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Student's solutions manual to accompany Jon Rogawski's Multivariable calculus, second edition PDF

711 Pages·2012·7.3 MB·English
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Preview Student's solutions manual to accompany Jon Rogawski's Multivariable calculus, second edition

This page intentionally left blank Student’s Solutions Manual to accompany Jon Rogawski’s Multivariable CALCULUS SECOND EDITION GREGORY P. DRESDEN Washington and Lee University JENNIFER BOWEN The College of Wooster RANDALL PAUL Oregon Institute of Technology W.H.FREEMANANDCOMPANY NEWYORK ©2012byW.H.FreemanandCompany ISBN-13:978-1-4292-5508-0 ISBN-10:1-4292-5508-0 Allrightsreserved PrintedintheUnitedStatesofAmerica First Printing W.H.FreemanandCompany,41MadisonAvenue,NewYork,NY10010 Houndmills,BasingstokeRG216XS,England www.whfreeman.com CONTENTS Chapter 10 INFINITE SERIES (LT Chapter 11) 1 Chapter 14 DIFFERENTIATION IN SEVERAL VARIABLES (LT Chapter 15) 336 10.1 Sequences (LTSection11.1) 1 10.2 SumminganInfiniteSeries (LTSection11.2) 13 14.1 FunctionsofTwoorMoreVariables 10.3 ConvergenceofSerieswithPositiveTerms (LTSection15.1) 336 (LTSection11.3) 23 14.2 LimitsandContinuityinSeveralVariables 10.4 AbsoluteandConditionalConvergence (LTSection15.2) 345 (LTSection11.4) 38 14.3 PartialDerivatives (LTSection15.3) 351 10.5 TheRatioandRootTests (LTSection11.5) 44 14.4 DifferentiabilityandTangentPlanes 10.6 PowerSeries (LTSection11.6) 52 (LTSection15.4) 363 10.7 TaylorSeries (LTSection11.7) 64 14.5 TheGradientandDirectionalDerivatives ChapterReviewExercises 81 (LTSection15.5) 374 14.6 TheChainRule (LTSection15.6) 386 14.7 OptimizationinSeveralVariables Chapter 11 PARAMETRIC EQUATIONS, (LTSection15.7) 399 POLAR COORDINATES, 14.8 LagrangeMultipliers:OptimizingwithaConstraint AND CONIC SECTIONS (LTSection15.8) 419 (LT Chapter 12) 96 ChapterReviewExercises 442 11.1 ParametricEquations (LTSection12.1) 96 Chapter 15 MULTIPLE INTEGRATION 11.2 ArcLengthandSpeed (LTSection12.2) 112 (LT Chapter 16) 458 11.3 PolarCoordinates (LTSection12.3) 120 11.4 AreaandArcLengthinPolarCoordinates 15.1 IntegrationinTwoVariables (LTSection16.1) 458 (LTSection12.4) 133 15.2 DoubleIntegralsoverMoreGeneralRegions 11.5 ConicSections (LTSection12.5) 143 (LTSection16.2) 471 ChapterReviewExercises 154 15.3 TripleIntegrals (LTSection16.3) 489 15.4 IntegrationinPolar,Cylindrical,andSpherical Chapter 12 VECTOR GEOMETRY Coordinates (LTSection16.4) 502 (LT Chapter 13) 166 15.5 ApplicationsofMultipleIntegrals (LTSection16.5) 521 12.1 VectorsinthePlane (LTSection13.1) 166 15.6 ChangeofVariables (LTSection16.6) 537 12.2 VectorsinThreeDimensions (LTSection13.2) 176 ChapterReviewExercises 554 12.3 DotProductandtheAnglebetweenTwoVectors (LTSection13.3) 184 Chapter 16 LINE AND