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Solution Manual for Calculus for Engineers, 4th Edition - Page 1

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Solution Manual for Calculus for Engineers, 4th Edition

Struggling with textbook problems? Solution Manual for Calculus for Engineers, 4th Edition offers a clear breakdown of every exercise for easy understanding.

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Solution Manual for Calculus for Engineers, 4th Edition - Page 1 preview imageSOLUTIONSMANUALCalculusforEngineersFourthEditionDonaldTrimUniversityofManitoba
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Solution Manual for Calculus for Engineers, 4th Edition - Page 2 preview imageDownloadedfromStudyXY.com®+StudyXYSdYe.o>\|iF’prE\3SStudyAnythingThisContentHasbeenPostedOnStudyXY.comassupplementarylearningmaterial.StudyXYdoesnotendroseanyuniversity,collegeorpublisher.Allmaterialspostedareundertheliabilityofthecontributors.wv8)www.studyxy.com
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Solution Manual for Calculus for Engineers, 4th Edition - Page 3 preview imageCONTENTSThismanualcontainsdetailedsolutionstoallexercisesinthetext.Wewouldappreciatebeingmadeawareofanyerrors.Chapter1CalculusPreparation1Chapter2LimitsandContinuity72Chapter3Differentiation99Chapter4ApplicationsofDifferentiation170Chapter5AntiderivativesandtheIndefiniteIntegral300Chapter6TheDefiniteIntegral332Chapter7ApplicationsoftheDefiniteIntegral360Chapter8TechniquesofIntegration443Chapter9ParametricEquationsandPolarCoordinates506Chapter10InfiniteSequencesandSeries553Chapter11VectorsandThree-dimensionalAnalyticGeometry644Chapter12DifferentialCalculusofMultivariableFunctions722Chapter13MultipleIntegrals816Chapter14VectorCalculus935Chapter15DifferentialEquations1012i
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Solution Manual for Calculus for Engineers, 4th Edition - Page 4 preview imageEXERCISES1.21CHAPTER1EXERCISES1.21.Thesolutionisz=2/3.2.Thesolutionis2=5/14.3.Since2%+2x—3=(z+3)(z—1),solutionsarez=—3andz=1.4.Since122%+11x—5=(3x—1)(4z+5),solutionsarez=1/3and&=—5/4.5.Sincethediscriminant52—4(2)(10)=—55isnegative,theequationhasnorealsolutions.—104TOO4(4)(9)-10+v316.Quadraticformula1.5givesz=0x10-440)=clove=EEN7.Since22+8z+16=(z+4)?,thesolutionis#=—4withmultiplicity2.8.Since422—36x+81=(2x—9)?thesolutionis==9/2withmultiplicity2.5+BHAD)51VIB9.Quadraticformula1.5givesz=SoxyBAAC=RENE10.Sincethediscriminant(—8)%—4(4)(9)=—80isnegative,theequationhasnorealsolutions.11.Since23—32%+32—1=(x—1)%,thesolutionis#~1withmultiplicity3.12.Possiblerationalsolutionsare+1,-£1/2,1/4,+1/8.Wefindthat#=—1/2isasolution.Wefactor22+1fromthecubic,82%+122”+62+1=(20+1)(da?+do+1)=(22+1)(2z+1)?