @(f)is acontinuous function with continuous derivativehavingan.inversefunction. Thendx=g’(f)df;weshallprovethat inthis case wehave the following equation: JFdemiow) ae. @ Here weassume that after integration wesubstitute, ontheright side, theexpression of¢interms ofxonthebasis of(1). Toestablish that the expressions tothe right and leit are the same inthe sense indicated above, itisnecessary toprove that their derivatives with respect tox'are equal. Find thederi- vative ofthe left side: (JF)de),=F). We differentiate theright side of(2)with respect toxasacom- posite function, wherefistheintermediate argument.Thede- pendence offonxisexpressed by(1);here,S=q'(t) andby the rule ofdifferentiating aninverse function, . “wo a-TO" We thus have (Sree eat)=(FFlec}e’(at),fm OeD =HeOleOem=lem)=1 0. Therefore, the derivatives, with respect tox,ofthe right and left side of(2)areequal, asrequired. The function x=@(4) should bechosen sothat onecanevaluate the indefinite integral onthe right side of(2). Note. When integrating, itissometimes better tochoose a change ofthevariable intheform oft=1p(x) and notx=@(0). Byway ofillustration, letitberequired tocalculate anintegral ofthe form sreevay” Here itisconvenient toput vayat 350 Indefinite Integrals ‘then W)dxmdt, Wide ca =SRBE= [Fe inje1¢e=inj pele. The following areanumber ofinstances ofintegration by substitution. Example 1.[V'sinx cosxde—? Wemakethesubstitution ¢=sin.x; then di=cosxdxand,consequently, [Vinkcosxde=[ VTa=fe"ar— athQo, =cmSaintetc. Example 2A=? Wepute=ipathendt=2edeand|AE= Lat1 apfFegimitcazinatsnse. Example3.Saray Weput(=;thendemadt,@ dx 1 Cad 1c at 1 1 xSatinhstenbftorn}weunpcntaetanEte. ae} ar x, Example4.Siew Saye Weputrmsthen@ az 1 _aat at Inadt, a 7 aneSvan)pide -Spieanmnte sarcsla£46 (tisassumed that2>0) The formulas 11” and 13° given inthe Table ofIntegrals (see above, Sec. 2) avederived inExamples 3and 4. Example&[narHerPuttminxsthenarmZt,Fanatm oda-fe dtmP+CmFlin $C, Examples. |tiem2Putraatthenattede,(Atmd(oo SFwetant$Cmpare tanx40. The method ofsubstitution isone ofthebasic methods forcalculat- ing indefinite integrals. Even when we integrate bysome other Integrals ofFunctions Containing aQuadratic Trinomial 361 method, weoften resort tosubstitution intheintermediate stages ofcalculation. The success ofintegration depends largely onhowappropriate thesubstitution isforsimplifying thegiven’integral. Essentially, the study ofmethodsofintegration reducestofinding outwhat kind ofsubstitution hastobeperformed foragiven ele- ment ofintegration, Most ofthis chapter isdevoted tothis problem, SEC, 5,INTEGRALS OF FUNCTIONS CONTAINING ‘AQUADRATIC TRINOMIAL I.Let usconsider the integral ae t= Samper Let usfirst transform thetrinomial inthe denominator byrep- resenting itinthe form ofasum ordifference ofsquares: axttoxtoma[e+ hatj- sober (b) ge (o)P =a[e+2ne+(s) +5—(z) 1]= b\t (eo 2) pe where oO asaag th The plus orminus istaken depending onwhether theexpression ontheleft ispositive ornegative, that is,onwhether the roots ofthe trinomial ax?+6x-+-c are complex orreal. Thus, theintegral 1,will take theform Aejded fae= (arporpe™ @ arn[(-+3)" Inthelatter integral letuschange the variable: xtpet, de=dt. We then get I=t)as =a) aE These aretabular integrals (see formulas 11’and 12), Example 1,Calculate the integral aeSaxpiea: 382 Indefinite Integrats Solution. ae 1 ae (=egere 7|epee =Jeet) Ci =9 )wypaqep—i- 7)ape Let usmake the substitution x+2—f, de=df. Putting itinto the integral weget the tabular Integral Led 14 ttad(Bedbmctanetc Substituting inplaceof£itsexpression intermsofz,weGnallyget d=.1aretan2t? 4.0, 2Vve ve HL.Let usconsider anintegral ofamore general form: Ax+B ho Santee Perform anidentical transformation oftheintegrand: rH fewto+ (0-32) t=)AttSem\22Ne yy a= )arpbrre arybere Represent thelatter integral inthe form ofasum oftwo inte- grals. Taking the constant factors outside theintegral sign, we get A2ax+o Ab dxhotsarti+(8%)ari The latter integral isthe integral J,,which we are able to evaluate. Inthe first integral make asubstitution: at+ox-+o=t, (Qax-+b)dx—dt. Thus,: SSPE =JF=in|1}+C=Injart+bx+0/C. And wefinally get 1=gplnjax'+bx+e]+(B—)1. Example 2.Evaluate the integral 43tm|eee Integrals ofFunctions Containing aQuadratic Trinomial 383 Applying the foregoing technique wehave 1 1 alaapes aOP aa—3JFaas _1p@r2ar de-7|SPS ae Ligiet dx afinpetaae 544ade 1 1nS aHinit215) 44en|VE=E—D) 5g, zint aan yeaah * UL.Letusconsider theintegral ae Bymeans oftransformations considered in.Item 1,this integral reduces (depending onthesignofa)totabular integrals ofthe jorm at at— 0, i}Tire(oFa>0oFjTanforot which have already been examined inthe Table olIntegrals (see formulas 13°and 14). IV.Anintegral oftheform AxtBratte” isevaluated bymeans ofthe following transformations, which are similar tothose considered inItem Ul: A Ab 5TDaxa{memeVax*tboxe Varybape-4fPorthde+-(8—4) (ax 2a)Vartporte ta)Varporpe” Applying substitution tothefirst oftheintegrals obtained, axt+oxto=t, (2ax+6)dx—dt, weget Cartode dtot, m9VTEESES\veRS Sve2Vi+C=2VaxFoxte+C. The second integral was considered inItem IIIofthis section, 123388 34 Indefinite Integrals Example 3. 5 5feh3fncee Vepar tl Vere +i =(te ee = TVraeeloVourare =5V FEF 10—7 Inx$24 VEFD FOl405 =5VPERFO—7nix24VEPRPC. SEC. 6.INTEGRATION BY PARTS Let uand vbe two differentiable functions of x.Then the differential oftheproduct uvisfound from thefollowing formula: d(uo)=udo=vdu, Whence, byintegration, wehave uo=fud+fodu or Sudv—uo—fodu. (a) This formula iscalled theformula ofintegration byparts. Itis most frequently used inthe integration ofexpressions that may berepresented intheform ofaproduct oftwo factors wand dv insuch away that thefinding ofthefunction ofrom itsdifferen- tialdv,andtheevaluation oftheintegral Sodu should, taken together, beasimpler problem than the direct evaluation ofthe integral [udv. Tobecome adept atbreaking upagiven element ofintegration into the factors uand dv, ore has tosolve problems; weshall show how this isdone inanumber ofcases. Example1,[rsinxdea?Weletuss, domsinedy, thendumds,v=—cosx, Hence, {xsinede—xcons+{cosxde—=—x cosx-+sinx +6. Note. When determining the function vfrom the differential dvwecan take any arbitrary constant, since itdoes not enter into thefinal result {this can beseen byputting the expression Integration byParts 355 v+C into (1)inplace ofo}.Itistherefore convenient tocon- sider this constant equal tozero. The rule forintegration byparts iswidely used. For example, integrals oftheform (xtsinaxdx, Jx*cosaxdx, Gxtesdx, “GxtInds, and certain integrals containing inverse trigonometric functions areevaluated bymeans ofintegration byparts. Example 2.Itisrequired toevauste Jaretan-as. Letting anacetans, ar dvmds,wehavedump2,vmx,Ths, zdx 1 ,Jactansdemsarctans— ftasarctans—y In|1bet14C. rample 8.Itisrequired toevaluate [2¥4i. Letusputwae,dometds; thendum2xdx,v-+e%,[retdemeer—o[setts ‘The last integral weagain Integrate byparts, letting was, duy=de, aede,Oe. Then xedese!—[etdeseorc, Finallywegettetdamate—2(xe08)+CmteDee420Ome(tODEC, Example 4.Iticequred toevaluate ((e425)con2sdx,Wetet ast}Te—§;domcos2xdx:then oedum(Qe+Dde, omit, a ;sind sia2 Se4728)condedems+758)2(oc47)MOde, Applytntegration bypartstothelatterintegral, lettinga=27,dopesin2dthen oecos duymds,oe SS, 247 247(_cos2 cos2 SpromaeaeBe(Se)f(a) em wnEETcose,sine 4 tte 1 6 Indefinite Integrate Therelore, wefinally get ftreoeasdeme7e—8)BEceEMG Example 6.J=[VaP arm? Perform Identical transformations. Multiply and divide the integrand by at dx xtdx Vimarm arma! —(fae 5 Syne’ \yace [ras motucain£—(te.a Va—e Integrate the latter Integral byparts, letting tan duns, tonHE,eeVIR then ‘ ‘idx xdx aoa = ult hetatrelIntheeater obtained expresion ofthegivenintegral JVaR aemataresn4xVIR [VRPae Tranporng thenegra fomrighttolettandperorming elementary trans formafionas wenally get JVtFam Fares45VIR. Example 6.Evaluate the Integrals yaGeconodeandtyeftindede Applying integration byparts tothe frst Integral, weget une, du=ae, domcottsds,mt-sinbs, * 1ax. 2 (poxfectcondearm-L etait[etede ‘Again apply themethod ofintegration byparts tothelast integral: ume, dumaet®, domsindxdx,om—condsfds,om—conde, Jertandem—hettconne[econdea Partial Rational Fractions and Their Integration 37 Patting into the preceding equation theexpression obtained gives us crcbrdemetafetconbef(code, From this equation tetusfind /y: (1462)fertcndndeoer(snortctx), whence 1cosbxdemtestSinbx4-0608ba) n=fe brdeaSOsaeeet c, Similarlywefindas, _e**(asinbx—8cosbx)tumfandemTHESEBosh, SEC, 7,RATIONAL FRACTIONS, PARTIAL RATIONAL FRACTIONS AND THEIR INTEGRATION ‘Aswe shall see below, not every elementary function byfar has anintegral expressed inelementary functions. For this reason, itisvery important toseparate out those classes offunctions whoseintegrals areexpressed intermsofelementary functions. The simplest ofthese classes istheclass ofrational functions. Every rational function may berepresented intheform ofa rational fraction, that istosay, asaratio oftwo polynomials: Qe)_Bex+B". BeTe)AgeFARA, Without restricting the generality ofour reasoning, weshall assume that these polynomials donot have common roots. Ifthedegree ofthenumerator islower than that ofthedenomina- tor, then the fraction iscalled proper, otherwise the fraction is called improper. Ifthe fraction isanimproper one, then by dividing the nu- merator bythe denominator (bythe rule ofdivision ofpolyno- mials), itispossible (orepresent thefraction asthesum ofa polynomial and aproper fraction:Qu Fo, Tay—MO)+FeyhereM(x)isapolynomial, and73isaproperfraction. Example 1.Given animproper rational fraction #3 eyphtT 388 Indefinite Integrate Dividing thenumerator bythe denominstor (by the cule. of.division of polyoma, wae -3 4x6 Bee aT Since integration ofpolynomials does not present any difficul- ties, thebasic barrier when integrating rational fractions isthe integration ofproper rational fractions. Definition. Proper rational fractions oftheform: 1,—AGp&isapositiveinteger2), ‘ActB mnwee (therootsofthedenominator arecomplex, that is,$-q<0), A+B gy ivei . IV,GttBin (kisapositive integer 52; theroots ofthe denominator arecomplex) are called partial fractions oftypes |, M1,TH, and 1V. Itwill beproved below (see Sec. 8)that every rational fraction may berepresented asasum ofpartial fractions. Weshall there- fore first consider integrals ofpartial fractions. ‘The integration ofpartial fractions oftypes 1,IandIITdoes notpresent any particular difficulties soweshail perform their integration without any remarks: 1.JAjde=Anix—al+c. A * (e—a)-*" UfAdee Afea)demALS Cm =—4 +c! (=k) (ay A ApmLfacestenfBet(9), Fepete FateAC 2kp Ap\ (__de A2, ip de =Antetertalt(8—%) (az z z P(+ $)'+(--4) A inixt 28—Ap Deke=Au BAParctan c zinixt+petaltee yet (see Sec. 5). Partial Rational Fractions and Their Integration 399 The integration ofpartial fractions oftype IVrequires more involved computations. Suppose wehave anintegral ofthis type: WV.eevee tal(t+px+a) Perform the transformations: A Ap) yute+ (BF [atte are2(2-#) peta) pa AC_2rbp Ap’ a pe dx+(B— .+Sape (8-4)lara The first integral istaken bysubstitution, x*-+px-+q=6; (2x+ p)dx—dt: 2tp dt = watta dem |= |i" dt=——+C=Sone Se] Et 1=——'_ic¢,aero Wewrite thesecond integral (letusdenote itby/,)intheform 1arm \ea eer b=a 7aE|ea Gttartar 2 i (emer[(+4)'+(-4)] assuming Po = Pintet$at, deadt, q—f=m' (itisassumedthattherootsofthedenominator arecomplex, andhence,q—2">0).Wethendoasfollows: af (etme yy 'lo =fme 4 Lea 1p_ea -3=|\a—ai ttmaa a)ea ® Transform thelast integral: 5eat=ffede (em) Oper 1 ¢pduttm’) H 1 ad(etter a _ig(_1__), 3erm man) (7a) 360 Indefinite Integrals Integrating byparts weget — ——igmer Smo] Saan 2k)LCem )em] Putting this expression into (1), wehave u(_# Lipa=lant alet 1 ‘ a _‘ ee 5a Ink)(Om Int(RD) (Em Ontheright side isanintegral ofthe same type as/,,but, the exponent ofthe denominator ofthe integrand isless’ byunity (e—1); wehave thus expressed /,interms of/y.,. Continuing inthe same manner wewill arrive atthefamiliar integral aw ot t t=ated arctandec. Thensubstituting everywhere inplaceof¢andmtheirvalues, we getthe expression ofintegral IVinterms ofxand thegiven numbers A,B,P,4. Example 2. j1 feeeneie weary) aEeo 1 d42 aratte? apie state ate DepETy") a ‘We apply thesubstitution x+1=/ tothe last integral: a ds a ene yySori Serie Stare) Gea ipa ie _#-3)ae-3Seppe at soaetan tt (fa.oehnos3Sarea Decomposition ofaRational Fraction into Partial Fractions 361 Let usconsider the last integral: fat 1c idet$2 1)Scetnne |Ge) «(aes)--1_t 1i)dt —~arete )apmage teinteTED 2VT ve (we donot yet write the arbitrary constant but will take itinto account in the nal rest Consequently, det aretan 21PFREHHTVE ve a 22es ee +[-awieataye Fa] Finally weget 5egg tH? VEtanttlig, eR A rep — TENS +e. SEC 8,DECOMPOSITION OF ARATIONAL FRACTION INTO PARTIAL FRACTIONS We shall now show that every proper rational fraction may be decomposed into asum ofpartial fractions. ‘Suppose wehave aproper rational fraction Fey Fa" Weshallassume thattheeoelficents ofthepolynomials arereal numbers and that the given fraction isnonreducible (this means that the numerator and denominator donot have common roots). Theorem 1.Let x=a bearoot ofthe denominator ofmulti- plicityk;thatisf(x)=(x—a)*f, (x)where[(a#0 (seeSec.6,Ch.VII).Thenthegivenproperfraction -&)mayberepresented intheform ofasum oftwo other proper fractions asfollows: Fi __A Fite)Tey~Gah eal," 0 where Aisaconstant notequal tozero, and F,(x) isapolyno- mialwhosedegree islessthanthedegree ofthedenominator(x—a)*"F, (x). Proof. Let uswrite the identity Fu) A Fiano 2Tey Gaokt Gare 8) 262 Indefinite Integrals (whichistrueforeveryA)andletusdefinetheconstantAsothat‘thepolynomial F(x)— Af,(x)can bedivided byx—a. Forthis, bytheremainder theorem’ "itisnecessary and sufficient that the following equality befulfilled: F(a)—Af,(a)=0. Sincef,(a)#0, F(a)#0,Aisuniquely defined by Anim OM For such an Awe shall have F(x)— Af,(2)=(ea)F,(x), where F,(x) isapolynomial ofdegree lessthan thatofthejolynomial (x—a)*-"f, (x).Cancelling (r—a) from the fraction inForce (2),weget(1). Corollary. Similar reasoning may beapplied tothe proper ra- tional fraction F,0) Gah)" inequation (1). Thus, ifthe denominator has aroot x=a of multiplicity k,one can'write : Fe)A Ae Anns4Fx) Ta)anat goat teat Ray where22)isaproper nonreducible fraction. Toitwecanapply thetheorem that has justbeen proved, provided f,(x) hasother real roots. Let us"now consider thecase ofcomplex roots ofthe denomi- nator. Recall that the complex roots ofapolynomial with real coefficients are always conjugate inpairs (see Sec. 8,Ch. VII). When factoring apolynomial into real factors, toeach pair of complex rootsofthepolyomial therecorresporids anexpression ofthe form x*+px-+g. But ifthe complex roots areofmulti- Dlicityp,theycorrespond totheexpression (xtpetgy.Theorem 2.Iff(x)=(x'+px-+g)', (x), where thepolynomial p(x) is_not divisible byx*+px-+q, then theproper rational fraction =may berepresented asasumoftwoother properTe)ypre Prope Iractions inthefollowing manner: Fa) Mrtn_, ___@) @)Te)er Fert Pe)" Decomposition ofaRational Fraction into Partial Fractions. 363 where ®,(x)isapolynomial ofdegree lessthanthatofthepoly-nomial (x*-+px-+q)'='9, (x). Proof. Let uswrite the identity Fi) F(x) aMEEN FUME MO) (4)Fey GFF prt oPon) textor Ptpx+P@(a)* which istrue forall M and N,and let usdefine Mand Nso that the polynomial F(x)—(Mx-+N)q,(x) isdivisible by x'+px+q. Todothis, itisnecessary ‘and sufficient that the equation F(x)—(Mx+N) @,(x)=0 have the same roots a-tiB asthe polynomial x*+px+q. Thus, F(@+i8)—[M(a+iB)+N]q,(@-+iB)=0 or i)4NwFoti) MetiB+N=earip) Butfeteisadefinitecomplexnumberwhichmaybewritten intheform K-+iL, where Kand Larecertain realnumbers. Thus, M(a+ip)+N=K+il; whence Ma+N=K, Mp=L or — =KBakaMa5, vaMeoe. With these values ofthe,coefficients Mand Nthe polynomial F(x)—(Mx+N)@,(x) has’ the number a4+-iB ‘for aroot, and, hence, also theconjugate number a—iB. But then thepolynomial can bedivided, without any remainder, by the differences x—(a+-iB) and x—(a—iB), and, therefore, by their product, which isx*+-px-+q. Denoting the quotient ofthis division by ©,(x), weget F(x)—(M+N)9,(2)=(2+px+9)®, (2). Cancelling x*+px-+-g from thelast fraction in(4), weget(3), and itisclear that thedegree of,(x) isless than that ofthe denominator, which iswhat wesetouttoprove. Nowapplyingtotheproperfraction7atheresultsofTheorems 1and 2,wecan obtain, successively, allthe partial fractions 364 Indefinite Integrals corresponding toall the roots ofthe denominator f(x). Thus, from theforegoing there follows the result that If F(x) =(ea) (xb... petar lett lee), thenthefractionaocanberepresented asfollows: Pov, acsFay aaa taap te HEE t B B, Bonstaaptaaoe ts tet Met |ayes” Mew mey|©Beaetartoepeor taepeee “peta ee |koetapesy tapes ttS The coefficients A,A,..., B,B,, ... may bedetermined by thefollowing reasoning. Thisequality isanidentity: andfor this reason, byreducing the fractions toacommon. denominator wegetidentical polynomials inthe numerators onthe right and left. Equating the coefficients ofthe same degrees ofx,weget a,system ofequations todetermine theunknown coefficients , een eed iinaddition, todetermine the coefficients we can take advan- tage ofthe following: since the polynomials obtained on the right and leftsides oftheequality must beidentically equal after reducing toacommon denominator, their values areequal forall particular values ofx.Assigning particular values tox,weget equations fordetermining thecoefficients. We thus see that every proper rational fraction may berepre- sented intheform ofasum ofpartial rational fractions. ample Lettberequted todecompo thetation @-o#"#2 ao partial fractions, From (6)wehave P42 AA A 8aos al a Reducing toacommon denominator and equating the numerators, we have wfDeeA(x—2)4+Ay(K+1)(H—2+A, (EFIDEBEHI,—6) “ Ap 2—(Ay B)8+(A,+38) 2+ IAA 8A, +38) x4(—2A 24,24, +Bp ‘Integration ofRational Fractions 365. Equating thecovificients ofx*,x4,x,x(absolute term,wegetasystem ofequations fordetermining theeoeffiients:OnA+B,1aai+38, 0=44,34, 438, 2m2A2A,2A,+B, Solving this system wefind Ae; Aad; 4y=—2; Bad f Ange Ang 7 Itmight also bepossible todetermine some ofthe coefficients ofthe equations. that result Tor’ some. particular values ofxfrom equality (6), which isanidentity Inx‘Thus,setting¥2—Iwehave$——3A orA=—I;setting#=2,venave62278;Ba? Iftothese two equations we add two equations that result from equating thecoefficients ofthesame powers ofx,weget four equations for deter mining thefour unknown coefficients. Asaresult, wehave thedecomposition 42 EF GBF TET REN TET SEC, 0,INTEGRATION OF RATIONAL FRACTIONS Let itberequired toevaluate the integral ofarational fraction Festhatis,theintegralge)Sfewae. Ifthe given fraction isimproper, we represent itasthe sum ofapolynomial M(x)andtheproperrationalfraction7a(see Sec. 7).This latter werepresent, applying formula (5),Sec. 8, asasum ofpartial fractions. Thus, the integration ofarational fraction reduces tothe integration ofapolynomial and several partial fractions. From theresults ofSec. 8itfollows that the form ofpartial fractions isdetermined bythe roots ofthe denominator f(x). Here, thefollowing cases are possible. Case LThe roots ofthe denominator are real and distinct, that is F(x)=(ea)(x6)... .(¢—d). Here,thefraction7aisdecomposable intopartialfractions oftypeI: FaA8 DTay eat z—et eetae 366 Indefinite Integrals and then Fu) A b DSFiaem) Aca[Pode 2.4)Pde =Aln|x—a|+BIn|x—6]+...+Din|x—d|+C, Case I,The roots ofthedenominator arereal, and some ofthem are multiple: F(x)=(x—a)*(x—bY..." Inthiscasethefraction FH}isdecomposable intopartial frac- tions oftypes Iand II. Example 1.(see example inSee. 8,Ch. X). w40 ax 1 ds 20 aeSartone--latorslates [it 2de11 12 2 $5) ad ee ea Hg eth cm 1 2) enesagt fate Case IIL. Among theroots o}thedenontinator there are. complex nonrepeating (that is,distinct) roots: F(x)=(x+pxtq)(ettlx+3)..(x—a)*.. (x—dy Inthiscasethefractionraisdecomposable intopartialfrac-tions oftypes I,II,and III. Example 2.Evaluate the integral Sensi@+DE=* De thefraction undertheintegral signintopartialfractions Becopogte tegral sign intopartial fractions [see 6), x Art, C@aDG=D TT te Consequently, eee(Ax$B)(x—1I)4C Ut+1), Settingx=,wegetI=26,Co; setting20,weget0=—B46, 1 sat. Integration ofRational Fractions 367 Equating thecoefficients ofxf,wegetO=A+C, whenceAm=—-y. Thus, xdx Leet yt taeSeeitay 3SageretsAeLesdeJ10de10de--rfBits) sete A Linear pad 1aad inpteiitdarctanct tintstite Case IV. Among theroots ofthedenominator there are complex ‘multiple roots Fx)=(x8+e+gy(xt+le+3)". (ea). (ed). Inthiscase,decomposition ofthefraction Fwillalsocontain partial fractions oftype IV. Example 8.Itisrequired toevaluate the integral Se as Fae FD 7 Solution. Decompose the fraction into partial fractions: SHAPE MAIS Arh. CetD yB(P+2cSPFT FLT FFB FHEFT whence pdt$Me128 (Ax+BY(x$1)+(Cx+D)(x842e+3)(x+I)EE(x?+De+3), Combining the above-indicated methods ofdeterinining coefficients, wefind A=l, Ba—1, C=0, D=0, E=1. Thus, weget AAP Naf 128 fan aeSaasraeir=areeraptet |e x42 3Bctantt!~—aetiery—F actsEipimetiiec. The first integral onthe right was considered inExample 2Sec. 7,Ch. X. The second integral istaken’ directly. From theforegoing itfollows that theintegral ofany rational function may beexpressed interms ofelementary functions in final’ form, namely, interms of: 1)logarithms in'the case ofpartial fractions oftype I; 2)rational functions inthecase ofpartial fractions oftype 11; 368 Indefinite Integrals 3)logarithms andarctangents inthecaseofpartial fractions oftype III; 4)rational functions and arctangents inthe case ofpartial fractions oftype 1V. SEC. 10.OSTROGRADSKY’S METHOD Inthecase ofmultiple roots inthe denominator, the integral ofarational function may beevaluated byadifferent method that leads tosimpler computations, This method permits separat- ing out the rational part ofthe integral without decomposing the fraction, into partial fractions, and then integrating therational fraction whose denominator has only simple roots. Itiseasy to integrate such afraction since itisdecomposable into partial fractions oftypes Iand III. This method belongs tothe noted Russian mathematician M. V.Ostrogradsky (1801-1862) and is based onthe following reasoning. Let itberequired tointegrate the proper rational fraction Fix)Fee, where F(x)= (xa) (xb)... +pxtay". Here, onthe basis of(5), Sec. 8,everything isreduced tointe- grating proper rational fractions offour types (see Sec. 7).Here, 1)theintegral ofafraction oftheformatisafraction of theformAT; MxtN. 2)theintegralofnefactionWeete isasumoffrac- tionsoftheform<4", where w*<p—l, andofanin- tegral oftheform weeSap Wewill not yetintegrate fractions oftypes Iand III. Combining the rational fractions obtained after integrating fractions oftypes IIand IV, wegetaproper fraction ofthe form£12,wherethepolynomial Q(x)isequalto Q(x) =(xa)? (x— bY. 8+pt gt... 8+etsy ¥(2)isapolynomial ofdegree one less than that ofthepoly- nomial Q. Ostrogradshy's Method 369 Combining the integrals ofallthefractions oftypes 1and III (including thoseintegralsoftheformJot whichareobtained byintegration offractions oftype IV), wegetanintegral ofaproperfractionoftheform38,wherethepolynomialP(x) is P(x) =(x—a)(x—6)...(2?+pxtg)... (+439). We thus find that Fads YO4¢Xu) Set =cit) aa a Here X(x) isapolynomial ofdegree one less than that ofthe polynomial P(2). Naw letusdetermine the polynomials X(x) and ¥(x) inthe numerators. Todothis, differentiate both sides of(1): Fu)g'—oy |x Tay or fey 10ev ,baxFy aie aa @ We shall show that theexpression ontheright isapolynomial. Noting that f(x)=PQ wecan rewrite (2)inthe form 7PQ" o F(y=py’"OF4ax, @) Whatremainsnowistoprovethattheexpression forisapolynomial orthat PQ’ isdivisible byQ.Wenote that Falla Ql==1)Ina)+G1)Ine—H+... weeFD) In+pxtant... HV1)Inet+lets) = a—!, B-1 =!)Rx+p) =)Art) sisatisete+epperebotpats The polynomial Pisthe common denominator ofthe fractions onthe right side. Inthe numerator there will beacertain poly- nomial ofdegree less than that ofP.Letusdenote itbyT.Then, Q_t gaz. Hence, the expression Q TpLy=phy=ty 30 Indefinite Integrals is@polynomial... Equation (2’) takes the form F(x)=PY'—TY+.QX. 6) Comparing the coefficients ofthe same powers ofthe variable in(3), weget asystem ofequations from which wefind the unknown coefficients ofthepolynomials Xand Y. Example.Evaluate Satey Solution. In this case, Fe)= (IP Otte, P(x)=(x—1) (P++==, we =e, Equation (I)has the form deARHBKEC,(EstRete i)[rsi +f[Te “ Ditterentiating both sides of(4)weget LPN) GA+B)—(Artt Be+6)3a"|BetFe[a wai oT Clearing fractions, wehave Lm(x?1)(2Ax+B)—(Ax* +Bx$C)3x8+(x?1)(Ex?+Fe+). Equating the coefficients ofidentical powers ofxondifferent sides ofthe ‘equation wegelasystem ofsix equations for delermining the coefficients, i ome, once 0=—2A—F, 1=£—B8-G. Solving thissystem wefind ' : B=0, Amd, CaO, Bat, Fad, G=—2. Putting thevalues ofthecoefficients thus found into (4), weget ol (os a 73 3 The denominator ofthe later integral has only simple roots, thus making itcasy{ocompute theIntegral. Wefinaly obtain “ 2 2,4Settpensine loeEtt|e (1381) aot}eypeet at?fines 2v3 241 Sa tgnatbek tlearetonFt4c. Integrats ofIrrational Functions sn SEC. 11, INTEGRALS OF IRRATIONAL FUNCTIONS Itisimpossible toexpress interms ofelementary functions the integral ofevery irrational function. Inthis and thefollowing sections weshall consider irrational functions whose integrals are reduced (bymeans ofsubstitution) tointegrals ofrational functions and, consequently, are integrated totheend. 1,Weconsider theintegral R(x,x*,...,x*) dxwhere R isarational function ofitsarguments.*) Let&beacommon denominator ofthefractions =,...,.. We make the substitution zat, de=kt"' dt. Theneachfractional powerofxwillbeexpresced imtermsof anintegral powerof¢andtheintegrand willthusbetransformedinto arational function of¢, Example 1.Itisrequired tocompute the integral dx atl 13 Solution,Thecommon denominator ofthefractions -L.2is4:andsowe substitute: x= dx=4i%dt; then adefe.aoe |fe-sebaimsJatpatma(Has)dtm ated 7 e #4 . nsfearns[oar Aineeiiec 41thin [et=$ [fmf] ee +1thenotation (2,7, ust) ten thatonlyctionoperations areperformed onthequantities x,2")..4.2°. hisisprecisely theway that theYollowing notations arehenceforward to beunderstood:(.(aea)--)- RG.VaFFOFO), Risins,cosny ele. For instance, the notation R(sin, cos.) indicates that rational operee tions aretobepefformed onsin and cos8. om Indefinite Integrals II.Now consider anintegral oftheform axtb)" fax+b\* Se[=(Sepa) "+(Ee) |ae ThisIntegral reduces totheintegral ofarational function by means ofsubstitution: ebyt ayaa! where &isthecommon denominator ofthefractions =,...,. Example 2.Itisrequired tocompute the integeal ipezzin Solution. We make the substitution x-44—0%, x=t*—4; de=2tdt: then Vea 8 4 a JBien styrene) emfarseste 1-2 a Vext-2=4210[13]4c—2Vert YEE|+0. SEC.12.INTEGRALS OFTHEFORM |R(x,Vax"pox-Fe) dx Let usconsider theintegral JR(x,Varbx+e)de. a) An integral ofthis kind reduces tothe integral ofarational function ofanew variable bymeans ofthe following Euler sub- stitutions. 1,First Euler substitution. fa>0, then weput Vax+bxFe=4Var+t. For the sake ofdefinitenesswetaketheplussigninfrontofVa. Then ax?bx+o=axt+2Vaxt +r, whence xisdetermined asarational function oft: toe *"2Vai IntegratsoftheForm{R(x,VarORFS)ax m3 (thus, dxwill also beexpressed rationally interms of¢).Therefore, VarFbape=Vax+0=Va +0, andVax"6x6isarational function of¢.Since Vax*+6x-+c, xanddxareexpressed rationally interms oft,the given integral (I)istransformed into anintegral ofa rational function of ¢. Example 1,Itfsrequired tocompute the integral deSri apeSelatlon, Sincehereamt>0,weputVFFCm —ahss en BECa set 4, whence tc nS. Consequently, aeEEat,TEC pteOg aE VPFC=—2 pt—ptES, Relurning tothe initial integral, wehave PHCy aePua Jefe|BeenJPomincomies verti2 (see formula 14inthe Table ofIntegrals). 2.Second Euler substitution. If¢>0, weput VaxFtbx-e=xttVG then axt+bx+o=x't? 42xtVere, (For the sake ofdefiniteness we took the plus sign infront of the radical.) Then xisdetermined asarational function of¢: ralVet Sincedx-andVax"6x6arealsoexpressed rationally intermsof¢,bysubstituting thevalues ofx,Vax'--6x-rc anddxinto 3m Indefinite: Integrals theintegral'{R(x,Vax*-Fbx-+¢) dx,wereduceittoan.integralofarational function of¢. Example 2Itisrequired tocompute the integral fsVIFEERYeV Solution, We setVTpapetext+1; then Meebsteaieadth emo: deeMay, Vite Pant1aPoth, a Putting theexpressions obtained Into the original integral, wefind ueVERE gem(MEO ED ge ewVite T= Oe =+2)pattem agin|eco =TEER5g[EEVTEEEP=I 0 ¥ r-Vitete +l =UTEPOD)inte2VTERHIHC. 3.Third Euler substitution. Let aand Bbethe real roots ofthetrinomial ax*+bx-+c,Weput Vax +bx+e=(x—a)t. Sinceax*+bx-+c=a(x—a)(x—B), wehave Va(x—a) (x—B)=(x—a)1,a(x—a)(x—B)=(x—0)",a(x—B)=(x—a)*, Whence we find xasarational function of¢: _op—atera Since dxandVax"+bx+¢ also rationally depend upon t,the given integral istransformed into anintegral ofarational function off. Integration ofBinomial Dierentials 8 Note 1.The third Euler substitution isapplicable’ notonly fora<0,butalsofora>0,provided thepolynomial ax‘+-6x-+chastwo real roots. Example 3.Itisrequired tocompute the integral deJV8Pu—a Solution. Since x*-+3x—4=(¢-+4)(e—1), weput VETOED=64465 then +) G—NSE+4P A,eletat, Lea te eat deme ope. fiat 51VerTe= [rea] apa. Returning tothe original integral, we have ae 10H) 2in| AceSpan oe =Spain| Ei ZI eeVecain|VEEVET],6 ioeryeleeeoel-V Note 2.Itwill benoted that toreduce integral (1)toanintegral ofarational function, the first and third Euler substitutions are sufficient. Let usconsider thetrinomial ax'+ bx+-c. If6*—4ac >0, then the roots ofthe’ trinomial are real, and, hence, thethird Euler substitution isapplicable. If6*—4ac<0," then inthis case ax+6x+o=pt[(2ax-+b)*+ dac—b*)} and therefore the trinomial has thesame sign asthat ofa.For Vax"+6x-+e toberealitisnecessary thatthetrinomial beposi-tive, and wemust have a>0. Inthis case, the first substitution isapplicable. SEC, 13, INTEGRATION OF BINOMIAL DIFFERENTIALS Anexpression ofthe form p x"(a+bx")Pdx, where m,n, p,a,6areconstants 1scalled abinomial differential. 36 Indefinite Integrals Theorem. The integral ofabinomial differential Sat(a+bx"de ifm,n,parerationalnumbers, isreducedtoanintegralofara-tional function and thus isexpressed interms ofelementary func- tions inthefollowing three cases: 1.pisaninteger (positive, negative orzero); 2,2!isaninteger (positive, negative orzero); 3.£14pisaninteger(positive,negativeorzero). Proof. Transform the given integral bysubstitution: xem, dealt "de. Then - 1pSt 1 Se(atbxFde=t fz*(a+beyde=tf24(a+b2)dz,a) where q=tttai. 1.Letpbeaninteger. Since qisarational number, wedenote itbyL.Integral (1)isthenoftheform SRE, ade. ‘AswaspointedoutinSec.11,Ch,X,itreducestoanintegralofarational function bythesubstitution 2—=7", : 2.Let“+!beaninteger. Theng=*!—1 isalsoaninteger. Thenumber pisrational, p=. Heretheintegral (1)isofthe form . JRL,(a+bz)]dx. This integral was considered inSec. 11,Ch. X.Itreduces toan integral ofarational function bythesubstitution a+bz=, 3.Let“#!4p beaninteger. Butthen"+14 p=g+p is aninteger. We transform integral (1): J(a+bay?dz—{ater(24%)"ae, Integration ofBinomial Dierentats a where q+pisaninteger andp= isarational number. The latter integral belongs totheclass ofintegrals t Jes,(2)"]a. This integral was considered inSec. 11,Ch. X.Itreduces toan integral ofarational function bythesubstitution24%1! Let usexamine examples ofintegration inallthe three cases. ax Sa anAty Example1.Satan )*(4x)dx.Herep=—I(integer). tsSram ’ Putting x?=2,wemake thequantity inparentheses linear in2: Sept) eeefora teedfeFagartae Now make thesubstitution 2*=f.Then 2=¢, d2=2dtand fete) seedfeFapertased(erated =9)Bamsaretan4.C=daretanVFC=Baretanj/F+C. # -t xample2.(dem (ete Fas.Here,m=3,nm, trample2.fide=(8) re,3n=, poh,MELaodntegen, Wesvbstutesta: thenxearmsas and Ed fa—a th ted “i,Sparen fue hae fennnte fekonters esta farenthesis rationalweput(1—2)?=¢;then Jtapedfeoteenfeneiaaefecnaen afittcet 9$05 (29aca Fong, a8 Indefinite Integrals Example3.race =fen$29YeHere,m=—2,n=,p=2 evita2 sndE14pi2(integer),Wereducetheexpressioninthepareltesesto 2linea?funetion: a neatezaaah;demye*de Jato panFae[tates pearmLeyfaaayture! feofLEE) Efe Fates arden EE) Fas he first factor isa rational function. In order tomake the second factor rational aswell, we make the substitution: then[eoee eToh may aa Thus, mtpayPde(gt(12) Fare Satatay Mdemy fat(EF) Farm Le yr_peet att (flap hace<afucuer ape foetae p46 LteyF_ (2) leet (tr =-(12)- (5 )F4c=-(LEt)t- (24)tec (4#)'-(ca)'+e--(F) (ea) Vis x-- ise.= Vise Note. The noted Russian mathematician P.L.Chebyshev proved that only inthe above three cases inanintegral ofbinomial differentials with rational exponents expressed interms ofelemen- tary functions (provided, ofcourse, that a#0 and 640). But if neither p,nor=+4, nor*+14p areintegers, thentheintegral cannot beexpressed interms ofelementary functions. SEC. 14. INTEGRATION OF CERTAIN CLASSES ‘OF TRIGONOMETRIC FUNCTIONS. Uptonow wehave made asystematic study only oftheinteg- rals ofalgebraic functions (rational and irrational). Inthis section we‘shall’ consider integrals ofcertain classes ofnonalgebraic Integration ofCertain Trigonometrie Functions 379 functions, primarily trigonometric. Let usconsider anintegral of the form JR(sinx, cosx)dx. a) We shall show that this integral, bythe substitution tang=t @) always reduces toanintegral ofarational function. Let usexpress sinxandcosx interms oftan5,andhence, interms of¢: Bsn eos—PsinEcos tan og,site eeeoeee satcost 1ptant x eX at res 2x cost—sintZ cost—sint itn | 008pam ipl ee cost+sint t+tant And eat x=2arctant, desi Inthis way, sinx, cosx and dxare expressed rationally in terms off.Since arational funetion ofrational functions isa rational function, bysubstituting theexpressions obtained into the integral (1)weget anintegral ofarational function: mae) ar JRisin, cosmde=lR[ Aa.TER]a. Example 1.Consider the integral aeSite On the basis ofthe foregoing formulas wehave 2atJin[EEStmnrconfinglse cog This substitution enables ustointegrate any function ofthe form R(cosx, sinx). For this reason itissometimes callea a“universal trigonometric substitution". However, inpractice it frequently leads to.extremely complex rational functions. Itis 380 Indefinite Integrals therefore convenient toknow some other substitutions (inaddi- tion tothe“universal” one) that sometimes lead more quickly to the desired end. 1)Ifanintegralisoftheform{R(sinx)cos.xdxthesubstitu- tion sinx=t, cosxdr=dt reduces this integral tothe form fRipat. 2)IftheintegralhastheformJR(cosx)sinxdx,itisreduced toanintegralofarational function bythesubstitution cosx=,sinxdx= —dt. 3)Ifthe integrand isdependent only ontan x,then the substitution tan x=¢,x—aretanf,dx=7Sreducesthisinte- gral toanintegral ofarational function: JRian)de=[ROSo. 4)Ifthe integrand has the form R(sinx,cosx),butsinxand cosx areinvolved only ineven powers, then the same substitu- tionisapplied: tanx=t, 2’) because sin*x andcos*x areexpressed rationally interms oftan.x: tue! 1 cosx=Tyiante TR? intye tate Rube Ttants “Te adem ne Alter: the substitution weobtain an integral ofarational function. Example 2Compute theinesratfyaeSolution.Thisintegralisreadilyreducedtotheform{R(cosx)sinxdx. Indeed, sitxsinxsinxde_¢1—costxSafar) Sppeea|apa Wemakethesubstitution: cosx2.Thensinxde—=—datsin? x 1-2 #1 3Saetbate~ JapeconfForen{setsps)om=Fre9in+2)+CaAS20924 1m(cose+2)+6. Integration ofCertain Trigonometrie Functions 381 ae Example8.Compute|-—2 Make the substitution tanx=t: [tin |pote ftieva rst tae \ PV Ve care 1 tan $mpgaretan(FEE)40 5)Now let usconsider one more integral ofthe form JRisinx, cos.x)dx, namelyanintegral underthesignofwhichistheproductsin*xcos"xdx (wheremandnareintegers). Herewe shall have toconsider three cases, a)sin"xcos"xde, wheremandnaresuchthatatleastone ofthem isodd. For definiteness let us assume that nisodd. Put n=2p+1 and transform the integral: Jsinxcost?** xd=fsin"xcos"?xc0s.xdx= =5sin™x(1—sin* x)?cosxdx, Change thevariable sinx=t, cosxdx=dt. Putting thenew variable into thegiven integral, weget §sin®xcos"xde= {e(l—eyrde, which isanintegral ofarational function of¢. Example 4. cost,_(costcosxdx_0(1_—sla?s)cosxdx SS] aS -f ees. Denoting sinz=t, cosxde—di, weget cote, CU—Mat cat ea ttJGitan [Uae ee te riot=santa sineaad byfsin”xcos"xdx, where mandnarenonnegative andeven numbers. Put m=2p, n=2g. Write thefamiliar trigonometric formulas: sintx=p—4 cos2x,costx=++4c0s2x, @ 382 Indefntte Integrals Putting them into the integral weget Jsin’?xcostxd=(+—7cos2x)’(5+cos2x)"de, Powering and opening brackets, wegetterms containing cos2x inodd and even powers. The terms with odd powers areintegra- tedasindicated inCase (a). We again reduce theeven exponents byformulas (3). Continuing inthis manner wearrive atterms of theform{coskxdx,whichcaneasilybeintegrated. Example 5. Jsrsacmfyfd—cos2otdem [12cos24-4cost28)dm 1 1 ys sindet[ecantetgfuteortnas]=p[Fanaa] 40, ©)Ifboth exponents are even, and atleast one ofthem is negative, thenthepreceding technique dossnotgivethedesired result. Here, one should make the substitution tanx=t* (or colx=t). Example 6. Sintxdx_(slotx(alax-+costx* JSE [ee eeeaeantetantads, .a Puttanserts thenxarctant, dxm7S'y andweget aint2sain fel ele Jen foaterta foatodafegece tants, fants ate. 6)Inconclusion letusconsider integrals oftheform Scosmxcosnxdx, [sinmcosnxdx, {sinmxsinnxdx, They are taken bymeans ofthe following*) formulas (m+n): cosmxCOSnx==+[COS(m+)x+COS(m—n)x1, *)These formulas are easily derived asfollows:os(m+n)x=c08mixcosnz—sinmxsinnx,os(m—n)x=c08mxcosaxsinmxsinnx. Combiningtheseequations termwise anddividing theminhall,wegetthe first ofthe theee formulas. Subtracting termwise and dividing inhalf, weget {he third formula. The second formula issimilarly derived ifwe write analo- gousequations forsan-haye andsin(m™n)s and’ten“combine “them jermwise, Integration ofCertain Irrational Functions 4383 sinmxcosnx=[sin(m--n)x-+ sin(m—n)x}, siimxsinnx=5[—cos (m+n)x-+008(m—n)x}. Substituting and integrating, weget Jcosmscosnxdx—-yf[cos(m+n)x-+c0s(m—n) x]d=_sia(m-tn)x 4sla(m—aye=Toneny tTmay FC The other two integrals areevaluated similarly. Example 7. finsesnseara{eoscotedemHEE4HEE, SEC. 15. INTEGRATION. OF CERTAIN IRRATIONAL FUNCTIONS BY MEANS OF TRIGONOMETRIC SUBSTITUTIONS Letusreturn totheintegral considered inSee. 12,Ch. X: JRVax$bx40) dx. Ww Here we shall give amethod oftransforming this integral into one ofthe form -5R(sinz,cosz)dz, 2) which was considered inthe preceding section,‘Transform thetrinomial undertheradical sign: ax'+bxtema(x+t)'+(c—B). Change thevariable, putting e+pat, dead. Then Vaepbspe= Vw+(e—8). Letusconsider allpossible cases.1,Leta>0, c—#>0. Weintroduce thedesignations: a=m', pe Acant. Inthiscaseweheve Varpbepea Vinpa 334 Indefinite Integrals 2.Leta>0, c—¥<o. Then Pee eens Thus, Vaxfbxe=Vm't—n*. 3.Leta<0, c—E>0. Then fo ee Hence, Vary bete= Vim, 4.Leta<0, c-¥<0. InthiscaseVar+bx46 isacom- plex number forevery value ofx. Inthis way, integral (1) isreduced toone olthe following types ofintegrals: L.srwViet$n)dt. @1) ul.sre.Vine) dt. (3.2) I.jr(t,VibmF)dt. (3.3) Obviously, integral (3.1) isreduced toanintegral ofthe form (2)bythesubstitution t=2tanz. Integral (3.2) isreduced totheform (2)bythesubstitution tafsece Integral (8.3) isreduced to(2)bythesubstitution tad sint. Example, Compute theintegral - ax nclaton, This anintegral oftype11,Makethesustitution ean xacossds, Integrals notExpressed inTerms ofElementary Functions 385 5ees =fete 1i4ManeyViera JVa—atsinnep )oeosts oF)costaat1 sing 1 sing 1One BE4Capitt aya te. SEC, 16, FUNCTIONS WHOSE INTEGRALS CANNOT BE EXPRESSED IN TERMS OF ELEMENTARY FUNCTIONS InSec. 1,Ch. X,we pointed out (without proof) that any function f(x) continuous onthe interval (a,6)has anantideriva~ tive onthis interval; inother words, there exists afunction F(x) such that F’(x)=/(x). However, not every antiderivative, even when itexists, isexpressible, infinal form, intermsofelemen- tary functions. For instance, wehave already pointed out that theantideriva- tives ofbinomial differentials that donot belong tothe three examined typescannot beexpressed interisofelementary fune- tions infinal form (Chebyshev's theorem). Such aretheantideriva- tivesexpressed bytheintegralsJertde, jeerde,jedx, SVi=esnrede,[AZandmanyothers. Inallsuch cases, the antiderivative isobviously some new function which does’ not reduce toacombination ofafinite number ofelementary functions. For example, that one oftheantiderivatives Jedx+e, which vanishes forx=0 iscalled the Gauss function and isdeno- tedby®(x). Thus, (x=Jer+C,, if ©()=0. This function has been studied indetail. Tables ofits values for various values ofxhave been compiled, We shall see how this isdone inSec. 21,Ch.XVI. Figs. 204 and205show the graph oftheintegrand y=e-** and thegraph oftheGauss func- tion y=@(x). That one ofthe antiderivatives \VIRPsmede+C (k<1), which vanishes forx=0 iscalled an“elliptic integral” and is 13-2388 36 + Indefinite Integrate denoted byE(x), E(x)=[VIRBSintde+-C,, if E()=0. 9 i, id7 7-200 yen fern (0 aern a a ¥ Fig. 204 Fig, 205, Tables ofthe values ofthis function have also been compiled for various values ofx. Exercises onChapter X 1.Compute theintegrals: 1.[atds.Ans.240.2[e+Vids. atVE 3VR amFene 8S(faEVE)ae,ameOVR— Leysade 2 1,4 ~Leyzee. 4 (24.ans.2evaee. 5.(L442) dx. 7Vee‘aa eeSGtast je Ans, —1——F— 4240. 6. sR. ans. tyete.2Vetet Sy amgVee yt #43array 2S(«+773)dx,Ans.£43BYF43E46. Integration bysubstitution: &[ettar. Ans.get+c. 9[cosdede, ans,S540. 10.Vsnards, ans,40. 1[Eas Lge ae cot3e ax ans.GZoteeci,PS. ans,—G,Poo.tanTe ae 1 a ans,Mec, aFHS. aneZiniae—ri¢c. 18.fH. Ans,In|1146.16.) 8Ans.—Fn|5— 2146.17. Vtandede. Ans,—Flalcose146. 18.fcotGe—7)dx.Ans.In|sin(Sx—T1 46. Exercises onChapter X 287 ay 1 x x 10§lhe.ane,Lintcnar46.2.footEte,anestaantc. a.ftmesetogs, AmSuunteec, —mSctene® Annjaneriec. 28.Q(tantsmot$)4s.Aneintents mtiofea S-c,26(arecsde, AmMC. 28Fontses ans,HE4C,8VFIeds,Ans,SVFPLC. fade 1 stds 2. vans. LV TEPC. 28 (EHane,2VRETHO. wm.[panied SYHcosxde 1 sindx 1 ae,(SEEM. An,—+c. 0.jae. Ans.yhte.ane tants cote ote aSEEeamEneaFaeaneEEge, ac— Ine) a (ams, Viena yc. (EHDay, Samar al fs Ans.wetec,38Te. Ans,WVUsinxFi+0. sin2cde H sindede (tiated ans, tigen p *[fear ce aed ans,VTHME, 38[VREge,ans,2VTaREETC, cosBede 1 sin3cdx 39, .Ans.—— Cc.40. + [aaa ~hapeee War 1 Intxde ints resindx Ansepee. fu. Ans.TEC.aPras: atesatx arctan xde aretants atecost smHE caF8SnEdsngEUGa,PH, anHEE 4sPEae,ane,EEG, ade Line z41 6[Ham pint gc, PAE ae,anhinarereg nec,aPRE ans,bintans+946. as ‘ (et aAnsarom! J2ecrtitds. Ans.eplne aPlatetsans,BEtote omPee ae dns intact46.8(ato. aneLint). 1B 2388 Indefinite Integrats 54jae ansBELG 55.Sota dns Injacensi46.st.rt2deame‘ini24sainel46. 1,Geostiny #.Anssin(insy$C.88contatx)de.” dnantic, afede AneLenn. oo.Sean Ans.%?46.61Gedesde.Ans.AEC 62,Sathede Ans.346.63.feaeAns.of46,4.[emntae. Ans,etc. oxforede ameEee. oe.Sertde Ameheme. orPiersamnan amL(erngtec). 05feterriennen smeoorrne, caSEHaeaneELEY orsme[sified binatiensc m(PTansbineeenste. 2.Se Ans.pun B46 mjr Ansparen VR+Carh Ans,aresin4-6. 1.pcg.amLactanSc.mfpHs.amein|ZEHL4c. 2.Srm Ans.InjxtVEFFOIEC. 80,_—- Ans.LinoeVORA+6,at.Svea Einfor +VFa+c.2(A. Ans.shin[A—E|4c.wnfA amepg[Stbelec [Alea amLacan. 85.pregAns.portZee6SeAns.atcsine’+C. Exarcses onChapter X 589 ae 1 S cosxdx_ a.(i qeaesin VExec. ete, Sree pare Gere donDein(M8)4000foe. anawcaminntos nccosxx i —_ wo, (MSAdeAns,—L(reconeVIRFHC. i)ViaAns—Plarecossi'+ VImM+6 seauctans \ \ : 1,SSAC ae,Ans.find+eh—ploretan et$C, (PERE aane2yTeW LCon[VETS4, AVaR. a ve Ans.AVOEV +E. lave Ans.SVieVE+e. ‘ VaVi4Veear conde var 95.SSSans.aretan$C.96,SFiae Ans.3}/IaF+0.0.fVTFCGsin2eds.Ans.—2VTFSC+C.98,(Hae 9 Vireo dns,2VTRmRHC. 9mPEaeAns,ttc, Yim3 yoo woo,\ax Ans.2imac. a a z 0Sraarercarg: Amepeuetan(Fins 40. integralsoftheform(—AL+8—gy, Integralsofthe{Satineax WctanSt! a weSapteps: amqactinttine mmPo. 1ant a 1jgEVE doepyneinigag+6.4Sarasran Moeare meeyete ts1 f=5 a aee a de 1 dentee (=a : (rae sonFySESDEE. Aaninatei4c,wonSySEDEE denSingr—seera eweatecnaFBRae tosFiwmeenegtqwcintec,afgtitloan, 300 Indefinite Integrats Ans,Fin@e—+F inrty+e. maPea. Aas.ineteb+rasaetanOTC nia,FSET ERge, dmotebigiaeeetiesshguctn sl Integralsoftheformjyatta « us[pt teFoemntittec me(pata, anstafeeheverrerTte.on(pe. ansnse¢ +VISES140.118,= Ans.yamine 110,\7- Ans.pains +54VORHS| +6. 120.(r= ‘Ans.aresinBate. 121.\reSs: Ans.PEO 1+VORP +6.122.Irate Ans,2VatPoepe+C. wm(ERS Ans.VFRshine verER. me[te dnsfyVERO +6.128,EEOAns—VTE Se a XIn(4x—1 4V5@EP—H))+C. M,Integration byparts: yor,[xetde. AnsN+, 198Gxtnede. Ans. patx x(Ine—4)40. tanGasncds dns.snxmecone tc10,[neds Ans.x(lax—46, 181,Garesinxds. Ans,xaresins+VI=F4C, Exercises onChapter X 391 132.Jina—aax. Ans.—x—(1—a)In(I—x) $C.138.ferineds, a n \ ansEE(memzty)4e. otfroctanede, AneLettrarctanr—rlbC. 185Sewesneds, AnsE(Qet—aesine + $eVTHALC 106,[In(eteide,Ans,xine¢)—242arctan$C, 137. JeetanVedx. Ans. (e+aretanVE—VE $C, sreanVF, THVT. = tanPUSSYEt,ane2VFacanVF-42VToe+6108freanV/Erte, ans,xaesinWEPVFarctanVF4C. 10,xcosteday#41 1 atesins aneadentesLeader.wh[EMEA AnnaT= saetans 41 rarest ua,SEREDGetas,gtothoctine —FMBGc.a,PrarctanVHTde.Ans.Faterctan VFT— HERAT 40.4[MSEae.Ans,nfVERE)Lesa0, 145.Smet ViFwdx. Ans. xlnjx+ VTFH|-VI FR+O, ade aesing|1),)1=¥ 6.ParesneAtta.aneHEBbE, Use trigonometric substitutions inthe following examples: ae Foe x wr,[YEEaeans,VEEP vesin4c,us.ftVIBae *_lyaseale yr ax Ans,2aesine—5x VIERtitVIRPSC, 109,Save an,YER en(VERBeeane,YFHP—earecor 40, de zou ws. (5. am. Sotto,Svea ®Yaga Integration ofrations fractions: Bet aeans,nfEEE ade weJapp AmemE|.seSeandeners: wm Indefinite Iitegras ee aeCe a xeeptgietnee se(ame ce static in(aA deamSy, wefFaeamHEtnecme(A. Ans,PEN in(EES)40.00,Ft. Ans.neato wor(EE teamInBH4vctanELC. wsSpA amEinEM etaBEC,veJape twnhatacunS40.msfe nwEpettint mp6.tor,PEE!aeanePEG+net?— tineMactan te160.Combe anenS! ~ppp eae 10. [ye ans.ALY10VBn]+e. m,[YEEBn an.BVP-BYR40.mnJME Ans.matytty eae Exercises onChapter X 393 24Ve beaSemis ya im.See as.88S BuyRR VeVi Vea BVRaVEY OYFGF910(I/F+1)+5In(/P41)H3aretanP/F+C, fest Vire+ Vite|_Vie meSVR mlerty eeeTHe ansamctanY/TeeinVERVE FYE. AnetatctanYV/E+Visie Vie im|PEEaaeYEE EVEY +hyBl4c.mjVaeAns.VETO Hex xin(224fe—Feat)46. Integra oftheform§Rix,Va UESae ax 1 ,|YBSRES—VT 1Ans. eta) MESES, oJs 730|EG alte Side sein|VERA HVT mse Osye!| ¥ tryate aa. Saxesn£2 VEER won[epee teHweesshec i (YEEBe PPH+Inle +14 VER EC. 2 (—_*_ . Ans.VFR+I|x+1VP+8Svar Ans.year 183.JVBRaeAas.$e)VE faresine—D)+C. 184[— ans.S42VFR 1 _ tae- FET+6. 185. ————yileevine J+e)Vitate xtVitepe (+1) COrere oe1 1Vipeee, b4s—2VTEETH = csr,(I=VEREERs,ans,n=2VTERE, ans—epOe Postale 188.jGE« Ans.yaa tine VPRRLEC. son IndefriteIntegra Integration ofbinomial diferentials: = .rae Le hoeee(Oo fosWEST Hem[ePOEATTen wo16),yh ax ae an,MEBo eee. i.fziAnsRate. aes wefete ctsHarttocindVaan ged? 2 aedans.SOVE—O04VH46.194,(GFe Yoo Yat russDe-Va ya Ans,2A4SY )O—V 195, SeVira. amB=Bagan®, Integration oftrigonometric funtion: 10,Gatteas, Ameotemeonesc. tm(sited, Anneedeat 40.wefeuteante, ambeatsbeatae. oo.[PrEax.Ans,exerfesctatc, 200.flcostde anEaLomtegc. meGontede, danDenBe84 . 1 sine 3 sm.fentsan am Sepeanar tltSnes)wosotscateae, dasry(seainars) sc,antfue ne Eines 46.20Seated,Am—Heategbeats Finisinxiec, m6,Peottads, Ans,—2%"F—inisina+0. wr Gntea, dm taptana Exercises onChapter X 2395 00,Ftantssecteds. Ans, BEERS C00,fA amtnstbianteee. 0SSE dmCmte cpaiatade 8gt oo au,Sus Ans,ZooFresco*4C.ate[sinrsindeds,sind,sin2x sintz,singe Ans,S84MPEG, 21a,[costrcosreds, Ans.MOEMI9, cos6_cos2e 1coed 21h,[cos2esintede, Ans,—2798S080. ais.[intxcon$ade tan—9| £08) aoe ae 1 2 an ae 1 x sinde a,Sgm$Eay. Amearetan|2tenZ]4c. aus,fSete 2 cosede x An, —24x40, cna ee A+tan Sita z sinde a as wan,[ARB deAns.sretan(2siote—1y4c. 201.[oa“tan£44tan? ae 1 Ans, tan$+ptant4c,omefet. an,—5[cote +pgaeten(FE)|+6mes,fBde, Ans.VBarctanx tanxfs) _rye,(FE) CHAPTER XI THE DEFINITE INTEGRAL SEC. 1.STATEMENT OF THE PROBLEM, THE LOWER ‘AND UPPER INTEGRAL SUMS The definite integral isone ofthe basic concepts ofmathemat- ical analysis and isapowerful research tool inmathematics, physics, mechanics, and other disciplines. Calculation ofareas bounded bycurves, ofarclengths, volumes, work, velocity, path length, moments ofinertia, and soforth reduce tothe evaluation ofadefinite integral. y CnSeonKAZE|No 7 eal4 VAR Aa OfPTXHeXygqrd x Oras Me% web Fig. 206. Fig. 207. Let acontinuous function y=f(x) begiven onthe interval fa,4](Pigs. 206and 207). Denote bymand Mitssmallest and largest values onthis interval. Divide the interval a,6]into n subintervals bypoints ofdivision: OmEyLysKyoesy Zuens FaOy so that BS <<< ty and put Rye Aa eA coe type, =Ae Then denote the smallest and greatest values ofthe lunction {(2) ontheinterval [x,,x,]bym,and M,ontheinterval [x,,x)bym,andM, ontheinterval [%,-y» ¥,]bYmyand M, Form the sums p=MAX, +MAX, +ee+MyAx,=DMAx, Ww & at Statement ofthe Problem. The Lower and Upper Integral Sums 397 5=MAx, +M,Ax,+...+Mabey=FM,Ax;. (2) a Thesum s,iscalled thelower (integral) sum, andthesum 5, iscalled theupper (integral) sum.Iff(x)>0,thenthelowersumisnumerically equaltothearea ofan“inscribed step-like figure”AC,N,C,N,... C,_,N,BA bound- edbyan“inscribed” broken line, the upper sum is'equal numer- ically tothe area of an “circumscribed step-like figure” AK,C,K,...C,-,Ky-C,BA bounded byan“circumscribed” bro- ken’ Tine! The following are some properties ofupper and lower sums. a)Since m;<M,foranyi(@=1,2,...,n),byformulas (1) and (2)wehave _ 55<5y (Theequalsignoccursonlywhenf(x)=const.)b)Since m, Sm, mm, ..., m,>m, : where misthesmallest value off(x) on(a,6),wehave Sy=m,Ax, +mAx, +... +m, Ax, >max, +mAx, +...4mAxy= * =m (Ax, +bt,+ ...-+Ax,)=m(b—a). Thus, ©)Since 5,>m(b—a). M,<M, M,<M, ..., M,<M, where Misthegreatest value off(x) on[a,6],wehave B= Mx, +M,Ax, +... +Mybty <MAx, +MAx, +...efMAd,=M(Ax,+Ax,£00+Ax,)=M(b—a). Thus, _5,<M(b—a,), neCombining theinequalities | itobtained,’ wehave 9 ll m(b—a) <s,<5,<M(b6—a). { 'Iff(x)>0,thenthelatterin- 1 equality has 'asimple geometric u meaning (Fig. 208), because the ibe products m(6—a) andM(6—a)are,respectively, numerically dl leequaltotheareaso{theinscribed” ‘; *rectangle AL,L,B and the“cir- cumscribed” ‘rectangle AL,L,B. Fig.208. 398 The Definite Integral SEC. 2,THE DEFINITE INTEGRAL Let uscontinue examining the question ofthe preceding sec- tion, Ineach ofthe intervals [x,t], [XsXp] --++ [ens Xn) take apoint and denote them by’, By,-.-s Ey(Fig. 209): BySES SES oy KaeSEShe Ateach ofthese points find the value ofthe function f(&,), F(Eas e++s F(Eq). Form asum: Sa=T(E)At,+PEs)Ay++++PEn)Ofna=UIE)Oeay This sum iscalled theintegral sum ofthe function f(x) onthe interval [a,6].Since foranarbitrary belonging totheinterval xj) 2)weWill have m;SFE) <M, and allAx;>0, itfollows that mAx; <f(5)Ax,<M;Ax, ‘and consequently %im,Ax;<%16)A<M, or SnSn<5, (2) The geometric meaning ofthelatter inequality forf(x)>0 con-sistsinthefactthatthefigurewhoseareaisequaltos,isyey bounded byabroken fine 4lying between the“inscribed* re broken line and the*circum- 4 scribed" broken line. Ab Thesum s,depends upon4, re thewayinwhichtheinterval fa,0}isarielintothesub- , xintervals [x;,, x]andalso 4%ESTartHD" yonthechoice ofpoints & Fig.209. insidetheresulting subinter- vals. Letusnow denote bymax [x;.,, x;]thelargest ofthelengths ofsubintervals [xyx], [tyX)o «++» Ergays a) Letusconsiderdifferent partitions ofthe‘interval {a,'6)" intosubintervals Uraai]suchthatmax[44]—-0. Obviously, thenumber ofsubintervals napproaches infinity here. Choosing theappro- The Definite Integrat 399 priate values of&,,itispossible, foreach partition, toform the integral sum Dre as. Wecan thus speak ofasequence ofpartitions and acorrespond- ing sequence ofintegral sums. Let this sum* approach the limit 7for some chosen sequence ofpartitions when max Ax, —0. itfor any partitions ofthe interval [a,6]such that max Ax;—+0 andforanychoiceofpoints§,thesumBreanapproaches thesame limitJ,wesaythatthefunction f(x)is integrable ontheinterval fa'B]:theVinitJiealledthedefiniteintegral ofthefunction f(x)ontheinterval (a,6].Itis denoted byJfeae andwewrite : ; , olf) Aa[10de The number aiscalled the lower limit ofthe integral, 6isthe upper limit. The interval (a,6]iscalled theinterval ofinte- gration, theletter xisthevariable ofintegration. Let ‘itbestated without proof that ifafunction y=f(x) iscontinuous ontheinterval (a,6),then itisintegrable onthis interval. Itisobvious that ifforsome sequence ofpartitions such that max Ax;—+0 weconsider thesequence oflower integral: sums 5, and ofupper integral sums 5,foracontinuous function f(x), then these sums will” tend towards the same limit /—the defi- nite integral ofthefunction f(x): ool, Bmdein [Tea .° cli, Sota frees Among discontinuous functions there are both integrable func- tions and nonintegrable ones. syinthiscasethesumIsanorderedvariablequantity, 400 TheDefiniteIntegrat Ifwe construct the graph oftheintegrand y=/(x), then in thecase of/(x)=0 theintegral A Sreyax willbenumerically equal totheareaofaso-called curvilinear trapezoid bounded bythegiven curve, thestraight lines x=a and x=b, and the x-axis (Fig. 210). For ‘this reason, ifitisrequired tocompute thearea ofacur- vilinear trapezoid bounded by the curve y=/(x), the straight lines x=a and x=6, and the x-axis, this area Qiscomputed bymeans ofthe inte- . gral 2 Q=fie)de. ® Cn * Note 1.Itwill benoted that thedefinite Fig.210. integral depends only ontheform ofthe function f(x)andthelimits ofintegration, andnotonthevariableofintegration, whichmaybedenotedby any letter. Thus, without changing the magnitude ofadefinite integral itispossible toreplace thelatter xbyany other letter: : ° ® Jrerde=[fydim =fpede. . > Whenintroducing theconcept ofthedefinite integral srede weassumed thata<b. Inthecasewhere 6<a wewill, by definition, have ’ A Srerdem—Jierar. “ Thus, for instance, Jetdr=—[ rae, Finally, inthecase ofa= weassume, bydefinition, that for The Definite Integral 40 any function /(x) wehave Sheyde=0. Oy This isnatural also from the geometric standpoint. Indeed, thebase ofacurvilinear trapezoid has alength equal tozero; consequently, itsareaiszerotoo, 5» Example .Compute theintegral faxde(>0) 3Solution. Geometrically, theproblem isequiva- yy lent tocomputing the area Qofatrapezoid bounded by the lines y=kx, r=a, r=, y=0 ig,210, 4 he{Unction y=ixunder the integral sign is continuous. Therefore, in.order tocompute. the definite integral wehave theright, aswas stated ayy x above, todivide ‘the interval (a, 6]inany way andchoosearbitrary inlermediate pointsB.The Fig.211. Fesult ofcomputing definite integral isindepend- ent ofthe way inwhich the integral sum isformed, provided that the subiaterval approaches 2270. Divide the interval {2,6]into mequal subintervals. Thelengthaxofeachsubinterval isar=°—*; thisnumber isthe subinterval (partition unit). The division points have coordinates: a=% aed OPI coonHOHMAL, For thepoints &take theleftend points ofeach subintervat: haa, Geetas Beeb GOH) de, Form the integral sum (1). Since f(&j)=AR;, wehave SpSAACEAEAEHAEAhaAx+[h(a+Axl]Ox+...+{2 la+(a—I) ax}}Oe Ra(G+Ax)+(+204)4...(8-H(8—1)Ax}x= =k{na+[Ar-+2Ae+... +(a—l)Axl}Ax sk{nat(+24... +20] ax}Ax, whereax=22. Taking intoaccount that “gt eyed (2sthe sum ofanarithmetic progression), ain—1)ba)ba napa sak[roteee] an[ers ]o—a. 40 The Definite Integrat Sincetim2=121,wehaveims=Q=k [a4P=2](baekOe dimwmdat[a42S2]0-0)aS, Thus, ¢ bat ferarease, ‘The area ofABba (Fig. 211) isreadily computed by the methods ofele-mentary”geometry.“refstwitbetgSameY Yyoxt example 2.Evaluate (s*dr. Solution.Thegivenintegral1sequaltotheareaQofa curvilinear’ (rapefold. bousded ‘by"a"parabola poosstheordigaexbandthestraightineyb(rig13 Divide the interval [a,b] into mequal’ parts by the points B20,HOE HEME coosHebe, bee Forthe&pointstaketherightextremities ofeachsubin- Forim the integral sum Wage be seextartelAckoobehamas act Fig. 212. (OAR Ax ees(aA) Ar=(Aa)(EE pat, As we know, omeVEEpeteOEDCOED| therefore a ome 0went (141)(242), lin=Q=Jtar= Fe ° Example2.Evaluate{mdx(m=const). Solution.. maze Ii mam lim mS Am =m tin Sarame—a. The Definite Integral 403, Here,$3axisthesumofthelengthsofthesubintervalsIntowhich theintervai{a,b]wasdivided.Nomatterwhatthemethodofpartition, thesumisequal tothelength ofthe'segment b—a. Example 4.Evaluate {etdr. Solution. Again divide theinterval (a,6]intonequal partst Take the letextremities asthe points &Then form thesum Semetax ettaed..pett— Mrare met(1petpete. pelt 84)ae, The expression ithe brackets isageometcic progression with common ratio e® and first term 1;therefore ie nas. ax sentGeaarmen) GT Then we have nge=b—a; imAX—aa, "ano OFT , z 1‘Hospital'sruletim2timLot.)Ths, (ByL’Hospita’s ruletim#5limJot.)Thus, lims,=Q—et(P-2—1)-1—0?—e8, that is, :(etdmet, Note 2.The foregoing examples show that the direct evaluae tion ofdefinite integrals asthe limits ofintegral sums involves great difficulties. Even when theintegrands arevery simple (kx, 2',e*), this method involves cumbersome computations. The find- ingofdefinite integrals ofmore complicated functions leads to still greater difficulties. The natural problem that. arises isto find some practically convenient way ofevaluating definite inte- grals. This method, which was discovered byNewton and Leibniz, utilises theprofound relationship that exists between integration and differentiation. The following sections ofthis chapter are devoted tothe exposition and substantiation ofthis method. 404 TheDefiniteIntegrat SEC. 3,BASIC PROPERTIES OF THE DEFINITE INTEGRAL Property 1.The constant factor may betaken outside the sign ofthedefinite integral: ifA=const, then ° r §arindx=Al(x)dx. () Prool. . ‘ . £ Aldx= li =JAr@ydr=_ tim|3are)dx, ’ =AlimSp)ax=ASf(yae.asares (al H Property 2.The definite integral ofanalgebraic sum ofseveralfunctions isequaltothealgebraic sumoftheintegrals ofthesum-mands. Thus, inthe case oftwo terms ob A A Si.@+hwlde= [fede +ffede. o) Proof. SUe4 ede tim 3,60+, Ga)dx= =lin13hGi)dat3hGdded= =obitn|DaGdOxihimBfGo)Axim ° , =Jhwadrt Jide The proof issimilar forany number ofterms.Properties 1and2,though’provedonlyforthecasea<b,holdalso for a>. However, the following property holds only fora<6: Property 3./fontheinterval la,6](a<6), thefunctions f(x)and@(x)satisfy thecondition f(x)<@(x), then : A Srupde< Spear. ) Basie Properties oftheDefinite Integrat 405 Proof. Let usconsider the difference * ° A Setdx—J/() de=S(9@)—fde =limS@-1€) ax: Here, each difference @(€,)—/(&,)>0, Ax;>0. Thus, each term ofthesum isnonnegative, theentire sum isnonnegative, and its limitisnonnegative; that’is, i> “ <2 we Stef) arao V(xy: or A ere, Soudse—Jfdeao, 7. . Fig.213. whencefollows inequality 8). IfF()>0 and g(x)>0, then this property isnicely illustrated geometrically (Fig. 213). Since @(x)>f(x), the area ofthecurvi- linear trapezoid @A,B,6 does notexceed thearea ofthecurvilinear trapezoid aA,B,b. Property 4.‘7fmandMarethesmallest andgreatest values ofthefunction f(x) ontheinterval (a,6)and a<6, then m(b—a)< J/(x)dr<M(b—a). “ Proof.Itisgiventhatmaf(x)<M. Onthebasis ofproperty (3)wehave > » » Smdx< ff(xydr< |Mae. 0) But ’ ° Jmar=m(b—a, |Mdx—=M(b—a) (see Example 3,Sec. 2,Ch. XI). Putting these expressions into inequality (4’), wegetinequality (4). 406 TheDefiniteIntegral Iff(x) 0, this property isclearly illustrated geometrically (Fig.214).The \j% area ofthe curvilinear trapezoid aABb 18 lies between theareas ofthe rectangles @A,B,b and aA,B,b. Property 5.‘(Mean-value theorem). Ifafunction f(x)iscontinuous onthe ANinterval (a,6},thenthereisapoint& Arl”onthisintervalsuchthatthefollowing “a |, equality holds: ao Bx i Fig.214 Si)dx=(6—a) f(8). (5) Proof. For definiteness leta<. Ifm and Mare, respectively, thesmallest and greatest values off(x) ‘on[a,6},then byvirtue of(4) 6 maAJi@decm. Whence , - AeSierdemn, wheremap<M. Since f(x) iscontinuous, ittakes on allintermediate values between mand M.Therefore, forsome value §(a<§<6) wewill have p=/(O, of : §f(x)dx=1(&) (6a). Property 6Forany three numbers a,b,¢theequality : ‘ ° Sitode=Jfemdet[forde, (3) 3 3 ? istrue, provided allthese three integrals exist. Proof, First suppose that a<c<b, and form the integral sum ofthefunction f(x) ontheinterval {a,6. Since thelimit ofthe integral sum isindependent oftheway inwhich theinterval (a,6]isdivided into subintervals, weshall divide [a,6]into subintervals such that thepoint ¢isthedivision point. Thenwepartition thesum»which corresponds tothe Evaluating aDefinite Integral 407 interval (a,6],intotwosums: 3§,which corresponds to(a,¢],and 3,whichcorresponds to(c,6].Then ® « ® DG) Ax=DIG) Axi+iG) dee Now, passing tothelimit asmax Ax,—+0, wegetrelation (6), Ifa<6<c, then onthe basis ofwhat has been proved wecan write ° p A : ‘ ‘ Srede=Jpirde+ S1(pdxorSf(e)dr= Jf(x)de—Jfear; ‘ a + 2 ° 3 but byformula (4), Sec. 2,wehave ¢ ’ Siadr=—Jfede. 3 ? Therefore, y 8 : F ’ 100%SF)demffcaydet[Feeae. A This property issimilarly proved for any other arrangement ofpoints a,6, and ¢. rr % Fig. 215 illustrates, geometrically, Fig,215. Property 6forthecase when f(x)>0 and a<c<b: thearea ofthe trapezoid aABb isequal tothe sum oftheareas ofthetrapezoids aACc and ¢CBb. SEC, 4,EVALUATING ADEFINITE INTEGRAL. NEWTON-LEIBNIZ FORMULA Inadefinite integral . Side letthelower limit abefixed, and lettheupper limit 6vary. Then the value ofthe integral will vary aswell: that is,the integral isafunction ofthe upper limit. Soastoretain customary notations, weshall denote theupper limit byx,and toavoid confusion we shall denote thevariable 408 TheDefiniteIntegral ofintegration by¢.(This change innotation does not change the valueoftheintegral.) Wegettheintegral |f(t)df.Forconstant a, thisintegral willbeafunction oftheupper limitx.Wedenote this function byD(x): (x)= Fat. a) Iff(6 isanonnegative function, thequantity ®(x) isnumeri- cally equal tothearea ofthecurvilinear trapezoid aAX« (Fig. 216). Itisobvious that this area varies with x. Let usfind the derivative of@(x) with respect tox,orthe derivative ofthe definite integral (1) with respect tothe upper limit. y >Theorem 1.Iff(x) isacontinuous De HyVV)4pfunctionand@(x)=f(t)dt,thenwe [3have theequality WY ©(=f(x). a XExedx x Inother words, thederivative ofa Fig.216. definite integral ‘with respect tothe upperlimitisequaltotheintegrand, inwhich the value ofthe upper limit replaces the variable of integration (provided that theintegrand iscontinuous). Proof. Let usgive the argument xapositive ornegative incre- ment Ax;then (taking into account Property 6ofadefinite integral) weget sean * sear Detan= Jhod=Sfind+ Jrae. The increment ofthe function (x) isequal to AO=O(4+4—O=J fats JFeOa—Sfide, that is, rene ao= Jp(pat. Evaluating aDefinite Integral 409 Applytothelatter’integral themean-value theorem (Property 5 ofadefinite integral): AD=f(8)(n+Ax—x)=f)Ax, where &liesbetween xand x+Ax. Find the ratio ofthe increment ofthe function tothe increment ofthe argument: a_/@ar_iene 7@ Hence, ©(x)=tim42—timf(. bre OE Aree But since E—-x asAx—+0, wehave limf(&)=limF(8),rr Sees and due tothecontinuity ofthefunction /(x), icyFG)=F(x). Thus, '(x)=/(x), and the theorem isproved. The geometric iliustration ofthis theorem (Fig. 216) issimple, theincrement AD=/(E) Ax isequal tothearea ofacurvilinear trapezoid with base Ax, and thederivative @’(x)=/(x) isequal tothe length ofthe interval xX. Note. One consequence ofthetheorem that has been proved is that every continuous function has anantiderivative. Indeed, if thefunction f(f) iscontinuous ontheinterval [a,x],then aswas pointed out inSec. 2,Ch. XI, inthis case the’ definite integral J1(Odeexists, which istosaythatthefollowing function exists: Ow=Sfioae. Bulfromwhathasalready beenproved, itistheantiderivative of f(x), Theorem 2/fF(x) issome antiderivative ofthecontinuous Junction f(x), then theformula ° SF de=F(6)—F (@) @) holds, 410 The Definite Integral This formula isknown astheNewton-Leibniz formula. *) Proof. Let F(x)besome antiderivative ofthefunction f(x). By Theorem 1,thefunction |f(¢)déisalsoanantiderivative off(x). Butanytwoantiderivatives ofagivenfunction differbythecon- stant C*, And sowe can write Sidt=Fy+cr. @) Within anappropriate choice ofC*this equality holds forall values ofx,that is,itisanidentity. Todetermine theconstant C* putx—a inthe identity; then Sidt=F@+cr, or O=F(a)+C*, whence Cta—F (a. Hence, SF@dt=F()—F @). Putting x=, weobtain theNewton-Leibniz formula: * Siigdt=F(o)—F(@), or,replacing thenotation ofthe variable ofintegration byx, Six)de=F(6)—F(a). Itwill benoted that thediflerence F(b)—F(a)is_independent ofthechoice ofantiderivative F,since allantiderivatives differby aconstant quantity, which disappears upon subtraction anyway. *)Itignecessary topoint out that the name offormula (2)Isnotexact, since neither Newton notLeibnia had.anyauch formula. Theimportant thing, however, isthat namely Leibniz and Newton were the first toestablish are- Tationship between integration and- differentiation, thus ‘making. possible the rule forevaluating definite, integrals. Evaluating @Definite Integral a Ifwe introduce the notation *) F()—F @)=F (oh, then formula (2)may berewritten asfollows: ’ §f(x)de=F(x)8=F(6)—F(a). The Newton-Leibniz formula yields apractical and convenient method forcomputing definite integrals incases where theantide- rivative ofthe integrand isknown. Only when this formula was discovered did thedefinite integral acquire itspresent significance inmathematics. Although the ancients (Archimedes) were familiar with aprocess similar tothecomputation ofadefinite integral as the limit ofanintegral sum, theapplications ofthis method were confined tothevery simple cases when thelimit ofthesumcould becomputed directly. TheNewton-Leibniz. formula greatly expanded the field ofapplication ofthedefinite integral, because mathemat cs obtained ageneral method forsolving various problems ofapar- ticular type and socould considerably extend the range ofappli- cations ofthedefinite integral totechnology, mechanics, astronomy, and so on. Example 1. : 1)bat Fran tpHoe Example 2. ° ep batJaden Paes Example 3. é heb pmet ametSeen (=e eox—. '*)Theexpression [*iscalled thesignofdouble substitution. Inthelite- rature we find two nofations: F()—F(a)=(F(ig or F()—F (a)=F @)8. We shall use both notations. a2 The Definite Integral Example 4. ° fetdemet teen Example 5.. §sined=—cosxl3"=—(cos2—cos0)=0. Example 6.. re yedt VFRAVE. jTrea Aha VI SEC. 5.CHANGING THE VARIABLE IN THE DEFINITE INTEGRAL Theorem. Let there beanintegral : Stadr, where thefunction f(x) iscontinuous ontheinterval [a,6]. Introduce anew variable ¢using theformula 2=9(0. It 1)@(@)=a, 9(8)=6, 2) g(t)and g’(¢) arecontinuous on[a,Bl,3)Teor isdefined andiscontinuous on{a,B),then 5 A Srerdr= ilemle’de. ) Proof. IfF(x) isanantiderivative ofthefunction f(x), wecan write thefollowing equations: Sfeide=F x)+0, @ Sele Wdt=F @@]+e. @) The truth ofthelatter equation ischecked bydifferentiation of both sides with respect to¢.(Itlikewise follows from formula (2), Sec. 4,Ch. X.] From (2)wehave $ Jrerdr=F wl,=F)—F@. Integration byParts 43 From (3)wehave 5 ° YVRSrewle ode=Fleole = =F[9B—F [ea] = i =F(6)—F (a). > The right sides ofthe lalter expressions areequal, and sotheleft sides areequal aswell, thus proving thetheorem. Note. Itwill be noted that when ‘com- puting thedefinite integral from formula (1) Fig,217. we do not return tothe old variable. Ifwe compute thesecond ofthedefinite integrals of(1), wegetacer- tain number; thefirst integral isalso equal tothis number. Example. Compute the integrat iyVF=Far. Solution. Change the variable: xersint, de—r cost dt. Determine the new limits: x=0 for t=0, x rer fort=, Consequently, GVPFacm|PSP econdtme|VTEcoea= aCcostedtart ((Lat aftst020)7_wt ar[costdeme(tyoor)amr(+57); ==. Geometrically, thecomputed integral istheareaof+ofthecirclebounded bythecircle x-+-yt=r* (Fig. 217). SEC, 6. INTEGRATION BY PARTS Let uand vbediferentiable functions ofx.Then (uo)’ =u'o +u0", a4 The Definite Integrat Integrating both sides ofthe identity from ato6,wehave ’ ’ A §(uoy’de=fu’vde+§uode. (ly e ie Since§(uvy’de=u0-+C, wehave{(uv)'dx=uo[};forthisreason, theequation canbe written intheform + ’ w=Jodu+Sudv, or,finally, . .?Sudo=uol, —Sdu. Example, Evaluate the tntgral y= side, Ipfstxte={attasnde—[shotdons: : tS Sa asiet cone] (00 [attconondem atofsattacontee =(=1)|stotFestat)de mina §settrde——0 fsxd Inthe notation chosen wecan write the latter equation as In= (0M) Ina) Integration byParts 415 whence we find In noe @ Using thesame technique wefind nasInes lnww and so noin—3Inepale Continuing inthesameway,wearriveatfyorJ,depending onwhether thenumber nisevenorodd. * Let us consider two cases! 1)nis even, n= 2m: eeilarmeeasBe BB 2)nisodd,n=2m+1:ee eeeecae cae a but since tonfuireaen farm, t=Ssinede=t, we have Com Qm—1 m3 538I ygfsatcdeIF STS, C einsest Qm_2m—2 64 Nanay=fstedem2AFORESAE, From these formulas there follows theWallis formula, which expresses the umber ©intheformofaninfinite product Indeed, from the latter two equations wefind, bymeans oftermwise dle vision, (246.0. 2m \F 1 aw§-(s tehweim) SETTemes ® 416 The Definite Integral WeshallnowprovethatliJimweTans Foralloheinterval(0,%)heingeities sin~*x>sin™x>sin™™**x hota, Integrating from0to3,weget Sam=hem>hams whence peste atoi. “ From (2)itfollows that Iypes_2m-41 Tass om Here, Jam=s jim2MEN PayereealMierie From inequality (4)wehave unweeTuen Passingtothelimitinformula(3),wegetWallis’formula(Wallis'product)for x u 2-4-6... 2m \F 1Le) (ee inn) mT]: This formula may bewritten tnthe form Bom (2.2,4,4,8. 2ma2, dm om mse \TO°S'S Wm Im—T “Im+Fi)* SEC. 7. IMPROPER INTEGRALS 1.Integrals with infinite limits. Letthefunction f(x) bedefined and continuous for all values ofxsuch that a<x<+oo, Consider theintegral . 1(0)=JF(x)de, This integral ismeaningful forany 6>a. The integral varies with band isacontinuous function of6(see Sec. 4,Ch. XI). Letus consider thebehaviour ofthis integral when 6—+-+00 (Fig. 218). Improper Integrals a7 Definition. Ifthere exists afinite limit ° liesSede, then this limit iscalled theimproper integral ofthefunction f(x) inthe interval {a,+o] and isdenoted bythesymbol §fear. Thus, bydefinition, wehave Jfear= timSpear. Inthiscaseitissaidthattheimproper integral {f(x)dx exists , : orconverges. If{(x)dx as6—+-+c0 doesnothaveafinitelimit, onesaysthat[/(x)dx déesnotexist’ 4 ordiverges.* > Itiseasy toseethegeometric Yyyemeaning ofanimproper integral WZ forthecase when [(x)50:ifthe~p-a@—% * integral|f(x)dx expresses thearea Fig.218. ofaregion bounded bythecurvey=/(x), thex-axis and theordinates x=-a, x=, itisnatural toconsider that the im- proper integral /(x)dx expresses theareaofanunbounded (in- finite) region tyingbetween thelinesy=/(x), x=a, andtheaxis ofabscissas. Wesimilarly define theimproper integrals ofother infinite in- tervals: §feydxm tim[f(x)dx, Jfeyde= Jfayaet Jpear. 142000 as The Definite Integral The latter equation should beunderstood asfollows: ifeach of the improper integrals on the right exists, then, by definition, the integral onthe left also exists (converges). F dx trample Evatt nega fay eeFay#0and yo a y ' 7 5 * 7 ¥ Vitte id Fig. 219. Fig. 220. Solution. Bythedefinition ofanImproper integral wefind ts ’ ax “ 5 * JrSemtimSra, dinretans|’,timactin=F, vari ete gxgrenes theareofanintecrvinartraperié cots: Example 2°Find outatwhich values ofa(Pig. 221) theintegral fFonF onverges and atwhich itdiverges.es TeSolution.Since(wheno¥1)y os , wwehave 7” te y ax Lgebs {Seam gop. a‘Consequently,ae te 5 amm Taps, then)Get andthe Fig. 221, Integral converges Improper Inteqrats a9 We<t, then|Sac,andtheintegraldiverges Whenamt,[minz["*=coy theintegral civrges. camp a.Beas {7. Solution. = ae ax as ‘Thesecond inlegral isequalto%(exeExample 1.Compute theAstintegral: i Ht..limfae limarctanx|°SrFenetefren inetons= =,lim(aretan0—aretana)=F Therefore, dxain i)Teer atye* Inmany cases itissufficient todetermine whether thegiven integral converges ordiverges, and toestimate itsvalue, The fol- lowing theorems, which wegive without proof, may beuseful in this respect. Weshall illustrate their application inafew cases. Theorem 1.Ifforallx(x>a) theinequality 0</(e) <9(2) isfulfilled andif{q(x)dx converges, then {f(x)dx also converges, and. . §fades §pide. Example 4,Investigate theintegral tng :5UF) for convergence. 1“ 40 The Definite Integral Solution, Itwill benoted that when La, 1 1 RU Se And e 1 Lite{fant on. Consequently, ‘.as?BFR) converges, and itsvalue isless than 1. Theorem 2.Ifforallx(x2a) theinequality 0<9(x) <S(®) isfulfilled, and{@(x)dx diverges, thentheintegral §|(x)dx also diverges~* . Example 5.Find out whether the following integral converges: 5xtl2VR” We notice that eeVa Va Ve" but + tolim2Vz|=+o0. SFennel a+ Consequently, the given integral also converges Inthelast two theorems weconsidered improper integrals of nonnegative functions. Forthecase ofafunction f(x)which changes itssign inan infinite interval wehave the following theorem. Theorem 3.Iftheintegral ||f(x)|dx converges, thenthein- tegral §f(x)dxalsoconverges. Inthiscase,thelatter integral iscalled anabsolutely conver gent integral. Improper Integrats at Example 6.Investigate theconvergence ofthe integral vasmae, Solution. Here, the integrand Isanalternating function. We note that sinx,1 Fae Lite|Se]slal 20S See[=r Therefore, theintegral [|S54|axconverges.Whenceitfollowsthatthe aiven integral alsoconverges 2The integral ofadiscontinuous function. Let the function [(2) bedefined and continuous when a<x<c, and forx=c let the function beeither not defined or let itbe discontinuous. In thiscase,onecannot speak oftheintegral |f(x)dx asoftheli- mitofintegral sums, because /(x)isnotcontinuous ontheinter- val(a,cl,and forthis reason thelimit may notexist. Theintegral {F(x)dxofthefunction f(x)discontinuous atthe point ¢isdefined asfollows: * * SP)demtimShae. Itthelimit ontheright exists, theintegral iscalled animpro- perconvergent integral, otherwise itisdivergent. Ithefunction /(x) tsdiscontinuous attheleftextremity otthe Interval (a,¢](that is,forx=a), then bydefinition Sfeode= timSfoade.} owarep Ifthefunction f(x) isdiscontiquous atsome point x—=x, inside the interval [a,cl],weput : * ‘ Grendx=(forde+[F(ayde, f z 5 ifboth Improper integrals ontheright side oftheequation exist. a The Definite Integral Example 7.Evaluate §ayVi Solution P * a aatti tim (He tim 2THEB= cateet =a,lim21VT=8-12. ax Example 6.Evaluate, theintegral (4. Solution. Since inside theintervalofintegration thereexistsapointx=0 where the lntegrand Tsdiscontinuous, the Integral must berepresented asthe Sum oftwo terms: dehax (ae i)gam fStimI a ot* ae Calculate each timit separately: Gaede tory_ anJSe-uae ttan ia) Thus, theintegral diverges onthe interval (—1, 0] dx 1 um(== am(“=ePe eae ee And this means that theintegral also diverges ontheinterval (0,1 Hence, thegiven integral diverges on'the entice interval [=et, 1. IWehould benoted thet Iwehad begun. toevaluate the glen” Integral without paying allention tothediscontinuity ofthe. integrand atthe peint “e=0,"theresult.would“have‘beenwrong. iyIndeed, ( de aye 11 §$--3|1,--(t-4)--2 9Hen tsimpose Fe 22 Note. Ifthe function /(x), defined ontheinterval[a,6},has,withinthis rr interval, afinitenumber ofpoints of 41 *discontinuity a,,a,,..., a,then the >Fig.222. integral ofthe’ function’ f(x) onthe Improper Integrals 2 interval [a,6}isdefined asfollows: ° os Ps . Sreode=Jfonde+ Sfedet... +)f@dx, ifeach oftheimproper integrals ontheright sideoftheequation converges. Butifeven oneofthese integrals diverges, then {F(x)dxiscalled divergent aswell. * For determining theconvergence ofimproper integrals ofdis- continuous functions and forestimating their values, one can frequently make use oftheorems similar tothose used toestimate integrals with infinite limits. Theorem 1’:Ifontheinterval |a,c|thefunctions {(x) and9(x) arediscontinuous atthe point c,and atall points ofthis interval theinequalities @(x)>[(x) >Oare fulfilledand§q(x)dxconverges, thenJf(x)dxalsoconverges. Theorem 2.Ifontheinterval [a,c]thefunctions [(x)and p(x) ‘are discontinuous atthepoint c,and atallpoints ofthis interval theinequalities f(x)>9(x)>0arefulfiltedand(q(x)dx diverges, then§f(x)dxalsodiverges. Theorem 3'.Iff(x)isanalternating function ontheinterval a,cland discontinuous only atthe point c,and the improper integrat §|f(x)|dx oftheabsolute valueofthisfunctionconverges, thentheintegrot \f(x)dx ofthefunction itself alsoconverges. Useisfrequently madeof<5. asfunctions withwhichitis convenient tocompare thefunctions under thesign oftheimproper integral,Itiseasytoverifythattad converges fora<l, and diverges fora> 1.* 424 TheDefiniteIntegral ThesameappliesalsototheintegralsSatedx, Example 9.Doestheintegtal=! axconverge? : el re c Solution. The integrand isdiscontinuous atthe left extremity ofthe in- terval[0,1].ComparingitwiththefunctionTFwehave 1 1 yeas <yge: Vitae Ve Theimproper integralfFEexists,Consequently, theimproper integral ( 1 Q ofalesserfunction,thatis,Syrre alsoexists, SEC. 8.APPROXIMATING DEFINITE INTEGRALS Attheend ofChapter Xitwas pointed outthat notforevery continuous function isitsantiderivative expressible interms of elementary functions. Inthese cases, computation ofdefinite in- tegrals bythe Newton-Leibniz formula isinvolved, and various methods ofapproximation are used toevaluate the definite inte- grals. The following areseveral methods ofapproximate integration based onthe concept ofadefinite integral asthe limit-ofasum. 1.Rectangular formula. Let acontinuous function y=/(x) be given onaninterval [a,6].Itisrequired toevaluate thedefinite integral ’ Srondx. Divide theinterval [a,6]bythepoints a=x,, x,,%,+... %,=5 into nequal parts oflength Ax: Avat=, Then denote bYYorYsYar«+++ Yass Yuthevalues ofthefunc- tion f(x) atthe points 't,,"x,xy,ss xqthat is, =F) WAL oi Ya Fn) Approximating Definite Integrals 45 Form the sums: YAK +YAK. AY AE, y,Ax+y,Ax+...-+y,Ax. Each ofthese sums isanintegral sum forf(x) onthe interval {a,6]and forthis reason approximately expresses the integral 6 Otel Ceeeeeee an a ca , Sedx=2=8y,tut... +40 co) This istherectangular formula. From Fig. 223 itisevidentthatiff(z)isapositiveandincreasing function, thenformula(1) expresses thearea ofthe step-like figure composed of“inside” rectangles, while formula (1’) yields thearea ofthestep-like figure composed of“outside” rectangles yt 6 ah4 AZ wel - WZZ "y 4«lage(O4] ve |4 Nae a ad Cee a al Fig. 223, Fig. 224 The error made when calculating integrals bytherectangular formula diminishes withincreasing m(thatis,thesmaller the oma divisionsabUi,The trapezoidal rule. Itisnatural toexpect ‘that we will obtain amore exact value ofthedefinite integral ifwereplace thecurve y=/(x) notbyastepped line, asinthe rectangular formula, but byaninscribed broken line (Fig. 224). Then the area ofthecurvilinear trapezoid aABbwill bereplaced bythesum oftheareas oftherectilinear trapezoids bounded from above by thechords AA,, 4,A,, ..., A,-,B. Since thearea ofthefirstof 26 The Definite Integral thesetrapezoids is“t%Ax,theareaofthesecondisU4Ax, and soforth, so Sitaydr=(BpHartSptart... Mattear) or ¢ b SFG)de=8(MEH yt tan): ® This isthe trapezoidal formula (trapezoidal rule). The choice ofnisarbitrary. The greater this number, thesmaller willbethedivision (subinterval) Ax=2=* andthegreater will bethe accuracy with which thesum, written onthe right side of the approximate equality (2), yields thevalue ofthe Integral. II. Parabolic formula (Simpson's rule). Divide the interval fa,6}into aneven number ofparts n=2m. Replace thearea of thecurvilinear trapezoid, corresponding tothefirst two subinter- vals [x,,x,]and [x,,x,]'and bounded bythegiven curve y=/ (x), bythe'area ofacurvilinear trapezoid such that isbounded bya quadratic parabola passing through three points: Mets Ys MyCY) Miler Wade and with anaxis parallel tothey-axis (Fig. 225). Weshall call thiskindofcurvilinear trapezoid aparabolic raped: The equation ofaparabola with anaxis parallel tothe y-axis isofthe form yaAd +Bete. The coefficients A,Band Careuniquely determined from the condition that the parabola passes through three specified points. Analogous parabolas areconstructed forother pairs ofintervals as well. The sum oftheareas oftheparabolic trapezoids will yield theapproximate value ofthe integral. Letusfirst compute theareas ofone parabolic trapezoid. Lemma. /facurvilinear trapezoid isbounded bythe parabola y=At+Bx+C, thex-axis and two ordinates separated byadistance 2h, then its area is S=hU+4y tH @) Approximating Definite Integrals er where y,and y,aretheextreme ordinates and y,istheordinate ofthecurve atthemidpoint ofthe interval. Proof. Arrange anauxiliary coordinate system asshown in Fig. 226. pmAx+BX4C y% MMe wy. yfoa /veMy OTK . oh a x Fig. 225. Fig. 226 Thecoefficients intheequation oftheparabola y=Ax*-+Bx+ +C aredetermined from thefollowing equations: ifx=—h, then y=Ah'—Bh+C;itx,=0,|theny= roi “ifxj=h, then yj=Ah?4Bh+C. Considering thecoefficients A,B,Cknown, wedetermine the area oftheparabolic trapezoid with theaidofadefinite integral: * SeScartBetCydrm[48498 4cx]"=5RAN+60). ah But from equalities (4)itfollows that Wt Ay,+y= 2AR +6C, Hence, S=4*U444,44) which iswhat had tobeproved. Letuscome back toourbasic problem (see Fig. 295). Using formula (3)we can write the following approximate equalities(=Ax): {peare2t tay, +00. cnn 28 The Definite Integral Jierde nsWA tude sme? ArJ1)deSFUmeatWames+Yen) Adding theleftand right sides, weget(ontheleft) thesought- forintegral and (ontheright) itsapproximate value: . SreydeEY+4,+UtANH etems +amas +Yam) 6) or ¢ boa Site)dem Sty +Yee 2ptYsboteed FAW Tt Fmd y This isSimpson's formula (rule). Here, the number ofdivision points 2m isarbitrary; but the more of them there are, the more accurately thesum onthe right side of(5)yields thevalue Aofthe integral. *) nx 9 Example. Evaluate approximately OMT ats 2X * y A m2(te,Fig. 27. j Solution. Divide theinterval {1,2 into 10equal parts (Fi. 227). Assuming 2-1a2ataon, 7)Tefindouthowmanydivision.pointsareneededtocomputeaninte es"ainedesitednumberpidecimalplaces,onecanematese"offormalsesimaing theerreaiting IromSnfoiting tenia Wedonot tive thee esfimates here: The reader wil fnd themin more sdvanced courses Si"nalysig sey" for example, Fikbtengolts, “Course “ofDiferential ond Integral Calculus, 1969, Vole Ii,Chr IXy'See” 6,(Rusean edition) Approsimating Definite Integrals 29 wemake table ofthe values ofthe integrand: ||| .got : vet aaio|arom |ais|4062500S=rt|fZogmm |SEF|prosefoi|flomme |Sore|foolsBrr3|fZozems |SI|fCosasemeld |y=0.71499 xn=20 |yy.