Solution Manual for Vector Calculus, 5th Edition

Preview (16 of 171 Pages)

100%

Solution Manual for Vector Calculus, 5th Edition - Page 1 preview image

Loading page ...

CONTENTSHowtoUseThis BookIVAcknowledgmentsI VCHAPTER 1The Geometry of Euclidean Space1CHAPTER 2Differentiation21CHAPTER 3Higher-Order Derivatives;Maxima and Minima43CHAPTER 4Vector-Valued Functions63CHAPTER 5Double and Triple Integrals77CHAPTER 6The Change of Variables Formula andApplications of Integration97CHAPTER 7Integrals over Paths and Surfaces17CHAPTER 8The IntegralTheorems of Vector Analysis143CHAPTER 9Sample Exams161APPENDIXAnswers to ChapterTets and Sample Exams167

Solution Manual for Vector Calculus, 5th Edition - Page 2 preview image

Loading page ...

Solution Manual for Vector Calculus, 5th Edition - Page 3 preview image

Loading page ...

1THEGEOMETRYOFEUCLIDEAN SPACE1.1:VECTORSINTWO-ANDT HREE-D IMENSION A L SPACEGOALS1.Be abletoperform the following operations on vectors:addition, subt raction , scalar multiplication.2.Given avector and a point, be able to writethe equation of the line passing through the pointin the directionofthe vector.3.Given two points, be able to write the equation of the line passing through them.STUDYHINTS1.Space notation.T hesymbol]Ror]RI refers to all points on the real number line or a onedimensional space.]R2refers to all ordered pairs(X,y)which lie in the plane, a two-dimensionalspace.]R3refers to all ordered triples(x,y,z)which lie in three-dimensional space. In general,the "exponent" in]R71tells you how many components there are in each vector.2.Vectors and scalars.Avector hasbothlength (magnitude)and direction.Scalars arejust~umbers.Sc~larsdo not have direction. Two vectors are equalIf and only If they both havethe same lengthandthe same~~direction .Pictorially,they do not needtooriginate from thesame starting point.Thevectors shown here are equal.3.Vector notation.Vectors are often denoted by boldface let ters , underlined letters, arrows overletters, or by an n-tuple(Xl,X2,... ,x71) .EachXiofthe n-tupleiscalled thelthcomponent.BEWARE thatthen-tuple may represent either a pointora vector. T he vector (0,0, ... ,0)isdenoted O. Your instructor orother textbooks may use other notations such as a squiggly lineunderneath a letter. A circumflex over a letterissometimes usedtorepresent aunitvector.4.Vector addition.Vectors may be added componentwise ,e.g.,in]R2(XI, yr)+(X2,Y2)=(Xl+YI,X2+Y2).Pictorially, two vectors may be thoughtofas the sides of aparallelogram.Star ting from the vertex formed by the twovectors,weform a newvectorwhich endsatthe opposite corner;;;?Jvof the parallelogram.Thisnew vectoristhesumofthe othertwo.Alternatively,one could simply translate v sothatthetail of v meets the headofu .Thevector joining the tailofuUItothe head of v isu+v .£?:jU+VI________1

Solution Manual for Vector Calculus, 5th Edition - Page 4 preview image

Loading page ...

CHAPTER 125 .Vectorsubtmction.Justas with addition, vectors may be su btracted componentwise.Thinkof this as adding a negative vector .Pictorially,thevectors a , babanda -bform a triangle. To determine the correct direction ,hb~you should Ibe able toaddabandbto geta.a--Thusab~go,"f<omthetipofbtothetipof8 .6 .Scalarmultiplication.Here, each component of a vectorismultipliedbythesame scalar,e.g.,inJR.2,r(x,y)=(rx, ry)for any real number r .Theeffect of multiplicationbya positive scalarist.ochangethe length by a factor.Ifthescalarisnegative, the lengthening occursinthe opposite direction.Multiplication of vectorswill be discussed in the next two sections .7.Standardbasis vectors.These are vectors whose components are all 0 exceptfor a single1.InJR.3,i,jandkdenote the vectors which lie on thex,yandzaxes.T hey are (1, 0, 0),(0, 1,0)and(0,0, 1), respectively.Thestandardbasis vectors inJR.2areiandj ,which are vectors lyingon thexandyaxes, and their respective components are(1,0)and (0,1).Sometimes, thesevectors are denotedby8.Lines.(a)Theline passing through a in the direction ofvisl (t)=a+tv.This is called thepoint-direction form of t he line because the only necessary informationisthe point a and thedirectionofv .(b)Theline passing through a andbisl(t)=a+t(b-a).Thisiscalled the point-pointformof the line. To seeifthe directioniscorrect , plug int=0andyou should get the first point.Plugint=1 and you should get the second point.9.Spanningaspace.Ifall points in a space can be written in the formA I V I+A2 V 2+ ...+AnV n ,whereAiare scalars, then the vectorsVI,...,Vnspanthatgiven space.For example, thevectorsiandjspan thexyplane.10.Geometricproofs.Theuseofvectors can often simplifya proof. Try to compare vector methodsandnon-vector methodsbydoing example10withoutvectors.SOLUTIONSTOSELECTEDE XERCISES1.Wemustsolve the following equations:-21-x-2523-6y.Wegetx=4andy=17,so(-21,23)-(4,6)=(-25,17).4.Convert-4i+3jto(- 4, 3,0),so(2,3,5) -4i+3j=(2,3,5)+(-4, 3,0)=(-2, 6, 5).

