工程电磁场第八版课后答案

CHAPTER11.1.GiventhevectorsM=�10ax+4ay�8azandN=8ax+7ay�2az,find:a)aunitvectorinthedirectionof�M+2N.�M+2N=10ax�4ay+8az+16ax+14ay�4az=(26,10,4)Thusa=(26,10,4)|(26,10,4)|=(0.92,0.36,0.14)b)themagnitudeof5ax+N�3M:(5,0,0)+(8,7,�2)�(�30,12,�24)=(43,�5,22),and|(43,�5,22)|=48.6.c)|M||2N|(M+N):|(�10,4,�8)||(16,14,�4)|(�2,11,�10)=(13.4)(21.6)(�2,11,�10)=(�580.5,3193,�2902)1.2.VectorAextendsfromtheoriginto(1,2,3)andvectorBfromtheoriginto(2,3,-2).a)Findtheunitvectorinthedirectionof(A�B):FirstA�B=(ax+2ay+3az)�(2ax+3ay�2az)=(�ax�ay+5az)whosemagnitudeis|A�B|=[(�ax�ay+5az)·(�ax�ay+5az)]1/2=p1+1+25=3p3=5.20.TheunitvectoristhereforeaAB=(�ax�ay+5az)/5.20b)findtheunitvectorinthedirectionofthelineextendingfromtheorigintothemidpointofthelinejoiningtheendsofAandB:ThemidpointislocatedatPmp=[1+(2�1)/2,2+(3�2)/2,3+(�2�3)/2)]=(1.5,2.5,0.5)Theunitvectoristhenamp=(1.5ax+2.5ay+0.5az)p(1.5)2+(2.5)2+(0.5)2=(1.5ax+2.5ay+0.5az)/2.961.3.ThevectorfromtheorigintothepointAisgivenas(6,�2,�4),andtheunitvectordirectedfromtheorigintowardpointBis(2,�2,1)/3.IfpointsAandBaretenunitsapart,findthecoordinatesofpointB.WithA=(6,�2,�4)andB=13B(2,�2,1),weusethefactthat|B�A|=10,or|(6�23B)ax�(2�23B)ay�(4+13B)az|=10Expanding,obtain36�8B+49B2+4�83B+49B2+16+83B+19B2=100orB2�8B�44=0.ThusB=8±p64�1762=11.75(takingpositiveoption)andsoB=23(11.75)ax�23(11.75)ay+13(11.75)az=7.83ax�7.83ay+3.92az11.4.Acircle,centeredattheoriginwitharadiusof2units,liesinthexyplane.Determinetheunitvectorinrectangularcomponentsthatliesinthexyplane,istangenttothecircleat(�p3,1,0),andisinthegeneraldirectionofincreasingvaluesofy:Aunitvectortangenttothiscircleinthegeneralincreasingydirectionist=�a�.Itsxandycomponentsaretx=�a�·ax=sin�,andty=�a�·ay=�cos�.Atthepoint(�p3,1),�=150�,andsot=sin150�ax�cos150�ay=0.5(ax+p3ay).1.5.AvectorfieldisspecifiedasG=24xyax+12(x2+2)ay+18z2az.Giventwopoints,P(1,2,�1)andQ(�2,1,3),find:a)GatP:G(1,2,�1)=(48,36,18)b)aunitvectorinthedirectionofGatQ:G(�2,1,3)=(�48,72,162),soaG=(�48,72,162)|(�48,72,162)|=(�0.26,0.39,0.88)c)aunitvectordirectedfromQtowardP:aQP=P�Q|P�Q|=(3,�1,4)p26=(0.59,0.20,�0.78)d)theequationofthesurfaceonwhich|G|=60:Wewrite60=|(24xy,12(x2+2),18z2)|,or10=|(4xy,2x2+4,3z2)|,sotheequationis100=16x2y2+4x4+16x2+16+9z41.6.FindtheacuteanglebetweenthetwovectorsA=2ax+ay+3azandB=ax�3ay+2azbyusingthedefinitionof:a)thedotproduct:First,A·B=2�3+6=5=ABcos✓,whereA=p22+12+32=p14,andwhereB=p12+32+22=p14.Thereforecos✓=5/14,sothat✓=69.1�.b)thecrossproduct:BeginwithA⇥B=������axayaz2131�32������=11ax�ay�7azandthen|A⇥B|=p112+12+72=p171.Sonow,with|A⇥B|=ABsin✓=p171,find✓=sin�1�p171/14�=69.1�1.7.GiventhevectorfieldE=4zy2cos2xax+2zysin2xay+y2sin2xazfortheregion|x|,|y|,and|z|lessthan2,find:a)thesurfacesonwhichEy=0.WithEy=2zysin2x=0,thesurfacesare1)theplanez=0,with|x|<2,|y|<2;2)theplaney=0,with|x|<2,|z|<2;3)theplanex=0,with|y|<2,|z|<2;4)theplanex=⇡/2,with|y|<2,|z|<2.b)theregioninwhichEy=Ez:Thisoccurswhen2zysin2x=y2sin2x,orontheplane2z=y,with|x|<2,|y|<2,|z|<1.c)theregioninwhichE=0:WewouldhaveEx=Ey=Ez=0,orzy2cos2x=zysin2x=y2sin2x=0.Thisconditionismetontheplaney=0,with|x|<2,|z|<2.21.8.DemonstratetheambiguitythatresultswhenthecrossproductisusedtofindtheanglebetweentwovectorsbyfindingtheanglebetweenA=3ax�2ay+4azandB=2ax+ay�2az.Doesthisambiguityexistwhenthedotproductisused?WeusetherelationA⇥B=|A||B|sin✓n.WiththegivenvectorswefindA⇥B=14ay+7az=7p52ay+azp5�|{z}±n=p9+4+16p4+1+4sin✓nwherenisidentifiedasshown;weseethatncanbepositiveornegative,assin✓canbepositiveornegative.Thisapparentsignambiguityisnottherealproblem,however,aswereallywantthemagnitudeoftheangleanyway.Choosingthepositivesign,weareleftwithsin✓=7p5/(p29p9)=0.969.Twovaluesof✓(75.7�and104.3�)satisfythisequation,andhencetherealambiguity.Inusingthedotproduct,wefindA·B=6�2�8=�4=|A||B|cos✓=3p29cos✓,orcos✓=�4/(3p29)=�0.248)✓=�75.7�.Again,theminussignisnotimportant,aswecareonlyabouttheanglemagnitude.Themainpointisthatonlyone✓valueresultswhenusingthedotproduct,sonoambiguity.1.9.AfieldisgivenasG=25(x2+y2)(xax+yay)Find:a)aunitvectorinthedirectionofGatP(3,4,�2):HaveGp=25/(9+16)⇥(3,4,0)=3ax+4ay,and|Gp|=5.ThusaG=(0.6,0.8,0).b)theanglebetweenGandaxatP:TheangleisfoundthroughaG·ax=cos✓.Socos✓=(0.6,0.8,0)·(1,0,0)=0.6.Thus✓=53�.c)thevalueofthefollowingdoubleintegralontheplaney=7:Z40Z20G·aydzdxZ40Z2025x2+y2(xax+yay)·aydzdx=Z40Z2025x2+49⇥7dzdx=Z40350x2+49dx=350⇥17tan�1✓47◆�0�=261.10.Byexpressingdiagonalsasvectorsandusingthedefinitionofthedotproduct,findthesmalleranglebetweenanytwodiagonalsofacube,whereeachdiagonalconnectsdiametricallyoppositecorners,andpassesthroughthecenterofthecube:Assumingasidelength,b,twodiagonalvectorswouldbeA=b(ax+ay+az)andB=b(ax�ay+az).NowuseA·B=|A||B|cos✓,orb2(1�1+1)=(p3b)(p3b)cos✓)cos✓=1/3)✓=70.53�.Thisresult(inmagnitude)isthesameforanytwodiagonalvectors.31.11.GiventhepointsM(0.1,�0.2,�0.1),N(�0.2,0.1,0.3),andP(0.4,0,