=0-50000NOUS|(ovgsbe 1Bythe frst rectangular formula (I)weget $4Sorustnt.tyeos-tusmmoneT, Bythesecond rectangular formula (1")weget §4%SonGhat.tndeotseamsno.8i, Itfollows directly fromFig.227thatinthiscasethefirstformula yields thevalue oftheinegral withtan excess, theseconds with adetec Il.Bythe trapezoidal rule (2), wehave f01 (42240.070)<a. Ui, BySimpson's rule (6), wehave (dx OLSESE etek eetoctet HAGbutotosOl =pospeamete443.4505) 0.0018. Aatualy, In2=(no.601e2 Gosevenleesofdina TosswenleisyGeeiteryalpoyto10partsnySineee’s nwo gcfegan decalbyheIapeotal rae,onlytreyabydhe FSetangulaf formula, weotesureonlyeftheAsdecimal 430 TheDefiniteIntegral SEC, 9,CHEBYSHEV'S FORMULA Inengineering computations, useisfrequently made ofCheby- shev's formula ofapproximate integration. Onceagain, letitberequired tocompute Gitex. Replace theintegrand bytheLagrange interpolation polynomial P(x) (Sec. 9,Ch. VII) and take certain nvalues ofthe junction ‘ontheinterval (a,b]:f(x,), F(x), «+++ F(%q) Where xy,Xy)005 Xq areany points ofthe interval [a,6]: (2) ey). le) POBSEDaaaay|Od+ (4) (ee)FSR ag)dE (=H)(R=).(Tyas) FtSGA Gann! de a ‘Wegetthefollowing approximate formula ofintegration: ° ’ {finde JPwax @) alter some computation ittakes the form JHa)deCf) +CM) +00$C,Ha) ) where thecoefficients C,arecalculated bythe formulas éct). )(J. ) Se Ce ere Yee &5GR)Ha weed Formula (3)iscumbersome and inconvenient forcomputation because thecoefficients C,areexpressed bycomplex fractions. Chebyshev posed theinverse problem: specify notthe abscissas Ay,Xyseer %qbutthecoefficients C,,C,,..., C,and determinetheabscissas X,,xy)see)Ape Chebysheo's Formate «at The coefficients C,arespecified sothat formula (3)should be assimple aspossible forcomputation. This will obviously occur when allthe coefficients C;areequal: C,=C,=...=C,. Ifwedenotethetotalvalueofthecoefficients C,,C,...,C, byC,, formula (3)will take the form . SFde Ce) thle) +. +hdk 6) Formula (5)is,generally speaking, anapproximate equality, but iff(x) isapolynomial ofdegree nothigher than n—I, then the equality will beexact. This circumstance iswhat permits determin- ingthequantities Cy.Xj.Xp,601 Lae Toobtain aformula that isconvenient forany interval ofin- tegration, letustransform the interval ofintegration [a,6]into theinterval [—1, 1].Todothis, put then forf=—1 wewill have x=a, forf=1, x=6. Hence, SreydemP52[1(2$242520) atmPs*Foat, where @(t) denotes thefunction of¢under theintegral sign. Thus, theproblem ofintegrating the given function f(x) onthe interval [a,6|can always bereduced tointegrating some other function p(x) onthe interval {—1, 1]. Tosummarise, then, theproblem has reduced tochoosing, in the formula SidraC,Ue)+1)++h (O) thenumbers C,,x,,X,,+++,%,$0thatthisformula willbeexactforany function f(x) ofthe form [@)=a,+axtax? +... +a, 2" 2) «2 The Definite Integral Itwill benoted that Siade= JQ+ax+ayet... +a,x")des .2(q+34+E4F4...+ 1),itnisodd; 2(q+$+...+ 24),imiseven, ®) Ontheother hand, thesum ontheright side of(6)will, onthe basis of(7), beequal to Cyla, +4,(yp yb eeba) HGP Pee EH) eee weg (EAT DL ) Equating expressions (8)and (9), weget anequation that should hold forallay,ay,ay,+++ Oya? 2(q+4S4G+..a =C,Ina,+a,(x,+y+2HQ) PORE AEEPAE)egy (EPRI bE Equate thecoefficients ofay,a,aj,Gy)+++) yy onthe left and right sides oftheequation: 2=C,norC,=2; Bytayb ee+e,2052a Beet. thea asi usx, (10) ee eee Abate taegeeg Fromthelatter.nequations wefindtheabscissas xy,xy)... X_-These solutions were found byChebyshev forvarious. values ofa. Chebyshev's Formula 433 The following solutions arethose that hefound forcases when the number ofintermediate points nisequal to3,4,5,6,7,9: 2 f=—y=0,707107 2 a r= 00187592 A y=—x,=0.832498, 5$ Ra—n=0.57454Ln=0 ii my—x4=0.866247 6 a Raa=0422519 3x= — =0.266635, ayax0.880862 A 2 soa=0.5296877 MaaH=0300012%=0 ns ns areA ==m=0.601019 9a soa#)0.528702 ' =—,=0.167906=o Thus, onthe interval [—1, 1],anintegral can beapproximated bythefollowing Chebyshev formula: A ; - Srey de=ZU +e) + +e where nisone ofthe numbers3,4,5,6,7or9,andx,...,ty arethenumbers given inthetable. Here, ncannot be&orany number exceeding 9;forthen thesystem ofequations (10) yields imaginary roots. : 4 The Definite Integral ‘When thegiven integral has limits ofintegration aand 6,the Chebyshev formula takes onthe form A Sle)de" XIX) +.FAX b+a)b—a whereX,=2$24°S"4,(/=1, 2,...,n)andx,havethevalues given inthetable. The following example illustrates theuseofChebyshev's approx- imation formula forcalculating anintegral. eeample, Evauate (4(—I02, Solution. First, bychanging variables, transform thisinfegral intoanew ene with Timts oftategratton —t and ele?2-1,3,¢ade sep ty ‘sgte- a a dent, Then fae (dtSoJstr- Compute thelalfer lateral, taking n=3, byChebyshev's formula: [email protected]+1+1(—o.707107). Since- 7 A To.TOON=3,qoroT=s7o7107=2, \ f=zhg=0.3388, 1(-0.0009 =5—shonrgy =aes=0408100, we have (ott2o.nesa40.3808 0.405130)= 34073: : =2.100015 =0.6928100.69, Integrals Dependent onaParameter 435 Comparing this result with theresults ofcomputation using the rectan- gular formulas, the trapezoidal rule, and Simpson's rule (see the example fhthe preceding section), wenote that the result (given byChebyshev's formula "with. three intermediate points) isin better ‘agreement with the true value ofthe integral than the result obtained byihe trapezoidal. rule (with nine intermediate points). The theory ofapproximating integrals was further developed in theworks ofAcademician A.N.Krylov (1863-1945). SEC. 10, INTEGRALS DEPENDENT ON APARAMETER Differentiating integrals dependent onaparameter. Let there be anintegral ; 1(a)=Sf(x,a)dx, 0) inwhich theintegrand isdependent upon some parameter a.If theparameteravaries,thenthevalueofthedefiniteintegralwill also vary. And thedefinite integral isafunction ofa;wecan therefore denote itby/(a). iSuppose thatf(x,a)andfa(x,@) arecontinuous functions when c<acd and acrad. 2) Find thederivative ofthe integral with respect tothe parame- ter a: jimL@tAM—1@) _7ay,rT) Infinding this derivative wenote that A Matda)={F(x,0-4+Aa) dx nd, consequently,aiquently, : . \1(a+a)—1(@)= 9f(x,0+Aa)dx—|f(x,a)drm =fUG,+4a)—f (,older, Hataa—!@)_ ¢1a.e+ aa)—I(04+ Aa)—! (a)_(1G,a+Aa)(2) uo i) a dx, 496 The Definite Integrat Applying the Lagrange theorem tothe integrand wehave [G.e+ 0019) _fie,a400a), where 0<0<1. Since fa(x,a)iscontinuous intheclosed domain (2), wehave fa(x,2+08a) =fa(X,a)+2, where thequantity e,which depends onx,a,Aa, approaches zero as Aa—+0. Thus, J(a+Aa)—/ (@)° é £ Heated Ufa(x,a)+ede={faleade+Vede. Passing tothe limit asAc—+0, wehave*) 5onLotbw— a)_fimLEER) 7,(ayeff,2)de or ® ’ [Jiea)dxj=iyfeleaydx. This formula iscalled the Leibniz formula. 2.Now suppose that inthe integral (1)the limits ofintegration aand6arefunctions ofa: ,0 1(a)=© (a,a(a),b(a)|=§f(x,a)dx. ayaio Ia, a(a), 6(a)] isacomposite function ofa,and aand 6are intermediate arguments. Tofind thederivative of/(a), apply the tule fordifferentiating acomposite function ofseveral variables (see Sec. 10,Ch. VIII):1(q)<042da,90dbV@)=9aGadat3da @) ’ *)Theintegrand intheintegral [edaapproaches zeroasAa—+ 0.From thetctatthellgrnd approach srdoesnot always followCal heintegral ssoapproaches zero.However, inthegivencase,J©dx approaches zero asAa—+0. Weaccept this fact without proof.2 Integrals Dependent onaParameter ar Bythe theorem for the differentiation ofadefinite integral with respect tothevariable upper limit {see formula (1), Sec. 5] weget , 2aSPH|fe,a)dx=/16 @),al, oo_a¢ acana|1Oadem—ayTea)dx=—fla(a), al. Finally, toevaluate $®usetheabove-derived Leibniz formula: ’ 20R=Sfetxa)dx. Substituting into (3)theexpressions obtained forthederivatives, we have 1 . a aFa(a)=Jfale,a)de+110(a),ulS—Fla(a), a)$2.(4)ate Using theLeibniz formula itispossible tocompute some defi- nite integrals. Example. Evaluate the integral fenwtintt ae Solution. First note that its impossibie tocompute the integral directly, because theatiderlvative ofthefonction e=#8824 iggotexpreaible in terms ofelementary functions. Tocompute this Integral weshall considet It 252function oftheparameter a: Hay=JeB ae Thea,adervative withrespecttoaisfundfomtheabovederived Lelie Wm{feeBe]ax=(e-*cosaras =)Leibnies formula was derived ontheassumption that the limits ofinte- gation aand bate finite. However, inthis case Leibnitlormula_ also holds, Even though one ‘ofthelimils ofiniegraion ‘isequal Toinalty. 408 The Definite Integral Batthelatterintegralisseadilevaluatedbymeansofelementaryfunctions:itisequal topoe Therefore, ; "Omar Integrating the identity obtained, wefind 1(a): J(a)=are tana+C. 6) We have Ctodetermine now. To dothis, we note that 10)f#08armfoaemo. What ismore, are tan 0=0. Substituting into (@)a=0, weget 1(Q)=are tan 0+C, whence C=O. Hence, forany value ofawehavetheequality 1(q)=are tana; that fs, JeeSBaemaretana, Exercises onChapter XI 1.Forming theintegral sum s,and passing tothe limit, compute the desinite integrals , (tae. Hint,Dividetheinterval_{a, 6]intonpartsbythepointsx;=ag!(/=0, 1,2oomwheregmJE.Ans,HSH, 52.[Senerocec,Ansin. Hint. Divide theinterval a,6]inthe same way asinthepreceding example, ‘ aJYFas.Ans,F(t—a", Hint. See Example 2. * 4Gstneas.Ans.cosa—cos8. Erercises onChapter XI 9 Hint, First establish the following identity: sina-tsin(a-+)-+sin a2)... -+sla(e-+(a1)hlh A -cos(a-4) cootan—*]ae Qin Todothis, multiply and divide alltheterm ofthe fet side bysin and replace the product ofsines bythedifference ofcosines .8.feosea Ans.sinb—sina, Using the Newon-Leibniz formula, compute the definite integrals: (eaname.£1fetedane 0[ancesAns tofrdnamb.1fecesAmentaGomeanans1 z s ¢ dx xi dx x¢ efit Baie [pptamBam[tonsaanetn v2FEtanrete(8AniontefondsdaseE15(ten: ; ? ye vaé ax¢ 2 x sas258. uwFftp.anenae,anfcosted,ansB. v8faieads ane, Evaluate the following integrals applying the indicated substitutions: ¢ 2 . 1i dx x x 10fanscotedeere tsdnt20fprfnattneoe, (saz avr ‘dx (AE apteettAns,VE, . setant, asSg beenam weafi esis 40 TheDefiniteIntegral x1 Cea * ans.S44. wy(PRs, rte ns,2(2—aretand, ¢ dz 1 3fcospdp reds ansind. a5,(80 sings, Ans in , ; Provethat26.[aQ—atdem [etd—ardeim>0, 0>0) n.frorde=[lato—nds. 28[rahdrmy[1ands, Evaluatethefollowingimproperintegra: 28(ELE am1 c Cae ES (ae x .Cereds,Ans1.at.(a4.Ans.Zia>0). 92.(2. Ans,% 30)Ans.18JatAns.E(a>0).82.Svsans. a$4.awd. 90finde dae198Fesineds ArmThedate 1sveges.36.FAn.Theintegraldiverges.8.foe. A eral diverges. 98.7 Ans. Theintegraldiverges.82.[eee—ey- Ans a(sh.amon(Sanetentgeaiveraen0.f2h 2v* aie pea Ans.Fe.ase‘Ans,Theintegraldiverges.42.fere*sin dede(a>0), ST eal OS EB Evatt theflowing Integrals approxinately: M4n=(SEbythe trapezoidal rule andbySimpson's rule(n=12).Ans.1.6182(bythetrapezoidal rule);1.6098(bySimpson's rule.45,[x"dxbythetrapezoids! ruleandby Exercises onChapter XI a Simpson'srule(n=10).Ans,2690;3660.48.[YT=Pdebytetrapezoidal (ax " o= rule(n=6).Ans.0.8109.47.PrabySimpson'srule(n=4).Ans.0.8111. 48Jtoesaebythetrapezoidal ruleandbySimpson's rule(n=10). Ans,6.0696;6.0596.4.Evaluate xfromtherelation =(72applying Simpson'srule(n=10)Ans.3.14159.60,f2dxbySimpson'srule(n=10) Ans.1.71.51.Evaluatee-#2"dxforintegralm>0byproceedingtromthe equalityPentdemZwherea>0.Ans.a!62.Proceedingfromtheequality fan tegral(<2, Ang,#1:3°5..-(22—1)Jpepengprp:evaluatetheintegral[aaSipeeyAnsSS2D 5%Evaluate theintegral (Y=ds,Ans.in(1+a)a>—1).64Utilising theequatity [xt-tde—Z, compute theintegral [x47Inaes, a: ans.(=. CHAPTER XII GEOMETRIC AND MECHANICAL APPLICATIONS OF THE DEFINITE INTEGRAL SEC. 1.COMPUTING AREAS IN RECTANGULAR COORDINATES Ifon the interval [a,6]the function f(x)>=0, then, aswe know from Sec. 2,Ch. XI, thearea ofacurvilinear trapezoid bounded bythecurve y=f(x), thex-axis, and thestraight lines x=a and x=b (Fig. 210) is ° Q=ficndr. a . Iff(x)<0on(a,6},thenthedefiniteintegral(f(x)dxisalso<0: IWisequal, inabsolute value, totheareaQcorresponding tothe curvilinear’ trapezoid: ’ —0=ff(xdx. Iff(x) changes sign ontheinterval [a,6]afinite number of times, then webreak uptheintegral throughout [a,6]into the y sum ofintegrals ofthesubintervals.The integral will bepositive onthose yf) subintervals where f(x)>0, and nega- tive where /(x)<0. The integral over theentire interval will yield thediffer- 4 %enceoftheareas above andbelow the qa-axis (Fig. 228). Tofind thesum ofthe Fig. 228. areas inthe ordinary sense, one has to find the sum ofthe absolute values ofthe integrals over theabove-indicated subintervals orcompute the integral Q=f[Fax Example.1.ComputethearesQboundedbythesinecurvey=sinxand tnenananforOSes ge229) * , Computing Areas inRectangular Coordinates 43 fret Sincesiax0whenO<x<x andsinx<0whena<x<2n, wehave Q={sincde-+|( sinxde|—C [staras, {sinsd=cosx|=—(connos)=—(—1—)=2, {siadra—cos|m=(C0822—c08a)=—2 Consequently, Q=2+|—2] =4. =f) y " Wy,yosing \S lo oe ox Fig.223. Fig.230. If_one needs tocompute thearea bounded bythecurves y=f, (x), 'y=F,(x) and theordinatesx=a,x=6,thenprovidedf,(x)>/,(x) wewill obviously have (Fig. 230) A * * Q=SAde—JF, de= ff,@)—F, (w))dx. 2) Example 2.Compute the area bounded bythecurves (Fig. 231) y=Vi andyas, Solution, Find the points ofintersection ofthe curves: Yeast, rast, whence x,=0, x=1. Therefore, Cys C Cre 2oft 2ot =| VFde—[ eralVEapaendeh2aPtL, Now letuscompute thearea ofthecurvilinear trapezoid boundedbyacurverepresented byequations inparametric form(Fig.232): z=9(1) Y=V(O, @) where a<t<p 444 Geometric andMechanical Applications oftheDefinite Integral and e(a)=a, @(B)=b. Let equations (3) define some function y=f(x) onthe interval [a,6]and, consequently, the area ofthecurvilinear trapezoid y may becomputed from thefor-yor? mula ove Q=Sfldx=fyde. iy BA) oN 1* Sse aa % Fig, 231. Fig. 292, Change the variable inthis integral: x=@(0; de=q’(dt, From (3) we have y=1) =H) =v. Consequently, ~ : Q=fwe Hat. a This isthe formula for computing the area ofacurvilinear trapezoid bounded byacurve represented parametrically. Example &Compute the area ofaregion bounded bythe ellipserasacost,ymslat,Solution.Computetheareaoftheupperhalloftheellipseanddoubleit Herat aries temo tortor and S0'% varies betweea #2040 m2conn(asinety——200 [sattm208fe Fecosa ftsin2e)ncaf1Sapts282"an, The Area of@Curvilinear Sector inPolar Coordinates 4 Example 4Compute the area bounded bythe x-axis and anareofthe eyelid rea(t—sing, y=a(l—cos 9 Solution, The variation of xfrom01028acorresponds tothevariation ot fiom Dto Bx. From ()we have =|aconna(hoon9deena?f(1costat= -[iaahentersea FarnamPeoncarer(costed§LEMgra Wefinaly get Qa!(20+m)=3na", SEC. 2,THE AREA OF ACURVILINEAR SECTOR IN POLAR COORDINATES Suppose inapolarcoordinate systemwehaveacurvegiven bytheequation e=/), where /(0) isacontinuous function when a<9<f. Let usdetermine thearea ofthesector OAB bounded bythe curve o=/() and bytheradius vectors {=a andb=B. Divide thegiven area byradius vectors 9,=a, =, ..., 6,—B into nparts. Denote byA,, A0,,..., A0,theangles between theradius vectors that wehave drawn (Fig.” 233), Denote bygthelength ofaradius vector corresponding tosome angle 0,between 0,., and 6,. _ Letusconsider thecircular sector with radiusqandcentralangle A0,. Its area will be _ AQ=FeAY. The sum 1.ot .2 =F LGA =FDVT AG, will yield thearea ofthe“step-like” sector. 416 Geometric and Mechanical Applications oftheDefinite Integrat Since,thissum_is anintegral sumofthefunction ot=[f()|* ontheinterval a<6<B, itslimit, asmax A0,—0, isthedefi- niteintegral . IfozSedo. Itisnotdependent onwhichradius vector Q;wetakeinsidethe 8 9-0) . ZS %YZ é. A.oZo 7 > Fig. 283. Fig. 234 angle A0,. It,isnatural toconsider this limit the sought-for area ofthe figure*). Thus, the area ofthe sector OAB is 8 Q=4fedd a) or ry Q=ZJUma. ay Example. Compute thearea bounded bythe lemniscate =a cos28.Fig.204 ° i, olution, The radius vector will describe afourth ofthe sought-for area it@variesbetweenOand: 1getPoeaoebarf otsin20[tT_at ond[eratedaron20abtm2(Fat, Hence. . Qaat ~7)Temight beshownthatthisdetermination oftheareadoesnotcontradict that given earlier. Inother words, if-one computes thearea of8curvilinear sectot bymeans ofcurvilinear trapezoids, theresult will bethesame. The Are Length ofaCurve a7 SEC. 3.THE ARC LENGTH OF ACURVE 1,The are length ofacurve inrectangular coordinates. Let ‘acurve begiven bythe equation y=/(x) inrectangular coordinates inaplane. Let usfind the length ofthe arc AB ofthis curve between the vertical straight lines x=a and x= (Fig. 235). The definition ofthelength ofanarewasgiven inChapter VI, Sec. 1.Let usrecall that definition.OnanarcABtakepointsA,M,, M,,«1.Mz... ,Bwithabscissas x=,X,,Xp)v0.0)isoo)Oy and draw thechords AM,, M,M,, .-.. M,.,Bwhoselengthsweshalldenoteby4¥m8feo AS, AS...5MSrespectively. This oteTiy,|* gives thebroken line AM,M, ...M,_,B ay; inscribed inthearcAB.Thelengthof|4the broken line is s=DAs. Olaxa 7Oe = iy The length, s,ofthearc AB isthe Fig. 285, limit which the length ofthe inscribed broken line approaches when the length ofitsgreatest segment approaches zero: =i 7 1seal2s ® Weshall now prove that ifontheinterval a<x<6thefunc tion f(x) and itsderivative /’(x) are continuous, then this limit exists. Atthesame time weshail specify atechnique forcomputing thelength ofthearc. Let us introduce the notation .by=f(xf(x24). Then 3=<VORFaa= V14(Seyan ByLagrange’s theorem wehave BiyaHedHtad)1(Ey, where HaySB<op 448 Geometric and Mechanical Applications oftheDefinite Integral Hence, As,=VIFU Ga An Thus, the length ofaninscribed broken line is 5=LVTFE GOOx. ro Itisgiventhat/’(x)iscontinuous; hence,thefunction V+[P(r isalso continuous. Therefore, this integral sum has alimit that isequal toadefinite integral:fi °s=limLVIFV CpOx,=SVIFFOFax. We thus have aformula forcomputing the are length: * + s=(VIFCIax=[14(Z)ax. @) Note. Using this formula, itispossible toobtain thederivative ofthe arc length with respect totheabscissa. Ifweconsider the upper limit ofintegration asvariable and denote itbyx(we shall notchange thevariable ofintegration), then thearelength swill beafunction ofx: sw=f V4(2)ax. Differentiating this integral with respect totheupper limit, we obtain as_/7aaS g=V1+(#)- @) This formula was derived inSec. 1,Ch. VI, oncertain other assumptions. Example 1.Determine thecircumference ofthe circle Beyer Solution. First compute the length ofafourth part ofthe circumference lying inthefirst quadrant. Then theequation oftheare.AB’ will be y=Vom, whence “wy xa oe The Are Length ofaCurce 49 Consequently, pee)Vopte|ptearmratesnd[ParB. The length ofthe circumference iss=2ar. Let usnow find the arc length ofacurve when the equation ofthecurve isrepresented inparametric form: x=9(), y=~l) (@<t<f), (4) where g(t) and 1p(é) arecontinuous functions with continuous de- tivatives, and @(t) does notvanish inthegiven interval. Inthis case, equations (4)define afunction y=/(x) which iscontinuous and has acontinuous derivative: ay84ae Let a=(a), 6=@(f). Then substituting inthe integral: (2) x=9(0, dx=q' (t)dt, we have 5 = FOP yy sSVi+ Polowat, or,finally, A s=\VEO FYOde. (6) Note 2.Itmay beproved that formula (5)holds also forcurves that are crossed byvertical lines inmore than one point (in particular, forclosed curves), provided that both derivatives @’(¢) and 4’(f)’arecontinuous atallpoints ofthecurve. Example 2.Compute the length ofthe hypoeycloid (astroid ecco!t,yeast. Solution, Since the curves symmetric about both coordinate axes, weshall first compuie thelength ofafourth part ofHtlocated inthefrst quadrent, Weta a .aacostfant dy3asin*fcos HYsaintcos 15 ase 450 Geometrie and Mechanical Applications ofthe Definite Integral Theparameter ¢willvaryfrom0to%..tence nn (SRT ETOPTEP nfSOTA = a inttefsintcoseat—3a P|,=%;560 Note 3.Ifaspace curve isrepresented bytheparametric equations x=, Y=¥O, 2=2(0 an) where a<ft<B (see Sec. 1,Ch. IX), then thelength ofitsarc isdefined (inthe same way asforaplane arc) asthelimit whichthelengthofaninscribed broken lineapproaches whenthelengthofthe greatest segment approaches zero. Ifthefunctions @(t), (6), and %(f) are continuous and have continuous derivatives on theinterval [a,BJ,then thecurve hasadefinite length (that is, ithas the above-mentioned limit) which iscomputed from the formula 8 s=\ VieOFF OFF OFat. (u) This result weaccept without proof. Example 3.Compute the are length ofthe helix smacut, poesin', reomt as{vaties from 0to2 Solution. Prom thegiven equations wehave drm—asint dt,dymacostdt, dzmamdt. Substituting into formula (7), wehave sa)Variaraterpataarma)VTFatt=IneVTA 2,The arc length ofacurve inpolar coordinates, Given (in polar coordinates) theequation ofthecurve e=/(0) (8) where gistheradius vector and 0isthe vectorial (polar) angle, The Are Length ofaCurve 451 Let uswrite the formulas forpassing from polar coordinates toCartesian coordinates: x=0050,y=esind. (| Ifinplaceof@weputitsexpression —— : (8)intermsof8,wegettheequations ( x=1(0)c0s®, y=F(0) sin®. pratsecost These equations may beregarded as the parametric equations ofthe curve Fig.236. and wecan apply formula (5)forcom- puting thearclength. Todothis, find thederivatives ofxandy with respect totheparameter 0: Sar(8)cos8—f(8)sind; S$=L(©)sind+7(0)cos8. Thenaz\*(4v\"_op@yt+weget (f6)'+(38)<0@r+V@r=c'+e. Hence, : s=Vere do. ‘ Example 4.Find the length ofthe cardioid @=a(l +cos 0) ig, 230). Vary te,vectorial ange©trom0tox,wegethalfthesoughtfor length. Here, g’=—a sin0.Hence, smo|VarFROROFATARTD do a2|VIFesddom ato[cos5dVtosin$|"ato, 1s 452 Geometric and Mechanical Applications oftheDefinite Integral Example 5.Compute the length ofthe ellipse - seacharen }O<tctn, assuming that a>, Seaton. We take advanige offormula (at eamputing theae Teng thats, the length ofthe are that corresponds toavariation ofthe parameter fromfa20totm: SafVaramrearcorT a= -JVatcorVFOFcota=VaF=(a— FFcostdt= nolVSP eonarma)VIRcstTat, woereb=VRE Hence, sata) Vimweatrae “Theonly thing that cemains isto,compute thelastintegral. Butweknow that 12 nol eapressible byelementary functions see Seertl6,_ Ch. X). “This Ilegra cabcmputed onybyapproximation methods (ySpies re, for exampleFortistance,ifthesem:major axisofanellipses equal(o8andthe semi-minor axisis4,thenA=, andthecircumference ofthe ellipe is saa.) V1-(B) corer. Computing thisintegralbySimpson's rule(bydividingtheinterval [0,3] into four parts) wegetanapproximate value ofthe integral: auey1acotdt=1.208, and sothelength ofthearcoftheentice ellipse, isapproximately equal to ‘25.96 unlls oflength Computing the Volume ofaSolid from the Areas ofParallel Sections 453 SEC. 4.COMPUTING THE VOLUME OF ASOLID FROM THE AREAS OFPARALLEL SECTIONS (VOLUMES BYSLICING) Suppose we have some solid T.Let usassume that weknow thearea ofany section ofthis solid made byaplane perpendic- ilar tothex-axis (Fig. 237). This area will depend ontheposi- tion ofthe cutting plane; that is, it will beafunction ofx: 1 Q=Q2). Wf WeassumethatQ(x)isacontinuous |\7||}[\{%?function ofxandcalculate thevolume |} J—}o"* N ofthe body 4 AK, Draw theplanes x=a, x=x,x—x, % 40% VBviaR=HA=b. rarThese-planes willcutthesolid up ee into layers (slices). Ineachsubinterval #,.,<x<, wechooseanarbitrary point &,and foreach value i=l, 2,..., nweconstruct acylindrical body, thegeneratrix ofwhichis parallel tothex-axis, while the directrix isthe boundary oftheslice ofthe solid 7made bythe plane x=. The volume ofsuch anelementary cylinder, the area ofthe base ofwhich is QE) Gi Sb SH) andthealtitude Ax,,isQE)Ax, The volume ofallthecylinders will be n=DOE) Axe The limit ofthis sum asmax Ax,—+0 (ifitexists) isthe volume ofthe given solid: . Since 9,isobviously the integral sum ofthecontinuous function Q(x) onthe interval acx<b, the indicated limit exists and is expressed bythedefinite integral : v= [Q(x)dx. ay 454. GeometricandMechanical Applications oftheDefiniteIntegral Example. Compute the volume ofthe (riaxial ellipsoid (Fig. 238), BottSafe gen abehed TAY a mer x Fig. 288. Solution. Inasection ofthe ellipsoid made byaplane paraliel tothe geplone and at&distance‘s from itywehave theellipse apr ae a. Bte-4 — =![5V1-43] cVi-= with semi-anes abWi-w aneV1. But the area ofsuch anellipse is6, (ee Example 8,Sec. 1) “Therefore, ewan (1-2). ‘The volume ofthe ellipsoid will be ” vyja 4 nate§(1-2)drone(eg)[*anode Inthe particular case, ambec, the ellipsoid turns into asphere, and we have 4pat ona The Volume ofaSolid ofRevolution 455 SEC. 5.THE VOLUME OF ASOLID OF REVOLUTION Let usconsider asolid generated bythe revolution, about the xaxis, ofacurvilinear trapezoid a4Bb bounded bythe curve y=I(x), the x-axis, and thelines xa, x=). a In this’ case, an arbitrary section ofthesolid made bya plane perpendicular tothex-axis isacircle ofarea Qaay'=nif (xl. +,Applying thegeneral formula pie forcomputing volume {(1), Sec. 4y,wegetaformula forcate: ae jating the volume ofasolid of 4 (elerevolution: ffsteve) ° : | vanlytdermal(oitdx. Fig.239. Example, Find thevolume ofasolid generated bythe revolution ofthe catenary n$(F4e*) about thex-axis onthe interval from x=0 tox= (Fig. 239). Solution. ee ot bo ns a a4 a, xa 2 =? venA(6+e")ante(>Ft*)drm = “, oo rataTyee))_nat(o_o), xa! aE[Gere gae (Fe)ee, SEC. 6.THE SURFACE OF ASOLID OF REVOLUTION Suppose wehave asurface generated bythe revolution ofa curve y=/(x) about thex-axis, Letusdetermine thearea ofthis surface ontheinterval acx<b. We take thefunction /(x) to becontinuous and tohave acontinuous derivative atallpoints ofthe interval (a,6). AsinSec. 3,draw thechords AM,, MyM,, ..., M,.,B, whose lengths aredenoted byAs,, AS, ..., As,(Fig. 240). 456 Geometric and Mechanical Applicatlons oftheDefinite Integral Each chord oflength As, (i=1, 2,..., 1)describes (inthe process ofrevolution) atruncated cone whose surface AP, is AP,=2n¥=tH! As, But = = 2 ayi\*As.=datagi=V1+(SH)ax 9 ty 8 Applying Lagrange’s theorem, weget ie19 Ay)PFny)op rain Ban PED aL) ley where aa dsl aed XpSE<p iHH hence, i! As,VETTE Ax, fig.0. AP=2aUY TEP) Ax,. The surface described bythe broken line will beequal tothesum woviet 7 P=onyVTE) On, orthe sum Pama dUle)+ +Fxd VIFF) Axe, ) extended toallsegments ofthe broken, line. The limit ofthis sum, when the largest segment As, approaches zero iscalled the area ofthe surface ofrevolution under consideration. The sum (1) isnot the integral sum ofthe function 2af(x) VIFF OP, 2) because theterm corresponding tothe interval [x;-,, x,involves several points ofthis interval x,_,, x,&.But itispossible to prove that the limit ofthe sum (i)isequal tothe limit ofthe Computing Work bytheDefinite Integral 457 integral sum offunction (2); that is, Pahima YU) +h VIFE GDda =lima |DE) VTFT GFOx or ’ P=2n\f(x)VIFF de. (3) Example. Determine thesurface ofaparaboloid generated byrevolution about thex-axis ofanareofthe parabola’ y*=2px, which corresponds tothe Yarlation of©from #'=0 tox=a! = Ve iearat 20. tetps-Vi, y=. VIR=Vivb= fe. Solution. By(3)wehave pata|VibeEEEanonV5[VFA rm =V5Seep $l teetor—s'h), SEC, 7,COMPUTING WORK BY THE DEFINITE INTEGRAL Suppose amaterial point Mismoving inastraight line Os under aforce F, and the direction ofthe force coincides with thedirection ofmotion. Itisrequired tofind thework performed bytheforce Fasthe point Mismoved from s=a tos=b. 1)IftheforceFisconstant, thentheworkAisexpressed bythe product ofthe force Fbythe path length: A=F(b—a), 2)Letusassume that theforce Fisconstantly varying, depend- ingontheposition ofthe material point; that istosay, itisafunction F(s)continuous ontheinterval as<b.Divide theinterval (a,b]into narbitrary parts oflength As, ASy soos ASpy then ineach subinterval {s,.., 5]choose anarbitrary point & and replace the work ofthe force F(s) along the path 488 Geometric'and Mechanical Applications oftheDefinite Integral As,(i=1, 2,..., m)bytheproduct F(&) As, This means that within the limits ofeach subinterval we take theforce Ftobeconstant: weassume F=F(&).. Here, theex-pression F(&;)As;willyieldanapproximate valueoftheworkdone bythe force Fover the path As;(for asufficiently small 4s,), and thesum An=DF(Ei)As; will bethe approximate expression ofthe work ofthe force F over the interval [a, 6]. Obviously, A,isanintegral sum ofthe function F=F(s) ontheinterval{a,6).Thelimitofthissumasmax(As;)—+0 existsandexpresses the.work oftheforceF(s) Aover the path from s=a los=6: ‘ |} >ql A=F(syds. a) s Example 1.Thecompression S$ofahelical spring. isproportional tothe applied. force -F. Compute thework atthe force Fehen thespring i im, wompresied 8em,itaforce, ofonekilogram 7 7feeulred tocompress em(Fig.2) Fig, 241 Solution. “itisgiven that theforceFandthe a.341. distance covered $“areconnected bytherelation FonS, where &isaconstant, Let usexpress Sinmetres and “Fin kilograms: Wien S=00L, Fe,thatis,1=2-0.01, whence k=100,F=100S.By(1)wehave AmFons4s=100$|"0.25sitegammetre Example 2,The force Fwith which anelectric charge e,repulses another charge ey(olihesame sign) ata"distance ofris expreséed byiheformula rant, where &isaconstant. * Determine thework done byaforce Finmoving thecharge ¢from the point A,(ata distance olrf,teom e)toAy(atadistance’ cffyfrom ,)Stuming haoelated afthepolitAyatheagin Solution. From formula (1)we have ¢ ees Lyn 1 1A-fateen tet[famn(t-4)- Coordinates oftheCentre ofGravity 459 When roo, wehave Whengat,A=kLL,Thisquantityiscalledthepotentiofthefeld generated bythe charge ¢- SEC. 8.COORDINATES OF THE CENTRE OF GRAVITY Suppose onanxy-plane wehave asystem ofmaterial points Pye W)i Palen Yadr- osPatan Yo) with masses im,my, «25 Myo Theproducts xm;andy,m;,arecalledthestaticmoments of themass m;relative tothey-and x-axes. Wedenote byx,andy.thecoordinates ofthecentreofgravity ofthe given system. Then, aswe know from mechanics, the coordinates ofthe centre ofgravity ofthis material system will be defined bythe formulas em Fg My an Ye MyMgtoeMy S We shall _use these formulas infinding the centres ofgravity of various figures and solids. 1,The centre ofgravity ofaplane line. Let there beacurve ABgiven bytheequation y=f(x), a<x<6, and letthis curve be amaterial line. Let the linear density*)ofsuchamaterialcurvebey.Divide the line into nparts oflength As,, As,,..., As,. The masses of these parts will beequal totheproduct oftheir’ lengths bythe (constant) density: Am,=yAs;. Oneach part ofthe are As;take *)Linear density isthemass ofunit length ofagiven line. Weassume that the linear density isthesame inallportions ofthecurve. 460 Geometric and Mechanical Applications oftheDefinite Integral anarbitrary point with abscissa —;. Now representing each part ofthearcAs;bythematerial point p;(E;, F(E,)] with mass yAs; and substituting into (1)and (2)§;inplace ofx;,/(E,) inplace ofy;,and thevalue ofyAs, (the mass oftheparts As,) inplace ofm;, weobtain approximate formulas fordetermining thecentre ofgravity ofthearc: neqe,y,weDLvas, yas Dros Ifthe function y=/(x) iscontinuous and has acontinuous deri- vative, the sums inthe numerator and denominator ofeach frac- tion have, asmax As;,—+0, limits equal tothelimits ofthecor- responding integral sums. Thus, thecoordinates ofthecentre of gravity ofthearcareexpressed bydefinite integrals: > > “ fea feVTE ae =p -=4——_. ay Ses VIFF aes . . Stemas (ie)VTFFFOVax y= == @’) fe SvTePmae Example 1.Find thecoordinates ofthecentre ofgravity ofthesemi-circlest4roa!situatedabovethex-axis.Solution. Determine theabscissa ofthecentre ofgravity: =Veoe, Hot ise WY"gy8ar, VOR, Hoh, aV4 dampoh, Coordinates ofthe Centre ofGravity 461 Find theordinate ofihe centre ofgravity: varsyaadea(de i~af_2at_2a tenae ee 2,The centre ofgravity ofaplane figure. Given afigure bounded bythelines y=f, (x), y=f,(x), #=a, x=6, which isa material plane figure. Wecon- sider constant the surface YA density, which isthemass iA aofunit’ area ofthe surface. \y-r00 Itisequal to6forallpartsoftheFigure. ol 00-Divide thegiven figure by. yhoo straight lines,x—a,x=,,... x=x,= into strips of‘width Ax, Dx, «+ AXq. The massoféachsirip willBeequalfothe product ofitsarea by i_—«thedensity 8.Ifeachstrip POGK ReeFe isreplaced byarectangle Fig.242. (Fig. 242) with base Ax; andaltitude f,(&,)—f, (6),where §,="=4*!, thenthemassofa strip will beapproximately equal to m= 91,GIA EM de G1, 2.041). Thecentreofgravity, ofthisstripwillbe situated approxi- mately inthecentre ofthe appropriate rectangle: (Dees Y=BOFEO, Now replacing each strip byamaterial point, whose mass is equal tothe mass ofthe corresponding strip and isconcentrated althe centre ofgravily ofthis strip, we find theapproximate value ofthecoordinates ofthecentre ofgravity oftheentire figure [by formulas (1)and (2)]: BO[fe(Bh Elax, _~§Ot—hEDan” FL G+hC19 GhGolan Uses Doh ea—henan . 462 Geometric andMechanical Applications oftheDefinite Integral Passing tothelimit asAx,—+0, weobtain theexact coordinates ofthecentre ofgravity ofthegiven figure: : ieJeteer—tenex $fVecofethdf,code 25 Oo > Juse—funee Jtee@—henae yiray These formulas holdforanyhomo.*geneous (thatis,having ‘constant density atallpoints) plane figure. = Weseethat the coordinates ofthe centre ofgravity areindependent of a-xthe density 6ofthe figure (§was cancelled out intheprocess ofcom- putation). Example 2.Determine thecoordinates of thecentre ofgravity ofasegment oftheFig.243. parabolagtdcuofbytheaaahTneyaa (Fig. Solution, inthiscase/,(x)= Vax, f,(x)=— Vax; therefore 2(eVard 2o oye 4_i} Zoya se,e+ -5_,—-43_-4.Tye hae afyma *Veae*k ge ye=0 (since thesegment issymmetric about thex-axis). Exercises onChapter XII Computing Areas 1.Findtheareaofafigureboundedbythelinesy*=9e,y=3r.Ans.+2.Findtheareaofafigureboundedbytheequilateral hyperbolaxy=a", theve-atis, andthelines#26,boo, Ans.aInt. yen Find teagenofgui’TyingBetween thecurey=4—s" andthe sanis, Ans. 105. 2.2 24.Findtheareaofafigurebounded bythehypocycloid x?-+y>=a®,Ans,0%, Exercises onChapter XII 463 5.Findtheareaof«figurebounded bythecatenary y=(«#40*), thesani, thepats, andthesraight line2m,Ans.1) 6,Findtheareaofafigurebounded bythecurve y=2%,theliney=8, andtheyratis. Ans.12. * , ’ 1,Fig’theareaofregion bounded byoneloopofasinewaveandthe 8.Findtheareaofaregion lyingbetween theparabolas y*=2px,x*=2py. Ans.$pt9.Findthetotalareaofafigureboundedbythelinesyaa,y=2x,yx.Ans.3. 10,Find theareaofaregionboundedbyoneareofthecycloidx=a(t—sin), y=a(l—cost) and the xaxis. Ans. Saat Tl,Find the atea of figire bounded bythehypocycloid x=a.cos'¢, y= masiat, Ans.$nat,antZfitd theaeaftheentireregionboundedbythelemniscategt=a cos2p.ns. at, 13.Compute thearea ofaregion bounded byoneloop ofthecurve @=asin2p. Ans. 14,Compute thetotalareaofareglonbounded bythecardioid ¢=a(1— e059) Ans, 0%, .15.Findtheareaoftheregionbounded bythecurve@=acos@. Ans.=. 16.Find theareaoftheregion bounded bythecurve gmacos2p. Ans, 32, 17,Findtheareaofthereonboundedbythecurve@=cos3,Ans. 18,Findthearenoftheregionbounded bythecurvegacond.Ans.2, Computing Volumes 19,Theellipse£54+-S5—1revolvesabouttheaxis,Findthevolumeof thesolidofrévolution. Ans.4nab*, 20,The segment ofaline connecting the origin with thepoint (a,6)ree volvesaboutthey-axis.Findthevolumeoftheresultingcone.Ans.--na%b 21.Find the, volume ofatorus generated by therevolution ofthecircle w#4(y—b)t=at about theeaxis (itis assumed that ba). Ans. Ona‘2!Theareaboundeddythelineshae,and=aFepovisboutthe xaris. Find the volume ofthe solid ofrevolution. Ans. spa" 464 Geometric and Mechanical Applications ofthe Definite Integral a2 2 2,Afigurebounded bythehypocycloid x?+-y®=aisrevolved aboutthexaxis.Findthevolumeofthesolidofrevolution, Ans.ST. 2%.Afigure bounded byone arcofthesine wave y=sin and thex-axis isrevolved about’ thea-axis, Find thevolume ofthesolid ofrevolution. ans. . 2.Aigurebounded bytheparabola gtx andthestraight lingx=revolvedaboutthex-axis,Findthevolumeofthesolidofrevolution. Ans.92m. 36igurebounded bythecurvey=ae* andihestraightLinesy=0,s=1, isrevolved about the x-axis. Find the volume of the solid. of revolution. Hog Ans.(1).2.Afigureboundedbyoneaeofacyloid=atala).ya(ene)and the x-axis isrevolved about the x-axis. Find the volume ofthe solid evolution. Ans. Sx’. 28.The same figure asinProblem 27isrevolved about they-axis. Find the volume ofthe solid ofrevolution. Ans. 6x%a*. 28:ThesamefigureasinProblem 27Isrevolved aboutastraight linethat isparallel tothegraxis andpasses through thevertex ofacycloid. Find thevolumeofthesolidofrevolution. Ans."2"(9u*—16).. go.ThesameigureasinProblem 27fsrevolved about¢straight, linepa rallel tothe.x-axis and passing through thevertex ofacycloid. Find thevo- ume ofthe solid ofrevolution, Ans. 7a'a*, Sl.Acylinder ofradius iscut byaplane that passes through the dia- meter ofthe base atanangle atothe plane ofthebase. Find the volume of thecut-offpart.Ans.3R*tana. 32.Find avolume that iscommon tothe two cylinders: 2*+y*=R4,y?-+ pateR* Ans.PRY 39,Thepointofintersection ofthediagonals ofasquareisinmotionalong the diameter ofacircle ofradius a;the plane inwhich thesquareliesremains perpendicular tothe plane ofthecircle,whilethetwooppositeverticesofthe Square move along the circle (asa tesult ofthis motion, the size ofthesquareobviously varies),Findthevolumeofthesolidgenerated bythismovingsquare.Ans.St.,,34,Compute thevolumeofasegmentcutofftheelliptical paraboloidSotGnbytheplanex=a.Ans.natV5} 36.Compete thevolumeaf»soldbounded bytheplanes20,40he@V2 cylindrical surfacesxt=2pyand2*=2prandtheplanex=a.Ans,@18 sind y 0 Pl rad (infirst octant).36.4straight lingisjnmotionparalleltotheyz-plane,andcutstwoel-rites+P, Ztfetlyinginthexy:andaa-planes, Compute thevo- lumeofthesolidthusobtained. Ans.abe Exercises onChapter XII 405, ‘Computing Arc Lengths 22 2 37.Find theentire length ofthehypocycloid x*+y* —a®. Ans. 6a. 38.Computethearclengthofthesemeusical parabolaay*=x*fromtheGrigintoapointwithabscissax=Sa,Ans.a, 40,Findthearelengthofthecatenary y=(e#be”#)fromtheorigin tothepoint(x,y)-Ans.$0<0®)=VpF=at 40,Findthelengthofoneaeofhecycloid=a¢—sin9,ya(Leos. 41.Find thelength ofanareofthecurve yin within the limits from reVBtoVE. AnsLyin 42, Find the arc length ofthe curve y=t—Incos between x=0 and rot. Ansinten3.43,FindthelengthofthespiralofArchimedes@—=apfromthepoletothe endofthefrstlop.Ans.xaVFI ©inOn+VIER, 44,Find the length ofthespiral g=e% from the pole tothe point (@,¢). ans,VEEge8THe45,Findtheentirelengthofthecureqesin“®.Ans.$e 46.Findthelengthoftheevoluteoftheellipsex=cost’, y=C'sintt. Ans129,42.Findthelengthofthecardoldo=a{1-+cos).Ans.48:Findthaelenginof theInvite oftbecicle*a(eos9+9ng) y=a(sing—9cos9)from@=0to=.Ans.Sagt Computing Areas ofSurfaces ofSolids ofRevolution 49.Find thearea ofasurface obtained byrevolving theparabola y*Aas sbaut the ani, fom the enginOtoapntwithabax—sa.Ansya £0.Find thearea ofthesurface ofacone generated bytherevolution of alinesegment y==2e from x=0 tox=2 a)About thex-axis, Ansr8x V5. byAbout they-axis. Ans xV5 51Find theaven ofthe surlace ofaforus obtained byrevolving thecircle aE G—o)teat about theeaxtae Ans. dntab 52! Find the area ofthesurface ofasolid generated byrevolving acar sioidabouttheSani Thecardio isrepresenedby thepaar equations£0(2.c0sg—c0s29),y=a(2sing—sin 29).Ans.+nat, 466 Geometrle and Mechinleal Applications oftheDefinite Integral 58,Findtheareaofthesurface ofasolid obtained byrevolving one,are ofaeycloidx=a(t—sint), y=a(l—cost) aboutthexaxis, Ans,“2, 54,The areofaeycloid (see Problem 53) isrevolved about they-axis. Find. the surface ofthe solid ofrevolution. Ans. 16xta", 85,The arc ofacycleid (see Problem 83)Isrevolved about atangent line parallel tothex-axis and passing through the vertex. Find thesurface ofthe‘sdna® ° solid ofrevolution, Ans,222" 56,Theastroid xmasin'f, yacos!! Isrevolved about thex-axis, Findthesurfaceofthesolidofrevolution. Ans.1°22", 57.Anareofthe sine wave y=sinx from x=0tox—2risrevolvedabout theraxis.Findthe,surfaceoftheslid.ofrevolution. Ans.4n(V2++n(V2EN. 58,Theellipse{+471(a>6)revolvesaboutthex-axis.Findthesur- faceofthesolidofrevolution. Ans.2n6*4-2nab SE88¢,whereemaLEP Varlous Applications ofthe Definite Integral 58,Findthecentreofgravity oftheareaofone-fourth oftheellipse HPie ins,42,4b Fthal a0.yao. aw.FF. 60,Find thecentre ofgravity ofthearea ofafigure bounded bythe pa- tabolax*4+4y—16=0 andthex-axis.Ans.(0,+). G1.Findthecentreofgravityofthevolumeofahemisphere. Ans.Ontheaxisofsymmetry atadistance -}Rfromthebase. 62,Find thecentre ofgravity ofthe-surface of hemisphere. Ans. Onthe axisofsymmetry atadistance %fromthebase. 63, Find the centre ofgravity ofthe surface ofacircular right cone, theradiusofthebaseofwhichisRandtheallitudeA:Ans.Ontheaxisofsym:metryatadistance -fromthebase.4.Thefigureisboundedbythe lines y=sinx(0<x<A), y=0. Find the centreofgravityoftheareaofthisfigure.ans.($,4):65.Findthecentreofgravityoftheareaofafigureboundedbythepa-rabols yes, 20.“ (8,9) i ‘66. Find the’ centre ofgravity of thearea ofacircular sector with central angle2aandradiusR.Ans.Ontheaxisofsymmetry atadistance2R28# from the vertex ofthe sector. 67.Findthepresse ofWateronarectangle vertically submerged Inwa: teratadepth‘ofSmifiisknownthatthebasefs8metres,thealtitude,12metres andtheupperaseisparallel tothefreesurace oftheWate. Ans, Exercises onChapter XII 467 68. The upper edge ofacanal lock has the shape ofasquare with asideof8mlying’onthesurfaceofthewater.Determinethepressureoneachpart af‘the,lockformed. bydividing the’square byoneaf“itsdiagonals ‘Ans, 85,835.33 hg 170,686.67 kg.‘2:Computetheworkneededtoarp,thewateroutof«hemispherical vessel ofdiameter 20metres. Ans. 2.9%10% kg-m. 70.Abodyisinrectilinear motion according {0thelawx—cl?, where= isthepath length traversed inlime t,e-=eonst: Thetesistanceofthemedium isproportional tothesauare ofthevelocity, and istheconstant ofpro-portionality. Findtheworkdonebytheresistancewhenthebodymovesfromthepointx=0tothepointx=a.Ans.2&j/@@ 71.Compute the work that has, tobedone inorder topump aliquid of densiiy yfrom areservoir having theshape ofacone with’ vertex peintingdown,altitude 1andradiusofbaseR.Ans.SVRH|72.Awoodeniloatofcylindrical shapewhosebasalarea.$=4,000 cm?andaltitudeH7==60cmisfloatingonthesurfaceofthewater,Whatworkmustbedone pulltheoatuptothe surface? (Specific weight oftheWood,0.8.Ans,PES292kgm. 73.Compute theforcewithwhichthewaterpresses onadamintheform ofanequilateral trapezoid (upper base a=6.4 m,lower base b=4.2 m,alli fade =Sm)Ans222m 74,Find the axial component Pkgoftotal pressureofsteamonthesphe- riealbottom ofaboiler Thediameter ofthecylindrical partoftheboilerisDmm,thepressureofthesteamintheboilerisPkglem’,Ans,PaxSPD" 75.The end ofavertical shaft ofradius rissupported byaMat thrust bearing. The weight oftheshalt Pisdistributed equally over theentire surefaceofthesupport.ComputethetotalworkofIrietioninonerotationofthe shalt,Coefficient offriction ts.Ans.+appr. 76._A vertical shaft ends inathrust pin having the shape ofatruncated cone, The specific pressure ofthe pin on the thrust bearing isconstant and equal toP.The upper diameter ofthe pin isD,the lower, d,and the angle sil'the vertex ofthecone ts28.Coefficient offriction, jt. Find the work affrictionforonerotationoftheshaft,Ans.2H(pray, 71.Aprismaticrodoflength1isslowlyextendedbyaforceIncreasing from0to'Ps0thatateachmomentthetensileforceisbalancedbythefore cesofelasticity oftherod. Compute thework Aexpended bytheforce onfension, assuming thatthetension occurred within thelimitsofelasticity. Flevtheronesetional arenoftherod,and£isthemodulusofelasticity”ofHint. Ifxfstheelongation oftherodandfisthecorresponding force, thenf=FEx, Theelongation duetotheforcePisequaltoalmte. Pal_Pi Ans,A=PAleBe 78.Aprismatic beamissuspended vertically andatensileforcePisap- pliedtoitsTowerend,Compute theelongation ofthebeamduetotheforce OFilsweight andfotheforce Pifitisgiven that theoriginal length ofthe 468 Geometric and Mechanical Applications oftheDefinite Integral beam isJ,the cross-sectional area F,the weight Qand the modulus ofelas- Willyofthematerial £.Ans,aim@t20) 19,Detecine thetimeduring whichsTiquidwilfowotof&prismatic vesel filed fo's height 1The erosrsecitonal area ofthe vessel isFthe silt oer thexl easy campetea rom. termo=uVUBR,wherepisthecoelfcient ofviscosity,gIstheacceleration of Gravity, caeA’iothedistance from:theaperture tofhelevel‘oftheliquids Ans,TPH /wiVaiwlVg 2,Delisming theisharge Qe quantityofwalerowinginnittine over aspilivay ofreclangular cross seclion- Height ofspilivag, hywidth, 6 Ans.Q=%wohV%h 8,Delerminethedischargeof,water@flowingfromaside rectangular opening ofheight andwih 0,iftheheight ofthepensurfoe ofthever terabovethelowersideoftheopeningisH.Ans.Qu=208-28?(a). CHAPTER XII DIFFERENTIAL EQUATIONS SEC. 1.STATEMENT OF THE PROBLEM. THE EQUATION OF MOTION OF ABODY WITH RESISTANC OF THE MEDIUM PROPORTIONAL TO THE VELOCITY. THE EQUATION OF ACATENARY Let the function y=/(x) reflect thequantitative aspect ofsome phenomenon. Frequently, itisnot possible toestablish directly the type ofdependence ofy’on x,butitispossible togive therela- tionship between xand yand thederivatives ofywith respect to ary’, y',...,y™. That is,we are able towrile adifferential ‘equation. From therelationship established between thevariable x,yand thederivatives itisrequired todetermine thedirect dependence ofyonx;that is,tofind y=f(x) or,aswesay, tointegrate the differential equation. Let usconsider two examples. Example I”Abodyofmais isdropped fromsomeheight. Itisquired toestablishthatlawaccordingfowhichthevelocityvwill-varyasthebodyfalls, if,inaddition tothe force ofgravity, the body isacted upon bythe decelerating force oftheaie, which isproportional tothevelocity (with cone stant ofproportionality A);inother words, itisrequired tofind o=f (0). Solution. ByNewton's second law do mag where48istheacceleration ofamoving body(thederivative ofthevelocity with respect totime) and Fistheforce acting onthebody inthedirection oI motion. This force isthe. resultant “of two forces:” the force of gravily mg’and the force ofairresistance, —ko, which hasthe minus sign be- Eause itis inthe opposite direction tothat ofthe velocity. And sowehave dvmGf—mg— ko. w Thisrelation connects theunknown function vanditsderivative 42,which isadifferential equation inthe unknown function v.Tosolve the differen {ial equation istofinda function v=[ (0)such. that identically satisfies the given differential equation. There isaninfinilude ofsuch funetions. The stu. dent can easily verify that any function ofthe form i ace! mBv=Ce + 2 satisfies equation (1)nomatter what the constant Cis.Which one ofthese 47 Diferentiat Equations functions yieldstheSoughtfordependence ofvon{?Tofinditwetakead- vanlage ofasupplementary condition: when thebody was dropped itwasim parted’aninivelocity ta(which may,beseroae8particular cae;wea Kime this initial velocity t9beknown. But then the unknown function v= [gi sehthatwhen(0(whenmation begin) thecondition v=ey isfulhited. Substituting ¢=0, vv, into formula (2), wefind me acatt, whence =o,cau. Thus, theconstant Cisfound, and thesought-for dependence ofvonfis mg) «om 8 7 om(mE. e Itwill benoted that ifk=O (the air resistance is.absent ornegligibly satYothat$ecandisregard), thenwehavearesultamar romys ies*): y r v=vytat. @ This function satisfies thedifferential equation (i)and theInitial condition: o=v, when f=0.vgBasile2Afeniblehomogeheots thread issuspended attwoends.Findtheequationnotfneteurethatiteaeibesunder,“Hesown ‘weight(itisthesameasanysuspended ropes,Thtchain,soforstancetheealerpiiareck 1a]ofafank between two supporting rollers). Solution, LetM,(0,8)bethelowestpoint as ofthe. thread, and M” an arbitrary. point(Fig2,Letconsiderpartofthehead,MyM.Thispartisinequilib, theestat 0 ‘% 1)thetension 7,acting along thetangentPig:oe fp.thepoint"Mandforminganangie@with 2)thetension HatM,actinghorizontally; 3)theweight ofthe thread ysacting vertically dowhwards, wheresisthe lengthoftheareMM and ye linear specific weight ofthethreadfreakingupthetensionFintohorizontal andvertiealcomponents, weget theequations ofequilibrium: Teosg=l, Tsing=ys Dividing the terms ofthe second equation bythe corresponding terms ofthefirst,weoblain " onan tangaTes. ° *)Formula (2") can beobtained from (2') bypassing tothe limits a im[(o,—™E) e4TE]mo, tis,[(oe—9F) FF] moot Statement ofthe Problem a7 Now suppose thatthe equation ofthe sought-or curve may bewrilten intheform'y=/(a),Here,)isan_unknownlunetion thalhistobe.found. Westibe’noted inate er tang=/'(x)=92. Hence, fy_1rota a wheretheratio4isdenotedintermsofa, Dierentiate Uh sides of(4)with respect tox: ay _1dssyn tae, © But, a8weknow (ee Sec, 1,Ch. VI), _/ im)4-f nf).Substituting thisexpressionintoequation(8),wegetthediferentatequ tionofthesought-for curve: “ as dyV(#): (6)m7 I+(a): (0) Wexpreses therelationship between thefistandsecond derivatives of thet utiewe function Wittout going into tiemethods ofsolving the equations, weshall note that any function oftheform +(44a), -(440) osGr), Yeo ) satisfies equation (6)foranyvaluesthatC,andC,mayassume. Thisisevi- dentifweputthefirstandsecond derivatives ofthegiven function into(6), Weshall indicates without prool, that thse Tunetlong Gor aierent CeOO Efthas alpie slulion ofequation (3) ‘Thegraphs ofallthefunctions thusoblained arecalled catenaries, ictdsnowfindoutbowoneshould choore theconstants CyandCysoas toSehttowndathowoneaboucotethe,constantsandCygo, ShaveorPSOE potnt athe cateney orcuples thelowest posse eation thetangent bersorzontal on.Alteornthatatthps the ordinate tsequal to6,yobsrom(7)wefind 4 1(d°6,-(4"6)) pot(ate . Pulling 290hee,weeanOmGS-e°S, Hee,Ca UEtheordinate ofthepoint M,isb,then ye=® when x=0. an Diperentiat Equations Fromequation (7)wegetb=2(141)4C, assuming #=0andCy=0, whence C,=0—a. Finally wehave Equation (7)assumes avery simple form ifwetake theordinate'of M,equal toa. Then the equation ofthe eatenary is SEC. 2. DEFINITIONS Definition 1.Adifferential equation isone which connects an independent variable, x,anunknown function, y=/(x), and its derivatives y’,y’, -.-. y". Symbolically, adifferential equation may bewritten asfollows: FO, WyY'sYseee YO or dyty ay) F(x, 9%, o8, w Gh)=0. Ifthesought-for function y=/(x) isafunction ofone indepen- dent variable, then the differential equation iscalled ordinary. Weshall deal’ only with ordinary differential equations *). Definition 2.The order ofadifferential equation isthe order ofthe highest derivative which appears. For example, the equation yf—2xy°+5=0 isanequation ofthefirst order. *)Inaddition toordinary differential equations, mathematical analysis makes astudy ofpartial differential equations. Such anequation isarelation Between ‘anunknown function 2(that is,dependent. upon two otseveralvariables%y,=.+thesevariables x,y,'..andthepartialderivativesPe csoe Seyaeee The following isanexample ofapartial differential equation with unknown function2(x,9):ety8" ItIseasy toverify that this equation issatisfied bythe function 2=x%y* {and also by2multitude ofother functions) ithis course weshall have little todowith partial differential equa- tions. First-Order Diferentiat Equations a3 Theequationy'+hy’—by—sinx=0 isanequation ofthesecond order, etc. The equation considered inthepreceding section inExample 1 isanequation ofthe first order, inExample 2,one ofthesecond order. Definition 3.The solution orintegral ofadifferential equation isany function y=/(x), which, when put into the equation, converts itinto anidentity. Example 1.Let there beanequation ayGity=o The functions y=sinx, y=2cosx, y=3sinx—cosx and, ingeneral, functions ofthe form y=C, sinx, y=Cyosx y=G,sinx$C,cose afesolutions ofthe given equation forany choice ofconstants C,and Cy this isevident ifweput these functions into the equation. ‘Example 2.Let usconsider the equation yx—s—y=0. Is solutions are all functions ofthe form ya24Cr phereCisanyconstant, Indeed, diferentiating thefunctions y=s*4+Cs, we yarx+e. Putting the expressions foryand y’into the Initial equation, wegetthe identity (40)2-2"2—Cx=0. Each ofthe equations considered inExamples 1and 2has aninfnitude of solutions. SEC. 8.FIRST-ORDER DIFFERENTIAL EQUATIONS (GENERAL NOTIONS) 1.Adifferential equation ofthefirst order isoftheform F(x, ysy')=0. ay Ifthisequation canbesolved fory’,itcanbewritten intheform ¥=Ts, y). ay Inthis case wesay that the differential equation issolved for thederivative. For such anequation thefollowing theorem, called thetheorem oftheunique existence ofsolution ofadifferential equation, holds. m DiGerentiat Equations Theorem. /fintheequation ¥=1e 9 thefunction f(x, y)and itspartial derivative with respect toy, t.farecontinuous insomeregionDinanxy-plane containing some point (x,, y,), then there isonly onesolution tothis equation y=Q(x) which satisfies thecondition x=x,, y=y,. The geometricmeaning ofthetheorem consists inthefactthatthere exists one and only one such function y=q(x), thegraph ofwhich passes through thepoint (x,, y,). Itfollows from this theorem that equation (1’) hasaninfinitude ‘ofvarious solutions [for example, asolution thegraph ofwhichpassesthrough (x,,y,);another solution whosegraphpassesthrough(x,,4); through “(x,, y,), ete., provided these points lieinthe region D). The condition that forx=x, thefunction ymust beequal tothegiven number y,iscalled theinitial condition. Itisfrequent- lywritten inthe form Ylxen=Yoe Definition 1.The general solution ofafirst-order differential equation isthefunction y=9(x, C), @) which depends onasingle arbitrary constant Cand satisfies the following conditions: a)Itsatisfies thedifferential equation forany specific value of the constant C. b)Nomatter what theinitial condition y=y, forr=x,, that is,Wes =Yur itispossible tofind avalue C=C, such that the function y=@(x, C,) satisfies the given initial condition, Itis assumed here that the values x,and y,belong tothe range of thevariables xand yinwhich the conditions ofthe existence theorem are fulfilled. 2.Insearching forthegeneral solution ofadifferential equation we often arrive atarelation like D(x, y,C)=0, (2) which Isnot solved fory.Solving this relationship fory,weget thegeneral solution. However, itisnotalways possible toexpress -ytrom (2) interms ofelementary functions; insuch cases, the general solution isleftinimplicit form, First-Order Differential Equations 475, Anequation oftheform (x, y,C)=0 which gives animplicit general solution iscalled thecomplete integral ofthedifferential equation. Definition 2,Aparticular solution isany function y=9(x, C,) which isobtained from thegeneral solution y=@(x, C),ifinthe latter weassign tothearbitrary constant Cadefinite value C=C,. Inthis case, the relation O(x, y,C,)=0 iscalled aparticular integral ofthe equation. - Example 1.For the first-order equation yi ans thegeneral solution isafamilyoffunctions y=; thiscanbechecked by simple aubsittion intheequation, elusfind aparticular solution that will satisty the following inital condition: yy=l when %=2. Putting thesevaluesintotheformula y=, wehave1=-ZorC2 Consequently, thefunction y=willbetheparticular solution weare secking From the geometric viewpoint, thegeneral solution (complete integral) isafamily ofcurves in'a coordinate plane, which family depends onasingle arbitrary constant C(or, asitiscommon tosay, onasingle parameter C).These curves arecalled integral curves'of thegiven differential equation. Aparticular integral is associated with one curve ofthis family that passes through a certain given point ofthe plane. Thus, inthe latter example, the complete integral isgeometri- callydepicted byafamily ofhyperbolas y= whilethepartic- ular integral defined bythegiven initial condition isdepicted byone ofthese hyperbolas passing through thepoint M,(2, 1). Fig. 245 shows the curves ofafamily that are associatéd with certainvaluesoftheparameter: C=¥,C=1, C=2,C=—1, ete. Tomake thereasoning still more pictorial, weshall from now ‘onsay that notonly thefunction y=@(x, C,) that satisfies the equation but also the associated integral curve isasolution of theequation. We will therefore speak ofasolutionpassingthrough the point (x,, y,). Note.Theequation $¥=—¥ hasnosolution passing through a point lying onthey-axis (see Fig. 245). This isbecause the right 176 Digerentiat Equations side oftheequation isnotdefined forx=0 and consequently is not continuous. Tosolve (oraswefrequently say, tointegrate) adifferential equation means: ‘a)tofind itsgeneral solution orcomplete integral (iftheinitial conditions are notspecified) or b)tofind aparticular solution oftheequation that will satisfy thegiven initial conditions (ifsuch exist). y Chel CnYe mall cat Onde x Cn-2 cxt)|(t=-tohhe Fig. 248. 3.Letusnow give ageometric interpretation ofafirst-order differential equation. Let there beadifferential equation solved forthe derivative y GBale 9 ay and lety=@(x, C)bethegeneral solution ofthis equation. This general solution determines the family ofintegral curves inthe xy-plane. For each point Mwith coordinates xand y,equation (1’) defines thevalueofthederivative 44,ortheslopeofthetangent line tothe integral curve passing through this point. Thus, the differential equation (1") yields acollection ofdirections or,as wesay, defines adirection-field inthexy-plane. FirstOrder Diflrentiat Equattons ar Consequently, fromthegeometric pointofview,theproblem ofintegrating adifferential equation consists infindingthecurves, ateyoty resSigs\\jy=2x yax,38fogyo" ‘kofXI a) YrVex 1 yerae | ling utr, i Glos, BETISE Se 'eoi Fig. 246. the direction ofthe tangents towhich coincides with the direc- tion-field atthecorresponding points. Fig. 246 shows adirection-field defined bythe differential equation woe deme 4.Let usnow consider the following problem. Let there begiven afamily offunctions that depends ona single parameter C: y=0(% CO) (2) and letonly one curve ofthis family pass through each point of the plane (orsome region inthe plane). For what differential equation isthis family offunctions acom- plete integral? From relation (2), differentiating with respect tox,wefind uy_g=a.% CO). 8) Sinceonlyonecurveofthefamily passesthrough eachpoint oftheplane, forevery number pair xand y,aunique value of Digerentiat Equations Cisdetermined from equation (2). Putting this value ofC-into (3)wefind44asafunction ofxandy.Thisiswhat’yields thedifferential equationthatis oop2on satisfied byeveryfunction of otscop thefamily(2). ota cof Hence, toestablish arela- tionship between x,yand“, oe - that is,towrite adifferential equation whose general solution £-0_, (complete integral) isgiven by %formula (2), one has toeli- minate C from relations (2) and(3). on,cnt Example 2.Findthedifferential Fig.247. equation’ofthefamilyofparabolas pace(Fig.47, Differentiating theequation ofthe family with respect tox,weget dy49ce, Putting thevalueC4, intothisequation fromtheequation ofthefam- ily, weobtain adifferentiable equation ofthegiven family: dydx x” This diferential equation ismeaningful when x0; which Istosay, in any region not containing points onthe y-axis. SEC. 4.EQUATIONS WITH SEPARATED AND SEPARABLE VARIABLES. THE PROBLEM OF THE DISINTEGRATION OF RADIUM Let usconsider adifferential equation ofthe form #=,OhW Om) where the right side isaproduct ofafunction dependent only ‘onxbya'function dependent only ony.Wetransform itinthefollowing manner assuming thatf,(y)40: 1 "Fuahwde, wy Consideringyaknownfunctionofx,equation(1’)mayberegard- edasthe equality oftwo differentials, while the. indefinite Equations with Separated and Separable Variables «9 integrals ofthem will differ byaconstant term. Integrating the leftside with respect toyand theright with respect tox,we obtain y SrmnSr (x)dx+Ckw ‘ La which isarelationship connecting |c/ the solution ofy,the independent aNvariable x,andanarbitrary con- Cy, 7stantC;wehavethusobtained a Regeneral solution (complete integral) ofequation (1). 1.Atype (1’) differential equa- tion M(x)de+N(y)dy=0 (2) Fig.248. iscalled anequation with separated variables. From what has been proved, itscomplete integral is JMG)de+[NWdy=C. Example 1.Given anequation with separated variables: xdxtydy=0. Integrating weget the general solution: ogS4hec, Since theleft side ofthis equation is nonnegative, theright sideI onmegatives Deneting26,inteofCh'wesillkav“Entde#0BHyeCt Thisistheequationofafamilyofconcentric circles(Fig.248)wit centatthecoordinate originandradius Go ta ata) 2.Anequation ofthe form M,(2)N,(y)dx+M,(x)N,(y)dy=0 @ iscalled anequation with variables separable. Itcanbereduced *) toanequation with separated -variables bydividing both sidesbytheexpression N,(y)M,(x): Male)NSW)gy4Mal)Nal) Myo)Me)Tye) mg)Y=? *)Thesetransformations arepermissible onlyinaregionwhereneither 1M, tor My) vanish, 480 DiGerentiat Equations 72 or M »18)eyNa)gy= ea YtHyY= that is,toanequation like (2). Example 2.Given theequation woe aoe Separating variables, wehave ceca vos Integrating wefind eewo (#46,§$--S3+ which Is * Can Infy|=—In]x1-41n]C]*) ofIn}yl=ta |S; c whencewegetthegeneral solution: y=, Example 3.Given the equation (taydetwxdy=0. Separatingvariables we have SEDdegStaymo, (F+1)xt(Z-1)dyno. Integrating we obtain In]x|-+2-4Inlyl—y=C orInxyl-+x—y=C. This relation, isthecomplete integral ofthegiven equation, Example 4.Itisknown that thedecay rate ofradium isdirectly propor- tional toitsquantity ateach given instant. Find thelaw ofvariation ofamassofradium asafunction ofthetimeifat£—-0themassofradiumwasmy ‘Thedecay rateisdetermined asfollows. Letthere bemass mattime f,andmassm-pamattime{++A0.DuringAfmassAmdecays.TheratioAFisthemeanrateofdecay.Thelimitofthisratioasat—r0 imSmaateodldt isthe rate ofdecay ofradium attime ¢. *)Having inview subsequent transformations, we denoted the arbitrary constant byIn|C], which ispermissible since In|C| (when C#0) can take onany value trom —e to+o. Equations with Separaied and Separable Variables 481 Itisgiven that dm at im, o where&istheconstantofproportionality (#>0).Weusetheminussignecause the mass ofradium diminishes with incfeasing time and therefore Heo. nm<0. Equation (4) isan equation with variables separable. Let us’separate the variables: kat. mo Solving theequation weobtain 0 Inm=— kt—In€ Fig. 239, whence in at, m=ce™, © Since at£0 themass of.radium, wasmy,Cmust satisfy therelationship male a6. Putting thevalue of©into (6)wegetthedesired mass ofradium asafune lionoftime(Fig.249): Me mame™. ) ‘The constant &isdetermined from observations asfollows. During time ty lela ofthe original mass ofradium decay. Hence, the following relation: shipisfulfilled: :a atte (iio)m=me whence —Hy=In (179) or aa=—} in(1—8=-i (1-75) - Thus, ithas been determined that for radium k=0.00044 (the unit of measureoftimeisoneyeu) Putting this value ofkinto (6)weobtain mame, Let usfind the radium half-life, which isthe Interval oftime during which’aifoftheoriginal massofradium decays. Putting*inplaceofm 16-9380 482 Diferentiat Equations inthe lat‘er formula, weget anequation fordetermining the half-life 71 m_ -emee Meme whence 0.000447 =—In? or In? Tmgigas=11590years. Note.Thesimplest differential equation withseparated variables isone ofthe’ form 4Geasiey ofdy=f(e)dx. iscomplete integral isofthe form y=Jferaetc. We dealt with thesolution ofequations ofthis kind inCh. X. SEC. 5. HOMOGENEOUS FIRST-ORDER EQUATIONS Definition 1.The function f(x, y)iscalled ahomogeneousJunction ofdegreeninthevariables xandy,ifforany&thefollowing identity istrue: Fx, dy)=O F(x, 9). Example 1.Thefunction f(x,9)—=}/3°FH isahomogeneous function ‘ofdegree one, since fds, 1)={/TPFOIAYPEPRM(8,9) Example 2.[(x, y)=xy—yt,is ahomogeneous function ofdegree two,since(hx)(hy)—(hy)*=9?[xy4). Example8./lxp=2=2 is«homogeneous function ofzerodegre, nceOA—Oy)PH thatis, =Fx, = since COR SEthat is,fs,Alix vor [0x9) =i Definition 2.Anequation ofthefirst order sale, v) w iscalled homogeneous inxand yifthefunction f(x, y)isaho- mogeneous function ofzero degree inxand y. Homogeneous First-Order Equations 483 Solution ofahomogeneous equation. Itisgiven that f(Ax, Ay)= =/(x, y).Putting det inthisidentity, wehave fe,n=r(1, 4). Thus, ahomogeneous function ofzero degree isdependent only ontheratio ofthearguments. Inthis case, equation (1)takes theformay y #-1(1, 4). a Making the substitution y u=2, ory=ux, weget yyanita Putting this expression ofthe derivative into equation (1’), we obtain ‘ weep, w). This isanequation with variables separable: au du___ae xGri Wu orpas. Integrating wefind aa ar Srna SF+e- Putting theratio4inplace ofwafter integration, weget theintegral ofequation (1’). Example 4,Given the equation ayy aap (Onthe right isazero-degree homogeneous function, which means that wo haveshomogeneous equation. Making thesubstitution exwwehave gous Beat efi duu, du wtwhat ima Xai 16° 484 Diferentiat Equations Separating variables weobtain Gowdaa;(41a, Whence, integrating, wefind 1 1—ganlul=injxi+iniCy or—=I]uxCy, Substituting u=, wegetthegenerat solution oftheoriginal equation: ~finicy| 1ispossible heretogat_asanexplicit function ofxinteamsofte mentary functions. Incidentally, itis very easy toexpress xinterms ofyi sayV—3CTHTCH Note. Anequation ofthetype M(x, y)dx+N(x, y)dy=0 will behomogeneous if,and only if,M(x, y)and N(x, y)are homogeneous functions ofthe same degree. This follows from the fact that the ratio oftwo homogeneous functions ofthesame de- gree isahomogeneous function ofdegree zero. Example 5.The equations *Qx43y) dx-+(x—2y) dy=0,(+y")dx—2xy dy=0 are homogeneous. SEC. 6.EQUATIONS REDUCIBLE TO HOMOGENEOUS EQUATIONS Equations ofthe following type are reducible tohomogeneous equations: dy artoyte wdx ax+ byte * Ifc,=c=0, then equation (1)isobviously homogeneous. Now let ¢andc,(oroneofthem) bedifferent from zero, Change theva- riables: xaxth yayth Then dy_dn 2aedy* ® Equations Reducible: toHomogeneous Equations 485 Putting into(2)theexpressions x,y,and’2,weobtain dy,any+bytoh-+oh-be @)Gr, Gx, FO FORE ORES Choose Aand &sothat the following equalities are fulfilled: ah-+bk-+0=0,ah+b,k+c,=0. } @ Inother words, define hand &assolutions ofasystem ofequa- tions (4). Equation (3)then becomes homogeneous: dy_any by, ano” Solving this equation and passing once again toxand yby formulas (2), weobtain the solution ofequation (1). The system (4)has nosolution if ab/Lla,o|=° ie.,ab=a,b. Buttate, thatis,a,=Aa, b,=46,. and, hence, equation (1)may betransformed to * dy_(ax+by)+edeat ta,” i) Then bysubstitution z=ax+by © and theequation isreduced toone with variables separable, Indeed, a ayFaaro, whence utaw deoata o° ca) Putting into (6)expressions (6)and (7), weget laa zte : VarFAiate,’ whichisanequation withvariables separable, Thedevice applied tointegrating equation (1)isalsoapplied totheintegration oftheequation toy (actite ) ae! \aepoy ta) where fisanarbitrary continuous function. 485 Differential Equations Example 1,Given the equation dyete—8&7y1" Toconvert itinto ahomogeneous equation, make the substitution x=x,++h;y=y,+k. Then 2 = dy stn thth—s a, Fh Solving theset,oftwo equations h+k—3=0; h—k—1=0, we find haa, bed ‘Asaresult weget the homogeneous equation dy,_ty 30" which wesolve bysubstitution: Bou; then du yyMt duttu otshatte ‘and weget anequation with variables separable: datut sietine: Separating thevariables, wehave daa 44% Tata Integrating wefind arctanuJIn(Itun, -+1nC, aretanu=In(V Tpatx,C) Putting 2inplaceof1,weobtain ~ me tant of Hae Passing tothevariables xandg,wefinally get arr. teentet CVG=IPOS ae OS, First-Order Linear Equations 487 Example2.Theequationya2etytey FS cannot besolved bythe substitution x—sy-+h, y—y-+h, since imthis casethesetofequationsthatservestodetermine #andiisinsolvable (here,thedeterminant [2A]ofthecoelteients ofthevariables teequalto10) This equation may bereduced toone with variables separable bythe substitution aeons, Then y’=2'—2 and theequation isreduced totheform bgt!niet 849Y=RTS Solving itwefind 22 2inise49l—e4¢.5 B ~ * Since224-4,weobtainthenalsolutionoftheinitalequationInthe2 7FOrwtygla]10+5y+91—2+C 10y—5x-+7 In|10x+-5y-+91—C,, that is,asanimplicit function yofx. SEC. 7. FIRST-ORDER LINEAR EQUATIONS Definition. Afirst-order linear equation isanequation that islinear inthe unknown function and itsderivative. Itisofthe form ay+P y=), 0) where P(x) and Q(x) aregiven continuous functions ofx(orare constants). Solution oflinear equation (1). Let usseek the solution of equation (1)intheform ofaproduct oftwo functions ofx: y=u(x) 0(x). (2) Oneofthesefunctions maybearbitrary, whiletheotherwill bedetermined from equation (1). Differentiating both sides of(2), wefind ayyA4y dem" Gat ae 488 Diferential Equations Putting theexpression obtained ofthe derivative into (I), we have w+ 0+Pw=Q or u(+Po) +0=a. @) Let uschoose the function vsuch that 4Pu=0. (4) Separating thevariables inthis differential equation inthefunc- tion v,wefind ee—Pdx. Integrating weobtain Inc, +Inv=—f Pdx or ° vacenS Pe, Since for usitissufficient tohave some nonzero solution of equation (4), wetake,-as thefunction (x), vipee SPA, Oo) where {Pdxissomeantiderivative. Obviously, v(x)#0. Putting thevalue ofo(x) which wehave found into (3),we get(noting that$24Pv=0): (t= Qi), or du_2)dx™ u(x)” whence = (ee=(8dx+C. Substituting: into formula (2),wefinally get y=ow[[SGar+c] First-Order Linear Equations 09 or y=00)(38ae+Co(x). () Note. Itisobvious that expression (6)will not change ifin place ofthe function v(x)defined by(5)wetake some function(2)=Co(a).Indeed,puttingv(x)In(6)Inplaceofo(4),weeel CG Qix) Ce =t 28 den .y=Cv(x) jBSdx=CC(2). TheC’sinthefirsttermcancel out;inthesecondtermtheproductCC isanarbitrary constant, which weshall denote byC,and we againarriveatexpression (6).IfwedenoteJo@ar—ow, thenexpression (6)will take theform y=0(x)p(x)+Co(x). 6’) Itisobvious that this isacomplete integral, since Cmay bechosen insuchmanner thattheinitial condition willbefulfilled: when x=, y=uy. The value ofCisdetermined from the equation Ye=9(X,)P(4,)+C0(x,). Example. Solve the equation peeratete Solution. Putting yaw we have Hautes Mo, Batting the expresion lato the eign eqution, weobtan de, du, 2 .oto Ewe tOF, do 2 a«(B-ye) +eGao to. a Todetermine ©wegetthe equation do 2a2oma, 490 Digerential Equations thatis, fots o7E+T! whence Inv=2in(e-41) oF om(e4I)% Puttingtheexpression ofthefunctionointoequation (7),wegetthefollowing equation for ut du 5 au ety Bawty ofHaut, whence aewnEENbc, ‘Thus, thecomplete integral ofthe given equation will beofthe form yn cute ‘The family obtained isthegeneral solution. No matter what the initial condition (ty. 2, where x#—1, itisalways’ possible. fochoose C.s0 that thecorrespondite particulir solution should satisly the given initial condi- tion. For example, the particular solution that salisfies the condition y=3whenxj—00isoundasfollows: tnt *; eescorns Caz. Consequently, the desired particular solution is pnZt Sete However, iftheinitial condition (xq,ya)ischosen sothat xy=—1, wewillnotfind‘theparticular solution thatsatisfies this condition. Thisisdueto thefactthatwhensj=—I thetwntion PU)——=2, iedacontinuous and, hence, the conditions ofthe theorem ofthe existence ofasolution are not ‘observed, SEC. 8. BERNOULLI'S EQUATION Weconsider anequation ofthe form* 4 P,BtPwy=Qay oy) ©This equation results. from. theproblem ofthemotion ofa.bodyprovided theresistance ofmedium Fdepends onthe velocity: P= Ayo-+hyo% Theequationofmotionwillthenassumetheformm4?=—2yo—Ayo" or 40 hey Beye44ombon, Bernoulli's Equation a where P(x) and Q(x) arecontinuous functions ofx(orconstants), and n#0 and ny1(otherwise wewould have alinear equation). This equation iscalled Bernoulli's equation and reduces to alinear equation bythe following transformation. Dividing allterms oftheequation byy*,weget yh Py =Q ® Making the substitution zay™", we have az andyBang yyB. Substituting into (2), weget 4(ong1)Pe=(—ntlQ This isalinear equation. Finding itscomplete integral and substituting the expression y-** forz,wegetthecomplete integral oftheBernoulli equation. Example. Solve theequation ay 9twas. @ Solution. Dividing allterms byg*,wehave rytaytast, Cy Introducing the new function ay, we getdz__gynd ae eae Substituting into equation (), weobtain atae—2e, © This isaTinear equation, et usfind itscomplete integral:_dt_do,du rau; HauseyMy, Putexpressions zand$into(9): wf4o—2euo— Bet 42 Digerential Equations or do duosu(G—200) +oftenae Equate tozero the expression inthebrackets: Sesrom0; Wonredx; Invest ome", For uwweget theequation du__ osotto, Separating variables, wehave dum—2e-w eds,umn2fenestdegc. Integrating byparts, wefind gare het EC, seuvet $14Ce-*, Consequently, the complete integral ofthe given equation is yetel4ce,ofyet ;Vetiece* Note. Just aswas done forlinear equations, itmay beshown that thesolution oftheBernoulli equation may besought inthe form ofaproduct oftwo functions: y=u(x)o(x), where v(x) issome nonzero function that -satisfies the equation v’+ Po=0. SEC. 9.EXACT DIFFERENTIAL EQUATIONS Definition. The equation. MQ, yde+N(x, ydy=0 « iscalled anexact diferential equation itM(x, y)and N(x, y)are continuous differentiable functions for which the following rela- tionship isfulfilled amt_an 2M ® andaand9%arecontinuous insomeregion. Integrating exact differential equations. Weshall prove that if thelelfside ofequation (1)isanexact differential, then condi- Exact Differential Equations 493 tion (2)isfulfilled, and, conversely, ifcondition (2) isfulfilled the left side ofequation (1)isanexact differential ofsome fun- ction u(x, y). That is,equation (1)isanequation ofthe form du(x,y)=0 @) and, ‘consequently, itscomplete integral is u(x, y=C. Let usfirst assume that the left side of(1)isanexact diffe: rential ofsome function u(x, y);that is, M(x, y)dx+N(x,dy=du=$ de+Sdy, ‘then mu yaMah NH. (4) Differentiating the first relationship with respect toy,and the second with respect tox,weobtain OM_Ou,ONOu“Gy~dxay*Ge~Bydx* Assuming. continuity ofthe second derivatives, wehave am _ontyi that is,(2)isanecessary condition fortheleftside of(1)tobe anexact differential ofsome ‘function u(x,y).Weshall show that this condition isalso sufficient: if(2)is’fulfilled then the left side of(1)isanexact differential ofsome function u(x, y). From the relation au Mx, 9) we find u=JMix,arto), Fa where x,Istheabscissa ofany point ofthedomain ofexistence ofthe solution, When integrating with respect toxweconsider yconstant, and therefore thearbitrary constant ofintegration may bedependent ony..Let uschoose afunction @(y) sothat the second ofthe 404 Diperentiat Equations relations (4)isfulfilled. Todothis, wedifferentiate*)bothsides ofthelatter equation with respect toyand equate theresult to N(x, y): .24_6OMaege EnJMeetoeW=NUevb 5 butsince349%|wecanwrite [ideteoW=M5 thatis,N(x, ee Y)=N(x, y) or _- NE DN D+9 W=NUeWe Hence, #W=NGu 9 or oO=\ Ne dy+C,. x Thus, the function u(x, y)will have theform u=lM, drt ING, dytC. Here P(x,, y,)isapoint-in the neighbourhood ofwhich there isasolution ofthedifferential equation (1). Equating this expression toanarbitrary constant C,wegetthe complete integral ofequation (1): A ' {Mi yar+) NG, Wdy=c. ) % x 4)Theintegral §M(x,»)dxIsdependent ony.Tofindthederivative of thisintegral withrespect toy,differentiate theintegrand withrespect toy: EfmesarmSMae.thisfotiowstromLeiba"theoremforderen: tinting adefinite itegral withrespect toaparameter (seeSec,10,Ch.XI), Integrating Factor 495 Example. Given theequation2ayBO8Eay Fae OO ayo. Let uscheck toseewhether this isanexact differential equation.Denoting oeet ae, yeastMnB vat, aM_6,ON_be Oy OE For y#0, condition (2)isfulfilled. Hence, the left side ofthis equation is anexect ‘differential ofsome unknown function u(x, y). Let us’ find this iunetion since$2—25,itfollowsBe .unlFartew=F+9u. whe isanasyetundefinedfunctionofy. BinkCAating theelationwithrespectto"yandnotingthat uy aunts!Zoe. we find ae toe3 of ets;Ftv oe: hence yyy=, ——FO=p. w= THCy =—), 4N= B-L4e. ‘Thus thecomplete integral ofthe initial equation is zt Foc.ey SEC. 10.INTEGRATING FACTOR Let the left side oftheequation M(x, y)dx+N(x,y)dy=0 a) notbeanexact differential. Itissometimes possible tochoose afunction w(x, y)such that after multiplying allterms ofthe equation byit'the leftside ofthe equation isconverted into an exact differential, The general solution oftheequation thus ob- tained coincides with thegeneral solution oftheoriginal equation; thefunction (x, y)iscalled theintegrating factor ofequation (1). 496 Diferentiat Equations Inorder tofind the integrating factor ,doasfollows. Mul- tiply both sides ofthe given equation by’the asyet unknown integrating factor p: BMdx+pWdy=0. For this equation tobeanexact differential equation, itisneces- sary and sufficient that the following relationship ‘befulfilled: 2(uM)__2UN),vyoe that is, a ua , M 1wt Man +N, or ony OH (2N_ aMMN=o(HF) After dividing both sides ofthe latter equation byp,weget dinp _y@inuaN_amme ySeNe (2) Itisobvious that any function w(x, y)that satisfies this equa- tion isthe integrating factor ofequation (1). Equation (2)is apartial differential equation inthe unknown function depen- dent onthetwo variables xand y.Itcan beproved that under - definite conditions ithas aninfinitude ofsolutions and that, con- sequently, equation (1)hasanintegrating factor. Butinthegene- ralcase, theproblem offinding (x, y)from equation (2)ishar- derthantheoriginal problem ofintegrating, equation (1).Only incertain particular cases does one manage tofind the function Be, y). For instance letequation (1)admit anintegrating factor depen- dent only ony.Then - @inpann0 and tofind weobtain anordinary differential equation on_aM QingOeyay from which wedetermine (byasingle quadrature) Inj, and, hence, as well. Itisclear that this may bedone only iftheexpressionan_oM2% isnotdependent onx. \,TheEnvelope ofaFamily ofCurves ro aw_am Similarly, iftheexpression => isnotdependent onybut only onx,then itiseasy tofind anintegrating factor that depends only onx. Example, Solve theequation +a? de—x dy=0 Solution. Here, M=y+xy%; N=—x; am an__y, OM oNorton han, Se, Thus, the left side ofthe equation ismot anexact differential. Let usscewhether'this equationallowsforanintegrating factordependent onlyonyoFnot.Notingthatan_aMTeWy—1-1-2 2 7 oe v wweconclude that the equation permits ofan integrating actor dependentonlyony.Wefindit:"bs a sa ainu__ 2,ay vy whence j ating ie, wad. Ingwhe nad ‘Aiter multiplying through bytheintegrating factor, weobtain theequation 1 x L4n)dx—Aady—0(f+) a—Say wsancrndierent equation (3M=2M——), Slvingtiegatin,wefinditscompleteintegral: S 2454c=0, ee ales ol SEC, 11, THE ENVELOPE OF AFAMILY OF CURVES Let there beanequation ofthe form D(x, y,C)=0, a where xand yarevariable Cartesian coordinates and Cisapara- meter that can take onavariety offixed values, 498 Diferentiat Equations For each given value ofthe parameter C,equation (1) defines some curve inthexy-plane. Assigning toCall possible values, weobtain afamily ofcurves dependent onasingle parameter, orusingthemorecommonterm,aone- iyparameter family ofcurves. Thus,equation (1)istheequation ofaone:parameter family ofcurves (because it contains only onearbitrary constant). y z ¥ 7 % “#Fee Fig. 250. Fig. 251. Definition. The line Liscalled theenvelope ofaone-parameter family oflines ifateach point ittouches some lineofthefamily, and different lines ofthe given family touch the line Latdiffer” ent points (Fig. 250). Example 1.Consider the family oftines C—OyeRt, where Ris8constant and Cisaparameter. This isafamily ofcircles ofradius Rwith centres on the x-axis. This {amily will obviously have asenvelopes thestraight lines y=R andy=—R Fig. 251). Finding the equation ofthe envelope ofagiven family. Let there begiven afamily ofcurves, (x, y,C)=0, 0) that depend ontheparameter C. Let-us assume that this family has anenvelope whose equation maybewritten in,theformy=@(x), where9(2)isacontinuous and differentiable function ofx. Weconsider some point M(i, y) lying onthe envelope. This point also lies onsome curve of‘the family (1). Tothis curve there corresponds adefinite value ofthe parameter C,which value isdetermined from equation (1), for given (x,y):C=C(x, y).Thus, forallpoints oftheenvelope the following equality isfulfilled: (x, y,C(x, y))=0. oO) Suppose that C(x, y)isadifferentiable function that isnotcon- stant inany interval ofthevalues ofxand yunder consideration, The Envelope ofaFamily ofCurves 499 From equation (2)oftheenvelope wefind the slope ofthe tan- gent totheenvelope atthepoint M(x, y).Differentiate (2)withfespect toxconsidering thet’ isafunction ofx: @,A0C|[aD,aD4C}|,B+teat[tao] Yo or O,4Oy+Oe[5+52y']0. @) The slope ofthe tangent tothe curve ofthe family (1) atthe point M(x, y)isfound from O,+ Oy’=0 ) (on this curve, Cisconstant). Weassume that®'y#0,otherwise wewould consider xasthe function andyastheargument. Sincetheslope&oftheenvelope isequal totheslope &ofthecurve ofthefamily, from (3)and (4) we obtain . fac, ac |o[2+2y]=0. But since ontheenvelope C(x, y)#const, itfollows that a, we,ete to and soforitspoints thefollowing equation holds: e(x, y,C)=0. 