Solution Manual for Vector Calculus, 5th Edition - Page 5 preview image

Loading page ...

3THEGEOMETRYOFEUCLIDEANSPACE7.To sketch v,tartatthe origin and move2unitsalong thexaxis, then move 3 units parallel totheyaxis, and then move-6units parallel to thezaxis.T he vectorwis sketched analogously. The vector- vhasthe same lengthasv,butit pointsin the opposite direction . To sketch v + w , t ranslate the tailof w to the head ofvand drawthe vector fromthe origin tothe head of the translated w .Thevectorv -w goesfrom theyhead ofwto the head of v.9.On theyaxis, points have the coordinats(0,y,0),sowemust restrictxandzto be0.On thezaxis, points havez(~~~'51_.(O,y,z)the coordinates(0,0,z),sowemust restrictxandyto be(x,O,z)/ /y0.In t hexzplane, points have the coordinates(:I:,O, z).so((o,y,o)wemust restrictyto be0.In theyzplane, points have thecoordinates(0,y ,z ),sowemust restrictxto beO.x12.Every point on theplane spannedbythe givenve torscan be writ tenasaVI+bV2,whereaandbare real numbers; therefore,the plane is described bya(3,-1,1)+b(O, 3,4 ).15.Giventwopoints a and b , a linethrough them isl(t)=a +t(b-a).In this case, a=(-1,- 1,-1)and b=(1,- 1, 2), sowegetl(t )=(-1,-1,-1)+t(2,0, 3)=(2t-1,-1,3t-1).19.Substitutev=(x,y ,z)=(2 +i,- 2+t,-1+i)into the equation forx,yandzand get2:1:-3y+z-22(2+t)-3(-2+t)+ (- 1 +t)-24+2t+6 -3t--1+t-2=7.Since7f.0,there are no points(:I:,y ,z)satisfying the equation and lying on v .23.Just asthe parallelogramof example17was described byv=sa+tbforsand t in[0, 1],theparalielpiped can be describedbyw=sa+ t b + rc,fors,tandrin[0,1].Leta ,bandcbe the sides of the triangle as shown , and letVijdenotethe vector from pointito poinj .We assumethateach medianisdividedinto a ratio of2 : 1bythe point ofintersection. ThenwehaveVI2-a/ 2=- (c -b)/2;V23=(1/ 3)(a / 2+b);V34(-2/ 3)(a+b/ 2);28 .1V45(a+b )/2.ThevectorV I Sshould be the sumV12+V23+V34+V45,orc -b1(a)2 (b )a+ bcV1 5=--2-+3 2+ b-3a+2'+-2-=b -2'whichisthemedian of the vectorthatends on c.Theother two median ' are analyzed thesame way.

Solution Manual for Vector Calculus, 5th Edition - Page 6 preview image

Loading page ...