=(20+1)°.Theonlysolutionisz=—1/2withmultiplicity3.13.Possiblerationalsolutionsare2—+1,£2,#5,+10.Wefindthatz=2isasolution.Wefactorz—2fromthecubic,28—20%+52—10=(z—2)(2®+5)=0.Theothertwosolutionsarecomplex.14.Possiblerationalsolutionsare2=+1,+3,£9,butitshouldalsobeclearthatnopositivevalueof=cansatisfytheequation.Wefindthat==—1isasolution.Wefactor2+1fromthecubic,Pde?+1224+9=(a+)?+3z+9)=0.Sincethediscriminantofthequadraticisnegative,theothertwosolutionsarecomplex.15.Possiblerationalsolutionsare+1,£2,+4,+8,116,1:32,+64,butitshouldalsobeclearthatnopos-itivevalueof2cansatisfytheequation.Wefindthatx=—4isasolution.Wefactor=+4fromthecubic,2741222+485+64=(x+4)(2%+82+16)=(z+A)(x+4)*=(z+4)*Theonlysolutionisx=~4withmultiplicity3.16.Possiblerationalsolutionsare+1,£2,£3,£4,£6,£9,£12,£18,4:36.Wefindthatz=~3isasolution.Wefactorz+3fromthequartic,2%T2®+92%—21a—36=(+3)(a®+42°—3x.—12).Possiblerationalzerosofthecubicare+1,+2,+3,44,+6,+12.Wefindthatx=—4isazero.Wefactorx+4fromthecubic,a+7%+90%—20a—36=(2+3)(x+4)(a?—3).Thesolutionsare&=—3,4,£V/3.17.Since2?—16=(22+4)(z®—4)=(a®+4)(z+2)(z—2),therealsolutionsarez=+2.+StudyXxy
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Solution Manual for Calculus for Engineers, 4th Edition - Page 5 preview image2EXERCISES1.218.Possiblerationalsolutionsare£1,+3,+5,£15,£1/2,£3/2,£5/2,£15/2.Wefindthatz=—5isasolution.Wefactorz+5fromthequartic,22+92%—62%—8x—15=(z+5)(22®—a?—2—3).Possiblerationalzerosofthecubicave1,£3,£1/2,£3/2.Wefindthatx=3/2isazero.Wefactor2a—3fromthecubic,2+92%—62%—8x—15=(zx+5)(2x—3)(@%+z+1).Sincethequadratichasanegativediscriminant,theonlyrealsolutionsarez=—5and©=3/2.19.Possiblerationalsolutionsare+1,438,£0,£1/2,£3/2,£9/2,+1/3,1/6.Wefindthat=—1/2isasolution.Wefactor2z+1fromthequartic,6a?42°+532%+92—9=(2x+1)(3z°—a?+272—9).Possiblerationalzerosofthecubicare+1,+3,+9,21/3.Wefindthat=1/3isazero.Wefactor3x—1fromthecubic,62+2°+532%+9x—9=(22+1)(3z—1)(2?+9).Sincethequadratichascomplexzeros,theonlyrealsolutionsare2=—1/2and&=1/3.20.Norealnumberscansatisfythisequation.21.Possiblerationalsolutionsare£1,£2,+3,+4,+6,£8,£9,+12,+18,24,+36,£72.Wefindthatx=24isasolution.Wefactorx—24fromthecubic,2%—232%—21a~72=(2—24)(2®+2+3).Sincethequadratichasanegativediscriminant,theonlyrealsolutionisx=24.22.Possiblerationalsolutionsare+1,+2,£3,£4,+6,£8,£9,+12,£16,+18,+24,4:32,+36,+48,+64,+72,+96,+144,+192,£192,£288,£576.Wefindthat2=—4isasolution.Wefactor2+4fromthequartic,2%dp?—442?4962+576=(x+4)(2®—82%—122+144).Possiblerationalzerosofthecubicare£1,£2,+3,+4,£6,18,£9,£12,+16,+18,+24,£36,£48,+72,+144.Wefindthatz=—4isazero.Wefactorz+4fromthecubic,2%—4a?—442+963+576=(z+4)(z+4)(2®—122+36)=(+4)*(z—6).Thus,=—4and2=6aresolutions,eachofmultiplicity2.23.Since32%+2°+52%=2*(3z®+x+5),andthequadratichasanegativediscriminant,theonlyrealsolutionis2=0withmultiplicity2.24.Possiblerationalsolutionsare£1,£2,+4,+5,+10,+20,£1/2,45/2,+1/3,£2/3,+4/3,£5/3,+10/3,£20/3,£1/6,£5/6.Wefindthat2=5/6isazero.Wefactor6—5fromthecubic,62°+2%+192—20=(62—5)(2?+2+4).Sincethequadratichasanegativediscriminant,z=5/6istheonlyrealsolution.25.Possiblerationalzerosare+1,£3,5,1:9,£15,+45.Wefindthat=—5isasolution.Wefactor2+5fromthepolynomial,20+5a—92—45=(z+5)(a!—9)=(2+5)(2®+3)(x?—3).Solutionsare2=~5and©=+3.+StudyXY
