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Table of Contents
Chapter 1 Vectors
1.1Vectors in Two and Three Dimensions
1.2More About Vectors
1.3The Dot Product
1.4The Cross Product
1.5Equations for Planes; Distance Problems
1.6Some n-dimensional Geometry
1.7New Coordinate Systems
Exercises for Chapter 1
Exercises for Chapter 1
Chapter 2 Differentiation in Several Variables
2.1Functions of Several Variables; Graphing Surfaces
2.2Limits
2.3The Derivative
2.4Properties; Higher-order Partial Derivatives
2.5The Chain Rule
2.6Directional Derivatives
2.7Newton’s Method
Exercises for Chapter 2
for Chapter 2
Chapter 3 Vector-Valued Functions
3.1Parametrized Curves and Kepler’s Laws
3.2Arclength and Differential Geometry
3.3Vector Fields: An Introduction
3.4Gradient, Divergence, Curl and the Del Operator
5. Westartwiththetwovectors a and b.Wecancompletetheparallelogramasinthe figureontheleft.Thevectorfromthe origintothisnewvertexisthevector a + b.Inthe figureontherightwehavetranslatedvector b sothatitstailistheheadof vector a.Thesum a + b isthedirectedthirdsideofthistriangle.
8. Thevectors a =(1, 2, 1), b =(0, 2, 3) and a + b =(1, 2, 1)+(0, 2, 3)=(1, 0, 4) aregraphedbelow. Againnotethat theoriginisatthetailsofthevectorsinthe figure
Also, 1(1, 2, 1)=( 1, 2, 1).Thiswouldbepicturedbydrawingthevector(1,2,1)intheoppositedirection. Finally, 4(1, 2, 1)=(4, 8, 4) whichisfourtimesvector a andsoisvector a stretchedfourtimesaslonginthesamedirection.
9. Sincethesumontheleftmustequalthevectorontherightcomponentwise: 12+ x =2, 9+7= y ,and z + 3=5.Therefore, x =14, y =16,and z =8
19. Weprovidetheproofsfor R3 : (1)(k + l )a =(k + l )(a1 ,a2 ,a3 )=((k + l )a1 , (k + l )a2 , (k + l )a3 ) =(ka1 + la1 ,ka2 + la2 ,ka3 + la3 )= k a + l a (2) k (a + b)= k ((a1 ,a2 ,a3 )+(b1 ,b2 ,b3 ))= k (a1 + b1 ,a2 + b2 ,a3 + b3 ) =(k (a1 + b1 ),k (a2 + b2 ),k (a3 + b3 ))=(ka1 + kb1 ,ka2 + kb2 ,ka3 + kb3 ) =(ka1 ,ka2 ,ka3 )+(kb1 ,kb2 ,kb3 )= k a + k b (3) k (l a)= k (l (a1 ,a2 ,a3 ))= k (la1 ,la2 ,la3 ) =(kla1 ,kla2 ,kla3 )=(lka1 ,lka2 ,lka3 ) = l (ka1 ,ka2 ,ka3 )= l (k a)
20.(a) 0a isthezerovector.Forexample,in R3 :
(b) 1a = a.Againin R3 : 1
21.(a) Theheadofthevector sa isonthe x-axisbetween0and2.Similarlytheheadofthevector tb liessomewhereonthe vector b.Usingthehead-to-tailmethod, sa + tb istheresultoftranslatingthevector tb,inthiscase,totherightby2s (representedinthe figureby tb*).Theresultisclearlyinsidetheparallelogramdeterminedby a and b (andisonlyonthe boundaryoftheparallelogramifeither t or s is0or1.
(b) Againthevectors a and b willdetermineaparallelogram(withverticesattheorigin,andattheheadsof a, b,and a + b Thevectors sa + tb willbethepositionvectorsforallpointsinthatparallelogramdeterminedby(2,2,1)and(0,3,2).
22. HerewearetranslatingthesituationinExercise21bythevector → OP0 .Thevectorswillallbeoftheform → OP0 + sa + tb for 0 ≤ s,t ≤ 1
(c) Wesolve (3, 2)+ t( 1, 2)=( 4, 12) for t andgetthat t =7 minutes.Notethat both 3 7= 4 and 2 14= 12
(d) Wecanseethisalgebraicallyorgeometrically:Solvingthe x partof (3, 2)+ t( 1, 2)=( 13, 27) wegetthat t =16.Butwhen t =16, y = 30 not 27.Alsointhe figurebelowweseethepathtakenbythe fleawillmissthe point ( 13, 27).