CHAPTER 1430.(a) Usingx,thenumberof C atoms;y,the numberofH atoms; andz,the numberof0atoms,as coordinates,wegetp(3,4,3)+q(O,0,2)=r(l,0,2)+s(O,2,1).{b) To find the smallest integer solution forp,q,rands,webalance the equation componentwise:3p=r(equatingx)2s=4pi.e.,s=2p(equatingy)2q+3p=2r+s~6p+2pi.e.,q=(5/2)p.(equatingz)Letp=2,then the smallest integer solu t ionisp=2,q=5,r=6,s=4.(c)In the diagram, Pis(6,8 ,6), Qis(0,0,10) , R is (6,0,12) andSis(0,8,4).Both sidesofthe equationaddup tothe vector(6,8,16).Rpyx1.2:THEINNERPRODUCT,LENGTHANDDISTANCEGOA LSl.Be able tocomputeadotproduct.2.Be able to explain the geometric significanceofthe dotproduct.3.Beable to normalize a vector.4.Be abl e tocomputetheprojectionofone vector onto another.STUDYHINTSl.Inne1' product.Thisisalso commonly called thedotproduct,anditisdenoted by a . b or(a,b).Thedotproductisthesum2.:::7=1ajbj,whereajandbjarethe~thcomponentsofaandb , respectively.For example, inlR2,a . b=alb1+a2b2.Notethatthedotproductisascalar.2.Lengthofavector.Thelength or thenormofavectorx=(x,y,z)isJx2+y2+z2.Itisdenoted byIlxllandisequal toVX-:-X.Thisisderivable from the factthatx .y=X1YI+X2Y2+X3Y3withx=y.3.Unit vector.These vectors have lengthl.Youcanmake any non-zero vector a unit vector bynormalizing it. To normalize a vector, divide the vector by its length,i.e.,computea/liali.4.Cauchy-Schwarzinequality.Knowingthatla·hi~Ilallllbllismost importantfor doing proofsin the optional sectionsofthistextand in more advanced cou rses..5.Importantgeometric properties.Knowthata·b=Ilallllbllcose,whereBis the angle betweenthe two vectors.As a consequence, a . b=0impliesthataand b are orthogonal.Thezerovectorisorthogonal to all vectors.6.Projections.Theorthogonal projectionofb onto a isthe "shadow" of b falling onto a .Theprojectionofbontoaisa vectoroflength(a·b)/llall,inthedirectionofa/liali.Thus, theprojectionofb onto ais(a' b))a(a·b)(-WW=~a....

Solution Manual for Vector Calculus, 5th Edition - Page 7 preview image

Loading page ...

THE GEOMETRY OF EUCLIDEANSPACE57.Problemsolving.Sincevectors have magnitude anddirection, they can be represented pictori ally.Itis often usefulto sketch a diagramto help you visualize a vector word problem .SOLUTIONSTOSELECTEDEXERCISES3.From t he definit ionoft hedotproduct,wegetu·v(07 19) · (- 2 - 10)-7cos(j==',"=~-01546IlullllvllV72+192V22+12V410V5..From ahandcalculator,wefind that(j~99°.7.Ifw=ai+bj+ck, thenIlwll=va2+b2+c2,and soIlull=vT+4=..)5;Ilvll=v1+l=h .Using t he formula for the dot product,wegetu·v=(-1)(1)+(2)( - 1)=- 3.10.Using t hesameformulas as in exercise 7,wegetIlull=VI+0+9=JiO;Ilvll=vO+16+0=4.Since u does nothaveajcomponent and v does not have anyior k co mponent, the vectorsare perpendicular; therefore n . v=O.12.A vectorwisnormalized by constructingthevectorw/llwll.Forthevectors in exercise7:u1 (.2')v1 ('')W=V5-)+J;IRI=V2]-J .15.T he projection of vontouisn·v_(-1)(2)+(1)(1) + (1)(- 3)(_,,k)-_i(_,,k)Ilul12u-(V1+1+1)2l + J +-3I+J +.16.For orthogonality,wewantthedotproductto beO.(a)Thedotproductis (2i+bj).(- 3i+2j+k )=-6+2b,sobmust be 3.(b)Thedot productis(2i+bj).k=0,sobcan be any real number.21.(a) Lookingatthexcomponen s, the pilot needs to getfrom 3 to23.Hisvelocity in thexdirection is theicomponent,400km/hr. T hus,Llt=~d=23-3=~.v40020The pilot flies over the airport(1/ 20)hour or 3 minutes later.The sam e answer couldhavebeen obtained by an alyzing theycomponents.(b)Weloo katthezcomponents and use the formula~d=vLlt,i.e.,h-5=(-1)( 1/ 20),soh=99/20.Thus, the pilotis4.95kmabove the airport when he passes over.24.(a)Itisconvenient to draw the diagram with A on the .r axis.

Solution Manual for Vector Calculus, 5th Edition - Page 8 preview image

Loading page ...