6) Thus, thefollowing twoequations serve todetermine theenvelope: O(,y,C)=0,} G(x, y,C)=0. ©) Conversely, if,byeliminating Cfrom these equations, wegetan equation y=@(x), where @(x) isadifferentiable function, and C#const’on this curve, then y=@(x) isthe equation ofthe envelope. Note 1.Ifforthefamily (1)acertain function y= (x)isthe equation ofthelocus ofsingular points, that is,ofpoints where ©,=0and®,=0, thenthecoordinates ofthesepointsalsosatisfyequations (6). Indeed, thecoordinates ofsingular points maybeexpressed interms ofthe parameter Cthat enters into equation (1): x=A(C), y=n(C). ) 500 Differential Equations Ifthese expressions aresubstituted inequation (1), weget an identity inC: OC), HC), C}=0. Differentiating this identity with respect toC,weobtain o,24oH+oe=0. Since foranypoints theequalities ©,=0, @,=0, arefulfilled, itfollows that forthem theequality ®c=0 isalso fulfilled, We have thus proved that the coordinate ofsingular points satisfy equations (6). Summarising, equations (6)define either the envelope orthe locus ofsingular points ofthe curves ofthe family (1), ora combination ofboth. Thus, after obtaining acurve that satisfies equations (6), one has further tofind outwhether itisanenvelope orthe locus ofsingular points. Example 2Find the envelope ofthefamily ofcircles (0+ y*—Rt=0, that are dependent onthe single parameter C.Saltion Differentiating theequationofthefamilywithrespecttoC, wege 2(x—C)=0. Eliminating ©from these two equations, weobtain theequation y—R=0 or yet R. Itisclear, by geometric reasoning, that the pair ofstraight lines istheenvelope (andnotThetocusofsingular’ points, sincethecircles ofafamily Gonot have singular points).Example 3.Findtheenvelope ofthefamily ofstraight lines xcosa+ysina—p=0 (@ where [email protected]. Differentiating thegivenequation ofthefamily withrespect toa,wehave —xsina+y cosa=0. (b) Toeliminate the parameter @from equations (a) and (b), multiply the terms ofthe first bycosa, and-of the ‘second, bysina, and then subtract the second from the first; wewill then have x=pose, Putting this expression into (b), wefind y=psina. Squaringthetermsofthetwolatterequations andaddingtermwise,weget ateyt= pt The Envelope ofaFamily ofCurves 501 This isacircle. Itistheenvelope ofthefamily (and notthe’locus ofsingu Tarpoints, sincestright inesdonathavesingular point) (Fig.252) imple 4.Find theenvelope ofthe trajectories. ofshells ffed from agun with velocity’ v,atdifferent angles ofIneli- nation ofthe barrel tothe horizon. We shall Sh consider that thegun islocated at’the Aa}, tr Sya aie x9 casa: a Fig, 252. Fig, 253. coordinate origin andthat thetrajectories oftheshells lieinthexy-plane (air resistance isdisregarded). ‘Solution. First find the equation ofthe trajectory ofthe shell for the case when the barrel makes an‘angle awith the positive x-axis. Inflight, the shellparticipates simultaneously in,two,mations: auniform ‘motion, "withvelocity v,inthedirection of thebarrel ‘and afalling motion due tothe Torce ofgravity. ‘Therefore, ateach instant oftime fthe position oftheshell Mf (Fig. 253} will bedefined bytheequations xaugcosa, y=otsina Theseareparametric equations ofthetajectory. (theparameter isthetime). Eliminating £,wegettheequation ofthetrajectory inthe form ee :=rtona—U Detcosta, Finally, introducing thenotation tana—k, f>—a, weget y=heart(1+8, CC) ‘This equation defines a parabola with vertical axis passing through the originand.with,branchesdownwards, Weoblainavarielyoltrajectories Torthe different valuesofk.Consequently, equation (8)isAbe"equation of#one. parameter “amily oiparabotas, “Which, arethetrajectory olashellTorGifferent angles @and foragiven initial velocity 2,(Fig. 258), Letusfind theenvelope ofthis family ofparabolas, Differentiating with respect to&both sides of(8), wehavex—2abst=0. © Eliminating &from equations (8)and (9). weget =1 a yaqre sen Diferentiat Equations Tisteeatin ofart hve ateptt(42) ais ofwhich coincides with thegars. Itisnot alocus ofsingular points {since parabola (8)do-not have singular pains). Thus, theparabela, fis yoda istheenvelope ofthe family of{rajetories. Itiscalled asafety parabola Because nopoint outside itichrreach ofasell red tom gives gan with ariven inital velocity‘, 9 NS.. . . a 7 Fig. 254. Example 5.Find theenvelope ofafamily ofsemieubical parabolas POF=0. Solution,Differentiate thegivenequation ofthefamily withrespest to the parameter C: 2(¢-€)=0 Eliminating theparameter Cfrom thetwo equations, weget y=0.ThesarisisaTocusofsingularpoints—acuspofthefrstkind(Fig.255)Indeed, tetusfind thesingular pointe ofthe curve P—u—o=0 forafixed value ofC.Differentiating with respect toxandg,wefind Fe=—2—0)=0; Fiaay=0. Solving the three foregoing equations simultaneously, we find thecoordi-natesofthesingularpoint:<x=C,y==0;thus,eachcurveof-thegivenfamily has2singular point”ofthe y ran For confinvoss.vatiation ofthe parameter C,thesingular points wil ithe entire ‘axis Exanole 6.Fdtheenvelope andloctsofsingularpointsefthefaily LocusoFangular paints 2Fig.255 Wer FeO =O. (10 The Envelope ofaFamily ofCurves 503 Solution, Differentiating both sides of(10) with respect toC,wefind 24-014254¢-O%=0 y—C—(x—0)'=0. i) Noweliminate theparameter €from(1)andfromtheequation (0)ofteamity: y-C=u—oy. Patting theexpression y—C into theequation ofthefamily, weget 2 GOF00 i 2)L ocr[9-2] -0, whence weobtain two possible values ofCand two solutions ofthe problem Corresponding tothem First Solution: Second Solution: 2 cms,car-2 and sofrom (11) wefind and sofrom (11) wefind| 2teeg2)! yrx—(e—t=0 wot3[—+4]=0 2 yon y=s-G- Wehaveobiaied trosrg nesgeaadyrs2.Theioa afsingular points, thesecond tsanenvelope (Fig. 256). 9 y Z Loe 1 es FySZ y d Py —Ky PUPSeS Lasis i 4 Curve, Fig.256 Fig.257. 504 Diferentiat Equations Note 2.InSec. 7,Ch. VI, itwas proved that thenormal toa curve serves asatangent toitsevolute. Hence, the family of normals toagiven curve isatthesame time afamily oftangents toitsevolute. Thus, the evolute ofthe curve istheenvelope of thefamily ofnormals ofthis curve (Fig. 257). This remark enables ustopoint outanother method forfinding evolutes: toobtain the equation ofanevolute, first find thefamily ofallnormals ofthegiven curve and then find the envelope of this family. SEC, 12. SINGULAR SOLUTIONS OF AFIRST-ORDER DIFFERENTIAL EQUATION Let thedifferential equation e F(x,y#)-0 a) have acomplete integral (x, y,C)=0. (2) Let usassume that thefamily ofintegral curves that corresponds toequation (2)hasanenvelope. Weshall prove that this envelope isalso anintegral curve ofthe differential equation (1). Indeed, ateach point theenvelope touches some curve ofthe family; that is,ithas acommon tangent with it.Thus, ateach common point theenvelope and thecurve ofthefamily’ have the same values ofx,y,y’. But foracurve ofthefamily, thenumbers x,y,and y’satisty equation (1). Consequently, thevery same equation issatisfied by the abscissa, the ordinate and the slope ofeach point ‘ofthe envelope. But this means that the envelope is_an integral curve and itsequation isasolution ofthe given differential equation. Since, generally speaking, theenvelope isnotthecurve ofthe family, itsequation cannot beobtained from the complete inte- gral(2)forany particular value ofC.Thesolution ofthedifferential equation which isnotobtained from thecomplete integral forany value ofCand which hasasitsgraph the envelope ofafamily ofintegral curvesentering intothegeneral solution, iscalleda singular solution ofthe differential equation. Letthecomplete integral beknown: (x,y,C)=0; eliminating Cfromthisequationandfromtheequation@¢(x,y,C)=0 wegel9(éy)=0.Ifthsfunctionsatisfiesthedifferential equation land does ‘notbelong tothefamily (2)], then itisasingular integral. Clairaut's Equation 505, Itshould benoted that atleast two integral curves pass through each point ofthecurve that describes asingular solution; that is, uniqueness ofsolution isviolated ateach point ofasingular solution. Example. Find asingular solution ofthe equation wey eRe Solution. Letusfind itscomplete integral. Wesolve theequation fory': dy_ VRP .gor. © Separating variables, weobtain Ht nay,£VR Whence, integrating, wefind thecomplete integral: (HO)ytRE ILiseasy tosee that the family ofintegral lines isafamily ofcirclesof radiuswithcentresonthex-axis.‘Thepairofstraightlinesy=ceRwill Betheenvelope otheamily ofcurve: The functions y= Rsatisty the differential equation (I). This, conse quently, tsasingular integral SEC. 13, CLAIRAUT'S EQUATION Letusconsider theso-called Clairaut equation: wr ayyurg+e(a). a Itisintegrated byintroducing anauxiliary parameter. PutaD then equation (1)will take theform Y=xP+Y(p). ay Differentiate, with respect tox,alltheterms ofthis equation, bearing inmindthatp= isafunction ofx: 4 +o)paxh+aty (p)se or 4s Le+W(Ze=0. Equating each factor tozero, weget dpan? @) 06 Digerentia! Equations and £49" (p)=0. @) 1)Integrating (2)weobtain p=C (C=const). Putting this value ofpinto (1’), wefind itscomplete integral: y=sC+¥(0), ) which,geometrically, isafamilyofstraight lines. 2)Iffrom(3)wefindpasafunction ofxandputitinto (1'), we obtain the function Y=) +P POL a) which may bereadily shown tobethe solution ofequation (1). Indeed, byvirtue of(3)wehave shaptiety (iBao. : And so,bysubstituting the function (1") into equation (1)weget the identity P+(Pp)=19+VP). Thesolution of(1”)isnotobtained fromthecomplete integral (4) forany value ofC.This isasingular solution; itisobtained by elimination ofthe parameter pfrom the equations 9=*P+900),\ +8 (p)=0, or,which isthesame thing, byeliminating Cfrom theequations y=sC +9(C), x+e (C)=0. Thus, the singular solution ofClairaut’s equation defines the envelope ofafamily ofstraight lines represented bythecomplete integral (4). Example. Find the general and singular solutions oftheequation dy yorty—tle ayV'+(4) Solution. Thegeneral solution isobtained bysubtitling Cfor2 ocanro Lagrange’s Equation 507 Toobtain thesingular solution, differentiate ytheTatler equation with respect {o.C: r4—2 00. aston? The,singular,solution(theequationoftheenvelope) is obtained in parametric form (whereiepanera r— = --—* Dcs = "> particulara+ey® “Weeoo \)yee. a) | ae cy* Eliminating ¢,we,getadirestrelationship Fig.258. between xand’ y.Raising both sides ofacl equation tothepower-Zandaddingtheresultantequationstermwise,we getthesingular solution inthefollowing form: Pty aa, This anaeeoid, However, theenvelope ofthefamily, (and,hence, the Singular solution) isnot theentire astrotd, but only iteleft half (since itisevident fromtheparametric equations thatx<0) (Fig.258). SEC, 14. LAGRANGE'S EQUATION The Lagrange equation isanequation oftheform y=xp(y+¥(y') ay where @andpareknown functions of$4. This equation islinear inyand x,Clairaut’s equation, which was considered inthe preceding section, isaparticular ‘case of theLagrange equation when @(y')=y'. The Lagrange equation, likeClairaut’s, tsintegrated bymeans ofintroducing anauxiliary parameter p.Put , y= then the initial equation iswritten inthe form - 9=29(0)+90). ay Differentiating with respect tox,weobtain . voy P=9(P)+ Le!(P)+¥(PGE 508 Differential Equations or . (py 22 P—9(P)=[x9"(p)+(p)Gee ay From this equation wecan straightway- find certain. solutions: namely, itbecomes anidentity forany constant value p=p, that salisfies the condition P.— (P)) =0. Indeed, foraconstant valuepthederivative 4250, andboth sides ofequation (1") vanish. Thesolution corresponding toeachvaluep==p,, thatis,teDs isalinearfunctionofx(sincethederivative $4isconstantonly inthecaseoflinearfantions) Tofind:thisfunctionitissuf- ficient toput into (1’) thevalue p=p,: Y=(P.)+P(P,) Ifitturns outthat this solution isnotobtainable from thegener- alsolution forany value ofthearbitrary constant, itwill bea singular solution. Letusnow find thegeneral solution, Write (I") inthe form az_y 8) __W)ap—*590)—7-90) and regard xasafunction ofp.Then theequation obtained will bealinear differential equation inthe function xofp. Solving it,wefind x=0(p, 0). @ Eliminating the parameter pfrom equations (1’) and (2),we getthecomplete integral (1)intheform D(x, y,C)=0. Example. Given theequation yay ty wo Putting=pwehave y=xptph. ay Differentiating with respect tox,weget - p=p'+iex0-+91 2. i) Let usfind thesingular solutions. Since p=p* forpp=0 andpy=I, the solutfons will belinear Tunetions [see (I)] y=x-0840%, that is,y=0, Orthogonal and Isogonal Trajectories 509 and gant Whenwefindthecomplete integrals wewillseewhether thesefunctions arc particular orsingular solutions. Tofind it,write equation (I")intheform de 2ap *p—pti—p and weshall regard xasa function oftheindependent variable p.Integrating iis Vinear (in2)equation, wefind rent gS. ay Eliminating pfrom equations (I") and (II), weget the complete integral y=(C+VEFI The singular integral ofthe initial equation is y=0 sincethissolution tsnotobtainable fromthegeneral solution forany’value i:iever, thefunctiony=z4-1inatsingularbuaparticule solution: itisoblained from thegeneral solution when C=0. SEC. 15.ORTHOGONAL AND ISOGQNAL TRAJECTORIES Suppose wehave aone-parameter family ofcurves D(x, y,C)=0. w Lines intersecting allthe curves ofthegiven family (1)ata constant angle are called isogonal’ trajectories. Ifthis angle isarightangle,theyareorthogonaltrajectories, Orthogonal’ trajectories. Let usfind the equation oforthogonal trajectories. Write thedifferential equation ofthegiven family of curves, eliminating theparameter Cfrom theequations (x,y, C)=0 and a0 abdy_ae+oyae Let this differential equation be ay F(x,y,$)=0. a’) Here,£4istheslopeofthetangent tosomemember ofthe family atthepoint M(x, y).Since anorthogonal trajectory pass- ingthrough thepoint M(x,y)isperpendicular tothecorrespond- ingcurveofthefamily, theslopeofthetangent toit,42,is 510 Diferentiat Equations connected with44bytherelationship (Fig.259) dyae" Tar ® ae Putting this expression into equation (1") and dropping the subscript T,wegetarelationship between thecoordinates ofan arbitrary point (x,y) and theslope oftheorthogonal trajectory atthis point, that is,adifferential y equation oforthogonal trajectories: Ltaeaa)=0@) tay,a . Thecomplete integral ofthisequa- tion ©,(x,4,C)=0 yields afamily oforthogonal trajec- 5tories. ‘Aconsideration oftheplanefow ofafluid involves orthogonal trajec- Pa tories.'e Letussuppose thatthefluidflowinaplane takes place insuch man- ner that ateach point ofthe xy-plane the velocity vector, (x,y), ofmotion isdefined. Ifthis vector depends solely on the‘position ofthepoint intheplane, but isindependent ofthe time, the motion iscalled stationary orsteady-state. We shall consider such motion. Inaddition, we shall assume that there exists apotential ofvelocities, that is,afunction u(x, y)such that the projections ofthevector (x, y)onthecoordinate axis, v,(x,y) andv,(x,y)areilspartial derivatives with respect tox andy: auaufino, Fav, , 4) The lines ofthefamily u(x, y)=C 6) are_called equipotential Lines (lines ofequal potential), The lines, thetangents towhich atallpoints coincide with the vector o(x,y) indirection, are called flow lines and yield the trajectories ‘ofmoving points. Orthogonal and Isogonal Trajectories sil We shall show that the flow lines aretheorthogonal trajectories ofafamily ofequipotential lines (Fig. 260). Let @beanangle formed bythevelocity vector owith the x-axis, Then byrelation (4) du (x, aus, ;BED—\9)cosgiMH=|9|sing, whence wefind theslope ofthetangent totheflow line dus, 9) __tang= a: 0) ae We obtain theslope ofthetangent totheequipotential line by differentiating, with respect tox,relation yy 8):au|dudy_etyas= ¥ whence du aynna o) oy Thus, inmagnitude and sign, the slope 7 ofthetangent totheequipotential line is Fig.260, the inverse oftheslope ofthe tangent to theflow line. Whence itfollows that equipotential lines and flow lines aremutually orthogonal. Inthe case ofanelectric ormagnetic field, the lines offorce ofthe field serve astheorthogonal trajectories ofthefamily of equipotential lines. Example 1.Find the orthogonal trajectories ofthe family ofparabolas y=ce, Solution. Write the differential equation ofthe family y=2x. Eliminating C,weget roe oe Substituting -7fory',weoblainadifferential equation ofthefamilyof orthogonal trajeatories oa Wr 512 Diferentiat Equations or xdx sdye—“> Itscomplete integral is Fav osa+gec. Hence, the orthogonal trajectories ofthegiven family ofparabolas will be represented byacertain family ofellipses with semi-axes a=2C, 6=C V2 Gig: 261). y WD.Sa(SSv"ey SF x Fig. 261. Isogonal trajectories. Let the trajectories cut the curves ofa given family atanangle a,where tana=&. y Theslope$=tang (Fig.262)ofthetan- @ gent toamember ofthefamily andtheslope \ dur je“2tanyptotheisogonaltrajectoryarecon- \nected bythe relationship 14 WeatanyotenaFig.262. tang=tan (Y—@)=TFianateny! Orthogonal and Isogonal Trajectories 513 thatis, y dur_ya_ee a LS OOaE Sab Astat Sosy | Substitutingthisexpression into/PFE) tT] equation (1')anddropping thesub- LTES script7,weobtainthedifferential LESS SZ¥equationofisogonaltrajectories. tise y=Cx, (8) that cut the lines ofthe given family 7 aManangleaythetangent ofwhich Fig,268. equals fanak Solution. Letuswrite thedifferential equation ofthegiven family. Diffee reniiating equation (@)with respect tox,wefind dyWoe, Onthe other hand, from the same equation wehave cat. Consequently, thedifferential equation ofthegiven family isoftheform Moe dx” Utilising relationship (2’) weget the differential equation ofisogonal trajectories dur Gay ira)"ae Whence, dropping thesubscript 7,wefind yayAte eee Integrating this homogeneous equation, weget thecomplete integral: toV8FreLarctanL416, o which defines the family ofisogonal trajectories. Tofind out precisely which 17008 54 Diferentiat Equations curves enter into this family, letuschange topolar coordinates: pte tangs VIF Substituting these expressions into (9)weobtain ine=teting or ence. Consequently, the family ofisogonal trajectories isafamily oflogarithmic spirals (Fig. 263). SEC. 16, HIGHER-ORDER DIFFERENTIAL EQUATIONS, (FUNDAMENTALS) Ashas already been indicated above (see Sec. 2),adifferential equation ofthenth order may bewritten symbolically intheform FY Yseny= () or, ifitcan besolved for the nth derivative, YOST HU Ysoor Ym). a’y Inthis chapter weshall consider only such equations ofhigher order that may besolved for ahigher derivative. For these equations wehave atheorem ontheexistence and uniqueness of asolution, similar tothecorresponding theorem onthesolution offirst-order equations. Theorem. Jfintheequation WAL YY ey) thefunction f(x,y, y’,«++.Y°"") and itspartial derivatives with respect tothearguments y,y’,.+., y"-" arecontinuous insome region containing the values x=X, Y=Yy Y'=Yiy veer y= ye", then there isoneand only onesolution, y=y(x), of the equation that satisfies theconditions Yana=YorYousSis ® ; Wid =ye, These conditions arecalled initial conditions. The, proof isbeyorid thescope ofthis’ book. Higher-Order Differential Equations ‘51S Ifweconsider asecond-order equation y’=f(x, y,y’), then the initial conditions forthesolution, when x=x,, will be YY Y=H, where x,,yyyjaregiven numbers, which have thefollowing geometric meaning: only one curve passes through agiven point ‘ofthe plane (x,,y,)with given tangent oftheangle ofinclination ofthe tangent’ line y;.From this itfollows that ifwewant to assign different -values ofy{forconstant x,andy,,wegetan infinitude ofintegral curves with different angles ofinclination passing through the given point. ‘Wenow introduce theconcept ofageneral solution ofanequa- tion ofthe nth order. Definition. The general solution ofadifferential equation ofthe nth order isthe function Y=OOEC, Cyor Cs which isdependent onnarbitrary constants C,,C,, ..., C,and such that: a)itsatisfies theequation foranyvalues oftheconstants byforspecified initial conditions Yenrs=YoprunsYo the constants C,, C,, ..., C,may bechosen sothat thefunc- tion y=9(x, C,,Cy,.+-,C,) Will satisfy these conditions (ontheassumption thatthe‘initial values x,,y..yj,»-+.y@-” belong tothe region where theconditions oftheexistence ofasolution arefulfilled). Arelationship ofthe form (x, y,C,,C,,..., C,)=0, which implicitly defines thegeneral solution,” is’called “thecomplete integral ofthedifferential equation. Any function obtained from thegeneral solution forspecific values ofthe constants C,, C,, ..., C,iscalled aparticular solution. The graph ofaparticular solution iscalled anintegral curve ofthegiven differential equation. ToSolve (integrate) adifferential equation ofthe’ nthordermeans: 7 ” a6 Digerentiat Equations 1)tofind itsgeneral solution (ifthe initial conditions are not given) of 2)tofind aparticular solution ofthe equation that satisfies the given initial conditions (ifthere are such). Inthe following sections weshall present methods ofsolving Various equations ofthe nth order. SEC, 17,AN EQUATION OF THE FORM yim =F (x) The simplest type ofequation ofthenthorder isoftheform y=1(). oy Let usfind thecomplete integral ofthis equation. Integrating the left and right sides with respect tox,and taking into account that y=(y"-)’, weobtain y= lfxyde+C, a"; where x,isany fixed value ofx,and C,istheconstant of integration. Integrating once more weget yrr=( (fr(ayde) de+0,(2—x,) +Cy. Continuing, wefinally get(after nintegrations) theexpression of the complete integral: yaJ..fords...de+AERP CERO Cy. ae Inorder tofind aparticular solution satisfying theinitial condi- tions Yrwse= eiYears=iriKEP=, itissufficient toput Co“ Yor Cars =Uee oe EU Example 1.Find thecomplete integral oftheequation yf=sin(ke) and 4partcular solution sallafying the initial conditions Yenr=0 Yrno=le AnEquation ofthe,Form y!"=(x) 817 Solution. oo[snkrdepe.SHON40, :Ub . ( (cos kx—1¢ oJ (Sac efoarte, or sine oe,ga FETC te This isthe complete integral, Tofind aparticular solution satisfying the giveninital condition, itTesuiicent Yo"determine thecorresponding valuesFromtheconditionYeay=0,wefindC,=0.From thecondition y,-4—=1, wefind C,—0. Thus, thedesired particular solution isoftheform sinks|(A oot (+1) Differential equations ofthis kind areencountered inthetheory ofthe bending ofgirders. ° Example. 2.Lel usconsider anelastic prismatic girder bending under the action ofexternal forces. both continuously distributed (weighty, load) and Concentrated. Let the x-axis be horizontal along theaxisofthegirder initsunderformed x n ateandite ants bedirected vertically 5downwards (Fig. t). ) x‘Each force acting on the girder (the load ofthe girder, and thereaction oftheSupports, ¥ forinstance) hasamoment, relative tosome t iecross section ofthe girder, equal tothe prod- tict ofthe force by‘the distance ofthe point e‘ofapplication ofthe‘forcetromthegiven—'Y‘ross section. Thesum, M(x), ofthemoments Pig.266.ofalltheforces applied tothat part ofthe bd firder situated toone side” ofthe’given cross section with abscissa xiscalled thebending moment ofthegirder relative tothegiven cross section. Incourses ofstrength ofmaterials, itisproved that thebending moment ofthegirder is El a. where Eisthe so-called modulus ofelasticity which depends onthematerial athe girder, Jis.the moment ofinertia ofthe cross-sectional area ofthe girder relative tothe horizontal line passing through the centre ofgravity of fhe cross-sectional area, and Risthe radius ofcurvature ofthe axis ofthe ent ‘girder, which radius isexpressed bythe formula (Sec. 6,Ch. Vi). paya 518 Differential Equations Thus, the differential equation ofthebent axis ofagirder hastheform ve Minayy Br” ° weconsider thatthedeformations are_small_and thatthetangents to the axis ofthegirder, when bent, form asm angle with thex-axis, weean disregard thesquare ofthesmall ‘quantity y™andconsider 1 rad. ‘Then the differential equation ofthebent girder will have the form M(x) of-ap ca) but this equation taofthe form of(I). Example 2AgirderiStedinpceattheexcemity 0andisubjected totegelion ofafconcenrated, veri (ncePapplied, fotheendotthegirder Latadistance 1Irom Q(Fig. 264). Theweight ofthegirder isignored. ‘We consider aeross section atthepolit N(a). The bending moment rela: tive fosection Afayinthegiven ease equal fo M()=(—2)P. ‘the ditetentiat equation (2) has thefoim ? vagy. Thentlcoins sr:or0hedefection ysequal oroandthe Tangent tothe bent axis ofthegirder colnetdes with the‘xeaxis; tha i Hens. Yea=O. Integrating theequation, wefind (Pt _? x).opr) (nam F(UZ): Pyomapy(Ht5). ® Inpaticuar, trom formula (3)we determine the deflection Aattheextre- rity ofthe girder Le heer SESent FET SEC. 18, SOME TYPES OF SECOND-ORDER DIFFERENTIAL EQUATIONS REDUCIBLE TO FIRST-ORDER EQUATIONS, I.Anequation of‘thetypero n Gh=1(x.4) 0) does notexplicitly contain theunknown function y. Some Types ofSecond-Order Diferential Equations a9 Solution. Letusdenote thederivative 44interms ofp,that is,weset$¢—p. Then£4—42, Putting these expressions ofthe derivatives into equation (1), wegetafirst-order equation, 4Pale, py inthe unknown function pofx.Integrating this equation, we find itsgeneral solution: p=p(x, C), andthenfromtherelation $= wegetthecomplete integral ofequation (1): y=Spl C)de+C,, Example 1,Letusconsider the differential equation ofacatenary (see See. 1): dy 1 dyrareV1+(2)o Set thao then d¥y dp ma andwegetafirst-order differential equation intheauxiliary function pofx1 dp_t .Gay Vie Separating variabies, wehave mPa tt View 2" whence np+VIF=E+C, 1/,546_,- (3-4)pag (oe ). Butsincep=42, thelatterrelation isadifferential equation inthesought- forfunction y.Integrating it,we obtain the equation ofacatenary (see 620 Differential Equations «. Sec. I) fac-(Fe) 9G(ete())40. Let usfind theparticular solution that satisfies thefollowing initial con- ditions: Yene=OsHrae=0. The frst condition yields C,=0, thesecond, Cy=0. Wefinally obtain o-G(et +e*) Note. We can similarly integrate theequation y= f(x,y). Setting y"- =p, we get for a.determination ofpthe first- order equation “i=I, p)- Fromherewegetpas.afunction ofx,andfromtherelation y=pwefindy(seeSec.17). : IL,Anequation ofthetype. a aGai(v2) @ does not contain the independent variable xexplicitly. To solve it,weagain set 4gmp @) but now weshall consider pasaJunction ofy(and not ofx, asbefore).Thenayap_dpdu_do Gt de Wydx dy? Putting into(2)theexpressions #and$4,wegetafirst- order equation intheauxiliary function p: peal. pd “) Integrating it,wefind pasafunction ofyand thearbitrary constant C,: p=py, C). Some Types ofSecond-Order Diflerential Equations sat Substituting this value in(3), weget afirst-order differential equation forthefunction yofx: =o, Cy). Separating variables, wehave. ay ru. Integrating this equation, we get the complete integral ofthe initial equation: (x, y,C,,C)=0. Example 2.Find the complete integral ofthe equation yey Solution. Putpatandconsider p-asafunction ofy.Thengare and weget afirst-order equation forthe auxiliary function p! dp _-F.any . . Integrating thisequation, wefiad . praG—y oope VG Butp=44; consequently, foradetermination ofywegettheequation ‘yerfynas,otwht dr, Cy EVCy—1 whence dy x+Q=t let, Tocompute the latter integral wemake the substitution Cyh—1 =F, Then emcee pntbh;Ph=OH0" oy: eye + y= 0 02 Diderential-Equations Consequently, "4 2 ayePete 3(MU gad(S41Ju -*VEGF Cy+2). Finally weget* a+ VogForcy"rea EVECy"b+2, Example 8.Letapoint move along the x-axis under theaction ofaforce that depends solely omthe position atthe point. ‘The differential equationofmotion will be Ps maa Fe. ax AttH0 tetraay Hany Mattipying bothsidesoftheequation by£41andinteraling rom0 tot, we have 1 (atta$m(2)'—L t=freee 1 dx\*( 39(#) +[-Jre ax]=Fmo}sconst. Thefrsttermofthisequation isthekinetic energy, thesecond term, ‘thepotential energy ofthemoving point.Fromthisequation iffollows thaty iheinhte Kinetic andpotential energy re mains constant’ throughout. the time. ofmotion, ‘Theproblem ofsimplependulum. Letthere be a"material point of mags. my which isination(bytheforeoferavity)angthectl Tying inthevertical plane. Let us fdtheequa: Konto! mlio olth!potneglecting resistanceforeetlon, aireason, ele 1 ating the origin atthe iowest point ofthe ice, WEputdiewcaxis slongtheTangent to thecircle (ig.263). Denote by/the radius ofthecircle, bysthe are lengih from theorigin Otothevariable point Biwhe temasmishosed; hislengthstokenwih"teappropiate sign(> 0."the Degsing FpointMisontherightofO;s<OilMison \ Theettof0). \ ‘Our problem consists inestablishing sas a ngt-S—tunction ohthetime f.et usdecompose the force ofgravity mginto Fig. 266, tangential “and formal ‘components. ‘The Tormer, Some Types ofSecond-Order Diferential. Equations 523, equal to—mg sing, produces motion, the latter iscancelled bythe reactionoftheurgealongbic‘themass'mismoving. * ‘Thus, theequation ofmotion isofthe form as mts —mgsing. Sincetheangle@=- foracircle, wegottheequation as : fro gsins. This isaType Ifdifferential equation (since itdoes not contain the inde- pendent variable {explicitly). ‘Let'usintegrate itintheappropriate fashion: ds_) ds_dp : an? Bra? Hence, or pdp=—esint ae, whence pratgont40, Let,usdenotebysythegreatestarelengthtowhichthepointMswingzy Fors5theveloety’of thepointiszero: ms ae 4sFer ea This enables ustodetermine Cyt omtetcor+0, whence C=—24tcos“ Theretore, ds)" 8 coal t=(G)'—20(cosf—cont) or,applying tothe latter expression theformula forthe difference ofcosines, as)" Sh ght(3)=tgtsinSFsin8, ) sm Digerentiat Equations o és ite he?ova Vanean © This isanequation with variables separable. Separating thevariables, weget Sypeeree ae 0Ysn8sin Weshall assume, forthe time. being, that s5,, then thedenominator ofthefractionisdiferentfromzero.i’we‘considet”thats=0Torf=0,thenfrom (7)weget 5stent. ® i%,,8=32safans ThisIstheequation thatyields. asafunction ‘oft.Theintegralonthe left cannot beexpressed interms ofelementary functions; neither can thefunction sof¢.Letusconsider thisproblem approximately. WeshalSand 4 eangles £2" ang SE assumethattheanglesStand4aresmall,Theangles£3and{7 willnotexceed $¢.In(6)letusreplace, approximately, thesinesofthe angles bytheangles as eastae ae “Em , sf f/faa. cy Separating variables, weget(assuming, forthetime being, that s¥s,) fun fa. “”Vo Again weconsider that 5-0 when ¢==0. Integrating thelatter equation, weget a iV si1/8, esa Sa Et, *)Weputtheplussigninfrontoftheroot.Fromthenoteattheendof the solution itfollows that there isno need toconsider the cate with the minus sige, Some Types ofSecond-Order Differential Equations 525 whence = zsayan VEe, o Note, When solving, weassumed that ss,. But itisclear, bydirect ree thatthefunction (9)isthesolution ofequation (6')forany value of f Let itbe recalled that the solution (9) isan approximate solution of equation (5),since equation (6)was replaced bytheapproximate equation (6). equation (a)shows, that“thepoint (hich maybergatded astheextremity ‘ofthe pendulum) performs: harmonic. oscillations ‘with aperiod TatViThisperiodisindependent oftheamplitude 5,.Example 4,Escape-velocity problem Determine ‘thesimallest velocity ‘with which abody must bethrown ver- tically upwards so"that itwill not return tothe earth. Air resistance is neglected Solution. Denote the mass ofthe earth and the mass ofthe body byM andmrespectively. ByNewton's lawofgravitation, theforceofattraction f actingonthe bodymis ,path, where risthe distance between the centre of the earth and the centre of sravily ofthe body, and-k isthe gravitational constant. The differential equation ofmotion ofthis body with mass mwill be momene “ a M Sane (10) The minus sign indicates that the acceleration isnegative. The differen-tial-equation (10)Isanequation oftype(2.Weshallsolveitforthefol. lowing initia! conditions: ra fortao reek, Smo, Here, Rtstheradius oftheearth and otsthe launching velocity. Wedenots dey eedo dodr doan” Bane ana where oisthe velocity ofmotion. Putting this into (10), weget onaanata: Separating variables, weobtain odo = a Integrating this equation, wetndFa 1 Gatem tec, ay 526 Differential Equations From thecondition that v=u, atthe earth's surface (for r=R), wedeter- mine Cy: a 1 Fate HG or AM GREAT Weputthevalue ofC,into(IN): , o 1_aM, of . at Rt or opm ta (22Mgamde(3-2). ay Itisgiven thatthebody should move sothatthevelocity isalways postive:hence,2>0.Sinceforaboundless increase of#thequantity #4becomes arbitrarily small, thecondition “5>0 willbefulfilled forany onlyfor the case vy aMZoitoo (3) o oeVE Hence, the lowest velocity will bedetermined by.theequation emonVE, a where £=6,66-10-* cim"/em-sec,R=63-10"cm.i Atthe earth's surface, forr=R, theacceleration ofgravity Isg(g=981 cm/sec). For this reason, from (10) weobtain M for o akLe Putting this value ofMInto (14) weobtain t=VIER=VERVE =11.2108 1.242, Graphical Method ofIntegration sar SEC. 19. GRAPHICAL METHOD OF INTEGRATING SECOND-ORDER DIFFERENTIAL EQUATIONS Let usfind out thegeometric meaning ofasecond-order differ- ential equation. Suppose wehave anequation F=f). 0) Denote by@the angle formed bythe positive x-axis and the tangent toacurve; then ayGistang. @ To find the geometric significance ofthe second derivative, recall the formula that determines the radius ofcurvature ofa curve atagiven point*)Rate" atte | ‘Whenceoe ; ye . But ye=tang; 14+y=14tantp=sectg; (I+y'h= pal =lsee’ol=iosrgr ‘therefore flf=mara ) Now putting into (1)the expressions obtained foryand y’,we have Baran hey.tang) or : Rarer tauaaa “ Itisthus evident that asecond-order differential equation deter- mines the magnitude ofthe radius ofcurvature ofanintegral curve ifthe coordinates ofthe point and the direction ofthe tangent tothis point arespecified. *)Up till we have als jidered theradius of curvature positive;inthisectionweshall'sonaider’R® numberthatcantake:onbothpoate andnegative values: ifthecurveisconvex (y’<0), Wweconsidér the-radius ‘ofcurvature negative (R<0); ifthecurve isconcave (y">0), itispositive(R>0). a -. 528 Differential Equations From the foregoing there follows amethod ofapproximate con- struction ofanintegral curve bymeans ofasmooth curve com- posed ofarcs ofcircles. *) Toillustrate, letitberequired tofind the solution ofequation (1)that satisfies thefollowing initial conditions: Yours =Yoi Your,=Yor Through the’point M,(x,,y,) draw arayM,T, with slope y’= =tang, —y, (Fig. 266). From equation (4)wefindthemagnitude ofR=R,. Lay offasegment M,C,, equal toR,,perpendicular toM,T,, and from thepoint C,(ascentre) strike anarcM,M7 with radius R,.Itshould benoted ‘ Tele thatifR,<0, thenthesegment M,C,Ta mustbe"drawninthatdirection’ soix 7,thatthearcofthecircleisconvexwre upwards, andforR,>0, convex downGK (seefootnote onpage527). Thenlet-x,,y,bethecoordinates ofthe point’ Af, which lies onthe constructed arc’ and issufficiently closetothepointM,whiletan9,is a% theslopeofthetangént M7,tothe Fig,266. circle drawn atM,. From equation (4) wefindthevalueofR=R,thatcor- responds to'M,. Draw thesegment M,C,, perpendicular‘toM,T,, equaltoR,,andfromC,(ascentre) strike anarcM,M, with radiusR,.‘ThenonthisarctakeapointM,(x,, y,)closetoM, andcontinue construction asbeforeuntilwe‘get'asufficiently large piece ofthe curve consisting ofthe arcs ofcircles. From theforegoing itisclear that thiscurve isapproximately anintegral curve that passes throught thepoint M,. Obviously, thesmaller theares M,M,, M,M,,..., thecloser the constructed curve will betothe Integral ‘curve. SEC. 20. HOMOGENEOUS LINEAR EQUATIONS. DEFINITIONS AND GENERAL PROPERTIES Definition 1.Annth-order differential equation iscalled linear ifitisofthe first degree inthe unknown function yand its *)Acurveiscalledsmooth ifithastangents atallpointsandtheangle ofinclination ofthe tangent isacontinuous [unction ofthe are length s. Homogeneous Linear Equations 529 derivatives y’,...,y™-", ym; that is,ifitisofthe 1orm ay +a,y"— +... any =F(x), 0) where @,,@,,@,,...,@, and f(x) are given functions ofxor constants, and ‘@,%0 forallvalues ofxfrom thedomain inwhich weconsider equation (1). From now onweshall presume that thefunctions a,,a,,...,a, and f(x) arecontinuous forallvalues ofxand that ‘the’ coefficient a,=1 (ifitisnot equal to|we can divide allterms oftheequation byit). The function f(x) ontheright side oftheequation iscalled theright-hand member ofthe equation. Iff(x)#0, then theequation iscalled nonhomogeneous linear oranequation with aright-hand member. But iff(x)=0 then theequation hasthe form yFaye"... -+4,y=0 C3) and iscalled homogeneous linear oranequation without. aright- hand member (the left. side ofthis equation isahomogeneous function ofthefirst degree iny,y’,y’, ....y%). Let usdetermine some ofthe basic properties ofhomogeneous linear equations,confining our proof tosecond-order equations. Theorem 1.Ify,and y,aretwoparticular solutions ofahumo- geneous linear equation ofthesecond order ¥tay’+ay=0, ) then y,+4, isalso asolution ofthis equation.Proof. Since y,andy,aresolutions oftheequation, wehave yitay, tay, =0 and. O) Yt ays+ay,=0. Putting into equation (3)thesum y,+y, and taking into account the identities (4), wewill have WitUO+a+9!$4(YU)= HWYtay tay) +Us+ays+ay,)=0+0=0, Thus, y,+4, isasolution oftheequation. Theorem 2.Ify,is@solution ofequation (3) and Cisacon stant, then Cy, is‘also asolution of(3). Proof. Substituting into (3)theexpression Cy,, weget (Cy) +4, (Cu) +4,(Cy) =Cly,+ay,+4444] =C-0=0; and the theorem isthus proved. 530, Dierential Equations Definition 2.The two solutions ofequation (3), y,and 4,arecalled linearly independent onaninterval [a,6)iftheirratio onthis interval isnot aconstant; that is,if fseconst. Otherwise the solutions are called linearly dependent. Inother words, two solutions, y,and y,,arecalled linearly dependent on aninterval [a,6]ifthere exisis aconstant number 4such that Bodwhenacxecb. Inthiscase,y,—=Ay Example 1.Lettherebeanequation y'—y=0. Itiseasytoverily that thefunctions e%e-*, Se". Sen are solutions ofthis equation. Here, the functions e*and e-*"are linearly independent onany interval because the ratioSpe doesnotremainconstant asxvaties. Butthefunctions e* and3e*arelinearly dependent, since3=3.—const Definition 3.Ify,and y,arefunctions ofx,thedeterminant =|" 4|—xu—viMow=| pl=semwn iscalled the Wronskian ofthegiven functions. Theorem 3.Ifthefunctions y,and y,arelinearly dependent on aaninteroal (a,6),then the Wronskian ‘onthis interval isidenti- cally zero. . Indeed, ify,=Ay,where 4=const, then y=Ay,and =|%4la|4 MlalhB= ronw=(% Bl-[e l-+[h fl-° Theorem 4.IftheWronskian W(y,,y,), formed forthesolutions y,and y,ofthehomogeneous linear equation (3), isnotzero for ‘some value x=x,onaninterval (a,b]where the coefficients of theequation are‘continuous, then itdoes not vanish forany value ofxwhatsoever onthis interval. Proof. Since y,and y,aretwo solutions ofequation (3), we have tayitay,=0, yebayetay,=0. Multiplying theterms ofthefirst equation byy,,theterms of thesecond equation by—y,, and adding, weget ‘ iY. 9.9) +4,YY,—YY)=0 ©) Homogeneous Linear Equations Ba The difference inthe second brackets istheWronskian’W(y,,y,). Theexpression inthefirstbrackets isaderivative oftheWrons- kianWW,44): WY) =YY)’ =9sFY IW=Ye—i> Thus, equation (5)assumes the form W'=—aW. 6) Separating variables (for W+0), weobtain va—o, Integrating, wefind InW=—(a,dx+InC or nga —Sade, whence * aie W=Ce® , 0) itisgiven that Ween =Ce= C40, But then from (7) itfollows that W#0 for any values ofx, because the exponential function does not vanish forany finite value ofthe argument. Note 1.Ifthe Wronskian iszero forsome value x=x,, then itisalso zero forany value xintheinterval under consideration. This follows directly from (7):. ifW=0 when x=x,, then Wiens,=C=0; consequently, W==0, nomatter what thevalue oftheupper limit ofxinformula (7). Theorem 5.[fthesolutions y,and y,ofequation (3)arelinearly independent onaninterval [a,'6), then theWronskian W,formed forthese solutions, does not. vanish atany point ofthegiven interoal.. We shall hint atthe proof ofthis theorem without giving it completely. 532 Differential Equations Suppose that W=0 atsome point ofthe interval; then, byTheorem 3,theWronskian willbezeroatall,pointsof[a,6]: w=0 or IY—99.=0. Let usfirst consider those subintervals in(a,6]where y,#0. Then AYVYwero a or ws)(y-°. Consequently, oneachofthesesubintervals Isaconstant fora 5B=1=const Taking advantage ofthe existence and uniqueness theorem, it may beshown that y,==Ay, forallpoints ofthe interval [a,6] including those where y,=0; but this isimpossible since itis given that y,and y,arelinearly independent. Thus, theWrons- kian does notvanish forany single point of[a,6). Theorem 6.Ify,andy,aretwolinearly independent solutions ofequation (3),then y=Cy, +Cyn (8) where C,and C,arearbitrary constants, isitsgeneral solution.Proof. From Theorems 1and2itfollows that thefunction Cut Cus isasolution ofequation (3)forany values ofC,and C,. Weshall now prove that nomatter what theinitial conditions Year,=YorYenxs=Ya,itispossibletochoosethevaluesofthearbit- raty’constants C,‘andC,so,thatthecorresponding particularsolution C,y,+C,y,shouldsatisfythegiveninitialconditions.Substituting the ‘initial conditions into (8), wehave Wy=Cy+Cer aC tCen a where we put dean=taiYdeos=YriUeany=SisiUadenny=Yane Homogeneous Linear Equations 533 From thesystem (9)wecan determine C,and G,,since thedeter- minant’of this system ° : fa|Oe isthe Wronskian forx=x, and, hence, isnot equal to0(by Virtue of.thelinearindependence ofthe,solutions wand ¥,) The particular solution obtained from the family (8) for the found values ofC,and C,satisfies thegiven initial conditions. Thus, thetheorem'is proved. Example 2The equation o+yV—pyH0 notcontain’thepoint#=0,permitsoftheparticular, solutions nen wet (Ihis isreadily verified bysubstitution). Heice, itsgeneral solution isof {he form yaOx+C4. : Note 2.There are nogeneral methods for finding (inGnite form) thegeneral solution ofalinear equation with variable coel- ficients, However, such amethod exists for anequation with constant coefficients. Itwill begiven inthe-next section. For the case ofequations with variable coefficients, certain devices will begiven inChapter XVI (Series) that will enable ustofind approximate solutions satisfying definite initial conditions. Here weshall prove atheorem that will enable ustofind the general solution ofasecond-order differential equation with variable coefficients ifone ofitsparticular solutions isknown. Since it issometimes possible tofind orguess one particular solution directly, this theorem will prove useful inmany cases. Theorem 7.Ifweknow one particular solution ofasecond-order homogencous linear equation, the finding ofthe general solution reduces tointegrating thefunctions. Proof. Let y,besome known particular solution oftheequation yf+a,y'+ay=0. Wefind another particular solution ofthe given equztion sothat y,and y,are linearly independent. Then thegeneral solution will beexpressed bythe formula y=C,y,-+C,y,, where C,and C,arearbitrary constants. Byvirtue offormula (7)(see proof of 54 Differential Equations Theorem 4),wecan write Soe YdsIasi,=Co55 Thus, foradetermination ofy,weobtain afirst-order linear equation, Integrate itasfollows. Divide allterms byyf: sinus 1nfoeae or4(u)_1ggSoe,a(#) we b whence SoePay ha. Hyi)Fde+C,. Since weare seeking aparticular solution, weget (by putting C,=0andC=1) irSede =y,(—nau [Goa (19) Itisobvious ‘that y,and y,are linearly independent solutions since##const, Thus,thegeneral‘solutionoftheaeequation isoftheformSea =Cy,+C,y, |——de. 1gala,+Cu,fe ay Example 3.Find thegeneral solution oftheequation (=a) of—2xy’+2y=0. Solution. Itisevident, bydirect verification, that this equation has a articular solution yy. Let usfind the second” particular solution yy#0fhat'y, andy, should belinearly independent. NotingthatInourcasea==", wehave,by(10), sees poefeedees(toes alge tot1 eeeTes? =(Stegca teres)oo[-Ft"|l]- Consequently, thegeneral solution tsoftheformTain[LEE paeue+6,(4en|!*2|-1), Second-Order Homogeneous Linear Equations 535 SEC. 21, SECOND-ORDER HOMOGENEOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS We have asecond-order homogeneous linear equation +py+qy=0, a) wherepandqarerealconstants, Tofindthecomplete integral ofthis equation, itissufficient (ashas already been proved) to find two linearly independent particular solutions. Let uslook forthe particular solutions inthe form y=e,wherek=const; @ then yak, yah, Substituting theexpressions ofthederivatives intoequation (1), wefind (ke!+pk-+9)=0. Since e*%0, itmeans that k+pk+q=0. @) Thus, if&satisfies equation (3), then ewill bea.solution of(1).Equation (3)iscalled anauxiliary equation with respect toequation (1). The auxiliary equation isaquadratic equation with two roots; letusdenote them by&,and &,.Then p a a a Arebet GG ba$V Go. The following cases are possible: 1,k,and &,arereal numbers and notequal (k,%2,); II.&,and &arecomplex numbers; III,&and &yarereal and equal numbers (&,=2,). Let _usconsider each case separately. I.The roots ofthe auxiliary equation are real and distinct, kyAk. Here, the particular solutions arethe functions ya, ya These solutions arelinearly independent because ahe arsBaSemele pconst. Hence, thecomplete integral hastheform y=Cet +e, 536. Diferentiat Equations Example 1.Given the equation 5 ¥+y'—%=0. ‘The auxiliary equation isoftheform fth—2=0. Wefind theroots oftheauxiliary equation: ighaVFas the completeintegrals TE ET?complete integral feecg, ; Il.The roots oftheauxiliary equation arecomplex. Since complex roots areconjugate inpairs, wewrite k=o+iB; kaif, where 2,as—$: BeVHF, The particular solutions may bewritten inthe form gaeerOe, “y,meloire, On) These arecomplex functions ofareal argument that satisfy the differential equation (1)(see Sec. 4,Ch. Vil). Itisobvious that ifsome complex function ofareal argument y=u(x)+10(x) 6) satisfies (1), then this equation issatisfied bythefunctions u(x) and o(2). Indeed, putting expression (5)into (1), wehave (ue) +"+7[4(2)+(X)]+9[u@)+0(2)=O or (e+pu’+44)+6(0"+po"+90)=0. But acomplex function isequal tozero if,and only if,thereal part and theimaginary part areequal tozero; that is, u"+pu+qu=0, of+po"+qu=0. Thus wehave proved that u(x) and v(x) are‘solutions ofthe equation. Letusrewrite thecomplex solutions (4)intheform ofasui ofthe real part and the imaginary part: y,=ecosBx+iet*sin Bx, y,=@*cosBx—is™sinBx. Second-Order Homogeneous Linear Equations 837 From what has been proved, the particular solutions of(1)are the real functions .9,=e"cosBx, @) 9,=e sinBr. ) The functions j,andJ,arelinearly independent, since smSegapeotBeaconst Consequently, the general solution ofequation (1)inthe case of complex roots oftheauxiliary equation isoftheform y=Ag,+By,=A&*cosBx+Be™*sinBx or y=e'*(Acosx-+BsinBx), ® where Aand Bare arbitrary constants, Example 2.Given theequation 6+2y' +5y=0, . Find thecomplete integral and aparticular solution that satisfies the initial conditions Yeag=0, Yeaq=1. Construct thegraph. Solution. 1)We welte the auxiliary equation AP42k-4+5=0 andfindilsroots: by142, y=1-2 Thus, the complete integral is . y=en* (Acos2x+Bsin2x). 2)We find particular solution that satisfies thegiven initial conditions and determine thecorresponding values ofAand. B. rom the first condition we find O=me-*(A cos2-0+8sin2-0),whenceA=0. Notingthat Y= 07728 c0s2—e-*B sin2 we obtain from the second condition 1 1-28,0B=4. Thus, the desired particular solution is iepatho sind Usgraph isshown inFig. 267. IIL The roots ofthe auxiliary equation are real and ‘equal. Here, &,=ky. 538 Digerentiat Equations One particular solution, y,=e%™, isobtained from earlier reasoning, Wemust find thesecond particular solution, which is N SS Se e-fersinte 7 Zonas Pa Fig. 267. linearly independent ofthe first (the function e* isidentically equal toe“*andtherefore cannot beregarded asthesecond par- ticular solution). Weshall seek’the second particular solution intheform y,=u(x)eh* where u(x) istheunknown function tobedetermined. Differentiating, wefind y,sue+hued=eh(u'+h,u), yeaule*+Qku'es* +ktuel*=eb(u"+2k,u'+kt). Putting theexpressions ofthederivatives into (1), weobtain eM[ul+(2k,+p)u’+(Ri+pk,+9)u]=0. Since k,isamultiple root oftheauxiliary equation, wehave A+pk,+9=0. Inaddition, 4,=k,=—% or2k,=—p, 2k,+p=0. Hence, inorder tofind u(x) we must solve the equationeu’=6oru’=0.Integrating, wegetu=Ax+B.Inparticular,we can set A=1 and B=0; then aan Homogeneous Linear Equations ofthe nthOrder 539 Thus, for.the second particular solution wecan take y,=xeh*, Thissolutionislinearlyindependent ofthefirst,since##=xconst,Therefore, thefollowing function isthecomplete integral: y=Cer +Cyxeh eh (C,+C,2). Example 3.Given the equation ofAy +4y=0. Write the auxiliary equation &*—4k-44=0. Find itsroots: j=#,=2. ‘The complete integral isthen y=Ce 4Cyue, SEC, 22. HOMOGENEOUS LINEAR EQUATIONS OF THE NTH ORDER WITH CONSTANT COEFFICIENTS Letusconsider ahomogeneoits linear equation ofthenthorder: y™tay"4...+4,y=0. a) We:shall assume that aj,a,, ..; a,areconstants. Before-giving amethod forsolving equation (1), weintroduce adefinition that will be needed later on. » Definition 1.Ifforallxofthe interval {a,6)wehave the equality Pn(2)=A,®,(4)+A,8)+ee+Anya where A,, A,,+.-,A, areconstants, not allequal tozero, then wesaythat p,(x)isexpressed linearly interms ofthefunctions DC, Pe(Xs oeosPnmy (X)- Definition 2.nfunctions @,(x),Ps(X)s +++» Pn (X), Pa(2)are called linearly independent ifnotone of‘thefunctions isexpressed linearly interms ofthe rest. Note 1.From the definitions itfollows that ifthe functions (2), 0). +1 G(x) are linearly dependent, there will be found constants C,,'C,, ..., C,, not allequal tozero, such that forallxoftheinterval [a,6)thefollowing identity will beful- filled: C®, (4)+Cp,(2)+++CuO,(4)=O. Examples: TThe functions y,—e, yy—e, yy—=Se® arelinearly dependent, since for Cyt, Cyn,Cyd wehavetheldetily CeeCet4Chemo, 540 <<.Digerentiat Equations... +. 2.The functions y—1, wu=x. ye=xt are linearly independent, since the ae CMEC REC wallnotbeMeatcally seroforanyCy,CyCythatare_not simultaneouslyPuifunctions yj—eh,yg—ehF,oo.dye, WhereysyyoonBs2 are different rumbets which are linearly ‘adependent. (This asseftion isgiven without proof) Letusnow solve equation (1). For this equation, thefollowing theorem holds. Theorem. Ifthefunctions yy,Yq,---+Yy arelinearly independent solutions ofequation (1), then’ itsgeneral solution is Y=Cyt Cet oo+Cavan @) where Cy, ..., C_arearbitrary constants, Ifthe’ coefficients ofequation (1)are constant, the general so- lution isfound inthe same way asinthecase ofsecond-order equations. 1)We form the auxiliary equation Rak tak", ta, 2)We find theroots ofthe auxiliary equation’ 3)From the character ofthe roots wewrite out the particular linearly independent solutions, taking note ofthe fact that: a)toevery real root &oforder one there corresponds aparti- cular solution e*; )toevery pair ofcomplex conjugate roots k””—a+i® andk®=a—if therecorrespond twoparticular solutions e*cosBxand e*sinBx; ©)toevery real root &ofmultiplicity rthere correspond r linearly independent particular solutions et xe, atte, 4)toeach pair ofcomplex conjugate roots &=a+iB, A*'—a—if ofmultiplicity pthere correspond 2uparticular so: lutions: e*cosBx, xecos, ..., a-'e*cosBx, e*sinBr, xe“sinBx, ..., 2*-'esinBr. Thenumberoftheseparticular solutions isexactlyequaltothe degreeoftheauxiliary equation (thatis,totheorderoftheglven lineardiferential equation). It'may beprovedthatthesesolutions arelinearly independent. : Nontiomogencous Second-Order Linear Equations sn 4)After finding nlinearly independent particular solutions YueYar+++ YqWeconstruct the general solution ofthe given linearequation: YC +Cytoo+Cad where C,,C,, -.., C,arearbitrary constants, Example 4.Find thegeneral solution oftheequation yao Solution, Form theauxiliary equation Ha1=0 Find the roots ofthe auxiliary equation: heal kerk kel kek Write thecomplete integral YRC Ce Acosxt Bsins, where C,,Cy, A,Barearbitrary constants Note 2.From theforegoing itfollows that thewhole difficulty insolving homogeneous linear differential equations with constant coefficients lies inthe solution oftheauxiliary equation. SEC, 23, NONHOMOGENEOUS SECOND-ORDER LINEAR EQUATIONS Let there beanonhomogeneous second-order linear equation ¥ tay’+ay=F(x). 0) The structure ofthe general solution ofsuch anequation is determined bythe following theorem. Theorem 1.The general solution ofthenonhomogeneous equation (1)isrepresented asthesum ofsome particular solution ofthe equation y*and thegeneral solution yofthecorresponding homo- geneous equation +49 +a,9=0. 2) Proof. Weneed toprove that thesum yaoty @) isthegeneral solution ofequation (1). Let usfirst prove that the function (3)isasolution of(1). Substituting thesum y+y* into (1)inplace ofy,weget Oty HQGry tag ty)=le) $0 Differential Equations or YUHay+ay)+Uy"+ay”+a,y*)=F(x). (@) Since 7isasolution of(2),theexpression inthefirstbrackets isidentically zero. Since y*isasolution of(1), the expression inthesecond brackets isequal tof(x). Consequently, (4)isan identity. Thus, the first part ofthe theorem isproved. We shall now prove that expression (3) isthe general solution ‘ofequation (1); inother words, weshall prove that thearbitrary constants that enter into the expression may bechosen sothat thefollowing initial conditions aresatisfied: Yrury=Yor vis 5Yr=ry=Yor} ® nomatter what thenumbers x,,y,andy,[provided thatx,is taken from the region where the functions a,,a,and f(x) ‘are continuous}. _ Noting that 7may begiven intheform 9=Cy,+ Cy where y,and_y, arelinearly independent solutions ofequation (2), and C,and C,‘arearbitrary constants, wecan rewrite (3)inthe form y=Cy, +Cy, ty* @) Then, bythe conditions (5), wewill have *) CetClan+H=Yor CuntCuntut=. From this system ofequations wehave todetermine C,and C,. Rewriting thesystem intheform CY.+CaYan=4,a yetCatan=YoHe 6Ciute+ Cue=Hoth © wenote that the determinant ofthis system isthe Wronskian forthefunctions y,and y,atthepoint x=x,. Since itisgiven that these functions are linearly independent, the Wronskian is notzero; consequently, system (6)has adefinite solution, C, *)Here, YioYanYooHierYowr98denote thenumerical values “otthe Functions yyyYmY%sYinYasYPwhen xy, Nonhomogeneous Second-Order Linear Equations 543 and C,;inother words, there exist values C,and C,such that formula (3)defines the solution ofequation’ (1) which satisfies thegiven initial conditions. The theorem iscompletely proved, Thus,- ifweknow the general solution yofthe homogeneous equation (2), the basic difficulty, when integrating the nonhomo- geneous equation (1), lies infinding some particular solution y*. We shall give general method forfinding the particular so- lutions ofanonhomogeneous equation. The method ofvariation ofarbitrary constants (parameters). We write thegeneral solution ofthe homogeneous equation (2): y=Cy, +Cy a We shall seek aparticular solution ofthe nonhomogeneous equation (1)intheform (7), considering C,and C,assome (as yet) undetermined functions ofx.Differentiate’ (7): re:y=CutCuntCy,+Cot Now choose theneeded functions C,and C,sothat thefollowing equation isfulfilled: = Cy, +Cy, =0. (8) Ifwetake note ofthis additional condition, thefirstderivative y’ will take the form ; yfHCW+Cy. Differentiating this expression, wefind y': YFRCHACTCin+Cys. Putting y,y’and y”into (1), weget Cy +CetCrys+Crysta,CytCys)+. +a, (Cy, +Cy,) =F(x) or -, woe ve LGAA+AaY)+L,atah+ays)+Cini+Civ=F(2). The expressions inthefirst two brackets vanish, since y,and y, aresolutions ofthe homogeneous equation. Hence, thelatter equa: tion takes’on ‘the form ts _ ;Cin +Cae =10). O} Thus, ihefunction (7)will beasolution ofthenonhomogeneous equation (1)provided thefunctions C,andC,satisfy thesystem ‘of,equations (8)and(9);that is,if CytCu=0 CtCyi=l(e). oa Differential Equations Since the determinant ofthis system isthe Wronskian forthe linearly independent functions y,andy,,itisnotequal tozero, Hence, insolving thesystem wewill find C;and C,asdefinite functions ofx: ; . C=9,(2),C=9,(x). Integrating, we obtain C=fSe@de+e; C,=Sq@)drtZ,, where C,andC,areconstants ofintegration.Substituting {heexpressions obtained ofC,andC,into(7),we find anintegral that isdependent onthetwo arbitrary constants G,andC,;that is,wefind thegeneral solution ofthenonhomo- geneous equation *). Example. Find thegeneral solution oftheequation Lae Solution. Letusfind thegeneral solution ofthehomogeneous equation v y—Lao. Since. gat wehaveIny’=Inx+InC; y'=Cx; and s0 yaCut+Cy. Forthelatterexpression tobeasolution ofthegivenequation, wehave todefine C,and C,asfunctions ofx[rom thesystem CPE C10, 2Clx+Ch0—n. Solving this system, wefind a oeGay. Gaps whence, alter integration, weget GQaZtt, G=-F4+t, Putting thefunctions obtained into theformula y=Cx?-+C,, we gettheteneral solution ofthenonhomogeneous equation |<" " ecient ory=Ct+C,4%, whereT,and7,arearbitrary constants, *)ItweputC,=T,=0, wegetaparticular solution ofequation (1). Nonhomogeneous Second-Order Linear Equations 545, When seeking particular solutions, itisuseful thetake advan- tage ofthe results ofthe following theorem. Theorem 2.Letthenonhomogeneous equation ¥+ay' +ay=F, (0)+h) (10) besuch that theright sideisasumoftwofunctions, f,(x)andf,(x). Ify,is@particular solution oftheequation ytay’ +ay=F, (2), ayy and y,isaparticular solution oftheequation tay’ +ay=f, (x), (12) then y,-+Y, isaparticular solution *)ofequation (10). Proof. Substituting theexpression y,+y, into (10), weget +H) +4, ty) +4, +H )=h +hOD or Vitay Fay) +Gitay+ay,)=h@)+h (0 (13) From equations (11) and (12) itfollows that equality (13) isan identity. And thetheorem isproved. SEC. 24. NONHOMOGENEOUS SECOND-ORDER LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS Suppose wehave the equation ¥teu +qy =F(a) a where pand qarereal numbers. ‘Ageneral method forfinding the solution ofanonhomogeneous equation was given inthe preceding section. Inthecase ofan equation with constant coefficients, itissometimes easier tofind aparticular solution without resorting tointegration. Letusconsi- derseveral suchpossibilities forequation (1) I.Let the right side of(1)bethe product ofanexponential function byapolynomial; that is,oftheform He)=P, (ee, 2) where P,(x).isapolynomial ofdegree n.Then thefollowing par- ticular cases are possible: *)Obviously, the appropriate the ‘ins true for any aumber oftedNltheappropriate theoremcemainstueforanyumber0 18-2088 56 Differential Equations a)The number aisnotaroot oftheauxiliary equation H+ pk+q=0. Inthis case, the particular solution must besought forinthe form aA +A +... +A e=Q, ()e @) Indeed, substituting y*into equation (1)and cancelling e**out ofall terms, we will have Qa(x)+(20+P)Qa(x)+(a?+patg)Q(x=Py(). (4) Q,(t) isapolynomial ofdegree n,Qn(x)isapolynomial ofde- gree n—1, and Qj(x) isapolynomial ofdegree n—2, Thus, n-degree polynomials arefound ontheleft and right oftheequa- lity sign, Equating thecoefficients ofthe same degrees. ofx(the number ofunknown coefficients isn-+1), wegetasystem ofn-+1 equations, fordetermining theunknown ‘coefficients Ay.Ay “b)The‘number aisasimple (single) rootoftheauxiliary equa- tion. Ifin this caseweshouldseektheparticular solutioninthe form (3),then ontheleftside of(4)wewould have apolynomial ofdegree n—1, since thecoefficient ofQ,(x),that is,a*+pa+q isequal tozero, andthepolynomials Q(x) andQz(x) have deg- rees less than n.Hence, (4)would not beanidentity, nomatter what theA,Aj,..., 4,Forthis reason, the particular solution inthis case’has'to betaken intheform ofapolynomial ofdegree n+l, butwithout the absolute term (since the absolute term of this polynomial vanishes upon differentiation) *): y= 2Q, (x). ©)The number aisadouble root ofthe auxiliary equation. Then, asaresult ofthesubstitution ofthefunction Q,(x)e*into thedifferential equation, thedegreeofthepolynomial isdiminished bytwo unils. Indeed, ifaisthe root ofthe auxiliary equation, then a’+pa+q=0; moreover, since aisadouble root, itfollows that 2a=—p (Since byafamiliar theorem ofelementary algebra, thesum oftheroots ofareduced quadratic equation isequal to thecoefficient ofthe unknown inthe first degree with sign rever- sed). And so2a+p=0. : +)Weremark thatalltheresultsgivenabovealsoholdfor,thecase whenaisacomplex number (thisfollows fromtherulesofdifferentiation of the function e*, where misany complex number; seeSec. 4.Ch. Vil). emetenne orate meer eee Pe Consequently, ontheleftside of(4)there remains Q;,(x), that is,apolynomial ofdegree n—2. To obtain apolynomial of degree nasaresult ofsubstitution, one should seek theparticular solution inthe form ofaproduct ofe*bythe(n+2)nd degree polynomial. Then the absolute term ofthis polynomial and the first-degree term will vanish upon differentiation; forthis reason, they need notbeincluded inthe particular solution. Thus, when aisadouble root ofthe auxiliary equation, the particular solution may betaken inthe form yt=xQ, (x)em. Example 1.Find thegeneral solution ofthe uation V+4y' +3y—x, fetetln. The goer! lationoftheresgoodlghomogesnes gunsFae +Ce. Since theright-hand side ofthegiven nonhomogeneous equation isofthe forsee ita ig ofeach eatoaeeea epee che equation k'+4k++-3=0, itfollows that weshould seek theparticular solution {nthe form y*=Q, (eM; inother words, weput plane Subelittng this exjenton tothe given easton, wewll nee Gtaaeries Eeguaing thecocina ofLenin deren of, weet 3A=1,44,434,=0, mee ; aets Amd. comyueny“ patedpatent, ‘Thegeneral solution ofy=yty* will be x. nary|4 yaletle +5es, Hzanple 2Fed thegeneral sltln ofhesquation ¥+9 =lt+ Ie, Saletan, The geet saluton ofthe homogeneous squiion lsrelfound: se idFHC,cos3+C,sin3x. The seh ie offongiven equation ce ba Snform Py(xye™, i 648 Dierential Equations Since thecoefficient 3intheexponent isnotaroot oftheauxiliary equa- tion, weseek the particular solution inthe form PH Qe of ya (Ast+Bx=Ce. Substituting this expression inthe differential equation, we will have (9(APH Bx+C)+6(2AL+B)+2A+9(Aat+BrCea(attNe. GarcetingouteMand equating the coefficients ofidentical degrees ofx,we obtain WAm1, 1244188=0, 24468-418C=1, 1 15 whenceAmik:B=—dh:C=S.Consequently, theparticularsolutionis (Mp ty5 w=(tna and thegeneral solution is 1 15)ae y=C, cos3e-4C,sin3x4(eeatm) eo, Example 3.Tosolve theequation PTY+6y=0—2)4. Solution. Here, theright side isoftheform P,(x)e™ and thecoeficient1 intheexponent iasimplefootoftheaunliary polynomial. Hence,weseek theparticular solution Intheform y*=20, (2)¢* or parce Be: putting this expression intheequation, weget(AstBi+A+28)4-27(AstBs)—7Ar+B)+F6(AP4Billee208 “ (—10Ax—5B-+24) F(xet Equating thecoefficients ofidentical degrees ofx,-we gel —l0A=1, —584+2A——2, 1 49 whenceA=—7,, B=. Consequently, theparticular solution is . 149vae(-prtg) and thegeneral solution ts eco 1gpacyteacete(—heeg)en Il.Let theright side have theform J(2)=P(x) ecosBx-+ Q(x) e*sinBx, © whereP(a)andQ()arepolynomials This case may beconsidered bythetechnique used inthepre-cedingcase,ifwepassfromtrigonometric functions toexponential Nonhomogencous Second-Order Linear Equations 9 functions. Replacing cosfx and sinfBx byexponential functions using Euler's formulas (see Sec. 5,Ch. VII), weobtain pxcibent pxnet Fay=P yeEHE™ 5QuyesPoem or ; fa=[FP+H) ]errs[pPM—gz QU]rm.(6) Here, thesquare brackets contain polynomials whose degrees are equal tothehighest degree ofthe polynomials P(x) and Q(x). WeLathusobtained therightsideoftheformconsidered in Case I. ~ Itisproved (we omit theproof) that itispossible tofind par- ticular solutions which donot contain complex numbers. Thus, iftheright side ofequation (1)isoftheform F(x)=P(x)e*cosBx+Q(x)e™sitiBx, @ where P(x) and Q(x) arepolynomials inx,then theform ofthe particular solution isdetermined asfollows: a)ifthe number a+-iB isnot aroot oftheauxiliary equation, then theparticular solution ofequation (1)should besoughtinthe formyt=U(x)&*cosBx-+V(x)e%*sinBx, (8) where U(x) and V(x) arepolynomials ofdegree equal tothehigh- estdegree ofthepolynomials P(x) and Q(x); b)ifthenumber otiBisarootoftheauxiliary equation, wethen write theparticular solution intheform yt=x[U (2)e**cosBx-+ V(x) esinBx]. O) Here, inorder toavoid mistakes wemust note that these forms ofparticular solutions, (8)and (9), are obviously retained when one ofthe polynomials P(x) and Q(x) onthe right side ofequa-tion(1)isidentically zero;thatis,whentherightsideisoftheorm P(x)e*cosBx orQ(x)e™ sinBx. . Let usfurther consider animportant special case. Lettheright side ofasecond-order linear equation have the form I(x)=McosBx+NsinBx, 7) where MandNareconstants.a)IfBiisnotaroot oftheauxiliary equation, theparticular solution should besought inthe form y*=Acos Bx+B sinBx. (8) 550 Differential Equations b)IfBiisaroot oftheauxiliary equation, then theparticular solution should besought inthe form yt=x(A cosBx+B sinBx). @) Weremark that thefunction (7') isaspecial case ofthefunc- tion (7)[P(x)=M, (Q)x=N, a=0); thefunctions (8°) and (9’) arespecial cases ofthe functions (8)and (9). siEZ#M0I€ 4Findthecomplete integral ofthenonhomogeneous linearequa VHD+5y=208x. Solution, Theauxiliary equation &t-+-2k+5=0 hasroots ky=—1-42i; Aye13. Therefore, thecomplete integral ofthecorresponding homoge: neous equation is Fae-*(C,cos2e-+-C,sin2s) Weseek theparticular solution ofthe nonhomogeneous equation intheform =Acosx+Bsinx, where Aand Bare constant coefficients tobedetermined. Putting y*into thegiven equation, wewill have AcosxBsinx+2(—Asinx-+Bcos2145(Acosx+Bsinx)—=2cosx, Equating the coefficients ofcosx and sin, we gettwo equations forde-termining AandB: « " -A42B45A=2; —B—2A458=0, 1. pat whenceAwd; Bat. ‘Thegeneral solution ofthegiven equation isy=7-+y*, that is, yaen™(C,c0s246,sin2x)+Ecosx7hsinx. Example 5.Tosolve the equation V+4ycos2,Solution.Theauxiliary,equationhasroots,=2/,y=——2%;therefore, ‘thegeneral solution ofthe homogeneous equation 1softheform FC,cos2x-+C,sin2e. We seek the particular solution ofthenonhomogeneous equation inthe form th yt=x(Acos2e-+Bsin2x).yt=2x(—Asin2x-+Bcos28)4-(Acos2x+Bsin2x),y=—4x(—A cos2e—Bsin2x)+4(—Asin2e+Bcos24), Putting these expressions of,the derivatives into the given equation andequalinghecoelclentsofcoseandsin2er'wegetasjsemafequations fordetermining AandB:Bal; —44=0, Higher-Order Nonhomogeneous Linear Equations 551 whenceA=0andB=-1.Thus,thecompleteintegralofthegivenequationis y=0,005246,sin2x4-1xsin2s, Example 6.Tosolve the equation y=Secons. Solution, The right side ofthe equation has the form I(x) =e (Mcos.x-+N six), andM=3,N=0.Theauxiliary equation &*—1—0 hasrootsk=1,ky=—I. ‘The general solution ofthe homogeneous equation is aCe +Ce-*. Since thenumber a-+iB=2-+i-1 isnot aroot ofthe auxiliary equation, we seek the particular solution inthe form y*=e* (Acosx+B sinx). Patting this expression into theequation, weget(ater collecting like terms) (2A+4B)e*cosx+(—4A +28)e™sinx=3e™cosx. Equating thecoefficients ofcos and sinx, weobtain 2A+4B=3, —444+2B=0. Whence A=B,andB=. Consequently, theparticular solution is mee (33sax gmet(jGeortgains), and thegeneral solution is ary (3 3, =CarpCenttet(coset Zsins). SEC. 25. HIGHER-ORDER NONHOMOGENEOUS LINEAR EQUATIONS Letusconsider theequation Yay"... ay=F(Xs 0) where a,,a,..., dq,{(x) arecontinuous functions ofx(orcon- stants). Suppose weknow thegeneral solution Y=Cw, +CU o+Calle @ ofthecorresponding homogeneous equation Fay Fay +...bay =0. ) Asinthecase ofasecond-order equation, thefollowing asser- tion holds forequation (1). 562 Digerentiat Equations Theorem. Ifyisthegeneral solution of-thehomogeneous equa- tion (3) and y*isaparticular solution ofthe nonhomogeneous equation (1), then f yaa+u" isthegeneral solution ofthenonhomogeneous equation. Thus,theproblem ofintegrating equation (1),asinthecase ofasecond-order equation, reduces tofindingaparticular solution ofthe nonhomogeneous equation. Asinthecase ofasecond-order equation, theparticular solution ofequation (1)may befound bythemethod ofvariation ofpara- meters, considering C,,Cy,++.»Cyinexpression (2)asfunetions of x Weform thesystem ofequations (ef.Sec. 23): Ciy,+Cy,+--+Cry,=0, Cy+Cige+++Cain=0, Cry4Cy +...+Cay=0, Cag Cay +oe+Cay?=f(x). Thissystemofequations withtheunknownfunctionsC;,C;,..., C,has very definite solutions. (The determinant ofthecoef- ficients ofC;,Ci,...,CxistheWronskian formed fortheparti- cular solutions ¥,,¥,,---, Yqofahomogeneous equation, and since these particular'solutions are, bydefinition, linearly ‘inde- pendent, theWronskian isnotzero.) Thus, thesystem (4)maybesolved forthefunctions Ci,C;,..., C,. Findiug them and integrating, weobtain CalCde4E; Cra[Cdr Gs---3C=SCrde+F,, where C,,C,,..., €,aretheconstants ofintegration. Weshall prove that insuch acase theexpression Yr=Cy, tC t--- +CY 6) isthegeneral solution ofthe nonhomogeneous equation (1).Differentiate expression (5)ntimes,eachtimetakingintoaccountequations (4); this yields HHCY, +O ACY, +oe+Caer yh=Cyi+Cys +Cyat o--+Cans gM =CyO $C +.Oye”, yO +CY + ACU +E(RD. Higher-Order Nonhomogencous Linear Equations 553 Multiplying theterms ofthefirst, second, ...and, finally, second tothelast equation bya,,a,-,, .-., a,respectively, andadding, weget ygFaye +...Hay=f(x), since Jy)Yy«++, Yqare particular solutions ofthe homogeneous equation; forthis reason, the sums oftheterms obtained inadding vertical columns areequal tozero. Hence, the function y*=C,y,+...-+Cyq [where Cy, ..., Cyarefunctions ofxdetermined from equations (4)|isasolution of the nonhomogeneous equation (1), and since this solution depends onthe arbitrary constants C,,C,,«.-, Cy itisthegeneral solution. The proposition isthus proved. Forthecase ofahigher-order nonhomogeneous equation with constant coefficients (cf. Sec. 24), theparticular solutions arefound more easily, namely:1.Lettherebeafunction ontherightsideofthedifferential equation: f(x)=P(x)e, where P(x) isapolynomial inx;then wehave todistinguish two cases: a)if@isnotaroot oftheauxiliary equation, then the parti- cular solution may besought inthe form wr=Qwe*, where Q(x) isapolynomial ofthe same degree asP(x), butwithUndetermined coefictents;b)ifaisarootofmultiplicity oftheauxiliary equation,then theparticular solution ofthenonhomogeneous equation may besoughtintheformyt=7Q(3)8, whereQ(2)isapolynomial ofthesamedegreas,P(a). IL.Lettheright side oftheequation have theform F(x)=MoosBx+-NsinBx, where Mand Nareconstants. Then the form ofthe particular solution will bedetermined asfollows: a)ifthe number iisnot aroot oftheauxiliary equation, then theparticular solution hastheform y*=AcosBx+BsinBx, where Aand:B are constant undetermined coefficients; b)ifthenumber Biisaroot oftheauxiliary equation ofmul- tiplicity p,thenyh=x"(AcosBx +BsinBx). 554 Differential Bquations IL. LetHx)=P(x)&*cosBx+Q(x)e*sinBx, where P(x) and Q(x) are polynomials inx.Then: a)ifthenumber «+i isnotarootoftheauxiliary polynomial, then weseek the particular solution inthe form yt=U(2)e%cosBx-+V(x)e*sinBx, where U(x) and V(x) are polynomials ofdegree equal tothehigh- estdegree ofthe polynomials P(x) and Q(x); b)ifthe number a-+i isaroot ofmultiplicity poftheauxiliary polynomial, then weseek.the particular solution intheform yt=x"(U(x)&*cosBx+V(x)e%sinBx], where U(x) and V(x) have thesame meaning asinCase a, General remarks onCases Itand Il,Even when the right side ofthe equation contains anexpression with only cosBx or“only sinBx, wemust seek thesolution inthe form indicated, that is,withsineandcosine.Inotherwords,fromthefactthattherightsidedoesnotcontain cose.orsin, itdoesmotintheleast follow that the particular solution oftheequation does notcontain these functions. This was evident when weconsidered Examples 4, 5,6ofthepreceding section, and also Example 2ofthe present section, Example 1.Find the general solution oftheequation oVmyaa th. Solution. The auxiliary equation A*—1=0 has the roots hel heh hel hank We find thegeneral solution ofthehomogeneous equation (see Example 4, See: 22}: GaCertCe$C,005x40,sin We seek the particular solution ofthe nonhomogeneous equation inthe form WmaAgiAtAy Differentiating y*four times and substituting the expressions obtained into the given equation, weget AgtaAtAgeAa$1. Equating thecoetficients ofidentical degrees ofx,wehave Asal —A=0; —A=0; —Ah Hence aera The Differential Equation ofMechanical Vibrations ‘555 Thecomplete integral ofthenonhomogeneous equation isfound from the formula y=y-+y"tY=Cek+Ce-*+0,cose+Cysinx—xt 1. Example 2.Tosolve theequation yy=Scoss. Solution.The auxiliary equation A—1—0 hastheroots =I, t=—I, besthaat Hene, thegeneral solution ofthecorresponding homégeneous equation‘isGaCet+Cye-*$C,cosx+C,sinx. Further, theright side ofthegiven nonhomogeneous equation has the formTeay=Mcosx-+Wsin, where M=5 and N=0. ‘Since {isasimple root oftheauxiliary equation, weseek the particular solution inthe form yt=x(A cosx+B sinx). Putling this expression into the equation, wefindhs 4Asins—4Bcosx5cosx, 4A=0, —4B=5 orA=0,B=—5. Consequently, theparticular solution ofthedifferential equation is os wa Ssing and thegeneral solution is YRC4CyeF4-6,cos+CysineSxsine, SEC. 28, THE DIFFERENTIAL EQUATION OF MECHANICAL VIBRATIONS Inthis and thefollowing sections weshall consider aproblem inapplied mechanics, and investigate and solve itbymeans of linear differential equations. Let aload ofmass Qbeatrest onanelastic spring (Fig. 268). We denote byy.the deviation ofthe load from the equilibrium position. We shall consider deviation downwards aspositive, upwards asnegative. Intheequilibrium position, theforceofthe weight isbalanced bytheelasticity ofthe spring. Letussuppose that theforce that tends toreturn the load toequilibrium (the so-called restoring force) isproportional tothedeflection, that is, equal toky, where &issome constant forthe given spring (the so-called “spring rigidity”)*). *)Springswhoserestoringforceisproportional tothedeflection arecalledsprings with a“linear characteristic’. 556 Digerential Equations Let ussuppose that the motion ofthe load Qisrestricted by aresistance force operating inadirection opposite tothat of motion and proportional tothe velocity ofmotion ofthe load relative tothe lower point ofthe spring; that is,aforce —ko= =a, where 4=const>0 (shock absorber). Write thedif- ferential equation ofthemotion ofthe load onthe spring. By Newton's second law we have a pOG 0) (here, &and 4are positive numbers). We thus have ahomoge- neous linear differential equation ofthe second order with con- stant coefficients. “a,sen Egutioriumpostion pt ll Epulibium postion—_ <; Z-plt)‘A A. A Fig. 268 Fig. 269, This equation may berewritten asfollows: ay dys Gat Pat y= ay where a & p=qi I=q- Let itfurther beassumed that the lower point ofthespring A executes vertical motions under the law z=@(é). This will occur, forinstance, ifthelower end ofthespring isattached toarol- ler, which moves over anuneven spot together with the spring and the load (Fig. 269). Inthis case the restoring force will beequal not to—&y, but to—k[y+(t)], theforce ofresistance will be—A[y’+9' ()], and inplace ofequation (1)wewill have the equation OTEASE+ky=—he(9%(0: @ ‘Free Osctilations 557. al d%diGetegetay=F). @) wherehettie') ——el+he'(0 f=Oto We thus have anonhomogeneous second-order differential equation. Equation (1’)iscalled anequation offreeoscillations, equation (2') isanequation offorced oscillations. SEC. 27, FREE OSCILLATIONS Let usfirst consider the equation of{ree oscillations ¥+py+qy=0. We write the corresponding auxiliary equation b+ pk+q=0 and find its roots: a Foy: pa? m a $4VERa:a$-VVFa. 1)Let#>q. Thentheroots&,and&,arerealnegative num- bers. The general solution isexpressed interms ofexponential functions: y=Ce +Ce! (k,<0, &,<0). ) From this formula itfollows that the deviation ofyfor any initial conditions approaches zero asymptotically if{—+oo. Inthe given case, there will benooscillations, since theforces ofresist- ance are great compared tothe coefficient ofrigidity ofthe spring k. 2)LetF=q; thentheroots#,and£areequal(andare alsoequaltothenegativenumber—4).Therefore, thegeneralsolution will be he Be att, y=Cet $C (C4000 7 @ Here thedeviation also approaches zero as¢—+00, but not so rapidly asinthepreceding ‘case (due tothefactor C,+C,!). 558 Differential Equations fp’),LttP=0(noresistance), Theauxiliary equation isofthe form #+q=0, anditsroots arek,=Bi; ky=—Bi, where B=V. The general solution is | y=C, cosBf+C, sinBr. @) Inthe latter formula, we replace the arbitrary constants C, and C,with others. We introduce the constants Aand @,,which areconnected with C,and C,bythe relations C=Asing, C,=Acosq,. Aand q,are defined asfollows interms ofC,and C,: AV GHG, 9,=arctan&. Substituting thevalues ofC,and C,into formula (3), weget y=A sing,cosBi+Acos@,sinBr or y=Asin Gf+9,). 6) These oscillations are called harmonic. The integral curves’ are sine curves. The time interval 7,during which the argument of thesinevariesby2a,is gyrAsin(Bt+ pe) calledtheperiodofoscil- AEsay lation;here,TaF.The % frequency isthenumber T-ofoscillations duringtime T-Qn; here, the frequency is Fig.270. B;Aisthe greatest de- viation fromequilibrium andiscalledtheamplitude; @,istheinitial phase.Thegraph ofthe function (3') isshown inFig. 270, 4)Letp#0 andF<q. Inthis case, theroots ofthe auxiliary equation are complex numbers: A=a+if, k,=a—ip, whereD jaea=—$<0, B=V/a—8. The complete integral hastheform y=e" (C,cosB+,sinBr) 4 Forced Oscillations 859 or y=Ae*sin(Bt+9). 4’) Here, fortheamplitude wehave toconsider the quantity Ae which depends onthetime, Since a<0, itapproaches zero as U <_/Drdettsin( ptt) 7 areas 2 Fig. 271. t—+00,whichmeansthatherewearedealingwithdampedoscil-lations: The graph ofdamped oscillations isshown inFig. 271. SEC. 28. FORCED OSCILLATIONS The equation offorced oscillations hasthe form ¥+py’+ay=F(0. Let usconsider animportant practical case when the disturb- ingexternal force isperiodic and varies under the law HO=asinot; then theequation will have the form ¥+py+qy=asinot. aw 1)Letusfirstpresume thatp40 and2<g, thatis,the roots oftheauxiliary equation are thecomplex numbers a+iB, Inthis case [see formulas (4)and (4’), Sec. 27], thegeneral solution ofthe homogeneous equation has the form y=Aesin(Bt-+9,). 2) ‘Weseek aparticular solution ofthe nonhomogeneous equation inthe form y*=Mcosot+Nsinot. @) 560 Digerentiat Equations Putting this expression ofy*into theoriginal differential equa- tion, we find the values ofMand N: = area yy 0aMaGrane N=Grange Before putting these values ofMand Ninto (8),letusintroduce the ‘new constants A*and g*, setting M=Atsing*, N=A* cosq*, that is At=VFN = a »tan gta,VIM ~Temoregey On Om Then theparticular solution ofthenonhomogeneous equation may be written inthe form y*=Atsin@*coswt+A*cosp*sinof=A*sin(wt+9"), or, finally, *=4___ sin(of+9"). P=amareaOl+9 Thecomplete integral ofequation (1)isy=y+y* or y=Aem™sinO40)+pai (of+9°). The first term ofthe sum onthe right side (the solutionofthe homogeneous equation) represents damped oscillations; itdimini- shes with increasing ¢and, consequently, after some interval of time thesecond term (which determines the forced oscillations) will acquire prime importance. The frequency @ofthese oscilla-tionsisequaltothefrequency oftheexternal forcef(¢);theamplitude oftheforced vibrations isthegreater, theless pand thecloser a*isto9. Let usinvestigate more closely thedependence oftheamplitude offorcedvibrations onthefrequency oforvarious valuesofp. For this, wedenote theamplitude offorced vibrations byD(a): Dw@=7—Vo-oF Fru Putting g=B} (for p=0, B,would beequal toitsnatural frequen- cy), wehave D@)=——.