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Solution Manual for Calculus for Engineers, 4th Edition - Page 6 preview imageEXERCISES1.2326.Possiblerationalsolutionsare+1,+2,+3,£4,15,16,£8,1:10,+12,+15,+20,£24,1:30,3:40,160,+120.Wefindthat2=1isasolution.Wefactorz—1fromthepolynomial,25—152%+852%—22522+27dw—120=(x—D(z?—14z®+71a?—154z+120).Weusethesamesetofrationalpossibiliticsforthequartic.Wefindthat#=2isazero.Whenwefactorx—2fromthequartic,2°—152%+852°—2262°+27d—120=(2—1)(z—2)(=®—120°+47a—60).Possiblerationalzerosofthecubicare+1,+2,+3,+4,£5,£6,£10,+12,+15,+20,£30,£60.Wefindthatz=3isazero.Wefactor2—3fromthecubic,25150%4852°—225%+274—120=(x—1)(z—2)(z—3)(2®—9a+20)=(z—1){z—2)(z—3)(z—4)(z—5).Solutionsaretherefore=1,2,3,4,5.27.Possiblerationalsolutionsare£1,+2,+4,+1/2,+1/4,butitshouldalsobeclearthatnopositivevalueof=cansatisfytheequation.Wefindthatz=—1/2isasolution.Wefactor2z'+1fromthequartic,dat440%+172%+162+4=(2x+1)(2°+2%+82+4).Possiblerationalzerosofthecubicare£1,£2,+4,+1/2,andonceagainwerejectthepositivevalues.Wefindthat©=—1/2isazero.Wefactor22+1fromthecubic,4a+de®+172%+162+4=(20+1)(2z+1)(2?+4).Thus,&=—1/2istheonlyrealsolutionandithasmultiplicity2.28.Possiblerationalsolutionsare£1,£2,4,1/5,£2/5,34/5,£1/25,+2/25,+4/25.Wefindthat»=2/5isasolution.Wefactor52—2fromthequartic,250—1200°+1092—362+4=(52—2)(bz®—22%+132—2).Possiblerationalzerosofthecubicare+1,+2,+1/5,£2/5.Wefindthat©=2/5isazero.WefactorSz—2fromthecubic,257—1202°+1092°—362+4=(5x—2)(52—2)(z*—dz+1).Thus,2=2/5isarealsolutionwithmultiplicity2andthequadraticformulagivestheremainingtwosolutionspodEVI64SARL—24v329.Possiblerationalsolutionsare£1,£2,£4,+5,£8,£10,£20,#25,+40,£50,£100,£200.Wefindthatx=1isazero.Wefactor2—1fromthepolynomial,2°+9a?+472%+1252%+18x—200=(z—1)(=*+102°+572%+1820+200).Weusethesamerationalnumbersforthequartic,butrejectthepositivevalues.Wefindthatz=2isazero.Wefactorz+2fromthequartic,25+9a+470°+12527+18—200=(&—1)(z+2)(z®+82%+41x+100).Forzerosofthecubicweuse—1,—2,—4,—5,—10,—20,—25,—-50,—100.Wefindthatx=—~4isazero.Whenwefactoritout,a®492+472%+1252°+18%—200=(x—1)(z+2)(x+4)(z*+4x+25).Sincethequadratichasanegativediscriminant,therealsolutionsare©=—4,-2,1.+StudyXY
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Solution Manual for Calculus for Engineers, 4th Edition - Page 7 preview image4EXERCISES1.380.Possiblerationalsolutionsare£1,£2,3,+4,+6,+8,+9,+12,+16,£18,£24,+27,-:36,+48,+54,+72,+81,+108,+144,+162,£216,£324,+432,1648,£1296.Wefindthatz=3isasolution.Whenwefactorz—3fromthepolynomial,284162%—812%—1296=(x—3)(2°+327+250%+7527+14x+432).Whenwenotethatnopositivevaluecansatisfythefifthdegreepolynomial,possiblerationalzerosarc—1,2,—3,—4,—6,—8,—9,—12,—16,—18,—24,27,—36,—48,—54,—72,—108,—144,—216,—432.Wefindthatz=—3isazero.Whenwefactor+3fromthepolynomial,28+162—81a?