CHAPTER16yAx(b)From the diagraminpart(a),wegetA=150iandB=(llOcos600)i+(llOsin600)j.A+B=(150 +llO cos600)i +(llOsin600)j=205i+ 55V3j.TheanglethatA+B makes with A is()=tan-1( ;)(5~:)~t an -1(0.4647)~250.=tan-1Alt.ernat.ively, the definition of the dot productgives us()=cos-1(A ·(A+B ) )=cos-1(150)(205)+ (0)(55V3)IIAIIIIA+BII(150)(J51l00)~cos-l(O.gO~g)~25°.27.(a)Geometrically,weseethattheicomponentofFisIIFIIcos().Similarly, thejcomponent ofFisIIFIIsinB.T hereyFfore,F=IIFIIcos()i+IIFIIsin()j,where()isthe angle from thexaxis. Since theangle from t heyaxisis7r/4,()isalso 7r/4, so F=3J2(i+j).(b)WecomputeD=4i+2j,soF·D=(3)2)(4)+(3)2)(2)=18"\/2.Also,IIFII=6 andIIDII=J20. From the definition ofthe dot product,F ·D18)23cos()=IIFIIIIDII=6J20=vT5~0.9487,i.e.,()~18°,or equi valently,()~0.322radians.(c)Frompart(b),wehad computedF·D=18)2.Knowing thatcos()=3/.JfO,wecalculateIIFIIliDIIcos()=(6J20)(3/JIO)=18)2,also .1.3:M ATRICES,DETERMINANTSANDTHECROSSPRODUCTGOALS1.Beable tocomputea cross product.2.Beable to explain the geometric significanceofthe cross product.3.Be abletowrite the equationofa plane from given information regarding points on t he planeor normals to the plane.x

Solution Manual for Vector Calculus, 5th Edition - Page 9 preview image

Loading page ...

7THE GEOMETRYOFEUCLIDEAN SPACESTUDYHINT S1.Ma trices anddetermi1wnts.A matrix isjusta rectangulararrayofnumbers.Thearrayiswrittenbetween a set ofbrackets.Thedeterminantofamatrixisa.number; amatrixhasno numerical value.Thedeterminant is defined only forsquarematricesanditisdenoted byvertical bars.2.Computing determinants.K nowthatI~!I=ad-be.Also knowthatabcdef9hNote theminus sign in frontofthe secondterm on theright-handside.Thegeneralmethodforcomputing determinantsis describednext.3.Computingnxndeterminants.Usethecheckerboardpatternshown here which beginswitha plus sign intheupperleft corner.Choose anycolumnorrow -usually pickingthe one wit,hthemost zeroes saves work.++Drawverticalandhorizontal linesthroughthefirstnumberof+t he row or column.Thenumbersremainingforman(n-1)x++(n-1)determ inant, whichshould bemultipliedbythenumber(with the sign det erm ined by the checkerboard)throughwhichbothlines aredrawn .Repeatfortheremainingnumbersoftherow or column . Finally,sumthe results.Thisprocess , calledexpansion by minors , works for any roworcolumn.Thebestwaytoremem bertheprocess is bypracticing. Be suretousethe correct signs.4.Simplifying determinants .Deter minants are easiesttocomputewhen zeroesarepresent.Adding a non-zero multipleofone row or one columntoanotherroworcolumndoesnotchangethevalueofthedeterminant,andthiscanoftensimplify thecomputations.See exa.mpIe 3.5.Computingacross product.Ifa=(aI,a 2 ,a3)andb=(b1 ,b2 ,b3),thenJka xb=a1a 2a3b1b2b3Theorder matters: a x b=-(bxa).The cross product isnotcommutative.Also,notethata x b is a vector ,nota scalar.6.Propertiesofthe cross product.T hevectors a , band aXb form aright-handedsystem(seefigure 1.3.2 ofthetext).The cross product a x b isorthogonaltobothaandb .Thelengthofa xbisIIallllbillsinOI, where0isthe angle between aandb.Notethattheformulaforthecrossprod uctinvolves sin0,whereas t hedotproductinvolves cosO.7.More properties.Ifthe crossproductis zero ,theneither: (i) t he lengthofoneofthevectorsmustbezero , or (ii) sinO=0,i.e.,0=0, sothevectorsmustbe parallel.8.Geometry.The absolute value ofthe determinantI~!Iis theareaoftheparallelogramspannedbythevectors(a,b)and(c,d)originating from thesamepoint.Theabsolutevalueofabcthedeterminantdefisthevolume oftheparallelpipedspanned bythevectors(a,b,c),ghi(d,e,J)and(g,h,i)originating fromthesamepoint.Thelengthofthecrossproduct11axhllistheareaofthe parallelogramspanned by the vectorsa andb.Thevector a x b gives avectornormaltothe plane spannedby a andb .

Solution Manual for Vector Calculus, 5th Edition - Page 10 preview image

Loading page ...