SURFACE INTEGRALS 12.4 TheCrossProduct (LTSection13.4) 198 (LT Chapter 17) 576 12.5 PlanesinThree-Space (LTSection13.5) 212 12.6 ASurveyofQuadricSurfaces (LTSection13.6) 223 16.1 VectorFields (LTSection17.1) 576 12.7 CylindricalandSphericalCoordinates 16.2 LineIntegrals (LTSection17.2) 581 (LTSection13.7) 230 16.3 ConservativeVectorFields (LTSection17.3) 599 ChapterReviewExercises 240 16.4 ParametrizedSurfacesandSurfaceIntegrals (LTSection17.4) 606 16.5 SurfaceIntegralsofVectorFields Chapter 13 CALCULUS OF VECTOR-VALUED (LTSection17.5) 622 FUNCTIONS (LT Chapter 14) 250 ChapterReviewExercises 636 13.1 Vector-ValuedFunctions (LTSection14.1) 250 13.2 CalculusofVector-ValuedFunctions Chapter 17 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS (LTSection14.2) 261 13.3 ArcLengthandSpeed (LTSection14.3) 273 (LT Chapter 18) 649 13.4 Curvature (LTSection14.4) 282 13.5 MotioninThree-Space (LTSection14.5) 303 17.1 Green’sTheorem (LTSection18.1) 649 13.6 PlanetaryMotionAccordingtoKeplerandNewton 17.2 Stokes’Theorem (LTSection18.2) 665 (LTSection14.6) 319 17.3 DivergenceTheorem (LTSection18.3) 678 ChapterReviewExercises 326 ChapterReviewExercises 693 1019763_FM_VOL-I.qxp 9/17/07 4:22 PM Page viii 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 S 50 R 51 1st Pass Pages 10 INFINITE SERIES 10.1 Sequences (LT Section 11.1) Preliminary Questions 1. Whatisa4forthesequencean=n2−n? solution Substitutingn=4intheexpressionforangives a4=42−4=12. 2. Whichofthefollowingsequencesconvergetozero? (cid:2) (cid:3) n2 −1 n (a) (b) 2n (c) n2+1 2 solution (a) Thissequencedoesnotconvergetozero: n2 x2 1 1 lim = lim = lim = =1. n→∞n2+1 x→∞x2+1 x→∞1+ 1 1+0 x2 (b) Thissequencedoesnotconvergetozero:thisisageometricsequencewithr =2>1;hence,thesequencediverges to∞. (c) Recallthatif|an|convergesto0,thenanmustalsoconvergetozero.Here, (cid:4)(cid:2) (cid:3) (cid:4) (cid:2) (cid:3) (cid:4)(cid:4)(cid:4) −1 n(cid:4)(cid:4)(cid:4)= 1 n, 2 2 whichisageometricsequencewith0<r <1;hence, (1)nconvergestozero.Itthereforefollowsthat(−1)nconverges 2 2 tozero. √ 3. Letanbethenthdecimalapproximationto 2.Thatis,a1=1,a2=1.4,a3=1.41,etc.Whatisnl→im∞an? √ solution nl→im∞an= 2. 4. Whichofthefollowingsequencesisdefinedrecursively? √ (cid:5) (a) an= 4+n (b) bn= 4+bn−1 solution (a) ancanbecomputeddirectly,sinceitdependsonnonlyandnotonprecedingterms.Thereforeanisdefinedexplicitly andnotrecursively. (b) bniscomputedintermsoftheprecedingtermbn−1,hencethesequence{bn}isdefinedrecursively. 5. Theorem5saysthateveryconvergentsequenceisbounded.Determineifthefollowingstatementsaretrueorfalse andiffalse,giveacounterexample. (a) If{an}isbounded,thenitconverges. (b) If{an}isnotbounded,thenitdiverges. (c) If{an}diverges,thenitisnotbounded. solution (a) Thisstatementisfalse.Thesequencean = cosπnisboundedsince−1 ≤ cosπn ≤ 1foralln,butitdoesnot converge:sincean=cosnπ =(−1)n,thetermsassumethetwovalues1and−1alternately,hencetheydonotapproach onevalue. (b) ByTheorem5,aconvergingsequencemustbebounded.Therefore,ifasequenceisnotbounded,itcertainlydoes notconverge. (c) Thestatementisfalse.Thesequencean=(−1)nisbounded,butitdoesnotapproachonelimit. 