—1206=(x—8)(x+3)(a+252%+144).Sincetherecanbenorealzerosofthequartic,therealsolutionsarex=£3.31.2(z+5)(xz—1)32.2x—(3+v05)/4)fx~(3-V65)/4]33.(z+2)(z—3)z10)34.24(x+5/8)(x~1/4)(x—1/2)35.(z41)(z—1)°36.16(z—1/2)%(z+1/2)?37.Onepolynomialis(x+41/3)(z—4/5)(z—3)(z—4)*.Thispolynomialcouldbemultipliedbyanyconstant.38.Accordingtotherationalroottheorem,possiblerationalzerosmustbedivisorsofapdividedbydivisorsof1.Thismeansthatpossiblerationalzerosaredivisorsofao.EXERCISES1.31.Withformula1.10,thedistanceisv/22+12=/5.2.Withformula1.10,thedistanceis1/62+(=3)Z=3/5.3.Withformula1.10,thedistanceis\/{—2)2+(—6)%=2V/T0.4.Withformula1.10,thedistanceis\/{—7)2+(3)?=v/58.5.Withformula1.11,themidpointis(259)—(=3)-6.Withformula1.11,themidpointis(5H155=(1-3)7.Withformula1.11,themidpointis(229)=(-2,-5).Noo(3-421118.Withformula1.11,themidpointis(32)=(-35)9.Withslope—2/4=—1/2,equation1.13gives10.Thelineishorizontal.1y-2=—le-1)=a+2=35.Ttsequationisy=—6.eo4%(1.2)%0G6)(5-6)+StudyXxy
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Solution Manual for Calculus for Engineers, 4th Edition - Page 8 preview imageEXERCISES1.3511.Withformula1.13,theequationis12.withformula1.13,theequationisy+3=3(x+2)==>y=238z13.yty=—3=1)=at+2+2=0yidx(32)x(2:3)13.Theequationofthey-axisis@=0.14.Theequationofthez-axisisy=0.15.Withm=(3—0)/(4+2)=1/2,equation1.1316.Withm=(4+2)/(0+1)=6,equation1.13givesy=(1/2)(z+2)=>2y=z+2.givesy—4=06(x—0)=>y=06x+4.yhy3)4€lx1-2)x17.Withslope(0+3)/(1—0)=3,equation1.1318.Themidpointofthelinesegment,is3-74+8givesy—0=3(x—1)=y=3z—3.(55-2Withm=(6—0)/(—2—0)=—3,equation1.13givesy—0=—3(z—0)=y=—3z.odY£78)fe)1£3pH19.Sinceslopes~1and1arenegativereciprocals,thelinesareperpendicular.20.Sinceslopesofbothlinesare—1/3,theyareparallel.21.Sinceslopesofbothlinesare1/3,theyareparallel.22.Sinceslopes—2/3and3/2arenegativereciprocals,thelinesareperpendicular.23.Sinceslopesare3and—1/2,thelinesareneitherparallelnorperpendicular.24.Sinceslopesarc1and—2/3,thelinesareneitherparallelnorperpendicular.25.Thelinesareperpendicular.26.Sinceslopesare—1and3,thelinesareneitherparallelnorperpendicular.27.Whenwesubtracttheequations,y+2y=0+3=>y=1.Thepointofintersectionis(=1,1).28.Thepointofintersectionis(1,2).29.Whenwesubtract3timesthesecondequationfromthefirst,4y+18y=6—9—>y=—3/22.Thepointofintersectionis(24/11,—3/22).30.Whenwesubstitutey=2a+6intothesecondequation,©=(22+6)+4=>x=—10.Thepointofintersectionis(10,14).|vv1StudyXY
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Solution Manual for Calculus for Engineers, 4th Edition - Page 9 preview image6EXERCISES1.331.Whenwesubstitutez=2—2/3intothesecondequation,2(2—2y/3)—y/4=15=>y=—132/19.