CHAPTER 189.Equationofaplane.Recall thatthe equationofthe planeisax+by+cz+d=O.Thevector(a,b,c)is orthogona] to the plane . Knowing twovectors in the plane ,wecan deter mine anorthogonal vector by using the cross product. Compare methods1and2of example11.10.Distance from pointtoplane.You should understand the derivationofthe equationinthebox preceding example12.Ifnecessary, review the geometric propertiesofthedotproduct insection1.2.SOLUTIONSTOSELECTEDEXERCISES2.(b)Wesubtract12times the third row from the first row andsubtract15timesthe third rowfrom the second row.'fhenweexpand along the first column:361817o-4241-42411. -.452420o-5150=3-5150=3(-2100+2091)=-27.135-235-25.Theareaofthe parallelogramis/la x hll .Wecomputejka x b=1-21=-3i+j+5k211andsothe areaofthe parallelogramislin x hll=v'9+1+25=55.8.Thevolumeofthe parallelpipedisthe absolute valueofthe 3x3 determinantmade upoft hevectors ' components. Expand along the first row:100_13-103-1-2'I=-l.-142-1Thus, the volumeis1.11.Wewanttofind the cross product and then normalize it .WecomputeSoJkv=-59-4I=113i+17j103k.789There are two orthogonal vectors in opposing directions; theyare given by±v/llvll=:±(113i+17j -103k)/v'23667.Since all vectors are orthogonal to0,the inclusion ofthatvector into the problem doesnotaffect the answer.15.(a)Theequationofa plane with normal vector(A,B,C)and passing through the point(xo,Yo,zo)isA(x-xo)+B(y-Yo)+C(z-zo)=O.In this case, the equationisl(x-1)+l(y-0)+l(z-0)=0orx+y+z=1.(d)Here, the normal vector is parallel to the line, so itis(-1, -2,3). Hence, t he equationofthe desired plane is-1(x-2)-2(y-4)+3(z+1)=0or-x-2y+3z+13=O.

Solution Manual for Vector Calculus, 5th Edition - Page 11 preview image

Loading page ...

9TH EGEOMETRY OFEUCLIDEAN SPACE16.(b) Two vectorsinthe desiredplane are v=(0-1, 1 -2,- 2 -0)=(-1,-1, -2)andw =(4 -0, 0 -1,1+2)=(4, - 1,3).Thecross pr duct v x w isorthogonal tobothvectors,and hence normal to the desired plane.Wecomputev xw=- 5i -5j+5k,so the desired equationis-5(x-1)-5(y-2)+5(z-0)=0or-x-y+z+3=O.22.(a) LetDbethe matrix with rowsu,v, w . T henUlU2U3det D =u·(vxw)=VIV2V3W IWzW3Usethefollowing property of determinants:v·(wxu ) correspondstotwo row exchanges ofthe matrixD ,80wehavv .(wx u ) =(-1)(-1)detD=detDandw · (uxv ) =(-1)(-1)detD=detD.To prove the otherthree,recallthatu x v =-(vxu).26. The line perpendicular to theplane is parallel tothe normal ofthe plane,sot he equation ofthelineisl (t)=(1,-2,- 3)+t (3,-1,-2).29.Let u be thevector normal tothe plane.Then u is perpendicular to3i+2j+4ksince vis on the plane.Alsu uisperpendicular to2i+j -3k,because the vectorAi+Bj+Ckiserpendicular toallvectors in t he planeAx+By+Cz=D .To findu,we take t he crossprodu t of 3i+2j+4kand2i+j -3k:Jku =324=-10i +17j -k.21- 3Whent=0,wefind that a point on the planeis(-1,1, 2),sothe equation of the plane is- 10(x+l)+17(y-l)-(z-2}=0or-10x +17y-z-25=0.30.First,find the normal ofthe plane. The norm al oftheplane is perpendicular0thelinepassingthrough (3,2,-1)and(1, - 1, 2).T he equationofthe lineisl(t)=(3,2,- 1)+t(2,3,- 3).Thenormaloft he plane is alsoperpendicular to v=(1, -1,0)+t(3,2,-2). T herefore,twovectorson t he desired plane are2i+3j -3kand3i+2j -2k, anthe normal is(2i+3j -3k) x (3i+2j -2k)=- 5j+5k.Now,weneed a point on the plane, say,(3,2,-1).Thus, the equationof t he planeisO(z -3)-5(y-2)+5(z+1)=0or-y+z=l.34.Theplane passi ng through the origin and perpendicular toi -2j+kisx-2y+z=O.Bythe dist ance formu la with(A, B ,C ,D )= (1, -2,1,0)and (Xl,Y1,zt)=(6,1,0),d=IAxl+BYI+C Zl+DI=6 -2_--±-_2V6VA2+B2+C2VI+4+1V63'37.Sinceall vectorsinthis exercise are unit vectors,liNxall= sin 01andIINxhll= sinO2 .FromSnell's law,n lsin01=n2sinO2 .Hence,nlllN x all=n211Nx hll.To establishthatNx a and N x h have the same direction, we assume t hatN,a andhalllie inthe same plane, andaandhare on the same side ofN.HenceNxaandNxhbothare perpendicular to this plane and paralleltoachother . ThusnlllNxalland n211NXhllare equal.