1 May18,2011 2 CHAPTER 10 INFINITESERIES (LTCHAPTER11) Exercises 1. Matcheachsequencewithitsgeneralterm: a1,a2,a3,a4,... Generalterm (a) 1,2,3,4,... (i)cosπn 2 3 4 5 n! (b)−1,1,−1,1,... (ii) 2n (c)1,−1,1,−1,... (iii)(−1)n+1 n (d) 1,2,6,24... (iv) 2 4 8 16 n+1 solution (a) Thenumeratorofeachtermisthesameastheindexoftheterm,andthedenominatorisonemorethanthenumerator; hencean= n+n1,n=1,2,3,.... (b) Thetermsofthissequencearealternatingbetween−1and1sothatthepositivetermsareintheevenplaces.Since cosπn=1forevennandcosπn=−1foroddn,wehavean=cosπn,n=1,2,.... (c) Thetermsanare1foroddnand−1forevenn.Hence,an=(−1)n+1,n=1,2,... (d) Thenumeratorofeachtermisn!,andthedenominatoris2n;hence,an= 2nn!,n=1,2,3,.... InExercises3–12,ca1lculatethefirstfourtermsofthesequence,startingwithn=1. Let3ann= 2n−1 forn=1,2,3,....Writeoutthefirstthreetermsofthefollowingsequences. 3. c(an)=bnn=! an+1 (b) cn=an+3 solu(ct)iodnn=Saetn2tingn=1,2,3,4intheformulaforcngives (d) en=2an−an+1 31 3 32 9 c1= 1! = 1 =3, c2= 2! = 2, 33 27 9 34 81 27 c = = = , c = = = . 3 3! 6 2 4 4! 24 8 5. a1=2, (a2nn+−1=1)!2an2−3 solutibonn=Fornn!=1,2,3wehave: a2=a1+1=2a12−3=2·4−3=5; a3=a2+1=2a22−3=2·25−3=47; a4=a3+1=2a32−3=2·2209−3=4415. Thefirstfourtermsof{an}are2,5,47,4415. 7. bn=5+cosπn 1 solutibo1n=F1o,rnb=n=1,b2n,−3,14+wehave bn−1 b1=5+cosπ =4; b2=5+cos2π =6; b3=5+cos3π =4; b4=5+cos4π =6. Thefirstfourtermsof{bn}are4,6,4,6. 1 1 1 9. cn=cn1=+(−21+)2n3++1 ···+ n solution c1=1; 1 3 c2=1+ = ; 2 2 1 1 3 1 11 c3=1+ + = + = ; 2 3 2 3 6 1 1 1 11 1 25 c4=1+ + + = + = . 2 3 4 6 4 12 May18,2011 SECTION 10.1 Sequences (LTSECTION11.1) 3 11. b1=an2=, nb+2=(n3+, 1)bn+=(n2+bn2−)1++·b·n·−+2(2n) solution Weneedtofindb3andb4.Settingn=3andn=4andusingthegivenvaluesforb1andb2weobtain: b3=2b3−1+b3−2=2b2+b1=2·3+2=8; b4=2b4−1+b4−2=2b3+b2=2·8+3=19. Thefirstfourtermsofthesequence{bn}are2,3,8,19. 13. Findaformulaforthenthtermofeachsequence. 1 c−n1=n1-placedecimalapproximationtoe 2 3 4 (a) , , ,... (b) , , ,... 1 8 27 6 7 8 solution (a) The denominators are the third powers of the positive integers starting withn=1.Also, the sign of the terms is alternatingwiththesignofthefirsttermbeingpositive.Thus, 1 (−1)1+1 1 (−1)2+1 1 (−1)3+1 a1= 13 = 13 ; a2=−23 = 23 ; a3= 33 = 33 . Thisruleleadstothefollowingformulaforthenthterm: (−1)n+1 an= . n3 (b) Assumingastartingindexofn=1,weseethateachnumeratorisonemorethantheindexandthedenominatoris fourmorethanthenumerator.Thus,thegeneraltermanis n+1 an= n+5. InExercSiusepsp1o5se–2th6a,tunsl→eimT∞heaonre=m41atnoddnel→tiemr∞mibnne=the7.liDmeitteorfmthinees:equenceorstatethatthesequencediverges. 