——= ~_ . V(BjoF+p*o*a/i-=eea Br) BEBE Forced Oscillations 56 Introducing the notation faa; 2aBoM BY where 4istheratio ofthefrequency ofthedisturbing force to the frequency offree oscillations ofthesystem, and theconstant y isindependent ofthedisturbing force, wefind that themagnitude ofthe amplitude will beexpressed by’the formula D0=—S 4)O-RyTSae i Letusfind themaximum ofthis function. Itwill obviously be forthat value of&forwhich the square ofthe denominator has aminimum, But the minimum ofthe function Va—ey rye (6) isreachedwhen —a :A=Vi-y and isequal to —a Hence, themaximum amplitude isequal to Dux=——2—. tyVie Thegraphs ofthefuriction D(A) forvarious values ofyareshown inFig.272(inconstructing thegraphs weputa=1, B,=1for thesake ofdefiniteness). These curves arecalled resonance curves. From formula (5)itfollows that for small ythe maximum value ofamplitude isattained forvalues of%close tounity, that is,when the frequency ofthe external force isclose tothe fre- quency offree oscillations. Ify=0 (thus, p=0), that is,ifthere isnoresistance tomotion, theamplitude offorced vibrations increases without bound asA—+1 ofaso—>B,=Vq: timDa)j=x.ao, Atot=q wehave resonance, we enats 2)Now letussuppose that p=0; that is,weconsider the equation ofelastic oscillations without resistance butwith a periodic external force:ae , “_LCEALL LTT“CCH eet ETCHAI7 i ANCE haOR NTT‘AT NC LECCE NETos} S—] ~CECE SS HHH: We The general solution ofthehomogeneous equation is G=C,cosBt+C, sinBt (B'=q). IfB#a, that is,ifthe frequency oftheexternal force isnot equal tothenatural frequency, then theparticular solution of thenonhomogeneous equation will have theform y*=M cosat+Nsinot. (6) Putting thisexpression intotheoriginal equation, wefind M=0, N=5 The general solution is y=Asint+)+7a—sinot. ‘Systems ofOrdinary Diferential Equations 363 Thus, motion results from the superposition ofa.natural oscilla tion with frequency Band aforced vibration with frequency w. IfB=o, that is,the natural frequency coincides with the frequency ofthe external force, then function (3)isnot asolu- tion ofequation (6). Inthis case, |, inaccord with the results ofSec. 44" 24,wehave toseek theparticular a solution intheform gfyreaspe 7y*=t(M coswt+N sinot). (7) Ss 7 Substituting thisexpression into|prcont,/ theequation, wefindMandN: wy, M=—z5; N=0. y Consequently, SNyaateaser. \\\ Thegeneral solution willhave Lthe form S y=AsinGt-+q,)—ft cospt. ‘NY The second term on the right S side shows that inthis case the Fig.273. amplitude increases without bound with the time f.This phenomenon, which occurs when the natural frequency ofthe system coincides with the frequency of the external force, iscalled resonance. ‘The graph ofthefunction y*isshown inFig. 273. SEC. 29, SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS Inthe solution ofmany problems itisrequired tofind the functions y,=¥, (2), Ys=Is(Xs«+++Gp=Yn(%),Whichsatisfya system ofdifferential equations containing’ the argument x,the unknown functions y,,¥4,+.» y,and their derivatives. Consider thefollowing’'system ‘offirst-order equations: uy,BEAN YoYrveeYad auFratgeoeee w Fate ee J 564 Digerential Equations where Yj,Yur+++» Yaare unknown functions and xistheargu- ment. Asystem ofthis kind, where the left sides ofthe equations contain first-order derivatives, while the right sides do not contain derivatives, iscalled normal. Tointegrate the system means todetermine the functions YorYar--~+ YorWhich satisly thesystem ofequations (1)and the given initial conditions: Widemse=YaarYadenss=Yaar+++Yudemee=Yo (2) Integration ofasystem like (1)isperformed asfollows. Differen- tiate the first equation of(1)with respect tox: ay,_ay4Ohdy af,diy Mah he. +e, Replacing thederivatives 4%,4,..., 4withtheirex- pressions f,,f,,+.+, f,from equations (1), weget theequation 4FEF Yue Ue) Differentiating this equation and then doing asbefore, weobtain ayGet=P YasYaroesUade Continuing inthe same fashion, wefinally get the equation a ee Wethus getthefollowing system: yyFel Ce ayetHPS, Yn) @) FeFsYarveyWade From thefirst n—1 equations wedetermine (ifthis ispossible) YoYooos+ Jnand express them interms ofx,y,and the ives dH, S44 amtyderivatives $2,59,...,Fatt r=PaErYasYineeeYOM Ye=P lkYeIY, ¥ ny Y= Val! YarGyover YOM) Systems ofOrdinary Differential Equations 565, Putting these expressions into thelast oftheequations (3), we getannth-order equation fordetermining y,: FE=OysYyyeresUm) ©) Solving this equation, wefind y,: Wa CyCyrey Cade (6) Difierentiating thelatter expression n—1 times, wefindthederivatives 24,$8,...,S78asfunctions ofx,CysCys.++Cue Substituting these functions into equations (4), we determine Yo Yor eres Yat M=Bs(%CyCys Cads Yn= alts CyCyy+++ Cn) For this solution tosatisfy the given initial conditions (2), it remains forustofind {from equations (6)and (7)] theappro- priate values oftheconstants C,,C,,..., Cy(like wedid inthe case ofasingle differential equation). Note 1.Ifthe system (1)islinear inthe unknown functions, then equation (5)isalso linear. Example 1.Integrate the system dy a 3Baytits, Bate @ with the initial conditions Weae=heOyae=0. ) Solution. 1)Differentiating the frst equation with respect tox,wehave dy_dy,dzaeaxtaeth Puttingtheexpressions $%and4fromequationsa)intothisequation, wwe getL ayGamWE8)+(—My32+24)41 or . PY ay23H © 2)From thefirst equation ofsystem (a)wefind ratty @ 566 »+Differential Equations ‘and put itinto theequation just obtained; weget a aGham—92(Hye) 43041 “ @cyey WYymdat? gyty=5e+1. (o) ‘The general solution ofthis equation is Y=(C+Cy)e-*+5x—9 © and from (4)wehave B=(Cy—2C, —2Cyx)e-*—6r$14. @ Choosing theconstants CyandCysothattheInitial conditions (b)are. eae=l Weoe=0, weget, from equations ()and (@), 1=6,-9, 0=¢,—20, +14, whence C,—10 and C,=6. tort ‘thesolution thatsatisfiesthegiveninitialconditions (b)hasthe y=(10-46x)e-*$529, =(—M128)e644, Note 2,Intheforegoing weassumed that from thefirst n—1 equations ofthe system (3) itispossible todetermine the functions y,, Y,-+-, Y,. Itmay happen that the variables Yys+++ Yqareeliminated notfrom n,butfrom asmaller numberofequations, Then todetermine y,wewillhave anequation oforder less than n, Example 2.Integrate thesystem trey Wig &Gath Batt Gaety. Solution. Dilferentiating thefrst equation with respect to¢,wefind ae_dy,de Seat Gout atito, aeSeetetyte Eliminating thevariables yand 2from theequations de ae Heyer Masetyts, wwegetasecond-order equation inx: de, de$5MEoem, Systems ofOrdinary Diferential Equations 567 Integrating this equation, weobtain itsgeneral solution: alel4Cet, @ Whence we find ae ot26andyetiseCet20gH GataCt+2CeMandy= Crem2Ce"#—2, (e) Putting into the third ofthegiven equations the expressions that have beenfoundfor'sandy,wegetanequation fordetermining #: a actHy emscyet, Integrating this equation, wefind2aCe-!4-Cet, ” But then, from equation (B), weget Y=— (C+ Cy)em! Cee, Equations (a), (B), and (y)give thegeneral solution ofthegiven system. The differential equations ofthesystem cancontain higher-order derivatives. This then yields asystem ofdifferential equations of higher order. For instance, the problem ofthe motion ofamaterial point under the action ofaforce Freduces toasystem ofthree second-order differential equations. Let’ Fe, Fy, Fzbetheprojections oftheforce Fonthecoordinateaxes. Theposition of”thepoint atanyinstant oftime#isdetermined by Uscoordinates x,y,and 2.Hence, x,9,2atefunctions of{.The projections ofthevelocity vectorofthepointontheaxeswillbe2,4¥,42 Suppose that theforce Fand, hence, its projections Fy, Fy, F,dependonthetimettheportionx,j,0hThepoint,andofth”velocityofix’ “dy dt motionofthepoint, thatis,on$¢, 44,4 Inthis problem thefollowing three functions arethesought-for functions; raz, yy, e=2(0. ‘These functions aredetermined from equations ofdynamics (Newton's law): ae dx dy dzmgine (innaeB), ay dx dy deCd Cee ee ® ae dx dy denGanFe(4nnnEH,#)- Wethus have asystem ofthree second-order differential equations. In.the case ofplane ‘motion, that is,motion” inwhich the ‘trajectory isaplane 568 Diflerentiat Equations curve (lying. forexample, inthexy-plane), wegetasystem oftwo equations fordetermining the functions x(2)and y(J: ae ae dymtn (naw EH) o ay de dymGhnt, (snFH). (io) Wiposable tosolveasystemofdiferentia equations. ofhigherorder byreducing ittoasystem offirst-order equations, Usingequations (®)and (Hb) asexamples, weshall show how this isdone. Weintroduce thenotation ar, woan” a Then dr_dud'y_dv aa ana The system oftwo second-order equations (9)and (10) with two unknown functions x(0)and () 1sreplaced byasystem offour Arstorder equations with four unknown functions x,4,u,0: fog, an dyean dumapaPalleHssO $F xyw,OD mah yl Hy uO) Weremark inconclusion that the general method that wehave considered atsolving. the‘system ‘mayinceriin spec cases, bereplaced bysome artificial technique that gets the result faster. Example 3.Tofind the general solution ofthe following system of differential equations: ty - poe a Say, Solution. Differentiate, with respect tox,both sides ofthefirst equationtwice a ae ButSay, andsowegetafourth-order equation: “iene lotegrating this equation, weobtain Itsgeneral solution (see Sec. 22 Example 4):paOyeh$Ce-*$0,00540,508, Systems ofLinear Diferential Equations 569 Finding 4%fromthisequation andputting itintothefirstequation, we find 2: 2aCe 4Ce-*—C, cos2—C,lax SEC. 30. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS Suppose wehave the following system ofdifferential equations: Brean, tattoo tate de eeci (ty aeSTFas,+Ogakyoo+danke where thecoefficients @,,areconstants. Here, ¢istheargument, and x,(f), x,(1),-.., 4(0) are the unknown functions. The system (1)is'a system ofhomogeneous Linear differential equations with constant coefficients. Aswaspointed outinthepreceding section, thissystem may besolved byreducing ittoasingle equation oforder n,which inthegiven instance will belinear (this was indicated inNote 1ofthe preceding section). But system (1) may besolved in another way, without reducing ittoanequation oftheathorder. This method makes itpossible toanalyse the character ofthe solutions more clearly. Weseek aparticular solution ofthe system inthefollowing form: Kaae, x=ae, 0... cece, Oy Itisrequired todetermine the constants a,, a,,..., a,and & insuch away that the functions ae“, ae", ..., a,e“” should satisfy thesystem ofequations (1).Putting them into System (1), weget: : hae =(a,,0, +-4,,0,4 ...+0,,0,) e", hea=(a0,+4,,0,+...+44,0,)2, hae! =(ag, +40 ++Ogg eM. 870 Diflerential Equations Cancel oute“,Transposing allterms toone side and collecting coefficients ofa,,a,,..., G,wegetasystem ofequations: (a,—8)a,+4,,0,+++040,=0, 44,0,+(y4— A)G+ 6+04,0, =0, 5,0, 45,04 «+ (Qqa— A)og 0. Choose ,,cy,..., a,and &such that will satisty thesystem (3). This isa’system oflinear algebraic equations ina,,a,,... Gy. Let usform the determinant ofthe system (3): Jak yy veeOy Gy,Oy—hoeOy Toia @) Gq, Ang +++(Gun—h) If&issuch that the determinant Aisdifferent from zero, then the system (3)has only trivial solutions a=a,—...—a,=0 and, hence, formulas (2)yield only trivial solutions: 4(Q=x, (N=... =4,(00. Thus, weobtain nontrivial solutions (2) only for&such ‘that the determinant (4)vanishes. We arrive atanequation oforder nfordetermining &: A,—k Oy ossyy Gy, O4,—k oes Oy yy Ogg sesygHe This equation iscalled thecharacteristic equation ofthesystem (1), and itsroots are theroots ofthecharacteristic equation. Let usconsider afew cases. I.The roots ofthe characteristic equation arereal and distinct.Denoteby&,,Ry++»Bqtherootsofthecharacteristic equation.For each root “k;write the system (3) and determine the coefficients ai), al,..., al, Itmay beshown that one ofthem isarbitrary; itmay be considered equal tounity. Thus weobtain: fortheroot-k, thefollowing solution ofthesystem (1) xaalret, xSade, ...,x=atdehts ‘Systems ofLinear Differential Equations sm forthe root ,the solution ofthesystem (1) xraairel!, xaired! ..., xmalteht, fortheroot &,the solution ofthe system (1)x10)airetat,x10amet, txalteta! Bydirect substitution into theequations weseethat thesystem offunctions x,=Care +Cares! +... +C,aiett, x=Cael +Cares 4...+Cyalmetst, a =Carel! +Careh! +... 4+C,airerst, where C,, C,,..., Cyare arbitrary constants, islikewise a solution ofthe system ofdifferential equations (1). This isthe general solution ofsystem (1). Itmay readily beshown that one can find values ofthe constants such that the solution will satisfy thegiven initial conditions. Example 1.Find thegeneral solution ofthesystem ofequations dry ty,Fiat eyGanon. Solution. Form thecharacteristic equation 2k 2Pat24-0 or—5k+4=0, Find itsroots: hal, had Seek the solution ofthesystem inthe form aad, Waae and aaa, mal ot Form thesystem (3)fortheroot&,=1 anddetermine a{”andaf (21)of+20"=0, tal"+—1)al"=0 or af?4-200”=,af?+20,=0, 5m Diflerentiat Equations whence af=—a,Putting af!=1,wegeta{!=—4.Thus,weoblain the solution ofthesystem: Mod, et,Mad, Wage. Now form thesystem (8)fortherootAy=4 anddetermine af"andof”: —2a{")+20)=0,af?—20!"=0, whence of?=a andaf=1, af—=1, Weobtain thesecond solution of the system: Peet, Poet The general solution ofthe system will be[see (6)] nae+cet, mam 7Cel+Cat I.The roots ofthe characteristic equation are distinct, but include complex roots. Among the roots ofthecharacteristic equa- tion letthere betwo complex conjugate roots: k=atip, k=a—ip. Tothese roots will correspond thesolutions xPmafMerrion Gal, 2... ,), ” xPaaf ett Gael, 2... 1). ® Thecoefficients ajandaf”aredetermined from thesystem of _equations (3). Just asinSec. 21,itmay beshown that thereal and imagi- nary parts ofthecomplex solution arealso solutions. Wethus obtain two particular solutions: x}=e(Af?cosBx+A)”sinBx),\ ® x}? =e"(A)?sinBx+4)”cosBx), whereaf”,A),4),2)"arerealnumbers: determined interms ofa”andaf”, Appropriate combinations offunctions (9)will enter into the general solution ofthesystem, Systems ofLinear Diferentiat Equations 573 Example 2.Find the general solution ofthe system dyGant 48onse, 326,58, Solution. Form the characteristic equation -7-k 1[Ze" sal or#4 12k-437=0 and find itsroots: h=—64i, abt Substituting Ay=—6-40intothesystem(8),wefind at, of=14. We write the solution (7): alert, (1 piel t, co) Putting k,=—6—i into system (3), wefind at,af=1—e Wegel asecond system ofsolutions (8): wWadotit, Magy e-tat, ® Rewrite the solution (7’): amen (cost+ising), HPO40e-(cost+iste) or ametcosttleMsin, a4)eM(608t—sint)4le"(cos+81a0). Rewrite the solution (8: afmenMcos(testat, 22)=e"(cos(—sinf)—le!(cost+inf). Forsystems ofparticular solutions wecantake therealparts andtheimagl- nary parts. separately: HMse-Heost, FY=e-"(cos¢—sint), eo, ‘ ” HY—e-Msint,. Hae (cosf+sin, The general solution ofthe system is 4,=Cye-"'cos (+Ce-“sint,y=Cee"(cost—sinf)+Ce(cos¢-+-sin1). 4 Digerentiat- Equations Bya’similarmethoditispossibletofindthesolutionofasystem Haadifferential equations ofhigherorderwithconstant coef- ficients, For instance, inmechanics and electric-circuit theory astudy ismade ofthe solution ofasystem ofsecond-order differential equations: a GEAttOya aa (10)GeAOnk+ayy. Again weseek thesolution intheform x=ae, y=pett, Putting these expressions into. system (10) andcancelling outeM wegetasystem ofequations fordetermining a,Band &: (@,,—H)a+a,,8=0, ana,,0-+(a,, —k*)P=0. Nonzero aand Baredetermined only when thedeterminant of thesystem isequal tozero: aka, This isthe characteristic equation ofsystem (10); itisafourth- order equation in&.Letk,,fy.k,.and k,beitsroots (we assumethattheroots aredistinct). For“each root&;ofsystem (11)we find the values ofaand f.The general solution, like (6), will have the form £=Cae+Care+Careh+Carer!, y=CPreN TCM! +Cpe +Cprent, Ifthere arecomplex roots, then toeach pair ofcomplex roots inthe general solution there will correspond expressions ofthe form (9). Example 3.Find the general solution ofthe following system of aierentiat equations as fae, fyftsty, Systems ofLinear Differential Equations 875 Solution. Write the characteristic equation (12) and find itsroots: Ine 4[ot i=. bak hank =VT =—VS We shall seek the solution inthe form Maalrel, —gapivelt,foramen, java pie-, Ama eh ymapne’t, MmQeVEt, yiapigYF", From system (11)wefindaandpu: ama, pond, wont, pad, omni, pmm—t, ovat, peed, Wewrite out thecomplex solutions: MaeHecosttisint, Y=t(ost-tising, #1e~Hecost —isint,=(Costin, ‘The real and imaginary parts separately form thesolutions Host, I=foost, Homsint, P=Pint We can now write thegeneral solution: X=Cy08£40,sin+6,07?!4.0,0°¥74,PRCeonlesCysint—C, oe!Che, Note. Inthissection wedidnotconsider thecase ofmultiple roots ofthecharacteristic equation, This question isdealt with in detail in“Lectures ontheTheory ofOrdinary Differential Equations” byLG. Petrovsky. 576 Diferentiat Equations SEC. 31. ON LYAPUNOV’S THEORY OF STABILITY Since thesolutions ofmost differential equations and systems ofequations arenot expressible interms ofelementary functions orquadratures, useismade (inthese cases when solving concrete differential equations) ofapproximate methods ofintegration. The elements ofthese methods were given inSec. 3;inaddition, some ofthese methods will beconsidered inSecs. 32'through 34andin Chapter XVI. The drawback ofthese methods liesinthefact that they yield only one particular solution; toobtain other particular solutions, one has tocarry out allthe calculations again. Knowing one par- ticular solution does not permit ustodraw conclusions about the character ofthe other solutions. Inmany problems ofmechanics and engineering itissometimes important to'know not thespecific values ofasolution forsome concrete value ofthe argument, but the type ofbehaviour for changes intheargument and, inparticular, foraboundlessincrease ofthe argument. For example, itissometimes important toknow whether thesolutions thatsatislythegiveninitialconditions are periodic, whether they approach some known function asymptoti- cally, etc. These arethequestions with which thequalitative theoryofdifferential equations deals. One ofthebasic problems ofthe qualitative theory isthat of the stability ofthe solution orofthe stability ofmotion; this problem was investigated indetail bythenoted Russian mathe- matician A.M.Lyapunov (1857-1918). Let there begiven asystem ofdifferential equations: ae waht» 7PA a)aahlt *y) Let x=x(t) and y=y(t) bethe solutions ofthis system that satisly theinitial conditions Stee %e .Yoe=Yor} “ Further, letx=x(f) and y=y(f) bethesolutions ofequation (1)that’ satisfy theinitial conditions Finehe\ a)Yine= Yor OnLyapunoo's Theory ofStability 87? Definition. Thesolutions x=x(t) andy=y/(é) thatsatisfy theequations (1)and theinitial conditions (I’) arecalled Lyapunov’s stable ast—+-o ifforevery arbitrarily small e>0 there isa5>0 such that forallvalues ¢>-0 the following ‘inequalities will be fulfilled: ko—xOl<e \aa ly@—yOl<e. aG-Grvett fe iftheinitial data satisfy thein- i raequalities willl® |z,—x,1<8, \— 3) xln—-vl<6 fig. Letusfigureoutthemeaning ig274. ofthis definition. From inequali- ties (2)and (3)itfollows that forsmall variations inthe initial conditions, thecorresponding solutions differ butlittle forallpositive values of¢.Ifthesystem ofdifferential equations isasystem that describes some motion, then inthe case ofstability ofsolutions, thenature ofthe motions changes but slightly forsmall changes inthe initial data, Letusanalyse anexample of9fst-order equation Lettherebegiven2differential equation: aay tt Oy The general solution ofthis equation isthefunction y=Ce-! 41. (b) Find-a particular solution that satisfies the initial condition Yiwo=l. ©) Itisobvious that this solution y= results when C=0 (Fig. 274). Then find the particular solution that satisfies the initial condition Yaado Find the value of©from equation (b): R=C+1, whence C=y—1. Putting this value ofCinto equation (), weget G=G—Net 41. The solution y=I Isobviously stable. 19 3388 578 Differential Equations Indeed, _ man 9—G=1Go— Nem" 1=G—Newt+0 THence, inequality @)will befulfilled foranarbitrary &ifthe following inequality isfulblled W—I)=b<e. Letusfurther consider thesystem ofequations Saatey, “Hemant by, assuming that thecoefficients a,b,c,gareconstant and g=0. Letusfind outwhat conditions thecoefficients must satisfy sothat thesolution x=0, y=0 ofsystem (4)should bestable. Differentiating the first equation and eliminating y,weget a second-order equation: a_deydy_ode as aeGame teGt—cF+e(ax+by)=eFF+age+b(Fe) or (b+) F—(ag— be)x=0. 6) Isauxiliary equation isofthe form M—(b+c)h—(ag—be)=0. 6 Letusdenote theroots oftheauxiliary equation by4,andA,. The following cases are possible. 1.The roots oftheauxiliary equation arereal, negative and distinct: A<0, 4,<0, AAD. Then xa +Cet, HAICO,—Oe+C,0,—c)eI 2. The solution that satisfies the initial conditions Xion Yt=Yor will be tenths gt4ue onF[Si a,get4aaa Goel]. OnLyapunov's Theory ofStability 579 From the latter formulas itfollows that forany e>0 itis possible tochoose x,andy,sosmall that forall¢>0 wewillhave lx®@l<e, ly@l<e since <1 and &'<i, Hence, inthis case thesolution x=0, y=0 isstable, 2.Let4,=0, 4,<0. Then x=C,+Cyer', y=PIC,Q—cje¥—eC,), and thesolution, asinthepreceding case, proves stable, 3.Let A,=4,<0. Then ee(C,+C,eM, Y=FOMIC A—I+C,(U4—el) Since te 0and &t—+0 when t—+00, itfollows that forsufficiently small C,and C,(that is,forsuffi ciently small x,and y,)wewill have |x(I'<e and |y(Q|<eforany£>0. Thesolution isstable. 4.Let, =4,=0. Then x=C,+Cf. YRFHC, +0, ,n1. Weseethat foranarbitrarily small C,0 both xandyapproachinfinity (as¢—+00), which means thalthesolution inthiscase isunstable. 5.Letatleast one ofthe roots 4,and A,bepositive; for instance, 2,>0. Fromformula (7)itfollowsthatnomatterhowsmallx,and Yyi or,+a,—Xb, £0, that is,ifC,40, then |x(¢)|—+00ast—+00, Hence, inthis case too the solution isunstable. 6.The roots oftheauxiliary equation arecomplex with nega- tive real part: a,=a+if,RItHh Nace 1 co) Diferential Equations Inthis case, x=Ce" sin(Bt-+8), =y=}Cet[(a—c)sin(Bt+8)+c0s(Bt+8)}®) Itisobvious that forany e>0 itispossible tochoose x,andy, suchthatwewillhave|C|<e andato <sand,conse- quently, [x(|<e and |y@|<e. The solution isstable. 7.The roots ofthe auxiliary equation are pure imaginaries: 2,=Bt, = Br. Inthis case,x=Csin(Bf+6),9=FCIp.cos(Be+6)—csin(Bf-+8)] which means that x(é) and y(f) are periodic functions oft.As inthe preceding case, weverify the solution and find itstable, 8.The roots of‘the auxiliary equation are complex with positive real part (a>0). From formulas (8)itfollows that here forarbitrarily small x, and y,(that is,for arbitrarily small C0) and forincreasing the quantities "|x()| and |y(f)| can take onarbitrarily large values, since e“Sos asf—+0o. Thesolution isunstable. Togive ageneral criterion ofthe stability ofsolution ofthe system (1), wedoasfollows. Wewrite theroots oftheauxiliary equation inthetorm ofcom- plex numbers: Wate, amt (inthecase ofreal roots, 4;°=0 andA;"=0). Letustaketheplaneofacomplex variable A++*anddisplay the roots ofthe auxiliary equation bypoints inthis plane. Then, onthebasis ofthe eight cases that have been considered, the condition ofstability ofsolution ofthe system (4)may befor- mulated asfollows. 5 Ifnotasingleoneoftherootsy,i,oftheauxiliary equation (6)liestotheright oftheaxis ofimaginaries, and atleast one Toot isnonzero, then the solution isstable; ifatleast one root Euler's Method ofSolution ofDifferential Equations 581 lies totheright oftheaxis ofimaginaries, orboth roots areequal tozero, then the solution isunstable. Let usnow consider amore general system ofequations: de Faetev+P(ev| dy a)Grartoy +e,w.| But forexceptional cases, thesolution ofthissystem isnotexpres- sible interms ofelementary functions and quadratures. Toestablish whether the solutions ofthis system are stable or unstable, the system iscompared with the solutions ofalinear system. ‘Suppose thalforz—-0andy—-0, thefunctions P(x,y) and Q(x, y)also approach zero and approach itfaster than @, where 9=Vz'+g%; inother words, lim2&9; timC4<0,one oe Then itmay beproved that, save fortheexceptional case, the solution ofthe system (4’) will bestable when thesolution of thesystem de_antey.|dyHoax +by, isstable, and unstable when the solution ofthe system (4)is unstable.’ The exception isthat case when both roots ofthe auxil- iary equation lieonthe axis ofimaginaries; inthis case, the question ofthe stability orinstability ofsolution ofthesystem (4') isconsiderably more involved Lyapunov *)investigated thequestion ofthestability ofsolutions ofsystems ofequations forrather general assumptions concerning the form ofthese equations. SEC, 32, EULER'S METHOD OF APPROXIMATE SOLUTION OF FIRST-ORDER DIFFERENTIAL EQUATIONS We shall consider two methods ofnumerical solution ofafirst- order differential equation. Inthis section, weconsider Euler's method. *)A.M. Lyapunov, TheGeneral Problem ofStability ofMotion, ONTI, 1935 (Russian edition), 582 Differential Equations Find (approximately) the solution ofthe equation Hass, ») 0) ontheinterval [x,, 6]that satisfies theinitial condition atx=x, y=vq- Divide the‘interval (x,,6]bythepoints x,,2,ty...) _= 0 into nmequal parts (here x,<x,<x,<...<x,). Denotex,—x,==x,—4, 2...b—4,.,= Arh;‘hence, nae, Lety=@(x) besome approximate solution ofequation (1)and Y= P(E) Y=PUdeveerYn=P(Eade Denote AY.=Ys—YorMY,=YaYares»MYnns=YnYame Ateachofthepointsx,,x,,..., x,inequation (1)we-replace the derivative with the ratio offinite differences: . Hale v e Ay=F(x, yydx @) When x=x,wehave SH=F(x0,),AYy=T(tyY)Ator ‘ Yy—Yo= Fas Yo)Ae Inthis equation, x,,y,,hareknown; thus wefind Hatley yh. When x=, equation (2') takes the form : Ay=F yh or Waly Ydhy Wy= IHF yeWhe Here, x,,y,,&areknown andy,isdetermined, Similarly, wefind Ya YatTay Ya)hy Yavs=Yatl(ayYa)ty Yn= Yams FESamay Inns)he Euler's Method ofSolution ofDiferentiat Equations 583 We have thus found the approximate values ofthe solution at thepoints x,,x,,..., X,-Connecting, inacoordinate plane, the oink tenSaUeBlooaeUaBY" Straight-line ‘segments, we“get abroken | line—an approximate integral curve (Fig. Grn) 275). This broken line iscalled Euler's lo: broken line. om Note. Wedenote byy= (x)anapprox- imate ‘solution ofequation (1), which |p,yycorresponds toEuler’s broken line when Ax=h. Itmay beproved®) that ifthere exists aunique solution y=@*(x) ofequa- 9 % GT tion (1)that satisfies the initial condi- Fig.5.tions and isdefined onthe interval [x,, 8},thentim], (&)—@* ()|=0 forany¥oftheinterval (xy,8) Example, Findtheapproximate value(forx=1) ofthesolution oftheequation Weete that satites the jitial condition ya=l forx40.Selon. Dividetheinterval(Bilni"T0artsbythepointsx40,OsOBee»10,Hence,A=0.1. Weseekthevaluesyy,Ys+--+Ynbyforeuence Aa=Uy4)ot tenia We thus get n=1+0-$0)-011401=11,e114 0.)-0.1—1.22, Tabulating theresults aswesolve,weget: <0 1.000|1,000 0.100Btn fo} 1200 0.1292203 ago|1490 o:142£203 1362|1620 162ete Ven|om O:19%5208 iis|aiates O:22162208 1:g980|6980 0:2598S209 arog|3ceol8 0.2810peer aurso|3:zn0 oarBo a:aoos|3:7008 0.3700Bato 3.1708 | 7)For the proof see, for example, 1,G.Petrovsky's “Lectures onthe Theory ofOrdinary Differential Equations”. 584 Differential Equations Wehavefoundtheapproximate valueyleaj=3.1703. Theexactsolution ofthis equation that satisfies the indicated tnitial conditions is pater. Hence, ear =2(e —1)=3.4365. 0.2662 ‘Theabsolute erroris0.2662 therelative erroris9-250"=0.07~8%. SEC, 33, ADIFFERENCE METHOD FOR APPROXIMATE SOLUTION OF DIFFERENTIAL EQUATIONS BASED ON TAYLOR'S FORMULA. ADAMS METHOD We once again seek the solution ofthe equation y= Fix, 9) a) ontheinterval [x,, 6],which solution satisfies theinitial condi-tiony=y,whenx=.x,.Weintroduce notation thatwillbeneededlater on.’ The approximate values ofthesolution atthe points PED onan will be The first differences, ordifferences ofthe first order, are AYe=Yy—YorBYy=YeYayoesBYnm4=Iu—Ynnr The second differences, ordifferences ofthesecond order, are Aty,= Ay,—AY,=Y4— 24,+Yor Ay, =A9,— AY,=4-2 +Hy Bp =AYnm1— AYns=Yn Ys FYnme Differences oftheseconddifferences arecalleddifferences ofthe third order, andsoforth. Wedenote byys,ys,«+++Untheapprox- imate values ofthe derivatives, and byYaUi)...» Yqthe approximate values ofthe second derivatives, etc. Similarly we determine the first differences ofthe derivatives: Ayo=Y:—Yor AYr=Y2— Yas++>AYnma=Yn—Yams the second differences ofthe derivatives: Atys=Ayi—Aya,Aty)=Ayi—Ay,+A*Ynna =AY Ayame and so on. ADifference Method forApproximate Solution ofDifferential Equations 585 Write Taylor's formula for solving anequation inthe neigh-bourhood ofthepointx=,[Ch.IVsSee.6.formula’ (6): ed ee et ) Inthisformula y,isknown, andthevalues ofys,gs,...ofthe derivatives are found from equation (1)asfollows. Putting theinitialvaluesx,andy,intotherightsideofequation (1),wefind yi: Yo=F(KerYa) Differentiating the terms of(1)with respect tox,weget a4 Ayv=Z4gy. @) Substituting into theright sidethevalues x,,y,,yswefind -_(a4a,»el(1Oncemoredifferentiating (3)with,respecttoxandsubstitutingthevalues x,,Y,,ys,yo,wefindy,”.Continuing inthisfashion.*.wecanfind‘the’valuesofthederivatives ofanyorderforx==x,)Allterms areknown, except theremainder R,,ontheright side of(2). Thus, neglecting the remainder, wecan"obtain anapprox- imation ofthe solution for any value ofx;their accuracy will depend upon the quantity |x—x,| and the number ofterms in the expansion Inthe method given below, wedetermine byformula (2)only thefirst fewvalues ofywhen |x—zx,| issmall. We determine thevalues y,and y,for x,—2,-+h and forx,—2,-+2h, taking four terms ofthe expansion (y,isknown from theinitial data): nantt atatk, i) Dh (2A) (BAY ee o=e EyeOT “ We thus consider known three values**) ofthe function: y, Y. Yx On the basis ofthese values and using equation (1), *)From now onweshall assume that the function f(x, 4)isdifferentiable witnreapect foandgas,manytimesasisrequired bythereasoning*+)'iPweweretoseckthesolution withgreater accuracy, wewould have tocompute more than the frst three values ofy.This fsdealt with Iadetailby.Ya.S._Betikovich in“Approximate Calculations” (Gostekhizdat, 1949) (Russ'an edition). *586 Diferential Equations we find , . Y=Fey Yad =EUs Ysdy Ye=Flay Yade Knowing y.,yi.yi,itis possible todetermine Ay, Ay., Atys. Tabulate the results ofthe computations: E foe |vw|w|iw * |4# al | | [an | nanth |ow 4% ay, | | [an | ee ee | eet OK|tae tha| | | | [avs | ee re [ati | [anes nant |om|ok| | Now suppose that weknow the values ofthe solution Yor Yar Yar ores Yor From these values wecan compute [using equation (1)] thevalues ofthe derivatives oy ; YorYayYaserryYo and, hence, _, , Ayo, AYiy ++++AYaas and : BY Yin eee MYkme Let usdetermine the value ofy,,, from Taylor’s formula (see Ch. V,Sec. 6),setting a=x,, x=%,,,—=x,+h: yg Be Hn anYur=YetTtpoet rage teta MtRee ADifference Method for Approximate Solution ofDifferential Equations £37 In our case we shall confine ourselves tofour terms ofthe expansion: hg Be BnYrs=YatTtpata Me 6) The unknowns inthis formula arey,andy{”, which weshall trytodetermine byusing theknown first-order and second-order differences. First, represent y;_, inTaylor's formula, putting a=.x,, x—a=—h: a a ra et aesSoa PASP © and yj_,, putting a=x,, x—a=—2h: yt (PN) pe (DA oeyaaIeSO9,+SOyp 0) From (6)wefind VpYoor=AY =TYe3 Hi (8) Subtracting theterms of(7)from those of(6), weget Yoes— Yann=Mena =TY Yas ® From (8)and: (9)-we obtain AY Apa =AY =" °r 1 Ye=Aay (10) Putting theexpression y;"into (8),weget Ahan Oaaaase. an Thus, yjandyj”have been found. Putting expressions (10) and (11) into theexpansion (5),weobtain bya Bay 45hage Yur =Yet TUF TAesth (12) This isthe so-called Adams formula with four terms. Formula (12)enables onetocompute vanwhenYasWaco,Yesateknown, hus,knowing y,,y,andy,we'can findy,and,further, ys,Yu... Note 1.We state without proof that ‘ifthere exists a'unique solution ofequation (I)ontheinterval [x,,6],which solution satisfies the initial conditions, then theerror’ oftheapproximate 588 Dierential Equations values determined from formula (12) donot exceed, inabsolute value, MA‘, where Misaconstant dependent onthe length of theinterval and theform ofthefunction f(x,y)and independent ofthemagnitude ofA. Note 2.Ifwewant toobtain greater accuracy inourcomputa- tions, wemust take more terms than inexpansion (5), and for- mula (12) will change accordingly. For instance, ifinplace of formula (5)wetake aformula containing five terms tothe right, that is,ifwecomplete itwith aterm oforder A,then inplace offormula (12) we, insimilar fashion, get the formula bahay Rasy 4BRaay Yas=YatTetFAY +GOYeatTYee Here, y,,, isdetermined bymeans ofthe values Y4,Yp-as Yrs and yy, Thus, inorder tobegin computation using ‘this formula wemust know ‘thefirst four values ofthesolution: 4,Y,.Uys¥. When calculating these values from formulas oftype (4), one should take five terms ofthe expansion, Example 1.Approximate the solution ofthe equation yoyte that satisfies the initial condition y= when x=0. Determine the values ofthe solution forx=0.1, 0.2, 0.3, 0.4. Solution. First wefind yyand yyusing formulas (4)and (4’). From the ‘equation and the initial data weget Y=WAeo= HotO=1+0—1. Differentiating thegiven equation, wehave vayth. Hence, KAW +Veww=1412 Differentiating once again, weget yay. Hence, tee Vy=H=2 Substituting into(4)thevalues ys,yj.vyandA=O.1, weget pate LOM2OMpeice Similarly, forh=0.2 wehave went422OM222ans ADifference Method forApproximate Solution ofDiferential Equations 589 Knowing yo,YuYayWefind (onthebasis oftheequation) Y=Ht0=1, Y=,+0411.110840.1=1.2108,9-=Ua+0.2=1.2426+0.2—1.4426, Ay,=0.2103, Ay,=0.2323,1%)=0.020. Tabulating the values obtained, wehave BE|se=com|yet| | | Iay)=0.2108| n=O| geat.2i09| |a,=0.0m0 02|n=1-208|detaee| |o%.=00m | | |dy,0.2551 n=03|gentoo|gate| | ny|a | From formula (12) we find yy _ on 0.1, 5-(0.1) =1.2426491.442642 «0.20204 2G).0.02201.3977Wethenfindthevaluesofy,Ay‘,A%y{.Againusingformula(12)wefindy*: 1.397742 1.6077491.0.2561+50.10.0228 1.5812. ‘Theexact expression ofthesolution ofthegiven equation is yatta. 590 Diferential Equations Hence, venagt2e'4—O4—I-= 1.5896.Theabsoluteeroris0.00%;therelax tiveerror,7-004=0.0015~0.157,. (InEuler'smethod,theabsoluteerror ‘ofy,is0.06, the relative error, 0.038=3.8%/,.) Example 2.Approximate thesolution oftheequation vawee that satisfies theinitial condition yp=0 forxy=0. Determine thevalues of the solution for z=0-1, 0.2, 0.3, 0-4. : Solution. Wefind.” =O +00, Few20+28)gag=0, Vene=24"+200"+2)cag2 By formulas (4)and (4) we have Gy 2 y=GpP=0.0003, yy3-20.0026. From theequation wefind¥=0,(=0.0100,y;=0.0400. Using these data, weconstruct thefirst rows ofthetable, and then deter- mine the Values ofjyandy, from formula (12), <0|n=0 |m0| | | | axi=oor00 | n=O|10.0008 |v=0.0100| |‘74,=0.0200 | | 4202|o=0.0026 |¥=0.0000 | |‘a'y;-=0.0201 | ‘Ay-=0.0501 | =03|s4=0.0089|¥-=0.0901 | nod|vnoom| | | ‘AnApproximate Method forIntegrating Digerential Equations 391 Thus, =0.0026-+20.0400+°:!.0.0300 45-0.1-0.0200 =0.0089, ,=0.0089+-2:1.0.0901 +-2:1.0.0501 +-5,.0.1.0.0201 =0.0204. Wenote that the first four correct decimals inyyare yy=0.0219. (This may beoblained byother, more accurate, methods with errot evaluation.) SEC, 34. AN APPROXIMATE METHOD FOR INTEGRATING SYSTEMS OF FIRST-ORDER DIFFERENTIAL EQUATIONS The methods ofapproximate integration ofdifferential equa- tions considered inSecs. 32and33arealso applicable forsolving systems offirst-order differential equations, Here, weconsider the difference method forsolving systems ofequations. Our reasoning willdealwithsystems oftwoequations intwounknown, function. Itisrequired tofind the solutions ofasystem ofequations 4Fahey 2) a de Eh. y2) @ that satisfy theinitial conditions y=y,, z=z, when x=x,. Wedetermine thevalues ofthefunction yand 2forvalues of the argument. x,Xj)X61aFagusory Oncemore,let KeepAeh(E=O,1,2,...,0=D, ) We denote the approximate values ofthe function as YorYar=+29YarYroreoeYn and Write the recurrence formulas oftype (12), Sec. 33:heahat4Sanne rss =etAGetEMeasFSAAYb “ hehe, Space Zee PeEET Aten +GAA tae (5) Tobegincomputations usingtheseformulas wemustknowyu4x2,,2,inaddition toy,andz,;wefind these values from formu-lasoftype(4)and(4’5,Sec.32: A he Ee W=YtTHtT WtGM» Dh OWE#AY =H Pe Oe yl, 592 Digerentiat Equations a ee ARAttAtTatges Dh oe (2h) = (A)? er 2,524 FtGE4 a", Toapply these formulas onehastoknow Yo,YorYouZorZarZa’ which weshall now determine, From (I)and (2)wefind Y= (erYor2s 201, CherYor20> Differentiating (1)and (2)and substituting thevaluesofx,,y,,2 y.andz,wefind HoWea (H+Fy+22Yee! £=Oenm (H+hy+Be) Differentiating once again, wefindys”and2”.Knowingy,,¥4. 2,2»wefind from thegiven equations (1)and (2), YasYou2s2pAyl, Aus, Oty, Aza, Azi, A%2, after which we can fill inthe first five rows ofthe table: IpeeSelae ||jt [ftfe Lf ft ftp fe| ||fw] fT fe slelél PTdelet | AnApproximate Method forIntegrating Diferential Equations 59 From formulas (4)and (5)wefind y,and 2,,and from equations (1)and(2)wefindy,and x,.Computing Ay:, ty, Az, Atz, wefind y,and y,,etc., byapplying formulas (4)and (8)once again. Example. Approximate the solutions ofthesystem yet, aay with initial conditions yy=0 and z4=1 forx=0. Compute the values ofthesolutions forr=0, 0.1,6.2,0.3,0.4. Solution. From'the given equations, wefind WaFeae=hs HaYene0. Differentiating thegiven equations, wefind WWxnoene=O, He eae= Weveby WWVena(2eneby 2° = Vane Wxae=0- Using formulas oftype (4)and (5), wefind Obey aONEgyON|mph 4OE9OM"0.1002, 0492.40.2"g4.2" 20492 1OF9 OF 10.2016, Ot9,O.N?,,(.F sateStoOM4".on1,0080, 0.290.2;40.2"g_ 2$22.94 OM5 OM =1.0200. Using thegiven equations, wefind j= 1.0080, y= 1.020, 7/=0.1002, #=0.2016, ‘ay,=0.0060, ‘82,=0.1002, Ay, =0.0150, az}=0.1014, ‘4%y,=0.0100, 6%=0.0012. so Digerential Equations Filling inthe frst five rows ofthe table, wehave reaa xno|nao|iat | ac |=0.1002|y=1.00850| |‘aty,=0.0100 | ay,=0.0150| x024=0.2016|v=1.0200 |exo. | | |ay)=0.0059| 203=0.3049|y=1.0459| | a0|yc04N17 | | 5 5 7 ow oe ano|nel|x0 | | |a=o.00| so|2,=1.0050|0.1002| |av,—o0n2 y=0.2|sy=1.0200|si=o2n6| |ax/=a0019 | | Teqaoroa_| na03|4=1069 |2,=0309 | anos|zo1.0817| | | Exercses onChapter XIII 55, From formulas (4)and ()wefind noanits 2.14% ons -01-d0m—020, sem.0m0424120164 2.001648,01.00012=1 089 and similarly a4 01 5aen0.3049-+2:1.1.0459 49.0.0280.45.0.1.0.0109=0.4117, 24=1.0459-4-2:.0,3049-+9:1.0.1033-+-5,.0.1-0.0019=1.0817.7 z @ Itisobvious that the exact solutions ofthe system of equations (the so-lutions satistying theintialconditions) willbe“7 Stations ( §tes), 2a =x)gabe, ca perten. And s0,solutions correct tothefourth decimal place are WAP ene04107,em(eten)1081, Note. Since equations ofhigher order and systems ofequations ofhigherorderimmanycasesreduce to.system offirstorder equations, themethod given above itsppiicsble tothesolution ofsuch problema, Exercises onChapter XIII Show that theindicated functions, which depend onarbitrary constants, satise fythecorresponding differentia) equations:Funetions Differential Equationssine ay 1 1.yesing14 Ce, Mycosz=sind ay\*_d bpecepc—ce (ie)tateenmo. * De 3.y=2Cecr, 9(34)+2Bpmo.tacet20 —(40)") acta lt trace 2. ssf(4)Jaen onlt acre lt fySo 5yaletS40, $442Stn, = ac oYoptspyo, &y= Cne +EH Thanh Rymet,suciar4cg-suniian yy an T.y=Cettrek4Cye-aarcsing, x9Fhe aty=0. 6 dy, 2dy 8yaSG, Spode. 596, Differential Equations Integrate the differential equations with variables separable 9%ydx—xdy=0. Ans. y=Cx. 10.(1+u) 0du+(I—v)udo=0. Ans. Inuo+ +u—v=C. ,MW(l+y)de—(l—x) dy=0. Ans. (1-+y)(I—x)=C.1atBetiteo.Ans.HpIn£6.18y—ayde-+stdy=0 1 a 5 sof=4 ae ae OT ade=0.Ans.@=C7T5. 1a eF = f ATER. anseefEE ttendt—VTAsm0, Ans, 207—arcans=C. 17.de+qtanOd0=0. Ans.p=Cos.18sinOcos@40— rosOiadg—0.Anscosp=Ccos.19.secOtangd0-+sec@ tan0dq—0. Ans.tanOtan@=C. 20.sec*Otan@dp-+sec*ptan@dd—0. Ans.sin*@+satg=C. 21. (L429 dy—VT—prdxn0. Ans, aresiny—aretanx=C. 2.Vimxtdy—VT—gtde=0. Ans.yVTat VT—pt. 8.Setanyx Xdx+(1—e%)sectydy=0.Ans.bragfotstan 24,(x—ytx)det +y—xy) dy=0.Ans.xPyxy Problems inForming Dilferential Equations 25. Prove that acurve having the slope ofthe tangent toany point pro- portionaltothe‘atscisaofthe"pointoffangeneyis&parabola”Ans mart C. G 26.Findacurvepassingthroughthepoint(0,—2)suchthattheslopeof thetangent atanypointofitisequalttheordinate ofthispointincreased bythree units, Ans yme—3, ErFund carvepassing through thepoet(14thattheslopeofthe tangent tothe curve'at any pointis proportional othe square af the ordi: nateolthispointans,EGEI)y yet.a:rind@cre far‘ich theHopeofthe tangent. atanygoatistimestheslopeofastraight lineconnecting thispointwiththe ong. Ans#78" Through thepoint(2,1)drawacurveforwhich thetangent atanypointcoincideswith'thedirectionoftheradiusvectordrawnfromtheorigintothesamepoint.Ans.y=-y-x. 30.Inpolar coordinates, find the equation ofacurve ateach point of wietheangen oftheanglebetween heradi vector andthe tangent Tine isequal fothe reciprocal ofthe radius vector with sign reversed. Ans reycmldibs polarcoordinates, tin,theequation ol,acurveateachpointof whieh the Tangent ofthe angle formed bythe radius vector and-the tangentTinetsequalfothesquareoltheradiusvector.Ans,==20+-C).S32! Prove that acurve with the property that allitsnormalepassthrough aconstant point isacircle 38, Find’a curve such that ateach point ofitthe length ofthesubtan gent isequal tothedoubled abscissa. Ans. y=C Vr. ‘0 Find acurve forwhich the radius vector sequal tothe length ofthe tangent belween the point oftangency and thex-axis Solution. Byhypothesis, LV]y%—VEG,whencem4 tn Exeréises onChapter XII 57 35, By Newton's law, the rate of cooling ofsome body inair is propor-tional{0'thediferencebetweenthetemperature ofthebodyandtheRempe- rature ofthe air. Ifthe temperature ofthe air is20°C and the body cools for20minutes from 100° to60"C, how long will ittake forits. temperature todrop to30°C? Solution. Thedifferential cquation oftheproblem is47=a(7—20), Inetegratingweind:T—20—Cet;T=100when=O;T=60when¢—=20;there-fore,C=80;40=Ce™*,e=(5)% consequently, T=m+00(7)%, AesumingT'=90,wefindt=60mia.36Duringwhattime7willthewaterfowoutofanopening0.5cmat thebottom ofaconicfunnel 10cmhighwiththevertex angled=60"? Solution. Intwo ways wecalculate thevolume ofwater that will flow out during the time between the instants ¢and ¢-+-At. Given aconstant rate ¢, during 1seeacylinder ofwater with base 0.5cm® and altitude Aflows out, and during time a¢theoutflow isthevolume ofwater dvequal to —dv=—050dt=—0.3 Vighat.") Onthe other hand, due totheoutflow, the height ofthe water receivesa negative “increment” dh, and the differential ofthe volume ofwater oulllow domartdh=F (h-+0.7)dh Thus, Eenpontdna—03 Vihar, whence = 0.0315 (10"—A") 40.0732(10°A") 40.078(V10—Vi). Settingh=0,wegetthetimeofoutflowT=12.5sec.37.The retarding action offriction onadisk rotating inaliquid isprow portional tothe angular velocity ofrotation w.Find the dependence ofthis Angular velocity onthetime ifitisknown that the disk begins. rotating at 100 revolutions per minute and, after theelapseof one minute, rotates at60 vl revolutionsperminute. Ans, @=100( 2).rpm. 35,Suppose thatinavertical colufnn’of alrtheir,pressure ateachevel isdue tothe pressure ofthe above-lying layers. Find thedependence ofthe pressure onthe height ifitisknown that atsea level this pressure is1kg Percm, while at900 mabove sealevel, 0.92 kgper cmt, Hint, Take advantage oftheBoyle-Mariotte law, byvirtue ofwhich the density ofthegas Isproportional tothe pressure. The differential equation oftheproblemisdp=—kpdh,whencep=e-"™™. Ans.pme-me™h *)Therateofoutflowvofwaterfromanopeningadistanceffromthe iseaa-tacaltaetroubytbeformeslal 0.)V%h,wheregistheacceleration ofgravity. 508 Diflerentiat Equations Integrate the following homogeneous differential equations:30.Yn)detrts)dynO.Ans.yetDymateC. M0(eby)dxbedy=O.Ans. xt2y=C. At.(ety)de+(y—2)dy=0. Ans. In(x+y! — —metanhaC, 4xdy—ydemVEPGde.Ans.142Cy—Ciet=0. 48.(8y+10x)dx-+(Sy+7x)dy=0. Ans.(x-+y)*(2e+y)"=C. 44.(2Vit—s)x xttds=o.Ans.teVTacorstint, 45.(—9dtttds=0. Ans. thacor satin, asytdymcrty ds.Ans.yarVOM.a1,eos(yde-tedy)—ysin(edy—yde,Ans,xycosLac. Integrate the differential equations that lead tohomogeneous equations: wgate) deeYd. Ans,(etyeymirc 49,(£29+1)dx—@r+4y-+3) dy=0.Ans.inthefy$8)+8—tec, 50. (e+2y-$1)de—@r—9)dy=0.Ans,Inx32, 51,Determinethecurvewhosesubnormalisthearithmetical meanbetween theabscissa and theordinate. Ans. (—y)*(e+-2y)—C. ‘52Determine thecurve inwhich the ratio ofthesegment cutoffbya tangent onthey-axistotheradiusvectorisequalto'sconstantGeer (3)'-(S)r=% Solution.Byhypothesis,apeaymmwhence&)-(¢)-%. 58. Determine thecurve inwhich theratio ofthesegment cut offbythe normal onthe x-axis tothe radius vector isequal toaconstant, dyeto Solution. Itteelventhatj7—EGam, whence xh+ytamteC) 54,Determine thecurveinwhich thesegment cutoffbyatangent onthe y-axis isequal {0@'see6, whore Oistheangle between theradius vector and fhe eens. Solution. SincetanO=% andbyhypothesis avprtmasecd, wwe obtain dyVEEP poeng VEEP whence fae -(200 55,Determine thecurve for which the segment cutoffonthey-axis by ‘2normal drawn tosome point ofthe curve isequal tothedistance ofthis point trom theorigin. Exercises onChapter XIII 599 Solution.Thesegmentcutoffbythenormalonthegaxisisy+;therefore, byhypothesis, wehave otha FTR whence xt=C(Qy+C). 56, Find the shape ofamirror such that all rays emerging from asinglepointOwould berellected parallel tothegivenditection.Solution.Forthez-axiswetakethegivendirection.and0a8theorigin, baom‘betheincident ray,MPthereflected ray,andMQthenormal to re desired curve. a=; OM=09, NM=y, NQ=NO+0Q—=— 14VFFPmycotBay, whence pdy=(—3+ VFRds, Integrating, we have yact+2cx, Integrate thefollowing linear differential equations: ye. Ans.2y=(x (x-+1)8.58,yaatt! Shy SETI, Ans.2y—E+YCOI.88,yo EET, Ans.y=Cetpe332, BRGY+Oy—ar=0. Ans.y= waryCx VIRH. 60,Seostspssintmt. Ans.smsint+Ccost. oh.Hp fscost=y sin2t.Ans,s=sint—14Ce-™!, 62.y’—Symete™ Ans, Pe4O),0.yey. Aneyar. tyme. Ans 12 + cyaetC.68Sy1a.Ansyas(14c07) Integrate theBernoulli equations: 68.yfexpat. Ans,Flt+1+Ce)=1, 67.(Lx)yxy—axy=0,Ans,(CVT==—a) y=1,68,y'y—ay*—z—1=0. Ans.oyCaenmaeti=t. myteyttanat. Ansxle—yye?* 4c]ae?70.(yinx—2)y de=xdy.Ans.y(Cx+Inx-41)=1. Tl.y—y’cosx=y*cosxx tan 24secx x(l=aingy,Ane,yoBEES, ‘oo Diferential Equations Integrate the following exact differential equations: 12.(ety)dx-(e—2y)dy=0.Ans.[rte—rac] 73.(—3ehdx— (ly)dy=0,Ans.Paypet=O.TA.WP—ayyey. Ans,yrodey+C. vot | vyee) Pe es Pe to ac. 1[tet [F-e] eneans, $276."2ty+28)de4-3Oty+97dy0. Ans.443444y'=C.sae+OeEDMYog,Ans,in(e+y)—— =C.7(3)pey 7SryemAns.niedae=6.7(Bt)dente dnsatgtecet 10ERLE An,A.Ondeydy 4. Ans.stot ParetanTC. sikDelermine thecarvethatas,fheproperty thattheproguct ofthe square ofthe distance ofany point ofit {rom the origin info the segment ‘ut ofonthe x-axis bythe normal atthis point isequal tothe cube ofthe abscissa ofthis, point.dns,QamC.2Find the envelope ofthe following families oflines: a)y=Cx-+Ct Ans.x-44y=0. b)y=Z-bC* Ans,2Txt=Ay", o)caeraed Ans.Bye.d)Cxp-Cy—1=0. Ans.y*44x=0. @)(EC —O}'=C. Ans.£20, a0. 1) OF -bylmdC. Ans. omarth.@)eCU—Cmd ‘ns. @—y)t=8 tyCoCyai.Ant,£p4y= "A straight line isinmotion sothat the sum ofthe segments itcuts ‘offonthe axes isaconstant a.Form the equation ofthe envelope ofal positions ofthestraight line. Ans. x'*-y'>—a' (parabola)84.Findtheenvelope ofafamily ofstraight linesonwhich thecoordi- nateantscutoffasegment ofconstant lengtha.Ang.=hy?=a" fs:Find the envelope ofafamily ofcircles whose diameters are the doubledordinates oftheparabola y*=2px. Ans.20at). 86,Find the envelope ofafamily ofcircles whose ctntres Heonthe pa- rabola g#—-2p; allthecircles ofthe family pass through the vertex of”this parabola, Ans. The elssold 2°-¢-y*(x+2)-20. '7.Find theenvelope ofaTamily ofcircles whose diameters arechords oltheeliseaye! pereniclr tothe eax Aasge ae 488,Find theevolute oftheellipse 6%"-+a'y*—a"H* astheenvelope ofits normals,Ans.(ax)?(by)?=(at—08)*- Integrate the following equations (Lagrange equations): oi c 20—p* ne 80.y=2ey’ty".Ans.reguppF. 00.yaytty"Ans.y=(VEFI+CP Singular solution: y=0, 81. yax(ty)tW). Ansee eyecDShethpe,ae!pe. Ans. Exercises on Chapter XIII ot 4Cx=AC*—y?, 93.Find acurve with constant normal. Ans. (e—C)-+y*=at, Singular soition: y="3nterate theagenCiiraut guano yee 4. gate ty Ans. y=Ce+C—CE_ Singular solution: 4y=(e+1) 95.yay VT Ans.ymCeIMC. Singular solution: yt—at= SNymayHY. Ans.ymCeO. .ymay’ ty.Ans.y=Cee. * 7 cet Singularsolution: yt=4x. 98.yay’ —Zp.Ans.y=Ce— 2p. Singular pa et solution:yt=—"at, 99,Theareaofatriangle formed bythetangent tothesought-lor curve and thecoordinate axes is@constant. Find the curve. Ans. ‘The equilateral hyperbola 4xy=+a%. Also, anystraight line ofthefamily y=CetaVC.Mio.Find’scurvesuchthatthesegmentofitstangentBetweentie”coor: falrsWolcott lngAneye8geSisston:shpyhhaa'h,W0i!Findacurvethetangents,towhichform,ontheaxes,segmentswhosesumis20,Ans.y=Cr—26,Singularsolution:(y—x—2a)*—8ox. 102.Findcurves forwhichtheproduct ofthedistance ofanytangent line totwo given points isconstant. Ans. Ellipses and hyperbolas. (Orthogonal and ‘sogonal trajectories.) 103, Find the orthogonal trajectories ofthe family ofcurves y=as" Ans. 4nyt=C. 04. Find theorthogonal trajectories ofthefamilyofparabolasy*=2p(x—a) (istheparameter ofthefamily). Ans.y=Ce? os.'Findtheorthogonal trajeeiries ofthefailyofcurvesxt—yt—a (aistheparameter. Ans. y=S. 106. Find theorthogonal trajectories ofthe family ofcircles x*4+-y*=2ax, Ans, Circles: y=(9%, " 07. Find theorthogonal trajectories ofequal parabolas tangent atthe verter ‘ofthegivenstright line,Ans.il2p"theparameter oftheparabo Jas. and the given straight line isonthey-axis, then the equation ofthe 24/27 trajectory willbeym VE xP. ; 108.Findtheorthogonal trajectories ofthecissoids yg. Ans. att yaliy+e (ina thebrthogonaltrajectoriesoftheemnlscates(2+y*=tty")at Ans.(aty")t=Cry. a110:Fldthegna! trajectories ofthefamilyofcurves:42ay—x3), whereaisavariable parameter Iftheconstant angleformed bythetrajec: Tories and the lines ofthe family is60". Solution. Wefindthedifferential equation ofthefamilyyavs_ivatano ono", andtorysubstitutetheexpression g=i4—"22, 1wmGr,then 2 Digerential Equations =LAE andwegetthediterential equation Y=VE 2 yy,SEV oe ® ee Ley VE‘hecompleteintegralytaxCx—yYT)yieldsthedesitedfamilyoftratelories. Hi, Find the isogonal trajectories ofthe family ofparabolas y*=4Cx when @=45", Ans.y*—1y+2tt=Ce? A 12.Findtheisogonaitrajectories ofthefamilyofstraightlinesy=CxforaonBYEaretan£ thecase@=30°,45°.Ans.Thelogarithmic spirals oie . tyme *.M13.y=GeF+Ce-*. Eliminate C,andC,.Ans.yf—y=0.tng,Wate tert eguatio’ aicleshinginoneplane.Ans Ley)yrayy OAceMRdtp ofalemoreau whose principal axescoincidewit iex-andy-axes. Ans.x(yy" +y")—y'y=0. solutiony=Gye"+-Cye-*+Cee TWfequired Yor)verifythatthe lvenfamilyofcurvesisindeed.the eneralaston;2Gnd'sparticular asutionifforx=r0.wehavey=ly Y=,mm.Ans,yyWerpeeheh), 117,Giventhedifferential equation vay anditsgeneral solution 2 2 pre ZUG) $y. Itisrequired fo:1)verify that thegiven family ofcurves isindeed the general solution; 2)find the integral eufve passing through. the point (1,2) W'the tangent atthis point forms with the positive x-difection inangle of 8.Ans.y=GZVBHS. Integrate some ofthe simpler types ofdifferential equations ofthe secondorderthateadtofistorderequations. *Vigy=. Ans.postngtGatGurs-Cy pick,outparticular sly. tion that satisfies the following initial conditions: x=1; y= 1;y’=1; y=3. 19,fare, Ans.pT CaM CattCee1Daly. Ans,oxen(ay+VaFEC)+C, oFymCpeGye". 121.ym.Ans.(Cyt C= Cy'—a. ‘inNos. 122-125 pick out aparticular solution that satisfies the followinginitialconditions: x=0,y=—1; (ans 122,ay"—y'=xte®. Ans.ym meeD+GatCy.Particularsolution:yer(ei).28.yi’ Foo) ats.” F4C;Iny=x4,. Barticdlar solution!” y=— 1A,tytanzeesinds. Ans.y=Cy+C,sinx—x—4sin2x.Particular solution: y=2sinx—siaxcosz—x—1.125.FF(y')t=at.Ant.y=Cy— tacos(eC).Particular solutions: y=a—l—acoss; ymacosx—(at i). Exercises onChapter XIII os (Hint,Parametric form:y’macoet, y/masin. 128.Ymply. Ant,y= HtZebCyBEGy127,omy", Ans.y=(C2 lla(G2)e+ +Cy. 128. y’y"—3y%=0. Ans. x=Cyy*+Cy+C,.Thtegrate theTolfowing linearsiterentialequatfons withconstantcoef.129.'=9y. Ans.y=Cye™4+Cye"™*, 130.y'+y—=0. Ans.y=Acosx+ PobingiatyOO.inkyacecen tan,findteans,ya=Ce™ 4Cye™,133.y’—4y'+4y—=0.Ans.y=(Cy+Cyx)e™. 134.y+ %2y"Wy. Ans.yore(A'cosSe-+Bsine).195.oO+3Y—2y=0. Ans. Lavin,anv yaCe * ‘$Ce * *.136. 4y'—12y'+9y=0. Ans. y= H+Cue,197.vtyt=O.ansyn[Acn(UE)4 +04 s)]- 138.Twoidentical loadsaresuspended fromtheendofaspring. Find the motion imparted toone load ifthe other breaks loose’ Ans, x= moc(Er}.Wwhere’aistheinereaseinlengthofthespringunder theaction ofone foad at rest.139,Amaterialpointofmassmisattractedbyeachoftwocentreswith 4foree proportional tothedistance, Thefactor ofproportionality is&.The distance’ between the centres Is2. ALthe initial Instant the point lies on the line connecting the centres atadistance afrom the middle. The initial velocityiszero.Findthelawofmotionofthepoint.Ans.x=acos(y2‘). 140.y!V—Sy+4y—0.Ans.y=Cye*Cre"Coe+Ce".141.yf—Ap ae MineeCPEEEeNeatPesaytial= aty=0. Ans. y=(C,+Cx +Cure. 143.yY—4y"=0. Ans.y=,+ FOR tCH Ce+Ce, M4.y!Y424"+9y-—0.Ans.g=(CycosVBE+ $C,sinVBeye~*+(C,cosVIE+C,sinVBE“. 148,y'V—By"+16y=0. Ans.y=Ce+Cye-™*+Cyxe™Cue,46."VtyO.Ans.y= Fr x x WF x x =e(6,cos$4C,sin H)+¢7(C0084-46,sin), (Cocopagum rip)+0(Creorghaan) 147.y'Y—aty=0. Findthegeneral solution andpickoutaparticular solutionthatsalisflestheinitialconditionsforxy—0,y=1.y'=0,y'=~at, KaoAns. General solution: y=Ce**+Cye~**+C,cosax+C,sinax. Par- 'etftegrate the(liowing nonhomogeneous lineardiferential equations(Hind integratethe(cllowingnonhomogeneous lineardifferentialequations (fin thegeneral solution) * y MB.ofTyF12yex, Ans.ymCeeCeEET149,¢oer+. Ans.saCyett4cetfE1,50,ytty—tymBsin2e. Ans.yee 604 Diferentiat Equations Oye —LGsin24200520).151.yyBx?Ans.ymCyet+Cem*— —5e—2,182.Das’patsme!(a#1).Ans.smepcyeltoy. 153.FLO45y—e%. Ans.ymCe"FCe" +ge,154.ofL9yEM Ans,y=,cosde+C,sin Septem, 185.f—3y=26x.Ans.ym,+ Pyat186.of—2y'+Syme"cose,Ans.yet(AcosExt BainVFn"(core—4sins). 157.yft4y—2sin2e, Ans.y= =Asn24.8on2—Fcos2e,158.fA454—2y—26-49. Ans.y= =(C,+Cyx)oF+Ce—x—4.159.y!V—atySatesinax.Ans.y=(C,— —sinas)eC**+0,cosax+Cysinar. 160.VY4208’+aty=8cosax.Ans.y=(Cy+Cye) cosax-+(C,+C,x) sinax—Zcosax. 161.Findtheintegralcurveoftheequation y'+4'y=0thatpassesthrough tnePolatAteeporaodTetangent atthdolaottheCurveproce Ans,ymyscosh(8a)+¢sink(ex). 162. Find asolution ofthe equation y'+2hy/-+aty=0 that satisles the conditions ya,y’=Cwhenx=0.Ans.Forh<ay=e7™ (cosVitae + +A VARa)forhanyee(CHab)xtah fork>a OtathVRE (Via s_CHa(h— VR) ens ”2Vita 2Vit—at :163,Findsolutionsoftheequationy/+-n'y=hsinpx(p#1)thatsatisty toenitionsumesmG”"for"emo.ans”“ymaconn C0POND inns+ymsinpr. PP sinxtoysn 164, Aload weighing 4kegissuspended from aspring and increases its length by|cm. Find thelawofmotion ofthis load ifweassume that the lpper. end of"the spring performs harmonic oscillations “under the lawomainVTODgE, whereyIsmeasured vertically Solution. Denoting byxthevertical coordinate oftheload reckoned from theposition ofrest, wehave AdieSGia—ke, where 1isthelength ofthespring inthefreestate andk=400, asIsevi- dentfromtheiltialconditions. Whence 4+100gx=100gsinVTOUgE+100ig. Wemust seek theparticular integral ofthis equation Intheform #(C,cosVTO0@t-+C, sinVTO0GH)+e, Exercises onChapter XIII 605. since the frst term onthe right enters into the solution ofthe homogeneous equation.‘%"{65.InProblem 139,theinitialvelocity isvyandthedirection isper- pendieular tothe straight Tine connecting the centres. Find the. trajectories, Solution. ifortheorigin wetakethemid-point between thecentres, thesiteretiat equations ofationwllbemf=rk(Cs)(C42) Zhe dys 1¢initial datafor¢=0ai mid2by,Theinitialdatafor=0are a “oe yoo Yoaa Hao, yao: Yan, Integrating, wefind% Fi % reaces(VB1), yonVpsa( VB). Whence $4224 ie 166.Ahorizontal tubeisinrotation about avertical axiswith constant angular Velocity «A sphere inside the tube slides along Itwithout frictionFindthelawofmotion ofthesphere itattheinitial instant ititeson.the fatsofrotation andhasvelocity (along thetubeMint.Theferential equation ofmotion Is©. Theini dat are:rat,Have for0Integrating, wend rapsletbe. Applying the method ofvariation ofparameters, integrate the followin,ditterential equations Wen fTFeymns,Ans,yetCtSHEETEE65yyy eckAns.vecconee tnxteslatoorxincose. 108.tymme Ans.y=,con2+0,sins—VCORTE cose Vtos y * Integrate thefollowing systems ofequations: 1m,fmptt, ort, Pickoutthepatcular slalom thatsataytheinitalconditionsx—==—2,y=0for(0.Ans.y=G,08/4,sin Fn (C,+Cy)cost-+(Cy—C) sind,Particularsolution:<*cost—tsint,geetcos tnt,ety May, Pickoattheputer slain titsatisfytheinitialconditions: =1,y=for£=0.Ans.y=C,cost-+C,sint, FmUC+C)c0st-41C,—CysintsParticular,solution:~e#Setost“sin yy?cost. A Adx_ dy, Ans, =Ge""+ Gem, imae Beeeyacont acSsymcost «06 Digerential Equations ey Ans.x=Cylt-Cye~!-+.Cy008£4C,sinfy {ans y=Cele" Cycos!Caine, 13. ay ona ae, dy 1[Gite deerncterec— gate, 74, deat 1 \atgee Y=CoC+20)=9(C—O Lemp dtneASO ge. ay Ans. y=(Cy-+Cqx)e-*, aah FEC—C—G) es, us.) Gr-y—e. 4 Ans, y=Ce4-Ge-*,7temo, 252(Ce—Cye-™), Spayao. 4y Ans. y=Cy4Cy42In5,in,|eteteaten 25726,76,Gri)—3sins—2cosx. 24y—temcose. a Ans. x=Cyentt Cet,(Ginves AFreeho ZEAE) e+Cet, ve,|Harts, ae Hearty. iBait, Ans.rach, 179. Ht yest getBera otag ee awo,|te Agp=Ee|eit a3ae,fad woGane, Integrate the following different types ofequations: _y" shpeltCoggCCEC).gp,Hy—vide 1h,yay. Ans. ymele + 1.102, AHF, Exercises onChapter XIII or Ans,bm.18,yay"0Ansy=(WEFT407Singlesltions:y=0; x+1=0. 184. y+y=secx. Ans. y=C,cosx+C,sinx+xsinx+ “Feobrlncoss, 185. ex)y—ay—a0. Ans,yarCVTER108,xeonManyconbnn,AnsselFc.18.fpAyersade Ans ymCye-* 4Cott an2420820). 188. ayy—stlnz=d Ans. (Inx+14-Cx)y=1.189.(2x-+-2y—1)dx-+(x+-y—2)dy=0. Ans.2x+ pitaee,ine,Beikaanyy20ata —“lnvestigate and determine whether thesolution x=0, y=0 isstable for tne! Toiowing sytens ofdierent! equations ar[Soe.om.|4 Ans.Unstable,"|Basten Stet192,dyAns. Stable. Mar—ty. farocttoy, 193.dyAns.Unstable. 44xy, 104, Approximate the solution ofthe equation y’=yt-tr that satintheitilPondiion:yest’whenx0.Findthevaluesofthesolution tort tal te0.10 0.0 02, 04, 05 Ans. ney 195. Approximate thevalue ofyza,, of solution oftheequation gtdyeet thatstses tetaleandins pathen1Compare the raul obtained with the exact solution 108 Fisd theapprorimate Values Opes anddjay. ofthe solutions of atenoheguntns Gfoges Maange thetalcon ditions #=0, y=1 when tal, Compare thevalues obtained with the exact ‘alse CHAPTER XIV MULTIPLE INTEGRALS SEC. 1,DOUBLE INTEGRALS Inanxy-plane weconsider aclosed*) region Dbounded bya line L. Inthis region Dletthere begiven acontinuous function z=f(s, 9) Using arbitrary lines wedivide theregion Dinto nparts As,, As, As, 2... AS, (Fig. 276) which weshall call subregions. Soasnottointroduce new symbols wewill denote byAs,, ..., As,both thesubregions and their areas. Ineach subregion ‘As, (itisimmaterial whether inthe interior oronthe boundary) take apoint P,;wewill then havenpoints: PyPuyceesPye Wedenote by/(P,), F(P.), +++» f(Pq) thevalues ofthe func- tions atthe chosen points ‘and then form thesum oftheproducts FP) As: Vg=F(P)B8, +F(P,)AS++HF(Pasa DT(PidA5,- () This istheintegral sum ofthefunction f(x, y)intheregion D. Iff>0 inD,then each term f(P,)As; may berepresented geometrically asthe volume ofasmall cylinder with base As; and altitude f(P;). The sum V,isthe sum ofthe volumes ofthe indicated ele- mentary cylinders, that is,the volume ofacertain “step-like” solid (Fig. 277). Consider an arbitrary sequence ofintegral sums formed by means ofthe function f(x, y)forthegiven region D, *)AregionDiscalledclosedifitisboundedby2closedline,andthepoints lying onthe boundary are considered asbelonging 0the region D. Double Integrats 609 fordifferent ways ofpartitioning Dinto subregions As;. Weshall assume that the maximum diameter ofthe subregions As;ap- proaches zeroasny—co, andthefollowing yy-proposition, which’wegivewithout proof, ctholds true.Theorem1.[fafunction(x,y)iscontinu- CER\‘ousinaclosedregionD,thenthereisaZTiqlimitofthesequence(2)ofintegralsums(1)if7themaximumdiameterofthesubregions As, (} approaches zeroasn—+0o, Thislimitisthe Gasameforanysequence oftype(2),thatis, QT iuisindependent either oftheway theregion Dispartitioned into subregions As, orofG 3 thechoice ofthepoint P,inside thesubre- Fig.276. gion As,. This limit iscalled thedouble integral ofthe function f(x, y) over theregion Dand isdenoted by SJrcPrds orSLFG,ydedy, a 3 that is, adit,BFPOBs=SSFea,y)dxdy. This region Discalled the domain (region) o}integration. IfF(x, y)=0, then the double integral off(x, y)over Disequalto'thevolumeofthesolidQbounded byasurfacez=F(x,y), z[=n FIRE) ‘ ZEBSRE)DB ASSB )Ae Myeea an|H (erSher | xNe El Fig. 277. Fig. 278. the plane z==0, and acylindrical surface whose generators are parallel tothez-axis, while thedirectrix istheboundary ofthe region D(Fig. 278). 20- s388 10 Muttipte Integrals Now consider the following theorems about thedouble integral. Theorem 2.The double integral ofasum oftwo functions (x, ¥)+¥(x, y)over theregion Disequal tothe sum ofthe double integrals over Dofeach ofthefunctions taken separately: Sloe, n+l, Mds=(loe, wds+[Fvlx, yds. 3 3 ° Theorem 3.Aconstant factor may betaken outside the double integral sign: ifa-const, then (Sage, yds=affocx,yds.3 3 The proof ofboth theorems isexactly the same asthat ofthecorresponding theorems forthedefinite integral (see.Sec.3,Ch.XI).Theorem 4./faregion Disdivided into tworegions D,andD,without common interior points, and thefunction f(x,y)is continuous atallpoints ofD,then Sie,wadsdy=VTFee,wdxdy+(SF(x,ydedy. (3) ° , o Proof. The integral sum over Dmay begiven inthe form (Fig. 279) BPAs=FP)As,+BHP)As, 4) where the first sum contains terms that correspond tothe subre- gions ofD,,thesecond, those corresponding tothesubregions of D,. Indeed, since the double integral does not depend onthe manner ofpartition, wedivide theregion Dsothat thecommon boundary ofthe regions D,and D,isaboundary ofthe subre-gions As;.Passing tothelimit im(4)asAs;—-0, weget(3). This theorem isobviously true forany number ofterms. SEC. 2,CALCULATING DOUBLE INTEGRALS Letaregion Dlying inthexy-plane besuch that anystraight line parallel toone ofthe coordinate axes (for example, the y-axis) and passing through aninterior*) point ofthe region, cuts theboundary oftheregion attwopoints N,andN,(Fig. 280). *)Aninterior pointofaregionisonethatdoesnotlieontheboun- dary oftheregion. Cateutating Double Integrals au Inthis case weassume that the region Disbounded bythe lines: y=9, (2), 9=4,(2),=a,x=6andthat (0) <9, (%), a<b while the functions @,(x)and @,(x)arecontinuous ontheintervalfa,6}.Weshallcallsucharegionregular inthey-direction. Thedefinition issimilar foraregion regular inthex-direction. y 4 egit_y,J Se NeOren |yu 7 ¥ wa ed Fig, 279. Fig. 280 Aregion that isregular inboth x-and y-directions we shall simply call aregular region. InFig. 280 we have aregular region D. Let the function f(x, y)becontinuous inD. Consider theexpression +o 00) Ip= (JSFenwrdy)de Baw which weshall call aniterated integral off(x, y)over D.Inthis expression wefirst calculate the integral intheparentheses (the integration isperformed with respect toy)while xisconsidered tobeconstant. The integration yields acontinuous *)function ofx: ote Om= Jflevay. aie Weintegrate this function with respect toxfrom ato6: * [p=J(x)dx. This yields acertaln constant. *)Wedonot here prove that the function (x) iscontinuous, oy 612 Multiple Integrals Example 1.Tocalculate the iterated integral tn=J(Jottvnay)ae. Solution. First calculate the inner integral (inbrackets): ©=[ote ay[evG]iawes Pawee Integrating the function obtained from 0to1,we find ¢Ea ceseee§(#49)a=[S+ér]i-s+anie- Determine the region D.Here, Disconsidered the region bounded bytheTines(Fig.281) y=0, £20, yout, cal. Itmay happen that theregion Dissuch that oneofthefunc- tions y=, (x), y=qa(x) cannot be represented by asingle y 4 9-900) | nai HH i i { 4Frill| + 3 Fig. 281. Fig. 282, analytic expression over theentire range ofx(irom x=a tox=).Forexample, leta<c<6,and 1(x)=1p(2) ontheinterval [a,c], (x) =7(x) ontheinterval [c,6]. where p(x) andx(x) areanalytic functions (Fig. 282). Then the Cateutating DoubleIntegrats 13 iterated integral will bewritten asfollows: 2 ene STS fenayjar— drat ¢940s) »oe =$[ Jfeny)aet iD)henayjar= tla feito eon 2 on =S[S fendyjde+ {0)fe,nay]ae. at vee eo ate The first ofthese equations iswritten onthe basis ofafamiliar property ofthedefinite integral, thesecond, due tothefactthat ontheinterval (a,c}wehave @,(x)= p(x), and ontheinterval Ic,6]wehave @,(x)=x(x). ‘Wewouldalsohaveasimilar notation fortheiterated integral ifthefunction ,(x) were defined bydifferent analytic expres- sions ondifferent subintervals ofthe interval |a,6). Let usestablish some properties ofaniterated integral. Property 1.Ifaregular y-direction region Disdivided into two regions D,and D,byastraight line parallel tothey-axis orthe x-axis, then theiterated integral Iover Dwill beequal tothe sum ofsuch integrals over D,and D,; that is, Ip=1,+lop 0) Proof. a)Let the straight line x=c (a<c<b) divide the region Dinto two regular y-direction regions *)D,and D,.Then § exe * F ° Jo=4(iyF(x:y)dy)dx={(x)de=(O(x)de+( D(x)dx=3 Sen a 3 2 «ot » oun =S(ffeway)det (fFydy)de=loy+ lowa ole 2 Seite *)ThefactthatapartoftheboundaryoftheregionD,(andofDy)iportlonoltheverticaleraightlinedoesGotstopifsegiohWromwelseat farinthe y-direction: foraregion toberegular, Itisonlynecessary” that gay,feted Straight ne“pasing Through, ‘ailerior point the.region should have nomore than two common points with the boundary (see foot- rote onpage 610). ou Multiple Integrals b)Let the straight line y=h divide the region D_into two regular y-direction regions D,and D,asshown inFig. 283. Denote byM,and M,the points’ ofintersection ofthe straight line y= with theboundary LofD.Denote theabscissas ofthese points bya,and 6,. F Theregion D,is'bounded bycon-Y=9rlr) tinuouslines: M, Me 1)y=,i -' 2)“the ‘curve A,M,M,B, whose 1g equation weshall conditionally write ! inthe form y=,(%), |having inview that 9f(1)—9,(#) LU, when a<x<a, andwhen6,<x<b ore a 3% and that eee O)<h when arc, 3)bythestraight lines x—a, x—b. The region D,isbounded by’the lines Y=, (0, 9=9, (2), where a,<x<b,. We write the identity byapplying tothe inner integral the theorem forpartitioning the interval ofintegration: 2 oun Ip={({ flewdy) dem rary win =S[ Jfeadr (fe, pdy]ar= »oye ©on =J(fe may)dx+{( Feay)ae. aSete aNye Webreakupthelatterintegral intothreeintegrals andapply totheouter integral the theorem for dividing the interval of Calculating Double Integrals 615, integration: eee) onSCYteswav)deeT(Tree,way)eetaSyia Bun boxe oe +I(fenay)ax+0(fFle,y)dy)dx: Cee 1ote since g;(x)=@,(x) onthe interval (a,a,] and on[b,, 6],it follows ‘that the first and third integrals are identically zero. Therefore, »oe oes To={( Jfewdy)dx+[( |fleway)de. aein atw Here, thefirst integral isaniterated integral over D,,thesecond, over 'D,.Consequently, Ip=I,+Io» The proof will besimilar forany positionofthecuttingstraight line M,M,. IfM,M, divides Dinto three oralarger_number of regions, We get'a telation similar to(1), inthe first part of which wewill have the appropriatenumber ofterms. \y Corollary. We can again divide yamineeachoftheregionsobtained (using [4FESastraight lineparallel tothey-axis [J][|Jarl Th orx-axis) into regular y-direction FROregions, andwecan apply tothem PSS Eat equation (2).Thus, Dmay bedivided =bystraightlinesparalleltotheHLT coordinate axes into any number of 1: regular regions we Be] Dy Dy Dy vs Dy Fig. 264. and the assertion that the iterated integral over Disequal tothe sum ofiterated integrals over subregions holds; that is(Fig. 284), Ip= lo,+10,+10+ ++++Lop @ Property 2(Evaluation ofaniterated integral). Letmand M betheleast and greatest values ofthe junction f(x, y)inthe 1s Multiple Integrats region D,Denote bySthearea ofD.Then wehave therelation oe) mS<f( 1Te,yy)dxMs. 8) a Sole Proof. Evaluate the inner integral denoting itby®(x): on on OM= 1fenays |Mdy=M(9,9,WH). oe oe We then have oon ® to=J (JMesydy)dr<lMig,@)—@,@)]dx=MS,Earn) 3 that is, In<MS. 6) Similarly oe oe D= )fe,ydy> |mdx=m[g,(o—9, oie sie A : Ip=J(x)dx[m[g(x—@,(&)]dx=ms, thatis, Ip=ms. By From the inequalities (3") and (3*) follows therelation (3): mS<Ip<MS. Inthenext section wewill determine the geometric meaning of this theorem. Property 3.(Mean-Value Theorem). An iterated integral Ipof acontinuous function }(x,g,overaregionDwithareaSisequal totheproduct oftheareaSbythevalueofthefunction atsomepoint Pinthe region D;that is, ein SCJfeenay)d=sys. (4) Proof. From (3) weobtain maglp<M. Cateutating Double Integrals (continued) 87 Thenumber-{J»liesbetween thegreatestandleastvaluesof Hix, y)inD.Due tothecontinuity ofthe function f(x, y),at somepointPofDittakesonavalueequaltothenumber Ip; thatis, j<slo=K(P), whence Ip=H(P)S. 3) SEC, 3,CALCULATING DOUBLE INTEGRALS (CONTINUED) Theorem. The double integral ofacontinuous function f(x, y) over aregular region Disequal tothe iterated integral ofthis Junction over D;that is,*) Sffuemdedy=(( Jfle,ydy)de. 8 a sew Proof. Partition the region Dwith straight lines parallel tothe coordinate axes into nregular (rectangular) subregions: AS, As, ..., As,. ByProperty 1{formula (2)] ofthepreceding section wehave Bomlanlant20Lamy=Zilone w Each ofthe terms ofthe right we transiorm by the mean- value theorem foraniterated integral: Tay=F(P))As, Then (1)takes the form p=FP.)M5,+f(P,)AS,++++P(Pg)AS,=BLP as.(2) where P,issome point ofthesubregion As; Ontheright isthe integral sum ofthefunction f(x, y)over theregion D.From the existence theorem ofadouble integral itfollows that the limit ofthis sum, asn+ coand asthegreatest diameter ofthe sub- regions As; approach zero, exists and isequal tothe double integral off(x, y)over D.The value ofthedouble integral [yon *)Here,weagainassume thattheregionDisregular inthey-direction andboundedbythelinesy=,(«),¥=y(n),ta,k=0. 618 Maltipte Integrats the right side of(2)does not depend onn.Thus, passing tothe limit in(2), weobtain = ii = 1» y)dxdi fomfimBHP) As,{fiey)dxdy or Spiesw)dxdy= Ip. @) Writing out infull theexpression ofthe iterated integral Ip, we finally get ac) {Sie wdedy=([ Jfeway]ae. 0) ° are Note 1.For thecase when f(x, y)=0, formula (4)has apic- torial geometric interpretation. Consider ‘asolid bounded bythe surface z=f(x, y), the plane z=0, andacylindrical surface whose generators areparallel tothe z-axis and the directrix of whichistheboundaryofthe zflay)regionD(Fig.285).Calculate Zof thevolumeofthissolidV. feIthasalreadybeenshown 0)thatthevolumeofthissolid AN isequaltothedouble integral %ut a ofthefunction f(x,y)overli~theregionD: jv V=(Sitxwdedy.(6) Ape f 2 $00Now let us calculate the ot ¥ volume ofthis solid using Fa.18. theresults ofSec. 4,Ch.7 XII, on the evaluation of the volume of asolid from theareas ofparallel sections (slices). Draw theplane x=const (a<x<6) that cuts the solid. Calculate thearea S(x) ofthe figure obtained’ bycutting x=const. This figure isacurvilinear trapezoid bounded bythelines z=(x,y)(x=const), 2=0, y=9,2), y=9, (x). Hence, this area can beexpressed bytheintegral oe s= 7fe,nay. ® oie Knowing the areas ofparallel sections, itigeasy tofind the Calculating Double: Integra (continued) sig volume ofthe solid: * V=fs@ dx; or,substituting expression (6), weget forthe area S(x) bone v=l[ Jflway)ax. 2) ah ee Informulas (5)and (7)the left sides areequal; and sothe right sides are equal too: 2 een SSiee, ydedy=([ Jfeeway]de. 3 ite Itisnow easy tofigure outthegeometric meaning oftheevalu- ation theorem ofaniterated integral (Property 2,Sec. 2):the volume Vofasolid bounded bythe surface z=f(x, y),the zzexeyt 2M) on ‘e ii) ; Hf ! RAL “ip“ i- |i ely) 4:il Kx9 Fig. 286. Fig. 287. lane z=0, and acylindrical surface whose directrix istheBoundary oftheregionD,exceedsthevolumeofacylinderwithbase area Sand altitude m,but isless than the volume ofa cylinder with base area Sand altitude M{where mand Mare the least and greatest values ofthe function z=f(x, y)inthe region D(Fig. 286)]. This follows from the fact that iterated in- tegral I,isequal tothevolume Vofthis solid. 0 Muttipte tntegrats Example 1.Evaluate thedoubleintegral ((4—x*—y" dedyiftheregion 3 tsbounded bythestraight Hines10,xml,yO,andyd. Solution. Bythe formula onf[fea ae]ane(fee2]tem ,‘e 1: vodi35 =f(1-5) v=(w—-$—v)|=B- Example2.Evaluatethedoubleintegralofthefunctionf(x,y=L+be-+yoveraregion bounded bythelines y=—x, x=Vy,y=2, 2=0 (Fig. 287). Solution. onfffaretunes]ayef[pene] arm =§[(Votev7+$)-(-9— +9)= =)[Vo+esV9-F] = ay*,Syt,y*_y?)2_44 5-(¥+443 ~§)-5 V+q. Note 2.Letaregular x-direction region D.be bounded bythe lines X=WW, =WW Y=O Y=, and let,(W)<¥,(y) (Fig. 288). Inthis case, obviously, é sun SSwdedy=l( 0Fx,wae)dy. @) To evaluate the double integral we must represent itasan iterated integral. Aswehave already seen, this may bedone in two different ways: either byformula (4)orbyformula (8). Depending upon thetype oftheregion Dor theintegrand ineach specific case, we choose one ofthe formulas tocalculate the double integral. Calculating Double Integrals (continued) eat Example 3,Change theorder ofintegration inthe integral vr taf([feemayor, Solution, The region ofintegration isbounded bythestraight line y= andthe parabola y=Vz.(Fig,280) Every straight line parallel tothe x-axis cuts the boundary ofthe region atnomore than two points; hence, weean compute the integral byformula @),'selting BW=H BW=y Osos then taf(fre nar)dy. ae Example 4.Evaluate (FedsIfthereglonDis»triangleboundedby 3 thestraight lines y=s, y=0, and com (Fig. 290) \y q a y y3 bySs s y x "|Z 7 0 x P U *¢ to Fig. 288. Fig. 259, Fig. 290. Solution. Replace this double integral byaniterated Integral using formula(4).fitweusedformula (8).wewould havetointegrate. theTonction €*withrespect tox;butthisintegral isnotexpressible intermsofelemen- tary functions): ove ot, edem e*dy|de=([ we‘|’axe fF fl f-Far]eenflPY (x]_e-1=Jxe—narme—nF [=ept0.880... Note 3. Ifthe region Disnot regular either inthex-direction orthe y-direction (that is,there exist vertical and horizontal straight lines which, while passing through interior points ofthe region, cuttheboundary oftheregion atmore than two points), then wecannot represent the double integral over this region in UF y qj“FH.1ot the form ofan iterated integral. Ifwe manage topartition the irregular region Dinto afinite number ofregular x-direction or y-direction regions D,,D,, -.-, D,,then, byevaluating thedouble integral over each ofthese subregions bymeans oftheiterated integral and adding the results obtained, wegetthesought-for integral over D. Fig. 291 isanexample ofhow anirregular region Dmay be divided into two regular subregions D,and: D,. ‘square isequal to2and that oftheouter square is4(Fig. 292), Sferrsenfferaes[fernaes[Perrars(feroran Cateulating Areas and Volumes 62s Repeating eachoftheeintegrals intheformofanHeated integra, Jfevran[[Pewa}acs ([fevar]acs ° AED any +f[fer a]oe EferaJere lee“ (ee $e) (Cem Hee CEE +(e?—e-) (e?—e)=(e—e-) (ee!) =4sinh3sinh1, Note 4.From now on, when writing the iterated integral bein To= SJMswd)ax, shale wewill drop the brackets containing the inner integral and will write beriIo=J Jfewdydx. dete Here,justasinthecasewhenwehavebrackets, wewillconsiderthat the first integration isperformed with respect tothevariable whose differential iswritten first, andthen with respect tothevariable whose differential iswritten second. [We note, however, that this isnot the generally accepted practice; insome books the reverse isdone: integration isperformed first with respect tothe variable whose differential islast.”’) SEC. 4, CALCULATING AREAS AND VOLUMES BY MEANS OF DOUBLE INTEGRALS 1.Volume. AswesawinSec.1,thevolume Vof asolid bounded bythesurface z=/(x, y), where f(x,y)isanonne- gative function, bythe plane z==0 and byacylindrical surface whose directrix isthe boundary ofthe regionDandthegenerators areparallel tothe z-axis, isequal tothe double integral ofthe function f(x, y)over the region D: V=(Ile, yas. 2 ”The following notation isalso sometimes used: fae tes to={[Jfenay]acm(arfFeeway. dha ma om Multiple Integrals Example1.Calculatethevolumeofasolidboundedbythesurfacesx=0, a0, rtgte—i, 20 (Fig. 299). Solution. va[Sons-navas weeDts(inFig.209)theshaded triangular regioninthexy-plane boundedbythestraightlinesx0,y=0,andr-+y=i.Pullingthelimitsinthedouble integral, wecalculate the volume val (0-endy arm(fang 9]!arfpuateng Thus.Vy cubieunits. Note 1.Ifasolid, the volume ofwhich isbeing sought, is bounded above bythesurface z=@, (x,y)>0, andbelow bythe surfacez=®,(x,y)=0,andtheregionD istheprojection ofboth surfaces onthe 42 z-@cxy) xy-plane, then thevolumeVofthissolidfy z|h\ aeyezt wo" h}2QeW) | . ot i f Z£LSS Fig. 293. Fig. 294. isequal tothedifference between thevolumes ofthetwo “cylindrical” bodies; thefirst ofthese cylindrical bodies has the region Das itslower base, and thesurface 2=®, (x,y)foritsupper base; thesecond body also has Dasitslower base, and the surface z=, (x,y)forits upper base (Fig. 294). Therefore, the volume Visequal tothe difference between the two double’ integrals v=SlOc,y)ds—S{®, (x,y)ds,8 3 or V={FIM, YO, (Was. a 3 Calculating Areas and Volumes 025 Further, itiseasy toprove that formula (I) holds true notonlyforthecasewhen®,(x,y)and@,(x,y)arenonnegative, butalso when ®,(x,y)and ®,(x, y)are ‘any continuous functionsthatsatisfy therelationship O,(x,y)>, (x,9). Note 2.Ifinthe region Dthe function f(x, y)changes sign, then wedivide theregion into two parts: 1)the subregion D, where f(x, y)=0; 2)thesubregion D,where f(x, y)<0. Suppose thesubregions D,and D,aresuch that thedouble integrals over them exist. Then theinfegral over D,will bepositive and equal tothe volume ofthe solid lying above thexy-plane. The integral overD,willbenegative andequal, inabsolute value,tothevolume ofthesolid lying below thexy-plane. Thus, the integral over Dwill beexpressed asthe difference between the corresponding volumes. 2.Calculating the area ofaplane region. Ifweform the inte- gral sum ofthefunction f(x, y)==1 over theregion D,then this sum will beequal tothe area S, S=$1-as, foranymethod ofpartition. Passing tothelimitontheright side oftheequation, weget S=[fdeay.i IfDisregular (see, forinstance, Fig. 280), then the area will beexpressed bythe double integral bone s=S[ Jajar. 3 teva Performing the integration inthe brackets, weobviously have ° S=Jle,9,Wide (cl. Sec. 1,Ch. XII). Example 2.Calculate the area ofaregion bounded bythe curves yatawt, yen. Solution. Determine thepointsofintersection ofthegivencurves(Fig.295). AAtthe point ofintersection’ theordinates areequal; that is, eels, Wegettwo.pointsofintersection: M(—2%—2,M,(1.0.Hence,the saf(f)ermfeenden [xSg]. Suppose that inapolar coordinate system 0,g,aregion Dis given such that each ray*) passing through aninterior point of the region cuts theboundary ofDatnomore than two points. yo SSp=Pye) AN a cet salla ame SSO|iyas \ (Zaaeel| )ASL Fig, 295. Fig. 296. Suppose thattheregionDisbounded bythecurves e=©,(6),e=©,(0)andtherays@=aand@=B,where®,(8)<@,(0)and a<f (Fig. 296). Again weshall callsuch aregion regular. Inthe region Dletthere begiven acontinuous function ofthe coordinates 8and g: z2=F(G, Q). Wedivide Dinsome way into subregions As,, As,,..., As,. The Double Integral inPolar Coordinates oar Form the integral sum Va=2FOP)AS a) where P,issome point inthesubregion Asy. From the existence theorem ofadouble integral itfollows that asthegreatest diameter ofthe subregion As, approaches zero, there exists alimit Vofthe integral sum (1). By definition, this limit Visthe double integral ofthe function F(6,g)over theregionD: valFO,ods. Oy 3 Let usnow evaluate this double integral. Since the limit ofthe sum isindependent ofthe manner of partitioning Dinto subregions As,, wecan divide the region in away that ismost convenient. This most convenient (for purposes ofcalculation) manner will betopartition the region bymeans oftherays O=6,, 0=0,, 0=6,,..., 0=0, (where 0,=a, 0,=,8,<6,, <0,<...<6,) andtheconcentric circles ¢=0,, ¢=0,, «is, =p {where Q,1sequal tothe least value ofthefunction©,(0),andQq,tothegreatest value ofthefunction ®,(0)in theinterval @<0<B, @<o,<...