Solution Manual for Vector Calculus, 5th Edition - Page 12 preview image

Loading page ...

10CHAPTER138.First,4 timesthefirst row issubtractedfromthesecond row.Next, 7 times the first row issubtractedfromthethirdrow .Thenextstepis expansion by minors along the first column.Finally,the2 x 2determinantiscomputed.1.4:CYLINDRICALANDSPHERICALCOORDINATESGOALSl.Be able to convert backandforth betweenthecylindrical, spherical and cartesian coordinatesystems.2.Be able to describe geometric objects with cylindricalandspherical coordinates.3.Be able to describe the geometric effectsofchanging a coordinate.STUDYHINTSl.Review.You should reviewpolarcoordinates in your one-variable calculus text ...~2.Cylindrical coordinates.Denoted(r,{},z),this isjustlike polar coordin ates exceptthatazcoordinatehas been added.Knowtheformulasx=rcos{},y=rsin{}, r=Jx2+y2and{}::::;tan-1(y/x).3.Spherical cooridinates.Denoted by(p,{},¢»,pisthedistance fromthe origin,¢>is the anglefrom the positivezaxisand{}isthesameas in cylindrical coordinates. Knowtheformulasx=pcos{}sin¢>,y=psin{}sin¢>andz=pcos¢>.Also know t.hatp=Jx2+y2+Z2,{}=tan-1(y/x)and¢>=cos-1(z/Jx2+y2+z2)=cos-1(z/p).4.Graphsofr,p=constant.Notethatr=constantin cylindrical coordinates describes acylinderandthatp=constantin spherical coordinates describes a sphere.Youmayhavesuspected this from thenameofthecoordinatesystem.5.Computing{}and¢>.Rememberthat,in this text,¢>takes values from 0 torrand{}rangesfrom 0 to2iT.Insome instances,itismore convenient to define{}intherange-rrtoiT .Youshould be very carefulaboutcomputing{}.Ifx=y=-1,t.hentan-1(y/x)=iT/4,butplottingthepoint(-1,-1)in thexyplane showsthat,inreality,{}=5rr/4.Thisis why theauthorsfuss withtan-1(y/x)inthedefinition.Plottingthexandycoordinates is very helpful fordetermining{}.6.Negativer,p.Notethatwehave defined randpto be non-negative.Ifthedistanceisgivenas a negative number,weneed to reflectthegiven point acrosstheorigin .7.Unit vector'sinsphericalandcylindrical coordinates.Theunitvectors in cylindrical coordinates areer ,eoandez .Thevectorerpointsalongthedirectionofr,whileeogoes inthedirection in which{}is measured ,andez=k.As onemightexpect., those three unit vectorsform an orthogonal basis, anderxeo=ez .Those vectors, however, lirenotfixedasisthecase with i,jand k ,thatis,ifyou changethepoint(7',{},z),thesetofunitvectorsrotates.For spherical coordinates, there is also a setofunitvectorsep ,eoande",.Those vectors , intermsofi,j,k and the cartesian coordinatesofthepointare workedoutin exercise 7 (seebelow)ofthis section.

Solution Manual for Vector Calculus, 5th Edition - Page 13 preview image

Loading page ...