0.1),find:a)thevectorRMN:RMN=(�0.2,0.1,0.3)�(0.1,�0.2,�0.1)=(�0.3,0.3,0.4).b)thedotproductRMN·RMP:RMP=(0.4,0,0.1)�(0.1,�0.2,�0.1)=(0.3,0.2,0.2).RMN·RMP=(�0.3,0.3,0.4)·(0.3,0.2,0.2)=�0.09+0.06+0.08=0.05.c)thescalarprojectionofRMNonRMP:RMN·aRMP=(�0.3,0.3,0.4)·(0.3,0.2,0.2)p0.09+0.04+0.04=0.05p0.17=0.12d)theanglebetweenRMNandRMP:✓M=cos�1✓RMN·RMP|RMN||RMP|◆=cos�1✓0.05p0.34p0.17◆=78�1.12.Writeanexpressioninrectangularcomponentsforthevectorthatextendsfrom(x1,y1,z1)to(x2,y2,z2)anddeterminethemagnitudeofthisvector.Thetwopointscanbewrittenasvectorsfromtheorigin:A1=x1ax+y1ay+z1azandA2=x2ax+y2ay+z2azThedesiredvectorwillnowbethedi↵erence:A12=A2�A1=(x2�x1)ax+(y2�y1)ay+(z2�z1)azwhosemagnitudeis|A12|=pA12·A12=⇥(x2�x1)2+(y2�y1)2+(z2�z1)2⇤1/21.13.a)FindthevectorcomponentofF=(10,�6,5)thatisparalleltoG=(0.1,0.2,0.3):F||G=F·G|G|2G=(10,�6,5)·(0.1,0.2,0.3)0.01+0.04+0.09(0.1,0.2,0.3)=(0.93,1.86,2.79)b)FindthevectorcomponentofFthatisperpendiculartoG:FpG=F�F||G=(10,�6,5)�(0.93,1.86,2.79)=(9.07,�7.86,2.21)c)FindthevectorcomponentofGthatisperpendiculartoF:GpF=G�G||F=G�G·F|F|2F=(0.1,0.2,0.3)�1.3100+36+25(10,�6,5)=(0.02,0.25,0.26)41.14.GiventhatA+B+C=0,wherethethreevectorsrepresentlinesegmentsandextendfromacommonorigin,a)mustthethreevectorsbecoplanar?Intermsofthecomponents,thevectorsumwillbeA+B+C=(Ax+Bx+Cx)ax+(Ay+By+Cy)ay+(Az+Bz+Cz)azwhichwerequiretobezero.SupposethecoordinatesystemisconfiguredsothatvectorsAandBlieinthex-yplane;inthiscaseAz=Bz=0.ThenCzhastobezeroinorderforthethreevectorstosumtozero.Therefore,thethreevectorsmustbecoplanar.b)IfA+B+C+D=0,arethefourvectorscoplanar?ThevectorsumisnowA+B+C+D=(Ax+Bx+Cx+Dx)ax+(Ay+By+Cy+Dy)ay+(Az+Bz+Cz+Dz)azNow,forexample,ifAandBlieinthex-yplane,CandDneednot,aslongasCz+Dz=0.Sothefourvectorsneednotbecoplanartohaveazerosum.1.15.Threevectorsextendingfromtheoriginaregivenasr1=(7,3,�2),r2=(�2,7,�3),andr3=(0,2,3).Find:a)aunitvectorperpendiculartobothr1andr2:ap12=r1⇥r2|r1⇥r2|=(5,25,55)60.6=(0.08,0.41,0.91)b)aunitvectorperpendiculartothevectorsr1�r2andr2�r3:r1�r2=(9,�4,1)andr2�r3=(�2,5,�6).Sor1�r2⇥r2�r3=(19,52,37).Thenap=(19,52,37)|(19,52,37)|=(19,52,37)66.6=(0.29,0.78,0.56)c)theareaofthetriangledefinedbyr1andr2:Area=12|r1⇥r2|=30.3d)theareaofthetriangledefinedbytheheadsofr1,r2,andr3:Area=12|(r2�r1)⇥(r2�r3)|=12|(�9,4,�1)⇥(�2,5,�6)|=33.351.16.IfArepresentsavectoroneunitinlengthdirecteddueeast,Brepresentsavectorthreeunitsinlengthdirectedduenorth,andA+B=2C�Dand2A�B=C+2D,determinethelengthanddirectionofC.