24.(a) Theplaneisclimbingatarateof4milesperhour.
(b) Tomakesurethattheaxesareorientedsothattheplanepassesoverthebuilding,thepositive x directioniseastandthe positive y directionisnorth.Thenweareheadingeastatarateof50milesperhouratthesametimewe’reheadingnorth atarateof100milesperhour.Wearedirectlyovertheskyscraperin1/10ofanhouror6minutes.
(c) Usingouranswerin(b),wehavetraveledfor1/10ofanhourandsowe’veclimbed4/10ofamileor2112feet.Theplane is 2112 1250 or862feetabouttheskyscraper.
(b) The z componentsofthetwovectorsalongtheropesmustbeequalandtheirsummustbeoppositeofthe z component inpart(a).Their y componentsmustalsobeoppositeeachother.Sincethevectorpointsinthedirection(0, ±2,1), the y componentwillbetwicethe z component.Togetherthismeansthatthevectorinthedirectionof (0, 2, 1) is (0, 50, 25) andthevectorinthedirection(0,2,1)is(0,50,25).
27. Theforce F duetogravityontheweightisgivenby F =(0, 0, 10).Theforcesalongtheropesareeachparalleltothe displacementvectorsfromtheweighttotherespectiveanchorpoints.Thatis,thetensionvectorsalongtheropesare
F1 = k ((3, 0, 4) (1, 2, 3))= k (2, 2, 1)
F2 = l ((0, 3, 5) (1, 2, 3))= l ( 1, 1, 2), where k and l areappropriatescalars.Fortheweighttoremaininequilibrium,wemusthave F1 + F2 + F = 0,or,equivalently, that k (2, 2, 1)+ l ( 1, 1, 2)+(0, 0, 10)=(0, 0, 0). Takingcomponents,weobtainasystemofthreeequations:
+2l =10
Solving,we findthat k =2and l =4,sothat F1 =(4, 4, 2) and F2 =( 4, 4, 8)
1.2MoreaboutVectors
ItmaybeusefultopointoutthattheanswerstoExercises1and5arethe“same”,butthatinExercise1, i =(1, 0) andinExercise 5, i =(1, 0, 0).ThiscomesupwhengoingtheotherdirectioninExercises9and10.Inotherwords,it’snotalwaysclearwhether theexercise“lives”in R2 or R3
(d) Thesymmetricformsare: x +1 2 =7 y = z 3 5 (for(a)) 5 x 5 = y +3 4 = z 4 5 (for(b)) x +1 2 = y 7= z 3 5 (forthevariationof(a)) x 5 = y 1 4 = z 9 5 (forthevariationof(b))
22. Solvefor t ineachoftheparametricequations.Thus
andthesymmetricformis
23. Solvingfor t ineachoftheparametricequationsgives t = x 7, t =(y +9)/3,and t =(z 6)/( 8),sothatthesymmetric formis
24. Seteachpieceoftheequationequalto t andsolve:
25. Let t =(x +5)/3.Then x =3t 5.Inviewofthesymmetricform,wealsohavethat t =(y 1)/7 and t =(z +10)/( 2) Henceasetofparametricequationsis x =3t 5, y =7t +1,and z = 2
30. Ifyoumakethesubstitution u = t3 ,theequationsbecome:
x =3u +7, y = u +2, and z =5u +1
Themap u = t3 isabijection.Theimportantfactisthat u takesonexactlythesamevaluesthat t does,justatdifferenttimes. Since u takesonallreals,theparametricequationsdodeterminealine(it’sjustthatthespeedalongthelineisnotconstant).
31. Thistimeifyoumakethesubstitution u = t2 ,theequationsbecome:
x =5u 1, y =2u +3, and z = u +1
Theproblemisthat u cannottakeonnegativevaluessotheseparametricequationsareforaraywithendpoint ( 1, 3, 1) and heading (5, 2, 1).
36. Wecouldshowthattwopointsonthelinearealsointheplaneorthatforpointsontheline: 2x y +4z =2(5 t) (2t 7)+4(t 3)=5,sotheyareintheplane.
37. Forpointsonthelineweseethat x 3y + z =(5 t) 3(2t 3)+(7t +1)=15,sothelinedoesnotintersecttheplane.
38. Firstweparametrizethelinebysetting t =(x 3)/6,whichgivesus x =6t +3, y =3t 2, z =5t.Pluggingthese parametricvaluesintotheequationfortheplanegives 2(6t +3) 5(3t 2)+3(5t)+8=0 ⇐⇒ 12t +24=0 ⇐⇒ t = 2.
39. We findparametricequationsforthelinebysetting t =(x 3)/( 2),sothat x =3 2t, y = t +5, z =3t 2.Plugging theseparametricvaluesintotheequationfortheplane,we findthat 3(3 2t)+3(t +5)+(3t 2)=9 6t +3t +15+3t 2=22 for all valuesof t.Hencethelineiscontainedintheplane.
40. Againwe findparametricequationsfortheline.Set t =(x +4)/3,sothat x =3t 4, y =2 t, z =1 9t.Plugging theseparametricvaluesintotheequationfortheplane,we findthat
(c) Thisisjustlikepart(b)exceptthe x and y coordinateshavebeenswitched.Thisisthesameasreflectingthecircleabout theline y = x andsothisisalsoacircleofradius5.Ifyoucare,thecirclein(b)wasdrawnstartingatthepoint(5,0) counterclockwisewhilethiscircleisdrawnstartingat(0,5)clockwise.
(d) Thisisanellipsewithmajoraxisalongthe x-axisintersectingitat (±5, 0) andminoraxisalongthe y -axisintersectingit at (0, ±3): x
46. Thediscussioninthetextofthecycloidlookedatthepathtracedbyapointonthecircumferenceofacircleofradius a asitis rolledwithoutslippingonthe x-axis.Thevectorfromtheorigintoourpoint P wassplitintotwopieces: → OA (thevectorfrom theorigintothecenterofthecircle)and → AP (thevectorfromthecenterofthecircleto P ).Thissplitremainsthesameinour problem.
Chapter1Vectors
Thecenterofthecircleisalways a abovethe x-axis,andafterthewheelhasrolledthroughacentralangleof t radiansthe x coordinateisjust at.So → OA =(at,a).Thisdoesnotchangeinourproblem.
Thevector → AP wascalculatedtobe ( a sin t, a cos t).Thedirectionofthevectorisstillcorrectbutthelengthisnot.If weare b unitsfromthecenterthen → AP = b(sin t, cos t)
Weconcludethenthattheparametricequationsare x = at b sin t,y = a b cos t.When a = b thisisthecaseofthe cycloiddescribedinthetext;when a>b wehavethecurtatecycloid;andwhen a<b wehavetheprolatecycloid.
Note:Theanswersto12and13arethesame.Youmaywanttoassignbothexercisesandaskyourstudentswhythisshouldbe true.Youmightthenwanttoaskwhatwouldhappenifvector a wasthesamebutvector b wasdividedby √2
12. proji+j (2i +3j k)= (i + j) (2i +3j k) (i + j) (i + j) (i + j)= 2+3 1+1 (1, 1, 0)= 5 2 , 5 2 , 0
13. proj i+j √2 (2i +3j k)= ⎛
i+j √2 (2i +3j k) i+j √2 · i+j √2
14. proj5k (i j +2k)= (5k) · (i j +2k) (5k) · (5k) (5k)=
18. Herewetakethenegativeofthevectordividedbyitslength: i 2k i 2k = 1 √5 (1, 0, 2).
19. SameideaasExercise17,butmultiplyby3: 3(i + j k) i + j k = 3 √3 (1, 1, 1)= √3(1, 1, 1).
20. Thereareawholeplanefullofperpendicularvectors.Theeasiestthreeto findarewhenwesetthecoefficientsofthecoordinate vectorsequaltozerointurn: i + j, j + k,and i + k