11THEGEOMETRY OF EUCLIDEANSPACESOLUTIONSTOSELECTEDEXERCISES1.(a) To conver t torectangularcoordinates, usex=r cos&andy=1'Sin11:x= 1cos 45°=v2/2andy= 1sin45°=V2/2.Next , usep=J x2+y2+z 2andrP=cos -1(z / p)togetthespherical coordinates:Ir.- 1 (1l'P:::;;VI2"+2"1+1=V2and¢=cos)21)="4'Hence,forthe cylindricalcoordinates(1,45°,1),therect angularcoordinates are given by(/2/ 2,)2/2,1)andhesph rical coordinatesare(V2,rr/ 4,1l'/ 4).(b)Toconvert to cylindrical coordinates,weusel'=..jX2+y2and(J=tan-1(y/x ):r=.)4+1=v'5andB=tan-1(1/ 2).Next, use thesameformulas asinpart (a) to get t hspherical coordina tes:p=V4+4+1 = 3and¢ = cos-1(-2/ 3).Hence, the rectangularcoordin ates (2,1,- 2) convert tothecorresponding cylindrical coordinats(/5,t an-1(1/2),- 2) and t othespherical coordinates(3,tan -1(1/2),cos-1(-2/ 3)).2.(b) T his mappin g takes a point a nd rotatesitby1l'radians about thezaxis .Thisisfollowedby a reflection acrossthexyplane.Theneteffectisthatthe pointisreflectedthrough theorigin.3.(b) Recallthatthe angle¢ismeasured fromthe"Nor th pole."If¢is7iradians , t hen t helocationisatthe"Soutlpole."Theeffect.of changing ¢ to1['-¢ istaking a pointandreflectingit across thex yplane.5.Letp~0, then(p,0,0)isthe positi vezaxis.Now, let ¢ vary from 0to1l'.Then(p,0,¢)isthehalfplane inthexzplanewit h z~0. By allowing&to vary from 0 to21l',werotatet hehalfplane described above.Therefore,p~0,°~(J<21l'and0~¢~1l'describes all pointsinJR3.Ifp<0 also , thecoordinates are not unique . Forexample,(x,y,z)= (1,1,0)has sphericalcoordinates(V2, 1l'/4,1l'/2)and(-V2,51l'/4,1l'/2).7.(a) First,episthe unit vector along the vector(x,y,z);t herefore, the formulaisxi+yj+zkep=.J x2+y2+z2zy-----k------~ryxx(x,y,O)Next,e oisparallel tothexyplaneanddenotes t he direction in which t he angleBismeasured .Itis perpendiculartor=(x ,y,0), so(ai+bj).(xi+yj ) =O.SincewewantBmeasuredcounterclockwise, a=-y(insteadofy)andb=x.Therefore,- yi+zjeo=J x 2+y2'

Solution Manual for Vector Calculus, 5th Edition - Page 14 preview image

Loading page ...

12CHAPTER 1Tofinde,p,wenotethatep ,eoande,pis a setoforthogonal vectors; they form a right-handedcoordinatesystem withepxeo=- e,p.So.(x, y, z )x(-y,x,0)x zi+yzj-(x2+y2)k~=-=.rpJ x2+y2Jx2+y2+Z29.(a)Thelength ofxi+yj+zk isJx2+y2+z2,which isthedefinitionofp.(b)NotethatIlvll=Jx2+y2+z2=Pandv·k=z;therefore,<p=cos~l(z/p)=cos-1(v.k/llvll)·(c)NotethatIlull=Jx2+y2,which isthecylindricalcoordinaterandu ·i=x;therefore,()=cos-1(x/r)=cos-1(u . i /llull).13.Notethat<pwill be between7r/ 2 and7rbecausetheregion liesinthe lower hemisphere.Fromthetriangle,weseethatcosa=(d/6)-;-(d/ 2)=1/ 3;therefore,wehave7r-a::;<p::;7ror7r-cos-1(1/3)::;¢::;7r. Now,pcanbeaslarge asd/2;however, aspgets smaller, its lowerlimitdepends on<p.Pick any¢ ,then¢+(3=7rand accordingtothediagramcos(3=(dj6)-;-p.Rearrangementgivesd/(6cos(3)=p=d/(6cos(7r-¢))=-d/(6cos¢).Therefore,-d/(6cos¢)::;p::;d/2.So far ,wehave describedthecross-section in onequadrant.Theentirevolume requires a revolutionaround thezaxis, soits description is1__d_<p<~0<()<27rand7r-cos-1(-3)<_A,<_7r.6 cos¢ --2 '--'i'1.5:n-DIMENSIONA LEUCLIDEANSPACEGOALS1.Be abletoextendthe ideasofthe previous sections tojRn.2.Be abletomultiplymatrices.STUDYHINTS1.The space]Rn.Mostofthistextbook deals withtheEuclidean spacesthatwecan visualize,jR2andjR3.Manyofthesameproperties hold injRn.Vector addition, scalarmultiplication,vector lengths,thedotproductand the triangle inequalityare defined similarly.2.No cross product analog.Thecrossproductin]R3doesnothave an easy analog injRn,n~4.3.Standard basis vectors.The analogsofi,j,k are defined e j.Thevectorejis(0,0, ...,1,...0)with 1 in thezthposition.The vectors ej andejareorthogonalifi=fj.

Solution Manual for Vector Calculus, 5th Edition - Page 15 preview image

Loading page ...