(di�culty1)Takenorthasthepositiveydirection,andtheneastasthepositivexdirection.ThenwemaywriteA+B=ax+3ay=2C�Dand2A�B=2ax�3ay=C+2DMultiplyingthefirstequationby2,andthenaddingtheresulttothesecondequationeliminatesD,andweget4ax+3ay=5C)C=45ax+35ayThelengthofCis|C|=⇥(4/5)2+(3/5)2⇤1/2=1Cliesinthex-yplaneatanglefromduenorth(theyaxis)givenby↵=tan�1(4/3)=53.1�(or36.9�fromthexaxis).Forthosehavingnauticalleanings,thisisveryclosetothecompasspointNE34E(notrequired).1.17.PointA(�4,2,5)andthetwovectors,RAM=(20,18,�10)andRAN=(�10,8,15),defineatriangle.a)Findaunitvectorperpendiculartothetriangle:Useap=RAM⇥RAN|RAM⇥RAN|=(350,�200,340)527.35=(0.664,�0.379,0.645)Thevectorintheoppositedirectiontothisoneisalsoavalidanswer.b)FindaunitvectorintheplaneofthetriangleandperpendiculartoRAN:aAN=(�10,8,15)p389=(�0.507,0.406,0.761)ThenapAN=ap⇥aAN=(0.664,�0.379,0.645)⇥(�0.507,0.406,0.761)=(�0.550,�0.832,0.077)Thevectorintheoppositedirectiontothisoneisalsoavalidanswer.c)FindaunitvectorintheplaneofthetrianglethatbisectstheinteriorangleatA:Anon-unitvectorintherequireddirectionis(1/2)(aAM+aAN),whereaAM=(20,18,�10)|(20,18,�10)|=(0.697,0.627,�0.348)Now12(aAM+aAN)=12[(0.697,0.627,�0.348)+(�0.507,0.406,0.761)]=(0.095,0.516,0.207)Finally,abis=(0.095,0.516,0.207)|(0.095,0.516,0.207)|=(0.168,0.915,0.367)61.18.AcertainvectorfieldisgivenasG=(y+1)ax+xay.a)DetermineGatthepoint(3,-2,4):G(3,�2,4)=�ax+3ay.b)obtainaunitvectordefiningthedirectionofGat(3,-2,4).|G(3,�2,4)|=[1+32]1/2=p10.SotheunitvectorisaG(3,�2,4)=�ax+3ayp101.19.a)ExpressthefieldD=(x2+y2)�1(xax+yay)incylindricalcomponentsandcylindricalvariables:Havex=⇢cos�,y=⇢sin�,andx2+y2=⇢2.ThereforeD=1⇢(cos�ax+sin�ay)ThenD⇢=D·a⇢=1⇢[cos�(ax·a⇢)+sin�(ay·a⇢)]=1⇢⇥cos2�+sin2�⇤=1⇢andD�=D·a�=1⇢[cos�(ax·a�)+sin�(ay·a�)]=1⇢[cos�(�sin�)+sin�cos�]=0ThereforeD=1⇢a⇢b)EvaluateDatthepointwhere⇢=2,�=0.2⇡,andz=5,expressingtheresultincylindricalandcartesiancoordinates:Atthegivenpoint,andincylindricalcoordinates,D=0.5a⇢.Toexpressthisincartesian,weuseD=0.5(a⇢·ax)ax+0.5(a⇢·ay)ay=0.5cos36�ax+0.5sin36�ay=0.41ax+0.29ay1.20.IfthethreesidesofatrianglearerepresentedbythevectorsA,B,andC,alldirectedcounter-clockwise,showthat|C|2=(A+B)·(A+B)andexpandtheproducttoobtainthelawofcosines.Withthevectorsdrawnasdescribedabove,wefindthatC=�(A+B)andso|C|2=C2=C·C=(A+B)·(A+B)Sofarsogood.Nowifweexpandtheproduct,obtain(A+B)·(A+B)=A2+B2+2A·BwhereA·B=ABcos(180