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

711 Pages·2012·7.3 MB·English

by  Gregory P Dresden

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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
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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