<Oq]- Denotebygytheshbregion bounded bythelinese=Q)-.,onTHesubregions As;willbeofthreekinds: 1)those that are not cut bythe boundary and lieinD; 2)those that arenot cut bythe boundary and lieoutside D; 3)those that are cut bythe boundary ofD. The sum ofthe terms corresponding tothecutsubregions have zero astheir limit when A®,—-0 and Ao;—+0 and forthis reason these terms will bedisregarded. The subregions As;, that lie outside Ddonot interest ussince they donotenter into thesum. Thus, the integral sum may bewritten asfollows: VaZUDF(Pu)Asal where P,, isanarbitrary point ofthe subregion Asi. The double summation sign here should beunderstood as meaning that wefirst perform the summation with respect to the index i,holding &fast (that is,wepick out allterms that correspond tothe subregions lying between two adjacent rays *). *)We sole that insumming over the index ¢this index will not run through ‘allvalues from Itom, because not allofthe subregions lying between therays O=0, and 0—6,,,, belong toD. 628 Multiple Integrals The outer summation sign signifies that wetake together allthe sums obtained inthe first summation (that is,wesum with respect tothe index k). Letusfind the expression ofthe area ofthesubregion As,, that isnotcut bythe boundary oftheregion. Itwill beequal tothe difference ofthe areas ofthe two sectors: 1 Panessq(ei+Ae)"40,—7-010,=(ce)+4f)AeA, or Asin=070A, where@<er<e+4Q- Thus, the integral sum will have the form*) Vi=ZLBFOs,oi)e440), where P(8;,0;)isapoint ofthesubregion Asi, Now take the factor A@, outside the sign ofthe inner sum (this ispermissible since itisacommon factor foralltheterms ofthis sum): Va=2CEPCie’cia)4%. Suppose that Ao,—-0 and AQ, remains constant. Then the expression inthe brackets will tend tothe integral (2) {FG,cede. (4) Now,assuming thatA@,—-0, wefinallyget**) 2oi v=S( JFO,eede)do. @) 2 Nein =)We can consider the Integral sum tnthis form because the limit ofthe sum does not depend onthe position of{the point inside the subregion. *6)Sut derivation offormats (3)4nocgorous in,derving thsformula vefist let‘Ag;approach zero, lenving40,constant,andonlythenmadeAO, approach zeioe THs, does nol exactly correspond to’the definition ofadouble Ingray shich werepard"as thehaiofandotegral aunasthediametersaltheSubregions prone ero(i thesane, approach t0zero GlAdy.andAo), However, though theproof lacks rigour, the.result lsiruefi&formula’ ta)istrue),ThisYormula couldbe‘igorously derived bythe ‘same’ method used when considering the double integral inrectangular fourdinates, We aiso note that this lormula ‘will bederived ‘once. again in ‘Sec.6withdifferent reasoning (asaparticular caseofthemoregeneral formula forteansforming coordinates inthe double integra). The Doubie Integral inPolar Coordinates o Formula (3) isused tocompute double integrals inpolar coordinates. Ifthe first integration isperformed over @and the second one over g,then weget the formula (Fig. 297) eo v=l(\ Fo.49)ede.(3')hy Letitberequired tocompute thedouble s integral ofafunction /(x, y)over aregion 3 Dgiven inrectangular coordinates: 4 SSree, gardy. a 8 IfDisregular inthepolar coordinates 8, Fig,297. @then thecomputation ofthegiven integral can bereduced tocomputing the iterated integral inpolar coordinates. Indeed, since x=ocos8, y=gsind, Fx, 9)=Flecos®, ¢sin0]=F(8,oy itfollows that 2om SSrex, wdedy=(( Jfigcos®, esinBiede)dd. (4) 3 ao Example 1.Compute thevolume ¥of«solidbounded vythespherical hyptatat an. the cylinderey—2ay=0. Solution. For the region ofintegration here we can take the base ofthe qld PHO Tn hele withcent (0a)aausa ieequationofthiscirclemaybewrittenin’theform #*+W—a)=at (Fig.298). * Wecalculate +oftherequired volume V,namely thatpartwhichis situated inthe first octant. Then forthe region ofintegration wewill have iotake thesemicircle whose boundaries aredelined bytheequations £=9,W)=0, 1=9,y)=V2ag—H, y=0, y=2a The integrand is 2a) ya Via oo MultipteIntegrats Consequently, 19Vine dyn VisBap ae)ay 7 Transform theintegral obtained tothe polar coordinates 8,¢: x=0050,y=osind Determine thelimits ofintegration. Todoso,write the equation ofthe : given" circle inpolar coordinates; 4 tyae seybeztadas> y=esin8, a [Wil RJ? Poy eeiy-apieat > Fig. 298, Fig. 299, itfollows that e209 sinB=0 or e=2esind, Hence, in polar coordinates (Fig. 299), the boundaries ofthe redefined bytheequations plas 0=9,0)=0,0=0,(0)—=2asin8, a=0, B=->, and the integrand has the form FQ, )=Via—e.‘Thus,wehave e yemaee 4 ,— (4at— gf's)s0408 AS (JVie ae)aom|[ME]. 1 - “HAFJtot40%sat)—dah]d= Ba .4 =AFSa cost)d=oFon—4, The Double Integral inPolar Coordinates ou Example 2.Evaluate the Poisson integral Fewae Solution. Fits evaluatetheintegral/p={e-#*-"dedy,whereteregion ofintegration Disthe circle att yteRt (Fig, 300). Bassing tothe polar coordinates 0,9,weobtain ar aR ta=|(Serede) a=4[e-# |do—na—e™ Now, ifweincrease theradius Rwithout bound (that is,ifweexpand without’ limit. the region” ofintegration, weget Ihe socalled improper Tterated integral: =< Par e a= e-Pede) a0— ean, IJede)40aes odo)d0—jim,xe an Weshalshowthattheintegral ([e--7%de dyapproachesthelimit 3 itthe region D’ofarbitrary form expands insuch manner that finally any Point ofthe plane gets into D'and remains there (we shall conditionallyFraicatesuch’anexpansionofD?bytherelationship DY—>e). _y Ry (7yet .y GZW) 74 1 LEA Fig. 300. Figs. Let and Rbethe least and greatest distances ofthe boundary ofD’fromtheorigin(Fig.201),‘Since thefunction e~**-¥* iseverywhere greater than zero, thefollowing inequalities hold: ty|e ay<i oz ‘MultipleIntegrals or =a Rt a(Ine) [enn rtardycn (Ie).(-e eff (a) Since for D’—coitisobviousthatRj—-2andRyo,itfollows thattheextreme parts oftheinequality tendfooneandthesame limit x. Hence, the median term also approaches this limit; that is, otenWTdxdy=n. © Asaparticular instance, letD’beasquare with side 2aand centre at the origin: then eortas dy{(etaedy= af[emevtaray=Q [femme ax]av. Now take thefactor e-”* outside thesign oftheinner integral (this is_per-imissible sincee~?*doesnotdepend onthevariable ofintegration 2).Then eMdxdym|em[[e-tar]ay. Jorreen| entire} Set{e-**dx=Bg.Thistsaconstant(dependent onlyona}:therelore, [femetacae |mescrassferan o fa co Butthelatterintegral tslikewise equaltoBy(because [e-**ax— =ergy): thus) [ler -¥dxdy=3,8,=8%. o We pass tothe limit inthis equation, bymaking @approach infinity (inthe process, D’expands without limit): i8-0aedy=limBEIiae]=[[eae]. ols,[fenereran. [here -s . iyeden VR Weremark thatwewould notbeableto.compute {hisIntegral directly (by means ofanindefinite integral) because thederivative ofe~** isnotexpres Inthe xy-plane letthere bearegion Dbounded by the line L.Suppose that the coordinates xand yare functions of new variables uand o: x=O(u, 0), y=Pu, Oo); () letthefunctions p(u, v)and *p(u, v)besingle-valued and con- tinuous, and letthem have continuous derivatives insome region D’, which will bedefined later on. Then byformulas (1)toeach pair ofvalues uand vthere corresponds aunique pair ofvalues y y % rya segs Minuail cet(Tr xandy. Further, suppose that the functions @and wpare such that ifwegive xand ydefinite values inD,then byformulas (1) we will find definite values ofuand v. Consider arectangular coordinate system Ouv (Fig. 302). From the foregoing itfollows that with each point P(x, y)inthe oot MultipleIntegrals xy-plane (Fig. 303) there isuniquely associated apoint P’(u, v) inthewv-plane with coordinates u,v,which are determined’ byformulas (1).Thenumbers uandoarecatledcurvilinear coordi-nates ofthe point P. Ifinthexy-plane apoint describes aclosed line Lbounding the region D,then inthe wv-plane acorresponding point will trace outaclosed line L’bounding acertain region D'; and to each point ofD’there will correspond apoint ofD. Thus, the formulas (1) establish aone-to-one correspondence between thepoints oftheregions Dand D',or,themapping, by formulas (1), oftheregion Donto region D'issaid fobeone-to-one.IntheregionD’letusconsideralineu=const.Byformulas(1) we find that inthe xy-plane there will, generally speaking, bea certain curve that will correspond toit.Inexactly thesame way, toeach straight line v=const oftheuo-plane there will correspond some line inthexy-plane. Let usdivide the region D'(using the straight lines u—const and v=const) into rectangular subregions (we shall disregard subregions that overlap the boundary ofthe region D’). Using suitable curved lines, divide Dinto certain curvilinear quadran- gles (Fig. 303). Consider, in’theuv-plane, therectangular subregion As’ bounded bythestraight lines u=const, u-+Au=const, v=const, 0+Av= =const, andconsider also thecurvilinear subregion Ascorresponding toitinthe xy-plane. We denote the areas ofthese subregions byAs’ and As, respectively. Then, obviously, As’ =AuAv. Generally speaking, theareas Asand As’aredifferent. Inthe region D,letthere beacontinuous function z=/ (x,y)- Toeach value ofthe function z=f(x, y)intheregion Dtherecorresponds theverysamevalueofthefunction z=F(u,v)intheregion D’,where F(u, =flo uo),Plu, 0D]. Consider theintegral sums ofthefunction zover D.Itisobvious that wehave the following equation: Die, y)As=DF(u!0)As. @ Letuscompute As,which isthearea ofthecurvilinear quad- rangle P,P,P,P, inthexy-plane (seeFig:303)." Changing Variables ina Double Integral (General Cast) 635 We determine the coordinates ofits vertices: PylyWade%=(HsYs Y= PU, vo), Peat meee Y=(U+Au,0), ®@Pili Ma =OU+AWV+A),Y=Y(U+Au,0+Av), PolenYdsX=(U,V+M0}, Y=Plu,0+AD). Whencomputing theareaofthecurvilinear quadransle Py,PyP,,P,weshall consider the lines P,P,, P,P,, P,P,, P,P, asparallelinpairs;weshallalsoreplacetheincrements ofthefunctionsbycorresponding differentials. We shall thus ignore infinitesi- mals oforder higher than theinfinitesimals Au, Av. Then formu las(3)will have theform X=OU, 0), W=V(4, 0), x=9(u,0)+52du, =9(us0)+32Au, Ky9+EAutLav,y=vu,+3autWao,8) 2=9(ts0)+38Ao, =P(th0)4-2dv. With these assumptions, the curvilinear quadrangle P,P,P,P, may beragarded asaparallelogram. ItsareaAsisapproximately equal tothedoubled area ofthetriangle P,P,P, and isfound by the following formula ofanalytic geometry: As=|(%,—%) W404 —¥)(Ys) = =|(SauSEav)BYav—2av(Baw43%dv)|— 2938 2909 292%_2929} =|FFauboEGEAuAo|=|SOStSESEAudu oe du 80 “1138Se]auao.ud *)Thedoubled tinesin:thedeterminant indicate thattheabsolute Value |of the determinant istaken. 636 MuttipleIntegrals We introduce the notation 292e|au d0| _ oy09|=!OuOv Thus, As=|I\As’. (4) The determinant Iiscalled thefunctional determinant ofthefunctions @(u,v)andwp(u,0).ItisalsocalledtheJacobian afterthe German mathematician Jacobi. The equality (4)isonly approximate, because inthe process ofcomputing thearea ofAsweneglected infinitesimals ofhigher order. However, thesmaller the dimensions ofthesubregions As and As’, themore exact will this equality be. And itbecomes absolutely exact inthelimit, when thediameters ofthesubregions ‘Msand As’approach zero: im 38 Mimi ae Let usnow apply the equation obtained toan evaluation of the double integral. From (2)wecan write Tiley) AsxDF(u,[1]As) (the integral sum onthe right isextended over the region D‘).PassingtothelimitasdiamAs’—+0,wegettheexactequation S$rteydedy=(fF,o)|1|dudo, 6) 8 This istheformula fortransformations ofcoordinates inadouble integral. Itpermits reducing theevaluation ofadouble integral over aregion Dtothe computation ofadouble integral over a region D’,which may simplify theproblem. Arigorous proof of this formula was first given bythenoted Russian mathematician M.V.Ostrogradsky. Note. The transformation from rectangular coordinates topolar coordinates considered inthepreceding section isaspecial case ofchange ofvariables inadouble integral. Here, u=8, v=@: x=0088, y=esind. The curve AB(g=g,) inthexy-plane (Fig. 304) istransformed into the straight line A’B’ inthe Og-plane (Fig. 305). ThecurveDC(g=0,)inthexy-plane istransformed intothestraightline D’C" inthe6g-plane. Changing Variables inaDouble Integral (General Case) 637 The straight lines AD and BC inthe xy-plane aretransformedintothestraightlinesA’D’andBC’intheSo-plane. The.curvesL,and L,are transformed into the curves L,and Li. ¢ lee . MMPIAS Cy etLOxS Hto/4NOON aaainKD, [oaee YYCEs 0 os(NIE We See an ia TeypA {UAL ae 7 ote ad Fig. 30 Fig. 306 Let uscalculate the Jacobian oftransformation ofthe Cartesian coordinates xand yinto thepolar coordinates ®and g: ax oe[303|_J-esin8cos 8)ianocost— 282 Hence, |/|=¢ and therefore bamJSre,mardy=S( |FC,ede)a0.° ahem This was theformula that wederived inthe preceding section. Example. Let itberequired tocompute thedouble integral (are 2 over theregion Dintheay-plane bounded bythestraight lines Fa anes pea, Itwould bedifficult tocompute this double integral directly: however, a simplehange ofvavabes pers reducing thsIntegral tooneover9etanglewhoseSides‘praTonecordateaxes wayns veytye © 638 MultipleIntegrals 0 ‘Thenthestraight linesy=x-+1, y=x—3 will ues ust betransformed, respectively, inio the straight Noesasia 3"in theyo-plane; andthestraightlinesy=a—ttt,ya—be ts +5 3*t53 Lywillbetransformed intothestraightlines Yes3 into "the rectangular region D’ shown in Fig. 906, Itremains tocompute the Jacobian interms ofwand o.Solving. the system of BS oot Wequations (6), weobtain Fig.$06, saSupdo yaturde Consequently, arar)|3.3eolayay|=| A3)=~Tee ajudo]|9 andtheabsolute valueoftheJacobian is|/]=2. Therelore, Lia 3,43,\)3 Sfurnerar= ff[(+40+ fe)-(—$erde)] feud= 3 ae) offfadudem ffGedcom, 2is SEC. 7,COMPUTING THE AREA OF ASURFACE Let itberequired tocompute thearea ofasurface bounded by the line [(Fig. 307), the surface isdefined by the equation z=f(x,y), where the function f(x, y)iscontinuous and has con- tinuous partial derivatives. Denote theprojection ofthe line fonthe xy-plane by&, Denote byDtheregiononthesy-plane bounded bythe,lineL. Inarbitrary fashion, divide Dinto nelementary subregionsAs,As,,...,As,.In'eachsubregion As,takeapointPiten).To'the point 'P,there will correspond, onthe surface, apoint MlbMeFB WL : Computing theArea ofaSurface 6 Through M,draw atangent plane tothe surface. Its equation isofthe form 27=fe(ByWEB) +hGeWY—m) ) (see Sec. 6,Ch. 1X). Inthis plane, pick out asubregion Ao, which isprojected onto thexy-plane inthe form ofasubregion As,. Consider thesum ofallthesubregions Ao,: ¥Ao, a ert »Sy)H 7 0 hk dA y Fig, 507, Fig. 08. ‘We shall call the limit oofthis sum, when thegreatest ofthe diameters ofthesubregions Ao, approaches zero, the area ofthe surface; that is,bydefinition weset = Ao,. 2)Crane een! ty Now letuscalculate thearea ofthe surface. Denote byy,the angle between the tangent plane and thexy-plane. Using afami- liar formula ofanalytic geometry wecan write (Fig. 308) As,=Aa,cosy; or abtdo=A. 8) The angle y,isatthesame time theangle between thez-axis and theperpendicular tothe plane (1). Therefore, byequation 640 MattipteIntegrats (1)and the formula ofanalytic geometry wehave 1 £08Yj=a,VitiG.wthend Hence, b0;=V 140nd)+he(1)As, Putting this expression into formula (2), weget o=lim Vith&, wythe wAs. Since thelimit oftheintegral sum ontheright side ofthe last equation is,bydefinition, the double integral 1+(5)+(3,)dxdy,wefinallyget iVi@y+Gy Teyale o=SSVis(S)'+(HYaxay. “)8 This isthe formula used tocompute the area ofthe surface z=](x9).Iftheequation ofthe surface isgiven inthe form x=p(y, 2)orintheform y=x(x, 2), then thecorresponding formulas forcalculating thesurface areof theformo=ffV1+(%)+(#)avez, @) Tuleey o~{fV+ +)wa, «By where D’andD”aretheregions inthexy-plane and thexz-plane inwhich thegiven surface isprojected. Example 1.Compute the surface oofthe sphere ae yttate RE Solution. Compute thesurface ofthe upper half ofthe sphere: inVR (Pig.90),tnthiscase &--VRoay Computing the Area ofaSurface el oy 9 VR Hence, 2)(8)=Veo yee VE +(5) ae ae The region ofintegration isdefined bythecondition eter. Thus, by formula (4)wewill have oa, er, dee oot wy)aereaSC[7ea)Rk-VR=By) Ls Tocompute thedouble integral obtained letusmake the transformation topolar coordinates. In.polar coordinates theboundary oftheregion of integration isdetermined bythe equation Q=R. Hence, mR ™ R on(fpbaose)d0=28flVRSw0= DArRHBee/ Om) iu =2R|Rddmdn RE Example 2.Find thearea ofthat part ofthesurface ofthecylinder Bypae which iscut out bythe cylinder atpateat, Solution. Fig. 810shows 1/8th ofthedesired surface. The equation ofthe surfacehastheformy=Vas Fae VRE _—=——_ cae ——,|A —«,14 I iH oi ey fll 17 y8D ii,7 “FS gy K hk xeeyea? Fig.309, Fig. 310, 2assee 6 ‘Multiple Integrals therefore, a a, +o a Vaca m8) Wa a V+4)+(%)=Veeevrs ancTis fiom ofIntegration isaquarter circle,thatIs,iisdetermined by sfsteat, 1B0; 250. Consequently, are ¢ vee @ oma", SEC, 8,THE DENSITY OF DISTRIBUTION OF MATTER ‘AND THE DOUBLE INTEGRAL Inaregion D,letacertain substance bedistributed insuch‘mannerthatthereisadefiniteamountofthissubstance perunitareaofD.We shall henceforward speak ofthe distribution ofmass, although ourreasoning willholdalsoforthe casewhen,speaking ofthe distribution ofelectriccharge,ofquantityofheat,andsoforth. Weconsider anarbitrary subregion Asof theregion D.Letthe mass ofsubstance ‘associated with this given subregion beAm. ThentheratioM¥iscalled themeansurface density ofthesub- stance inthesubregion As. Now letthe subregion Asdecrease and contract tothepoint P(x,y).Consider thelimitJim35Ifthislimitexists,then, generally speaking, itwilldepend ontheposition ofthepointP, that is,upon itscoordinates xand y,and will besome function 1(P) ofthepoint P.Weshall call this limit the surface density ofthesubstance atthepoint P: lim87=f(P)=1(x,9). Q ane Thus, thesurface density isafunction f(x, y).of the coordi- nates ofthepoint oftheregion, Conversely, letthere begiven, inaregion D,thesurface den- sity ofsome’ substance assome continuous function /(P)=/(x,y) The Moment ofInertia ofthe Area ofaPlane Figure 643 and letitberequired todetermine thetotal quantity ofsubstance ‘Mcontained intheregion D.Divide Dinto subregions As,(i= =1,2,...,) and ineach subregion take apoint P;;then [(P,) isthe surface density inthe point P;,. Towithin higher-order infinitesimals, the product /(P,)As, givesusthequantity ofsubstance contained inthesubregion As, and the sum 2F(P,)As; expressesapproximately the total quantity ofsubstance distribu- ted inthe region D.But this isthe integral sum ofthe function F(P) inthe region D.The exact value isobtained. inthe limit asAs,—0. Thus, *) M=limY(Pdds= JCFP)ds—= [67%vdedy, 2) sete 3 ‘3 orthe total quantity ofsubstance inthe region Disequal tothe double integral (over D)ofthedensity [(P)=/(x, y)ofthis sub- stance. Example. Determine themass ofscircular plate ofradius &ifthe sure tneney [ep ofheatc tieplatasec patBs po.portional 18tedtancect hepato) fromiheeneotthece, Gai, Te,nak VEER Solution. Byformula (2)wehave Ma((AVF paras, 3 wheretheregionofintegration Disthecirclex74ycRE Passing tepolarcoordinates, weobtain! =a R mae (foods)ao—in® |=2ene SEC, 8.THE MOMENT OF INERTIA OF THE AREA OF APLANE FIGURE The moment ofinertia /ofamaterial point Mofmass mre- lative tosome point Oisthe product ofthe mass mbythe *)The relationship As;—+0 istobeunderstood inthesense that thedia- meter ofthe subregion %approaches tro, Fo ou MuttipteIntegrals square ofitsdisiance rfrom thepoint 0: T=mr, The moment ofinertia ofasystem ofmaterial points m,, m,, seus m,Telative toOisthesum ofmoments ofinertia ofthe yindividual points ofthe system: 1=3mr. a nw Letusdetermine themoment ofinertia ofamaterial plane figure D. LetDbelocatedinanxy-coordinate 7)7 %plane. Let usdetermine the moment of inertia ofthisfigurerelative totheorigin, Fig,3th, assuming that the surface density is everywhere equaltounity. DividetheregionDintoelementary subregions As,—-(i=1, 2,...,n)(Fig.311).Ineachsubregion takeapointP,with coordinates&,1;.Letuscalltheproduct ofthemassofthe subregion As;bythesquare ofthedistance r}=E+n} anele- mentary moment ofinertia AJ,ofthesubregion As: Al=(E+n)As, and let usform the sum ofsuch moments: >(b+ ni)As. This istheintegral sum ofthefunction f(x,y)—=2*-+y* over the region D.fedefinethemoment ofinertia ofthefigureDasthelimit ofthis integral sum when the diameter ofeach elementary subre- gion As;approaches zero: = lis As,. Butthedoubleintegral({(x*-+y*)dedyisthelimitofthissum.3 Thus, the moment ofinertia ofthe figure Drelative tothe originis 1=Sfatsy)dedy, 0)3 where Disaregion which coincides with thegiven plane figure The Moment ofInertia ofthe Area ofaPlane Figurt 645 The integrals IneSfy'dedy, ) v Tyy=Sfatedy (3)3 arecalled, respectively, the moments ofinertia ofthe figure D relative tothe x-axis and y-axis. Example 1.Compute the moment ofinertia ofthe area ofacircle Dof radius Rrelative tothe centre 0. Solution. Byformula (1)wehave ton[foreeravae Toevaluate this integral wetransform tothe polar coordinates 8,9.The equation ofthe circle inpolar coordinates is@=R. ‘Therefore ne =f(Serese) dom2R, Note. Ifthesurface densityyisnotequaltounity,butissome function ofxand y,i.e.,y=y (x,y),then themass ofthesub- region AS,, will, towithin infinitesimals of‘higher order, beequal to ‘y(& ,)As;and, forthis reason, the moment ofinertia ofthe plane figure relative tothe origin will be 1=fvMatty) dedy. a3 Example2.Compute themomentofinertiaofaplanematerial ‘gureD bounded bythe lines yt—1—x; 2-0, y—Orelative fothe yranis ifthe sur- face density ateach point isequal toy(Fig. 312). Solution, 1vine ty’FE ae :ty=J( Jota)emFAP |aayft—9deny. Ellipse ofinertia. Let usdetermine the moment ofinertia of thearea ofaplane figure Drelative tosome axis OLthat passes through the point 0,which weshall take asthe coordinate ori- gin. Denote by@the angle formed bythe straight line OLwith. the positive x-axis (Fig. 313). The normal equation ofOL is xsinp--y cos@=0. 646 ‘Multiple Integrals The distance rofsome point’ M(x, y)from this line is r=|xsing—y cos|. y yeotx A oo 1G alo Uy * Fig. 812, Fig. 318. The moment ofinertia Jof the area ofDrelative to OL is expressed, bydefinition, bytheintegral 1=S{rdedy=[f (xsing—y cosq)*dxdy= 8 3 =sintg[fx*dxdy—2 singcos@[fxydxdy+cost@S$ytdxdy. 8 Therefore T=1,ySin@—2gySi9C089+IxqCOS"G O) here, Iyy={{x*dedy isthemoment ofinertia ofthefigure 3 relative tothey-axis, I,,=({y'dxdy isthemoment ofinertia ° relative tothex-axis, andJ,y=({ xydxdy. Dividing allterms a ofthe latter equation by1,weget =1., (s988)*_. sin) /c089sing)* tae(FF)lo(FE)(FE)WAGE) © Onthe line OL take apoint A(X, Y)such that OA=yp 5 Tothe various directions oftheOL-axis, that is,tovarious values The Moment ofInertia ofthe Area ofaPlane Figure oT oftheangle ,there correspond different values Jand different points A.Letusfind thelocus ofthepoints A.Obviously, 1 Ls Xeappeose, Yawesing. Byvirtue of(5), the quantities Xand Yare connected bythe relationship 1a1,,X*—2 XY+1¥*. 6 Thus, the locus ofpoints A(X, Y)isasecond-degree curve (6). ‘We shall prove that this curve isanellipse. The following inequality established bytheRussian mathema- tician Bunyakovsky *)holdstrue: ({feudeai)'< (65x'deay)({5wares) r3 2 or Iealyy—y>0. *)To, prove Bunyakovsky's (also spelt Buniakowski) inequality, we con-sideribetOllowingobviousinequality.” )InequalitySfVeret, sitacdyao, % where &isaconstant. The equality sign ispossible only when f(r, y)— Ap (xyy=0;thatis,iff(x,y=Ap(x,y).Ifweassumethat[eda Aconst=’, then there will always bethe inequality sign. Thus, removing brackets’ under the integral sign, weobtain SSG,saxdy—2niN}FeMele,dedy-+¥V{@t(e,ydedy>0.° 3 Consider the expression on the left as afunction ofA. This isaseconddegree polynomial thatnevervanishes; hence, itsrootsarecomplex, andthis Will oceur when. the discriminant. formed ofthecoefficients ofthequadratle Polynomial isnegative, that is, (SShoaray )—SfFaxdy{{gtaxdy<0 3 8 or (Sftearan)'< Jfmacay[fotteas 2 3 This isBunyakovsky's inequality. Inourease,fle,=xOleN=.THconst. Bunyakovsky's inequality iswidely used invarious, lelds ofmathema- tics. Inmany textbooks itisincorrectly called Schwars* inequality. Bunya: Kovaty’ abled 1(among etherimpartant equalities) in1858.Sehwars published biswork 16years later, in’1875. 18 Multiple Integyals Thus, thediscriminant ofthecurve (6)ispositive and, con- sequently, thecurve isanellipse (Fig. 314). This ellipse iscalled theellipse ofinertia. Thenotion ofan yellipse ofinertia isvery important inme- <x chanics. } Wenote that the lengths ofthe axes of }theellipse ofinertia and itsposition in 4 “’ the plane depend onthe shape ofthe 4 f+kivenplanefigure.Sincethedistancefrom 7S32 theorigintosomepointAoftheellipse :VeisequaltorawhereIisthemoment Vana ofinertia ofthe figure relative tothe OA-axis, itfollows that,afterconstructing Fig,314. theellipse, wecanreadily calculate the moment ofinertia ofthefigureDrelative tosome straight line passing through the coordinate origin. In particular, itiseasy tosee that themoment ofinertia ofthefigure will beleast relative tothemajor axis ofthe ellipse ofinertia and greatest relative tothe minor axis ofthis ellipse. SEC. 10, THE COORDINATES OF THE CENTRE OF GRAVITY OF THE AREA OF APLANE FIGURE InSec. 8,Ch. XII, itwas stated that the coordinates ofthe centre ofgravity ofasystem ofmaterial points P,,Py,w+)Py with masses m,,m,,...m,aredefined bytheformulas Demy yyUe :1 . Oy Let usnow determine thecoordinates ofthe centre ofgravity of aplane figure D.Divide this figure into very small elementary subregions AS,. Ifthesurface density istaken asequal tounity, then themass ofthesubregion will beequal toitsarea. Ifitis approximately considered that the entire mass ofanelementary subregion AS; isconcentrated insome point ofitP;(E;, n,), the figure Dmay beregarded asasystem ofmaterial points. Then, byformulas (1),thecoordinates ofthecentre ofgravity ofthis figure will beapproximately determined bythe equations ase as The Coordinates ofCentre ofGravity ofaPlane Figure 619 Inthe limit, asAS;—+0, theintegral sums inthe numerators and denominators ofthefractions will pass into double integrals, and weobtain exact formulas for compu- ting the coordinates ofthecentre ofgra- vity ofaplane figure: wards ded5yae ee@*e*dxdy‘°° dedy" 0] Fp jaw : These formulas, which have been derived Fig. 315, foraplane figure with surface density 1, obviously, hold true also forafigure with any other density constant’ atallpoints. If,however, thesurface density isvariable, y=V(% y), then the corresponding formulas will have the form Sve, nxardy Sve, nydedy = =a——_—.. freacay S$VO,Waxdy 8 Theexpressions M,=sfye,y)xdedy andM,={\y(x,y) > ’ ydrdy arecalled static moments oftheplane figure Drelative tothe y-axis and x-axis. Theintegral |{y(x,y)dedy expresses thequantity ofmass ofthe figure inquestion. Example. Determine the coordinates ofthe centre ofgravity ofa quarter oftheellipse (Pig. 315) ,54K =, wuming that the surface density atall points isequal to1,sssugptution. Byformulas (2)wehave” * SlJsayJax2)Vermacae oe pat 4 $= 8en [Eve anal ta afa ratad realadne 650 Multiple Integrals \ 3 4 Tee 7 SEC, 11, TRIPLE INTEGRALS Let there begiven, inspace, acertain region Vbounded by aclosed surface S.Let some continuous function f(x, y,2),where %,y,2arethe rectangular coordinates ofapoint of’the region, begiven inthe region Vand onitsboundary. For clarity, if F(x, y,2)=0, wecanregard this function asthedensity ofdis- tribution ofsome substance inthe region V. Divide V,inarbitrary fashion, into subregions Av,; thesym- bol Av, will’ denote not only the region itself, but itsvolume aswell.Withinthelimitsofeachsubregion Av,,chooseanarbitrarypoint P,and denote byf(P;) thevalue ofthe function fatthis point. Form anintegral sum ofthe type Di) Av, 0) and increase without bound the number ofsubregions Av; sothat the largest diameter ofAv; should approach zero." Ifthefunction f(x, y,2)iscontinuous, there will bealimit ofthe integral ‘sums oftype (1), where the limit ofintegral sums istobeun- derstood inthe same sense asfor the definition ofthe double in- tegral.**) Thislimitisnotdependent eitheronthemanner ofpar-titioning the region Voronthe choice ofpoints P;; itisdesig- natedbythesymbol {({f(P)dv andiscalled atriple integral. ¥ Thus, bydefinition, li P,)Av,=P)d ali, (Pe:=SFP)do or S§feyao=S (fre, y,2)dxdydz. (2)iu : *)The diameter ofasubregion Avy isthe maximum distance betweenpoint) iyingentheboundary ofthesubregions75)This: theorem. oftheexistence ofalimit ofintegral sums (that is,of theexistence ofatriple integral) forany. function continuous ina closed region V(including theboundary) Isaccepted without rool. . Evaluating aTriple Integral 61 Iff(x,y,2)isconsidered thevolume density ofdistribution ofasubstance over theregion V,then theintegral (2)yields the mass ofthe entire substance contained inV. SEC. 12, EVALUATING ATRIPLE INTEGRAL Suppose thatthespatial(three-dimensional) regionVbounded by‘the closed surface Spossesses the following properties: 1)every straight line parallel tothez-axis anddrawn through aninterior (that is,not lying on the boundary S) point ofthe e pan region Vcuts the surface Sattwo points; f,\.-”--\\ aail By, ly Kea 99400 90200 Fig. 316. Fig. 317. 2)the entire region Visprojected on the xy-plane into a regular (two-dimensional) region D; 3)any part oftheregion Vcutoffbyaplane parallel toany fone ofthecoordinate planes (Oxy, Oxz, Oyz) likewise possesses Properties |and 2. Weshall call theregion Vthat possesses theindicated proper- ties aregular three-dimensional region. Toillustrate, anellipsoid, arectangular parallelepiped, atet- rahedron, and soonare examples ofregular three-dimensional regions. Aninstance ofanirregular three-dimensional region isgiven inFig. 316. Inthis section wewill consider only regular regions. Let the surface bounding the region Vbelow have theequa- tion z=x(x, y),and thesurface bounding this region above, the equation z=p(x, y)(Fig. 317). Weintroduce’ theconcept ofathreefold iterated integral Jy, over theregion V,ofafunction ofthree variables f(t, y,2) defined and continuous inV.Suppose that the region Disthe projection ofthe region V‘onto the xy-plane bounded bythe sa Maltipte Integra lines Y=.) Y=, (X), ¥=a, y=. Thenathreefold iterated integral ofthefunction f(x,y,2)over the region Visdefined asfollows: bows (x) OC) w=SE J{JSfew dz}dy]ae. 0) a tete earn We note that asaresult ofintegration with respect tozand substitution oflimits inthebraces (inner brackets) wegetafunc- y tionofxandy.Wethen compute thedouble integral ofthis function over the > region Dashasalready been done. Thefollowing isanexample oftheevalua- °4% tionofathreefolditeratedintegral. 2 + Example 1.Compute theiterated integral of the function f(x, y,2)=xyz over the region V bounded bythe planes abo ~ x=0, y=0, 2=0, xtyt2=l.Solution.ThisregionIsregular,itIsbounded Fig.8. above anilowOY"the’planes20"anda siereMind”te”projectedon“thexyplane Into»regular planeregonDyich tare bounded bye aight iesSO Ea ym2PigSi)“thereloe,the:tvelldieraed Integral Tyis'computed asfollows 7 wallinee] BL? Setting upthe limits inthe twofold HMerated integral over the region D,we cota wef{f[fwootJeyhaonf{[[abeen ~$[Youmaran fiemarernas- Letusnow consider some oftheproperties ofathreefold iterated integral. Property 1.IfaregionVisdivided intotworegionsV,and V,byaplane parallel tosome ofthe coordinate planes, then the ihreejold iterated integral over Visequal tothesum ofthe three- fold iterated integrals over theregions V,and V,. Evaluating aTriple Integral a3 Theproofofthispropertyisexactlythesameasthatfor twofold iterated integrals. Weshall not repeat it. Corollary. For any kind ofpartition ofthe region Vinto a finite number ofsubregions V,, ...,V,byplanes parallel tothe coordinate planes, wehave theequality Iya lytly tetlige Property 2(Theorem oftheevaluation ofathreefold iterated integral). /[mand Mare, respectively, thesmallest and largest values ofthefunction f(x, y,2)inthe region V,wehave the inequality mV<ly<MV, where Visthevolume ofthegiven region and Iyisathreefold iterated integral ofthefunction f(x, y,2)over theregion V. Proof. Let usfirst evaluate the inside integral inthe iterated vw integralofslJfeyaea wld va va vay ven, Srey ad< [Ma—M [da=Mz |= Pros) Pca) 1a) PCa =MO xe DL Thus, the inside integral does not exceed the expression M(p(x, y)—x(*, y)|. Therefore, byvirtue ofthe theorem of Sec. 1fordouble integrals, weget(denoting byDtheprojection ofthe region Vonthexy-plane) CawiliceadeJoosmints,N—xle,yido=Olean 3 =MIS Ie, xe W)ldo. 3 But thelatter iterated integral isequal tothedouble integral of thefunction w(x, y)—x (xy)and, consequently, isequal tothe volume oftheregion which liesbetween the surface z—x(x, y) az=p(x, y),thatis,tothevolume oftheregion V.There- fore, ly<MV. Itissimilarly provedthatJy>mV.Property 2isthusproved. Property 3(Mean-Value Theorem). Thethreefold iterated integ- ral[yof@continuous function f(x, y,2)overaregionVisequal totheproduct ofitsvolume Vbythevalue ofthe function at st MattipteIntegrats some point PofV;that is, OPws(x)(W(x,w)weil§{Vrwae}ay]de=[(P)V.— Q) tla@lian The proof ofthis property iscarried outinthesame way asthat foratwofold iterated integral [see Sec. 2,Property 3,formula (4)]. ‘We can now prove thetheorem for evaluating atriple integral. Theorem. The triple integral ofafunction f(x, y,2)over a regular region Visequal toathreefold iterated integral over the same region; that is, bpese (cenSSSfeyaae=f[ iy{§feyoathaos, 7 tLeto led o Proof. Divide the region Vbyplanes parallel tothecoordinate planes into nmregular subregions: Av, +Av, +... +:A0,. Asdone above,;denote by /ythe threefold iterated integral of thefunction f(x, y,2)over theregion V,and byJs,thethree- fold iterated integral ofthis function over thesubregion A,,. Then bythecorollary ofProperty |wecan write the equation TyIso,Lao02++Lange @) Wetransform each oftheterms ontheright byformula (2): Ty=H(P)M0,+4(P)A0,++sFI(Py)A0yn® whereP,issomepointofthesubregion Av;. Ontheright side ofthis equation isanintegral sum. Itis assumed that the function f(x, y,z)iscontinuous inthe region V; and forthis réason thelimit ofthis sum, asthelargest diameter ofAv, approaches zero, exists and isequal tothe triple integral ofthefunction f(x, y,2)over V.Thus, passing tothe limit in (4), asdiam Av,—+0, weget y=JSSfe,y,2)d0, ? or,finally, interchanging theexpressions ontheright and left, Brexn(er.0) S$renaaemf[ FTreaaahay|ax ? Floto lao Thus, thetheorem isproved. Evaluating aTriple Integrat 655 Here, z=x(x, y)and z=w(x, y)are theequations ofthesur- faces bounding the regular region Vbelow and above. The lines y=9,(*), Y=, (x), x=a, x=6 bound the region D,which is the projection ofV’onto the xy-plane. Note. Like inthe case ofthe double integral, wecan form a threefold iterated integral with adifferent order ofintegration with respect tothe variables and with other limits, if,ofcourse, theshape oftheregion Vpermits this. Computing the volume ofasolid by means ofathreefold iterated integral. Ifthe integrand f(x, y,2)=1, then thetriple integral over theregion VexpressestheVolume oftheregion V: yee va§{fdxdyde. 6) 7 cto Example 2.Compute thevolume of HoT }theellipsoid " Nar€7?0 eee KA - Solution. Theellipsoid (Fig. 319) Fig.319, is Bounded below by the surface sane 1EE, andabovebythesurface2=0V1 hewrojection ofthiselipsid onthexpplane (region D)tsanelise,rt+Eal. Hence, reducing toathreefold iterated integral, weobtain v-f]5 {a“|e te) Soe\ a/Fe - iseef {Virsa la. When computing the inside integral, xisheld constant. Make the substitu tion: gab Feaint,ayn1Fycostt. Thevariableyvariesfrom—bY/1—2to6Y/1—ApiAnereore& 656 Multiple Integrals variesfrom—S-to 4,Putting newlimitstntheintegral, weeet riffV(-3)-(-SmVEmale= f =20f[ialcostdtJansfie—aydraae, Hence, vanade MWa=b=e, wegetthevolume ofthesphere: Vedna4 at, SEC. 13,CHANGE OFVARIABLES INATPIPLE INTEGRAL 1,Triple integral incylindrical coordinates. Inthecase ofcylin- drical coordinates, theposition ofapointPinspaceisdetermined bythethreenumbers @,9,z,where©andgarepolarcoordinates Ofthe. projection ofthe point Pon theay-plane and2isthe z-coordinateofP,thatis,thedistance ofthepointtothexy- plane—with theplus sign ifthe point lies above the xy-plane, and with theminus sign ifbelow thexy-plane (Fig. 320). Inthis case, wedivide the given three-dimensional region V intoelementary volumes bythecoordinate surfaces @=0,, o=e,,2=2, (half-planes adjoining thez-axis, circular cylinders whose axis Coincides with the z-axis, planes perpendicular tothe z-axis). The curvilinear “prism” shown inFig. 321 will beavolume ele- ment. The base area ofthis prism isequal, towithin infinitesi- mals ofhigher order, togA@Ag, the altitude isAz (to simplify notation wedrop theindices i,j,&).Thus, Av=eA@AQAz, Hence, the triple integral ofthe function F(8,9,2)over theregion Vhas the form 1=$§ F@@2)Qd0dedz. Ww 7 The limits ofintegration aredetermined bythe shape ofthe region V. Change ofVariables ina Triple Integrat er 2 = az|SNeyKet 692)hiAA D42th SPR? aS=a x dp Fig. 320. Fig, 921 Ifatriple integral ofthe function f(x, y,2)isgiven inrectangular coordinates, itcanreadily bechanged toatriplein-tegral incylindrical coordinates. Indeed, noting that x=qcos®; y=osind®; z=z, we have SSSFeey,2)dedyde= (VFO, @,2)ed0dode, 7 ¥ where F(qcos®, gsin®, 2)=F(®, g,2). Example. Determine the mass Mofahemisphere ofradius Rwith centre attherorigin, ithedensity oftssubstance ateachpoix2)ispsPortionalfothedistanceofthispointfromthebase,thatis,F'—keSolution. ‘The equation ofthe upper part ofthe hemisphere a=VR incylindrical coordinates has the form 22VR Hence, aac /VRR=Gn=[finesoaeae=]j({sedeJea]- ark vR=e arebat k-J[hs|oa]o0-j[f$eee ae ReRe aR kakotf[F-T]onpf anSE, ose Mattptetnegrats 2,Atriple integral inspherical coordinates. Inspherical coor- dinates, theposition ofapoint Pinspace isdetermined bythree numbers, 0,r,@,where ris thedistance ofthepoint from 2 theorigin, theso-called radius vector ofthe point, @isthe PA wheY 9 Ae LS7 as ae K K ar* Fig. 92. Fig. 923. angle between theradius vector andthez-axis, 0istheangle between theprojection oftheradius vector onthexy-plane and thex-axis reckoned from this axis inapositive sense (counterclockwise) (Fig. 322). For any point ofspace wehave O<r<o, 0O<gan; 0<0<2n. Divide this region Vinto volume elements Avbythe coordi- nate surfaces r=const (sphere), p=const (conic surfaces with vertices atorigin), @-—const (half-planes passing through the z-axis). To within infinitesimals ofhigher ofder, thevolume element. Avmay beconsidered aparallelepiped with edges oflength Ar,rAg, rsingA@. Then thevolume element isequal (see Fig. 323) 'to Av=?singArA0Ag. The triple integral ofafunction F(®, r,q)over the region V has the form 1=$J[FO7,@rtsingdrddag. ; §§3 ;w The limits ofintegration are determined by the shape of theregion V.From Fig. 322 itiseasy toestablish theexpressi- ons ofCarlesian coordinates interms ofspherical coordinates:x=rsin9030,y=rsing sind, z=rcosy. Change ofVariables ina Triple Integral 69 For this reason, theformula fortransforming thetriple integral from Cartesian coordinates tospherical coordinates has the form SSShesy,2)dedyde= 7 =J[Jslrsingcosd, rsingsind,rcosq]r* singdrdid. 7 3.General change ofvariables inthe triple integral. Transformations from Cartesian coordinates tocylindrical and spherical coordinates inthe triple integral represent special cases ofthegeneral transformation ofcoordinates inspace. Let the functions x=@lu, t,w), y=9lu, t,w), z=1(u, t,w) map, inone-to-one manner, the region VinCartesian coordi- natesx,y,zontotheregionV’incurvilinear coordinates u,f,w. LettheVolumeelementAvoftheregionVbecarriedovertothevolume element Av’ ofV’and let limAo=|7|. dinae=Ih Then SSSfe,y,2)dedyde= 7 HSISilo4w,Blefw)xltw)]|Idedtdeo, ?- Asinthecase ofthe double integral, /iscalled theJacobian; and asinthe case ofdouble integrals, itmay beproved that theJacobian isnumerically equal toadeterminant oforder three: axaed Bu56a [242088 utdw|+220202 aiaoe Thus, inthecase ofcylindrical coordinates wehave x=Qc0s0, y=esing, z=2 (Q=u, O=1, =u); cos) —gsin90J=|sin6 teonta|me0 ol 60 MuttipleIntegrals : Inthecase ofspherical coordinates: x=rsingcos), y=rsing sind, 2=rcos@ (r=u, g=t, 0=w); sing cos rcos@ cosh —rsing sind|T=|sing sinrcos@sind singcos6|=r* sing.cosp|—rsing 0) SEC. 14,THE MOMENT OF INERTIA AND THE COORDINATES OF THE CENTRE OF GRAVITY OF ASOLID 1.The moment ofinertia ofasolid. The moments ofinertia ofapoint M(x, y,2)ofmass mrelative tothe coordinate axes Ox,Oy,andOz(Fig. 324) areexpressed, : respectively, bytheformulas Leg=(y'+24)m, a=Lyy=(8zt)Lee(ey). Hl 2 H Hh LP p beataeAl(4.2) 7gyA h ‘s Fig. 324. Fig. 925, The moments ofinertia ofasolid are expressed bythe corre- sponding integrals. For instance, themoment ofinertia ofasolid relative tothe z-axis isexpressed by the integral ,,= =JfS et+y)ve,y2)dedydz, where y(x,y,2)istheden- fj sityofthesubstance. Example J.Compute themoment ofInertia ofrightcircular eylinder ofaltitude2%andradiusRrelativetothediameterofitsmediansection,considering the density constant and equal to¥,Solon, "Chvse'scoriie sytemawsdetthezalsalong the“ansof,thepliner, andpattheoriginofcoordinates atitscentrea symmetry (Fig. 328). Moment ofInertia and Coordinates ofCentre ofGravity ofaSolid Gat Then the problem reduces tocomputing the moment ofinertia ofthe cylinder relative tothe x-axis: leaJSJUtevseduds7 Changing tocylindrical coordinates, weobtain marktaal{5[Sirtersmrora] eae}ao wR = = anRe nn{§PEtoerseo] eachmyf(aAeobat 200RE 2hae[AEonAE|net[3or]. 2.The coordinates ofthe centre of-gravity ofasolid. Like what wehad inSec. 8,Ch. XII forplane figures, thecoordinates ofthecentre ofgravity ofasolid areexpressed bytheformulas SSlavt anardedyde (Vaiss addrdyaey=———____ gsSpfvewaydedyds ~°VU\y(ey,2)dedyde : 7 SSfare nnarayas 28 — —“Thenaewe’ 7 where y(x, y,2)isthe density. Example 2.Determine the coordinates ofthecentre ofgravity ofthe upper halfofasphere ofradius Rwith centre attheorigin, considering the density ysconstant. SotutlSn, The hemisphere Isbounded bythe surfaces 22VRoF=P, 220. Thez-coordinate ofitscentre ofgravity isgiven bytheformula [fever ade TSSreteay ae” 7 ea Mattipte Integrals Changing tospherical coordinates, weget are nfKy({reetesneer) ]asogRLaa T2_3 tea nr. PR cavel[J(6sing.”)4]a Obviously, byvirtue ofthesymmetry ofthehemisphere, x.=y-=0. SEC. 15. COMPUTING INTEGRALS DEPENDENT ON APARAMETER Consider theintegral dependent ontheparameter a. A 1(a)={f(x,a)de. (We examined such integrals inSec. 10,Ch. XI.) Westate with- out proof that ifafunction f(x, a)iscontinuous with respect toxover theinterval [a,6)and with respect toaover thein- terval (a,,@,], then thefunction A 1(a)=Sf(,0)de isacontinuous function on[o,,a,].Consequently, thefunction1(a)maybeintegrated withrespect toaontheinterval (a,,a,): ,Greerdam [1c«)de]da. The expression onthe right isaniterated integral ofthefunc- tion f(x, a)with respect toarectangle situated intheplane xOa, Wecan change theorder ofintegration inthis integral: aoe tet §[Jre.a)as]aa=(Fe,a)aa]dx, This formula shows that forintegration ofanintegral depen- dent onaparameter a,itissufficient tointegrate theelement ofintegration with respect totheparameter a.This formula isalsousefulwhencomputing definiteintegrals. Exercises onChapter XIV ous Example. Compute the integral en poke{ae This integral isnot expressible interms ofelementary functions. Toevaluate ityweconsider another integral that may bereadily computed: femacn Lao Integrating this equation belween the limits a=a and a=, weget ie A f[Jerse]aoafttant. Changing theorder ofintegration inthe Mist integral, werewrite this equafionfatheTolloving form * ee S[fers]ecomt, whenee, computing the inner integral, weget Ce-at_e beee az * Exercises onChapter XIV oye aasray Boo Cfdude Evaluatetheintegrals%:1.SSespy)dedy.Ans.$2.S$oe 3we 4 ; fon8FPapcanane Baa oath afat ue 4)theIntegra iswtonas |fFpied then8basaed BR been slated, we can consider that thefirst integration isperformed with Fespeet tothevariable whose differential occupies the frst place; thal is ra a Sree.paste (Srods)ay. wR a 654 MattipteIntegrals ¢(xdyde xa L (% Ja* 8PPEe.ans,Smaaretonbea Yxydedy,Ans.Oe. oes Fyre rs 32Jfedde. Ans.jabs Be Determinethelimitsofintegration fortheintegral(f(x,y)dxdywhere3 the region ofintegration isbounded bythelines: &x=2, x=3, y=—l, 2 ice y=5.Ans.SfIte,9)dydz.®y=0,y=1—x*,Ans.iyfHey)dydx.= aVanaa woattgteatAn.Yfemdyds. Heverea, vetAns. Jteemayas. 1290,90,ys,yaa.Ans.FYfee,aay a8 MG Changetheorderofintegration intheintegrals: 18.(F(x,yldyds.A. a v5 We Jfresmaray WVUfemdvds, Ans.SyHewdedy. Myis : Via2 4 wie 150 tenpasa.ans.(Jtenavact §Fpendyar.¥3 3oR ae od rice aViss AnsJfewaedy tr.)[feesyddedy.Ans.§[f(xyayareOise ie 6 +5Jremayer. ‘Compute thefollowing integrals bychanging topolar coordinates: 18.fJVa=RaPay ds,ans,|VaReteadeneot a ui Exercises onChapter XIV 568 vam a artwf(teen dray.Ans.[(erdednn2.20,[Pe-rermay de, i. ava Eracossx ans,SVemedeamt. zt.ffdyarans.[[edodo =m, Transform thedouble integrals byUntroducing newvariables uandvcon-pectedwithandybstheformulassmu—uo,yuo:22.{§Mspda berepre ee Ans.J[fu—uo, uyududs. 23.|U7,ayar. oe *Feiss 1¢ ans.J[fluuo, :e)udude+ |[pu—wo, woududo. ae Caleulating Surfaces byMeans ofDouble Integrals 2.Compute theareaoffigurebounded bytheparabola y*—=2r andthe straight linegx. Ans.3. 25,Compute thearea ofafigure bounded bythelines y*=4ax, x-+y=S0, y=0.Ans.at. 28.Compute’ theareaof2figurebounded bythelinesxfy?ma, styna An.= 21,Compute the area ofaAgure bounded bythelines y=sins, y=cos.s, a0. Ans, VI=1 . 28.Compute thearea of@loop ofthecurve g=asin20.Ans.SE anzigomputethe entire area bounded bythe lemnisate tates29, ; Bt) oy 30,Computetheareaofaloopofthecurve(7445) =2.. Hint. Change tonew variables x=ga.cos0andy=gbsind.Ans.2, 666 Multiple Integrals Calculating Volumes 31.Compute the volumes ofsolids bounded bythe following surfaces:Sybya,x0,yd,200,Ans,ME,92200,ettytd,byt+2=3.Ans.3m.33.uae ceetexy=2,2=0. Ans.x.34,x8y8§——2ax=0, 220,ttyteetAns,Bat.35yest,xmyl,220,25124 sty—xt, Ans,3 36.Compute the volumes ofsolids bounded bythe coordinate planes, theplone2e+3y-—12=0 andthecylinder2my?,Ans.16 97.Compute thevolumes of,solidsbounded byacreular cylinder of radius a,whose axis coincides with the z-axis, the coordinate planes and. the pineF4Z—1. ans.(4-1) 38,Compute thevolumes ofsolidsbounded bythecylinders x-+y!=at,stesteat Ans.Wak30gttsten, cay,290,Ans.Zo.M0.ahhh pete PEA=R a>ReAns,Salt—(VERM. Mh.armay's 220,xt4yfe2ar, Ans.Saat.42,gtaateos2,attyttrtmat, 220. (Compute the volume that isinterior with respect tothe cylinder.) Ans. FoBa+-2-16VD. Calculating Surface Areas 43,Compute thearea ofthat part ofthesurface ofthecone x*p-y*=zt which igcutoutbythecylinder 2*-+-y*—=2ar, Ans, 2na?V3, 44.Compute theareaofthatpartoftheplane x-+y-+2=5a, which, lies inthefirstoctant andisbounded bythecylinder 2t4+yt= at,Ans,5V3. 45,Compute thesurface area ofaspherical segment (minor) ifthe radius ofthe spiiere isa,while the radius ofthe base ofthe segment is0. Ans, 2x(a'—a Va), .48,Findtheareaofthatpartofthesurfaceofthespherex*-+-y?-+2'= a?whichiscutoutbythesurfaceofthecylinder *-+Han1(a>6).Ans anat—tat—are sinVEE 41.Find the surface area ofasolid that isthe common part oftwocylinders#4mat,yf-t2'—at,Ans.I6at 48,Compute the‘area. ofthat part ofthesurface ofthe, cylinder stseghe2an, whichtiebetween theplanem0andthecones¥y=2 ‘inCompute tegreaoftatpartoffhesacoftheopin xgtaat ‘which liesbetween theplane 2—mx andtheplane z=0, Ans.nat, Exercises onChapter XIV oor 50, Compute the ares ofthat part ofthe surface ofthe paraboloid yt22S an, which lies belween the’ parabolic cylinder y*=ar and the plane rea,Ans,x0@V3-1), Computing theMass, theCoordinates ofthe Centre ofGravity, and the Moment ofInertia ofPlane Solids (inProblems 51-64 weconsider thesurface density constant and equal tounity) 51,Determine themassofalah theshapeofacircleofradius if the density atany point Pisinversely proportional tothe distance: ofP fromtheaxisofthecylinder(theproportionality factorisK).Ans.naK. '2.Compute the coordinates ofthe’ centre. ofgravity ofamequilateral triangie ifwetake itsaltitude forthex-axis andthevertex ofthe’ triangle forthecoordinate origin.Ans.x=23;yao 58,Find thecoordinates ofthe centre ofgravity ofacircular sector ofradits6,Takingtheisectorofteangleasthevans.Theangeofspread ofthesectoris2a.Ans.xeS88, ya ‘54,Find thecoordinates ofthecentre ofgravity oftheupper half ofthe ctcest4ytmet Anttents yen 155.Find thecoordinates ofthecentre ofgravity ofthe area, ofone are oftheeyelolds=a(t—sint), y=a(l—cos), Ans.sean, ye=2 56.Find thecoordinates ofthecentre ofgravity ofthe area bounded byma? aloopofthecurve Q*=a*cos20. Ans. xe He0. 57,Findthecoordinates ofthecentreofgravityoftheareaoftheear dioidg=a(1+cos®). Ans. x=, ye=0. 58.Computethemomentofinertiaoftheareaofarectangleboundedby thearefines#20,x20,40,yorbrelativetotheorigin.Ans aot U ° 2 2 58,Compute themoment ofinertiaoftheellipse24.atrelativefotheyaxist)relativetotheorigin.Ans.a)2%;pySAary.on, 60,Compute themoment ofinertia ofthearea ofthecircle @=2acos0 lative tothepole.Ant2 61, Compute the moment of inertia of, the area of the cardioid enedl—eond) rlatve tothepoe,Ane.S24, \ 62.Compute themoment ofinertia of.thearea ofthecircle (x—a)*-. Uber daFrelativetothey-axis.Ans.Sr. 668 MaltipleIntegrats 63.Thedensity atanyplotofsquare slawithsideaisaproportion- altothedistance ‘ofthis point from one ofthe vertices of {hesquare, Compute’ the ‘moment ofinertia ofthe slab relative totheside. passing through thisverter. Ans.2kat(7VB4+9In(VB1)whereksthe proportionalityfactor 4,Compule themoment ofinertia oftheareaofafigure, bounded by theparabola gar andthestraight linex—o, relative t0thestraight line gana. Ans, Bot Triple Integrals 65.Compute (UT24245, ithereglonofintegration isboundedarate hg bythecoordinate planesandtheplanex-ty-tz—. Ans.%2—5, 6s,Evawuate {[(({xsede) ay]dx.Anese 67.Computethevolumeof2solidboundedbythespherex*-+-y-+2t=4 andthesurfaceoftheparaboloid xt-+yt=Sz. Ans.12x. 68.*) Compute thecoordinates ofthecentre ofgravity and the moments ofinertia ofapyramid bounded bytheplanesx=0,g=0,zme0;244.4z a>ce.,athe» blac,_ctabpee Ans seeds neds tes lee, eR, Se, 1M(atote.69.Compute the moment ofinertia ofacircular right cone relative toits axis,Ans.qhabrtwherehisthealtitudeandeistheradiusofthebaseof the cone, 70.Compute thevolume ofasolid bounded byasurfac with equation Gttytetteats, Ans.Saat, 71,Compute the moment ofinertia ofacircular cone relative tothe diameter ofthebase.Ans.On*43°9), 72,Compute thecoordinates ofthecentre ofgravity ofsolidtying between asphere ofradius aand conic surface with angle althevertex 2a, iIthevertex ofthecone coincides with thecentre ofthesphore. Ans. x,=0, 4.=9,72=-$a(1+e0sa) (thez-axisistheaxisofthecone,andthever- tex lies atthe origin). *)InProblems 68,69and 71to73weconsider th:density constant and equal {0unity. Exercises onChapter XIV 669 73.Compute thecoordinates ofthecentreofgravity ofasolidbounded byaaphere ofradiusaandbytwoplanes pang through thecentreofthei sphere andforming anangle of60°.Ans. e=75, O=0, g=-Z (theline ofintersection ofthe planes istalen for the z-axis, the centre ofthe sphere fortheorigin; @,0,garespherical coordinates). 12 74. Using ‘he equation —==——| e~“*da (a>0) compute theeavtrail )comp Peosxde 4,(siaxdx VzVz integrals (S288 and[Seam VFVF CHAPTER XV LINE INTEGRALS AND SURFACE INTEGRALS SEC, 1,LINE INTEGRALS Let the point P(x, y)beinmotion along some plane line L {rom thepoint Mtothepoint N.ToPisapplied aforce F whichvariesinmagnitude and Fydirection with the motion of P; Maz itisthussomefunctionofthe Gres [77 coordinates ofP:‘ae F=F(P). MyLet us compute the work A % ofthe force Fasthe point Pis translated from MtoN(Fig. 326). 7 Todothis, wedivide the curve MN into narbitrary parts bythepoints M=M, M,My... My=N in a % _%#04_*F the direction fromMtoN'andwe Fig.326. denote byAs, the vector MiMra. WedenotebyF,themagnitude of theforée Fatthepoint M,. Then thescalar product F,As; may beregarded asanapproximate expression ofthe work ofthe force Falong the areM)M,,,: A,©F,AS,. Let F=X(x,ylt¥ (xy where X(x,y) and Y(x,y)are the projections ofthevector Fon thex-and y-axes. Denoting byAx,and Ay,theincrements ofthe coordinates x;and y,when changing from thepoint M,tothe pointM,,,,wegetAs,=Ax,t+Ay,J.Hence,FAAS,=X(i4)AtFYCty4)Adie The approximate value ofthe work Aofthe force Fover the entire curve MN will be AmDRA =BXGywbx+Y(814)duh ay Line Integrats on Without making any precise statements, we shall say that if there exists alimit ofthe expression on’theright asAs,—-0 (here, obviously, Ax;—+0 and Ay;—+0), then this limit expresses thework ofthe force Fover the curve Lfrom the point Mto ‘the point N: A=limD(X(x,y) Ax+(x,yi)Ayjil- (2)dacait The limit *)onthe right iscalled the line integral ofX(x,y) and Y(x,y) over thecurve Land isdenoted by A=[X(x, y)de+¥ %vdy @) i or “ A=\X(xy)de+¥(x,y)dy. @) do Limitsofsumsofpe(2)frequently occurinmathematics and mechanics; here, X(x,y) and Y(x,y)areregarded asfunctions oftwo variables’ insome region D. The letters Mand N,which take the place ofthe limits of integration, are inbrackets tosignify that they arenot numbers butsymbols ofthe end points ofthe line over which theline integral istaken. The direction ofthe curve Lfrom MtoNis called the sense ofintegration. Ifthe curve Lisaspace curve, then the line integral ofthree functions X(x, 92), ¥(x, 2),Z(% 2)isdefined similarly: SX (eyde+Y(x,y2)dy+Z(x,y,2)d2— i =,lim2X(asYu»Ze)MEY (XpeYarZe)MYAZ(KerYarZa)AZye gate The letter Lunder theintegral sign indicates that theintegration isperformed along the curve L. We note two properties ofaline integral. Property 1.Aline integral isdetermined bythe element of integration, the form ofthe curve ofintegration, and the sense ofintegration. *)Here, thelimit ofthe integral sum istobeunderstood inthesame sense asinthecase ofthe definite integral, seeSec. 2,Ch. XI. on Line Integrals and Surface Integrals Aline integral changes sign when the sense ofintegration is reversed, since inthat case the vector As, and hence itsproje- ctions Axand Ay, changes sign. Property 2.Divide thecurve Lbythepoint Kinto pieces L, andL,sothatMN=MK+KN(Fig.327).Then,fromformula (1)itfollows directly that on) 9 on{Xdr+Vdy= |Xdx+Y¥dy+ |Xdx+Vdy. iy ao & This relationship holds forany number ofterms. Itwill further benoted that the definition ofaline integral holds true also for the case when the curve Lisclosed. Inthis case, the initial and terminal points ofthecurve coin- cide. Therefore, inthe case ofaclosed curve wecannot write on ‘ 7|Xdx-+Y¥dy, butonly[Xdx+Y¥ dy;andwe do i havetoindicatethedirection ofcirculation by (sense ofdescription) over the closed curve L. The line integral over aclosed contour Lis ” frequently denotedalsobythesymbolfXdx. fig.527. jrequently denotedalsobythesymbolf+ +¥dy. Note. Wearrived attheconcept ofaline integral while consi- dering theproblem ofthework ofaforce Fonacurved path L. Here, atall points ofthe curve Lthe force Fwas given as avector function Fofthe coordinates ofthepoint ofapplica- tion (x,y);theprojections ofthevariable vector Fonthe coor- dinate ‘axes areequal tothescalar (numerical, that is)functions X(x,y) and ¥(x,y). For this reason, line integral ofthe form \Xdv+Ydy mayberegarded asanintegral ofthevector functionFgivenbytheProjections XandY. The integral ofavector function Fover the curve Lisdeno- ted bythe symbol \Fas. z Ifthevector Fisdefined byitsprojections X,Y,Zthen this integral isequal tothe line integral [Xa Vdyt2de, Evaluating aLine Integral 73 Asaparticular instance, ifthe vector Flies inthexy-plane, then the integral ofthis vector isequal to 5Xde+Vdy. WhenthelineintegralofavectorfunctionFistakenslong aclosed curve L,this line integral isalso ealled acirculation ol the vector Fover the closed contour L. SEC. 2EVALUATING ALINE INTEGRAL Inthis section weshall make more precise the concept ofthe limit ofthe sum (1) ofSec. 1and inthis connection weshall make more precise the concept of the line integral and indicate a N method forcalculating it. ihLetacurve Lberepresented by4/7wy) equationsinparametric form: f,x=90, Y=9). iy Consider thearcofthecurve MN a ie(Fig.328).LetthepointsMandNOl ‘ae *correspond tothevalues ofthepara- Fig.328. meter aand f.Divide the arc MN into subarcs As, bythe points M,(x, ,), MyUy Yur soo My(XpYa)andputx,=@(f;), y=Plt).Consider the line integral SX, detV(x,yay 0) i defined inthepreceding section. Wegive without proof theexist- ence theorem ofaline integral. Ifthefunctions p(t) and %¥(t) arecontinuous and have continuous derivatives '(t)and ‘(t), and also continuous arethefunctions X(q(t), ‘p(t)] and ¥[p(¢), @(¢)) asfunctions oftontheinterval [ap], then the following limits exist: limYXGG) dx,=4X(x,wax, aan’ @ limDYGH)Au=S¥ (way, where ¥;and Yjare the coordinates ofsome point lying onthe arc As;. These limits donot depend onway thearc Lisdivided 22—ass on Line Integrats ondSurface Integrals into subarcs As;, provided that As,—+0 anddonotdepend onthe choice ofthepoint M,(%;, %;)onthesubarc As;; they arecalled line integrals and are denoted as limBXGW Ax=[Xewae, limBYGT) dm=T¥ Ceway. Note. From this theorem itfollows that the sums defined in ‘thepreceding section, where thepoints M;(%;, 9)arethe extremi- ties ofthesubare As; and the manner ofpartition ofthe arcL into subarcs As, isarbitrary, approach thesame limit—the line integral. This theorem makes itpossible todevelop amethod forcomput- ing aline integral. Thus, bydefinition, wehave ow a {X(x,yde= limDX&,7)dx, @) ihane teh where Ax=¥)—*}-.=OL)—9(t-1)- Transform this latter difference bytheLagrange formula An=O) OG-= 9H(t =0"CH)Als where t;issome value of¢that liesbetween thevalues f,—1 and #;,Since thepoint %,,%;onthe subarc As,may bechosen atpleasure, weshall choose itsothat itscoordinates correspond tothe value ofthe parameter 1: H=OH), =P Substituting into (3)thevalues of%,,7and Ax,that wehave found, weget “ A {XC,gdem lim3Xt9(e) (eI9(HdBt oy *suse fh Onthe right isthe limit ofthe integral sum forthecontinuous function ofasingle variable X(p(t), p(t)] ¢'(f) onthe interval fe,B). Evaluating aLine fategral ors Hence, this limit isequal tothe definite integral ofthis function: A {X(,yde=JX19,vOledt. cn 3 Inanalogous fashion wegetthe formula w 2 Vy,pdy=SV 19,violy(oat.io 2 Adding these equations term byterm, weobtain wy A XGndery dy|(XIV Olt (inH +Y19, VOIY (hat. “ This isthe desired formula forcomputing aline integral. Insimilar manner wecompute the line integral §Xdet+Ydy+Zde over thespace curve defined bytheequations x=9(f), y=‘p(), z=4(0). Example 1.Compute thelineintegral ofthreefunctions: 2%,3zy’,—x*y (or, which isthesame thing, ofthe vector function x*+43zy*f—x*yk) along Segment ofsstraight line” lssuing from thepoint 1(G21) tothe porat N(O,0,O)(Fig.329).Solution 16findtheparametric equations ofthelineMN,alongwhich theintegration Isfobe:performed, wewrite theequation ofthe atraigat line that pases through thegiven two! points: ci g727T? and denote allthese relations by&single letter th the equatiaftestraightline’inparametric forme, nn”We6*tIheeauations rel, yet, rat. Here,obviously, totheoriginofthesegmentMNcorrespondsthevaluetheparameterf=sl,andtotheterminusofthesegment;thevalueYO.The derivatives ofxy.'2 with cespect tothe parameter f(which will beneeded forevaluating iheline Integra) areeasily ound: 53, ype gel a ors LineIntegralsandSurfaceIntegrals Now the desired line integral may becomputed byformula (4): ow °(8drtSedy—atyde=[1007-3494 (0.2—GNF} Go ? . Cos 87=oa=—%. Example 2.Evaluate theline integral ofapair offunctions: 6x%y, 10xy* hlongaplanecurveyeratfromthe’pointM(1,1)tothepoint(2,8) ‘Figs330).‘Solution. Tocompute therequired yintegral Wwy {oxydettory?ay ny wemust have the parametric equations of the ‘given curve. However, the explicitly fefined equation ofthecurve y=? is& xspecial case ofthe parametric’ equation: A 7 12 ® ul yen hc Fig. 929. Fig. 880. here, the abscissa xofthepoint ofthecurve serves astheparameter, and the parametric equations ofthe curve are een, gaat The parameter xvaries from x=1 tox=2. The derivatives with respect totheparameter are readily evaluated: el, yaa, Hence, “: {ortydx10xy*dy—{[6x41410ee"3]de joi =f(+00) a=potaet 1008, Evaluating @Line Integral on Wenow indicate certain applica- yy tions ofaline integral. Pp 1.The expression ofthearea ofa GO) region bounded byacurve interms G ofalineintegral. Inanxy-plane let athere begiven aregion D(bounded bythe“contour L)such that Kany straight line parallel toone of |} the coordinate axes and passing through aninterior point ofthereg- ioncuts theboundary Lofthere-G7 5 gion innomore than two points (which means that the region Dis Fig.931. regular) (Fig. 331). Suppose that the region Disprojected on the x-axis inthe interval [a,6],and itisbounded below bythecurve (I,): Y=" and above bythecurve (I,): Y= Y@)<s.@)]. Then thearea oftheregion Dis > ® S=Jy()de—Jy, (@)de. Butthefirstintegral isalineintegral overthecurve’!,(MPN),since y=y, (x)istheequation ofthis curve; hence, ° Sy@de= |yde. 3 iow _ Thesecond integral isalineintegral overthecurve1,(MQN), that is, 5 Syide= Jyde. 3 aw ByProperty 1oftheline integral wehave §yde=— {yd. bw wba Hence, S=— fydx— §ydx——Syde. ® we Man fs om LineIntegralsandSurfaceIntegrals oe Here, the curve Listraced inacounterclockwise direction. Ifpartoftheboundary Listhesegment M,M, parallel tothe y-axis, then |ydx=0, andequation (5)holds trueinthis giocaseaswell(Fig,332). Similarly, itmay beshown that S=i)xdy, 6) Adding (5)and (6)term byterm and dividing by2,weget another formula forcomputing thearea S: 1Say]ray—ude. 2) Example 3.Compute thearea oftheellipse xeacos!, y=bsiat, Solution. Byformula (7)wefind SahJtacos:hcos—bsin (asin) dt=aab. Wenote that formula (7)and formulas (5)and (6)aswell hold true also forareas whose boundaries are cut bycoordinate lines inmore than twopoints (Fig. 333). To prove this, we divide the yy P given region (Fig. 333) into two regular regions by the line /*, y yoy) Mm W oyaeare aed al ¥ Fig. $32. Fig. 833. Formula. (7)holds foreach ofthese regions. Adding theleft and right sides, weget(on the left) thearea ofthegiven region, ontheright,alineintegral(withcoefficient "/,)takenovertheentireboundary, since thelineintegral overthedivision line/*istaken twice: in'the direct and reverse senses; hence, itisequal tozero. Green's Formuta oro 2.Computing the work ofavariable force Fonsome curved path L.Aswas shown atthe beginning ofSec. 1,thework donebyaforceF=X(x,y,2)i+Y (x,y,2)/+Z (x,y,2)&alongalineL=MN isequal totheline integral ry A=\X(x,y,2)de+¥ (x,y,2)dy+Z(0,y,2)dz. i Let usconsider aninstance that Ma(Or- shows how to calculate the work of the force inconcrete cases. iqeEttmole 4,Determine theworkA,ofthe slatedfromthepoate a)fethe ireBeitMelee ace)alongamaeiiry pathElaye)‘Solution. Theprojections oftheforceof 7 gravity Fon thecoordinate anes are =, X=0, Y=0, Z=—mg. Fig. 834. Hence, the desired work Is uy % Aa|arty ty424e=| me)demmele.—a). uly a Consequently, tinthis case theline integral Isindependent ofthepath of Integration anddependent onlyonthe inal andteil points. Nae:re cisely, thework ofthe force ‘ofgravity isdependent “only onthe dilference Setween theheights oftheTerminal and initia! points ofthepath. SEC. 3,GREEN'S FORMULA Let usestablish aconnection between adouble integral over some plane region Dand the line integral around theboundary L ofthis region. Inanxy-plane, letthere begiven aregion D,which isregu- larboth inthe direction ofthe x-axis and they-axis, bounded byaclosed contour L.Let this region bebounded below bythe curve y=y,(x), and above bythecurve y=y, (x), 9,(x)<y,(x) (a<x<b) (Fig. 331). Together, both these curves represent theclosed contour L.Let there begiven, inthe region D,continuous functions X(x,y)and Y(x,y) that have continuous partial derivatives. Weconsider the integral BritonyJSetaxdy. 620 LineIntegralsandSurfaceIntegrals Representing itintheform ofaniterated integral, wefind ee) A00 {iehecarn|['P gai]temfixe[aeBact 2 ne » =JSXG OY=X(x,yoo]de. O) Wenote that the integral A Sx wide isnumerically equal tothe line integral §X@wde ibm taken along the curve MPN, whose equations, inparametric form, are eax, y=y,(%), where xisaparameter. Thus ’ SX@wnde= [Xieyde @ 3 aw Similarly, theintegral ’ SX, ye)de isnumerically equal tothe line integral along the arc MQN: * Sx ynde= [Xx,yde @) H waa Substituting expressions (2)and (3)into formula (1), weobtain SSaeay— )X(x,y)dx—JX(x,y)de. @ 3 ab aiew But §X@yde=— [X(x,yde aw wew Conditions foraLine Integral Being Independent ofthePath 681 (see Sec. 1,Property 1).And soformula (4)may bewritten thus: SSShaedy—)X(x,y)de+5X(xy)dx. 3 bw Mew But the sum ofthe line integrals onthe right isequal totheline integral taken along the entire closed curve Lintheclockwise direction. Hence, the last equation can bereduced tothe form Sizdxdy= § X(x,yde. ©)773 (tnthecides senses Ifpart oftheboundary isthesegment /,parallel tothey-axis, then|X(x,y)de=0, andequation (6)holds trueinthiscaseas 4well. Analogously, wefind . Eg§Soeaedy=— iy ¥(y)dy. O)o Lan the eotkwis sens Subtracting (6)from (5), weobtain ax_a SS(HF) aeay § Xdx+Vdy. ° tn the clockwise sense) Ifthe contour istraversed inthecounterclockwise sense, then *) av_axSS(B—Sp)aedam[Xdvay, This isGreen's formula, named after the English physicist and mathematician D,Green (1793-1841)**). We assumed that the region Disregular. But, asinthe area problem (see Sec. 2),itmay beshown that this formula holds true forany region that may bedivided into regular regions. SEC. 4.CONDITIONS FOR ALINE INTEGRAL BEING INDEPENDENT OF THE PATH OF INTEGRATION Consider the line: integral o \Xde+¥dy, i 7)IfinaTine integral along aclosed contour the direction ofcirculation isnot indicated, itis-assumed that itisinthe counterclockwise sense. Ifthe direction ofcirculation isclockwise, this must bespecified **)Thisformula Isaspecial caseof2moregeneral formula discovered by the Russian mathematician M.V.Ostrogradsky. 682 LineIntegralsandSurfaceIntegrals taken around some plane curve Lconnecting thepointsMandN. Weassume thatthefunctions X(x,y) and¥(x,y) have conti- nuous partial derivatives intheregion D 2Wunder consideration. Let us find out under M what conditions the line integral above is 3 independent oftheshape ofthecurve Land isdependent onlyontheposition oftheini- Fig335. tialandterminal points MandN. Consider twoarbitrary curves MPNand MQNlyinginthegivenregionDandconnecting thepointsM and N(Fig. 335). Let {xXde+¥dy= |Xde+¥dy, () wen an that is, {Xde+¥dy— §Xdx+V¥dy=0. aby Man Then, onthebasis ofProperties 1and 2oflineintegrals (Sec. 1), we have {Xdr+¥dyt+ §Xdx+¥dy=0, aby wan which isaline integral around theclosed contour L: §Xdx+¥dy=0. r Inthis formula, the line integral istaken around theclosed con- tour £,which is‘made upofthe curves MPN and NQM. This contour Lmay obviously beconsidered arbitrary. Thus, from the condition that forany two points Mand Nthe lineintegral isindependent oftheshapeofthecurveconnecting them and isdependent only onthe position ofthese points, it follows that the line integral along any closed contour isequal to zero, The converse conclusion isalso true: ifaline integral around any closed contour isequal tozero, then this line integral isinde- pendent oftheshape ofthe curve connecting the two points, and depends only upon theposition ofthese puints. Indeed, equation (1) follows from equation (2). InExample 4ofSec. 2,theline integral isindependent ofthe path ofintegration; inExample 3theline integral depends onthe path ofintegration because here the integral around theclosed contour isnotequal tozero, but yields anarea bounded bythe Conditions for aLine Integral Being Independent ofthe Path 683 contour inquestion; inExamples 1and2thelineintegrals are likewise dependent onthe path ofintegration. The natural question arises: what conditions must thefunctions X(x,y) and Y(x,y) satisfy inorder that the line integral §Xdx+yYdyalonganyclosedcontourbeequaltozero.The answer isgiven bythe following theorem.Theorem.AtallpointsofsomeregionD,letthefunctions.X(xy), Y(x,y),together withtheirpartialderivatives en angG8) becontinuous. Then, fortheline integral along any closed contour L lying inthis region’ tobezero, that is,for [XGdertyee,ydy=0, @ itisnecessary and sufficient tofulfil theequation ox_ayx-F ) atallpoints oftheregion D. Proof. Consider an arbitrary closed contour Linaregion D and write Green's formula for it: ay_ax SS(GE—3)dedXayay. Ifcondition (3)isfulfilled, then thedouble, integral ontheleft isidentically zero and, hence, |Xdx+Vdy=0. L This proves thesufficiency ofcondition (3).Nowweprovethenecessity ofthiscondition; thatis,weprovethat if(2)isfulfilled for any closed curve L’inthe ‘region D, then condition (3)isalso fulfilled ateach point ofthis region. Letusassume, onthecontrary, that equation (2)isfulfilled, that is, Xdr+¥dy=0, L and that condition (3)isnot fulfilled; ay _ax aeayFO atleast inonepoint. For example, atsome point P(x,,y,) let ost LineIntegralsandSurfaceIntegrals there bethe inequality OY ax eo Since there isacontinuous function ontheleft, itwill bepositive andgreater than some number 8>0 atallpoints ofsome sufficiently smallregion D’containing thepointP(x,y,).Takethedoubleintegral ofthedifference xi overthisregion. Itwillhavea positive value. Indeed, ay_aX 7 SS(e—3) dxdy>({ddedy=0{ {dedy=a0’>0. ” But byGreen's formula theleft side ofthe last inequality is equal toaline integral along the boundary L’oftheregion D’, which, byassumption, iszero. Hence, the last inequality contra: dictscondition (2)andtherefore theassumption that$¢—$* is different from zero inatleast one point isnot correct. Whence it follows that w_oX_4oe og atallpoints ofthe given region D. The theorem isthus proved completely. InSec. 9,Ch. XIII, itwas proved that fulfillment ofthecon- dition Wey _aXe oF oy istantamount tothefact that the expression Xdx-+Ydyis.an exact differential ofsome function u(x, y),oF Xdx+¥ dy=du(x, ») and au au Xe,=F, YsMaze But inthis case the vector du42H FaXit Viale By isthegradient ofthefunction u(x, y);thefunction u(x, y),the gradient ofwhich isequal tothe ‘vector Xi+Yj, iscalled the potential ofthis vector. Conditions foraLine Integral Being Independent ofthePath 685. oy Weshallprovethatinthiscasethelineintegral I=|Xdx-+Y¥dy (inalong anycurve Lconnecting thepoints MandNisequal tothe difference between thevalues ofthefunction uatthese points: “ a {Xdx+¥dy= Jdu(x, y)=u(N)—u(M). a 7 Proof. IfXdx+Ydyistheexactdifferential ofthefunctionu(x,y),thenXai yap andthelineintegral takesonthe form ©ouou t=|Bde+3dy, on Toevaluate thisintegral wewritetheparametric equations of thecurve Lconnecting thepoints Mand N: «=e, y= old). Weshall consider that to,the value ofthe parameter ¢—t, there corresponds the point M,and to¢=T, the point N.Thea the line integral reduces tothe following definite integrals ¢ ‘duOx,Ou mf [te920 The expression inthe brackets isafunction off,and this func- tion isthe total derivative ofthe function wlp(), ()] with respect tot.Therefore fa1=SSidt=alo, VONE=uI9, OL —4le(t), p(.)] =4(N)—4(M). Aswesee, thelineintegral ofanexact differential isindependent oftheshape ofthecurve along which theintegration isperformed. Wehave asimilar assertion foraline integral over aspace curve (see below, Sec, 7). ee Line Integrals and Surface Integrals Note. Itissometimes necessary toconsider line integrals of some function X(x,y)along the length ofanare L: [XO dsmtimSXCay4)ds “4 where dsisthe differential ofthearc. Such integrals areevaluat- edinsimilar fashion tothe line integrals considered above. Let thecurve Lberepresented bytheparametric equations x=91), v=), where @(t), p(t), 9"(O,‘y'(8)arecontinuous functions oft. Let @and Bbevalues oftheparameter ¢corresponding tothe origin and terminus ofthe arc L. Since ds—Ve Fv Oat, weget aformula forevaluating integral (4): p $x, yds=\X lo. vOlVEO+¥Oat. 2 2 ‘We can consider the line integral along the arc ofthe space curve x=9(0), Y=), 2=4(0 : [XG¥2d5=JX6O, VO.LOWTOT OFTOat. Bytheuse ofline integrals along anarc wecan determine, for example, thecoordinates ofthecentre ofgravity oflines. Reasoning asinSec. 8,Ch. XII, we obtain aformula for evaluating thecoordinates ofthe centre ofgravity ofaspace curve. Kem) Yee eR 6)Je fe a z z pcfuamole Findthecourts ofthecent ofgravity ofeeturnofthe enact, yaasint, 2=bt O<t <n), ifitslinear density isconstant. Surface Integrals esr Solution. Applying formula (6), wefind [coneVaFaTCORTTD?at ehee [Varmarrareae Troat YaconsVarEDtae a atVEEEO5iz ‘OnVarpoF \Vara * Similarly, ye=0, fonVareaarcoTae i betaVEOan =PAVE Ea, onVapor anVapor ‘Thus, thecoordinates ofthe centre ofgravity ofoneturn ofthehelix are we=0, yenO, 2mad, SEC. 5.SURFACE INTEGRALS Let aregion Vbegiven inanxy2-coordinate system. Let a surface @bounded byacertain space line %begiven inV. With respect tothe surface oweshall assume that ateach point Pofitthepositive direction ofthenormal isdetermined bytheunit vector m(P), thedirection cosines ofwhich arecon- tinuous functions ofthe coordinates ofthe surface points. Ateach point ofthesurface letthere bedefined avector, FHaX eyDIFV (Hy.DI+Z( y2h, where X,Y,Zare continuous functions ofthe coordinates. Divide thesurface insome way into subregions Aq;. Ineach subregion take anarbitrary point P,and consider thesum PEPya(P)doy, (O) where F(P,) isthevalue ofthevector Fatthepoint P;ofthe subregion Agj; n(P;) istheunit normal vector atthis point and Fa isthe scalar product ofthese vectors. The limit ofthesum (1)extended over allsubregions Ag,as thediameters ofallsuch subregions approach zero iscalled ‘thie 688 LineIntegralsondSurfaceIntegrals surface integral and isdenoted bythesymbol SSFnac, Thus, bydefinition *) . audit SFin,b0,=§§Fndo. @ Each term ofthe sum (1) Fn,Ao,=F,Ao;cos(n;,F;) 3) may beinterpreted mechanically asfollows: this product isequal tothevolume ofacylinder with base Aq;and altitude F,cos(m;,F;). Ifthe vector Fisthe rate offlow ofaliquid through the’ sur- face o,then the product (3) isequal tothe quantity ofliquid flowing through thesubregion Ag, inunit time inthedirection ofthevector n;(Fig. 336). {> is [LZYo re) a as pda | 3 Fig, 586. Fig. 387. Theexpression {{Fndo yields thetotalquantity ofliquid flowing inunittimethrough thesurface ointhepositive direc-tionifbythevector Fweassume theflow-rate ‘vector ofthe liquid atthegiven point. Therefore, the surface integral (2)is called theflux ofthevector field Fthrough thesurface 0. From the definition ofasurface integral itfollows that ifthe surface @isdivided into theparts o,,0,..., o,,then SJFndo=S{ Fndo+S\ Fado+...+){ Fras. Dit thesurface gissuch that ateach point ofitthere exists-2 tangent plane that constantly varies asthepoint. Pistranslated over’ thesurface, Snd ifthe vector function Fiscontinuous onthis surface, then this limit exists (weaccept thisexistence theorem ofasurface integral without rool). Evaluating Surface Integrals 689 Let usexpress theunit vector minterms ofitsprojections on the coordinate axes: n= cos(n, x)i+cos(n, y)j+cos(n, z)R. Substituting intotheintegral (2)theexpressions ofthevectors Fand nminterms oftheir projections, weget $fFado=ff[Xcos(n,2)+Ycos(n,y)+Zcos(n, 2)}do.(2") The product Accos(n, 2)isthe projection ofsubregion Aoon thexy-plane (Fig. 337); ananalogous assertion holds true forthe following products aswell: Agcos(n, x)= Ady:, Aocos(n, y)= Adz, Agcos(n, 2)= Ady, (4) Where Ady, Ax, Adz atethe projections ofthe subregion Ao ‘ontheappropriate coordinate planes. Onthis basis, integral (2’) can also bewritten inthe form $fFado=Jf[Xcos(n,x)-+Ycos(n,y)+Zcos(n, z)]do= =f)Xdyde+¥ dzde+Zdxdy. @) 7 SEC, 6EVALUATING SURFACE INTEGRALS Computing theintegral over acurved surface reduces toeva- luating adouble integral over aplane region. Toillustrate, the following isamethod ofcomputing the integral §§Zcos(n,2)do. Let the surface obesuch that any straight line parallel tothe z-axis cuts itinone point. Then the equation ofthe surface may bewritten intheform z=1(, y). Denoting byDthe projection ofthe surface oonthe xy-plane, weget(by thedefinition ofasurface integral) SJZee,yz)cos(n, 2)do=, timXZinYin%)C08(ns2)Aye, am801-0fot 690 LineIntegralsandSurfaceIntegrats Noting, further, the last offormulas (4), Sec. 5,weobtain JfZeos(n, 2ydo— timS12(einYiFen4)(Qoan)= ? jamSoyot aylimOZGiveMe4D)Aeoles the last expression isthe integral sum foradouble integral of thefunction Z(x, y,F(x, y))over theregion D,Therefore, SJZcos(n, 2\domSSZ(x,yfle,y))dedy. < 3 Theplus sign infront ofthedouble integral istaken ifcos(n, 2)>0, the minus sign, ifcos(n, z)<0. Ifthesurface odoes not satisfy the condition indicated atthe beginning ofthis section, then itisdivided into parts that satisfy this condition, and the integral iscomputed over each part separately. The following integrals are computed insimilar fashion: SJXcos(n, edo; Sf¥cos(n, y)do. The foregoing proof justifies thenotation ofasurface integral inthe form of(2"), Sec. 5. Here, the right side of(2°) may beregarded asthesum of double ‘integrals over theappropriate projections ofthe region o and the signs ofthese double integrals (or, otherwise stated, the signs ofthe products dydz, drdz, dxdy) are taken in accord with the foregoing rule. Example 1.Letaclosed surface obesuch that any straight line parallel tothezaxis cuts itinno more than two points, ‘Consider the integral Secon, nde Wehallclltheouternorma thepositive direction ofthenormal, Inthis case, thesurlace may bedivided Info two parts: lower and upper: their equations’ ace, respectively, reh(e yand rmhy(e, Denote byDtheprojection @onthezy-plane (Fig. 338); then Gfzcostn, sydo—fFtatesvaray—(Cf(xw)deay. ° ’ o odwt ‘n d ai? o ae4ys Fig. 338. Fig. 339. vaffzee, ado Feber Sfesreneonfs [Jone satbea tm itnfem tae 2 Line Integrals and Surface Integrals SEC. 7,STOKES’ FORMULA Let there beasurface osuch that any straight line parallel tothe z-axis cuts itinone point. Denote by4the boundary of the surface o.Take the positive direction ofthe normal mso that itforms anacute angle with the positive z-axis (Fig. 340). Let the equation ofthe surface bez=f(x, y).The direction cosines ofthe normal are expressed bythe formulas (see Sec. 6, z Ch. TX): a oF x i . 008(1, 8)=oa i V+(%)+(%) >| —iay : cos(n,y)=——— 5 ¢(I) ih /— Yo(ay+(@)cos(a, 2)=——————————— . +7 V+(2) +) Fig. 540. We shall assume that the surface olies entirely insome region V.Let there be afunction X(x, y,z)given inVthat iscontinuous together with first-order partial’ derivatives. Consider the line integral along the curve A: fxey2)dx. Onthe line A,z==f(x, y),where x,yare the coordinates of ‘thepoints oftheline L,which isaprojection oftheline &on thexy-plane (Fig. 340). Thus, wecanwrite theequation JXteuwddee|Xte,wsHeMae @ z ‘The last integral isaline integral along L.Transform this integ- talbyGreen's formula, putting Xe HHe M=X(% yy O=Ve y) Substituting intoGreen's formula theexpressions ofXandY, we obtain a)ste)dydy=SKC,wedx@) Stokes? Formula 603 where the region Disbounded bythe line L.On the basis of the derivative ofthe composite function X(x,y,f(x, y)). where yenters both directly and interms ofthe function’ 2=/(x, 9), we find aXe vsHes W)_OXUewe2)4OX,we2)eV) a a Ree 4 eee ne) Substituting expression (4)into the left side of(3), weobtain ‘OX(x,y2),OX(x,y,2),OF(x,yy -Sf(Pp eeSOdedy =JXG wHemde. Taking into account (2), the last equation may berewritten as (CX gegy06OXa frey2)de=Ssdxdy-SSarSjaed 6) The last two integrals can betransformed into surface integrals. Indeed, from formula (2"), Sec. 5,itfollows that ifwe have some function A(x, y,2),thefollowing equation istrue: SJAlsy,2)c0s(n, 2)do=ffAddy. 3 5 Onthe basis ofthis equation, the integrals ontheright side of(6)are transformed asfollows: ax ax i}Sedxdy(0$*cos(n,2)d0, OXaf oxaf 6) i}Soaxay=SfFFcos(n,z)do. Transform the last integral using formulas (1)ofthis section: dividing the second ofthese equations bythethird termwise, we find cos(n,y)__oF‘cos(a,2)~~ ay or Leos(n,2)=—cos(n,y). Hence,OXoF ax SSifaxdy—S07cosn,y)do. m 8 3 oot LineIntegrateandSurfaceIntegrats Substituting expressions (6)and (7)into equation (6), weget {xuwade=— [1%cos(n,ado+[fFcos(n,y)do.(8) The direction ofcirculation ofthe contour 4must agree with the chosen direction ofthe positive normal n.Namely, ifan observer looks from the end ofthe normal, hesees the circula- tion along the curve 4asbeing counterclockwise. Formula (8) holds true for any surface ifthis surface can be divided into parts whose equations have the form z=f(x, y). Similarly, wecan write the formulas fy mady=Sf[HH05«,+Hcos(n,a]do,@’) az oz :" {20»adeff[—320sn,D+F008(n,»]do.6) Adding the left and right sides of(8), 8’), and (@), weget the formula av_ax’ §xdetVlyLaemff(¢-%cos(n,2)+ +(5Z—$) cosin,01+(52—$2)cosa,y)]do.