��↵)=�ABcos↵where↵istheinteriorangleatthejunctionofAandB.Usingthis,wehaveC2=A2+B2�2ABcos↵,whichisthelawofcosines.71.21.Expressincylindricalcomponents:a)thevectorfromC(3,2,�7)toD(�1,�4,2):C(3,2,�7)!C(⇢=3.61,�=33.7�,z=�7)andD(�1,�4,2)!D(⇢=4.12,�=�104.0�,z=2).NowRCD=(�4,�6,9)andR⇢=RCD·a⇢=�4cos(33.7)�6sin(33.7)=�6.66.ThenR�=RCD·a�=4sin(33.7)�6cos(33.7)=�2.77.SoRCD=�6.66a⇢�2.77a�+9azb)aunitvectoratDdirectedtowardC:RCD=(4,6,�9)andR⇢=RDC·a⇢=4cos(�104.0)+6sin(�104.0)=�6.79.ThenR�=RDC·a�=4[�sin(�104.0)]+6cos(�104.0)=2.43.SoRDC=�6.79a⇢+2.43a��9azThusaDC=�0.59a⇢+0.21a��0.78azc)aunitvectoratDdirectedtowardtheorigin:StartwithrD=(�1,�4,2),andsothevectortowardtheoriginwillbe�rD=(1,4,�2).Thusincartesiantheunitvectorisa=(0.22,0.87,�0.44).Converttocylindrical:a⇢=(0.22,0.87,�0.44)·a⇢=0.22cos(�104.0)+0.87sin(�104.0)=�0.90,anda�=(0.22,0.87,�0.44)·a�=0.22[�sin(�104.0)]+0.87cos(�104.0)=0,sothatfinally,a=�0.90a⇢�0.44az.1.22.Asphereofradiusa,centeredattheorigin,rotatesaboutthezaxisatangularvelocity⌦rad/s.Therotationdirectionisclockwisewhenoneislookinginthepositivezdirection.a)Usingsphericalcomponents,writeanexpressionforthevelocityfield,v,whichgivesthetan-gentialvelocityatanypointwithinthesphere:Asinproblem1.20,wefindthetangentialvelocityastheproductoftheangularvelocityandtheperperdiculardistancefromtherotationaxis.Withclockwiserotation,weobtainv(r,✓)=⌦rsin✓a�(r<a)b)Converttorectangularcomponents:Fromhere,theproblemisthesameaspartcinProblem1.20,excepttherotationdirectionisreversed.Theanswerisv(x,y)=⌦[�yax+xay],where(x2+y2+z2)1/2<a.1.23.Thesurfaces⇢=3,⇢=5,�=100�,�=130�,z=3,andz=4.5defineaclosedsurface.a)Findtheenclosedvolume:Vol=Z4.53Z130�100�Z53⇢d⇢d�dz=6.28NOTE:Thelimitsonthe�integrationmustbeconvertedtoradians(aswasdonehere,butnotshown).b)Findthetotalareaoftheenclosingsurface:Area=2Z130�100�Z53⇢d⇢d�+Z4.53Z130�100�3d�dz+Z4.53Z130�100�5d�dz+2Z4.53Z53d⇢dz=20.781.23c)Findthetotallengthofthetwelveedgesofthesurfaces:Length=4⇥1.5+4⇥2+2⇥30�360�⇥2⇡⇥3+30�360�⇥2⇡⇥5�=22.4d)Findthelengthofthelongeststraightlinethatliesentirelywithinthevolume:ThiswillbebetweenthepointsA(⇢=3,�=100�,z=3)andB(⇢=5,�=130�,z=4.5).Performingpointtransformationstocartesiancoordinates,thesebecomeA(x=�0.52,y=2.95,z=3)andB(x=�3.21,y=3.83,z=4.5).TakingAandBasvectorsdirectedfromtheorigin,therequestedlengthisLength=|B�A|=|(�2.69,0.88,1.5)|=3.211.24.Twounitvectors,a1anda2lieinthexyplaneandpassthroughtheorigin.Theymakeangles�1