21. Wehavetwocasestoconsider.
Ifeitheroftheprojectionsiszero:proja b = 0 ⇔ a
projb a = 0 Ifneitheroftheprojectionsiszero,thenthedirectionsmustbethesame.Thismeansthat a mustbeamultipleof b.Let a = cb,thenontheonehand
Ontheotherhand
Theseareequalonlywhen c =1
Inotherwords,proja b = projb a
22. Property2:
23. Wehave
24. Thefollowingdiagramsmightbehelpful: a
To find F1 ,thecomponentof F inthedirectionof a,weproject F onto a:
To find F2 ,thecomponentof F inthedirectionperpendicularto a,wecanjustsubtract F1 from F: F2 =(1, 2) 2 17 (4, 1)= 9 17 , 36 17 = 9 17 (1, 4)
Notethat F1 isamultipleof a sothat F1 doespointinthedirectionof a andthat F2 a =0 so F2 isperpendicularto a.
25.(a) Theworkdonebytheforceisgiventobetheproductofthelengthofthedisplacement( → PQ )andthecomponentof forceinthedirectionofthedisplacement(± proj → PQ F orinthecasepicturedinthetext, F cos θ ).Thatis,
28. Notethat i, j,and k eachpointalongthepositive x-, y -,and z -axes.Therefore,wemayuseTheorem3.3tocalculatethat
29. Asinthepreviousproblem,weuse a =3i +4k to findthat
α = (3i +4k) · i 3i +4k (1) = 3 5 ;
β = (3i +4k) · j 3i +4k (1) =0;
γ = (3i +4k) · k 3i +4k (1) = 4 5
30. Youcouldeitherusethethreerighttrianglesdeterminedbythevector a andthethreecoordinateaxes,oryoucoulduse Theorem3.3.Bythattheorem, cos α = a i a i = a1 a2 1 + a2 2 + a2 3 .Similarly, cos β = a2 a2 1 + a2 2 + a2 3 and cos γ = a3 a2 1 + a2 2 + a2 3
31. Considerthe figure:
If P1 isthepointon AB located r timesthedistancefrom A to B ,thenthevector → AP1 = r → AB.Similarly,since P2 isthepoint on AC located r timesthedistancefrom A to C ,thenthevector → AP2 = r → AC.Sonowwecanlookatthelinesegment P1 P2 usingvectors.
Thetwoconclusionsnowfollow.Because → P1 P2 isascalarmultipleof → BC,theyareparallel.Alsothepositivescalar r pulls outofthenormso → P1 P2 = r → BC = r → BC .
If M1 isthemidpointof AB and M2 isthemidpointof BC ,we’vejustshownthat M1 M2 isparallelto AC andhas halfitslength.Similarly,considertriangle DAC where M3 isthemidpointof CD and M4 isthemidpointof DA.Wesee that M3 M4 isparallelto AC andhashalfitslength.The firstconclusionisthat M1 M2 and M3 M4 havethesamelengthand areparallel.Repeatthisprocessfortriangles ABD and CBD toconcludethat M1 M4 and M2 M3 havethesamelengthand areparallel.Weconcludethat M1 M2 M3 M4 isaparallelogram. Forkicks—haveyourstudentsdrawthe figureforABCDa non-convexquadrilateral.Theargumentandtheconclusionstillholdeventhoughoneofthe“diagonals”isnotinsideofthe quadrilateral.
33. Inthediagraminthetext,thediagonalrunningfromthebottomlefttothetoprightis a + b andthediagonalrunningfrom thebottomrighttothetopleftis b a.
35.(a) Let’sstartwiththetwocircleswithcentersat W1 and W2 .Assumethatinadditiontotheirintersectionatpoint O that theyalsointersectatpoint C asshownbelow.
(b) Let’susetheresultsofpart(a)togetherwiththehint.Weneedtoshowthatthedistancefromeachofthepoints A, B , and C to P is r .Let’sshow,forexample,that → CP is r :
Theargumentsfortheothertwopointsareanalogous.