(2) This formula iscalled Stokes’ formula after the English physicist and mathematician D.Stokes (1819-1903). Itestablishes a.rela- tionship between the integral over the surface oand the lineintegral alongtheboundary Aofthissurface, thecirculation about thecurve %being performed according tothesame rule as that given earlier. The vector B,defined bytheprojections @_#7,p_oXa. ay_oxBema) Bye} Bay iscalledthecurlorrotationofthevectorfunctionF==X#+Yj+ Zkand isdenoted bythesymbol rotF. Thus, invector notation, formula (9)will have theform Fds={ nrotFao, @) i 3 andStokes’ theorem isformulated thus: Stokes* Formula 695 The circulation ofavector around thecontour ofsome surface isequal totheflux ofthecurl through this surface. Note. Ifthe surface oisapiece ofplane parallel tothe xy-plane, then Az=0, and weget Green's formula asaspecial case ofStokes’ formuia. From formula (9)itfollows that if ay _aX_y o%_a@_9 oX_dz HHao, Zao, Hd, (19) then theline integral along any closed space curve %iszero: §Xdx-+¥dy+Zdz=0. qd): Whence itfollows that the line integral isindependent ofthe shape ofthecurve ofintegration. Asinthecase ofaplane curve, itmay beshown that the indicated conditions are not only sufficient butalso necessary. Inthe fulfillment ofthese conditions, the expression under the integral sign isanexact differential of’some function u(x, y,2): Xde+V¥dy+Zde=du(x, y,2) and, consequently, o ry |Xde+¥ dy+Zde= |du=u(N)—u(M).Go Gib This isproved exactly like the corresponding formula for a function oftwo variables (see Sec. 4). Example 1.Write the basic equations ofthe dynamics ofamaterial point: to,_y, ay domtsax, moray, moins. Here, mis themass ofthe point, X,¥,Zatethe projections ofaforce, tetngontnpint,entthecocinale ant:y=, gm, opelwe thepotions ofvelocity theaxes fitiply the left and” right sides ofthese equations bythe expressions ogdtmds, o,dt=dy, o,dt—dz, ‘Adding the given equations term byterm, weobtain M(¥g d0g+0,d0y +0,40)=Xdx+dy+243; mbaotpopbodaX detYdy42d, : 696 LineIntegralsandSurfaceIntegrals Since vf+o}-+of=0% wecanwrite a(tmo!)=XdebYdy-+Zdz ‘Take theintegral along thetrajectory connecting thepoints My and My: 1 1 Merola ymofXde+¥ dy+Zdz, city where 9,and varethevelocities atthepoints M,and My. 4 ‘This last equation expretses. the theoremm, ofliveforces:theincreaseinkineticenergywhen passing from one point foanother is ggSquat forthe! work’oftheorceacting onthe Example2. Determine theworkoftheforce ofNewtonian attraction toafixed centre of % fps theIaplation ofuitmaefrom 9 1(24e Bus)tOMaly.byc)- 7Shatin LettheStigBinthexed centteofattraction, Denoteby7the.radius hc Yeclorofthepoint (ig34)corresponding Fig.341. tganarbitrary position ofunit mass, andby#theunitvectordirectedalongthevector ThenF=—M'r*, where&istheconstant ofgravitation. Theprojections of the force Fon the coordinate axes will be Ls 1 Xerhn eS;Ya—km at, Zante, Then thework oftheforce Fover thepath MyM, is MyAnnis|stebydyteds cA UG)" ty =ki)orate|4(4) city aly ince rast ytteh rdrmxdx-+ydytzde). Ilwedenote byr,and rtAefengths oftheradiusvectors of{nepoints'M, and’Methen” Anim(4-2). Thus,hereagainthetineintegral, doesnotdepend on.theshapeofthe curve ofintegration, but only on. the position ofthe initial and’ terminal points.Thefunctionw=AZiscalledthepotentialofthegravitational eld Ostrogradsky's Formula 697 generated bythe mass m. Inthe given ease, au yu yaw xt, vat, 2a, A=u(M)—u(M,). Thatis,theworkdoneinmovingunitmassisequaltothedifference be- tween the values ofthe potentiaf atthe terminal and inital points SEC. 8OSTROGRADSKY'S FORMULA Let there begiven, inspace, aregular three-dimensional region Vbounded byaclosed surface oand projected onanxy-plane into aregular two-dimensional region D.Weshall assume that thesurface omay bedivided into three parts 0,,0,and o, such that theequations ofthefirst two have theform 2=f(% y)and 2=1,(% 9), where f,(x, y)and f,(x, y)are functions continuous isthe regionDandthe’third parto,isacylindrical surface withgenerator parallel tothe z-axis. Consider the integral ImSSP2aaya. First perform theintegration with respect tozz her=f$(iGPa)aay= i) H=S$Ze% whe, mdedy—(S 20,»ileMdedy. ? 3 Onthenormal tothe surface, choose adefinite direction, name- lythat which coincides with thedirection oftheouter normal tothesurface 0.Then cos(n, 2)will bepositive onthesurface 9,and negative onthe surface o,;onthe surface 9,itwill be zero. The double integrals ontheright of(1)areequal tothecor- responding surface integrals: SZ, wheWdrdy=S)Z(x, y,2)cos(n, 2)do, 2") 2 oy SSze,HhWdedy= [5Z(x,y,2)(—cos(n,z))d0, 08 Line Integrals and Surface Indegrats Inthelast integral wewrote {—cos(n, z)]because theelements ofsurface 6,and o,and theelement ofarea Asoftheregion D are connected bythe relation As=Ao {—cos(n, 2)], since the angle (n,2)isobtuse. Thus, $f202wsfoespdedum—Sf 208,wfetesweos(ns2940.2) Substituting (2') and (2') into (1), weobtain Stee Dardydem =JJZe,yzcosin, 2do+ffZ(x,y,z2)cos(n, 2)do. Forthesake ofconvenience insubsequent formulas, weshall rewrite thelastequation asfollows [adding §$Z0, y,2)cos(n, 2)do=0, sincetheequation’ cos(n,2)=0isfulfilled onthesurface 0]: le.vs2) SISGt?aeayte =JfZeos(n, 2)do-+{fZcos(n, z)do+[f Zcos(n, z)do. But thesum ofintegrals ontheright ofthis equation isanin- tegral over theentire closed surface o;therefore, SSaearaude—(020,y,2)cos(n,2)do, Analogously, wecanobtain therelations SfaparaydemSvcsy,2)cos(n,y)do, S{SBetedyaemPFxcy,2)c08(n,*)do. Ostrogradsky's Formula 699 Adding together thelast three equations term byterm, weget Ostrogradsky’s formula*’: Ox,OY,atSSS(GE+35+32)tyae= =f(Xcos(n,x)-+Y¥cos(n,y)+Zcos(n, 2)do. (2) Theexpression 5°+5"+32iscalledthedivergenceofthevec- tor (or the divergence ofthe vector function): FoXt+Yj+Zk and: isdenoted bythesymbol divF: fayFenOX4.OY482- divFaR +t H We note that this formula holds good forany region which may bedivided into subregions that satisfy theconditions indi- cated atthe beginning ofthis section. Let usexamine ahydromechanical interpretation ofthis formula. Let the vector F=Xi+Y/+Zk bethe velocity vector ofa liquid flowing through theregion V.Then thesurface integral in formula (2) isan integral ofthe projection ofthe vector Fon the outer normal m;ityields thequantity ofliquid Mowing out ofthe region Vthrough the surface oinunit time (orflowing into Vifthis integral isnegative). This quantity isexpressed interms ofthe triple integral ofdiv F. IfdivF==0, then the double integral over any closed surface isequal tozero, that is,the quantity ofliquid flowing out of (or into) something through any closed surface owill bezero(nosources). Moreprecisely, thequantity ofliquidflowingintoaregion isequal tothequantity ofliquid flowing outofthis region, iinvector notation, Ostrogradsky’s formula has the form SffavFaof{Fads ay *)This formula (sometimes called the Ostrogradsky-Gauss formula) was discovered bythenoted Russian mathematician MV, Ostrogradsky (1801-1861) fand published in1628 inanarticle enlilled “ANote onthe Theory ofHeat". 700 LineIntegrals andSurface Integrals and isread: the integral ofthedivergence ofavector field F extended over some volume isequal tothevector flux through the surface bounding thegiven volume. SEC. 9,THE HAMILTONIAN OPERATOR AND CERTAIN ‘APPLICATIONS. OF IT Suppose we have afunction u=u(x, y2).Ateach point of the region inwhich thefunction u(x, y,2)isdefined and diffe- rentiable, the following gradient isdetermined: ou Ou ow gradu +S ERS. a The gradient ofthe function u(x, y,2)issometimes denoted asfollows: bude.aWatt STSs @ The symbol yisread “del”. 1)Itisconvenient towrite equation (2)symbolically as O4,9 a ) yun(iZtgthg)u 2’) and toconsider the symbol aa a vai tigteg ® asa“symbolic vector”. This symbolic vector iscalled theHamil- tonian operator ordel operator (y-operator). From formulas (2) and (2’) itfollows that “multiplication” ofthesymbolic vector vy bythescalar function wgives thegradient ofthis function: yu=erad a. “ 2)We can form thescalar product ofthesymbolic vector yby the vector F=iX+jY +kZ: aa a VF=(iSdthE) UXT +82)= a a aOX|OY,azabxt grt ga BF 4Zeaive (see Sec. 8).Thus, yF=divF. 6) The Hamiltonian Operator and Certain Applications ofIt 701 3)Form the vector product ofthe symbolic vector yby the vector F=iX+j¥+kZ: ao pe VXP= (IZ+E+R)xUXEI+42)= ik) ja) jaa) jaa29.0|_,|%ae|__ lara]|lara=laeayae|="|y z|—4|xz|+*ws XYZ 2_ov)_,(9Z_aX) ,4(2¥_ox =1(G—)—d(S—-e)+(= {(%_av),,(aX_az) |,(a¥_ax 13d) +4(eR)+4(teFp)OF (seeSec.7).Thus, yXxF=rotF. (6) From the foregoing itfollows that vector operations may be greatly condensed bytheuseofthesymbolic vector y.Let usConsiderseveralmoreformulas.4)The vector field F(x, y,z2)=iX+jY+4k2Ziscalledapoten- tial vector field ifthe vector Fisthe gradient ofsome scalar function u(x, y,2): Feagradu vr euOuouFaiR+ Ete. Inthis case theprojections ofthevector Fwill be X=%, vat, 22%aH, vee, 20%, From these equations itfollows (see Ch. VIII, Sec, 12)thatax_ay a_ozaX_azOy“ie?Oy? FeOe or OX oY oY az OX OZF-Hao, F-FZao, KZao, Hence, forthe vector Funder consideration, rotF=0. Thus, weget rot(grad u)=0. @ 702 LineIntegralsandSurfaceIntegrals Applying thedeloperator y,wecan write (7)asfollows fonthe basis of(4)and (5)]: (yx yu) =0. (7) Taking advantage ofthe property that for multiplication ofa vector product byascalar itissufficient tomultiply this scalar byone ofthe factors, wewrite (yxy)u=0, a) Here, the del operator again hastheproperties ofanordinary vector; thevector product ofavector into itself iszero. The vector field F(t, y,2),forwhich rotF=0, iscalled irro- tational. From (7)itfollows that every potential’ field isirrota- tional. The converse also holds: ifsome vector field Fisirrotational, then itispotential, The truth ofthis statement follows from reasoning given atthe end ofSec. 7. 5)Avector field F(x, y,2)forwhich divF=0, that is,avector field inwhich there are nosources (see Sec. 8) iscalled solenoidal. We shall prove that div(rotF)=0 ® orthat the rotational field isfree ofsources.Indeed,ifF=iX-+JY +2,thenaz_av),,(0X_az)4(2¥_ax rotF=i(23)+4(5-32) +4(—$) and therefore a(az_ov) ,9/aX_0z), a/a¥_ax div(otF)=3(Z—-H) +5(HF) +5(H—-H) =o. Using thedeloperator, wecan write equation (8)as V(vxF)=0. @) The left side ofthis equation may beregarded asavector-scalar (mixed) product ofthree vectors: Vy,Vy,F,ofwhich two are the same. This product isobviously equal to‘zero. 6)Let there beascalar field u—u (x,y,2).Determine the gradient field: du, 52Hgraduize+ip+kg The Hamiltonian Operator and Certain Applications ofIt 703. Then find : a (mu), 2 (du), 9 (dudiv(grad=5(%)+35(5)+(B) or ' Ou, Ou, tudiv(gradw)=34+34454. ® The right side ofthis expression iscalled theLaplacian ope- rator ofthe function wand isdenoted by uaF 4 10)aatapt aa ¢ Hence, (9)may bewritten as div(gradu)=Au. ay Using thedeloperator ywecan write (II) as (yya)= Au. a’) We note that the equation Ou|Ou,Oe+t Mao (12) or Au=0 (zy iscalled Laplace's equation. The function that satisfies the Lap- lace equation iscalled aharmonic function. Exercises onChapter XV Compute thefollowing line integrals: 1[ytacsh2aydy overthecircumference x=acos’, y=asint, Ans. 0. 2Jyde—xdy overanarcoftheellipse x=acost, y=bsint. Ans, —2nab, . 2 w 2(schetegett)overaclewthcateaheein Ans. 5, 4.§(CEEEEY) over«segment ofthestrat tneyefromz=tory a2. Ans, in’ 5.[yede-tredytay dzoveronarcofthehelixx=acost,y=sint, zh astvaries from 0to2x. Ans. 0. 704 LineIntegratsandSurfaceIntegrals 6.§xdy—yde overanarcofthehypocyclold x=acost, ymasintt. Ans.4nat(thedoubleareaofthehypocycloid). Eai)dy—ydsovertheloopofthefoliumofDescartesx=i224,3 opeAns.3a*(thedoubleareaoftheregionboundedbytheindicated 1o0p). 8.[xdy—yds overthecurver=a(t—sint), y=a(1—cos 10t<2). ‘Ans.—6na*(thedoubleareaoftheregion-bounded byoneareofacycloid and the axis).rovethatadwhere¢ksant. 8,grad(cp)=cerad@wherecisaconstant, Tograd Gyeay)-egrad gtegrad'p wheie¢isaconstant. 1,grad (9H) =Parad D+ Eradg, 12,Findgradr,grade,grad+,gradf(q)wherer =VIFFERAns, ton -Srot.13,Provethatdiv(A+B)=divA+divB.U4Computedvr,whetereelU/-+28,Ans. 3. 15.Compute div(Aq), where Aisavector function and @isascalar function.Ans.«divA-+(gradgA).16.Computediv(r-c),where¢Isaconstantvector.Ans.(7), 12.Compute divBUrA). Ans.AB. 18,rot(A,-+6A;) —6,fotAy+c,fotAywhere ¢,andc,areconstants,Ie,fodgtadancewhere'sTeaeSnotantvector20,rotrolAmgraddivA—VA.21,Axrotp=rot(pA). Surtace Integrals 22,Provethat[{cos(n,2)da=0 IfoIsaclosedsurfaceandntsanor-mal to it. 23.Find the'moment ofinertia ofthesurface ofasegment ofasphere with equation efy'f2t— Recutolfbytheplanez=/frelativetothesaris,Ans=34@Rt—3RtH +H".24,Find the moment ofinertia ofthe surface ofthe paraboloid ofrevo- lution xt-yf=der cut olf bytheplane eo relative tothez-axis. Ans. 55+9V3 aae ie25.Computé thecoordinates ofthecentre ofgravity ofapartofthesurefaceofthecone2*-+y"=Ry2tcutoffbytheplane,z=Hl.Ans.0,0,5H. Exercises onChapter XV 708 26.Compute thecoordinates ofthecentre ofgravity ofasegment ofthesurface ty ateRE = R+H ofthesphere2°y*-+24=R*cutoffbytheplane2=H.Ans.(0,0,AH), a7,Find((xcos(nx)-ycos (ny)+2608(nz)]do,whereoiso closedsurface. Ans.3V,whereVIsthevolume ofthesolidbounded bythe surface o. 2a,Find sdcdy whereSistheexternal sideofsaphere styt4st= ’ =PAns.Sart, 29,Find|{x*dyde-tySdede-tz4dx dywhereStstheexternalsideof s thesurface ofasphere et+y'p24—=R¥ Ans, xRt, 30,Find((Vx*Fy*ds whereSisthelateral surface ofacone S HGHao,ocect,Ans,MOVEER 31,UsingtheStokesformula, transform theintegral [ydz-fzdy+xdz.i Ans.—fftosetosB+cosy)ds, Find the line integrals, applying the Stokes formula and directly: 32, Jutaatetadtatia whereListhecirclex*4y*+zt= =a,xyt2=0.Ans.0,33,jxiy'ds-+dy-+edz whereLtsthecircle pant no,Aan28 Applying the Ostrogradsky formula, transform thesurface Integrals into volume integrals: 34.[{(xcosa+ycosB-+2cosy)ds.Ans.iN)Baxdydz. 3 a[Juteremaydetdedearann AmSITwcbetsndeaydtsau au 38,[Vayaedytyedydeterdzds, Ans0.37.SfSeva dedetsau PuOu,tw+Hdxdy.Ans.S{S(Gaesp+a8) dxdydz,Using the Ostrogradsky formula compute the following integrals: 38, Jfteeosa-tycosp-+zcosypds where Sisthesurface oftheellipsoid 5 23—sass 706 Line Integrals and Surface Integrals HBrtat.Ans,dnabe,39,[fetcosaty*cosB+2%cosydswhereS3 isthesurfaceofthespheresty"b2t—=R* Ans,Baa40,(0stayde ‘s videdx+2tdxdy whereSIsthesurfaceofthecone44-20 oct). Ans.Pat. [Cedydetydrdetededy whereSisthe ‘s surface ofthe cylinder s*-+yt=at, —H<ecH. Ans. Snot, 42,Provetheidentity |((S542%) dedy— (Meds,whereCi 7 weta)dx4u=\54s, isacon- tourboundingtheregionD,and3isthedirectional derivative oftheouter normal. Solution. §S(G+g) a-[-¥aerxaynf [H¥cos(s,+Xsin(s,2]ds, 3 where(2) theangebetween thetangent lingtothecontour ©andthe axis. Ifwedenote by(n,2)the angle between the normal and thex-axis, then sin(5,x)eos(m,x),£08(,#)-—=—sin(n,4).Hence, S\E+%) dea=[xenX+Ysin(n,2))ds. ‘8 inex! yadltSettingX=Gz, YaST, weget SS(3B+5#) dxdy0[3cosos214-34sncn,aes© * a,OW ‘au Sf[Fa+3h]a{inas, Theexpression 34+5% iscalledtheLaplacian operator.48,Provetheidentity (called Green's formula) au__,2 S{Stosu—wanrandy temOf(o$5—efteo ‘where aand 9are continuous functions with continuous derivatives tothe Second order inthe region D. ‘The symbols Auand AodenoteBu,Fu,Oa do,do,atosunFetgate homme otoa. These expressions arecalled Laplacian operators inspace, Exercises onChapter XV 707 Solution. In the formula x, oY, a2)SSS +82)dryde((1Xcos(n,DEYcos(n,y)+2cos(n,2)do we putXeon,—u0', Yaou,—wi, 2mont—uor Then OXY Oat gah but) ule, hot, otBR ietyba)—tleeeOhy+)0auUbe, Xeos(a, 2)+Ycosa,y-+Zeos(n, 2)—= =0(u,08neu,cosny+1,08nz)—u(0,C08nx,£08ny+07608n2)= ao3H 28 aoHt, Hence, sfs(4u—uao) dedydem(((0aao, 44,Prove the identity ffavaranaen(00, v ° ou,4,tu wheresum$454+55(Laplacian. Solution.InGreen's formula, whichwasderived inthepreceding section, pulosl, Then So=0, and wegetthe desied Identity. 18.Iw(e.9.2) is harmonie function insome region, that is,afunce tion which at¢very point ofthls region satislies theLaplace equation atu, uu fatFatsano. then auSfdeo=0 where oisaclosed surface. Solution. This follows ditectly from the formula ofProblem 44. 4B.Letw(zy 2)beaharmonic function insome region Vandletthero be,inV,asphere owith centre atthe pointM(xj, yy2)and with radiusProve thal LLCs) 8Gwe=a ffedo. a 708 Line Integrals and Surface Integrals Solution. Consider theregion @bunded bytwospheres 0,6ofradius Rand@(@<R)withcentresatthepointM(x,,y,,2).ApplyGreen'sformula{aad Se“Prchln oo tue selon aking Wr dhe!sbowe acess lanchon, Sor the Tuneton ee TV ea Ua +E: ou, Bydiet diferentation andsubttation weareconvinced that244.2%4. +S40. consequently, 1 at Law427 IfGa)erno or 1 12) o6ffbe2) va2G va, 2¢r ; Ly 1ad2 ©thesulacesandothequantity2tsconstant(hand2)andean betakenoutside theintegral sign.By’viru,ofthefautobted inProbe Jem 45,wehave 1002 genoAfSeone 16paggESSao=o eC ca) Nees VuSedo, but~ Ly 4(2a(t) a(tG)a=)a aa ‘Therefore, 1do 1do=0affere {fare or Affaaoms, (fads.o[fuena Exercises onChapter XV 709 Apply the theorem ofthe mean tothe integral onthe right: LOCdont2D0deFaSfcossa ® winea(nDs,polatonthesuraceofasphereofradian@withWemake¢approachser!thenwm,+4(tnYnas@ appr 1 Ang?Ay((domf ae, eSfong ain Hence, as q-+0 weget 1(doubt ou2)4x. ral) Cn) Further, since the left side of(1)isindependent of@,itfollows that a q-r0 wefinally get u= EA pel)edenanu(ryYu2) or 1 Guto2gapeffs. CHAPTER XVI SERIES SEC. 1,SERIES. SUM OF ASERIES pebefinition 1,Lettherebegivenaninfinite sequence ofnum- ers*) The expression Ub Uyboybe buy beee a iscalled anumerical series. Here, thenumbers tly,tly,«++ Uqy«++ are called the terms oftheseries. Definition 2.The sum ofafinite number ofterms (the first a terms) ofaseries iscalled the nth partial sum oftheseries: Sp= UyHU ee tly Consider thepartial sums S=uy SM, tly SU, +H, +dy Spy tly Fytootye Ifthere exists afinite limit s=lims,, itiscalled thesum oftheseries (1)and wesay that theseries converges.IfTim's, does not exist (for example, s,—+00 asn—+oo), then wesay that the series (1)diverges and has nosum. Example. Consider the seriesabagbagtbag. ® This Isageomeicte progression with first term aand ratio.g (a#0) *)Asequence isconsidered specified ifweknow the law bywhich itis possible fodetermine any term upforagiven . Series, Sum ofaSeries m The sum ofthefirst nterms ofthegeometric progression is(when 9 #1) te airs or =2"== i=" 1)Itig]-<1, then g"-+0 a8+ coand, consequently, imsy=lim(220")a cis,=in,(55-1) =e Hence, inthecase of|q|<1, the series (2)converges and itssum is pty 2)If[gi>1, then|q"|+00asn+ooandthenSHA+scoasn+o, that is,lims,does notexist. Thus, when |g|>1, the series (2)diverges. 3)ItGa, then the series (2)has theform atatat... In this ease Sq=na, lim S400, and the series diverges, 4)Ifg=—L, then the series (2)has the form a—a+a—a+... In this case :-{0whenniseven, =@whenaisodd. Thus, 5has nolimit and theseries diverges. ‘Thus, ageometric progression (with first term different {rom zero) conver gesonlywientherationoftheprogression isTessthanunityinabsolute Theorem 1.Ifaseries obtained from agiven series (1)bysup- pression ofsome ofitsterms converges, then thegiven series itself converges. Conversely, ifagiven series converges, then aseries obtained [rom thegiven series bysuppressionofseveraltermsalsoconverges. Inother words, theconvergence ofaseries isnotaffected bythe suppression ofafinite number ofitsterms. Proof. Lets,bethesum ofthefirst nterms oftheseries (1), Cy,thesum of'& suppressed terms (wenote that forasufficiently large n,allsuppressed terms arecontained inthesum s,), and G,-, isthesum ofthe terms ofthe series that enter into the na Sertes sum s,but donot enter into c,.Then wehave 5p=CptOnan where cyisaconstant that isindependent ofn. From’the last relationship itfollows that iflim o,_, exists, thenlims,exists aswell;iflims,exists, thenlim0,_,also exists; which proves thetheorem. _ We’conclude this section with two simple properties ofseries. Theorem 2./faseries a+ a,+... 3) converges and itssum iss,then theseries 6a,+00+... “ where ¢issome fixed number, also converges, and its sum iscs. Proof. Denote the nth partial sum ofthe series (3)bys,, and that oftheseries (4), byo,.Then 0,=C0,++...+00,=C(0,+...+4,)=C5,. Whence itisclear that thelimit ofthe nthpartial sum ofthe series (4)exists, since limo,= lim(¢s,)=¢ lims,=cs. Thus, the series (4)converges and itssum isequal tocs. Theorem 3./ftheseries G,+G,+.-+ 6) and +b te. © converge andtheir sums, respectively, are$and§,then theseries @,+6) +, +6) +... (a) and (2,—6,) +(a,—b,) +0 ®) alsoconverge andtheirsumsareS45and$—S,respectively. Proof. Weprove the convergence ofthe series (7). Denoting itsnthpartial sum byo,and thethpartial sums oftheseries (5) and(6)by3,andSq,respectively, weget 6,=(0,+6.) +... +(d,+b,)= yt eeaH tee$FOQ)=SnFSue Necessary Condition forConvergence ofaSeries 13 Passing tothelimitinthisequation asn—oo,weget limo,=lim&,+5,)= limS,+ lim5,=3+s. Thus, theseries (7)converges anditssumis5+5. Itisanalogously proved that theseries (8)also. converges and itssum isequal tos—s. Ofthe series (7)and (8)itissaid that they were obtained by means oftermwise addition or,respectively, termwise subtraction ofthe series (5) and (6). SEC. 2,NECESSARY CONDITION FOR CONVERGENCE OF ASERIES One ofthe basic questions, when investigating series, isthat ofwhether the given series converges ordiverges. We shall establish sufficient conditions forone todecide this question. We shall also examine the necessary condition for convergence ofa series; inother words, we shall establish acondition forwhich theseries will diverge ifitisnot fulfilled. Theorem. /faseries converges, itsnth term approaches zero asn becomes infinite. Proof. Let the series UyFU, FU, +... tug tone converge; that is,letushave the equality lims,=s, where sisthe sum ofthe series (afinite fixed number). But then wealso have the equation lims,.=5, since (n—1) also tends toinfinity asn—+oo. Subtracting the second equation from thefirst termwise, weobtain lims,— lims,.,=0 or lim(s,—5,-,)=0. But Sp—Sqos =Uae m4 Series Hence, limu,=0, which iswhat was tobeproved. Corollary. Ifthenth term ofaseries does not tend tozero as n—= 00,then theseries diverges. Example. The series 12,3 2 gtetete tte. diverges, since =tim(2_)1 peekHamin(stu)—y#0: Westress thefact that this condition isonly anecessary con- dition, but not asufficient condition; inother words, from the fact that thenth term approaches zero, itdoes not follow that the series converges, forthe series may diverge. For example, the so-called harmonic series legtgttte tte... diverges, although limu,=lim1=0. Toprove this, write the harmonic series inmore detail: Vytadyayr ty ltgtgtatgtetrtet I LE tetera iyiyr yi iyttototntatetatetietit a Wealso write theauxiliary series lag¢gtptetetatet 16 terme TT 1,1,1,1,1,1, 1,7 7 tutetetetetetetetat: tat. @ The series (2)isconstructed asfollows: itsfirst term isequal tounity, itssecond is*/,,itsthird and fourth are'/,,thefifth totheeighth terms areequal to'/,, the terms 9to16areequal to"hy theterms 17to32areequal to"/,,, etc. Necessary Condition for Convergence ofaSeries 15 Denote bys?thesum ofthefirst mterms oftheharmonicseries(1)andbysithesumofthefirstmtermsoftheseries(2).Sinceeachtermoftheseries(1)isgreater thanthecorrespond-ing term ofthe series (2)orequal toit,then forn>2 30>. oy Wecompute thepartial sums oftheseries (2)forvalues ofequal to2,24,2%,24,2: gal¢tad, salt ge(Zea) altptgale2-g, Leda d)e(tatatad 1 ale ge(Z4q)t(Gtatate)al+og, Lada dya(a wwe 1sucltgt(ata)t(gt- ta)tette)= —e ee i al444, Liga 1a 1sults t(Ftgt(gtotat(ptetie)+ee 1 1 1+(gt---tg) <145-g1Tetame inthesamewaywefindthatse=1+6-4, se=1+7-4 and, generally, 5¢=1+2-y. Thus, forsufficiently large &,thepartial sums oftheseries (2) can bemade greater than any positive number; that is, lim3=00, but then from the relation (3)italso follows that ims(?)=00 which means that the harmonic series (1)diverges. 76 Series SEC. 8.COMPARING SERIES WITH POSITIVE TERMS ‘Suppose wehave two series with positive terms: Cteeee ee? (a) eS ee ees (2) For them the following assertions hold true, Theorem 1.Iftheterms oftheseries (1)are not greater than thecorresponding terms oftheseries (2); that is, u,<0, (N=1, 2...) @) and theseries (2)converges, then theseries (1)also converges. Proof. Denote bys,and’o,, respectively, the partial sums of the first and second series: 3-3uy,o=3oo From the condition (3)itfollows that S,<o,. “ Since theseries (2)converges, itspartial sum has alimit o: lim6,=0. From thefact that theterms oftheseries (1)and (2)areposi- tive, itfollows that 6,<o, and then byvirtue of(4) 54<8. We have thus proved that the partial sums s,are bounded. We note that asnincreases, the partial sum s,increases, and from the fact that the sequence ofpartial sums "isbounded and increases, itfollows that ithas alimit *) lim s,=5, and itisobvious that s<o. Using Theorem 1,wecanjudge oftheconvergence ofcertain series. *)Toconvince ourselves that thevariable sqhas alimit, Jetusrecall a condition forthe existence oflimitafseaience (eeCh.1;"it'avar able isbounded and increases, ithas alimit. Here, the sequence ofsums 5, isbounded and increases. Hence ithas alimit, I.eythe series converges. Comparing Series with Positive Terms 17 Example 1.The series legtpt at thtBtyte Mtoe converges because itsterms aresmaller than the corresponding terms ofthe ligt tet tpt. But the last series converges because itsterms, beginning with the second, fom2gometrte proven wthcommon railot=Thesumofthosete Santa uaatoy eames varere|ens patos a oimcneae ean Theorem 2.Iftheterms oftheseries (1) are not smaller than thecorresponding terms oftheseries (2); that is, PED, © land theseries (2)diverges, then theseries (1)also diverges. Proof. From condition (5)itfollows that 5,505. 6 Since the terms ofthe series (2)are positive, itspartial sum o, increases with increasing m,and since itdiverges, itfollows that iimo,=00. But then, byvirtue of(6), lim s,=00, the series (1)diverges. Example 2,The series rd 1lepgtpatetpgte diverges because its terms (from thesecond on) are greater than the corre: sponding terms ofthe harmonic series 11 1 lege ptetttes which, asweknow, diverges Note. Both theconditions that we have proved (Theorems 1 and 2)hold only forseries with positive terms. They also hold 78 Series true when some ofthe terms ofthe first orsecond series are zero. But these conditions do not hold ifsome of the terms of the series are negative numbers. SEC. 4,D'ALEMBERT’S TEST Theorem (d’Alembert’s Test). /finaseries with positive terms MyAug+Uyt+oetgtees a) theratioofthe(n+1)st termtothenthterm,asn—co,hasa(finite) limit 1,that is, lim#241=1, (2) then: 1)theseries converges for1<1, 2)theseries diverges for1>1. (For 1=1, the theorem cannot determine the convergence or divergence ofthe series.) Proof. 1)Let /<1. Consider anumber qthat: satisfies the relationship 1<q<1 (Fig. 342). From the definition ofalimit and relation (2)itfollows that forallvalues ofnafter acertain integer N, that is,forn>WN, wewill have the inequality eat<q, e) Indeed, sincethequantity “«*tends tothelimit /,thedif- ference between thequantity “2andthenumber /may(aftera certain N) bemade less (in “absolute value) than any positive number, inparticular less than g—J; that is, |s2-<a—t. Inequality (2)follows from,thislastinequality. Writing this inequality forvarious values ofn,from Nonwards, weget Uns, SI» Hyg,SMe Sea @Uys <MUves<Ttiyw D'Alembert's Test 79 Now consider the two series WyPU by hoe bly yyy blige bey qa) Uyt qty gut os ay The series (1')isageometric progression with positive common ratio q<1. Hence, this series converges. The terms ofthe gl at tta ———— of 7Setg 01ae Fig. 342. Fig. 348. series (1), after uy,,, areless than the terms ofthe series (1"). ByTheorem 1,See.’ 3,and Theorem 1,Sec. 1,itfollows that the series (1)converges. 2)Let />1. Then fromtheequation lim“s*t=1 (where >1)itfollows that, after acertain N, that isfor nN, we will have the inequality Sati1 (Fig. 343), oru,4,>-u, foralln>N. But this means that the terms ofthe series increase after the term N-+1, and forthis reason thegeneral term oftheseries does nottend tozero. Hence, the series diverges. Note 1.Theseries willalsodiverge when lim“#4100. This follows fromthefactthatiflim“+100, thenafferacertain n=N wewillhavetheinequality “11, oruy,>Uy Example 1.Test the following series forconvergence: ri 1 Solutton. Here, epee, tg e “Taal! TR) wr Moy ont dyWET 720 Series Hence, imett imCe Td ‘The series converges Example 2.Test forconvergence the series 2.0 2 Teepe tte. Solution. Here, 2 2 ayy tg Matt im 2Ameng ey! Bet limpetejim2epae> h ‘The series diverges and Itsgeneral term ugapproaches infinity. Note 2.DAlembert’s test tells uswhether agiven positive seriesconverges; butitdoessoonlywhenlimSateexistsandis different from 1.But ifthis limit does not “exist orifitdoes existandlim4s#=1, thend’Alembert’s testdoesnotenable us totell whether theseries converges ordiverges, because inthis case theseries may prove tobeboth convergent and divergent. Some other test isneeded todetermine the convergence ofsuch series. Itwillbenoted, however, thatiflim“%#=1, buttheratio 281foralln(after acertain one)isgreater thanunity, these riesdiverges. Thisfollows fromthefactthatif“4>1, thenUns>tqandthegeneraltermdoesnotapproach zeroasn—+o0,‘Toillustrate, letusexamine some examples. Example 8.Test forconvergence the series 1,2,3 A dtgsde ttt Solution. Here, aH nts—timB42LtimBHMt1 a ieae abanohai Inthiscasetheseriesdiverges because 2222>1forallmi fins mt+204“iewean Cauchy's Test m2 Example 4.Using thed'Alembert test, examine the harmonic series tit 1 lege gett 1 1 Wenotethatp=, tga=tyandyconsequently, Jim“ett jim 2meetgncaapT Thus, d’Alembert’s test does not allow ustodetermine the convergence or aivergence ‘of‘thegiven‘series. Butweeatier found outby’different expedient that ahatmonic series diverges. Example 5.Test forconvergence theseries roid 1 Tatagtgat--taeent Solution. Here, ee eea “aah GEN @r" Himfatto tim200) tim 8ntoUyTaeFIL aeWT D'Alembert’s test does not permitustoinferthattheseriesconverges; butbyother reasoning wecan’ establish thefact that this series converges, Notingthat Ee autD ne a+T" wwecan write the given series inthe form Lay t_ ayaa 1(4-4)4(4-4)4(S-H 4t(degh+ sngIte atial sumofthefirstmterms,aterremoving brackets andcancel ng, is. aet-ay IT Hence, 1 limsomim(1)=. That fs,the series converges and itssum is1. SEC. 5. CAUCHY'S TEST Theorem (Cauchy's Test). Jfforaseries with positive terms heeeee a thequantity {/u, hasafinite limit |asn—+0o, thatts, lim/u,=1, m2 Series then: 1)for1<1, theseries converges; 2)for1>1, theseries diverges. Proof. 1)Let1<1. Consider thenumber qthatsatisfies therelation [<q<1. Aiter some n=N we will have the relation \Wa—<g— whence itfollows that Vun<4 or a,<4" for all n>N. Now consider two series: WyHl yt eethy tne tye toes a) QN+Qh+ght... a) The series (1') converges since itsterms form adecreasing geometric progression. The terms oftheseries (1), after uy, are less than the terms ofthe series (1'). Consequently, theseries (1) converges. 2)Let1>1. Then, after some n=N, wewill have Vus>1 or u,>i1. But ifallthe terms ofthis series, after uy, exceed 1,then the series diverges, since itsgeneral term does nottend tozero. Example. Test forconvergence the series 11/2), (8)" ayxt(z)+(4) +--+(aa) + Solution. Apply theCauchy test: y— 7) jim82 lin,WantimV(sez)limargee<! ‘The series converges. Note, Asinthe d'Alembert test, the case lim/u,=1=1 requires further investigation. Among theseries that satisfy this condition areconvergent and divergent series. Thus, forthehar- The Integral Test jorConvergence ofa Series m3 monic series (which isknown tobedivergent) imYaretimYLstimiz,imVie. min /E Tobesure,weshallprovethatliminV/L=0. Indeed, Jimin/L=lim=", Here, the numerator and denominator ofthe fraction approach infinity. Applying I"Hospital's rule, wefind 14/Tetim=!tim2 timinF=tim=24=tim=P=0. Thus,InJ/£—0, butthen//2—1, i.e., limVii. For the series tit 1. peptgt thts wealso have theequality ae VT=timVEYTEsinFimtimVmtinVVGah but this series converges, since ifwesuppress the first term, the terms oftheremaining series will beless than the corresponding terms ofthe converging series on 1 Tatesttagep te (see Example 5,Sec. 4). SEC. 6,THE INTEGRAL TEST FOR CONVERGENCE OF ASERIES Theorem. Let theterms oftheseries Wybey bey bene bly bone Oy bepositive and not increasing, that is, 4,2, 2D, mu Series and letf(x) beacontinuous nonincreasing function such that HO)=u5 FQ)=us tf) =uy @ Then thefollowing assertions hold true. 1)iftheimproper integral Siade converges (see Sec. 7,Ch. XI), then the series (1)converges 100; 2)ifthegiven integral diverges, then the series (I)diverges as well. Proof. Depict theterms oftheseries geometrically byplottingonthex-axis thenumbers 1,2,3,--.1,nl, ...oftheterms oftheseries, andonthey-axis, thecorresponding Values oftheterms Oftheseries ty,yysay tgs =»(Fig. 344). Inthesame ‘coordinate system plot thegraph ofthecontinuous noninereasing function y=1%) which satisfies condition (2). Anexamination ofFig.344showsthatthefirstoftheconstruct- edrectangles hasbase equal to1and altitude f(J)=u,. The area ofthis rectangle isthus u,.The area ofthesecond oneisu,, and soon; finally, the area of"the last (nth) ofthe constructedrectangles isw,,Thesumoftheareasoftheconstructed rectanglesisequal tothe sum s,ofthe first mterms ofthe series. On theother hand, the step-like figure formed bythese rectangles embraces aregion bounded. bythecurve y=f(x) and thestraight lines x=1, x=n-+1, y=0; the area ofthis region isequal to JI(x)dx, Hence, >Jfx)de. @) Let usnow consider Fig. 345. Here thefirst oftheconstructed rectangles onthe left has altitude u,; and soilsarea isu,. The area ofthesecond rectangle isu,, and soforth. The area ofthelast oftheconstructed rectangles isw,,,. Hence, thesum oftheareas ofallconstructed rectangles iséqual tothe sum of allterms oftheseries beginning from thesecond tothe(n+1)st The Integral Test forConvergence ofaSeries 725 ors,4,—t,. Ontheother hand, itisreadily seen that thestep- like figure formed bythese rectangles iscontained within the iy NI ;4= NXte,afox) lyetoi Ek GTR) Tas at Fig. 344. Fig. 345, curvilinear figure bounded bythecurve y=/(x) and the straight lines x=1, x=n-+1, y=0. The area ofthis curvilinear figure isequal to|f(x)dx. Hence, 1 Sent §Fade, whence San< JFa)detuy. C) Let usnow consider both cases. 1,We assume that the integral f(x)dxconverges, thatis, hasafinitevalue. ‘Since Spade <Stwae, itfollows, byvirtue ofinequality (4), that Sn<Siar<IPQ)deuy. Thus, the partial sum s,remains bounded forallvalues ofa. But itincreases with insreasing n,since alltheterms u,are 76 Series positive. Consequently, s,(asn—+eo) hasthefinite limit lims,s and the series converges. 2.Assume, further, that|f(x)dr—oo. Thismeansthat §F(x)dxincreases without boundasnincreases. Butthen,by virtue ofinequality (3), s,likewise increases indefinitely with n; the series diverges. The theorem isthus proved completely. Example. Test for convergence the series tot 1 btptpte tote Solution. Apply the integral test, putting to=3. innTyfelonsaisesslltheconditionsofthetheorem.Considerthe integral Lapfg|eaeinrspurr—ameva <.?Ingffinv whenp=1. Allow Ntoapproach infinity and determine whether the improper Integral converges invarious cases iff Thembeposable fojudgeabouttheconvergence ordivergence of theseries forvarious values ofp. or>t,F825, theino isfileand,bene,theseries converges; Cae ; tforp<l.SGq«. theintegral isinfinite, andtheseriesdiverges; Cae in i; forpat, (mes, theintegral isinfinite; andtheseries diverges. Alternating Series, Leibnis? Theorent m1 We note that neither the d'Alembert test nor the Cauchy test, which were considered -earier, decide whether theseries isconvergent oFna, since on =e (ghy'-» _ 7 aadtinVig=inGa im(VT) ava. SEC. 7.ALTERNATING SERIES. LEIBNIZ’ THEOREM Sofar we have been considering series whose terms are all positive. Inthis section weconsider series whose terms have alternating signs, that is,series ofthe form Cetet ee O) where UW,Uy,..., Ug+++ are positive. Leibniz’ Theorem. /finthe alternating series CeeeeC) 0) the terms are such that 4>u >Use ®) and limu,=0, @) then the series (1) converges, itssum ispositive and does not exceed the first term. Proof. Consider the sum ofthe first n=2m terms of the series (1): Sym (=U) +(y=) vos ym —Ham From condition (2)itfollows that theexpression ineach of thebrackets ispositive. Hence, thesum s,,ispositive, Sim>0, and increases with increasing m. Now write this sum asfollows: Sm=1(y=) —(4) =lanes)Haae Byvirtue ofcondition (2), each oftheparentheses ispositive. Therefore, subtracting these parentheses from u,wegetanumber less than u,,or Sim<tye 7 Series We have thus established that s,» increases with increasing mandisbounded above. Whence itfollows thats,,hasthelimit s: lim54>, and 0<s<u,. However, wehave not yet proved the convergence ofthe series; wehave only proved that asequence of“even” partial sums has asitslimit thenumber s.Wenow prove that “odd” partial sums also approach the limits.Consider thesumofthefirstm=2m-41termsoftheseries(1): Semi =Sint Mamas Since, bycondition (3), lim u,_4,=0, itfollows that Fim Syuey tim Syqt limyng, lim 5,_=8. We have thus proved that lim s,=s both foreven nand foroddn.Hence, theseries(1)converges. ; Note 1.The Leibniz theorem may beillustrated geometrically asfollows. Plot the following partial sums onanumber line (Fig. 346): 5,=Uy, $,=U, —U=5,—ly,5,=SpAV=S—y S=HEM, ete, Thepoints corresponding topartial sumswillapproach, acertain point s,which depicts thesum oftheseries. Here, the points correspondingtothe 4)evenpartialsumslieon % the left ofs,and those yy corresponding ‘tooddsums, usonthe right ofs. Us Note 2.Ifanalternating . =seriessatisfies thestatement Oe“4 ofthe Leibniz theorem, Fig. 346. then itiseasy toevaluate the error that results if we replace itssum, s,bythe partial sum s,.Inthis substi- tution wesuppress ailterms after u,,,. But these numbers form bythemselves analternating series, Whose sum (inabsolute value) islessthan thefirst term ofthis series (that is,less than u,,,). Thus, theerror obtained when replacing sbys,does notexceed (inabsolute value) the first ofthesuppressed terms. Plus-and-Minus Series. Absolute and Conditional Convergence 729 Example 1,Theseries baa lagtg—gte converges, since roi,Di>gogoes 1 ‘The sum ofthe first mterms ofthis series e1-tyt_t naeQalogt gogt Hey , 1 differs fromthesums oftheseries byaquantity lessthan=. Example2.Theseriesodytidyare ae converges byvirtue ofthe Leibniz theorem. SEC. 8,PLUS-AND-MINUS SERIES. ABSOLUTE AND CONDITIONAL CONVERGENCE Wegive the name plus-and-minus series toaseries that has both positive and negative terms. ‘Obviously, the alternating series considered inSec. 7isaspecialcase‘ofplus-and-minus series*).Weshall consider some properties ofalternating series. Incontrast totheagreement made inthepreceding section we will now assume that the numbers uj, uy..., U,..-can be both positive and negative. First, let usgive an important sufficient condition for the convergence ofanalternating series. Theorem 1.Ifthealternating series UyUy bee Py bee ay issuch that aseries made upofthe absolute values ofitsterms, CARACAReeeaCeee) @) converges, thenthefemalternating seriesalsoconverges. Proof. ‘Let s,and o,bethe sums ofthe first nterms ofthe series (1)and (2). *)Inthis English edition we shall use the term alternating series for both types.— Te. 70 Series Also, letsi,bethesum ofallthepositive terms, ands", the sum ofthe absolute values ofallthe negative terms ofthe first nnterms ofthegiven series; then Sp=5,—S5 =5,+5, Byhypothesis, ©,hasthelimit o;sandsare positive in- creasing quantities “less than a,Consequently, they have the limits s’and s’.From therelationship 's,=s,—s, itfollows that s,also hasalimit and that this limit isequal fos’—s', which means that the alternating series (1)converges. Theaboveproved theorem enables onetoJudgeaboutthe convergence ofsome alternating series. Inthis case, the test for convergence ofthe alternating series reduces toinvestigating a series with positive terms. Consider two examples. Example 1.Test forconvergence theseriessing|sin2asin3a, sinaSeee Cc) where aisany number. Solution. Also consider the series ina|sin2a)|sina sina[ene+[252|+[ae]+++. ” and tii 1 Pty tte tet. ) “The series (8) converges (see Sec. 6), The terms ofthe series (4)arenot greater than thecorresponding. terms ofthe series (8); hence, the series i) fiso ‘converges, But then, invirtue ofthe theorem just proved, thegiven series ()likewise converges. Example 2.Test forconvergence the series we cas® cas% coteSeaa nha nae © Solution. Inaddition tothis serles, consider the series Ltt 1 : dade e. tht. o Thisseriesconvergesbecauseitisadecreasing geometricprogression with Pera yg nepray yaes Cara RO ofitsterms are less than those ofthe corresponding terms oftheseries (7). Plus-and-Minus Series, Absolute and Conditional Convergence 731 We note that the convergence condition that was proved earlier isonly asufficient condition forconvergence ofanalternating series, but not anecessary condition: there arealternating series which converge, but series formed from the absolute values of their terms diverge. Inthis connection, itisuseful tointroduce the concepts ofabsolute and conditional convergence ofan alternating series and, onthe basis ofthese concepts, toclassify alternating series. Definition. The alternating series Ut Up Ut eeebute. qa) iscalled absolutely convergent ifaseries made upoftheabsolute values ofitsterms converges: CAPSCA Raa aee ) Ifthe alternating series (1) converges, while the series (2) composed ofthe absolute values ofitsterms diverges, then the given alternating series (1) iscalled aconditionally convergent series. Example 8.The alternating series aiid lagtgogte: 1sconditionally convergent, since aseries composed ofthe absolute values ofitsterms isaharmonic’ series, tit legtgtgtens whichdiverges. Theseries itseltconverges (thiseanbereadily veried byExample 4,Thealternating series ria Iy+g-qte Jsabsolutely convergent, since »series made upofthe absolute values of its terms, tid legty tate converges, asestablished inSec. 4. Theorem 1isfrequently stated (withthehelpoftheconcept ofabsolute convergence) asfollows: every absolutely convergent series isaconvergent series. Inconclusion, wenote(without proof)thefollowing properties ofabsolutely convergent andconditionally convergent series. 732 Series Theorem 2.Ifaseries converges absolutely, itremains abso- lutely convergent forany rearrangement ofitsterms. The sum of theseries isindependent oftheorder ofitsterms. This property does nothold forconditionally convergent series. Theorem 3./faseries converges conditionally, then nomatter what number Aisgiven, theterms ofthisseries canberearranged insuch manner that itssum isexactly equal toA.What ismore, itispossible sotorearrange the terms ofaconditionally conver- gent series that theseries resulting after therearrangement is divergent. The proofs ofthese theorems arebeyond thescope ofthis course. Toillustrate thefactthatthesumofaconditionally convergentseries can change upon rearrangement ofitsterms, consider the following example. Example 5.The alternating series 114 l-gtg-qte ® converges conditionally. Denote its sum by s.Itisobvious that s>0. Rearrange ‘thetermsoftheseries()$0°that twonegative termsfollowone positive term: 1ouyi1a 4 14 \-g-atg-o ratteaaat o enSiI, ee We shall prove that the resultant series converges, but that itssum sis halfthesumoftheseries(8):anDenotebys,andsi,thepartialsumsof ‘the series (8)and (9). Consider the sum of3kterms ofthe series (9): 1_tyy(t_ ad roa 4 sua(19-4) +(g-3-8) ++(gira)riya 14=(3-4)+(e-3)+--+(ia-a)=t ty fat ae=F[(\-4)+($-4)+--+(eh-a)] >i/; ttt 1 ltyt ericgtycate tein) a Consequently,11 Alyta=UEsee Further, imsings=lim(Sytap=p tmSines=lim(Setog) aes ; eer eeea Bmsnes (ato aees)-E* Functional Series 733 ‘And we obtain liig=y5 Thus, inthis case the sum ofthe series changed after itsterms were rearranged (itdiminished byafactor of2). SEC. 8 FUNCTIONAL SERIES Theseriesu,+u,+:...+u,+... iscalledafunctional series ifits terms are functions ofx. Consider the functional series 4,(44,(FayOe bag(Eo Oy Assigning toxdefinite numerical values, weget different numerical series, which may prove tobeconvergent ordivergent. The set ofall’ those values ofxfor which the functional series converges iscalled thedomain ofconvergence ofthe series. Obviously, inthedomain ofconvergence ofaseries itssum is some function ofx.Therefore, the sum ofafunctional series is denoted bys(x). Example. Consider the functional series Teta tate. Thisseriesconverges forall valuesof#intheinterval (1, thatIs for allxthat satisfy the condition [xj-<i. For each value’ of’ inthe interval (—1,1),thesumoftheseriesisequalto7-4(thesumofa decreasing geometric progression with ratio x).Thus, inthe interval (—1, 1}thegivenseriesdefinesthefunction GADU) sma which isthe sum ofthe series; that is, 1 Aptpops tetette ti Denote bys,(x)thesumofthefirstmtermsoftheseries(1), Ifthis series converges and itssum isequal tos(x), then 5) =5, (0+1 (2), where r,(x) isthesum oftheseries u,4, (x)+Uays(X)-+ --4) bee, Fy2)=teas (8)4Ugg DE oo 14 Series Here, thequantity r,(x) iscalled the remainder oftheseries (1). For allvalues ofxinthe domain ofconvergence oftheseries we have therelation lims,(x)=s(x); therefore, limr,(x)= lim[s(x)—s, (2)]=0, which means that theremainder rq(x)ofaconvergent series ap- proaches zero asn—+oo. SEC. 10, DOMINATED SERIES Definition. The functional series My(1)+Hy(2)ty(0) oehla(4)+ 0) iscalled dominated insome range ofxifthere exists aconver- gent numerical series GFatapebat... @) with positive terms such that forallvalues ofxfrom this range thefollowing relations are fulfilled: [Se 1)Seyvoy[eeO|<Styo-—B) Inother words, aseries iscalled dominated ifeach ofits terms does notexceed, inabsolute value, thecorresponding term ofsome convergent numerical series with positive terms. For example, theseries SOE4SOESE pO, isaseries majorised ontheentire x-axis. Indeed, for all values ofx,therelation|S2|<% (n=1,2.) isfulfilled and the series tata Ttatgyte as_we know, converges. From thedefinition itfollows straightway that aseries domina- tedinsomerangeconverges absolutely atallpoints ofthisrange (see Sec. 8).Also, adominated series hasthe following important property. Theorem. Let thefunctional series 4,(2)FigAoebelaFoo Dominated Series 735 bedominated ontheinterval (a,b].Lets(x) bethesum ofthis series and s,(x) thesum ofthefirst nterms ofthis series. Then foreach arbitrarily small number e>0 there will beapositive integer Nsuch that foralln>N thefollowing inequality will be fulfitted, Is@)—s.(x)<e, nomatter what thexoftheinterval {a,6}. Proof. Denote byothe sum ofthe series (2): Ceee then o=0, +e, where ,isthesum ofthefirst nterms oftheseries (2),and e, isthesim ofthe remaining terms ofthis series; that is, Fn Oner tongsboos Since this series converges, itfollows that lim o,=0 and, consequently, lime,=0. Letusnow represent thesum ofthefunctional series (1)inthe form 8(X)= 5,(1)+0 (2), where Sy(4)=4,(X) t+ta (*)s Ta(2)=Une (X)tenes (X)tage (2)+ From condition (3)itfollows that Ines @)1Stneae [Maes|Sasa oor and therefore InGlen forallxofthe range under consideration. Thus, Is)—s, (4)|<n forallxoftheinterval (a,6],and e,—+0 asn—+oo. NoteI.Thisresultmay‘berepresented geometrically asfollows. Consider thegraph ofthe function y=s(x). About this curve construct aband ofwidth 2e,; inother words, construct the 736 Series curves y=s(x)-+e, andy=s(x)—e, (Fig. 347). Then forany e,the graph ofthefunction s,(x)will liecompletely intheband under consideration. The graphs ofallsuccessive partial sums will like- wise lie within this band. . Note 2.Not every function- aaa alseriesconvergent onthe vA / interval [a,6]hasthepro- ye Bypertyindleatedintheforego-ing theorem. However, there are nondominated series such that possess this property. Aseries that possesses this xproperty iscalledauniform- oe o ly convergent series onthe 7 interval [a,6). Fig.347. Thus, thefunctional series; ty(2)-H'u,(2)4.Etty(2) +... iscalled auniformly convergent seriesontheinterval(a,b] ifforanyarbitrarily smalle>OthereisanintegerNsuchthat foralln>N theinequality Is—s,@)|<e will befulfilled forany xoftheinterval (a,6}. From thetheorem that has been proved itfollows that adomi- nated series isaseries that uniformly converges. SEC, 11, THE CONTINUITY OF THE SUM OF ASERIES Let there beaseries made upofcontinuous functions Hy)FeyVFveebtlg(2)Eves convergent onsome interval (a,6]. InChapter Ilweproved atheorem which stated that the sum ofafinite number ofcontinuous functions isacontinuous func- tion. This property does nothold forthesum ofaseries (consist- ingofaninfinite number ofterms). Some functional series with continuous terms have for the sum acontinuous function, while inthe case ofother functional series with continuous terms, the sum isadiscontinuous function. Example. Consider the series gfe ete? 8). HEL, The Continuity oftheSum ofaSeries 131 The terms ofthis series (each term isbracketed) are continuous functions for allvalues ofx.We shall prove that this series converges and that its-sum isadiscontinuous function. ‘We find the sum ofthe first terms of the series: Find the sum of the series:* it'z>0, then =a=,5(=as,GPF eta, ifx<0, then $l ()—tima ix=0, then 5,0, and sos= lim s,=0. Thus, wehave s@)=l—x for 2>0, s@)=—1—x for2<0, s@)=0 for#=0. And sothesum ofthegiven series isadiscontinuous function. Itsgraph is shown inFig. 348 slong with the graphs ofthe partial sums (2h 8(3), and 54(8). The following theorem holds true fordominated series. Theorem. The sum ofaseries ofcontinuous functions dominated ‘onsome interval [a,6)isafunction continuous onthis interval. WcS \ S NSS 4 ‘ DNpS 7H ¥ S ybeoreo 5) WN0forx=0 SSteforx<0 NS ONG Fig. 348. 24—s308 738 Series Proof. Let there be aseries of continuous functions dominated ontheinterval {a,6): 4,(4)+4, (4)+4, (2)+--+ Q Let usrepresent itssum inthe form SQ) =5,(*)+70(2), where 8,(2)=u,(4)+bug(2) and Tal)=Hanes(X)FUnes(8)Foe Ontheinterval [a,6]take anarbitrary value oftheargument xand give itanincrease Axsuch that the point x-+Ax should algolieontheinterval [a,6]. We introduce the notations As=s(x+Ax)—s (x);As,=5,(4-+4x)—S,(x); then As=As,+r,(e-+Ax)—r,(2), from which we have [As|<<]As,|-+]rq4(¢ +42)|+174(2). @) This inequality istrue forany integer n. Toprove thecontinuity ofs(x), wehave toshow that forany reassigned and arbitrarily small'e>O there will beanumberBSo°such thatforall|Ax|<5 wewillhave[As|<e. Since thegiven series (1)isdominated, itfollows that forany preassigned e>0 there will befound an‘integer Nsuch that for allnN (and asaparticular case, n=N) the inequality Iw@)<$ @) willbefulfilled foranyxoftheinterval [a,).ThevaluexAx liesontheinterval (a,6]and therefore the following inequality isfulfilled: Irn(e+Ax)|<3. @) Further, forthe chosen Wthe partial sum sy(x)isacontinuous function (the sum ofafinite number ofcontinuous functions) and, consequently, apositive number 6may bechosen such that for every Ax that satisfies the condition |Ax|<6 the following inequalityisfulfilled: ldswi<$. (a) Integration and Diferentiation ofSeries 739 Byinequalities (2), (3),(3'), and (4),wehave ldsl<Zt+5tgee that is, JAs|<e for |Ax|<6 which means that s(x) isacontinuous function atthe point x (and, consequently, atany point oftheinterval (a,b}). Note. From this’ theorem itfollows that ifthe sum of aseries isdiscontinuous onsome interval (a,6],then the series isnot dominated onthis interval. Inparticular, the series given inthe example isnotdominated (onany interval containing the point x=0, that istosay, apoint ofdiscontinuity ofthe sum oftheseries). :Wenote, finally, that theconverse statement isnottrue: there areseries, not dominated onaninterval, which, however, converge on this interval toacontinuous function. Forinstance,every series uniformly convergent ontheinterval {a,6](even ifitisnot dominated) has acontinuous function forits’sum (if,ofcourse, allterms ofthe series are continuous). SEC. 12 INTEGRATION AND DIFFERENTIATION OF SERIES Theorem 1.Let there beaseries ofcontinuous functions Hy(A)AM(A)A ooHUA) $eee qa dominated ontheinterval [a,4andlets(x)bethesumofthis series. Then theintegral of$(x) between thelimits from atox, which limits belong fotheinterval [a,6),isequal tothesum of such integrals oftheterms ofthegiven series; that is, Ss@yde=Su,(ddxt fu,Q)det...t Ja,Q)det... Proof. The function s(x) may berepresented inthe form 8)=5,(2)+05(*) or 8(x)=4,(X)+H,(8)+eee+g()+n(XD Then : 5foedemfuycpdet fuerte t. : 5 +Sue)detVrQ(a)de e) ue 740 Series (the integral ofthe sum ofafinite number ofterms isequal to thesum ofthe integrals ofthese terms). Since the original series (1) isdominated, itfollows that. foreveryxwehave|r,(x)|<e,, where¢,—+0asn—-oo.Therefore, Sra)de|<<|r,(2)de<fe,dee,(xa)<e,(b—a). Since e,—+0, itfollows that lim[r,(x)de=0. But from equation (2)wehave Srate)de=[s(pdx—[Ju,@odet...+ u(r]. Hence tim{fs(x)de—[Ju, det... +a,(2)dx]}=0, or lim[a@)det... +Ju,(0)dx]=Js(n)de. @) The sum inthe brackets isapartial sum oftheseries Ja@ddet +a,(det... @) Since thepartial sums ofthis series have alimit, this series converges and itssum, byvirtue ofequation (3), isequal to Ssixydz, ie, Sseeydr=Ju,(xdrtfu,(x)det...4JuQde+..., this isthe equation that had tobeproved. Note 1.Ifaseries isnot dominated, term-by-term integration ofitisnotalways possible. This isto’beunderstood inthesense thattheintegral |s(x)dxofthesumoftheseries(1)isnotalways Integration and Diferentiation ofSeries ™m equal tothesum oftheintegrals ofitsterms [that is,tothesum oftheseries(4). Theorem 2./faseries, 4,(2)+H, Fo bug) oo 6) made upoffunctions having continuous derivatives ontheinterval [a,6]converges (on this interval) tothesum s(x) and the series Wy(2)Us(2)Fe tan(X) Eve © made upofthederivatives ofitsterms isdominated onthesame interval, then thesum ofthe series ofderivatives isequal tothe derivative ofthesum oftheoriginal series; that. is, SMD EU)HU)Htut Proof. Denote byF(x) the sum ofthe series (6): F(x)aa (tux) +...tu)H..., andprovethat F(x)=s'(x). Since theseries (6)isdominated, itfollows, bythepreceding theorem, that GFadem Sui(det Jui(pdt... +Jun(art... Performing the integration, weget [Pip de=lu,(a, (a+ +[4,0 —4,(@)] +...+[in(4)—4, (@)]+. But, byhypothesis, S(x)=a,(x)+4,(2)+...+u,(t)+..-,s(a)—=4,(a)+4, (a)+...+4,(a)+..., nomatter what the numbers xand aontheinterval {a,6]. Therefore, SF(yar=s(y—s(a). Differentiating both sides ofthis equation with respect tox, we obtain F(=s'(x). m Series ‘We have thus proved that when the conditions ofthe theorem are fulfilled, the derivative ofthe sum ofthe series isequal to the sum ofthe derivatives ofthe terms ofthe series. Note 2.The requirementofdominance (majorisation) ofaseries ofderivatives isextremely essential, and ifnot fulfilled itcan make term-by-term differentiation ofthe series impossible. This is illustrated byadominated series that does not admit term- by-term differentiation. Consider the series sitesage4sateygte This series converges toacontinuous function because itis dominated, Indeed, forevery xitsterms are (inabsolute value) less than’ the terms ofthenumerical convergent series with positive terms ; dahtyte thee. Write aseries composed ofthe derivatives ofthe terms ofthe original series: cosx-+2*cos2*x-+...-+n*cosn*x+... This series diverges. Thus, forinstance, forx=0 itturns into the series 14243. bathe. (Itmay beshown that itdiverges notonly forx=0.) SEC. 18. POWER SERIES. INTERVAL OF CONVERGENCE Definition 1.Apower series isafunctional series ofthe form G,+0,X-+4,0*+...+4,0+..., qa whereay,a,dy+++,dy,+++areconstants calledcoefficients of the series. The domain ofconvergence ofapower series isalways some interval, which, inaparticular ease,candegenerate intoapoint. Toconvince ourselves ofthis, letusfirst prove thefollowing theorem, which isvery important forthewhole theory ofpower series. Theorem 1(Abel’s Theorem). 1)/fapower series converges for some nonzero value x,,then itconverges absolutely forany value ofx,forwhich lel<layh 2)ifaseries diverges forsome value x,,then itdiverges forevery xforwhich . Ix1> [xh Power Series. Interval ofConvergence 143 Proof. 1)Since, byassumption, the numerical series 4G,+0,%,+00+...+O,Xtoe ay converges, itfollows that itscommon term a,x—+0 asn—oo, and this means that there exists apositive number Msuch that all the terms ofthe series are less than Minabsolute value. Rewrite theseries (1)inthe form and consider aseries ofthe absolute values ofits terms: lal+ax|Z|+laagle[ +.tlaagi[ef+e. The terms ofthis series areless than thecorresponding terms ofthe series * m+M|zlem[z[+...+u]Zf +... oy For |x|<|x,| the latter series isageometric progression with ratio\é|<1 and,consequently, converges. Sincethetermsoftheseries(3)arelessthanthecorresponding termsoftheseries(3), the series (2)also converges, and this means that theseries (la) or(1)converges absolutely. 2)Itisnow easy toprove the second part ofthe theorem: let the series (I)diverge atsome point x,Then itwill diverge at any point xthat satisfies thecondition |xl>[x,|. Indeed, ifat some point xthat satisfies this condition theseries converged, then byvirtue ofthe first part (just proved) ofthe theorem, itshould converge atthepoint x,aswell, since |x,|<|x| Butthiscon- tradicts thecondition that atthepoint x,the series diverges. Hence the series diverges atthepoint xaswell. The theorem is thus completely proved. Abel's theorem makes itpossible tojudge the position ofthe points ofconvergence and divergence ofapower series. Indeed, ifx,isapoint ofconvergence, then theentire interval (—|x,|, |x,[) isfilled with points ofabsolute convergence. Ifxisapoint of'divergence, then thewhole infinite hall-line tothe tight ofthe point |x,| and thewhole hali-line totheleftofthe point —|x;| consist ofpoints ofdivergence. From this itmay beconcluded that there exists anumber R such that for|x|<R wehave points ofabsolute convergence and for[x|>R, points ofdivergence. ™ Series We thus have the following theorem onthe structure ofthe domain ofconvergence ofapower series: Theorem 2.The domain ofconvergence ofapower series isan interval with centre atthecoordinate origin. Definition 2.Theinteoal of,convergence ofapowerseries.isan interval from —R to+R such that forany point xlying inside Seriesconverges . ~% a - 0 ‘Baresdiverges Seriesdiverges Fig. 349. this interval, theseries converges and converges absolutely, while forpoints x'lying outside it,the series diverges (Fig. 349). The number Riscalled theradius ofconvergence ofthe.power series, ‘Attheend points oftheinterval (atx—R and atx——R) the question ofthe convergence ordivergence ofagiven series is decided separately foreach specific series. Wenote that insome series theinterval ofconvergence dege- erates into apoint (R—0), while inothers itencompasses theentirex-axis(R=0).Wegive amethod fordetermining theradius ofconvergence of apower series. Let there be aseries a+ axtattoo. +a,x"+ 00, (a) Consider aseries made upofthe absolute values ofitsterms: NayJa[lx]+lay|leltaylL+“elagilelt+.-Flay|e... O) Todetermine theconvergence ofthisseries (with positive termsl), apply thed'Alembert test Let usassume that there exists alimit: iim22im|2ae282|tim|ae2Ain,gettim|225|fim[85]x1—Ce Then bythed'Alembert testtheseries (4)converges, itL|xI<1; thatis,if[x|<p, anddiverges ifL[x|>1, thatis, if[x1>}. Consequently, series(1)converges absolutely when|x|<7. But if[z[>, thenlim“s%—|x|L>1 andseries(4)diverges, and Power Series. Interval ofConvergence 45, itsgeneral term does not tend tozero.*) But then neither does thegeneral term ofthe given power series (1) tend tozero, and this means that (on the basis ofthe necessary condition ofcon- vergence) thispowerseriesdiverges (whenlxl>)- Fromtheforegoing itfollows thattheinterval (—2,t)is the interval ofconvergence ofthe power series (1): 1tim|eeRepo iin|| Similarly, todetermine the interval ofconvergence wecanmake useoftheCauchy test, and then R=—1__Tim{7Teal” Example 1.Todetermine the interval ofconvergence ofthe series Dpapatpat tbat... Solution. Applying d'Alembert's test directly, weget lim,|r|=tet. Thus,theseriesconverges when|x|<1 anddiverges whenlle1At theextremities ofthe Interval (1, 1)Itisimpossible to.investigate the series bymeans ofd’Alembert’s test. However, itis immediately apparent that when *—--—1 and when x=—1 the series. diverges Example 2.Determine theinterval ofconvergence oftheseries 2_ Qa, @otT2ty Solution. We apply the d’Alembert test: (ays im|EE) tim|"|i2e1= ois,|ase|=[arg]2112007 ‘Theseriesconverges if|e]<i,thatis,if[x|<2ys whenx=theseries converges; whenxo—dthesetsdiverees. *)Itwill berecalled that inproving d’Alembert’s test (see Sec. 4)we found that iftim 22>1,then thegeneral term oftheseries increases and, consequently, doesriottendtozero. 16 Series Example 3.Determine the interval ofconvergence ofthe series aEStes Solution. Applying the d'Alembert test weget sam [@meml=as[lees Since the limit iindependent ofxand isless than unity, theseries con Seppe or fltutes os Example 4.Theseries1-+x+-(2e)*+(3x)'+...+(nx)"+... diverges for ailvaluesofrexcept £20bechuse (aor weatNeewhomale wel the ite tong asi's diferent from eee Theorem 3.The power series B+ OE+ORE bd boo Oy isdominated onanyinterval [—@, Q]that lies completely inside theinterval ofconvergence. IntervalofconvergenceSSS SOO7 a (ntervatofmajorisation, Fig. 50, Proof. Itisgiven that e<R (Fig. 350) and therefore thenum- berseries (with positive terms) 1a,|-+14,le-+ 1a,lett---+layle © converges. But when |x|<o, the terms oftheseries (1)donot exceed, inabsolute value, the corresponding terms ofseries (5). Hence, series (1)isdominated ontheinterval [—g, @]- “Rk +0 FR Fig. 61. Corollary 1.Onevery interval lying entirely within theinterval ofconvergence, thesum ofapower series isacontinuous function. Indeed, the series onthis interval ismajorised, and itsterms are continuous functions ofx.Consequently, onthebasis ofTheorem 1, Sec. 11, the sum ofthis series isacontinuous function. Differentiation ofPower Series 147 Corollary 2./fthelimitsofintegration a,Bliewithintheintervalof convergence ofapower series, then theintegral ofthesum oftheseries isequal tothesum oftheintegrals oftheterms oftheseries, be- cause the region ofintegration may betaken inthe interval (=e. 0],where theseries isdominated (Fig. 351) (seeTheorem 2, Sec. 12,onthepossibility ofterm-by-term integration ofadomi- nated series). SEC. 14. DIFFERENTIATION OF POWER SERIES Theorem 1.Jfapower series sSQ)=atarctas tarpaste taste. (I) has aninterval ofconvergence (—R, R), then theseries (4) =4,+2a,x+ 34x"... fray + (2) obtained bytermwise differentiation oftheseries (1) hasthesame interval ofconvergence (—R, R); here, @(t)=s' (x), ifZ1<R, i.e., inside theinterval ofconvergence thederivative ofthesum ofthepower series (1)isequal tothesum oftheseries obtained by termwise differentiation ofthe series (1). “RkOP oO x9tkah Fig. 352. Proof. We shall prove that theseries (2)ismajorised onany interval [—g, @]that lies completely within the interval ofcon- vergence. Takeapoint&suchthato<E<R Fe352).Theseries(1) converges atthis point, hence lima,8*=0; itistherefore possible toindicate aconstant number’ Msuch that 1a,8"|<M (n=1, 2,...). Ii|x|<e, then 1 1 nat "-" Mnat,Ina,x"-|<|na,e"-"|]—=n]a,8°-"|| ¢[""<nZar, where q=f<i. 8 Series Thus, inabsolute value, the terms ofthe series (2), when x|<Q, areless than theferms ofapositive number series with constant terms: EU+29-4394... tngt+...). But this latter series converges, aswill beevident ifweapply the d'Alembert test: ngsimann ash Hence, theseries(2)ismajorised ontheInterval {2o],andby Theorem 2,Sec. 12, itssum isaderivative ofthe sum ofthe given series ontheinterval [—, Q],i.e., @l(x)=s' (x). Since every interior point ofthe interval (—R,R)maybe included insome interval [—g, g],itfollows that theseries (2) converges atevery interior point oftheinterval (—R, R). We shall prove that outside the interval (—R, R)the series (2) diverges. Assume that the series (2) converges when x,>R. Integrating ittermwise intheinterval (0,x,), where R<x,<x,, wewould find that theseries (1)converges atthepoint x,,but this contradicts the hypotheses ofthe theorem. Thus, the interval (—R,R)istheinterval ofconvergence ofseries’(2). Andthetheorem isproved completely. Series (2)‘mayagainbedifferentiated termbyterm,andthis may becontinued asmany times asone pleases. We‘thus have the conclusion: Theorem 2.Ifapower series converges inaninterval (—R, R), itssum isafunction which has, inside theinterval ofconvergence, derivatives ofany order, each ofwhich isthe sum ofaseries re- sulting from term-by-term differentiation ofthe given series an appropriate number oftimes; here, the interval ofconvergence of each series obtained by differentiation isthe same interval (-R, R). SEC. 15. SERIES IN POWERS OF x—a Also called apower series isafunctional series ofthe form 4,+4,(x—a) +4,(x—a)*+... +4,(x—a)"+..4, (I) where theconstants a,,@,,...5 dy) ++.arelikewise termed coeffi-cients oftheseries. Thisisapower series arranged inpowers of the binomial x—a. Series inPowers ofx—a 49 When a=0, we have apower series inpowers ofx,which, consequently, isaspecial case ofseries (1). Todetermine theregion ofconvergence ofseries (1), substitute the variable x—a=X, Series (1)then takes onthe form 4,44,X +4,X8 aX" oy @) wethus have apower series inpowers ofX. Lettheinterval —R<X<R betheinterval ofconvergence ofthe series (2)(Fig. 353, a).itthus follows that series (1)will converge forvalues ofxthat satisfy theinequality —R<x—a<R ora—R<x<a+R. Since series (2)diverges for|X|>R the series (1) will diverge for |x—a|>R, that is,itwill diverge outside theinterval a—R<x<a+R (Fig. 353, B). orf prthat ee Fig. 958. Fig. 354. And sothe interval (a—R, a+R) with centre atthe point a will bethe interval ofconvergence ofseries (1). Alltheproperties ofaseries inpowers ofxinside the interval ofconvergence (—R, +R) are retained completely for aseries inpowers of x—d inside theinterval ofconvergence (a—R, a+R). Forexample, after term-by-term integration ofthepower series (I), ifthe limitsofintegration liewithintheintervalofconvergence (a—R,a+R),wegetaserieswhosesumisequaltothecorresponding integral ofthesum ofthegiven series (1). Inthe case oftermwise diffe- rentiation ofthe power series (1), forallxlying inside the inter- val ofconvergence (a—R, a-+R) weobtain aseries whose sum isequal tothederivative’ ofthesum ofthegiven series (1). Example. Find the region ofconvergence ofthe series DEG DELERM Solution. Putting x—2=X, weget the series XXXRAM This series converges when —1< X<.H1, Hence, the given series converges forallxthat salisly theinequality —1<x—2< 1,that is,when bax <3 (Fig. 354). 150 Series SEC. 16, TAYLOR'S SERIES AND MACLAURIN'S SERIES InSec. 6,Ch. IV,itwas shown that forafunction f(x) that has allderivatives uptothe (n+l)st order inclusive, Taylor's formula holds intheneighbourhood ofthe point x=a (that is, insome interval containing the point x=a); 1@)=f@+547@+ FSP at. ASE MER O) where theso-called remainder term R,(x) iscomputed from the formula R=SOTPY[atOa), O<O<1. Ifthe function f(x) has derivatives ofallorders inthe neigh- bourhood ofthepoint x=a, then inTaylor's formula thenumber nmay betaken aslarge asweplease. Suppose that inthe neighbourhood under consideration the remainder term R,tends tozeroasn—co; limR,=0. Then, passing tothe limit informula (1)asn—+oo, wegetan infinite series onthe right which iscalled the Taylor series: 1)=F)4227@+...42S"Mt... —Q) Thisequation isvalidonlywhenR,,(x)—+0 asn—»oo,Thentheseries ontheright converges and itssum isequal tothegiven function f(x). Letusprove that this isindeed thecase: 14)=P,(2)+R,(2), where oeP,Q)=f(a)+22PF@+e.+2SIm. Since itisgiven that limR,=0, wehave F)=limP,(x). But P,(x) isthe nth partial sum oftheseries (2); itslimit isequal ‘tothesum oftheseries ontheright sideof(2).Hence, (2) istrue: 10=1@+ 7@+4s2r@+...+ 57w@t... Examples ofExpansion ofFunctions inSeries ‘73h From the foregoing itfollows that theTaylor series isagivenfunction f(x)onlywhenlimR,=0.IflimR,+0, thentheseriesisnot the given function, although itmay converge (toadifferent function). Ifinthe Taylor series weput a=0, wegetaspecial case of »this series known asMaclaurin’s series: FO)=FO+E/OFFO+..-+FMO+--. @) Iffor some function we have aformally written Taylor's series, then inorder toprove that this series isagiven function itiseither necessary toprove that the remainder term approaches zero, ortobeconvinced insome way that this series converges tothe given function. We note that foreach oftheelementary functions defined in Sec, 8,Ch. I,there exists anaand anRsuch that inthe inter- val(a—R, a+)itmaybeexpanded intoaTaylor's seriesor (ifa=0) into aMaclaurin’s series. SEC. 17. EXAMPLES OF EXPANSION OF FUNCTIONS IN SERIES 1.Expanding thefunction f(x)==sin.x inaMaclaurin’s series. InSec. 7,Ch. IV, weobtained the formula sine xB HI Et Rae Since itwas proved that limR,,=0, itfollows, bywhat hasbeen saidinthepreceding section, thatwegetanexpansion ofsinx inaMaclaurin’s series: singax—Ht Gt (I Goats a) Since theremainder term approaches zero forany x,thegiven series converges and, for itssum, has the function sin xfor any x. Fig. 355 shows the graphs ofthe function sinx and ofthe first three partial sums ofthe series (1). This series isused tocompute the values ofsinx for different values ofx. Toillustrate, letuscompute sin10°tothefifth decimal place. Since 10°==0.174533, wehave ee A/a)", 1(n\*_1 (nysin10°=h—3 (js)+81(is)—n(i)te 152 Series Confining ourselves tothe first two terms, wegetthefollowing approximate equality: intat 1 (ny.sink 8-5(%)+ here, weare in‘error by5,which inabsolute value isless than thefirst ofthesuppressed terms; that is, <5(fh)<p(0.2<4-10". . yy 1“ol Hlrn\ Sex4 ! \/I \4 i \y ! \ JOspgd \Z / \ x \ x \o v f a \ /oo\ Ha \ ioe \ I“ lal iY serge \ 17 ‘ i \ Fig. 356, Ifeachtermintheexpression forsin7giscomputedtosix decimal places, weget sinf=0.173647, We can besure ofthe first four decimals. 2.Expanding the function f(x)=e* inaMaclaurin’s series. On the basis ofSec. 7,Ch. IV, we have alte gtgtt+gts @) Euler's Formula 753 since itwas proved that limR,(x)=0 forany x.Hence, the series converges forallvalues ofxandisthefunction e*. 3,Expanding thefunction f(x)==cosxinaMaclaurin’s series. From Sec. 7,Ch. IV, we have aotot coseIF FG 8) forallvalues ofxtheseries converges and represents thefunction cos x. SEG. 18, EULER'S FORMULA Up till now wehave considered only series with real terms and have notdealt with series with complex terms. Weshall notgive thecomplete theory ofseries with complex terms, forthis goes beyond thescope ofthistext. Weshall consider only oneimportant example inthis field. InChapter VII wedefined the function e**” bythe equation ett)=6*(cosy+isiny). When x=0, weget Euler's formula: em cosy-+ising. Ifwedetermine theexponential function ¢”with imaginary exponent bymeans offormula (2), Sec. 17,which represents the function e*intheform ofapower series, "wewill get theverysameEuler equation. Indeed, determine e”byputting theexpres- sion iyinplace ofxinequation (2), Sec. 17: iy, iy (iy ww"Oe ee ee ee a) Taking into account that *=—1, @=—i, =I, =i, *=—1, and soforth, we transform formula (1) tothe form wei tt Clee ee ee Separating inthis series thereals from the imaginaries, wefind viv yey om(14g...) 46(f-G48—...)- The parentheses contain power series whose sums are equal to cosy and siny, respectively [see formulas (3)and (1)ofthepre- ceding section}. Consequently, e”=cosy+isiny. Thus, wehave again arrived atEuler's formula. 754 Series SEC. 19. THE BINOMIAL SERIES 1.Letusexpand thefollowing function inaMaclaurin’s series: Fx) =(1+2)", where misanarbitrary constant number. Here theevaluation oftheremainder term presents certain dif- ficulties and sowe shall approach the series expansion ofthis function somewhat differently. Noting that thefunction f(x) =(1+x)" satisfies thedifferential equation +2)F (2)=mf(2) (ay and the condition FO)=1, wefind apower series whose sum s(x) satisfies equation (1)and thecondition s(0)=1: sQ)altaxtast+... pax...) (2) Putting this series into equation (1), weget (1+2)(a,+2a,x+3a,x*+ 26.ax"+...)= Sm(LFa,xbaet +...bat...) Equating thecoefficients ofidentical powers ofxindifferent parts oftheequation, wefind =m; a,+2a,—=ma, ...5 na,+(n+1)a,,,= mays... Whence forthe coefficients ofthe series wegetthe expressions a= aam aad min=) ,pal a=m; a= 20> cand a(m—2) _mim—Vim—2), 2,=RD mm); (m=)... m—n-41), These are binomial coefficients. Putting them into formula (2), weobtain s(x)altmetBGSary veep SDMON go, @ *)Wetook theabsolute term equal tounity byvirtue oftheinitial con- dition s(Q)=1. The Binomial Series 785, Ifmisapositive integer, then beginning with theterm con- taining x**" allcoefficients ‘are equal tozero, and theseries is converted into apolynomial. For mfractional oranegative integer, wehave aninfinite series. Let’ usdetermine the radius ofconvergence ofseries (3): thyggA, Mm 1).[mA$2)naeGe inast|=sim|Seneae [==im|P=24|]x=. Thus, series (3)converges for|x|<1, In’the interval (—i, 1),series (3) isafunction s(x) that satisfies thedifferential equation (1)and thecondition sQ)=1. Since the differential equation (1) and the condition s(0)=1 aresatisfied byaunique function, ‘itfollows that the sum ofthe series (3)isidentically equal tothe function (1+2)", and we obtain the expansion (yt lpm SEYtyMONONA Gy Fortheparticular casem=—1,wehave Petlorttiete. (4) Form= weget — Lipid gag 19 yy 19-5ViFesltgr—gyt' tage reat te (6) Form=—4 wehave 1 Lh og 1-3-5 4 1.35.7poet icptee rete © 2,Weapply thebinomial expansion tothe expansion ofother functions. Expand thefollowing function inaMaclaurin’s series: f(x)=aresinx. 756 Series Putting into equation (6)the expression —x* inplace ofx,we get 1 Lg hdyrs tat tat 13-5 13-5... (201) vanFe FEOe, By the theorem ofintegration ofpower series wehave, for [x|<k: ¢ a " eed 23-5x7 [piprecsineeet Stagtaete 1.3-5...2n—1) x44 soot2-4-6...2n tyite This series converges inthe interval (—1, 1).One could prove that the series converges for x=-+1 aswell asthat forthese values thesum ofthe series islikewise equal toarcsinx. Then, setting x=1, wegetaformula forcomputing x: nla Lbs 1135 1aresinl=S=l+y-gtge gtpee pte ‘SEC. 20.EXPANSION OF THE FUNCTION In(14x) IN APOWER SERIES. COMPUTING LOGARITHMS Integrating equation (4), Sec. 19, from 0tox(when |x|<1), we obtain Sen Jdette ede or 2gt . Ind+x)=x—-F +FF4H.HHI EE, ( This equation holds true intheinterval (—1, 1). Ifinthis formula xisreplaced by—x, then wegetthe series att ot In(l—x)=—2—-2-2_¥_., @ which converges inthe interval (—1, 1). Using theseries (1)and (2)wecan compute thelogarithms of numbers lying between zero and two. We note, without proof, that forx=1 theexpansion (1)also holds true. Wewill now derive aformula forcomputing. the natural loga- rithms ofallintegers. Expansion oftheFunction in(1-+3) inaPower Series 187 Since intheterm-by-term subtraction oftwo convergent series wwegetaconvergent series (see Sec. 1,Theorem 3),then bysub- tracting equation (2)from equation (1)term byterm, wefind In+)—In(1—2) in}emo[x4F4E4...]. Nowputpeat; thenxegin.For any n>0 wehave O<x<1]; therefore lee j,atl_of 1 1 1injan 2(at saeeh +seapipt ‘}: whence 1 1 1 Inn+)—Inn=2[ tage teat |:@) For n=1 we then obtain in2—2[+aat stn2=2lratzat eet}: Tocompute In2 toagiven degree ofaccuracy 6,one has to compute thepartial sum s,,choosing thenumber pofitsterms such that thesum oftheSuppressed terms (that is,theerror R, committed when replacing sbys,)islessthan theadmissible error 6.Todothis, letusevaluate theerror Ry _ 1 1 1 2layer tT tT wets]. Since the numbers 2p-+3, 2p-+5, ... are greater than 2p-+1, it follows that byreplacing them by2p+1 weincrease each fraction. Therefore, 1 1 1 2<2[aAT attETE +EST +]: or Mpa yayaR<glwat watgent.--]- The series inthe brackets isageometric progression with ratio }+Computing thesumofthisprogression wefind 1 2 WH 1 RySgpete TFT ®3 Ifwenow want tocompute In2 to, forexample, seven decimal places, wemust choose psuch that ’R,<0.0000001. This eanbe done byselecting psothat theright Side ofinequality (4)isless 758 Serles than 0.000001. Bydirect choice we find that itissufficient to take p=8. Toseven-decimal accuracy wehave Vedat ya adin2ws=2[y+getsetretept it 1 +giantypige]=0.6991471. Thus, In2—0.6931471. These seven digits are significant digits. ‘Assuming n=2 informula @), weobtain Ins—in2+2[f-+satggt ]=1.098612, andsoforth. Inthis way weobtain thenatural logarithms ofany integer. Toget the common logarithms ofnumbers, use the following relationship (see Sec. 8,Ch. II) logN=MInN, whereM=0.434294, Then,forexample,wegetIn2=0.6931472, Jog2=0.30103. SEC. 21. INTEGRATION BY USE OF SERIES (CALCULATING DEFINITE INTEGRALS) InChapters Xand XIitwas noted that there exist definite integrals, which, asfunctions ofthe superior limit, arenot, in final form, expressible interms ofelementary functions. Itis sometimes’ convenient tocompute such integrals bymeans of series, Let usconsider several examples. 1.Let itberequired tocompute the integral Se-**dx. Here, theantiderivative ofe-*" isnot anelementary function. Toevaluate this integral weexpand the integrand inaseries, replacing xby—x* intheexpansion ofe*[see formula (2), Sec.17]: eo oe »etal eet tI + Integrating both sides ofthis equality from 0toa,weobtain ¢ J xe sot lo fedx=(FEtis—it) = a@ la at=Tcintas at Integration byUse ofSerles 159 Using this equation, wecan calculate the given integral toany degree ofaccuracy forany a. 2,Itisrequired toevaluate theintegral Expand the integrand inaseries: from theequation sincaa 4F4, weget sine. eoSeI-F+h Ft thelatter series converges forallvalues ofx,Integrating term byterm, weobtain Csinx aotaJiao Rteat The sum oftheseries isreadily computed toany degree of accuracy forany a. 3,Evaluate theelliptic integral {VISainap<0). Expand theintegrand inabinomial series, putting m=4, x=—F' sin’ [see formula (6), Sec. 19]: ViRFsing=1—FH!sintg—4tk!sintg—41 3pisintg—... This series converges forallvalues of@and admits term-by-term integration because itmajorises onany interval. Therefore, ° iat 11\Vizes Sin*Gdp=G—zk [sin*edp—7 4Asin*@dg— 113 uarTe" fsiedg—... 160 Series The integrals ontheright arecomputed inelementary fashion. Forp= wehave é int, 1:3...(2n—1) JsinodoaT (see Sec. 6,Ch. XI) and, hence, a x L\*p2_ (1:3\tR (1-3-5) 84"JvI=e anedom5[1—-(7)'#-(75)'5— (248) s—~]- SEC, 22, INTEGRATING DIFFERENTIAL EQUATIONS BY MEANS OF SERIES Iftheintegration ofadifferential equation does not reduce to quadratures, oneresorts toapproximate methods ofintegrating theequation. One ofthese methods isrepresenting theequation inaTaylor's series; thesum ofafinite number ofterms ofthis series will beapproximately equal tothe desired particular solution. Totake anexample, letitberequired tofind thesolution of asecond-order differential equation, =Flo ws 0) that satisfies the initial conditions enn =YoWear =Yer @) Suppose that thesolution y=f(x) exists and may begiven intheformofaTaylor'sseries(wewillnotdiscusstheconditionsunder which this occurs): 9=F)=f)+2SEPH+EPwt @) Wehave tofind f(x,), f’(x,), P(x,),--., ie., thevalues ofthe derivatives oftheparticular solution when x=x,. But this can bedone bymeans ofequation (1)and conditions (2). Indeed, from conditions (2) itfollows that Fd =e Fa) =H from equation (1)wehave PF) =aang =FerYorYodo Integrating Diferential Equations byMeans ofSeries 76 Differentiating both sides of(1)with respect tox,weget y= Filey VI+Fu HwYY+FelHYY(A and substituting thevalue x=x, into the right side, wefind PG)=Ueasy Differentiating the relationship (4)once again, wefind PY(8)=ODenny and soon. We put these values ofthe derivatives into (3). For those values ofxforwhich this series converges, this series represents the solution ofthe equation. Example 1.Find thesolution oftheequation y=-e, which satisfies the initial conditions Weao=l, Wieme=0. Solution. We have FO=ymls O=y,=0.Fromthegivenequationwefind(yzas=P" (0)=0:further,yamy's!204, W)eav=!" 0)=0,gmstyAry—2,gw—2 and, generally, differentiating &times both sides ofthe equation by the Leibnte formula, wefind (See. 22,Ch. Til) ptt M2 PN gk (1) Putting x=0, wehave yt? am—bh) ff? or,setting A-+2—n, =(n—3)(n—B yf, Whence WY=A gf=—5-645"=(—11-2)(6-6), 48=—9-104{" =(—IF1-2)(6-6)(9-10), ft=(—8(1-2)6-6)(9-10)...144-3)(44-3. In addition, i=0, yl=O, ..., tt <0, =o, yf!—0, ...,oto, A=0, gM=0, ..,, gt a0. Thus, only those derivatives whose order isamultiple offour donot become zero, 162 Series Putting the values ofthe derivatives that we have found into aMacla- uria's series, weget thesolution ofthe equation ot ”ymLaGl2+gy(1-2)6-6)—Fpj(1-2)6-6)(9-10)+... st—1)" (1.2)(6+ —'—' seHED 6-0...1449k=O)+... Bymeansofd’Alembert's fetwecan.verity that,thsseriesconverges for allvalues of%ence, itis the solution of theequation, Itthe equation islinear, itismore convenient toseek the coefficients ofexpansion of‘the particular solution bythe method ofundetermined coefficients. Todothis, weput the series Y=a,tartar +...bart into the differential equation and equate the coefficients of identical powers ofxondifferent sides oftheequation. Example 2.Find thesolution oftheequation oeAy that satisfes the initial conditions Wrens,(W'Vene=. Solution. We set YO,baxpagepagpo.page On the basis ofthe initial conditions we find 4=0, a,=1. Hence, YSEbORpat batbeYa14aye+Sayx*+...nage".Y=24,43-20... fa(n—l)aye” Putting these expressions into the given equation and equating thecoelicients ofidentical powers of*,weobtain eating22,0, whencea,=0;32q,=244, whence =I;4-30,—4a,+404,whence4,0: A(ANdg(02)DoggtAdganswhenceay=228=8, Consequently, a 211Zot 1. gated: gaat: gat: Bessel’s Equation 169 1aot way_1ce rT 4490; m0; 04=0. Substituting thecoefficients which wehave found, weget the desired solution _ tte, et garde tpt tate The series thus obtained converges for allvalues of+. ‘M'Will benotedthatthisparticular solution maybeexpressed interms oftheelementary functions: taking xoutside the brackets weget(inside the brackets) anexpansion ofthefunction e*.Hence, yan, SEC. 29, BESSEL'S EQUATION Bessel's equation isadifferential equation ofthe form ey tay’ +(t—p')y=0 (p=const). a The solution ofthis equation (asalso ofcertain other equations with variable coefficients) should besought not inthe form of &powerseries,butintheformofaproduct ofsomepowerof xbyapower series: y=xDew". (2) The coefficient a,may beconsidered nonzero due tothe indefiniteness ofthe’exponent r. Werewrite theexpression (2)intheform y=Bee and find its derivatives: y=Bert hag, =DetWrtka, Put these expressions into equation (1): BBCP +hDag+ HeDerbay’! +(tpt) Baye’=0. 