‘Thepointofintersectionis(126/19,—132/19).32.Ifwemultiplythefirstequationby5andaddtheresulttothesocondequation,weobtain73x=37=2=37/73.Thepointofintersectionis(37/73,153/146).CCIB+4-16 -6-3833.Formula1.16Ee—aN34.Formula1.16———==‘ormulagivesSixt7‘ormulagivesNie755-1-34-1435.Formula1.16givesbold=0.86.Formula1.16givesBed~%5"2V2.4436737.Thedistanceis7.38.Formula1.16givesmo-Ta39.Sincethepointofintersectionofthegivenlinesis(3,1),andtheslopeofz+2y=15is—1/2,therequiredequationisy—1=—(1/2)(z—3)=x+2y=5.40.Sincetheslopeofz—y=4is1,andthepointofintersectionof2+3y=3andz—y=4is(3,-1),therequiredequationisy+1=—1(z—3)=z+y=2.41.Sincetheslopeofthelinethrough(1,2)and(3,0)is2/4,therequiredequationisy—6=(1/2)(z—5)=2y=x+T.42,Sincetheslopeofthelinethrough(—3,4)and(1,2)is(4+2)/(~3—1)=—3/2,theequationoftherequiredlineisy+2=(2/3)(z+3)=»22=3y.43.Since[|OA|=[|OB||,coordinatesofAandB44.WhenweequateslopesoflinesegmentsACand5are(a,0)and(0,a),respectively.SincetheAB,2===b==areaofAOABis8,itfollowsthatCombinethiswiththefactthatthearea(1/2)a?=8=>a=4.TheequationofAQAB=(1/2)ab=30,andweobtaina=6oflineABisy=—(z—4)=z-+y=4andb=10.Theequationofthelineisy=—(5/3)(z—6)=Sz+3y=30.woofconfBG.5)olAw%Aap)x45.SincetheslopeoflinesegmentABis2,46.Iftheequationisy=¢,thenA=(¢,¢)andanditslengthis3,itfollowsthatB=(4—¢,c).SincetriangleABDhasarea9,—bfa=2andVa?+b?=3.Thesecanitfollowsthat9=le—)(4—2c),besolvedfora=—3/v/5andb=6/5.solutionsofwhichare5,—1.TheTheequationofABisy=22+6/V5.requiredequationisthereforey=—1.24odBOYANFr=4=x>D(2,2)TN[oepo)%1]a”47.(a)TheconversionequationisTir=5(Tr—32)/9.(b)TheconversionequationisTr=97¢/5+32.(¢)TheyareoneandthesamelineifweplotTpalongthehorizontalaxisandTcalongtheverticalaxis(leftfigurebelow).IfweplotTralongtheverticalaxisandTe:alongthehorizontalaxisinpart+StudyXY
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Solution Manual for Calculus for Engineers, 4th Edition - Page 10 preview imageiEXERCISES1.37(b),weobtaintherightfigurebelow.100%nal*32,32227,wo(d)IfwesetTr=Tt:intheequationfrompart(a),weobtainTg=5(Tc—32)/9=>4T¢=—160=Te=—40.48.(a)TheconversionequationisTx=5(Tr—32)/9+273.16.(1b)TheconversionequationisTp=9(Tj~273.16)/5+32.(c)TheyareoneandthesamelineifweplotTralongthehorizontalaxisandTcalongtheverticalaxis(leftfigurebelow).IfweplotTwalongtheverticalaxisandTxalongthehorizontalaxisinpart{b),weobtaintherightfigurebelow.TeTF255.3455.38T459.69T459.69)(d)IfwesetTp=Txintheequationfrompart(a),weobtainTx=5(Tx—32)/9+278.16==4Tx=2208.44==>Tx=574.61.49.(a)IfthetemperatureischangedtoT,thechangeIinthelengthofthebarisaLo(T—Tp),andPo—thereforeitslengthisLo1-oT)L=Lo+aLo(T—To)=Lo[l+o(T—To).{b)TheendswillbeincontactwhentherailshavelengthL=10.003m.Thisoccurswhen10.003=10[L+1.17x1075(T~20)]==T'=45.6.ET50.CoordinatesofE,themidpointofsideAC,are(2,3).SincetheslopeofBEis—1/2,theequationof4DESmedianBEisy—4=—(1/2)2—2+2y=8.BOASimilarly,equationsformediansADandCFare%7x—y=6andy=x+2,respectively.LinesegmentsrFADandCFintersectinthepoint(4/3,10/3),{andthispointalsosatisfiesx+2y—8.Thus,theALDthreemediansintersectatthepoint(4/3,10/3).x51.CoordinatesofP,Q,R,and§areYCladPl(@1+2)/2,(11+92)/2),Qa+23)/2,(v2+y3)/2),)R((zs+4)/2,(ya+y4)/2),andS((1+24)/2(51+y4)/2).®BGs.)SlopesofthelinesegmentsPS,RQ,PQ,andRSDeerare,respectively,owfo@r—w)/2(w-w)/2(uw)(@-y)/2§®(wz—@4)/2"(za—x0)/2"(za—11)/2"(23—21)/2AG)SincePSandRQareparallel,asarcPQandRS,PQRSisaparallelogram.+StudyXxy