and�2withthexaxisrespectively.a)Expresseachvectorinrectangularcomponents;Havea1=Ax1ax+Ay1ay,sothatAx1=a1·ax=cos�1.Then,Ay1=a1·ay=cos(90��1)=sin�1.Therefore,a1=cos�1ax+sin�1ayandsimilarly,a2=cos�2ax+sin�2ayb)takethedotproductandverifythetrigonometricidentity,cos(�1��2)=cos�1cos�2+sin�1sin�2:Fromthedefinitionofthedotproduct,a1·a2=(1)(1)cos(�1��2)=(cos�1ax+sin�1ay)·(cos�2ax+sin�2ay)=cos�1cos�2+sin�1sin�2c)takethecrossproductandverifythetrigonometricidentitysin(�2��1)=sin�2cos�1�cos�2sin�1:Fromthedefinitionofthecrossproduct,andsincea1anda2bothlieinthex-yplane,a1⇥a2=(1)(1)sin(�1��2)az=������axayazcos�1sin�10cos�2sin�20������=[sin�2cos�1�cos�2sin�1]azthusverified.1.25.GivenpointP(r=0.8,✓=30�,�=45�),andE=1r2✓cos�ar+sin�sin✓a�◆a)FindEatP:E=1.10a⇢+2.21a�.b)Find|E|atP:|E|=p1.102+2.212=2.47.c)FindaunitvectorinthedirectionofEatP:aE=E|E|=0.45ar+0.89a�91.26.Expresstheuniformvectorfield,F=5axina)cylindricalcomponents:F⇢=5ax·a⇢=5cos�,andF�=5ax·a�=�5sin�.Combining,weobtainF(⇢,�)=5(cos�a⇢�sin�a�).b)sphericalcomponents:Fr=5ax·ar=5sin✓cos�;F✓=5ax·a✓=5cos✓cos�;F�=5ax·a�=�5sin�.Combining,weobtainF(r,✓,�)=5[sin✓cos�ar+cos✓cos�a✓�sin�a�].1.27.Thesurfacesr=2and4,✓=30�and50�,and�=20�and60�identifyaclosedsurface.a)Findtheenclosedvolume:ThiswillbeVol=Z60�20�Z50�30�Z42r2sin✓drd✓d�=2.91wheredegreeshavebeenconvertedtoradians.b)Findthetotalareaoftheenclosingsurface:Area=Z60�20�Z50�30�(42+22)sin✓d✓d�+Z42Z60�20�r(sin30�+sin50�)drd�+2Z50�30�Z42rdrd✓=12.61c)Findthetotallengthofthetwelveedgesofthesurface:Length=4Z42dr+2Z50�30�(4+2)d✓+Z60�20�(4sin50�+4sin30�+2sin50�+2sin30�)d�=17.49d)Findthelengthofthelongeststraightlinethatliesentirelywithinthesurface:ThiswillbefromA(r=2,✓=50�,�=20�)toB(r=4,✓=30�,�=60�)orA(x=2sin50�cos20�,y=2sin50�sin20�,z=2cos50�)toB(x=4sin30�cos60�,y=4sin30�sin60�,z=4cos30�)orfinallyA(1.44,0.52,1.29)toB(1.00,1.73,3.46).ThusB�A=(�0.44,1.21,2.18)andLength=|B�A|=2.531.28.StatewhetherornotA=Band,ifnot,whatconditionsareimposedonAandBwhena)A·ax=B·ax:Forthistobetrue,bothAandBmustbeorientedatthesameangle,✓,fromthexaxis.Butthiswouldalloweithervectortolieanywherealongaconicalsurfaceofangle✓aboutthexaxis.Therefore,AcanbeequaltoB,butnotnecessarily.b)A⇥ax=B⇥ax:Thisisamorerestrictiveconditionbecausethecrossproductgivesavector.Forbothcrossproductstolieinthesamedirection,A,B,andaxmustbecoplanar.ButifAliesatangle✓tothexaxis,Bcouldlieat✓orat180��✓togivethesamecrossproduct.Soagain,AcanbeequaltoB,butnotnecessarily.101.28c)A·ax=B·axandA⇥ax=B⇥ax:Inthiscase,weneedtosatisfybothrequirementsinpartsaandb–thatis,A,B,andaxmustbecoplanar,andAandBmustlieatthesameangle,✓,toax.Withcoplanarvectors,thislatterconditionmightimplythatboth+✓and�✓wouldthereforework.Butthenegativeanglereversesthedirectionofthecrossproductdirection.Thereforebothvectorsmustlieinthesameplaneandlieatthesameangletox;i.e.,AmustbeequaltoB.d)A·C=B·CandA⇥C=B⇥CwhereCisanyvectorexceptC=0:Thisisjustthegeneralcaseofpartc.Sincewecanorientourcoordinatesysteminanymannerwechoose,wecanarrangeitsothatthexaxiscoincideswiththedirectionofvectorC.Thusalltheargumentsofpartcapply,andagainweconcludethatAmustbeequaltoB.1.29.Expresstheunitvectoraxinsphericalcomponentsatthepoint:a)r=2,✓=1rad,�=0.8rad:Useax=(ax·ar)ar+(ax·a✓)a✓+(ax·a�)a�=sin(1)cos(0.8)ar+cos(1)cos(0.8)a✓+(�sin(0.8))a�=0.59ar+0.38a✓�0.72a�b)x=3,y=2,z=�1:First,transformthepointtosphericalcoordinates.Haver=p14,✓=cos�1(�1/p14)=105.5�,and�=tan�1(2/3)=33.7�.Thenax=sin(105.5�)cos(33.7�)ar+cos(105.5�)cos(33.7�)a✓+(�sin(33.7�))a�=0.80ar�0.22a✓�0.55a�c)⇢=2.5,�=0.7rad,z=1.5:Again,convertthepointtosphericalcoordinates.r=p⇢2+z2=p8.5,✓=cos�1(z/r)=cos�1(1.5/p8.5)=59.0�,and�=0.7rad=40.1�.Nowax=sin(59�)cos(40.1�)ar+cos(59�)cos(40.1�)a✓+(�sin(40.1�))a�=0.66ar+0.39a✓�0.64a�1.30.Consideraproblemanalogoustothevaryingwindvelocitiesencounteredbytranscontinentalaircraft.Weassumeaconstantaltitude,aplaneearth,aflightalongthexaxisfrom0to10units,noverticalvelocitycomponent,andnochangeinwindvelocitywithtime.Assumeaxtobedirectedtotheeastandaytothenorth.Thewindvelocityattheoperatingaltitudeisassumedtobe:v(x,y)=(0.01x2�0.08x+0.66)ax�(0.05x�0.4)ay1+0.5y2a)Determinethelocationandmagnitudeofthemaximumtailwindencountered:Tailwindwouldbex-directed,andsowelookatthexcomponentonly.Overtheflightrange,thisfunctionmaximizesatavalueof0.86/(1+0.5y2)atx=10(attheendofthetrip).Itreachesalocalminimumof0.50/(1+0.5y2)atx=4,andhasanotherlocalmaximumof0.66/(1+0.5y2)atthetripstart,x=0.b)Repeatforheadwind:Thexcomponentisalwayspositive,andsothereforenoheadwindexistsoverthetravelrange.c)Repeatforcrosswind:Crosswindwillbefoundfromtheycomponent,whichisseentomaximizeovertheflightrangeatavalueof0.4/(1+0.5y2)atthetripstart(x=0).d)Wouldmorefavorabletailwindsbeavailableatsomeotherlatitude?Ifso,where?Minimizingthedenominatoraccomplishesthis;inparticular,thelattitudeassociatedwithy=0givesthestrongesttailwind.11