764 Series Equating tozero the coefficients ofxtothe powers r,r+1, r-F2,..., r+, weget asystem ofequations: Ir(r—1)+r—p'la,=0 or(?—p']a,=0, ) (+Dr+(+1)—p'}a,=0 of{(r+1)*—p']a, =0, [r+2)(+I)+(-+2)—p']a,+4,=0 oF[(r-+2)"—p']a,+a,=0, |(3) UFC FRDC+E)—play$ay4=0 oF |(r-FRYpila,+a,..=0.) Let usconsider the latter equation: (+2)—p']ay+a,.,=0- (3’) Itmay berewritten asfollows: (r+k—p)(r +k+p)]a,+a,.,=0. Itisgiven that a,#0; hence, . P—pt=0, therefore, r,—p orr,=—p. Letusfirst consider thesolution forr,=p>0.From thesystem ofequations (3)wedetermine allthecoeffi- cients a,,dy,...insuccession; a,remains arbitrary. For instance, puta,=1. Then MRO Assigning various values to&,wefind a,=0, a,=0 and, generally, a,.,,—=0; SSOo esFmHA EDO ED 4 we H 4(—VW"EER TDDTDBED* Putting thecoefficients found into (2), weobtain =. 2 x — nowt Wet TTA wT ¢—meraerreret): ©) Allthecoefficients a,,will bedetermined, since forevery &the coefficient ofa,in(3), (+k, will bedifferent from zero, Bessel's Equation 105, Thus, y,isaparticular solution ofequation (1).Letusfurther establish theconditions under which allthecoef- ficients a,will bedetermined forthesecond root r,=—p aswell.Thiswilloccur ifforanyevenintegral positive &thefollowing inequalities are fulfilled: (+k —pt #0 © or nthe. But p=r,; hence, ntken. Thus, condition (6)isinthis case equivalent tothefollowing none where isapositive even integer. But =P =—P hence ror, =2p. Thus, ifpisnotequal toaninteger, itispossible towrite asecond particular solution that isobtained from expression (5) bysubstituting —p forp: ax f1—,* 4, bs = w=[1esaaeeeees) # ; TEE TI— PTB TOset: ] 6) The power series (5)and (6!) converge forallvalues of2;this isreadily found byd’Alembert’s test. Itislikewise obvious that y,and y,arelinearly independent.*) The solution y,multiplied byacertain constant iscalled aBessel function ‘ofthe first kind oforder pand isdesignated bythesymbol J,.Thesolution y,isdenoted bythesymbol J_,. *)The linear independence offunctions isverified asfollows. Consider the relation tpt ee eeTED EHTS wT % ye‘345 tTTawa This relation ismotconstant, since forx-+0 itapproaches infinity. Hence ‘the functions y,and y,are linearly independent. 766 Series Thus, forpnot equal toaninteger, the general solution of equation (1)has theform Y=Culp+Cylge Forinstance, whenp= theseries(6)willhavetheform i at xt at ps ee SWi[e-E+8—-7+ +]: Thissolution multiplied bytheconstant factorJ/= is.called Bessel’s function Ji;wenote that the brackets contain aseries whose sumisequal tosinx. Hence, Ji@=VWZsinz Inexactly thesame way, using formula (5’), weobtain Laiw=YV Zee. Thegeneral integral of(1)forp= is 9=CJs (FCI (2s Now letpbeaninteger which weshall denote byn(n>0). The solution of(5)will inthis case bemeaningful and isthe first particular solution of(1). But thesolution of(5’) will not bemeaningful because one of thefactors ofthedenominator will become zero upon expansion, For positive integral p=n theBessel function J,isdetermined bytheseries (5)multiplied intotheconstant factor gray(when n=0 wemultiply by1): a x x 4.0)=$m[|—raeey$5+rapa DOP at +] TET mth m+ hme or _Sew payJne=LoanGan(z) o Best's Equation 167 Itmay beshown that thesecond particular solution should in this case besought intheform K,(x)=J, (x)Inxe-® 2byt. Putting this expression into (1),wedetermine thecoefficients by. The function K,(x), with the coefficients thus determined, muf- tiplied byacertain constant iscalled Bessel's function ofthe second kind oforder n. This isthe second solution of(1), which with the first one forms alinearly independent system. The general integral will beofthe form YAOI(8)+C.Ky(2). ) We note that limK,,(x)=00. Hence, ifwewant toconsider the final solutions forx=0, then wemust put C,=0 into formula (8). Example. Find thesolution ofBessel's equation, forp=0, vty tno that satistes the intial conditions: for#=0, y=2, f=0. Solution. From (7)wefind one particular solution: Su(=Ox)", 1(x), tbe)t_1fet1600S (4)"=!~a(4)am(4)ae(Z)+ soni, isolation, weeanwriteasolution thatsate thegiven ath, Note. Ifwehad tofind the general integral ofthis given equation we would sek the second particular solution Inthe form Kyepadsinet Soya, rs Without giving all the computations, we indicate that the second particular solitions whieh wedenate by1a), isafthe fora Bev ty 1ayy, ad Kom2anatin $)'(14)+a($)(ued) This function multiplied bysome constant factor iscalled Bessel’s function Gite 'econd kind otordee ae. 18 Series Exercises onChapter XVI Write thefrst several terms oftheseries according tothegiven general a et age eeeee ee a a ee 5.ug=j/mF1—Var. ; ve balay * Tatieolloring atsfrcomme: SASH ttn1 oneVerve Vs"tit AmDies a2 kase ttle... an Diverse. 8. ghetto...ytgttat ree vityet + 1 1 2\* 3\" n\n soyigyt dewDass4(3)'+(2) ent(GE) 142.3 )4 5 Oe Watstiotnt tapite:Ans.Diverges. WBpty tothe tpt++Ans.Converges. Tatforconvergence theellowing serieswithgivengeneraltems 1aya: AteConver, Metym Ant.Die, Ite iatE8AnsDiverges.1%earch Ans,Diveraes.8.tqtE%,Ans,Diverse.1%uyeraretory.Ans Converges. 18.nF ‘Ans.Diverges, 19.Provetheinequality Lyd 1 1a 1ltgtgttyeMtn>tate tear: 20.Isthe Leibniz theorem applicable tothe series StVari Vari Va-1 Van} Vani Vag ‘Ans. Ikisnot applicable because the terms ofthe series donot decrease mogtonically inabgolute velue, The sets diverges, How mary frst terns oust befaken Inthesei] sothat their sum should not differ bymore than 10-* ofthe sum ofthe corresponding series: Loayt 14 lait abade dddtht Amsammo, ozLot yd 1 ont tag 1 mbt anno bed eh \ oo, 4 i mt Amenet dobar egtth H mite Anene, Exerctses onChapter XVI 10 Findoutwhichofthefollowing seesconverges absolutely: %ophacntot. HOU aptonAnsConversabso-wey EE gape. dnsComersfelee an . absolutely. 27.gpg ympegted Wet. Ams.Converges cnn 2Lgl ataaACi verges conditionally. ae 1 Fnhewmofee yahtae an.2, For what values of do the following series converge: meee eB. ancnc oo Eee typolareptSRCad 8mata eeeDMEceAns,CREEE EMM1 100k,10,000x* ,1,000,000x* Ans.|<. 88.LpEpOEa Ansme<a<a, Wsnettandem pbong AnMacao. 5 a Z 8. tt. vee An. Leech.TeytiayetSagat3 a, 2 BENGARE tnditomcece,Mant tae tet bese Ans. e<e<e Bat Ee+ + eeHOAe AnAeA Fldtheuml tesi epRE bast. (Le]<M Hint? Write the'sefies in the form teeth. Beebe :Rhee ans. 7a.epi4=e Determine whichofthefollowing series Ismajored ontheindeatedintervals: 40. +E G+. OSSD. Ans. Majorised. atteSet tlt... ceed. AnsNotmajorised, sine sine sine, sian a,0OE SEEH. 10,2a].Ans,Majorised. 25-3388 7 Series Expanding Functions InSertes (3,Expand sglzginpowersofanddetermine theinterval ofconver gence. Ans. The series converges for—10 <x <10, 14Expand coxinpowersof(s—).Ans.glete(2-4)— 1 x\t, 1 x)~rye(*-4) taye(*-4) + 45,Expand e-*inpowers ofx,Ans.aeEEEaa.46,Expandefinpowersof(#2,Ans.ebete—D-+E 2+ +hamt... AT. Expand x*—2x*45x—7 inpowers of(x—l). Ans. —34+4(r—1)+ +(x1?+=1).448, Expand thepolynomial 2"42<t—3e"—Grt4.3r¢4.61"—x—2 inaTay- lors series inpowers of(2-1); check tosee that this polynomial has the number Iforatriple root, Ans, {(2)=01 (e—I)¥-+270 (xIe342(2D +330(x—1)"$186(x1)?+6321)"+12(21)+(e1)", Pi49,Expand cos(x-+a) inpowers ofx.Ans. cosa—rsina—¥ cosa+ cad at +7sina+7cosa—... 50,Expand tnxinpowersof(r—1). Ans,(1) (e—I+-5 (eI— 1"7 teat. riesofpowersof ins.=? 42" SI,Expande*inaseriesofpowersof(x-+2). Ans.[DS] . x 52.Expand costsinaseriesofpowersof(x—4). n(n"7 Ans.44-1 eco. 58,Expand4inaseriesofpowersof(+1).Ans.Sntneeti" (-2<%<0).= 54.Expandtanxinaseriesopowersot(x—) Ans.142(2—4E)4+2(1-4) sheeWritetheretfourtermsoftheseriesexpansion, im,powers,ofxthe following funetions: 65.tans, Ans,x44 HE... Exercises onChapter XVI ™m yy wat sist 5.4,Ans,(1F4FT...) cctan ateit 57. Ans. Lbt5bGaytee Sinden. Ans.nS 4, In eos80,24, Ans. het EAE. e48 60.(4a Ans. ete to, 61see.Ans1 wove 2,Incos, Ans. —E BS 63.Expandsinkx.inpowersofx.Ans.kx—Ws},(est(kel 4,Expand sintsJnpowers ofanddelgyming, theinterval ofconverencesAns,OEESEpmtt. Theseriesconver.es for allvalues of 65,Expandaanaseriesinpowersofx,Ans.Lmxttat—atpe.68, Expand arefanz inaseties inpowers ofx. nl,Takeadvantage ofthe formulaarctan Ppa Ants2-5 soe 4E-Fte. Clercn. 62,Expand tay isees ofpower An,MBean Gl<r<p.sngtheformulasforexpansionoftheinctionef,sin,com,i(-+2)and(1-f)™ into power series and applying various procedures, expang theictiowikgfunctionsinpowerseriesaeddetermineihefaervalsofconvergence! eye # 08,sinhx, Ans EEHEbo. (eo<x<c).6coshe,AnsLE a eeht Cecece, Meconte dn14hSeem (—@<<o),Th.(12)In(142).Ans.ALCva (kt<D. 7.ene Ans14D(ust at(ew<r<e). %Gh. 25° m2 Series ans.SA x<VB. HSE. ans.eee pe-Lars . atatetat (-@<r<2), 1%.gaAns.Satie" (zl<).18.esins, Ans,xt AYRsine (wercey I 12 Lae gt1e3..-20—1) 7.e+VIF8 Ans. t—3a trate HOIay amet In(l+2) SetX#pites Clerc. geata dx.Ans.Sms ielen, 7.(221%G,an,Swe tceen 5‘)etAne,STR cost Se 2 80.ja Ans.Coins NoGur (8<*<0and (ax oyata 7 O<xr<o).BtSs. Ans.JS—5. 62Provetheequations sin(a+x)=sinacosx+-cosa sinx,cos(a-+-x)=cosacosx—sina sinx byexpanding the left sides inpowers ofx Gtllising “appropriate series,“compute: 83.cos10°tofourdecimals.Ans. 0.9848. 84, sin1°tofour decimals. Ans. 0.0175.85,sin16totinesdecimals.Ans.0.909.88,sin“tofourdecimals. fsa27 2acctan tooardnclinh, An.97%,08tnt ts decimals, Ans. 1.609. 89,log,, 5tothree decimals. Ans. 0.699. 90.arcsin1to within 0.0001. Ans. 1.5708. 91.Vetowithin0,001.Ans.1.6487.92.logeto Within O.0000i.-Ans. 0.43429, 93, cos towithin 0.00001.” Ams. 0.5403, Using aMaclaurin series expansion ofthefunction f(x)=/a"Fx, compute towithin 0001: 04.j/30. Ans.3.107. 98,V7. Ans. 4.121,96.3/80,Ans.7.997.97.§/FHD.Ans,3.017,98.VB.Ans.9.165.99.3/FAns.” 1.2508 Expanding the integrand inaseries, compute the integrals: 100,[82dstofivedecimalplaces.Ans.0.94608.101.[e-*dxtofour decimal, Ans.07468, 102,[sage)detofourdecimals,Ans.087i Exercises onChapter XVI 713 ae (arctanx 103,[e¥Fadx totwodecimals. Ans.0.81.104|“F"*aetothreedeci- malplaces. Ans.0.487.105.[cosVzde towithin 0.001. Ans.0.764, x 1of 106.fin(1+Vx)dxtowithin0,001.Ans.0.071,107.fedxtowithin sing oom, Ans.ona. woe.(GATZae towithin00001. Am.0.021, 10s,|fastowithin0.001,Ans,0494.110,(OEDae,Ans. Nole.Whensolvingthisexerciseandthetwofollowingonesitiswellto ‘s fonsWe PSL At bearinmindtheequations:LaF5Loan Lens which will beestablished inSec. 2,Ch. XVit, 1,fM=Bae, an8, Itede x ma,fingteS. ans. Integrating Differential Equations byMeans ofSeries 143. Find thesolution ofthe equation y'=xy that satisfies the initial conditions forx=0, y=, y’=0- . Hint,Lookforthesolution Intheformofaserie, AniLig * a tryst tease cept 114.Findthesolution oftheequation yf-+y'-+y-=0 thatgps theInitiatconditions forr=0,y=0,yaleAns.xmStSam _epttent “155. ay)" 118. Find the general solution ofthe equation eyta+(#—t) mo Hint. Seek the solution inthe form YaALEAREAREoe ™ Series AnsGet|atat|Hee atansingyoone cE4c,RE, Ve Va. .118.Findtheolutionoftheequationxy'-+u'x90,thatstisestheInitialconditionfore=0,yal,90.AneIraRTIRE tb.UteetHint.Thetwolatterdifferential equations areparticular casesofthe Bestel equation 2 Hy+ty=0 fornodandnet. 117. Find thegeneral solution oftheequation fay+24+y=0. Hint, Seek the solution inthe form of@series 2?(a,-a,-+a,x"4...). Ans. C,cosVx-+C, sinVx,118.'Findthesofutionoftheequation(29of—sy=Oshat,satisfiesCer oo xe0andym.Ans.cttSytSey t+3aeTte11.Findthesolutionoftheequation(I++y'-+2xy'=0thatsatisfies theinitialconditions y’=1whenx=0andy=0.Ans.2545-44...120, Find thesolution oftheequation y”—xyy" that salisfies the initial conditions y'=1whenx=0andyol.Ans.1teth42hSy 121,Findthesolutionoftheequation(I—)y'=1-t#—y ‘that‘satises tne! inital 'conditons y=0when"0,and.indicateteialrvalofconver: genceoftheseriesobtained, Ans.e+25455+3G+... (Clerc, 122, Find thesolution oftheequation ay/+y=0 that satisfies the initial conditions y=1when,2=0andg=0,andindicate theinterval ofconver ence, Ans. eT taS Gat (Ce<*<e) 12.Findthesolutionoftheequationy+y’-+y=0 thatsatisfiesthe initialconditions y/=1whenx=0,y=.Ans.SE, 124.Findthesolutionoftheequationy'-++y'-+y==0 thatsatisfiestheinitialconditions y’=0whenx=0andy=1,andstleatetheintervalofconvergence oftheseriesobtained, Ans,IS-tghagra...(xi<@). Exercises onChapter XVI 1 Find the fits thre terms ofthe expansion Inapower series ofthe solu-totofthefollowingdiferentequationsfortheRivenntl”coniiens125.y'aattys forx0,yal.Ans.tettee, 128gtaHebefor20,yal,a0.AnsEE. 127yaysins a ee Find several terms oftheseries expansion ofsolutions ofdifferential equa- tonsinde thendieated ni conor ahyfyf atwhen£0, PaOenfatdmee tet tTgmat1 a4, a whenenandyoohAmspHEAgeetLis oe mee when xn0andyoo,Ans.Latagl eg atm. 3,Yexyf—1 whenx=0andyal,An,IreS42, :mound ye #28) ie 132.y'=e+xywhen x=0 andy=0, Ans, x+:at3tagate CHAPTER XVIL FOURIER SERIES SEC. 1.DEFINITION. STATEMENT OF THE PROBLEM Afunctional series ofthe form $a,cosx+b, sinx-+a, cos2x-+b,sin2x+..., ‘or,more compactly, aseries ofthe form 3+Lee,cosm+6,sinnx), a) iscalled atrigonometric series. The constants a,,a,and 6, (n=l, 2...) are called coefficients ofthe trigonometric series. Ifseries (1)converges, then itssum isaperiodic function f(x) with aperiod 2x,since’sinnx and cosnx areperiodic functions with period 2x. Thus, F(x)=F(e+2n). Let uspose the following problem. Givenafunction epwhichisperiodic andhasaperiod2x. Under what conditions forf(x) isitpossible tofind atrigonomet- ricseries convergent tothegiven function? That istheproblem that weshall solve inthis chapter. Determining the coefficients ofaserles from Fourier's formulas. Let theperiodic function f(x) with period 2abesuch that itmay berepresented asatrigonometric series convergent toagiven function inthe interval (—2, x); i.e., that itisthe sum ofthis series: 1)=$+ DG,cosnx+6,sinnx), 2) Suppose that theintegral ofthefunction onthe left-hand side ofthis equation isequal tothe sum ofthe integrals oftheterms ofthe series (2). This will bethecase, forexample, ifweassume that the numerical series made upofthe coefficients ofthegiven Definition. Statement oftheProblem m7 trigonometric series converges absolutely; that is,that the follow- ing positive number series converges: [S[tlalt lal+lalelolt +tlaltloalte @) Then series (1)ismajorised and, consequently, itmay beinte- grated termwise inthe interval from —x tox,Letustake advan- tage ofthis forcomputing thecoefficient a,. Integrate both sides of(2)from —ax to+x: Jrerde= [Sart (Jaycosnxde-+ [b,sinnxde). Evaluate separately each integral ontheright side: %sdeena,J$dranay Ja,cosnxdea,{cosmxdraS824" 0; Somsinnede= 6,{sinnxde—b, |"=0. Consequently, Sie)de=na,, whence“ a=+JFeayde. CO) To calculate the other coefficients ofthe series we shall need certain definite integrals, which wewill consider first. In and &areintegers, then wehave thefollowing equations: ifnk, then . Jcosnxcoskxdx=0; §cosnxsinkxdx=0; © §sinnxsinkxdx=0; 778 FourierSeries but ifn=&, then : §costkxde=x; §sinkxcoskxdy=0; ay fsintardxaa, Totake anexample, evaluate the first integral ofgroup (1). Since 0snxcoska=+[cos(n+4).x-+C08(n—E)x], itfollows that [cosnecostede— 7{cos(n-+h)xdx+{cos(n—b)xdx—0, The other formulas of(1)*) areobtained insimilar. fashion, The integrals ofgroup (II) arecomputed directly (see Ch. X). Now wecancompute the coefficients a,and 6,ofseries (2). Tofind thecoefficient a,forsome definite value &#0, mul-tiplybothsidesof(2)bycoskx: F(3)coskx=%coskx+D>(a,cosnxcoskx-+6,cosnxcoskx).(2') ‘The resulting series ontheright may bemajorised, since itsterms donot exceed (in absolute value) the terms ofthe convergent positive series (3). We can therefore integrate ittermwise onany interval. Integrate (2') from —x toa: JF(a)c0skxde=4[coskedet +3L(a,|cosnxcoskxde-+b,|sinnxcoskede). *)Bymeans oftheformulas cosnxsinkx="/,[sin(n+&)x—sin(n—k)x],‘sinnxsinkx="/, [—cos(n-+&)x4cos(n—A)x). Definition. Statement oftheProblem 19 Taking into account formulas (11) and (I),wesee that allthe integrals onthe right areequal tozero, with theexception ofthe integral with coefficient a,.Hence, JF(x)coskxdx=a,|costkxde=a,x, whence” ~ a=t))F(x)coskxdx. © Multiplying both sides of(2)bysingx and again integrating from —x tox,we find Jf(2)sinkxdx=6,§sin*kxdx=6,7, whence~ ~ b=fF(e)sinbeds, © The coefficients determined from formulas (4), (5)and (6)are called Fourier coefficients ofthe function f(x), and the trigono- metric series (1)with such coefficients iscalled aFourier series ofthefunction (x). Letusnowreverttothequestion posed.atthebeginning, of this section: What properties must afunction have sothat the Fourier series constructed for itshould converge and sothat the sum oftheconstructed Fourier series should equal thevalues of thegiven function atcorresponding points? We shall here state atheorem that will yield sufficient conditions forrepresenting afunction f(x) byaFourier series. Definition. Afunction f(x) iscalled piecewise monotonic onthe interval a,6]ifthis interval may bedivided byafinite number ofpoints x,,X,,..-, X,-, into subintervals (a,x,), (X. X)reeey (Xq-y, 6)such that thefunction ismonotonic (that is,either nonin- creasing ornondecreasing) oneach ofthe subintervals. From thedefinition itfollows that ifthefunction f(x) ispiece- wise monotonic and bounded onthe interval (a,6],then itcan have only discontinuities ofthe first kind. Indeed, ifx=c is apoint ofdiscontinuity ofthefunction f(x), then byvirtue ofthe monotonicity ofthe function there exist the limits Jimf(@)=/(e—0), limf@)=f(€+0), i.e, the point ¢isadiscontinuity ofthe first kind (Fig. 356). 780 FourierSeries Wenow state thefollowing theorem. Theorem. Ifaperiodic function f(x) with period 2nispiecewise monotonic and bounded ontheinterval |—x, x},then theFourier series constructed forthis function converges atallpoints. The sum oftheresultant series s(x) isequal tothevatue off(x) atthe discontinuities ofthe function. At the “aj discontinuities off(x), thesum oftheseries isequaltothearithmetical meanofthe (0dlimitsoff(x)ontherightandontheleft; foro)that is,ifx=c isadiscontinuity ofthe few function f(x), then 5(2)goq=LO=OENELD, al& x From this theorem itfollows that the Fig,356. class offunctions that may berepresented byFourier seriesisratherbroad.Thatiswhy Fourierseries have found extensive applications invarious divi- sions ofmathematics. Particularly effective useismade ofFourier series inmathematical physics and itsapplications tospecific problems ofmechanics and physics (see Ch. XVIII). Wegive this theorem without proof. InSecs. 8-10 wewill prove another sufficient condition fortheexpandability ofafunc- tion inaFourier series, which condition inacertain sense deals with anarrower class offunctions. SEC 2.EXPANSIONS OF FUNCTIONS IN FOURIER SERIES ‘The following aresome instances oftheexpansion offunctions inFourier “Example 1.Aperiodic function f(x)withperiod 2isdefined asfollows: Iwas, —acren. This function ispiecewise monotonic and bounded (Fig. 357). Hence, it admits expansion ina Fourier series. Byformula (4), Sec. 1,wefind if atinanyfedeEfno. y im me Fig. 357. parsons ofFurtions inFeuer Serie rat Applying formula (5), Sec. 1,.and integrating byparts, wefind 1 1 sin kx |= 1a malfscsmecen dE §sieteao Byformula (6),Sec. 1,wehave L¢ _a coske #1 ean2 nookPramtearnt[eetatfsteaeJaco2. Thus, weget the series wu[sts _siu2e 3e__ppen sinks Heya[Se SE(teEE, This equation occurs atall points except points ofdiscontinuity. Ateach Theothnatgr esa otheSee eet Mee etna ean oF fini ORE dha et whic ke Example 2.Aperiodic function (x) with period 2xisdefined asfollows: [@enr when wcrc, f@)ex when 0<zcm lor/(2)=II](Pig.858)Thisfunctionisalsopiecewisemonotonicandbound- 1ealRaf29it _y aaa,Se eae ar Fig. 298 Let usdetermine is Fourler oetctents: 7 mre Aantfroant[ f(-nartfete]on, cred]ficneostceet scot]= 1xsinkx|1¢ xsinkx|x1%wd-SeEEtdJaneane Janne] Lf cose cosaaa[-S" [Lt] = : afer &even,wdeaman-{_ ont Ey aa!‘Akodd; m Faure See netfcmaneees feumearae We tus tan thesnedfoneOEgcotbegycoltedie 1-5-4 [Tg tgtetpi tee| gant twace Ta)=1 for Ocean. This unin Fe. 358 Iplsele mootne andDonde ote Ital mone y | ——- a me 98, Let uscompute tsouter costs wot[romnt[fiversee]ae ife ¢ sinkx 0sinkx|rovefcrenneees fener]1stefan fe ¢‘e 1[coskxj*_coskxjrsootfrearaes fannie] (Pf2]=: 0traeen==;[1—cosx)-{4 i A,ark e twat (Ree Ee ee... This equation hols aallpnt with teexception ofdonna ge aie Tear Se Shar SA Ng re and more accurately the function f(x) asn—+co. aye SA Gla Wa [0 eis pvod 2m eed ows: Tees onerecn Oe oh ! f sefonn S a. Fie 0 atfadeLEP oteang fatter Sfa: ir, |xtsinarje26 antfscortex[antfang[etneae]= 2 xcos kx|n1fa =-3[SEPeffeoteu|=gine ‘-|iAtor&even, A Atr boa a1fsanneaen 4|—zteostee 12fFtookFartenraenst|2pfFronteee|= ~a(ss" rafsnteae|o re FourteSere Thus, theFourier series ofthegiven function hastheform waGna(SpE ote.) tion isfulfilled atallpoints. |“bn -4n -3n -2n i a fn on 4m Sk Fig. 361. Putting =a inthe equality obtained, wegel toyedsx a Example 5.Aperiodic function f(x) with period: 2xIsdefined asfollows: f)=0 for —mex<0, Pab=x for Ocxecn (Fig. 362, ne an in oe x on an an ome ig. 562, Determine the Fourier collet i¢ fe i Latoxcannyroad carr}eeetgs i¢L|xsinkx it svePecsncent]255244Fateae| 2 Hate-(eforhkodd, aeEle 0for&even; ARemark onthe Expansion ofaPeriodic Function inaFourier Series 7&5 tootfeomenaont [ft feewae 1 ree ee a =A for&even.z ‘The Fourier series will thus have the form HR_2(cosx,cos3x|cosSx sinsin2x|sin3xlearmales(SeeSpeSpe+.)+(SEe.) Atthediscontinuities ofthefunction f(2), the sum oftheseries 1sequal to thearithmetical meanofitslimitsontherightandleft(inthisease,tothe number), Putting 2=0 inthe equality obtained, we get t_y 1 SEC. 3.REMARK ON THE EXPANSION OF APERIODIC FUNCTION IN AFOURIER SERIES ‘We note the following property ofaperiodic function y(x) with period 2n: * Aasn Svcde= |pide, os cs no matter what the number A. Indeed, since E—2n)=H(8) itfollows that, putting x=§—2n, wecan write (for allcandd): 4 doan dasn donSvwmde= fye—2md= |v@at= [pear? even chin etn Inparticular, taking c=—x, d=A, weget” A dean Jverde= |year, 786 FourierSeries therefore, hese -s 5 dean§va@de= Jyeydet fp(det [yede= zx on 2 on 2 a * 5sv(odet fplaydet |pleyde= Jpaydx. Thisproperty meansthattheintegral ofaperiodic function 1(x) over any interval whose length isequal totheperiod always has thesame value. This fact isreadily illustrated geometrically: the cross-hatched areas inFig. 363 areequal. wa” aPiear Bane oT Fig. 263, From the property that has been proved itfollows that when computing Fourier coefficients wecan replace the interval ofin- tegration (—z, x)bythe interval ofintegration (A,4-+2n), that is,wecanputdean asmaati)fide,agetJF(2)cosnxdx, rea‘ 0 <4i)Fla)sinnxdx, |x J where 4isany number. This follows from thefact that thefunction [(x) is,byhypothe- sis, periodic with period 2x; hence, both thefunctionsf(x)cosnx and f(x)sinnx are periodic functions with period 2n.Wenow illustrate how this property simplifies theprocess offinding coef- ficients incertain cases. Example. Let itberequired toexpand inaFourler series thefunction ta) with period 2x, which isgiven ontheintervalOcxxbytheequation fay=s, ‘Thegraph off(2)isshown inFig. 364,Ontheinterval (—x, x)thisfunc.tion isrepresented bytwo formulas: f(x)=x+2n onthe interval (—z, 0]andj(z)=x ontheinterval (0,x].Yet,on(0,22)itisfarmoresimply Fourier Series forEven and Odd Functions rer epresented byasingleformulaf(«)—x.Therefore,oexpandthisfunctionTeRourerdriesitsellertoloususeofTomi (ipsetting Nos onbfp09deeLfcent an1cosedemEVaconmedem2[EMEA4SRE]Ao, reedfronsaardem tfsamacaent [teem] 2. Consequently, [avaa—2sins—2 sn2x2snae—2sinte2ax. nO 9onanonon6xmmOnx Fig. 964. This series yields thegiven function atallpoints with theexception ofpoints ofdiscontinuity (i,e.,except thepoints x=0, 2x,4x,...). Atthese points the Sum ofthe series isequal {0°the half_sin ofthe itmiting values ofthe function f(x) ontheright andontheleft(tothenumber i,inthiscase). SEC. 4,FOURIER SERIES FOR EVEN AND ODD FUNCTIONS From thedefinition ofaneven and odd function itfollows that ifp(x) isaneven function, then Swlaydx=2[y(xyar. Indeed,” . Jpde= Jvenar+|peae=(p(—aydet fy@odr= =Jve)det[peydr=2 [w(ayds, since bythedefinition ofaneven function p(—x)=1p(x). 138 FourierSeries Itmay similarly beproved that if@(x) isanodd function, then §@(e)dx=( @(—x)dx+{p(yde=—f o(x)de+[ p)de=0. Ifanodd function f(x) isexpanded inaFourier series, thentheproductf(x)coskxisalsoanoddfunction, whilef(x)sinkxisaneven function; hence, a=ti)F(x)dx=0; gatFrercoserar=o ay * } b=ESMx)sinkxde=2 (f(x)sinbede. Thus the Fourier series oranodd function contains “only sines” (see Example 1,Sec. 2). Ifaneven function isexpanded inaFourier series, thepro- duct f(x)sin kxisanodd function, while f(x)coskx isaneven function and, hence, = ) a,=2 |f(a)de, a=2Jf(x)coskedx, @) b=Jfe)sinaxde=o.on J Thus, the Fourier series ofaneven function contains “only cosines” (see Example 2,Sec. 2). The formulas obtained permit simplifying computations when seeking Fourier coefficients incases when the given function is even orodd. Itisobvious that notevery periodic function is even orodd (see Example 5,Sec. 2). Example. Let it,berequired toexpand inaFourler series the evenfunction} ()whichhasaperiod of22andontheinterval [0,)isgivenby the equation yes, The Fourier Series foraFunction with Period 2 789 WehavealreadyexpandedthisfunctioninaFourierseriaipExample2, Sec. 2.(Fig. 358), Letusagain compute the Fourier series ofthis function, faking advantage oftheTact that thegiven function iseven. Byvirtue offormulas (2)6,=0 forany &; ak(xdeam, a=2(xcoshdr 0forbeven, 2[xsinkbe cos kx] 2 . a We,oblained thesamecoeffiients asinExample 2,Sec.2,butthistimeby SEC. 5.THE FOURIER SERIES FOR AFUNCTION WITH PERIOD 2 Let f(x) beaperiodic function ‘with period 2i,generally speaking, different from 2x.Expand itinaFourier series. ‘Make asubstitution bythe formula xoit Thenthefunction f(4) willbeaperiodic function of#with period 2n. Itmay beexpanded inaFourier series onthe interval —nexen: fies)=$+©(aycoskt-+54sinkf), a) a where 1f u if ua=t (F(Lt)da—t JF(£4)coseeae, if ia=t JF(£4) sinaeae, Now letusreturn tothe original variable x: 1 x x xapt text, dt=Fdx 7% FourterSeries We will then have t 1 ant)Teds,ayng|Te)skFeds,a i. ®b=FfMe)sinkFede. 5} Formula (1)takes the form ay. Ae Afo=3+ (%cos“*x-+6,sinwr), 8) where thecoefficients a,,a,6,arecomputed from formulas (2). This istheFourier serits foraperiodic function with period 2, We note that all the theorems that hold for Fourier series of periodic functions with period 2n hold also forFourier series of Fig. 96. periodic functions with some other period 2/.Inparticular, the sufficient condition forexpansion ofafunction inaFourierseries (seeendofSec,1)holdstrue,asdoalsotheremarkonthepossibi- lity ofcomputing coefficients oftheseries byintegrating over any interval whose length isequal tothe period (see Sec. 3),and the remark onthepossibility ofsimplifying computation ofcoefficients oftheseries ifthe function iseven orodd (Sec. 4). Example. Expand inaFourier seriestheperiodic function £2)withperiod 21whtch'on theinterval [is 7)fsgiven bytheequation f(2)=11(Pay96) Solution, Stnce the funtion atHand iseven, ifellows that net a2frtens 1 x ©for&even, eBfronMaeeenteaee (atey (OntheExpansion of@Nonperiodie Function inaFourier Series 791 Hence, theexpansion isoftheform x da Cr+nxmegafte. oe lal tt tape SEC. 6.ON THE EXPANSION OF ANONPERIODIC FUNCTION INA FOURIER SERIES , Let there be given, onsome interval [a,6)apiecewise mono-toniefunction. f(x)(Fig,366).Weshallshowthatthisfunction F(x) may berepresented inthe form ofasum ofaFourier series atthe points ofitsdiscontinuity. Todothis, letusconsider an arbitrary periodic piecewise monotonic function f,(x) with period 24>|b—al, which coincides with thefunction /(x) ontheinter- val {a,6}.[We have redefined the function f(x).) y i 10), t ' a Aa DR wa Fig. 366. Expand f,(x)inaFourier series. Atallpoints oftheinterval {a,6](with' the exception ofpoints ofdiscontinuity) thesum of this series coincides with thegiven function f(x); inother words, weexpanded the function f(x) inaFourier series onthe interval la,6}. Let’ usnow consider the following important case. Let afunc tion {(x) begiven onthe interval (0,/).Redefining this function inarbitrary fashion onthe interval [—/, 0](retaining piecewise monotonicity), wecan expand itinaFourier series. Inparticular, ifweredefine thisfunction sothat when —/<x<0, [(x)=/(—x), wewill getaneven function (Fig. 367). {Inthis case wesaythat the function f(x) is“continued ineven fashion”.| This function isexpanded inaFourier series that contains only cosines. Thus, we,haveexpanded incosines thefuneton (x)givenontheinter val 0, 4. 792 FourierSertes Butifweredefine thefunction f(2)when—Le¢x<0 asfollows 1()= —F(—2), then wegetanodd function which may beexpan- ded insines (Fig. 368). [The function f(x) is“continued inodd fashion”. u 4 100, N4MN H Tp i i i a4 Fig. 367. Fig. 968. Thus, ifontheinterval (0,{]there isgiven some piecewise monotonic function f(x), itmay beexpanded inaFourier series both incosines and inSines. Example1.Letitberequired10expandthefunction f(x)exinaseri insinesontheinterval (Qn) ti) . Solution. Continuing ‘this function inodd fashion (Fig. 967), wegetthe seriesD ap[saz_sia2e sindean2[SS | (sce Example 1,Sec. 2). wecEtspnte 2;"Expand thefunction f(a)—x In9sviesIncosines onthe erval 10,3. igoitloe. Continuing thisfunction Inevenfashion, wegeh Hajelel, —a<xca (Fig. 358), Expanding itinaseries wefind _A[eos,cos3e,cos5x formZ—4[SAey] (ee,Example 2See.2Andsoontheinterval (,a]wehavetheeque- ion_A[eos,cosSx,cosSe xngt (S24e+Se+...] . SEC. 7.MEAN APPROXIMATION OF AGIVEN FUNCTION BY ATRIGONOMETRIC POLYNOMIAL Representing afunction byaninfinite series (Fourier’s, Tay- lor's and soforth) has the following meaning inpractice: the finite sum obtained interminating theseries with thenth term Mean Approximation ofaGiven Function 793 isan approximate expression ofthe function being expanded. This approximate expression may bemade asaccurate asdesired bychoosing asufficiently large value ofn,However, the charac- ter_of theapproximate representation may differ. For instance, thesum ofthefirst terms ofaTaylor's series s,coincides with the function athand atone point, and atthis point has derivatives uptothe nth order that coincide with the derivatives ofthe function under consideration. Annth degree Lagrange polynomial (see Sec. 9,Ch. VII) coincides with the function under consideration atn+-1 points. Let ussee what thecharacter isofanapproximate represen- tation ofaperiodic function f(x) bytrigonometric polynomials ofthe form 5,(1)=B+Ya,coskx+bysinkx,cot where dy,a,54,dy,by+» GyOyareFourier coefficients; thatis,bythesum'of the’firstnterms"of aFourier series. Wefirst make several remarks. Suppose weregard some func-tiony=/(x)ontheinterval(a,6] oo @and want toevaluate theerror <a’ when replacing this function by another function @(x). For the measure oferror wecan, for in- stance, take max |f()—g()| >‘ontheinterval [a,6],which is vv theso-called mazimum devia- Fig.369. tion of@(x) from f(x). But itis sometimes more natural totake for the measure oferror the so-called roof mean square deviation 6,which isdefined bythe ‘equation ief=GagsUe—ecotar. Fig, 369illustrates thedifference between theroot mean square deviation and the maximum deviation. Let thesolid line depict thefunction y=f(x), thedashed linestheapproximations 9,(x)and9,(x).Themaximum deviation ofthecurve y=9, (x)isless than ofthecurve y=, (x),butthe root mean square deviation ofthefirst curve isgreater ‘than the second because thecurve y=, (x)isconsiderably different from 74 FourierSeries the curve y=f(x) only onanarrow section and forthis reason characterises the curve y=f(x) better than the first. Now letusreturn toour problem. Let there begiven aperiodic function f(x) with period 2x. From among allthetrigonometric polynomials oforder n $+LX(a,cosex+B,sinkx) a itisrequired tofind (by choice ofthecoefficients ayand ,)that polynomial forwhich theroot mean square deviation defined by theequation aaa)[ro-9-Eew coskx-+B,sin9]dx, has the smallest value. ‘The problem reduces tofinding theminimum ofthe function 2n+1 ofthevariables a,,a,,...,GyByBy-+++Bar Expanding the square’ under theintegral sign and integrating termwise, weget atl{reo—see[ 3+Boscortr+sina| +[5+Zteemieeasintn |Vem =H)Pod—E)perde—PLen! M2)costedet a« att wo +7d HG)sinkedet+ a7Jdx+72dai)cos"kxdx+fd, 2,|Mitde +7DBSsintkxd+, Yaycoskedx+ md, tad, tundoe)sinkxde+YeSow,coskxcosjxde+~ neh ~ +£4DLDaB,fcoskesinjxdx+4)1,6,sinkssinjede,hes ies a het jen ey tei ‘Mean Approximation ofaGiven Function 795: We note that zsfx)de=a, +)F(x)coskxde=ay; x)f(x)sinkxdx=b, are the Fourier coefficients ofthe function f(x). Further, byformulas (I)and (II), Sec. 1,wehave: fork=j i)costhxdx=a, {sin*kxde=n, i)sinkxcosjxdx=0; forkj Jc0skxcosjxdx—=0, Jsinkssinjrde=0. Thus, we obtain soit p »_¥ s Fad 4-83ag) Pedde— -z(at+Bibdta +4=(ai-+-Bi). ‘Adding and subtracting thesum Z+TD GG+04), we will have sma) Fedde 2-3DetODF(00+ +3Dilla +Gx60) a mt Thefirstthreetermsofthissumareindependent ofthechoice ofcoefficients a,,@,,..-, GyBy,---»ByThe remaining terms Fa) TEOra) +Bab] 796 FourierSeries are nonnegative. Their sum reaches the least value (equal tozero) ifweput a=a,, a=a, ...,@,=Ay By=dy --.»Bye Oe With this choice ofcoefficients a,,@,,-+-,Gy»By«++»Bythe trigonometric polynomial B+Zi(ascosx+B,sinkx) will least ofalldiffer from the function f(x) inthe sense that in such achoice ofcoefficients thesquare deviation 63will beleast. We have thus proved the theorem: Ofalltrigonometric polynomials oforder n,that polynomialhas theleast root mean square deviation from thefunction f(x), the coefficients ofwhich polynomial aretheFourier coefficients ofthe function f(x). The least square deviation is rod tay ae eediagJFedeF—zLeto. 2) Since 6;>0, itfollows that forany nwehave (p LD ataptHSP@de>F445Dal+oH. plim Hence,theseriesontherightconverges (whenn—oo),andwecan write x)Podez+Dealtoh, ® This relation iscalled Bessel's inequality. We note without proof that forany bounded and piecewise monotonic function theroot mean square deviation obtained upon replacing thegiven function bythenth partial sum oftheFourier series tends tozero asn—oo, that is,64-0 asn—eo. But then from formula (2)there follows theequation at bt Cp .T+HDarw=z) Fide, co) which iscalled the Lyapunov equation. (We note that A.M. Lyapunov proved this equation even forabroader class offunc- tion than that which wehere consider.) ‘Mean Approximation of@Given Function 70 From what has been proved itfollows that for afunction which satisfies the Lyapunov equation (in particular, for any bounded piecewise monotonic function), the corresponding Fourierseriesyieldsarootmeansquare deviation equaltozero. Note. Let usestablish aproperty ofFourier coefficients that will beneeded inthe future. We first introduce adefinition. Afunction f(x) iscalled piecewise continuous onthe interval [a,6]ifithasadefinite number ofdiscontinuities ofthefirst kind onthis interval (oriseverywhere continuous). Weshall prove thefollowing proposition. Ifafunction f(x) ispiecewise continuous onthe interval[—x,x],thenitsFouriercoefficients approach zeroasn—00;that is, lima,=0, limb,=0. Oy Proof, Ifthe function [(x) ispiecewise continuous onthe in- terval [—x, x], then thefunction f*(x) tooispiecewise conti- nnuousonthisinterval. Then{/*(x)de exists andisafinite number). Inthiscase,fromtheBessel inequality (8)itfollows thattheseries $1(az +04) converges. Butiftheseries conver- gesthen itsgeneral term approaches zero; inthis case,lim(a3+63)=0.Whencewegetequations (4)directly. Thus,the foliowing equations are valid for apiecewise continuous and bounded function: timJF(x)cosnxdx=0, limFf(2)sinneds0. Ifafunction f(x) isperiodic with period 2a, then thelatter equations may bewritten asfollows (for any a): limJf(x)cosnxdx=0; limJf()sinnxde=0. qT lnkcral maybepresented asthesumofdefinite integrals ofconetinuous,tunctionsoverthesubintervals intowhichtheinterval-=a,a)Is 798 FourierSerle Wenote that these equations continue tohold ifintheintegrals wetake any arbitrary interval ofintegration (a,6],which isto say that the integrals A : Jie)cosnxdx and§f(x)sinnede approach zero when nincreases without: bound if[(x) isabound-edvand piecewise continuous function. Indeed, taking 6—a<2n fordefiniteness, weconsider theauxi- liary function g(x) with period 2ndefined’ asfollows: a) =f) when acx<b 9%) =0 when b<x<a+2a, Then . oom SiG)cosnrdr= J@(x)cosnedx, ° asin SiG)sinnzde= {p(x)sinnxde, Since @(x) isabounded and piecewise continuous function, theintegrals ontherightapproach zeroasn—oo.Hence,thein-tegrals ontheleft approach zero aswell. Thus, theproposition is proved; that is, : ’ limJf(x)cosnedx=0; lim{f(x)sinnxde=0 6) forany numbers aand 6and any piecewise continuous function F(x) bounded on[a,5). SEC, 8,THE DIRICHLET INTEGRAL Inthis section weshall derive aformula that expresses the nthpartial sumofaFourier seriesintermsofacertain integral. This formula will beneeded inthe subsequent sections, Consider the nth partial sum ofaFourier series fortheperi- odie function f(x) with period 2n: 540)=B+Si(aycoshrtbysinkx), where a=) J(costdt,ats F()sinkt dt, on os The Dirichlet Integrat 9 Putting these expressions into the formula fors,(x), weobtain VE 5.)=35|F)dt+ FD[EYreocosted+82) posinaat], Fo d, oe orbringing coskx and sinkx under the integral sign (which is possible since coskx and sinkx areindependent ofthe variable ofintegration and, hence, can beregarded asconstants), weget 5.)=)10dt+ +42[JF(t)coskxcosktasf10}sinkesinktdt). Nowtaking +outside thebrackets andreplacing thesumofine tegrals bytheintegral ofthesum, weobtain s@=25 {+E [F(0coskxcoskt+f(t)sinkxsin}dt, dnEo or S(t)=+5100]$+EcoAtcoskx+sinkfsina= pem -t)10[$4389 ]a, w Transform theexpression inthebrackets. Let 6,(2)=+0082-4cos22-+...+005nz; then 2o,,(2)cosz=cos2+2coszcosz+2coszcos22-+«27+12e082c08nz=cosz-+(I+c0s22)+(cos2-+c0s 32)+++(cos22+-c0s42)+...+[cos(n—1)z+cos(n-+1)2]==142cosz-+2cos 2+... +2cos(n—1)z-+cos nz+cos (n+1)z 0 Fourier Series or20,(2)cosz=26,(2)—cosnz+-c0s(n+1)2, ont ee, But cosnz—cos(n +1)2—=2sin(2n+1) 5sinZ, 1—cosz=2sin*Z Hence, sin20-1) 9,@)=——*. Paine Thus, equation (1)may berewritten as tx Hi‘sin(2n+1) 5,@)=+) f(0——*at. ahi 2sin * z Since the integrand isperiodic (with period 2n), itfollows that the integral retains itsvalue on,any interval ofintegration oflength 2x,Wecan therefore write tox odsin(2a+1)= s)=4 |Q)—— at. maaa Introducing anew variable a,weput t—x=a, t=x+a, Then weget the formula Fa sin2a)s)=2) He+e)—— da. @py 2sin The integral onthe right isDirichlet's integral. Inthis formula put f(x)==1; then a,—2, a,—0, b—0 when k>0; hence, s,(x)=1 forany nand wegettheidentity &sina) 1-4) —— aa, ®ad2sa> which we will need later on, The Convergence ofaFourier Series ataGiven Point ot SEC. 9,THE CONVERGENCE OF AFOURIER SERIES AT AGIVEN POINT Assume that the function f(x) ispiecewise continuous onthe interval [—z, =] Multiplying both sides ofthe identity (3)ofthepreceding section byf(x) and bringing f(x) under theintegral sign, weget theequation zsin@ntS 1z 10)=4f1) aa, o. Bind Subtract theterms ofthelatter equation from the corresponding terms of(2)ofthepreceding section; weget 1% sinQn-$1)F 58)—10) =)Ufe+«)—f)]—— aa, ae zane ‘Thus, the convergence ofaFourier series tothe value ofafunc- tion 'f(x) atagiven point depends onwhether the integral ontherightapproaches zeroasn—co.Letusbreak upthis integral into two integrals: if ot s.)—10) =z)[Fe+a)—f(2)] —Ssinnada+ 22ain +f [f(e+a)—F (x)}cosnada, 2 taking advantage ofthefactthatsin(2n+1)$==sinna cos$4 +cos nasin5.Break upthefirstoftheintegrals ontheright ofthelatter equation into three integrals: ie cogs@)—1=z JUe+e)—f(s)] —*sinnadaos 2sin ~ cos 1z +4)Ue+9—/09]—2+sinnaday os2sinz 1g cos ue 1tyUero—1e sinnada+2{ [f(x+0)—I()}e0snada,sint on 26 3388 a2 FourierSeries Put®,@=ete=te |Sincef(x)isaboundedpiecewise con- tinuous function, itfollows that ©,(a)isalso abounded and piecewise continuous periodic function of«.Hence, the latterintegralapproaches zeroasn—-oo,sinceitisaFouriercoeffi-cient ofthis function. The function cos ,(@)=f+0)—f()]—230 isbounded when —x<a<—6 and 6<a<q and Jo,@)<(M+mj—,, Pain where Misthe upper limit ofthe quantity |/(x)|. Also, the function ®,(a) islikewise piecewise continuous. Hence, by’for- mulas (5)ofSec. 7,the second and third integrals approach zero as n— oo. We can thus write ’ ae im[s,(2)—f (@)]=lim+)Ve+o)—F (9)—* sinnada.(1) mene), Dome Inthe expression on the right, the integration isperformed over the interval —b<a<6; consequently, theintegral isdepen- dent onthe values ofthe function f(x) only inthe interval from x—6 tox+6. An important proposition thus follows from the lalter equation: the convergence ofaFourier series atagiven point xdepends only onthebehaviour ofthefunction f(x) inan arbitrarily small neighbourhood ofthis point. Therein lies the so-called principle oflocalisation inthestudy ofFourier series. Iftwofunctions f,(x) and f,(x) coincide inthe neighbourhood ofsome point x,then their Fourier series simulta- neously either converge ordiverge atthis point. SEC, 10, CERTAIN SUFFICIENT CONDITIONS FOR THE CONVERGENCE OF AFOURIER SERIES Inthepreceding section itwas shown that ifthe function f(x) ispiecewise continuous inthe interval [—a, x],then theconver- gence ofaFourier series atthegiven point x,toavalue ofthefunetion f(x) depends onthebehaviour ofthefunction ina Certain Sufficient Conditions for the Convergence ofaFourier Series 803 certain arbitrary small neighbourhood {x,—6, x,-+6] with centre atthe point x,. Let usnow’ prove that ifintheneighbourhood ofthepoint x, the function f(x) issuch that there exist finite limits tim(tO) w ViLet10) ® while thefunction iscontinuous atthevery point x,(Fig. 370), then theFourier series converges atthis point toacorrespond: ingvalue ofthe function f(x)*). Proof. Let usconsider the fune- 9} tion ®,(a) defined inthe preced- ingsection: oe, MOVto—MelTesSg——y7 since thefunction /(x)ispiece- LD wise continuous on the iaterval [—s, a]and iscontinuous atthepoint x,,itistherefore con- tinugus insome neighbourhood (x,—8, x,+] ofthepoint x,,Forthisreason, thefunction ®,(a) iscontinuous atallpoints where 0340 and |a|<. When'a=0 the function ®,(a) isnot defined. Let usfind the limits lim@®,(a) and lim®,(a), making us> ofconditions (1)and(2): . cof lim©,(a)=tim(f(x,+)—/(e)I—> = one eee 2m timMeetO=10)Fogg own Pa =HimL169) timFim cosShy1-1hye oeose ingoes =)conditions (1)and(2)arefulfilled,thenwesaythatthefuncti(@asatthepoint2,9Gervative'on therightandaderivate “oythe leit.“Fig.370°shows ‘atunction where AyangyAy==tangarheh.th y=hy,thatis,ifthederivatives ontherightandfeft‘areequal,then'theftnetifa will Bedifferentiable atthegiven point. w 208 FourierSeries Thus, ifweredefine thefunction ,(a) byputting®,(0)=é,, thenit’willbecontinuous ontheinterval [—6, 0],and,hence, bounded aswell. Similarly weprove that lim®,(@)=hy. Consequently, thefunction ®,(a) isbounded and continuous onthe interval (0,4]. Thus, onthe interval [—8, 8]the func- tion ®,(a) isbounded and’ piecewise continuous. Now letusreturn toequation (1),Sec.9(denoting xinterms ofx,), it cos lim(sa(¢)FGM=lim+)FG,+e)—F«)] —sinnada oeore ©, dan or 2 lim[5,(%,)—F)]=lim}f©,(a)sinnada,ped nen dy From formulas (5)ofSec. 7weconclude that the limit onthe right isequal tozero, and therefore lim [s,(*)—F(,)] =0 or lims,(x)=F(%)- The theorem isproved. This theorem differs from the theorem stated inSec. 1inthat inthe latter case itwas required, forconvergence oftheFou- rier series atapoint x,tothevalue ofthefunction f(x,), that the point x,should beapoint ofcontinuity onthe interval [—x, 1], whereas thefunction should bepiecewise monotonic; here, however, itisrequired that the function atthe point x,should beapoint ofcontinuity andthattheconditions (I)and (2)befulfilled, while throughout theinterval [—zx, x]thefunc- tion should bepiecewise continuous and bounded. Itisobvious that these conditions are different. Note 1.Ifa piecewise continuous function isdifferentiableat tnepointx,itisobviousthatconditions (1)and(2)are.ful- filled’ Here’” k,ky. Hence, atpoints where thefunction f(s) isdifferentiable, the Fourier series converges toavalue ofthe function atthecorresponding point. Note 2:a)The function considered inExample 2,Sec. 2 (Fig. 358), satisfies conditions (1)and (2)atthepoints 0,+2n, 4x, ...Atalltheother points itisdifferentiable, Consequent- Practical Harmonic Analysis 805, ly,aFourier series constructed foritconverges tothevalue of this function ateach point. b)The function considered inExample 4,Sec. 2(Fig. 361), satisfies conditions (1)and (2)atthe points tx, 3x, -L5x, Itisdifferentiable atallpoints. Itisrepresented byaFourier series ateach point. ©)The function considered inExample 1,Sec. 2(Fig. 357), isdiscontinuous atthe points 2, +3n, ‘6m. Atallother pointsitisdifferentiable. Hence, atallpoints, withtheexcep- ionofpoints ofdiscontinuity, theFourier series corresponding toitconverges tothe value ofthe function atthecorresponding points. Atthediscontinuities, thesum oftheFourier series is equal tothe arithmetical mean limit ofthe function ontheright and onthe left (inthis case, zero). SEC. 11, PRACTICAL HARMONIC ANALYSIS The theory ofexpanding functions inFourier series iscalled harmonic analysis. We shall now make several remarks about approximate computation ofthecoefficients ofaFourier series, that istosay, about practical harmonic analysis. ‘Aswe know, the Fourier coefficients ofafunction f(x) with period 2xaredefined bytheformulas =z) F(x)dx;a=) F(x)coskxdx; b=ESFl)sinkeds, Inmany practical cases, thefunction f(x) isrepresented either intabular form (when the functional relation isobtained by experiment) orintheform ofacurve which isplotted bysome kind ofinstrument, Inthese cases the Fourier coefficients are calculated bymeans ofapproximate methods ofintegration (see Sec. 8,Ch. XI). Let ‘usconsider the interval —x<x<a oflength 2x, This can always bedone byproper choice ofscale onthex-axis. Divide the interval (—2, x]into nequal parts bythe points a a en oes Then the subinterval will be ara, 206 FourierSeries We denote the values ofthefunction f(x) atthepoints x,,x, Xy sss Xq(tespectively) interms of These values are determined either from atable or from the graph ofthegiven function (by measuring the correspondingordinates). .Then, taking advantage, forexample, ofthefectangular for- mula [see formula (1), Sec.’ 8,Ch. XI], wedetermine theFourier coefficients: i Fo ms Diagrams have been devised that simplify computation ofFou- rier coefficients (see, for instance, V.I.Smirnov, “Course of Higher Mathematics", Vol. Il;A.M.Lopshits, “Models forHar- monic Analysis”). Wecannot deal here with thedetails butwe can note that there are instruments (harmonic analysers) which permit approximating thevalues ofFourier coefficients from the graph ofthe function. SEC. 12, FOURIER INTEGRAL Let afunction {(x) bedefined inaninfinite interval (—0o, co) and absolutely integrable over it;that is,there exists an integral . fe@lara. ) Further, let the function f(x) besuch that itisexpandable into aFourier series inany interval (—!, +1): 1e)=3+Ziaycos"txtbysinFx, @) where ( 1( Ae 44-4) H(cosA# tdt,b=7)Hsin+de.—@) * * Fourier Integral 807 Putting into series (2)theexpressions ofthecoefficients a,and b,from formulas (3), wecan write t on 1 bn ax fe=za) fod+s 1(0)cos*¢dt\cos8%x+ uf,1ES1oonteateos 1 +(J)10sindt)sinxm 1 =! mafforeEsio[costiecosAx+sin’¢sin4x]at or 1( i;¢ hs Fey=qf10d+aEN110cos=A a) Let us investigate what form expansion (4) will take whenpassing tothelimitas!—oo,We introduce the following notation: a=5, ga, ga, ...andAgee. 6) Substituting into (4), weget F oy HermaJfOdt+E(S10cosoytat)bay6) As [—+co, the first term on the right approaches zero. Indeed, 1 1 5 11 |xJratl<gJiolde <aJif@lat=ze—o. 4 a eS For any fixed 1,theexpression intheparentheses isafunction ofa,(seeformula (S)],which takes onvalues fromtooo.We will show, without proof, that ifthefunction f(x) ispiecewise monotonic onevery finite interval, isbounded onaninfinite inter- valand satisfies condition (1), then as/—»+0o formula (6)takes the form ra=t)( Jrcosa(t—xydt) da, a 208 FourterSertes The expression ontheright isknown astheFourier integral of the function f(x). Equation (7)occurs forall points where the function iscontinuous. At points ofdiscontinuity wehave the equation +S(JFcosa(¢—2)4x)H=etOFTe—9 (7) Let ustransform the integral ontheright of(7)byexpanding cosa(t—x): cosa(t{—x)=cosafcosax+sinasina. Putting this expession into formula (7)and taking cosax and sinax outside the integral signs, where theintegration isperformed with respect tothevariable t,weget 1m=£5 (free0satdt)cosada+ +45( JFwsinatat)sinaxda, 8) Each ofthe integrals inbrackets with respect to¢exists, since thefunction f(t) isabsolutely integrable intheinterval (—0, 00), andtherefore thefunctions f(t)cosat and f(t)sinat arealso abso- lutely integrable. Letusconsider particular cases offormula (8). I.Letf(x) beeven. Then f(f)cosaf isaneven function, while f(Osinat isoddandwehave JF()cosatdt=2 {F(t)cosatat, JFsinatdt=0. Formula (8)inthis case takes theform 1=2)(Srcosatat) cosaurda, © Fourier tntegrat #9 2,Letf(x) beodd. Analysing thecharacter oftheintegrals in formula (8)inthis case, weobtain 1a)=2(((7sinatdt) sinaxda, (10) IfF() isdefined only intheinterval (0,oo), then forx>0 itmay berepresented byeither formula (9)or(10). Inthefirst case weredefine itinthe interval (—oo, 0)ineven fashion; in the latter case, inodd fashion. Let itbenoted once again that atthepoints ofdiscontinuity weshould write thefollowing expression inplace off(x) inthe left-hand members of(9)and (10): 1+4+1¢—0) LetOtTeo, Let usreturn toformula (8). The integrals inbrackets arefunc tions ofa.We introduce thefollowing notation: A@=+ frocosatdt, B@=t JFWsinatdt. Then formula (8)may berewritten asfollows: Fy=[[A@cosax +B(a)sina]da, aly Wesay theformula (11) yields anexpansion ofthefunctionf(x) intoharmonics withafrequency athatcontinuously variesfrom 0tooo.The law ofdistribution ofamplitudes and initial phases asdependent upon the frequency aisexpressed interms ofthe functions A(a) and B(a). Let usreturn toformula (9).Weset Fay=WVZIFOcosatas; (12) then formula (Q)takes the form 1e=VES F@cosaxda, 3) 810 Fourier Series The function F(a) iscalled theFourier cosine transform ofthe funetion f(x). Ifin(12) weconsider F(a) asgiven and f(0) asthe unknown function, then itisaninéegral equation ofthe function f(t). Formula (13) gives thesolution ofthis equation (Onthebasis offormula (10) wecanwrite thefollowing equations: o@=Y FSresinatat, (14) 10)=V2)0@)sinaxda, (15) The function (a) iscalled theFourier sine transform, Example. Let fene™ (>0, x20. From (12) wedetermine the Fourier cosine transform: eae Ts F@=Vas cosatdt=Vipta From (I4) wedetermine the Fourier sine transform: zt Za o@=V3setsnatdn V2pea. From formulas (19) and (16) wefind thereciprocal relationships 20eeedane «=o, 2fasnar2fgittiacre ya, =SEC, 13. THE FOURIER INTEGRAL IN COMPLEX FORM IntheFourier integral {formula (7),Sec. 12], thebrackets con- tain aneven function ofa;hence, itisdefined fornegative values of@aswell. Onthebasis ofthe foregoing, formula (7)can be rewritten asfollows: Hed=ayJ(J1cosa¢—x) dt)da, a) The Fourier Integrat inComplex Form on Letusnow consider thefollowing expression, which isidentically equal tozero: Moe f(frosina(t—x)dt) da=0.tu Se The expression onthe left isidentically equal tozero because the function ofainthebrackets isanodd function, and anin- tegral. ofanodd function from —M to+M isequal tozero. It isobvious that Moe lim|({F@sina(@—x)dt) da=0 useyd or §(J#@sinat@—xpat) da~0. (2) Note. Itisnecessary topoint tothe following. Aconvergent integral with infinite limits isdefined asfollows: Jeda fg¢arda+f9(a)da— Eaa Ee =lis da+ti ee a aim,J9fa+im{ota)la (o) : provided that each ofthe limits tothe right exists (see Sec. 7, Ch. XI). But inequation (2)wewrote * " =lis da, i) Je@damtim|o(a)da o Obviously, itmay happen that the limit (**) exists, while the limits onthe right side ofequation (*)donotexist. The expres- sion onthe right of(**) iscalled the principal value ofthe in- tegral. Thus, inequation (2)weconsider the principal value of theimproper (outer) integral. The subsequent integrals ofthissection will bewritten inthis sense. Letusmultiply thetermsof(2)by4andaddthemtothe corresponding terms of(1); wethen get ra=z J[J10(cosa(¢—x) +isina(¢—x)at]da a2 Fourier Series or 1o=% J[JFere-nae] da. @) This istheFourier integral incomplex form. Formula (3)may be rewritten asfollows: a=LfF(LF pipemar)em tomva |(7m[foeat)eda, Onthebasis ofthis latter equation wecan write Fr@= refreat, “ aofh pee 1O=7R jr(@e~*da, ©) The function F*(a)defined byformula (4)iscalled theFourier transform ofthe function f(t). The function f(x) defined byfor- mula (5)iscalled the Fourier inverse transform ofthe function F*(q) (the transforms differ inthe sign infront ofi). Exercises onChapter XVIL 1.Expand the following function inaFoutier series intheinterval( —z, ) I(x)=2xforOGxen, fx)=x for —a<x <0. 12 (cos,cos3x,cosSe sinx_sinde aeqed(SiSere...) 49 +42-...)2,Takingadvantageoftheexpansionofthefunction7(x)=1intheinter- vat(0,3) ithe sitesofmoltfple ares,calculate thesumoftheserie dy ans.2, 3.Utilising theexpansion ofthefunction /(x)=x* inaFourier series,compute thesumofthesresya LyAns3 4.Expand thefaction e)=f2—2 inaFourier sereintheinterval (<n,m.Ans.conx—S0524S08E_COBEY Exercises onChapler XVII 813 5.ExpandthefollowingfunctioninaFourierseriesintheinterval(—a,=)Hye"$9or—nce0, Fa)=} ns)for<x<e 1 1 Ans.sinxt-sinDetsinged... 6.Expand inaFourier series, inthe interval (—n, m), the function 1Q)=—x for—x<x<0, 1@)=0 forOcxca, H_ 2FV costs pgsian aF-35 SSLy aaa 7.Expand inaFourier series, intheinterval (—x, m),thefunction f@)=1 for —1<x<0, He)=—2 for0<rece. 1_6y sin(20-1) x ae-2-4h Wael . 8,Expand thefunction /(x)=x%, intheinterval (0,2),inaseriesofsines. . jx?2 aos,2 {4-3(ormai}nas,9.Expandthefunctiony=cos2xinaseriesofsinesintheinterval(0,7).4[sinx,Seine,SsinSx as—t[Fee]. 10,Expand thefunction y-—sin sinaseriesofcosinesintheinterval(0). 4°Fcos2x,costy nssheets].11,Expand the function y—e* inaFourier series intheinterval (—1, J. tt =(—1)"sia 2ene a 7 Ans,SPee)erage + nx 2(=1)""nsin tent r Se 12,Expand thefunction f(x)=2r inaseries ofsines intheinterval (0,1). ans,12Soe, ’ a Fourer See »& sin22anEar 2, af, tfrocrehwom{Fie PSESY inthe interval (0, 2): a)inaseries ofsincs; b)inaseries ofcosines, 8S ys 14 Sresanttnas inoSowa yg SH, CHAPTER Xxvitl EQUATIONS OF MATHEMATICAL PHYSICS SEC. 1.BASIC TYPES OF EQUATIONS OF MATHEMATICAL PHYSICS The basic equations ofmathematical- physics (for the case of functions oftwo independent variables) arethe following second- order partial differential equations. 1.Wave Equation: Haag. Oy This equation isinvoked inthestudy ofprocesses oftransversal vibrations ofastring, the longitudinal vibrationsofrods,electric oscillations inconductors, the torsional oscillations ofshafts, gas vibrations, and soforth. This equation isthesimplest oftheclass ofAyperbolic equations. Il.Fourier Equation forHeat Conduction: dtupeo a (2) This equation isinvoked inthe study ofprocesses ofthepropa- gation ofheat, the filtration ofliquids and gases inaporous medium (for example, the filtration ofoiland gasinsubterranean sandstones), some problems inprobability theory, etc. This equation isthesimplest oftheclass ofparabolic equation. Ill. Laplace’s Equation: fitSamo. @) This equation isinvoked inthe study ofproblems dealing with electric and magnetic fields, stationary thermal states, problems inhydrodynamics, diffusion, andsoon.This equation isthesimplest intheclass ofelliptic equations. Inequations (1),(2), and (3),theunknown function udepends ‘ontwo variables. Also considered are appropriate equations of functions with alarger number ofvariables. Thus, the wave 816 Equations ofMathematical Physics equation inthree independent variables isofthe form au_a(4,e Gsina(s+5h) to) the heat-conduction equation inthree independent variables isof the form oe gpa asHoa(+5). ca) the Laplace equation inthree independent variables has the form fu, Ot, Oe ;FatFatGan0. a) SEC. 2.DERIVATION OFTHE EQUATION OFOSCILLATION OF ASTRING. FORMULATION OF THE BOUNDARY-VALUE PROBLEM. DERIVATION OF EQUATIONS OF ELECTRIC OSCILLATIONS IN WIRES Inmathematical physics astring isunderstood tobeaflexible and elastic thread. The tensions that arise inastring atany instant oftime aredirected along atangent toitsprofile. Let a string oflength /be,atthe initial instant, directed along aseg- ment of the x-axis from 0to J. Assume that the ends ofthe string arefixedatthepoints x=0 4and x=I. Ifthe string isdeflected fromitsoriginal position andthen 14Mpletloose;orif‘withoutdeflectingthe luo TK string weimpart toitspoints acer- of —*¥— yy TF tain velocity attheinitial time, or ; ifwedeflect the string and impart Fig.S71. avelocity toitspoints, then thepoints of.thestring willperform certain motions; we say that the string isset into oscillation, The problem istodetermine theshape ofthestring atany instant oftime and todetermine thelaw ofmotion ofevery point ofthe string asafunction oftime. Letusconsider small deflections ofthepoints ofthestring from theinitial position. Wemay suppose that themotion ofthepoints ofthestring isperpendicular tothex-axis and inasingle plane. Onthis assumption, the process ofoscillation ofthe string is described byasingle function u(x, ¢),which yields the amount that apoint ofthe string with abscissa xhas moved attime¢ (Fig. 371). Since weconsider small deflections ofthestring inthe(x,u)- plane, weshall assume that the length ofan.element ofstring Derivation oftheFouation ofOscillations ofaString ar MM, isequal toitsprojection onthex-axis, that is,*) M\M,— ‘='r,2-x,. Wealso assume that thetension ofthestring ‘at’all points isthesame; wedenote it by7.‘Consider anelementofthestring fe apMM’ (Fig. 372). Forces7’actatthe ends ofthis element along tangents tothestring. Letthetangents form 4/17 5with thex-axis angles@and@+Ag. Then theprojection ontheu-axis of Fig. $72. forces acting ontheelement MM’ will beequal toTsin (p-+Ag)—T sing.Sincetheangle@issmall, wecan put tang=sing, and wewill have Tsin(@+Aq)—T sing STtan(g-+Ag)—T tang—T7[MEME O_O) Sule OAx 1), pFale, y) a7THETA DayeHEDAe, 0<t<1 there,weapplied theLagrange theorem totheexpression inthe square brackets). Inorder toobtain the equation ofmotion, we must equate to the force ofinertia the extemal forces applied totheelement. Let @bethe linear density ofthe string. Then themass ofthe element ofthestring will begAx. The acceleration oftheelement is24.Hence, byd’Alembert's principle wewillhave edeZt7MHas, Cancelling outAxanddenotingTact,wegettheequationofmotion: Yu suFeagtMH ) This isthewave equation, theequation ofvibrations ofastring. Equation (1)byitself ishot sufficient for acomplete definition *)This assumption Isequivalent toneglecting ui?ascompared’ with 1. Indeed, Maen|Viraae§(14pa.) des[deena 818 Equations ofMathematical Physics ofthemotion ofastring. The desired function w(x, f)must also satisfy boundary conditions that indicate what occurs atthe ends ofthe string (x=0 and x=1) and initial conditions, which describe thestate ofthestringattheinitialtime(f—0).Thebound- ary and initial conditions arereferred tocollectively asboundary- value conditions. For example, asweassumed, lettheends ofthestring atx—0 Ranaee befixed.Thenforany¢thefollowing equalities must old: uO, )=0, 2) u(l, )=0. (2) These equations arethe boundary conditions forour problem. Atf=0 thestring has adefinite shape, that which wegave it. [etthisshapebedefined byafunction f(x).Weshould then ave u(x, =| rao=F (2). 6) Further, atthe initial instant the velocity ateach point ofthe string must begiven; itisdefined bythe function @(x). Thus, weshouldhave au . Flea 9 co) The conditions (3) and (3°) arethe initial conditions. Note. For aspecial case we may have f(x)==0 or@(x)=0. But iff(x)=0 and g(x)=0, then the string will beinastate ofrest; hence, u(x, t)=0. Ashasalready been pointed out, theproblem ofelectric oscit- lations inwires likewise leads toequation (1), Let usshow this tobethecase. The electric current inawire ischaracterised by the current flow i(x, f)and thevoltage v(x, ¢),which aredepen- dent onthe coordinate xofthe point ofthewire and onthe time ¢.Regarding anelement ofwire Ax, wecan write that the voltage dropontheelement Axisequaltov(x,f)—v(r+Ax,f=sR AeThisvoltage dropconsists oftheohmicdrop,equal toiRAx, andtheinductive drop,equalto#LAx. Thus, —8bemiRAx+HLAs, “ where Rand Laretheresistance and thecoefficient ofself-induc- tion reckoned perunit length ofwire, The minus sign indicates Derivation oftheEquation ofOscillations ofaString 819 that thecurrent flow isinadirection opposite tothe build-up ofv.Cancelling out Ax, wegetthe equation S+iR+LF =, 3} Further, the difference between the current leaving element Ax and entering itduring time A¢will be i(e,Niet Ax,N=—Fdxdt, Itistakenupincharging theelement (thisisequaltoCAxg?at) and inleakage through the lateral surface ofthe wire due to imperfect insulation, equal toAvAxAt (here Aisthe leak coeffi- cient). Equating these expressions and cancelling out AxAt, we gettheequation a oa$4+C2 +Av=0. (6) Equations (5) and (6) are generally called telegraph equations. From the system ofequations (5)and (6)wecan obtain an equation that contains only the desired function i(x, f),and an equation containing onlythedesired function ox).Dillerentate the terms ofequation (6)with respect tox;differentiate theterms of(5)with respect to¢and multiply them byC.Subtracting, weget#42 cptop% FtaZ—crE— Closm0. Substituting intothelatterequation theexpression $2from(6), wegeta a at_oy ita (—iR—L2)—cr¥—cr Mano or Hop Ht aFanClFat(CR+AL)+ARI. @ Similarly, weobtain anequation fordetermining v(x, #): oe 2 SomCL$24(CR+AL)4ARD. ®) Ifweneglect the leakage through the insulation (40) and ‘theresistance (R=0), then equations (7)and (8)pass into the 20 Equations ofMathematical Physis waveequations 20Ot 200Oto aaa, oe, where a'=7.Thephysicalconditions dictatetheformulation oftheboundary and initial conditions oftheproblem. SEC. 3.SOLUTION OF THE EQUATION OF OSCILLATIONS OF ASTRING BY THE METHOD OF SEPARATION OF VARIABLES (THE FOURIER METHOD) The method ofseparation ofvariables (ortheFourier method), which weshall now discuss, istypical ofthesolution ofmany problems ofmathematical physics. Let itberequired tofind the solution oftheequation ouPane oe () which satisfies theboundary-value conditions 40, )=0, @ u(t, )=0, @) u(x, =F), ® aFluo 3) Weshall seek aparticular solution (not identically equal tozero) ofequation (1)that satisfies the boundary conditions (2)and (3), intheform ofaproduct oftwo functions X(x) and T(t), of which theformer isdependent only onx,and thelatter, only on ft: u(x, )=X()TO). 6) Substituting into equation (1), weget X(x)T"()=a'X"(x)T(1), and dividing theterms oftheequation bya*XT, rox Z=%. CO) The left member ofthis equation isafunction that does not depend onx,theright member isafunction that does not depend on¢,Equation (7)ispossible only when theleftand right mem-bersarenotdependent eitheronxoron¢,thatis,areequalto aconstant number. We denote itby—A, where A>0 (later on wewill consider the case 4<0). Thus, TX 4ar>x=—* Solution ofthe Equation ofOscillations ofaString 821 From these equations weget two equations: X'42X=0, an)T'+a'AT=0. @) The general solutions ofthese equations are(see Ch. XIII, Sec. 21) X(x)=AcosVix+BsinVix, (10) T(x)=CcosaVit+D sinaVit, ay where A,B,C,and Dare arbitrary constants. Substituting the expressions X(x) and T(t) into (6),weget u(x,t)=(AcosVix+B sinVXx)(CcosaV'At+DsinaVit). Now choose theconstants Aand Bsothat theconditions (2)and (3)aresatisfied. Since T(t)40 (otherwise wewould have u(x, t)=0, which contradicts thehypothesis), thefunction X(x)must satisfytheconditions (2)and(3);thatis,wemusthaveX(0)=0, X(1)=0.Putting the values x=0 and x= into (10), weobtain, onthe basisof(2)and(3), 0=A-148-0, 0=AcosVN+B sinVi =0. From the first equation wefind A=0. From thesecond itfollows that e BsinV=0, B+O, since otherwise we would have X==0 and u=0, which contradicts the hypothesis. Consequently, wemust have sinVil=0, whence Vi== (n=1,2,...) (12) (we donot take the value n=0, since then wewould have X=0 and w=0). And sowe have X=Bsin Fx, (13) These values ofAarecalled eigenvalues of‘thegiven boundary- value problem. The functions X(x) corresponding tothem are called eigenfunctions. Note. Ifinplace of—A wetook the expression +A=A%, then equation (8)would take theform XX =0, a2 Equations ofMathematical Physics The general solution ofthis equation is X= Ae™ +Be-™, Anonzero solution inthis form cannot satisfy the boundary conditions (2)and(3). Knowing V%wecan[utilising (11)] write T()=Ccos “44Dsin“ (n=1,2...). (4) Foreachvalueofn,henceforeveryteweputtheexpressions (13) and (14) into (6)and obtain asolution ofequation (1)that satisfies the boundary conditions (2)and (3). We denote this so- lution byu,(x,2): .ug(%,t=sinx(C,cosS44+D,sin), (15) For each value ofnwe can take the constants ©and Dand thus write C,and D, (the constant Bisincluded inC,and D,). Since equation (1) islinear and homogeneous, the sum ofthe solutions isalso asolution, and therefore the function represent- edbythe series u(x, N= Bale or u(x,0=$(C,cosS1-+D, sin“)sinFx(16) will likewise beasolution ofthedifferential equation (1), which will satisfy theboundary conditions (2)and (3).Series (16) will obviously beasolution ofequation (1)only ifthe coefficients C,and D,aresuch that this series converges and that theseries resulting from adouble term-by-term differentiation with respect toxand tofconverge aswell. The solution (16) should also satisfy theinitial conditions (4) and (5). We shall trytodothis bychoosing the constants C, and D,.Substituting into (16) £=0, weget [see condition (4)]: 1)=LG,sinFx. (17) Iithe function f(x) issuch that inthe interval (0,1)itmay be expanded inaFourier series (see Sec. 1,Ch. XVII), thecondition (17) will befulfilled ifweput 1 C.=FJFo)sinSExde, (18) The Equation forPropagation ofHeat inaRod #23 We then differentiate the terms of(16) with respect to and substitute 4=0. From condition (5)wegetthe equality 9X)=D,DF sinFx, We define the Fourier coefficients ofthis series: fi nx_2 ax D,MF=5|9(@)sinFxde or odD,-aJowsin!xdx. (19) Thus,wehaveprovedthattheseries(16),wherethecoefficients C,and’ D,are defined byformulas (18) and (19) [ifit“admits double termwise differentiation], isafunction u(x, f),which is the solution ofequation (1)and satisfies theboundary and initial conditions (2)to(5). Note. Solving the problem athand forthe wave equation by adifferent method, wecan prove that theseries (16) isasolution even when itdoes'not admit termwise differentiation. Inthis case thefunction f(x) must betwice differentiable and @(x) must be once difierentiable*). SEC. 4.THE EQUATION FOR PROPAGATION OF HEAT INAROD. FORMULATION OF THE BOUNDARY-VALUE PROBLEM Let usconsider ahomogeneous rod oflength !.We assume that thelateral surface oftherod isimpenetrable toheat transfer and that the temperature isthe same atallpoints ofany cross-sectional area (———1-.-—, ofthe rod. Let usstudy the process of He 1 propagation ofheat intherod. :Weplacethex-axis, sothatone ee end of the rod coincides with the point x=0, the other with the point x=/ (Fig. 373). Let u(x, f)bethetemperature inthe cross section oftherod with abscissa xattime ¢,Experiment tells usthat the rate ofpropa- *These conditions are dealt with indetail in“Equations ofMathematicalPhysics"ACN,Tikhonovsad’A.A,Samarshy, Gostekiedat, 1954(Russian om Equations ofMathematical Physics gation ofheat (that is,the quantity ofheat passing through a cross section with abscissa xinunit time) isgiven bytheformula q=ks ) where Sisthe cross-sectional area ofthe rod and &isthe coef- ficient ofheat conduction*). Let us examine an element of rod contained between cross sections with abscissas x,and x,(x,—x,=Az).Thequantityof heatpassing through thecrosssection ‘withabscissa x,during time Afwill beequal to au AQ,=al She, @) and thesame forthecross section with abscissa x,t 4Q,=—AG]Sas. @) The influx ofheat AQ,—AQ, into the rodelement during time Atwill be shOEAxSAt @ (weappliedtheLagrange theoremtothedifference Hl._——Hl,..,): ThisinfluxofheatduringtimeAfwasspentinraising thetemperature oftherodelement byAu: AQ,—AQ,=ogAxSAu or 8Q,—AQ,xcAxS HAt, 6) where ¢isthe thermal capacity ofthesubstance ofthe rod and @isthe density ofthe substance (gAxS isthemass ofanelement ofrod). *)The rate ofpropagation ofheat, orthe rate ofthethermal ux, is determined by AQ q=tm42, vuhere AQIsthequantity ofheat that haspassed through aeross section S during atime AC. Heat Propagation inSpace 825 Equating expressions (4)and (5)ofone and thesame quantity ofheat AQ,—AQ,, weget eu ou kgetAXSAt=cQAxSZF‘At or Qu_ kat wear Denoting k=a.,wefinallyget duuMma Sh, ©) This isthe equation forthe propagation ofheat (the equation of heat conduction) inahomogeneous rod. Forthesolution ofequation (6)tobedefinite, thefunction u(x,t) must satisfy the boundary-value conditions corresponding tothe physical conditions oftheproblem. Forthesolution ofequation (6), the boundary-value conditions may differ. The conditions which correspond totheso-called first boundary-value problem for0<t<T are as follows: 4(, )=90), a 40, N=, (9, 8) u(t, N=). () Physically, condition (7)(the initial condition) corresponds tothefactthatforf0atemperature isgiveninvarious cross sections ofthe rod equal to@(x). Conditions (8) and (9)(the Boundary conditions) correspond totheTactthatattheendsof the rod, x=0 and x=J, atemperature ismaintained equal to (0)anda(t),respectively, ttisproved that theequation (6)hasonly one solution inthe region 0<x</, 0</<T, which satisfies theconditions (7),(8), and (9). SEC. 5.HEAT PROPAGATION IN SPACE Let usfurther consider the process ofpropagation ofheat in three-dimensional space. Let u(x, y,2,f)bethe temperature at apoint with coordinates (x,y,2)attime ¢.Experiment states that the rate ofheat passage ‘through anarea As, that is,the quantity ofheat passing through inunit time isgoverned bythe formula [similar toformula (1)ofthepreceding section] AQ=—k As, ) 225 EquationsofMathematical Physics where &isthe coefficient ofheat conductivity ofthe medium under consideration, which weregard ashomogeneous and isotro- pic, mistheunit vector directed normally tothearea Asinthe direction ofmotion ofthe heat. Taking advantage ofSec. 14, Ch. VIII, wecan write Safcosa+Hcosp+54cosy, wherecosa,cos,cosyarethedirection cosinesofthevectorm,or $=gradu. Substituting theexpression %intoformula (1),weget AQ=—kngradu As. The quantity ofheat passing intime Atthrough the elementary area As will be AQMt=— kngraduAAs. Now letusreturn totheproblem posed atthe beginning of thesection. Inthemedium athand wepick outasmall volume V bounded bythe surface S.The quantity ofheat passing through the surface Swill be Q=—AtlS kmgraduds, 2)? where aisthe unit vector directed along the external normal to thesurface S.Itisobvious that formula (2)yields the quantity ofheat entering the volume V(orleaving the volume V)during time Af.The quantity ofheat entering Visspent inraising the temperature ofthesubstance ofthis volume, Let usconsider anelementary volume Av. Let itstemperature rise byAuintime At.Obviously, thequantity ofheat expended onraising the temperature oftheelement Avwill be chogAuxchug% At, wherecistheheatcapacity ofthesubstance andgisthedensity.The total quantity ofheat consumed inraising the temperature inthe volume Vduring time A¢will be a0$FcoSao,7 Heat Propagation inSpace wr But this isthe heat that has entered thevolume Vduring the time Af; itisdefined byformula (2). Thus, wehave theequality at[Vengraduds=at{(fooSed.Ss v Cancelling out Af, weget $femeraduds=ffo9$¢dv. @)8 7 The surface integral onthe left-hand side ofthis equation we transform bythe Ostrogradsky formula (see Sec. 8,Ch. XV), assuming F=kgradu: §§(egraduymds—Jdiv(egradujdv.‘8 ? Replacingthedoubleintegral ontheleftof(3)byatripleinte- gral, weget $fdivgradu)do=0009%dv ¥ fa or S$[aivegradu)—c9i]dv=0. “ ‘Applying the mean-value theorem ‘tothetriple integral onthe left (See Sec. 12,Ch. XIV), weget au [aiveegradue) yayysans" (6) where the point P(x, y,2)issome point ofthe volume V. Since wecan pick out anarbitrary volume Vinthree-dimen- sional space where propagation ofheat istaking place, and since weassume that the integrand in(4)iscontinuous, equality (5) will befulfilled ateach point ofthespace. Thus, cg=div(kgrad u). (O} But bgradum bithejthMh (see Sec. 14,Ch, VIII) and' mw)42,aediv(egradu)3.(#u)+a(#+h (#) 28 Equations ofMathematical Physics (see Sec. 9,Ch. XV). Substituting into (6), weobtaindu_2(428)42(4du),2/,au :oot=ae(#Be)ay(5p)Fae(FE)-a If&isaconstant, then div(kegradu)=kdiv(gradu)=k(S455455) and equation (6)then yields ou dtu, tu, Owwae(Satieta) or,putting a="duge(2%4Ou,Ot ane(S++aa)- ® Equation (8)isbriefly written Bmathu, oaOu, , : whereAu=gataataeistheLaplaceoperator. Equation (8) istheequation ofheat conduction inspace. Tofind itsunique solution that corresponds totheproblem posed here, itisnecessary tospecify the boundary-value conditions. Let there beabody Qwith asurface o.Inthis body wecon- sider theprocess ofpropagation ofheat. Atthe initial time the temperature ofthebody isspecified, which means that thesolu- tion isknown forf=0 (the initial condition): u(x, Y2,O=@(x, Y2). @ Inaddition tothat wemust know thetemperature atany point M ofthesurface oofthebody atany time f(the boundary condi- tion): u(M, )=9(M, (19) (Other boundary conditions arepossible too.)Ifthedesired function u(x,y,z,¢)isindependent ofz,which corresponds tothe temperature being independent ofz,weobtain the equation a eeHad!(+35). (uy which isthe equation ofheat propagation inaplane, Ifweconsider heat propagation inaflatregion Dwith bound- ary C,then the boundary conditions, like (9)and (10), are The First Boundary-Value Problem fortheHeat-Conductivity Equation 829 formulated asfollows: u(x, yN=O(% ¥), u(M, t)=9(M, ¢), where @and ware specified functions and Misapoint onthe boundary C.Butitthefunction udoesnotdependeitheronzorony, thenwegettheequation a a an"ae which isthe equation ofheat propagation inarod. SEC. 6.SOLUTION OF THE FIRST BOUNDARY-VALUE PROBLEM FOR THE HEAT-CONDUCTIVITY EQUATION BYTHE METHOD OFFINITE DIFFERENCES When wesolve partial differential equations bythemethod of finite differences, the derivatives, asinthe case ofordinary differential equations, are replaced by appropriate differences (seeFig.374):Que,uth N—ule, aox LJ ’ Fale,1{ft(eh,ule,)a(z,O—we—h, 5} ata Ly i or ue) _wleth,Ole, O-+ue—h OD,1ute, O—BleN-tut D @ similarly, u(x, t)u(x, t+)—ule, t)a T @) The first boundary-value problem fortheheat-conductivity equa- tion isstated (see Sec. 4)asfollows. Itisrequired tofind the solution oftheequation a amatOH Oy that satisfies the boundary-value conditions u(x, =x), O<x<L, 6) 40 )=¥,0, O<F<T, (6) a, D=¥,0, O<tar, @ that is,wehave tofind thesolution u(x, ¢)inarectangle boun- ded bythe straight lines ¢=0, x=0, x=L, ¢<T, ifthevalues 800 EquationsofMathematical Physice ofthedesired function are given onthree ofitssides: t=0, x=0, x=L (Fig. 375). We cover our region with agrid formed bythestraight lines xsth, G=1 Qe t=Al, k=l, 2wee, and approximate thevalues atthenodes ofthegrid, that is,at the points ofintersection ofthese lines. Introducing the notation it a A a a Cty 77a oe: OATOtCh.t) fe fifad aa a ¥ D 7 Fig. 874, Fig. 975. u(ih,kl)=u;,y,wewrite[inplaceofequation (4)]acorrespondingdilference equation for the point (ih, &l). Inaccord with (3) and (2), weget Eat et ac ® Wedetermine 1),pss? tynan(LF) teaFaas aMinaae O) From (9)itfollows that ifwe know three values intheAth TOW: Uj,4Us,ay“ins,»WEcandetermine thevalueuj,.4,in the(e-{1)st row. We'know allthevalues onthestraight line #=0 [see formula (5)]. Byformula (9)determine the values at alltheinterior points ofthesegment ¢=1, Weknow the values ‘oftheend points ofthis segment byvirtue of(6)and (7). Inthis way, row byrow, wedetermine thevalues ofthedesired solution atall nodes ofthe grid. Itisproved that from formula (9)wecan obtain anapproxi- matevalue ofthesolution notforanarbitrary relationship between thestepsAand1,butonlyif<j. Formula (9)isgreatly sim- plified ifthestep /along thef-axis ischosen sothat 20411—>r=0 Propagation ofHeat inanUnbounded Rot 831 or ie i=. Inthis case, (9)takes theform 1 Hepo Cron abMm a (10) This formula isparticularly convenient forcomputations (Fig. 376). This method gives thesolution atthe nodes ofthegrid. Solutions between thenodes may beobtained, forexam- ple,byextrapolation, bydrawingaplane[Jomo] _| through every three points inthespace (x,¢,u). cao) Let usdenote byu,(x,#)asolution obtained byformula (10) and this extrapolation. Itis proved that (i) >i Fimule, D=a(e, O, Fig.576 where w(x, f)isthe solution ofour problem. Itisalso proved *) that Ian Q)—u(e, )I<MA', where Misaconstant independent ofA. SEC. 7.PROPAGATION OF HEAT IN AN UNBOUNDED ROD Let the temperature begiven atvarious sections ofanunboun- ded rod ataninitial instant oftime. Itisrequired todetermine thetemperature distribution intherodatsubsequent instants of time. (Physical problems reduce tothat ofheat propagation in anunbounded rod when the rod issolong that the temperature inthe interior points ofthe rod atthe instants oftime under consideration are but slightly dependent onthe conditions at the ends ofthe rod.) Ifthe rod coincides with the x-axis, the problem isstated mathematically asfollows. Find the solution totheequation du_gadGn~* Sa ry *)This question isdealt with inmore detail inD.Yu, Panov's “Rel- erence onNumerical Solution ofPartial, Differential Equations", Gostekhizdat, 1961:Lothar "Gollatz,""Numerisehe Behandlung. vonDiffeentalgluchungea", 832 Equations ofMathematical Physics inthe region —oo<x<oo, 0<¢ which satisfies the initial condition u(x, Y=o(x). @ Tofind thesolution, we apply the method ofseparation of variables (seeSec.3);thatis,weshallseekaparticular solution ofequation (1)intheform ofaproduct oftwo functions: u(x, =X (x)T(t). @) Putting this into equation (1)wehave X(x)T’()=a'X"(x)T() or rox 'wage. O) Neither ofthese relations can bedependent either onxor onf;therefore, weequate them toaconstant, *)—A*, From (4) wegettwo equations: T'+a'MT =0, 6) XE MX=0, O} Solving them wefind T=Ce-om, X= Acoshx+Bsinhx, Substituting into (3), weobtain u,(x,t=ent [A(A)cosd.x+B(A)sinAx] ” [the constant Cisincluded inA(A) and inBQ). For each value of4we obtain asolution ofthe form (7). For each value of4thearbitrary constants Aand Bhave defi- nite values. We can therefore consider Aand Bfunctions of2. The sum ofthe solutions ofform (7) islikewise asolution {since equation (1)islinear}: yeaa) cosAx+B(A)sinAx]. Integrating expression (7)with respect totheparameter 4between 0 and co, we also get asolution u(x,t=Jere[AA)cosAx+B(A)sinAx|dh, (8) *)Since from the meaning ofthe problem T(t) must bebounded for anyt,if@(3)isbounded, itfollows that[>mustbenegative, Andsowe write — Propagation ofHeat inanUnbounded Rod 3 ifA(Q) and B(Q) are such that this integral, itsderivative with respect to¢and the second derivative with respect toxexist and areobtained bydifferentiation oftheintegral with respect to¢and x.Wechoose A(A)and B(A)such that thesolution u(x,¢) satisfies the condition (2). Putting ¢=0 in(8), weget [on the basis ofcondition (2)]: u(x,0)=@(x)= J[A@)cosAx-+B(A)SinAx]dh, @) Suppose that the function @(x) issuch that itmay berepresented bytheFourier integral (see Sec. 12,Ch. XVII): @G)=25 (Je(@)cosh(a—x)da)am orso o)=2[(J@(cosada)costxt +(i)(a)sindada)sindx]da.