13THEGEOMETRYOFEUCLIDEAN SPACE4.Matrices.Amatrixisa rect angulararray of numbers .Unlike adeterminant,amatrixhas nonumerical value. You should rememberthat annx m matrix hasnrowsandm columns.The(i,j)entryisthe number located in rowi,columnj.5.Mat rix m ultiplication.You should practice until matrix multiplication becomes secondnaturetoyou. Letthe components ofAbea i jand let those ofBbebkl,whereAis an m xpmatrixandBisapxnmatrix.Thenthe components ofA Barep(A B)mn=Lamjbjn .j=lWecan only multiply an m xpmat rixwith apxnmatrix,i.e.,[m xp][Pxn].Notethatthe numberofcolumns ofAand the number of rowsofBmustbe equal(pin this case).Theresult is an m xnmat rix ("cancelling"thep).6.Non-commutativityofm atrixmultiplication.In general,AB#B A .In fact,ABmay be definedwhenBAisundefined . However , matrix multiplication is associat ive;i.e.,(AB)C=A(BC)ifthe productAB Cis defined .7.Matrices and mappings.An m xnmatrix can represent amapping from]Rnto]Rm.To seethis, letAbe thematrix and let x be a vectorin]Rn,represented as annx 1matrix,andybe a vector in]Rm,an m x 1 matrix. ThenthematrixAtakes a point in]Rnto a point in]Rmbythe equationAx=y.SOLUTIONSTOSELE CTEDEXERCI SES2.(a) Use the properties of leng hsanddot products:(x+y) .(x+y)+(x-y).(x-y)x .x+2x .y+y .y+x .x -2x.y+y .y2x·x+2y·y=211xl12+211Y112.Thefigureatthe left depicts the equation geometrically.Bythe lawofcosines,wehave12<tlyandWealso notethata+j3=1r,soj3=1r -a .T herefore, cosj3=cos(1r -a)= -cosa.Addingthe two equations fromthe lawofcosines yields211xl12+211Yl12;:Ilx-Yl12+I'lx+Y112.4.To verify theCauchy-Schwarz inequality,wecomputeIx·yl1(1)(3)+(0)(8)+(2)(4)+(6)(1)1=17.Ilxll'1'1+0+4+36=J41.Il,yllJ9+64+16+1=V90.Thus,weindeed haveIx.yl=v'I7v'I7<v4fJ90=II'xlillyll.For the triangle inequality,wecomputex+y=(4,8,6,7)andIIx+yll=J16+64+36+49=v'l65<13.Indeed,wehaveIlx+yll<13<15=6+9<AT+v'9O=Ilxllllyll.

Solution Manual for Vector Calculus, 5th Edition - Page 16 preview image

Loading page ...

14CHAPTER18.Wecompute1nAB~r:-1~31andA+B~[:0211-11Expandingby minors across the first row givesdetA3:12-1I11~II=6 -2=4,01+1 1detE1101~1-11~~1= -3,det(AE)1-111_115and31-11~11-3151-111=-12det(A+B)-41~~I=8.11.(a) Forn=2,I'xalldet('xA)='xa21Forn=3,'xallAal2'xa13det(AA)Aa21'xa22'xa23Aa31'xa32'xa33'xalldet('xAd-'xa12det('xA2 )+'xa13det(AA3),X.,X2(a11detAl-al2detA2+a13detA3),X3detA,where AI ,A2andA3are2x2matricesobtainedbyexpandingacrossthefirst row .Assumethatforn=k,det('xA)=AkdetA.Theforn=k+1,det(AA)can be foundbya process analogoustothe3x3case:det(AA)'xalldet('xAd-'xa12det('xA2 )+ ... +(_l)k'xal,k+ldet('xAk+d,Xk+ldetA,whereAI,A2 ,...,Ak+larekxkmatricesobtainedbyexpandingacrossthefirst row .Byinduction,det('xA)=AndetAforannxnmatrixA.14.Assume, as inthebook,thatdet(AE)=(detA)(detE).Thendet(AEC)=det[(AB)C]=det(AE)detC=(detA)(detE)(detC).17.Multiplythetwomatricestogettheidentitymatrix:[ab]1[d-b]edad-be-eaSimilarly,wecanshowthat

Preview Mode

This document has 171 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

Precalculus - Polynomial and Rational Functions

Precalculus - Functions

Precalculus - Exponential and Logarithmic Functions

Math Word Problems - Translating Expressions

Math Word Problems - Equations

Algebra II – Word Problems

Algebra II – Sequences and Series

Algebra II - Segments Lines and Inequalities

Algebra II - Relations and Functions

Algebra II - Rational Expressions

Company

Explore

Study Tools