(c) Whatwereallyneedtoshowisthateachofthelinespassingthrough O andoneofthepoints A, B ,or C isperpendicularto thelinecontainingthetwootherpoints.Usingvectorswewillshowthat
36.(a) ThisfollowsimmediatelyfromExercise34ifyounoticethatthevectorsarethediagonalsoftherhombuswithtwosides b a and a b
Orwecanproceedwiththecalculation: ( b a + a ) · ( b a − a b).Theonlybitofgoodnewshereisthatthe crosstermsclearlycanceleachotheroutandwe’releftwith: b
=0 (b) Asin(a),theslickerwayistorecall(orreprovegeometrically)thatthediagonalsofarhombusbisectthevertexangles. Thennotethat ( b a + a b) isthediagonaloftherhombuswithsides b a and a b andsobisectstheanglebetween themwhichisthesameastheanglebetween a and b
Anotherwayistolet θ1 betheanglebetween a and b a + a b,andlet θ2 betheanglebetween b and b a + a b Then
Also
So b a + a b bisectstheanglebetweenthevectors a and b
19. Youneedto figureoutausefulorderingofthevertices.Youcaneitherplotthembyhandoruseacomputerpackagetohelp oryoucanmakesomeobservationsaboutthem.Firstlookatthe z coordinates.Twopointshave z = 1 andtwohave z =0 Theseformyourbottomface.Oftheremainingpointstwohave z =5—thesewillmatchupwiththebottompointswith z = 1,andtwohave z =6—thesewillmatchupwiththebottompointswith z =0.Theparallelepipedisshownbelow. We’llusethehighlightededgesasourthreevectors a, b,and c.Youcouldhavebasedthecalculationatanyvertex.Ihave chosen (4, 2, 1).Thethreevectorsare:
Finally,Volume = |(a × b) c| =53
Note:TheproofsofExercises20and28areeasierifyourememberthatifmatrix A isjustmatrix B withanytworows interchangedthenthedeterminantof A isthenegativeofthedeterminantof B .Ifyoudon’tusethisfact(whichisexploredin exerciseslaterinthischapter),youcanprovethiswithalongcomputation.Thatiswhytheauthorofthetextsuggeststhata computeralgebrasystemcouldbehelpful—andthiswouldbeagreatplacetouseitinaclassdemonstration.
27. Exercise25(f)showsusthat (a × b) × c isintheplanedeterminedby a and b andsoweexpectthesolutiontobeoftheform k1 a + k2 b forscalars k1 and k2 Usingformula(3)fromthetextfor a × b:
Themostimportantpairis a (b × c)= c (a × b).Becauseofthecommutativepropertyofthedotproduct c (a × b)= (a × b) c andsoweareshowingthat a (b × c)=(a × b) c c (a × b)=(a × b) c
(Forexample,the (a c)b cancelswiththe (c a)b becauseofthecommutativepropertyforthedotproduct.)
31. Ifyourstudentsareusingacomputeralgebrasystem,theymaynotnoticethatthisis exactly thesameproblemasExercise27. Justreplace c with (c × d) onbothsidesoftheequationinExercise27toobtaintheresulthere.
32. FirstapplyExercise29tothedotproducttoget
Youcaneitherobservethattwoofthesequantitiesmustbe0,oryoucanapplyExercise28tosee a · (c × a)= c · (a × a)=0 Exercise28alsoshowsthat b (c × a)= a (b × c).Theresultfollows.
33. WedidthisaboveinExercise29.
34. Theamountoftorqueistheproductofthelengthofthe“wrench”andthecomponentoftheforceperpendiculartothe “wrench”.Inthiscase,thewrenchisthedoor—sothelengthisfourfeet.The20lbforceisappliedperpendiculartothe planeofthedoorwayandthedoorisopen 45◦ .Sofromthetext,theamountoftorque = a F sin θ =(4)(20)(√2/2)= 40√2 ft-lb.
35.(a) Herethelengthof a is1foot,theforce F = 40poundsandangle θ = 120degrees.So
40.(a) v = ω × r =(0, 0, 12) × (2, 1, 3)=(12, 24, 0)=12i +24j
(b) Theheightofthepointdoesn’tchangesowecanviewthisasifitwereaproblemin R2 .When x =2 and y = 1,we can findthecentralanglebytaking tan 1 ( 1/2).Inonesecondtheanglehasmoved12radianssothenewpointis (√5cos(tan 1 ( 1/2)+12), √5sin(tan 1 ( 1/2)+12), 3) ≈ (1 15, 1 92, 3)
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