15. a(an)=nl→1im2∞(an+bn) (b) nl→im∞an3 solu(ct)ionnl→im∞Wceohs(aπvebna)n=f(n)wheref(x)=12;thus, (d) nl→im∞(an2−2anbn) nl→im∞an=xl→im∞f(x)=xl→im∞12=12. 5n−1 17. bn=an1=2n20+−9 4 n2 5x−1 solution Wehavebn=f(n)wheref(x)= 12x+9;thus, 5n−1 5x−1 5 lim = lim = . n→∞12n+9 x→∞12x+9 12 19. cn=−2−4n+n−3n2 solutiaonn=We4hna2v+ec1n=f(n)wheref(x)=−2−x;thus, lim (cid:6)−2−n(cid:7)= lim −2−x = lim − 1 =0. n→∞ x→∞ x→∞ 2x 21. cn=9n (cid:2)1(cid:3)n solutizonn=We3havecn=f(n)wheref(x)=9x;thus, lim 9n= lim 9x =∞ n→∞ x→∞ Thus,thesequence9ndiverges. n 23. an=zn(cid:5)=n120+−11/n x solution Wehavean=f(n)wheref(x)= (cid:5) ;thus, x2+1 x n x 1 1 1 lim (cid:5) = lim (cid:5) = lim √x = lim (cid:8) = lim (cid:8) = √ =1. n→∞ n2+1 x→∞ x2+1 x→∞ x2+1 x→∞ x2+1 x→∞ 1+ 1 1+0 x x2 x2 May18,2011 4 CHAPTER 10 INFINITESERIES (LTCHAPTER11) (cid:2) (cid:3) 12n+2 25. an=anl=n (cid:5)−9n+4n n3+1 (cid:2) (cid:3) 12x+2 solution Wehavean=f(n)wheref(x)=ln −9+4x ;thus, (cid:2) (cid:3) (cid:2) (cid:3) (cid:2) (cid:3) 12n+2 12x+2 12x+2 lim ln = lim ln =ln lim =ln3 n→∞ −9+4n x→∞ −9+4x x→∞ −9+4x InExercrinse(cid:9)=s2ln7–n3−0,luns(enT2h+eo1r)em4todeterminethelimitofthesequence. 1 27. an= 4+ n solution Wehave 1 1 lim 4+ = lim 4+ =4 n→∞ n x→∞ x √ Since xisacontinuousfunctionforx >0,Theorem4tellsusthat (cid:9) (cid:9) √ 1 1 lim 4+ = lim 4+ = 4=2 n→∞ n n→∞ n (cid:10) (cid:11) 29. an=anc=os−e41n/(32nn+3n93+) 1 solution Wehave n3 1 lim = n→∞2n3+1 2 Sincecos−1(x)iscontinuousforallx,Theorem4tellsusthat (cid:10) (cid:11) (cid:10) (cid:11) lim cos−1 n3 =cos−1 lim n3 =cos−1(1/2)= π n→∞ 2n3+1 n→∞2n3+1 3 n 31. Letaann==tann+−11(e.−Finn)danumberMsuchthat: (a) |an−1|≤0.001for n≥M. (b) |an−1|≤0.00001for n≥M. Thenusethelimitdefinitiontoprovethatnl→im∞an=1. solution (a) Wehave (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) (cid:4) |an−1|=(cid:4)(cid:4)(cid:4)n+n 1 −1(cid:4)(cid:4)(cid:4)=(cid:4)(cid:4)(cid:4)n−n(+n+1 1)(cid:4)(cid:4)(cid:4)=(cid:4)(cid:4)(cid:4)n−+11(cid:4)(cid:4)(cid:4)= n+1 1. Therefore|an−1|≤0.001provided n+11 ≤0.001,thatis,n≥999.ItfollowsthatwecantakeM =999. (b) Bypart(a),|an−1|≤0.00001provided n+11 ≤0.00001,thatis,n≥99999.ItfollowsthatwecantakeM =99999. Wenowproveformallythatnl→im∞an=1.Usingpart(a),weknowthat 1 |an−1|= n+1 <(cid:3), providedn> 1 −1.Thus,Let(cid:3)>0andtakeM = 1 −1.Then,forn>M,wehave (cid:3) (cid:3) 1 1 |an−1|= n+1 < M+1 =(cid:3). 33. UseLtehteblnim=it(cid:6)d13e(cid:7)finn.itiontoprovethatnl→im∞n−2=0. solu(at)ioFnindWaevasleueethoaftMsuchthat|bn|≤10−5forn≥M(cid:4) . (cid:4) (b) Usethelimitdefinitiontoprovethat|nnl→−im2∞−bn0|==0(cid:4)(cid:4)(cid:4). 1 (cid:4)(cid:4)(cid:4)= 1 <(cid:3) n2 n2 provided 1 n> √ . (cid:3) May18,2011

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