(10) Comparing the right sides of(9)and (10), weget A@=t Je(@)costada, ~ (dy Bay=t 59(a)sinkada, Putting the expressions thus found ofA(A) and B(A) into (8), we obtain 4,Q=Efee{(fo@coshada) coshe+ +(f(a)sindada)sindx]dha =the [§@(@)(cosdacosAx +sindasindx)aa]d= 1¢tat fenae (Je0ecoshca—syaa)an 27 3388 en EquationsofMathematical Physies or,changing theorder ofintegration, wefinally get ue,O=2f[a(fercos(a—a)on)]da,(12) This fsthesolution ofthe problem. Let ustransform formula (12). Compute the integral inthe parentheses: Facer yeeJecosh(a—x)di,aicosBzdz,(13) The integral istransformed bysubstitution: 2, OkeVing, Soh—B. (4) We denote K(p)=Se~**cosBzdz. (5) Differentiating, *)weget K'(@)=—Ser* zsinBzdz, Integrating byparts, wefind K’()=Fle*sinBal$fe-*cosBede or K’@=—5K@). Integrating this differential equation, weobtain _# K(B)=Ce*, (168) Determine theconstant C,From (15) itfollows that K@=fetan le *)Differentiation here iseasily justified, Propagation ofHeat inanUnbounded Rod 835 (see Sec, 5,Ch, XIV). Hence, in(16) wemust have Vacals, And so vi# Kp= Yee a7 Put the value (17) ofthe integral (15) into (13): . _o# _ dhe VEE fe08h(a—#)dhTeVes, Inplace ofBwesubstitute itsexpression (14) and finally getthe value ofthe integral (13): *— lent 0 1 a, feMteosh(a—a)dh—= zeHe7 (as) Putting this expression ofthe integral into thesolution (12), we finally get 1% tans 6%Dr ig@e#da, (19) This formula, called the Poisson integral, isthe solution to the problem ofheat propagation inanunbounded rod. Note. Itmay beproved that the function u(x, ‘),defined by integral (19), isasolution ofequation (1)and satisfies condition (2) ifthe function g(x) isbounded on an infinite interval (0, 00). Let usestablish the physical meaning offormula (19). We consider the function 0forwo<x<x,, gt=] els) for,cxce,+Ax, 2) 0 forx,+Ax<e<oo. Then the function 12 ato weN=ai)g@e“da (1) isthe solution toequation (1), which solution takes onthevalue a 836 Equations ofMathematical Physics 9*(x)when f=0. Taking (20) into consideration, wecan write mothe mea 1fom *(x, )=—1_ “a do,WO,Dare \e@e” da. Applying the mean-value theorem tothelatter integral, weget Ge,yaoeaeSa A 29)We, D=Tae +<bean. (22) Formula (22) gives the value oftemperature atapoint inthe rodatanytimeiffort=0thetemperature intherodisevery- where u*=0, with the exception ofthe interval [x,, x,+Ax], where itis@(x). The sum oftemperatures ofform (22) iswhatyieldsthesolution of(19).Itwillbenotedthatifgisthelineardensity oftherod, ¢the heat capacity ofthematerial, then the quantity ofheat ‘inthe element [x,,x,+Ax] for£=0 will be AQwo(E) Axe. (23) Let usnow consider the function ga 1ar ava (24) Comparing itwith the right side of(22) and taking into ac- count (23), wemay say that ityields the temperature atany point ofthe rod atany instant oftime ¢ifforf=0 there was aninstantaneous heat source with quantity ofheat Q=cg inthe cross section &(the limiting case asAx—0). SEC. 8.PROBLEMS THAT REDUCE TO INVESTIGATING SOLUTIONS OF THE LAPLACE EQUATION. STATING BOUNDARY-VALUE PROBLEMS Inthis section weshall consider certain problems that reduce tothe solution ofthe Laplace equation: Gu, Hu, Futip aes O) Asalready pointed out, theleftside ofequation (1), Fu Fu, FuSatgtge=eiscalledtheLaplacian operator. Thefunctions uwhichsatistytheLaplace equation are called harmonic functions. I.Astationary (steady-state) distribution oftemperature ina homogeneous body. Let there beahomogeneous body 2bounded The Laplace Equation aa byasurface o.InSec. 7itwas shown that the temperature at various points ofthebody statisfies equation (8):du_(2aOu,Ouane(+R+He): Ifthe process issteady-state, that is,ifthe temperature isnot dependent onthetime, butonly onthecoordinates ofthepoints ofthebody, then%=0 and,consequently, thetemperature satisfies the Laplace equation Fu, ou, Fugatgatgan. (1) Todetermine the temperature inthe body uniquely from this equation, one hastoknow thetemperature ofthesurface o.Thus, forequation (1), the boundary-value problem isformulated as follows. Tofind the function u(x, y,2)that satisfies equation (1)inside the volume @and that takes onspecified values ateach point M of the surface o: ul,=(M). @) This problem iscalled theDirichlet problem orthefirst boundary- value problem ofequation (1). Ifthe temperature onthe surface ofthe body isnot known, but the heat flux atevery point ofthe surface is,which ispro: portional to%(seeSec.5),theninplaceoftheboundary-value condition (2)onthe surface owewill have the condition a)_ye lav . ® The problem offinding thesolution to(1)that satisfies theboun- dary-value condition (8)iscalled the Neumann problem orthe second boundary-value problem. Ifweconsider the temperature distribution inatwo-dimensi- onal region Dbounded byacontour C,then the function wwill depend on two variables xand yand’ will satisfy the equation ou, Fufatga “ which iscalled theLaplace equation inaplane. The boundary- value conditions (2)and (3)must befulfilled onthe contour C. IL,The potential flow ofafluid. Equation-of continuity. Let there beaflow ofliquid inside avolume Qbounded byasur- face o(inaparticular case, @may also beunbounded). Letg 838 Equations ofMathematical Physics bethedensity oftheliquid. Wedenote thevelocity oftheliquid by vao,d+0,j+0,k, (6) wherev,,0,,0,aretheprojections ofthevector9onthecoor- dinate axes.”In the body @pick out asmall volume @,bounded bythesurface S.The foliowing quantity ofliquid will pass through each element As ofthe surface $inatime At: AQ=on Aso At, where aisthe unit vector directed along the outer normal to thesurface S.The total quantity ofliquid Qenteringthevolumeo (orflowing outofthevolume w)isexpressed bytheintegral Q=At{{eonds (} ‘s (see Secs. 5and 6,Ch. XV). The quantity ofliquid inthe volume @attime ¢was Set During time Afthe quantity ofliquid will change (due to changes indensity) bytheamount 2 Q=SSSAcdomAt SSSao. ) Assuming that there arenosources inthe volume @,wecon- clude that this change isbrought about byaninflux ofliquid to anamount that isdetermined byequation (6).Equatingtheright sides of(6)and (7)and cancelling out Af, weget a -Sfeonds= +(0(Pde, ® Wetransform theiterated integral ontheleft byOstrogradsky's formula (Sec, 8Ch. XV). Then (8)will assume the form -SSfdiv(@o)do=ff$Pao or. . §sf(B+4iv(ee)do=0. Since thevolume wisarbitrary and theintegrand iscontinuous we obtain ;+div(eo)=0 @ The Laplace Equation 839 or a a a ;REZCodt+H(ce,+Z(v,)=0. @) This isthe equation ofcontinuous flow ofacompressible liquid. Note. Incertain problems, for instance when considering the movement ofoilorgas inasubterranean porous medium toa well, itmay betaken that *grad om—Z gradp, where pisthe pressure and &isthecoefficient ofpermeability and 20, 5,20a22, A=const, Substituting into the continuity equation (9), weget 4.2—div(kgradp)=0 or 02/4 9)12(420)42/,aRASPm=5e(e52)+a;(#32)+35(#$6). (10) Ifkisa constant, then this equation takes onthe form an_(ap,Bp,ao B=(a+58+%). ay and wearrive atFourier’s equation. Let usreturn toequation (9). Ifthe liquid isnoncompres- sible,then=const, 220, and(9)becomes div(2)=0. (12) Ifthemotion ispotential, that is,ifthe vector 9isagradient ofsome function @: v=gradq, then equation (12) takes the form div(grad)=0 °rawFa+Fh+58=0; (13) that is,thepotential function ofthevelocity @must satisfy the Laplace equation. 840 Equations ofMathematical Physics Inmany problems, as,forexample, those dealing with filtra~ tion, wecan put o=—h, grad p, where pisthepressure and&,isaconstant; wethen getthe Laplace equation forthe determination ofthe pressure: Op Pp, Fp "Fe4ohSEO. (13') The boundary-value conditions forequation (13) or(13") may bethe following: 1.On the surface oare specified the values ofthe desired function p—pressure {condition (2)]. ThisistheDirichletproblem. 2.On the surface oare specified the values ofthenormal derivative 22;theflowthroughthesurfaceisspecified{condition(3)]. This isthe Neumann problem. 3.On parts ofthe surface oarespecified thevalues ofthe desired function p—pressure, and onparts ofthesurface are specified thevaluesofthenormalderivative $2—theflowthrough thesurface. This istheDirichlet-Neumann problem. Ifthemotion istwo-dimensional-parallel—that.is,thefunc- tion @(or p)does not depend onz—then weget the Laplace equation inatwo-dimensional region Dwith boundary C: aB+FG=0. (14) Boundary-value conditions oftype (2), the Dirichlet problem, oroftype(3),theNeumann problem, “arespecified onthecon: tour C. Ill. The potential ofasteady-state electric current. Let aho- mogeneous medium fillsome volume V,and letanelectric cur- rent pass through itwhose density ateach point isgiven bythe vector J(x,y,2)=Jqi+J,j-+J,k. Suppose that thecurrent den- sity isindependent ofthé time ¢,Further assume that there are nocurrent sources inthe volume under consideration. Thus, the flux oravector Jthrough any closed surface Slying inside the volume Vwill beequal tozero: §fJnas—0, ‘s where misaunit vector directed along theouter normal tothe surface, The Laplace Equation inCylindrical Coordinates sit From Ostrogradsky’s formula weconclude that divsJ=0. (15) The electric force Eintheconducting medium athand is,onthe basis ofOhm's generalised law, 4 E=t (16) or J=2E, where 2isthe conductivity ofthe medium, which weshall con- sider constant. From the general electromagnetic-field equations it,follows that iftheprocess isstationary, then thevector field Eisirro- tational, that is,rot E==0. Then, like the case wehad when considering the velocity field ofaliquid, the vector field ispo- tential (see Sec. 9,Ch. XV). There is'a function such that E=gradg. (17) From(16)weget J=Agradg. (18) From (15) and (18) wehave Adiv(grad@)=0 cala, Op, a‘? reSetaptae=O (a9) We get the Laplace equation. Solving this equation forappropriate boundary-value conditions, wefind thefunction @,and from formulas (18) and (17) wefind the current Jand the electric force E. SEC. 9,THE LAPLACE EQUATION INCYLINDRICAL COORDINATES. SOLUTION OF THE DIRICHLET PROBLEM FOR ARING WITH CONSTANT VALUES OF THE DESIRED FUNCTION ON THE INNER AND OUTER CIRCUMFERENCES Let u(x, y,2)beaharmonic function ofthree variables. Then Pu, Pu, HusatSethe. Oy We introduce the cylindrical coordinates (r,@,2): rercosg, r=rsing, 7=2, whence raVEFH, pmarctant, 222 @ 802 Equations ofMathematical Physics Replacing theindependent variables x,y,andzbyr,@,andz, we arrive at the function u*: u(x, y,z=u*(r, @,2). Let usfind theequation that will besatisfied byu*(%,@,2) asafunction ofthearguments r,@,and z;wehave du _dutOrdu"3 5&GetByoe?Gu_Gut(dr)*|Outdtr,9Dutdrd9,u*(d9\*,du"9, sanoe(5)+gatewopaccetae(se)Hagaes similarly, ~aut (ae)4dategDutdey4But(08%4due saner(5)+Sat? aeaes+oer(ae)tapoe besides,a_dur R=%. ©. We find theexpressions for : oa trOrtyoyoeHeBes 5p ae BP ee ee ae! from equations (2). Adding the right sides of(3), (4)and (5), and equating thesum tozero [since the sum ofthe left-hand sides ofthese equations arezero byvirtue of(1)}, weget BehreteaeGeO © This isthe Laplace equation incylindrical coordinates. Ifthe function wisindependent of2and isdependent onx and y,then the function u*,dependent only onrand @,satisfies theequation Hut Laut, 1utStee taaemo ) where rand @are polar coordinates inaplane. Now letusfind thesolution toLaplace's equation intheregionD(ring)boundedbythecirclesC,:x"-+y*= RYandCy:xt-+yt= RE with thefollowing boundary values imposed: UulCy=uy ®ulCy=u, @ where u,andu,areconstants. The Solution ofDirichtet’s Problem foraCirce as Wewill solve the problem inpolar coordinates. Obviously, it isdesirable toseek asolution that isindependent of.Equation (7)inthis case takes the form SyMno. Integrating this equation wefind u=C,Inr+C,. (io) Wedetermine C,and C,from conditions (8)and (9): u,=C,InR, +C,, u,=C,InR, +C,. Whence we find oy on WRCte C=, (4y—a) R Ry Substituting thevalues ofC,and C,thus found into (10), we finally get ne amu,t+—E(uu). aynk Note. We have actually solved the following problem. Tofind thefunction uthat satisfies the Laplace equation inafegion bounded bythesurfaces (incylindrical coordinates) aR, 1=R, 2=0, 2H, and that satisfies the following boundary conditions: Wak =H, UlR=Hy, au auFlees Flea™? (the Dirichlet-Neumann problem). Itisobvious that thedesired solution does not depend either onzoron@and isgiven by formula (11). SEC. 10, THE SOLUTION OF DIRICHLET'S PROBLEM FOR ACIRCLE Inanxy-plane, let there beacircle ofradius Rwith centre attheorigin and let there beacertain function f(g), where @ isthepolar angle, begiven onitscircumference. Itisrequired tofind thefunction u(r, @)continuous inthecircle (including eu Equations ofMathematical Physics theboundary) and satisfying (inside thecircle) theLaplace equa- tion aFuFe5e=0 ay and, onthecircumference, assuming thespecified values ular=F (9). (2) We shall solve theproblem inpolar coordinates, Rewrite equation (1)inthese coordinates: uy lou, 1ew ateatag? °r wt Ou,Oot EtRea ay Weshall seek thesolution bythemethod ofseparation ofvariables, placing u=O()R(N. @) Substituting into equation (1’), weget POQ)R ()+rOGR’D+O"'@RN=0 or ©) _ARR psoe)~ Ray OE “” Since the left side ofthis equation isindependent ofrand the tight isindependent ofg,itfollows that they are equal toaconstant whichwedenoteby—A*.Thus,equation (4)yieldstwo equations: ©"(9)+RO(p)=0, 6) PRY+rRBR=0. @) The complete integral of(5)will be O=A coskp+B sinkg. 6) We seek the solution of(5’) inthe form R=r™. Substituting R=r* into (6’), weget ; Pm(m—1) 9? +re"! —hr =0 or m—k=0. Wecanwrite two particular linearly independent solutions r*and r-* The general solution ofequation (5') is R=Cr*+Dr-*, @ We substitute expressions (6)and (7)into (3): Uy=(A,cosko+Bysinkg)(Cyr*-+ Dy"). @®) The Solution ofDirichlet's Problem foraCircle ‘845, Function (8)willbethesolutionof(1’)foranyvalueof&differentfrom zero. Ifk=0, then equations (5) and (5') take the form ©=0, rR'+R'=0, and, consequently, 4,=(4,+B,9) (C+D, Inv). @) The solution must beaperiodic function ofg,since forone and the same value ofrfor and @-+2n we must have the same solution, because oneandthesame point ofthecircle isconsidered. Itistherefore obvious that informula (8’) wemust have B,=0. Tocontinue, we seek asolution that iscontinuous and finite in the circle. Hence, inthe centre ofthe circle the solution must befinal forr==0, and forthat reason wemust have D,=0in(8)and D,=0 in8). Thus, theright side of(8’) becomes the product A,C,, which wedenote byA,/2. Thus, ; Ay .wade, 6) We shall form the.solution toourproblem asasum ofsolutions ofthe form (8), since asum ofsolutions isasolution, The sum must beaperiodic function ofg.This will bethe case ifeach term isaperiodic function [email protected], &must take onintegral values. (We note that ifweequated thesides of(4)tothenumber +k, wewould notobtain aperiodic solution.} Weshall confine ourselves only topositive values: a rr because theconstants A,B,C,Darearbitrary and therefore the negative values of&do’not yield new particular solutions. Thus, . u(r,—)=B+D (A,cosnp+B,sin.ng) (O} (the constant C,isincluded inA,and B,). Let usnow choose arbitrary constants A,and B,soastosatisfy theboundary-valuecondition (2).Putting into(§)r=R, weget,fromcondition (2), i@=P+Layersng-+B,sinng)R". (10 For ustohave equality (10), itisnecessary that the function should beexpandable inaFourier series intheinterval (—z, x) and that A,R" and B,R" should beitsFourier coefficients, Hence, a6 Equations ofMathematical Physics A,and B,must bedefined bythe formulas AnmapaSf(O)cosntdt, a (il) B,=aa|F(Osinntdt. Thus, the series (9)with coefficients defined byformulas (11) will bethe solution ofourproblem ifitadmits termwise iterated Giferentiation withrespet torandp(butwehavenotproved this), Let ustransform formula (9). Putting, inplace of A,and Bu, their expressions (11) and performing thetrigonometric frans- formations, weget u(r,o=%fhodted Ji@cosn(t—q) at(§)"= =tiyHe)[42d(4)cosne—o] dt.(2) Let ustransform the expression inthesquare brackets: * 1423 (g)'cosm¢—@ =14D(g)Teemerietoome -4E (gee) +(Gere-n)'] = ft Lgatit-w) tnext=wa=ouohe Ife en) a a7 TF >Ra WRreost—g Fr apet—or(e) +)Inthe derivation wedetermine thesum ofan,infinite geometric prog- ression whose ratio is&complex number the modulus ofwhich Isfess than Unity. This formula ofthesum ofageometric progression is.derived. inthe same way asinthecase ofreal numbers. Itisalso necessary fotake into Account thedeci othe Timiofthecomplex function ofrealgue iment. Here, the argument isn(ace See. 4yCh. Vil}. Solution ofDirichlet's Problem byMethod ofFinite Diferences 847 Replacing theexpression insquare brackets in(12) byexpres sion (13), weget 1% Rae Formula (14)icalledPoisson's integral. Byananalysis ofthis formula itispossible toprovethatifthefunction Te)iscon+ tinuous, then thefunction u(r, q)defined bythe integral (14) also satisfies equation (1')and'u(r, @)—+/(g) asr—+R. That is, itisasolution ofthe Dirichlet problem foracircle, SEC. 11, SOLUTION OF THE DIRICHLET PROBLEM BY THE METHOD OF FINITE DIFFERENCES Inanxy-plane, let there begiven aregion Dbounded by acontour C.Let ‘there begiven acontinuous function fonthe contour C.Itisrequired tofind anapproximate solution to theLaplaceequation Mu4Mo a oe TOF that satisfies theboundary condition ale=f @) We draw two families ofstraight lines: x=ih and y=kh, @) where Aisthe given number, and éand kassume successive integral values. We shall say that the region Discovered with agrid. Wecall thepoints ofintersection ofthe straight lines nodes ofthegrid. We denote byu;,, the approximate value ofthe desired function atthe point x=ih, y=kh; that is,w(ih, kh)=u,y. We approximate the region 'Dby the grid region’ D*, which consists ofallthe squares that liecompletely inDand'of some that arecrossed bytheboundary C(these may bedisregarded). Here,thecontour Cisapproximated bythecontour C*,which consists ofsegments ofstraight lines oftype (3). Ineach node lying onthecontour C*wespecify thevalue /*,which isequal{0thevalueofthefunction fattheclosest pointofthecon- tour C(Fig. 377). The values ofthe desired function will beconsidered only at the nodes ofthegrid. Ashasalready been pointed out inSec, 6, 848, Equations ofMathematical Physics thederivatives inthis approximate method arereplaced byfinite differences: | tena2,attra Ox|xmih,yath ee 'ou Mye120) thbet The differential equation (1)isreplaced byadifference equation (after cancelling out A"): Higa e—2p, aEGinny eM, ne2AM, way=O or(Fig. 378) MieT Miers aAMi,nastinnyaEMi,aad (a) For each node ofthegrid lying inside D*(and not lying onthe boundary C*), we form an’equation (4). Ifthe point (x= ih, y=kh) isadjacent tothe point ofthe contour C*, then the right side of(4) will contain known values of/*. Thus, we obtain anonhomogeneous system ofNequations inNunknowns, .whereWisthenumberofnodes 4ee ofthe grid lying inside the HEE] "Sve"chaitprovethattha feshallprovethatthesys- EHHA-EHSAKEH]tem(4)hasone,andonlyone, SGS08 SERS pops ia) CORE TT[[I] o8SSeSelCo dal ae CONSere) HR CCOTTrrrrryrrr d 7 a * Fig, 877. Fig. 578. solution. This isasystem ofNlinear equations inNunknowns. Ithas aunique solution ifthe determinant ofthe system isnot zero. The determinant ofthesystem isnonzero ifthe homoge- neous system has only atrivial solution. The system will be homogeneous ifj*=0 atthe nodes onthe boundary ofthe contour C*, Weshall prove that inthis case allthevalues u;,, atallinterior nodes ofthegrid areequal tozero. Inside the region, letthere bew;,, different from zero. For thesake of definiteness, wesuppose’ that thegreatest ofthem ispositive. Letusdesignate itbyu;,,>0. Solution ofDirichlet’s Problem byMethod ofFinite Diferences 849 By(4)wewrite Cr Ce er ee ee ) This equation ispossible only ifallthevalues ofuonthe right areequal tothegreatest u;,».Wenow have five points atwhich thevalues ofthedesired function areu;,,. Ifnone of these points isaboundary point, then, taking one ofthem and writing for itthe equation (4), we will prove that atcertain other points thevalue ofthedesired function will beequal to %;,y Continuing inthis fashion, wewill reach theboundary andwill’provethatattheboundary pointthevalueofthefunction will beequal toi;,,, which iscontrary tothe fact that /*=0 atboundary points. Assuming that inside the region there isaleast negative value, we will prove that onthe boundary the value ofthe function isnegative, which contradicts the hypothesis. ‘And sosystem (4)hasasolution which isunique. The values ;,, defined from thesystem (4) areapproximate values ofthesoliition oftheDirichlet problem’ formulated above. Itwas proved that ifthesolution ofthe Dirichlet problem for agiven region Dand agiven function fexists [we denote itby u(x, y)]and ifu,,, isthesolution of(4),then wehave therelation [ue 9)4,al<An 6) where Aisaconstant independent ofA. Note. Itissometimes justified (though this has not been rigorously proved) tousethefollowing procedure forevaluating theerror oftheapproximate solution. Letu{*} beanapproximate solution forastep 2h,uf", anapproximate solution forastep h, andletE,(x, y)betheerror ofthesolution u!",, Then wehave the approximate equality E(t, MoHWut) inthe common nodes ofthe grids. Thus, inorder todetermine theerror oftheapproximate solution forastep h,itisnecessary tofind thesolution forastep 2h. One third ofthe difference ofthese approximate solutions isthe error evaluation ofthe solution forastep (mesh-length) of4.This remark also refers tothe solution oftheheat-conduction equation bythefinite- difference method, 0 Equations ofMathematical Physes Exercises onChapter XVII 4,Derive anequation oftorsional oscillations of@homogeneous eylin- rica rod Tint “The torque'tn aeross section oftherodwith abscissaxisdeterminedBa eee chao Nar ena aT section with abscisa xattime f,@istheshear modulus, and1isthe polar tment ofete alcrowelonofHeTod Ans.arntawhereaaand&isthemomentofinerfiaofunit length ofthe rod, ;Pind scien oftheejuaton 22-28 tatsates thecon ditions0,)=0,84,H=0,8x,me(a),PEM0,where ye 1 a Be forOcred, ounFX4m,forParas. Give amechanical interpretation ofthe problem. ce,paBnSSD, Ott)egOktIynat Ans.O(a,=SeSYepotAOAcogCA seidBate2equation ofIonia! osciaion of»homogeneous ylin- sical 10 Hint. Ifw(x, f)isthe franslation ofacross section ofrod with abscissa x aHime't, then theHenle stress Tin across" sestion isdefined. bythe teu 792%, wher 3sUnelects motain offsmaa ea3 isUhe cross-sectional, area ofthe rod, aeBinet PEwheeoe, andinthednbrdmati 4.Ahomogeneous rodoflengthSxwasshortedby2kundertheaction offecesape oindsEt=oWisreeofHarescling exfertally. Betermine theduplacrtnt 4 ata cross tston ofther with abc See atime t(ihe tid-polnt Ofthe ants ofthe rod hes abeciva =O) SAI eIae||etN)nat doeOoBSG ayBEDEee 8.Oneendofa04oflng1sxdTheotherendisaceduponby _atenafle force P.Find the longitudinal oscillations ofthe rod ifthe force P SPL (=1)"gyQn)ax|.(2nt1)nat testoperatewen0,das.BELSM DA on (Eand SasinProblem 3). Exercises onChapter XVIIT 51 6.Findasolutionfotheequation$Yaa?2%thatsatisestheconsditions 40, H=0, u(t, N=Asinat, eo,aule.0) ale,)=0, —S—=0. Give amechanical interpretation oftheproblem. Asin® xinot-nt Ans.u(e)=——* 42aYee Sia ~sin21 ot() uu @ a T Hint. Seek the solution inthe form ofasum oftwo solutfons: Asin2xsinot mote, wherew=——2sin21 1sthe solution that satisfies the conditions 20, H=0, off, H=0 Cao G0(x,0)__ dw(x,0) :=o, 0,AIR SE (tsassumed thatsin-2140.)\ au_9pOu 1.Findasolution totheequation 2mat4.thatsatistes thecon- ditions 40. H=0, a(t, H=0, t>0, xwhenO<rct a(x,O= ji tox when bexcs ° (nse ins. Oe,yet EE On1)me Ans.hee,9=AET sin24Dne| Hint. Solve the problem bythe method ofseparation ofvariables, 8Findasolution totheequation 4enat2%thatsatisfiesthecon- ditions 40,=u, N=0,a(x,AGED, . ansrart =5 Ban COLIRe Ans.u(x,o-3Lo ‘sin 852 Equations ofMathematical Physics 8.Findasolution totheequation Sat2%thatsatisfiesthecon- ar"ast ditions aElem HEDaHywee=9(0). Point out the physical mesning ofthe problem. ee (on Ans.w(t,tug Age”cosEEy, 2¢io+1) ( wherey=fo(0)eosCOEDgyA Hint. Seekthesolution intheform u=uy-0(x,‘). 10,Findasolutiontotheequation24=a-2%thatsatisfiesthecon- ditions a=0, 4) au]. ule D= 9040,9=0, SE) aia, aeO=ee Point out the physicel meaning ofthe problem. - . ee .Ans.ute,023A,—OEMAFyBat, a PO+D+H, u ‘ whereAn=%J9epsinBEa,PHL,thyHyeesHinatepositiveroots oftheequation tanp=a— Hint, AUthe end ofthe fod (when x—=0) aheat exchange occurs with the environment, which has atemperature of220,TeFind{byformula(0),Se.6,puttingh=02)anapproximatesolution totheequation 4=22% thatsatisfies theconditions a,Q=x(F-#), uOm0,wdomz,ORteat 12,Findasolution totheLaplace equation S5-+-S4—0, inastrip 0<x<a, 0<y <a that satisfies theconditions 4O.N=0 ula,y=0,ue,Q—A(1-Z), we,y=0. 2A Me ane ans.ute,DmPAS LEYinBE Hint, Use the method ofthe separation ofvariables. Exercises onChapter XVIII 3 , au, uo , 18,FindasolutiontotheLaplaceequation2%424220intheree tangle O<x<a, O<y<b thal salisies the conditions 4(6m0, ult, 60, 40, =Ay(O—y), Ula, ¥)=0. gayeoyseAOENO=H)gyOat Ans. at, 98487 warir Gartae pa onaCED 1H,Findasolution totheequation -S44-+$%=0insidearingbounded bythecircles s+y=RE,444 REthat satisties theconditions | ats - a Give ahydrodynamic interpretation ofthe problem. Tint. Solve the problem inpolar coordinates.dinetygeein2amt Big 16. The function a(e, ymer¥sinx isasolution ofthe equation f+genointhesquare0<x<1,O<y<Ithatsatisfiestheconditions 40, Y=, ul, Y=eFsinl, ule, =sinx, ule, Nets x InProblems 12-15 solve the Laplace equations forgiven boundary condi- tions bythefiniledifference method forh=0.25, Compare the approximate golution with the exact solution CHAPTER XIX OPERATIONAL CALCULUS AND CERTAIN OF ITS APPLICATIONS Operational calculus isan, important branch of mathematical analysis‘Themethods ofoperational calculusareusedinphysics,mechanics,clectseai fngineering andelsewhere. Operational caleulus findsespecially broad. appli cations inautomation and lelemechanics. Inthis chapter wegive. (on the Dasis ofthe foregoing material ofthis text) the fundamental concepts of operational calculus and operational methods ofsolving ordinary differential equations. SEC. 1,THE INITIAL FUNCTION AND ITS TRANSFORM Letthere begiven thefunction of real variable {defined for1250 {we shall sometimes® consider that the function /() isdefined. onaninfiniteinterval —oo<¢<oo,but/(!)=0 when¢<0].Weshallassume thatthefunction f(t) ispiecewise continuous, that is,such that inany finite intervaliI'hasafitenumberofdiscontinuities oftheMestkind.(eeSee.9,Ch.II). TTovensure the existence ofcertain integrals inthe infinite interval Ox<t <0 weImpose anSdditional restriction onthefunction /(0): namely, weSuppose That There exist constant positive numbers Mand ssuch that IN| <Mert 0) forany value¢intheinterval0<f<eo. JLetusconsider theproduct ofthefunction /(t)bythecomplex function e-¥'ot areal variable?) 1,where p=a-+b issome complex nome: ett). @ Function (2)isalso acomplex function ofareal variable f: enPAY(1)meeib1tf()meHt}(t)e~ibtaeAtf(1)cosbt—ie~"4f(t)sinbt. Let usfurther consider theimproper integral Ferrara feretincormidt—ifererainvea ——@) Weshall,showthatifthefunction/({)satisescondition(1)anda>ty then ‘the integrals onthe right of(3)exit andthe convergence ofthe nls: grals isabsolute. Let usbegin byevaluating the first ofthese integrals: |Serst7cycosoeat|<{Jemety(cosoe[ar< <MVerse atcmfene-ntarma: 2 a *)SeeSec. 4,Ch. VI, concerning complex functions ofareal variable, Transforms oftheFunctions 0,(t),Sint, Cos 855 Insimilar fashion we evaluate the second integral. Thus, the integral Fe-Ptycde exits 1definesacertainfunctionofp,whichwedenote) byFp): . Fea lerrinat. Oy The function F(p) iscalled. the Laplace transform. oFthe L-fransform, or simply thetransform ofthe function /(t). The function f(t) isknown asthe initial function, ortheoriginal. IF(p) isthe transform off(t), then we~ write FOF. Cy or 1H =F ) o LAFO}=F(P). u) Aswoshallpresenty se,themeaning oftransforms consists inthefact ‘Yhat with their help itispossible tosimplify thesolution ofmany problems,forinstance, toreduce thesolution ofdifferential equations to”simple algeb: fale operations infinding atransform. Knowing the transform, one can find theoriginal either from specially prepared “original-transform” tables orby methods that will begiven below. Certain natural questions arise Uatherebegivenacertainfuntion Fp)Doesthereerst function 1(0)forwhich Fp) isa transform? Ilthere does, then is.this function the‘onlyone?Theanswerisyestobothquestions, givencertaindefiniteassump-fons with-respect to.Fip) and /(0). For example, the following. theorem, Which wegive without proof, esfablishes that the transform isunique: Uniqueness Theorem. Iftwo conlinuous functions p(t) and (1) have ont and thesame L-transform F(p), then these functions are identically cqual. Thstheorem wilplayantinportant tolethrughost {hesubsequent fest Indeed, ifinthe solution ofsome practical problem wehave determined, In someny,thefansorm of"desired function andromthetransform the ‘original function, then onthe ‘basis ofthe foregoing’ theorem we conclude that thefunetion wehave found isthe solution of‘the given. problem. and that no other solutions exist. SEC, 2,TRANSFORMS OFTHE FUNCTIONS o4(t), SIN f,COS ¢ 1,The funetion /(0), defined as I()=1 fort>0, H()=0 for <0, iscalled theHeaviside unit function and isdenoted byoy(t). The graph ofthi'funetion igiveninFig.378.et'usfindtheL-ttansfom ofTheHess ° erat (0,=(e-Plat =— L{oto}§a= fet 1),Thefunction F(),forp#0, thefunction ofacomplex. variable (lor example, seeV. LeSmirnov's “Course ofHigher Mathematics’, Vol. Ill,Part’) (ussian edition). 856 Operational Calculus andCertain ofItsApplications Thus,*) q1] eh 8) 5 ® or, more precisely, 2 G yetFig.879. CD? Insome books onoperational caleulus the following expression iscalled the transform ofthe funetion [(0): Fr(ppJe-Pth(at. With this definition we have o,(!)+1 and, consequently, C=C, more eacty, GanSC Tl,Let }(2)=slat; then e ePt(—psinx—cosx)|@_ 1 int}={e-Ptsintgtpsneos) And so 1 singe teaaa O} TIL Leff (f)=cos#; then Cee e-Pt(tsin t—pcost)|*__p. costh=(e-#aettsint—peos jee L{cost}icostd FT \?reas And so +h costa (9) SEC, 3,THE TRANSFORM OF AFUNCTION WITH CHANGED SCALE OF THE INDEPENDENT VARIABLE. TRANSFORMS OF THE FUNCTIONS SINat, COS at Letusconsider thetransform ofthefunction /(af), where a>0: L{fay}= fe7P4f(at)dt. Wechange the variable inthe latter integral, putting 2=al; hence, dz=a df; then weget ss Lran}=ife* “teas *)Incomputing theintegral Seretae ‘onemightrepresent itasthesumo integralsofrealfunctions; thesameresultwouldbeobtained. Thisalsoholdsforthetwo subsequent integrals, The Linearity Property ofaTransform 857 1p(2Lyan}=te(2). Thus,it FOFIO then ,Ee(2)+100. ay Example 1.From (9),by(11), westraightway get 21d sinal £3(ye sinat+a rte) Example 2.From (10), by(11), weobtain 2 cosat + teorn or 2? cosat Pa. ) sate (a) SEC. 4.THE LINEARITY PROPERTY OF ATRANSFORM, Theorem. The transform ofasum ofseveral functions multiplied byconstants tsequal tothesum ofthe transforms ofthese functions multiplied bythe corresponding constants, that fs.if 1O=D Cif ay Fy (Cjateconstants) and‘hes POEM, FMFhO, FO=Z CF) a) Proof, Multiplying alltheterms of(14) bye~?! and integrating with res- peel to!from0toc(taking thefactorsC;outside theintegral sign),weHanple 1,Findthetransform ofthefunction F(t)=35in4t—2cos5t. Solution. Applying formulas (12), (19), and (15), wehave 4 p__ 12a LY Ohad pte? PomHeWOb@3pete?pre”PEEFEB 858 Operational CalculusondCertainofItsApplications Example 2.Find the original function whose transform isexpressed by the formula em5%POTEES Solution, Werepresent F(p) as Fo=3 a tO,OD PFET PFO Hence, by(12), (19, and (14) wehave{Ee Sort, From theuniqueness theorem, Sec. 1,Itfollows that this Istheonly original function that corresponds tothe given F(2) SEC, 5,THE SHIFT THEOREM Theorem, /fF(p) isthe transform ofthe Junction [(t), then F(p-ba) istheansorn ofIRDRenetion’@ eFCines aG0) UFO) st HFH iD thenF(p+0)+e" [IsassmaderethatRepta)> Proot. Find iheranstorm ofie tunctlan em" (0, Leto} rrp9timfemPe9*FCO Thus, Lle“fO}=F 040). This theorem makes itpossible toexpand considerably the class oftrans- forms forwhiel itiseasy tofind theoriginal Tunelion®. SEC. 6.TRANSFORMS OF THE FUNCTIONS eth SIN uhCOSH af,e°" SIN af, e- COS at From (8, onthebasts of(15), westraightway get eywate 09) Similarly, pate (syaa Substracting from theterms of(161) thecorresponding terms of(16)anddivide ingtherests bywo, weeet V/A) gen(sha) +ze ) Transforms oftheFunctions e~*!,Sinat,Cosat,e~Sinat, e~*'Cos.at859 peatsinnat, an Similarly, byadding (16) and (161), weobtain espeatcoshat. 08) From (12), by(18), wehave praee tetsinat. 9 Using formula (15) wegetfrom (13) PHO omstcosa orate? o en tonEzAmle 1Findtheorginal funtion whose transform isgivenbythe 1 POET TD Solution. Transform F(p)totheformofexpression ontheleft-hand side of(19): en er PHFCFOFlo4OFTE Thus 74 $5DEE Hence, byformula (19) wewill have Ae -Fe petsinat, Example 2.Find the original function whose transform 4sgiven bytheformula ‘eo-_?POFIFO" Solution. Transform the function F(p): apPt8_ Otte ptt, 2 PHF OFFS” OFFER OF IET =Pty? 38O+IF FETT FIFE using formulas (18) and (20) wefind theoriginal function: F(py3erteosttZe!sinst, 380 Operational CalculusandCertainofItsApplications SEC, 7,DIFFERENTIATION OF TRANSFORMS Theorem. 1]F(p) F(t), then (0gpFHT00. en Proof. We first prove that iff(t) satisles condition (1), then the integral Jert(—onrmat 22) exists. °Byhypothesis [7(0)|<Me'!, p=a+t-ib, a>si;anda>0; %>0, Obvie ously, there will beane>0suchthatthe’inequality a<s,-+e‘willbetule fled? AsinSec. 1,itis proved that the following integral exists: fervor f(alat. We then evaluate the integral (22) GJereterrcn |ae=fe-e-ote-stry (|at. Since the function e~“¢" isbounded and, inabsolute value, isless than some number NVforany value ¢>0, wecan write Flereter faecaGeena co]armn Geen" |e[at<e. Itisthus proved that the integral (22) exists. But this integral may be fegatded abannilvorder derivative withrespect totheparameter *)polthe integral Se-Ptr(nat. And so, from formula F(=(ertF(nat wwegelthe formula SenetorpdtaginJerrtrinat. *)Earlier wefound aformula for differentiating adefinite integral with respect toareal parameter (see Sec. 10,Ch. XI), Here, the parameter pisay complex number, but the differentiation formula holds true The Transforms ofDerivatives 861 From these two equations wehave a”cr Corgroalenter, which isformula (21).. Let ususe (22) tofind the transform ofapower function. We write the formula (8); 1 dan 3 Using formula (21), from this formula weget aaog(F)+ or : Ase Simian “ Ase. pee For any awe have ; pte 5 (23) Example 1.From theformula [see (12) Fan)erinae, bydifferentiating thelef€andright sides with respect totheparameter p, we get 2a sinaeit tinal m% Example 2.From (13), onthebasis of(21), wehave —GEE Theosat, %) Example 3.From (16), by(12), wehave 1 aedort! 25) SEC. 8,THE TRANSFORMS OF DERIVATIVES Theorem. 1/Fp)+1(0,then PF(e)—1 0)+1(O @ Proot. From thedefinition ofatransform wecan write uy(op=Serr (at. (28) 862 Operational CateulusandCertainofItsApplications We shall assume that all the derivatives f°(0, f(D, «21» J2(®) which wetncounter Sateythe\coniton (ieaneconse, iheata G3)and similar integrals forsubsequent derivatives exist. Computing byparts thein- legral onthe right of(28), wefind LAr(ahJerr(dimer P(O|?+0e-AtTDat, But bycondition (1)d dimer10=0 and “Jetty dt=ron. Therefore BLAFO}=—1O+ oF(n). Thetheorem Isproved. Letusnowconsider thetransforms ofderivatives ofanyorder.Substituting into(27)theexpression pF(e)—1()inplaceof F(p)andtheexpression ’(t)inplaceofTb:weget PLO (FOF +P or, removing brackets, BF(e)—PfOF(0)+P(0. co) ‘The transform for aderivative oforder nwill be BF (BLP(OE PHO)ooPI—*(41° (OF(D-(80) Note. Formulas (27), (29), and (30) are simplified if =f =...aneRTSTanstasewegetTeminedFLOL) FO) +10, PF(p)+P (Oe BF (9)+1 (0. SEC, 9,TABLE OF TRANSFORMS For convenience, theIransforms which weobfained arehere given inthe form ofatable ‘Note. Formulas 13and 15ofthis table will bederived later on, Note. Il'for the transform ofthe function /()we Lake Fan lerfiat, then inthe formulas 1-19 ofthe able the expressions Inthe fist column must bemultiplied byp,and formulas 14and 15will take onthefollowing form. Since F*(p)=pF(p),itfollowsthat‘bysubstituting intotheleftside Table ofTransforms es Table 1 1 14 1 2 a sinaPra ae 3 ee 0sat H 4 a aPta . 5 —, al ple . sinhat 6 P ata cos hat 7 — e~"sinat@+aF+at 8 Pa — Darra ac 9 a ~ aa ' 10o_o inoem fsinat n aetoe ore teosat 12 1 et oe 13, 1 i!(sinat—atcosaf)wr Es iCrm “10 5 Fup)FAP) Shemhe—nae of14theexpression neinplaceofF(p)andmultiplying byp,weget ; 2ptt (EOD) 5uw. reg (5O) se70. 864 Operational CalculusandCertainofItsApplications Substituting into the left side of15 Fr FL)RO=+ 2,AO ‘(p) P () P and multiplying this product byp,wehave ‘ B phe Ros) AOAG—vae, SEC. 10, AN AUXILIARY EQUATION FOR AGIVEN DIFFERENTIAL EQUATION Suppose wehave alinear differential equation oforder mwith constantcoedelentsyyOywoesOpnax tx dx OyGatasTarantoeFOnaGetMa=P(O. @n Itisrequired fofind asolution ofthis equation x= (t)for120 that sati- sfles the initial ‘conditions KO, OR, ayxRORAM, 32) Before, weused tosolve this problem asfollows: wefound the general solu- ion ofequation (31) containing xarbitrary constants; then wedetermined the constants sothat hey should satisly the initial conditions (32). Herewegive simpler method ofsolving thisproblem ‘singoperational calculus. We seek the L-transform ofthe solution x(¢) of(31)_satisfying the conditions (32).Wedesignate thisL-transform by‘¥(p);thus,¥(p)~x(#).TLetussuppose thatthereexisttransforms ofihesolution of(31)andof itsderivatives oforder m(alter finding the solution wecan test the truth of Thisassumption), Wemultiply alltermsof(31)bye-Pt, wherep=a-+ib, and integrate with respect 10ffrom 0toat f-ngte Fanatics han ReanyeneThatta, |eneTFat...ta,je-Mx(ndt=e-rtf(oat.(83) On the leftshand sideoftheequationaretheL-transforms ofthefunctionx(0) and itsderivatives, onthe right, the L-transform ofthe function f(f), which wedenote byF(p). Hence, equation (33) may berewritten as as) ane ab{i}tab{Gri}+...agh{e}=LO}. Substituting into this equation the expressions (27), (29), and (30) inplaceof ihe transforms ofthe function and ofitsderivatives, weget 840°E(9)Loe Otbat ba ay{0"="F(p)[Bg Dm HAY tains{0%P)—Leal}+4%(2)=F(0). oy AnAuxillary Equation 865 Equation (34) 1sknown asthe auxilfary equation, orthe transform, equation, ‘The unknown inthis equation isthe transform X(p), which isdetermined from it. Transform itleaving onthe left the terms that contain ¥(p): ¥(P)[agp"+ayp"—*+...dgPtag)=0, (Oey EOE ah ELT Hay[PTEytOT oePRE Hagas[PtybEaayLed+F(0). on ‘The coefficient ofX(p) ontheleftof(34’) isanmth-degree polynomial inp, Which ‘results. when inplace ofthe derivatives. we put the corresponding degrees ofpinto theleft-hand member ofequation (31). Wedenote thepoly: nomial by((Pl Pn(P=AQP"OPE hgDtOye (35) The right-hand side of(34’) Isformed asfollows: ‘tecoefficient a, 1smultiplied by%, thecoefficient ay, ismultiplied by’pry-+2y, thecoefficient a,ismultiplied byp*#x,-+p""¥, 4...-ba{""™, thecoefficient a,ismultiplied byp%* +p%=Fx,-4...40°". All these products are combined. Tothis 1salso added the transform oftheFightsideofthedifferential equationF(p).Alltermsoftherightsideof(4’), with the exception ofF(p), form, after collecting like terms, apolynomial inpofdegree n—1 with known coefficients. Wedenote itbyY_—.(p). And soequation (34") can bewritten asfollows: (P) Pn(P)=Vn-s (P+F(P)- From this equation wedetermine ¥(p): Fpaden4FoFy abe FO 36OmGR) ea) oe Determined inthisway,x(p)isthetransformofthesolutionx(t)ofthe equation(31),whichsolution satisfies theinitialconditions (32).Ivwenow findthefunction x*()whosetransform isthefunction x(p)determined byequation (36),thenby‘theuniqueness theorem formulated inSec.1itwillfollowthatx*(¢)isthesolution ofequation (3)thatsatisfies theconditions(2), that is, 2(Q=x(D. Ifweseekthesolutionof(81)forzeroInitialconditions:xy=x,=35...— x{"-=0, then in(36) wewill have ‘p,-(p)=0 and theequation will take the form xin FoFOF 28 sate 866 Operational CalculusandCertainofItsApplications or sw< Fp) .LOGE Fin CH) Example 1.Find the solution oftheequation de Fart satistying the conditions x=0 for f=0.Solution. Formtheauxiliary equation - hoe aFWG+N-0-4 ofFW=THE Decomposing thefraction ontheright into partial fractions, weget sett FO=4- ser Using formulas 1and 4ofTable 1,wefind thesolution: x(Q=1—et, Example 2.Find the solution oftheequation ateoepoet thatsatisfles theinitial conditions: 24=:2)=0 fort=0. Solution. Write the auxiliary equation (34’)=, — 1 z y= or Z(—)= B)+9=oFFO=spEEHy Decomposing this fraction into partial fractions, weget 14 Fen?=aET tp Using formulas 1and 3ofTable 1wefind the solution: 1 1 xapcos WHE. Example 3.Find thesolution oftheequation ates ade $53Hporet thatsatisfies theinitial conditions j=, =0fort=0. Solution. Write the suxiliary equation (24’) E(p)Oren d or ro=-452.—7- "> 5OpERED —PFOFNOTD * Decompesition Theorem” sr Decomposing this {ration into partial fractions bythemethod ofundetermined Coelfictents we obtain soett_3tyt 1 *O-apateeTOFD” From formulas 9,1.and 4ofTable 1wefind the solution: Vy Sygtdee, x=gt—gteti tes, Example 4.Find thesolution oftheequation afhp9aSena salistying theconditions xy=1, x,=2 fort=0 Solution. Write the auxiliary equation (34°): Fp) +2p-48)—pl+242-14L4 sin} o spp cgn _ 1FOC +%+)=pt4t ots. whence wefind(9): setts 1*O~ Fm TSTHEN OTTO” Decomposing the latter fraction onthe right into partial fractions, wecan write _ett —hetFw=7 tePEIEST—pFT Fells etloy et at *O=19°GEFETISWHIFF 10"PITS aE" Applying formulas 8,7,3,and 2ofTable 1,weget thesolutionMWtrop9142e-taintrconta! 2O=Hec082+sesin2—75cost+sint or, finaly,xipnent(entBomat)=contdat SEC. 11, DECOMPOSITION THEOREM. Fromformula 9,oftheprevious section itfollows thatthetransform of thesolution ofalinear differential equation consists oftwo terms: thefirst term isaproper rational fraction inp,thesecond term ts. fraction whose numerator isthetransform oftherightsideoftheequation Fe).whilethedenominator” isthepolynomial "9,().:lfF(p)isavrational. fraction,thenthe second term will also bearaffonal traction. 1islus necessary 10be Able fofind the original. Tunction ‘whose fransform isthe proper rational a 0 Operational Calculus and Certain ofIts Applications fraction, We shall deal with this question inthe present section. Let the [transform ofsome Tunetion beaproper rational fraction inp: Yom (0). PnP) 11isrequired tofind theoriginal function. InSec. 7,Ch. X,_1t was shown That any proper rational fraction may berepresented” inihe'form of&sum ofelementary actions offour types: i ae ia U.a(p—ayF* AILAg+ wheretherootsinthedenominator arecomplex, thatPrapra," nomi plex, .fegeo, W.ABE, whereAi=2,theensthedemmintr atecomptesLet“tsfind“iheoriginal functions fortheseelementary fractions, For fraclion type 1weget(orthebasis offormula 4ofTable lyALeae a For atype Ifraction, byformulas 9and 4ofTable 1,wehave Ag tpgcoor 4G en Let usnow consider the type IILfraction, We perform identical transforma- Ap+B Ap+B 7Papta 7 Vay rs (+9) +(Va-#)4)4(p—A% a A(o+$)+(2-44) A+ i(Vat) *(Vty * (e+$)'+(V a-3) (0+ $)'+(Va3) (eaye( Va) Denotingthefist_and_second termsbyMandWrespectively, weget(from formulas 8and 7ofTable 1) -te —> pAe * xMxhecosV7Pacis Examples ofSolutions ofDiGerential Equations 869 = (p—A% L ~! on/ Ssws(2-4) 54 myoF 4-7 And, finally, Arto.PP+ap+a,” ef aona a=eF4AcosVuk ateVwi +8)V:an? We shall notconsider thecase ofelementary fraction LV,since itwould in- volve considerable calculations. Weshall consider certain special cases below. SEC, 12. EXAMPLES OF SOLUTIONS OF DIFFERENTIAL EQUATIONS AND SYSTEMS OF DIFFERENTIAL EQUATIONS BY THE OPERATIONAL METHOD Example 1.Find the solution ofthe equation axaatsemsinde thatsatisfies theinitial conditions %y=0, x5—0 when ¢=0. Solution. Form the auxiliary equation (34) 7 =p0+0ta2—, T=?FOC+)=PI0t sy. Wap a3 = _ a 13 3. 2 IO=ayst ETASPESTO PT whence wegetthesolution f x(o=fsin2—F sins, Example 2.Find thesolution ofthe equation axfeeno thatsatisfies theintial conditions x4—1,3)—3, 25=8 when f=0. Solution. Form the auxiliary equation (34°) ZOtap+p3+8, we find FpaZt32t8___P+30+8etl C+DG pth) ‘870 Operational Calculus andCertain of11sApplications Decomposing therational fraction obtained into partial fractions, weget B4Spt8 2, pt6 GINE=eFy PHTtHpsi1 v3 io ema ade omOepet Ty" syVs" ly ieC-ay+(Va) 9Oa) Using Table 1,we write the solution: Sons v¥3,, 1. V3e(Qedetpe (euBeaPainGr). Example 3.Find the solution oftheequation asapt*=600s2 ‘that satisfies the{initial conditions x=0, x,—0 when £=0, Solution. Write the auxiliary equation (34) <motsye ,FOU = GTR ~ 5.1.5 1.8 4 O=— oR OFT TT Consequently, 5 5 wal =~FantFynet-5(7snk—¢cos2”) Obviously, theoperational method may also beused tosolve systems of linear differential equations. The Tollowing isanillustration. Example 4.Find the solutions ofthesetofequations a sF+m+ Ha, de, dyFtsrw—0 that satlaly the initial conditions #=0, y=0 when (=0. Solution. Wedenote x(t) =x (p),y(f)++y(p)andwritethesystemof auxiliary equations: O+2F +7O=2, Px(—)+Up+3)¥(p)=0. Solving this system, wefind Fo= ets ek=FPF NUIpTO 2SFI) WUlp+e* Fw aprbeen-t (sr-ms) ‘ UpFOO+H~ 5\p+t—Tipe) ~ The Convolution Theorem a Fromthetransforms wefindtheoriginalfunctions—thesought-forsolutionsofthesystem: aotsM=g-geta pee 1 ote y= 5lef-e * ), Linear systems ofhigher orders aresolved insimilar fashion. SEC 18, THE CONVOLUTION THEOREM The following convolution theorem isfrequently useful when solving differential equations bythe operational, method. Convolution Theorem. IfF,(p) and F,(p) arethetransformsofthefunctions AW and f,(t), that is, Fy(p)+f,(0)andFy(p)+h then F,(p)-Fa(p) tsthetransform ofthefunction Sn@ne—n de, thatis, ‘ A FOF 0(het ae. co) Proof. We find the transform ofthe funetion fi Shonen fom the definition ofatransform: L{Jnione—oae b=fer[Crean—merJat, ‘TheIntegra onthe rightisadouble integral oftheform({@(x,#dtde, which is(aken overaregion bounded bythestraight ines <=0, =e (Fig. 980). Changing theorder ofIntegration inthisdouble Integral, weget A * ¢ 2{Sno@ne—nae}=( [romfern naeJas. Changing thevariable t—t=z intheinner integral, weobtain Jerrimndtmerep,(9deme[ert(deer Fi. er Operational CalculusandCertainofItsApplications Hence, ‘ . - £{Sheone—oar} (hier Ad=Fre)[e-Phdem= =F, (p)Fy(p). And so ' Shemht vrs FrFrio). Thisisformula 15inTable 1. Note1,Theexpression [f,(s)fx((—t) dsiscalled theconvolution y‘4 (Faitung,resultant) oftwofunctionsf,(f)and[4(0 ¥ The operation ofobtaining itisalso known asthe tr convolution oftwo functions; here ‘ ‘ Sno@nt—odr=J eonids. at t 3 3 Fig.$60. Thatthisequation istrueisevident ifwechange the variable ¢—taxz inthe right-hand integral. Example. Find the solution fothe equation ee gato thatsatisfles theinitial conditions x,=¥,==0 for=0. Solution. Write theauxiliary equation (34°) FOU+N=F Oo) 1 whereF(p)isthetransform ofthefunction f(f).Hence, ¥(p)=peT FO. 1. ‘ tatgrraeCandFU)1(.AnpingIheexavoltionformuleG2 anddenoting j= Fs)Fi)=Fi we get ‘ x(=JF(9)sia(¢—w)de. (40) Note 2.Onthe basis ofthe convolution theorem itiseasy tofind the transform ofthe integral ofthegiven function Ifweknow thelransform of this funetion; namely, ifF(p)21(0,then A harppros free. ay The DiGerential ‘Equations efMechanical Oscillations 373 Indeed, ifwe denote AO=O, (=H, thenF,e)=FO),Frl)=2. Putting these functions into (39), we get formula (41). SEC. 14, THE DIFFERENTIAL EQUATIONS OF MECHANICAL OSCILLATIONS. THE DIFFERENTIAL EQUATIONS OF ELECTRIC-CIRCUIT THEORY From mechanics weknow that theoscillations ofamaterial point ofmass mare described bythe equation *) @r, Ade kd tS ptr tyo, cr) where xisthedeflection ofthe point from acertain position and&isthe Tigidity ofthe, elastic system, Torinstance, aspring (aesr spring), the force oFresitance to.motion Is.proportional. (the proportionality constant. 163)fotheAstpowerofthevelocity, and(0thebuter” (ordisturbing) force. L Equations oftype(42)describe smallvibrations ofother mechanical systems with one degree ofIree: dom, forexample, thetorsional oscillations ofafly-wheel onanelastic shaft, ixlstheangle ofrotation & e Oftheflywheel, mis themoment ofinertia ofthe flywheel” kis thetorsional rigidity oftheshafl, and mmf), 1sthe moment oftheouter forces relative toUhesieofrotation,Equationsoftype(2)describe Ge hot only ‘mechanical vibrations. butalso. phenomenathatoccur inelectric circuits. Fig.381. Suppose. wehave ‘an_electric circuit. consisting ofaninductance L, resistance Rand acapacitance C,towhich isapplied anem.t. (Fig. 381), We denote by1the current inthe cireuit, byQthe charge ofthe capacitor:then,asweknow fromelectrical engineering, 7andQsatisly thefollowingequations: a°LatRi+ GE (43) 4aRau. a4) From (44) we get(4)we@ rao weana u Substituting (44) and (44°) into (43), wegetforQanequation oftype (42): #Q, dQ, boLoeRBLone. 4s) *)See, forexample, Ch. XIII, Sec. 26,where such anequation isderived inconsidering the oscillation of4weight ona car spring. am Operational Calculus and Certain ofItsApplications Differentiating both sides of(43) and utilising (44), we obtain anequation for determining the current i: i pt 1) ae Rg tin, 6) Equations (45) and (46) are type (42) equations. SEC, 15, SOLUTION OF THE DIFFERENTIAL OSCILLATION EQUATION Let uswrite the oscillation equation intheform ae Gata Gear l(O, an where the mechanical and physical meaning ofthe desired function x,oftheCcelficients ay,oy,andofthefunction|(0)isreadilyestablished bycomparingthis equation’ with equations (42), (48), (46). Let usfind thesolution to equation (47)thatsalisles theinitial conditions x=, x’, when (=0. ‘We form the auxiliary equation for equation (47): FP)ot+a,pay)=x09+4,404%,+F(P), (48) where F(p)isthetransform ofthe function f(t). From (48) wefind z AHR EON Fo)O=prapra toPapta,” Cy Thus, forasolution Q(1)ofequation (45)that satisfies theinitial conditions Q=Qs, Q’=Q, when £0, thetransform will have theform Top atLOHQIRG. E Fyer Ew.Lp?+Ret Let+Ro+e The type ofsolution issignificantly dependent onwhether the roots oftheIrinomlal p*-+a,p4-0, are-complex, ortealanddistinct, orrealandequal.Letsexamine'In'detal theceshenterootsoftbeirinomialarecomples, Anatis,when($)'—a,<0.Theothercasesareconsideredinsiafashion,Since the transform ofasum oftwo functions isequal tothesum oftheir transforms, itfollows from formula (38) that the original function forthe first fraction ontheright of(49) will have theform sentonEleet Prapra * s a fay — aea a 60) +asinYo|. V a> Investigating Free Oscillations tis Let usthen find theoriginal function corresponding tothefraction F(p) Prap +a,” Here, wetake advantage oftheconvolution theorem, fest noting that Prapta,”V:a ‘.meet Hence, from (39) weget peesfrpPOtatVo, roeVa\te siat—0/g—hae.6H afi And so,from (49), faking into account (60) and (6), weget syne*|ayeostTa at|e pect 1b tune a +:freesin@—9YYy—zae 62) Vay UItheexternalforce((()m0,whichmeansthatifwehavefree,mechanical orelectrical ocilations: then the solution igiven bythe fst: term onhe right-hand sideofexpression (62). Iftheinitial data areequal fozero, i.e., it'sy=x,=0, then thesolution isgiven bythesecond term ontheright side of(52). Let usconsider these cases inmore detail. SEC. 16, INVESTIGATING FREE OSCILLATIONS Let equation (47) describe tree oscillations, that 1s,/(Q)=0. For con- venlence inwriting weIntroduce the notation ,=2, a,—h#, Af=i—nt, Then (47) will have the form Fi+2He=, Co) Thesolution ofhisequation x,that satistes theintial conditions x= a8, forf=0 iselven bytheformula (60)orbythefrst Term of(62) sp(omen™[xscos1 sa]: on «6 Operational Caleulus and Certain ofItsApplications Wedenote0,ty, 1isobvious thatforany@and8wecan select Mand8suchthatthefollowing equalities willbefulfilled: =Msin8,baMcos8, here, Miaat+o% tan d= >. We rewrite formula (54) as y=en"[Mcosky!sin8-4Msinbytc0s3), or, infinal form, the solution may bewritten thus: paVOPReH"sin(byt+0). 65) Solution(5)corespond to,damped ocilations: Idna)=s0, that 1s,ifthere Isnointemal friction, then the solutionwillbeoftheform p= VEPBsin(yt-+8).Inthiseaseharmonicoscillations occur.(InCh.XIIL,Sec.27,Figs.270and211 give graphs ofharmonic and- damped coscilations.j SEC. 17. INVESTIGATING MECHANICAL AND ELECTRICAL OSCILLATIONS IN THE CASE OF APERIODIC EXTERNAL FORCE Whenstudyingelasticoscillations ofmechanicalsystemsand,inparticular, whenstudying electrical ‘oscillations, onehastoconsider diferent typesof External force f(®). Let us.consider indetail the cave ofaperiodic external force. Let equation (4?) have theform @x dxget gythem sinat. (66) To determine the nature ofthe motion itissufficient toconsider the case when x,=x/=0. Onecould obtain thesolution oftheequation byformula (62), butpedagogically speaking, itismore convenient toobtain thesolutionbycarrying outalltheIntermediate calculations. Lat tswrite the transform equation FW OrMptharqe,OO pt=APS from which weget =m Ao5O=GEER EFO CoWeconsiderthecasewhen2n#0(1#<2),Decompose thefractionon theright info partial Trections AwaNO+B 4Co+D a PEt) OyPMR por We determine the constants B,C,Dbythe method ofundetermined coefficients. Using formula (38), we find the original function from its Investigating Mechanical and Electrical Oscillations sm Laransform (67): A A=rathpaaar {meant tnacon+ tentt[(2mt—at-bon sianyt+-200conht]bs 6 here again, j=Vat, This isthesolution ofequation (66)thatsatisfies theinitial conditions ay=x;=0 when f=0. Let usconsider aspecial case when Zn=0. This corresponds toamecha. nical systems wilh nointernal fesitaties, offoanelectric circuit whste R=rO (no internal resistance inthe circuit). Equation (56) then takes the form osEEemAstnot, co andwegefthesolution ofthisequation satisfying theconditions x,=x,=0 for¢=0'if in9) weput n=O: " % 0—prtgre lest tsinatl.(6 Here wahave thesumofIwoharmonic cecttss OTT11|a Hens:naturaloscillations withtreqocsey’ Patt Ao aaphitttiyy irs Fatt=—apyHO, Hit ie forced oscillations with frequency «: wnmat ne AM irs& Aye)=pagsinot. He ofoscillationsforthecaseA>is showninFig,382. > HittLetusagainreturn toformula (69). 3 7 Utan0(whieh‘occurstnthemechanical andthecal forcesunderconigeaten, thenthe termcontaining thefactor e—"", which represents Fig.382, imped tart xlations orfneetig Tapidly decreases. For fsufficiently targercite character ofthe oscillations will bedetermined bytheterm that does notcontain thefactor e~"; that is, bytheterm A 7 #0=eaten ((H—0%snot—2nacost}. @ We introduce the notations AW) _cosg; A2noeae Mets Aap Mand, 6) where A a VBoanat 878 Operational Calculus and Certain ofItsApplications The solution (62) may berewritten asfollows: #1)=A not+0, 6 From formula (64itfollows that the frequeney offorced oscillations coineices with: that of‘the external fore, Ifthe Infernal resistance, characterised Dy the number'n, ismall ‘and thefrequency oftheexternal Tofce atsnot very different trom that ofthe natural otilations hythen the amplitude ofori Iations may'be made asreat atone. plesses, since thedenominator may be arbitrarily small. For 'm=0, a=, fhe solution Isnot expressed byTor- mula (64). SEC, 18, SOLVING THE OSCILLATION EQUATION IN THE CASE OF RESONANCE Let usconsider the special case when a,—2n—0, that is,when there is noresistance and the Treguency ofthe external force coincides with that of ihe natural osellations ‘ova. The equation then takes the form Beem Asia © Weshall seekthesolution thatsatisfies theinitial conditions x,=0, x,=0 for (0. The auxiliary equation will be . yeaE(p)(P+)AnyE, whence a OGRE GS) Wehave properrational action oftypeV,whichwehavenotconsidered In'the general form. To. find theoriginal lunclion forthe transform ol(66), welake advantage ofthefollowing procedure, Wewrite theIdentity formula 3 ofTable 1) kopie (oP sinatdt.PreJsinktdt ry Wesiteenate bthsideofthisequation withespe fo(theintegral onIheightmay fereened nthe formoasun ioVterate ot variable, each ofwhich depends onthe parameter A): apt emenet cosaFreTRF Jfeosktdt. ‘Utilising (67) wecan rewrife this equation as Det i 1aa(et[reeead at, rey i[°s& The Detay Theorem 89 Whence ifollows directly that Ak. ASABa a(gett—teonte) (Grom thisformula wehaveformula 13,Table 1).Thus, thesolution ofequa Hon (65)satisfying theinitial conditions x4—x,—0 for¢=0 willbe (0=f(palaht—teoe ht) 68) Let usstudy thesecond term ofthis equation: Abcoshts , (QS—FEfoshts (68) This quantity isnot bounded as¢increases. The amplitude ofoscillations that correspond {oformula (68') inereases. ‘without bound astincreases without Bound: Hence, the amplitude ofoscillations corresponding toformula (68) also Increases without bound. ‘This is.resonaner; itoeeurs when the frequency of the natural. oseilations coincides ‘with that ofthe external force (see also Ch, XIII, See. 29,Fig. 273}. SEC. 19. THE DELAY THEOREM Latthefunetion f),fr4-<0,beidentically equalfoaero(Fig.383, Then thefunction /@—f,) will be’identically. zero for f'<ty (Fig. 983, 6). We shall prove atheorem which isknown asthe delay theorem. 4 as { H 0 t 0 to t @ o Fig, 383. ‘Theorem. IfF(p) isthetransform ofthefunction f(t), then e~°"F(p)is thefanfor ofthefunction {cei that fF2Uaethen F(t) +“PF(p). Proof. Bythe definition ofatransform wehave . 4 . Leto =ferrr—tya=|eneptydt|e-FtUay dt,3 3 a 880 Operational CalculusandCertainof14s.Applications ‘The ‘irst Integral onthe right ofthe equation iszero since [(¢—t)=0 forreAtthelastIntegralwechangetheverlable,pulling(4,23) LAY—typ=femPEHf(a)deme|-FY(a)demeMPMFp), Thus,[tt e7PF(9). Example. InSec. 2ifwas established forthe anHeaviside unit function that at aes. Itfollows, from the theorem that has just been a F proved, that forthefunction 0,(¢—A) depicted InFig. 384, theC-transform is 1 Fig.548, der, boP that is, 1 tanya hem, ottt Exercises onChapter XIX Find solutions tothe following equations forthe indicated initial condi- Hons: ax dx fe - mate-t gent 1SpoBporeo,eel,x2fort=0.Ans.xmden!—3eH, aya P 2OEGeo rer,x=,Felfor(=0,Ans.r=l—tte, a ro 3TEaECapoxm0,rmayYee,forf=0.Ans.rm ==[xbcosbf+(x,—xya)sin64]. x jde = , ad et 4Ea tet, rel, ee? fortO. Ans. rebely La 4attpeape, ae : a 5Tetmemacosnt, xmzyax,fortO.Ansreat x x(ornconm0-+4,6081-4conmts 6.FEW at, 0, x0 for1-0, Ans.2=3e'—1e213, aeants=9 aed 3 : < Bxtetits onChapter XIX : a1" a_ pe 2Gitepee,newaxa0fora0,Ans,rad(Poort 3)tidetdfeos(1ya)—Lyayle +h)eaggertg fom(gy78)—Van(7H) he (a =s,=4,=0 fort= a aSpel, yexaa a0fortm0.Ansxet—t 23143 Fe cos 2 de ode a tteoS oF temsint, ymaym eng 0 for fm0.Ans. x= =FO—Hsint—S toss, 10,Find solutions tothe system ofdifferential equations ax ayGetynl Fats=0, thatsatisfy theinitial conditions xy=y—s,—y,—0 forf=0. Ans. (= apeosttttttetymcosttebe. iwpex A Auxitary equation, 836,865 Average wceetertion, 128 bets theorem, 142, 748Abssute constant 18 Average curvature, 31,927 Abwaute value TS eeAbsolutely convergent integral, 420, Sfimogineries, 293 Abvoltely convergent serie,71 me as”\cceleration: 4 atagiven instant, 125 2 Gflinear mation 125 . Adon formula $87 Aigare Bernoulli's equation, 480492 fondamental theorem of, 248 Bernstein, S.Na282 Agebrale eguaion, 225, 28, Beratts polyoma, 252 AgebrateTunetions, 26; 28 Bel funeion ofthe fst ind, 785Alferatingele,727 Beselfonctionofthesecondkind,767Amplitdeota complex number), 294 Bese's equation, 738, Tot Amplitode (otsoulfetion, 858" Besel's equally, 7Analyse Binomialdierent, 875harmonic, 808 Binomial sere, 84780 Anayttea expression, 21 Bina, 53 Anas ofcontingence (olanare), 210° Boundary conditions, 818, 826, 628 Antiderivative, 2 Boundary of«domeia, 286 Arc length o's curve, 47482 Boundary-vatue conitons,_818 archimeds Boundary-elue_pobtem spiel of,29 tint, 825,857 sequent 19 trond, 807 Str complex number, 234 Bounded function, 40, 41 Intermedlate, 88 Bounded variebe, 18 Aste 107 Briggs 56 Asymptote, 169 Broker tne Tec, 191 Euler $83 vertical, 190 Bunyetovsky, 647 Subject Index 883 Bunyakovsky's inequality, 647 powers of,237 Birgi, 56 roots of, 238 subtraction of, 235 c trigonometric form of,234 Complex plane, 240 CalculusComplex roots,248,249 operational,854 Complex variable, 240 Catenary,471 Composite exponential function, 93 Cauchy'stest,721,722 Composite function, 25 Cauchy'stheorem, 143, Concave curve,18 Centreofcurvature, 217 Cancavity (ofacurve),183 Centre ofaneighbourhood, 18 Conditions Change ofvariable, 348 boundary, 818,825,628 Characteristic equation, 570 boundary-value, 818 Chebyshev, P.L.,253 initial, 474, 514, 818, 825, 828 Chebyshev polynomials, 253 Conditional extremum, 200 Chebyshev's formula, 430-435 Conditionally convergent series, 731 Circle ofcurvature, 217 Conjugate complex numbers, 233 Cireulation (ofavector), 673 Conjugate pairs (ofcomplex roots), 249 Clairaut's equation, 505-507 Constant, 16 Glosed contour, 673 ‘sbeolete, 16 Closed domain, 256 Continuous function, 67,58Closed interval, 17 Contour Closed region, 608 closed, 672 Coefficient, 27 Convergence ofaseriesCoetficients necessarycondition for,713Fourier, 779 Convergent integral, 421 ofatrigonometric series, 776 Convergent. series Combined method, 229, AESnisarat ‘Common logarithms, 758 condltlonslly, 73t Complete integral (ofadifferential Convey curve. 183 equation), 475,515 Convex down (downwards), 183 ‘Complex function of@realvariable, Convex up(upwards), 183 242 Convexity (ofacurve), 183 Complex number Convolution, 872 imaginary part of,233 Convolution formula, 872 realpartof,235 Convolution theorem, 871 Complex numbers, 233 Coordinate ‘addition of,234 polar, 28 conjugate, 238 Coordinate system division of,236 polar, 28 exponential form of,243 Correspondance geometric representation of, 283 one-to-one, 634 ‘multiplication of,235 Critical points (values), 168, 294 884 SubjectIndex Curl (ofavector function), 694 of@function defined implicitly, Curve 276, 27 concave, 183 logarithmic, 94 conver, 183 ofalogarithmic function, 84 convex downwards (upwards) 183 Derivative Gaussian, 187 ofnthorder, 119 sinooth, 528 partial, 263-265 space, 450 ofaproduct, 82Curves second,119integral, 475, 515 ofsecond order, 119 resonance, 561 ofasum, 81 Curvature, 211, 212, 215, 216, 327 symbols of,71 average, 211, 327 third, 119 centre of,217 total, 275 circle of, 217 Determinant atapoint, 211, 212 functional, 636 radius of,217 Deviation Curvilinear trapezoid, 400 maximum, 793 Cusp root-mean-square, 793 double, 309 Diameter ofasubregion, 650 Cusp ofthefirst kind, 308 Differentiable function, 74 Cusp ofthesecond kind, 309 Diflerentiable atapoint, 267 Cyeloid, 106, 107 Differential, 113, 114, 116, 117, 118, 267 > ofanare, 210 D’Alembert’s test, 718 binomial, 375 Decomposition (ofarational fraction ath, 121 into partial fractions), 361 second (second-order), 121 Decomposition theorem, 867 third (third-order), 121 Decreasing function, 20 total, 267 Decreasing variable, 18 Differential equation, 469, 472 Definite integral, 396, 398, 399 exact, 492 Degree ofapolynomial, 27,244, 246 first-order, 473-478 Deloperator, 700 higher-order, 514-516 Delay theorem, 879, 880 Mnear, 628, 629 DeMoivre's formula, 237 ordinary, 472Density DifferentiaisTinear, 459 error approximation by, 270 surface, 642 Differentiation, 71 Derivative, 71,72,78,79,80,114 Direction ofcirculation, 672 ‘ofacomposite function, 85,8 Direction-field, 476, 477 directional, 284-286 Directional derivative, 284-286 discontinuous, 168 Dirichlet-Neumann problem, 840, 843 ofatraction, 8% Dirichlet problem, 837 Subject Index 885, Dirichlet’s integral, 800 Fourier (for heat conduction), 815 Discontinuity (see point of.) heat-conduction, 815, 816, 825, 828 Discontinuous derivative, 168 Equation (cont.) Discontinuous function, 60 ‘ofheatpropagation Divergence (ofavector,orofavectorinaplane,828 function),699 higher-order differential, 514-516 Divergentintegral, 421 homogeneous, 482 Domainhomogeneous linear,629 closed,256 hyperbolic, 815 ofconvergence (ofaseries)733 Lagrange's, 507-509 ofdefinition (ofafunction),19,256Laplace's703,815,836 natural, 21,22 incylindrical coordinates, 842 open, 256 linear, 487 Dominated series, 734-736 linear’ differential, 528, 529 Double cusp, 309 Lyapunov's, 796 Double integral, 609 nonhomogeneous linear, 529 Double root, 546 ofanormal, 126 ordinary differential, 472 E parabolic, 815.Pariaditerental, 472 igenfunctions, Parabolic, peepartial differential, 472 Eigenvalues,821 orytangent 126 Element of‘integration, 343 transform, 865 Elementary function, 26 vector, 314eles ene with'a right-hand member, 529 Elliptic equations, 815 withseparated variables, 479Elliptic integral, 385 with variables separable, 479Endpointsofaninterval, 17 toutatightcheed member, 529Envelope (of@familyoflines),498,Winout9ie! "pF Equations Equation parametric, 103,104,314 algebraic, 225,245 telegraph, 819 auxiliary, 535, 865 Equipotential lines, 510 Bernoulli's 490-492 Equivalent infinitesimals, 64, 65 Bessel's, 763, 764 Error characteristic, 870 maximum absolute, 270 Clairaut's 605-507 maximum relative, 272 ofcontinuity, 837 relative, 272 ofcontinuous flow ofacompressible Euler substitution quid, 839 first, 372 differential, 469, 472 second, 373 elliptic, 815 third, ‘374, 375 exact differential, 492 Euler's broken line, 583 first-order linear, 487 Euler's formula, 243, 753 rs SubjectIndex Euler's method (of approximate Formula (cont.) ‘olution offist-ordardiferential Green's, 670-681 tqualion), 681-584 oftateraton Bypars, 354 volute, 219 Lagrange's interpolation, 260, 251 Evolver, 219 Leb, 120, 6 Execl ferential equation, 492 -Maclaurin’s, 188 Exltence theorem of line integral, Newton-Letbnly 410, 411 on Ostogradshy’s 697700 Expansion (of function), 186-159 In parable, 28 iaylors sels, 186 Festangutar, 424, 425 Expl fonction, 90, 91 Simpoon'y 428 Exponential form (ol complex num- Stoker's 692697 ens 203 Toylo 182, 188 Exponential function, 22,24,99, 102 for transformations ofcoordinates ropeties of,241 ma double integral, 638 edpofential powee tention, 98 eosotaattese Expression Wallis’, 415,416anatytca, 21 Formulasrtvectevals(of«fonction,168“SeatPeo,336Eriremum (esbrems} (oa fooeloeh, pooni setncte, 779 166, 292 Fourier cosine transform, 810 conditional, 300 Fourer equation forheatconduction, ais r Fourler Integral, 96-808 incomplet form,810-812 FactorFourer tavetetansorm, 812 integrating, 495-497 Fourier series, 776-812 Faltung, 872 definition of,779 Fill cures, 475 Fourer sietransform, 810Family ofTunctions, 343 Fourier transform, 812 Family oforlhogonal iajectoris, 510. Fraction Tae, SPradients, 267 partial, 358 First Euler substitution, 372 Proper, 357 Flow fins, 810 Frectonalratfonal unetlon, 27 Flux (ola vector field through asur. Free vosllatons, S57, 876876 tac), 688 Frenet ice Serel-Prenet formulas, Foret’ oscillations, 657, 59569 38) Formula Frequency, 858 ‘Adams, 557 Fonction 19 Gheopiev's, 490435, aigerale, 26,28 convelution, 872 tnutytcaleepresentation of,2 De Moivre, 237 Baste elementary, 22 Euler, 23, 188 Bessel, ofthe ft kind, 765 Subject Index 887 Bessel, of the second kind, 767 Function (cont.) Bounded, 40,41 piecewise continuous, 797 composite, 25 piecewise monotonic, 779 composite exponential, 9 power, 22, 23, $3, 102 continued in.even fashion, 791 power-exponential, 93 continued inodd fashion, 792 Quadratic, 27 continuous, 57, 58 Fational integral, 27, 244 continuous in‘ domain, 261 represented parametrically, 104, 128 Continuous over an inierval, 59 ofeeveral varlables, 255 Continuous onthe left, 59 single-valued, 20 continuous atapoint, 261 tabular representation of,20 continuous onthe ight, 59 transcendental, 28 decrease of,163 trigonometric, 22, 24, 102 decreasing, 20 unbounded, 41 differentiable, 74 Functional determinant, 636 Function (cont) Functional relation, 19 differentiable at»point, 267 Functional series, 738, discontinuous, 60 Functions elementary, 26 hyperbolic, 110, 111 explicit, $0,91 linearly dependent, 539 explicitiy defined, 90 linearly independent, 699 exponential, 22,24,93, 102 rational, 357 exponential-power, 93 Fundamental theorem ofalgebra, 245 fractional rational, 27 ofafunction, 25 « frapbcal representation of,21tatfemeton, 385harmonic, 703,836 eee Gate i7Heaviside unit, 855 General solution (of@differential een equation), 474,475,515 genous,apheroataretcer Geometric mean,905increase of,163° ‘Ceometric progression, 710 ‘‘ Gradient,286,287 aie° Graph,21 ainaenienehed Greatestvalue(ofafunction), 61 initial,Green, D.,681 inverse,95, Green's formula, 679-681 inverse trigonometric, 22, 102 investigation of,194-198 fn ferational naz) Hamiltontan operator, 700 Iinear, 27 Harmonie analysis, 605 logarithmic, 22,24,103 Harmonic function, 703,836 mltiple-valueé, 20 Harmonie oscillations, 558 perlodie, 26 Harmonic series, 714, 715 888 SubjectIndex Heat-conduction equation, 815, 816, Infinitesimal oflower order, 64 825, 828 Infinitesimal quantity, 45 Heaviside unit function, 855 Infiitesimals Helicoid, 316 equivalent, 64, 65 Helix, 315, 316 ofsame order, 63 Hodograph, 314 Inflection (point ofinflection), 185 Homogeneous equation, 482 Initial condition, 474, 514 Homogeneous function, 482 Initial conditions, 8,18, 825, 828 Homogeneous linear equation, 529 Initial function, 855 Hyperbolic equations, 815 Initial phase, 558 Hyperbolic functions (sine, cosine, Integrable (said of@function), 399 tangent, cotangent), 110, 111 Integral, 473, Hypocyctoid, 449 absolutely convergent, 420 complete, 475, 515 1 convergent, 421 Identity, 364 definite, 396,398,399 Imaginary Dirichlet's, 800pure, 233 divergent, 421 Imaginary axis, 233 double, 609 Imaginary part’ofcomplex number, elliptic, 385 233, Fourier, 806-808Implicit function, 90,91,122 improper, 416,417 Improper fraction, 357 improper’ iterated, 631 Improper integral, 416,417 indefinite, 243 Improper iterated integral, 631 iterated, 611 Inclined asymptotes, 191 line,671,674 Increasing function, 20 particular, 475 Increasing variable, 18 Poisson's, 835,846Increment three-fold iterated,651-655partial (ofafunction), 259 triple, 650,654,656,658,659 fotal (ofafunction), 259, 265 Integral curves, 475,515 Indefinite integral, 343 Integral sign, 343, Independent variable, 19 Integral sum, 398,608 Indeterminate forms, 144-147, 150-152 Integral test(lorconvergence), 723-725 Inequality Integrals Bessel’s, 796 Table of,345 Bunyakovsky's, 647 Integrals ofirrational functions, 371, Schwarz’, 647 383 Infinitely large quantity, 39 Integrand, 243 Infinitely large variable, 34 Integrate (adifferential equation), 476 Infinitesimal, 42-45 Integrating factor, 495-497 Infinitesimal’ function, 42,44,45 Integration (ofafunction), 344 Infinitesimal ofhigher order, 64 Integration ofbinomial differentials, Infinitesimal ofkth order, 64 315 Subject Index 889 Integration byparts, 354-356, 413-416 Laplace transform, 855 Integration ofrational fractions, 365 Laplace's equation, 507-509, 703, 815, Integration bysubstitution, 348-351 836. Integrationoftrigonometricfunctions,Laplactanoperator,703,836 378-383,Least value (ofafunction), 61 Interior point (ofaregion), 610 Leibniz (see Newton-Leibniz formulay Interior points (of adomain), 256 410, 411) Intermediate argument, 85 Leibniz’ formula, 436 Interpolation, 250 Leibniz’ rule (formula), 120 Interval, 17 Leibniz’ theorem, 727, 728 closed, 17 Length of ofintegration, 399 ‘anarc, 208 open, 17 8normal, 127 Invariance (ofform ofdifferential), 117 asubnormal, 127 Inverse function, 95 fasubtangent, 127 Inverse trigonometric function, 22,102 atangent, 127 Investigation ofafunction, 194-198 Level lines, 283 Involute, 219 Level surfaces, 283 Irrational function, 27 L'Hospital’s theorem (rule), 145 Irrational numbers, 13 Limit Irrotational vector’ field, 702 lower (ofanintegral), 399 Isogonal trajectories, 609, 512-514 upper (ofanintegral), 399 Isolated singular point, 310 Limit of Iterated integral, 611 analgebraic sum ofvariables, 46 evaluation of, 615, 616 afunction, 35, 261 improper, 631 @product, 46 three-fold, 651-655 aquotient, 46 avariable, 32 4 Line Jacobi, 636 secant, 265 ‘Jacobian, 636, 659 Line integral, 671, 674 Line tangent, 73 K Linear densify, 459 Linear differential equation, 528,529 LeBOL) Linear equation, 487 D Linear function, 27 Linearity property (ofatransform), L-transform, 855 857 Lagrange form ofremainder, 155 Linearly dependent functions, 539Lagrange’s interpolation 'formula, Linearly dependent solutions, 30250, 251 Linearly independent functions, 539 Langrange’s theorem, 142 Linearly independent solutions, 530 Laplace equation incylindrical coor- Lines dinates, 642 flow, 510 89 SubjectIndex level, 283, Minimum (ofafunction), 165, 169, equipotential, 510 178, 292, 297 Logarithm Modulus, 18 common, 258 ofacomplex number, 234 Napierian, 56 oflogarithms, 56 natural, 58, 758 Moments Logarithmic derivative, 94 static, 649 Logarithmic function, 22,24, 103 Monotonicity, 226, 227Lopshits,A.M.,806 Multipleroots(ofapolynomial),Lower limit (ofanintegral), 399 7 Lower (integral) sum, 397 Multiple-value function, 20 Lyapunov, A.M.,576, B81 Multiplicity (ofroots), 247-249 Lyapunov stable (about solutions, conditions), 577 N Lyapunov equation, 796 ;semov's theory ‘ofstability, 576 Nthpartial sumofaseries, 710Lyspunov's theory yongNIHpartial” Napierian logarithms, 56 Maclaurin's formula, 158 Natural logarithms, 86,758 Maclaurin's series, 751-753 Necessary condition (forexistence of Mapping extremum), 166 one-fo-one, 634 Necessary conditions ofanextremum, Maxima (see maximum) 208 Maximum (ofafunction), 164, 169, Neumann problem, 837 178, 292, 297 Newton-Leibniz formula, 410, 411 Maximum ‘absolute error, 270 Newton's method, 227 Maximum deviation, 793 Neighbourhood (of point), 17,260 Maximum relative error, 272 centre of, 18 Mean radius of, 18 geometric, 305 Nodal point, 307 Mean-value’ theorem, 406, 616, 653 Nonhomogencous linear equation, 529Member Normal,221,320right-hand (ofanequation), 529 principal (ofacurve), 328Method Normaltoacurve,126ofchords, 225 Normal toasurface, 339 combined, 29 Normal plane, 320 Euler's, 581-584 Normal system ofequations, 54 Newton's, 227 Number Ostrogradsky's, 368 complex, 233 oftangents, 227 ,51, 53 ofvariation ofarbitrary Irrational, 13 constants (parameters), 543 rational, "13 Minima (see Minimum) Number (cont.) ‘Minimax, 297, 299 real, 13 Subject Index 891 Number pair, 256 Parameter, 103, Number quadruple, 258 Parametric, 103 Number scale, 13 equations, 103, 104, 314 Number triple, 257 Part Numerical series, 710 principal (of an increment), 113 Partial derivative, 263-265 ° Partialderivatives Operatorofdifferent orders, 279-283, v-operator, 700 Partial differential equations, 472 del, 700 Partial fractions, 358 Hamiltonian, 700 Partial increment (ofafunction),259 Laplacian, 703, 836 Particular integral, 475 One-to-one correspondence (mapping), Particular solution, 475, 515634 Partition unit,401One-parameter family ofcurves, 498 Period, 24 Open domain, 256 ‘ofoscillation, 558 Open interval, 17 Periodic function, 24 Operational calculus, 854 Piecewise continuous function, 797 Ofder ofadifferential equation, 472 Piecewise monotonic function, 779 Ordered variable quantity, 18 Phase Ordinary differential equation, 472 ofacomplex number, 234 Ordinary point, 305, 336, initial,558 Origin (ofavector), 314 Plane Original, 855 complex, 240 Original-transform tables, 855 normal, 320 Orthogonal trajectories, 509-512 osculating, 331Oscillations Langent,338forced, 557, 559-563 Plus-and-minus series, 729 free, 557, 875-876 Point Oscillations (cont.) critical, 168, 204 harmonic, 558 ‘ofdiscontinuity, 60,75 Osculating plane, 331 ‘ofinflection (of2curve), 186 Osculation (see Point ofosculation309) interior (of aregion), 610 Ostrogradsky, M.V., 368, 636, 681, isolated singular, 310699 nodal,307Ostrogradsky's formula, 697-700 ordinary, 305, 336 Ostrogradsky's method, 368, ‘ofosculation, 309 singular,306,322,336 PpPoints Parabola interior (of adomain), 256 safety, 502 Poisson's integral, 835, 847 Parabolic equations, 815 Polar axis, 28 Parabolic formula, 426 Polar coordinate system, 28 Parabolic trapezoid, 426 Polar coordinates, 28 02 Subject des Pote, 28 Radius oftorsion (of@curve), 333 Polynomial, 27,244 Radius vector, 314 Bernstein's, 252 Range ofavariable, 17 Chebyshev, 253 Rate ofmotion, 70 Potential offield, 459 Ratio (ofageometricprogression), 710 Potentialofagravitational field,696Rational functions, 367 Potentialofavector,684 Rational integral function, 27,244 Potential vector field, 701 Rational aumbers, 13 Powerexponential function, 93° Ray G06 Power function, 22,25,$3,102 Re'l gxis, 289 Power series, 742 Real number, 13 Principal vormal (ofacurve), 328 Real partofcomplex umber, 253 Principal part(olanIncrement), 13 Resafeatar formedes Anh 498 Principal value (ofamIntegral), 811 peerPrinciple oflocalisation, 802 ‘Slosed, 608 Problem ofintegration, 609 Dirichlet, 857 regular, 64, 611, 626 Dirichet-Newmenn 640, 843 regular inthe ‘direction, 611 First boundary-value, 825, 837 regular inthe y-direction, 611 ofinterpolating @function, 250° perstive error, 72 Neumann, 837 peas etersecond boundary-value, 857 lation py mole pendulum, Un5 ema a Lagrange formof,155 Remainder theorem,244 Progression Resonance, 563,879 geomet,20 fe ren beeperin,36 Resonancecurves, ropertyrd.member(ofanequation) eyotawanarm, sorRighhend member(ooein, Pureimaginary, 233, Rolle’s theorem, 140 a Root Quedeatie funeton, 27 double, 546 Quadratic trinomial, 351 ofanequation, 244 Quantity Ieytuple, 247 aesaneet ofmultiplicity i,247-249 infinitesimal, 45 ofpolynomial, 244 ath simple (ingle), H6 rdered. variable, 18 Root-mean-square deviation, 793 R Roots Complex,248,249 Radiusofconvergence,744, tmultple(of@polynomial),247 Radius ofaneighbourhood, 18 Rotation (ofavector function), 694 Subject Index 803 Rule Single-valued function, 20 Leibniz, 120 Singular point, 306, 322, 336 L'Hospital’s, 145 isolated, 310 Simpson's, 426, 428 Singular solution (ofdifferential equa- trapezoidal, 425, 426 tion), 504 ‘Smallest value (ofafunction), 61 s ‘Smitnov, V. 1.,806 Smooth ‘curve, 528 Safety parabola, 502 Solenoidal vector field, 702 Scalar field, 283 Solid ofrevolution, 455Scale Solutionnumber, 13 ofadifferential equation, 473 Schwar2’ inequality, 647 general, 474, 475, 515 Secant line, 265 particular, 475, 515 Second derivative singular, 504 mechanical significance of, 124 stable, 577, 579, 580 Second Euler substitution, 373 unstable, 579, 580 Sense ofdescription, 672 Solutions Sense ofintegration, 671 linearly dependent, 530 Separated variables, 479 linearly independent, 530Series Solve(adifferential equation), 476absolutely convergent, 731 Space curve, 450 alternating, 727 Spiral ofArchimedes, 29 binomial, 754-756 Stable conditionally convergent, 731 Lyapunov (about solutions, condi- dominated, 734-736 tions), 57 Fourier, 776-812 Stable solution, 577, 579, 580 definition of, 779 Static moments, 649 functional, 733 Stokes, D., 694 harmonic, 714, 715 Stokes" formula, 692-697 ‘Maclaurin’s, 751-753 Stokes’ theorem, 694-695 numerical, 710 Subinterval, 401 plus-and-minus, 729 Subnormal, 127 power, 742 Subregions, 608 Taylor's, 750, 751 Substitution trigonometric, 776 Euler, 372-375 Serret-Frenet formulas, 335 universal trigonometric, 379 Shift theorem, 858 Subtangenf, 127 Sign Sufficient conditions (for existence ‘ofdouble substitution, 411 ofanextremum), 169 integral, 343 Sum Simple (single) root, 546 integral, 398, 608 Simpson's formula, 428 lower (integral), 397 Simpson's rule, 426, 428 upper (integral), 397 Single (simple)' root, 546 ‘Sum ofaseries, 710 ou Subject Index nth partial, 710 Stokes’, 694, 695 Surface density, 642 uniqueness, 855,Surfaces Weierstrass’ approximation, 252level, 283 Theory ofstability Symbolic veetor, 700 Lyapunov's, 576 Third Euler ‘substitution, 374, 975 T Threefold iterated integral, 651-655 Table ofintegrals, 345 Torsion (ofacurve), 333 Table oftransforms, 862, 863 radius of,333,Tables Totalderivative, 275original-transform, 855 Total differential, 267 Tacnode, 309 Total differentials Tangent, 73,896 approximation by,268, 269 line, 73 Total increment (ofafunction),259,265 Tangent plane, 338 Trajectories Taylor's formula, 152,155 ‘sogonal, 509, 512-514 forafunction oftwovariables, 290 orthogonal, 509-512 Taylor's series, 750,751 Transcendental function, 28 Telegraph equations, 819 ‘Transform (L-transform), 855 Terminus (ofavector), 314 Transform, 855, 856 Terms ofaseries, 710 Fourier, 812 Tet Fourier cosine, 810 Cauchy's, 721, 722 Fourier inverse, 812 @'Alembert’s, ‘718 Fourier sine, 810 integral (forconvergence), 723-726 Laplace, 855‘Theorem Transform’ equation,865Abel's, 742, 743 Transforms Cauchy's, 143 ofderivatives, 861, 862 convolution, 871 differentiation of,860, 861 decomposition, 867 ‘Trapezoid delay, 879,880 curvilinear, 400 existence (ofalineintegral), 673 parabolic, 426 Theorem (cont.) Traperoidal formula, 426 fonfinite increments, 142 Traperoidal rule, 425, 426 fundamental (ofalgebra), 245 ‘Trigonometric function, 22,24,102 Hospital's, 145 Trigonometric series, 776 Lagrange's, ‘142 Trinomial Leibniz’, 727, 728 quadratic, 351 ‘mean-value, 406,616,653 Tripleintegral, 650,654,656,658,659fonratio ofincrements oftwo fun- Triple product (ofvectors), 933, 334 ctions, 143 remainder, 244 uv Rolle’s, 140 Unbounded function, 41 shift, 8658 Uniqueness theorem, 855 Upper limit (ofanintegral), 399 Variables Value Vectorequation, 314 critical, 168 ofgradients, 287 least (ofafunction), 61 potential, 701 extreme (ofafunction), 166 rymptotes, Variable, 16 w bounded, 18 Wallis'formula, 415, 416 infinitely large, 34 Wronkskian, 530-533, 542, 544, £52.