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Solution Manual for Calculus for Engineers, 4th Edition - Page 11 preview image8EXERCISES1.352.Themidpointofthelinesegmentis(1,~1),anditsslopeis—3/2.Theequationoftheperpendicularbisectorisy+1=(2/3}(z—1)=22=3y+5.53.Ifcoordinatesofthepointare(x,y),then(z—1)2+(y—2)?=(z+1)*+(y—4)%and(z—1)*+(y—2)*=(2+3)?+(y—1)%.Whencxpandedandsimplified,thesereduceto —y=—3and82+2y=5,thesolutionofwhichis(—11/10,19/10).54.Supposetwononvertical,parallellineshaveslopesm)andmy.Theirequationscanbeexpressedintheformy=myz+b;andy=maz+by.Tofindtheirpointofintersectionwesetmiz+by=max+b.Sincethelinesdonotintersect,theremustbenosolutionofthisequation.Thishappensonlyifmy=ms;thatis,thelineshavethesameslope.Conversely,iftwononverticallineshavethesameslope,theirequationsmustbeoftheformy=mz+byandy=ma+by,whereby#ba.Whenweattempttofindtheirpointofintersectionbysettingma:+by=ma+by,wefindthatby=bz;acontradiction.Inotherwords,thelinesdonotintersect.55.BecausetrianglesPTQandPSRaresimilar,ratiosofcorrespondingsidesareequal,i.by1PQl_IPT)_22—21Qari)IPR]IPS]2—aRG)Ifwesubtract1fromeachsideofthisequation,[25]x2—LAIR2SURtBL.IPR]pr.1Py)EmTle)IPQPal_@a—2ra_wyeyET=F2.2xIPR]TaOma-mThus,rox—ro@y=712—ML—>&=hme,Asimilarproofgivesthecorrespondingformula2forthey-coordinateofR.*56.No.Definition1.1definesparallelismonlyfordifferentlines.57.Wechooseacoordinatesystemwiththeoriginatonevertexofthetriangle,andthepositivez-axisalong\yBooneside.Thecoordinatesoftheverticesarethen0(0,0),A(a,0),andB(b,c).Usingequation1.11,coordinatesofthemidpointsofthesidesare|FPa+bcbea(55)e(3s)#Go)Thesumofthesquaresofthelengthsofthemediansisol®Aledxb\?ce)?b\?cy?a?Pl?2ir?=|(2Hc_b_c_a2IOPIP+AI+|BR)VG)(eg)+(5)|+|(b-5)+e3=5+64ab).Three-quartersofthesumofthesquaresofthelengthsofthesidesis3337(fOAI+ABI?+[0B]?=i{a®+[b=a)?+c?+(B®+A}=5(@®+62+c*—ab).58.Ifwechoosethecoordinatesysteminthefigure,equationsofACandABarey=—v/3(z—h/V/3)hom—V3z+y—h=0andy=V3(z+h/V3)—>Bx—y+h=0.IfP(d,e)isanypointinteriortothetriangle,wecanuseformula1.16to»5>findthesumofthedistancesfromPtothethreesidesIPEI+12D]+[PE]oTGases+StudyXY
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Solution Manual for Calculus for Engineers, 4th Edition - Page 12 preview imageEXERCISES1.4A9[V3d+e—h| |V3d—c+h|11—eteepo=e+|V3—hl+z|vV3d—e+h|.e+AToTetplVidteil+513e+hlBecauseP(d,e)isbelowandtotheleftoflineAC,itfollowsthat3d+e—h<0.BecausePisbelowandtotherightofAB,3d—e+h>0.Hence,11[|PF|+|PDI|+|IPE|=e+(=V3d—eth)+5(v3d—e+h)=hEXERCISES1.4A1.Thisistheparabolay=222shifted2.Factoredintheformy=—(z—3)(z—1),1unitdownward.the2-interceptsoftheparabolaare1and3.Itsmaximumisat©=2.4od7%-11xJS)3.Factoredintheformy=(x—1)%,4.Thisistheparabola2=4y%/3shiftedtheparabolahasitsminimumat(1,0).1/3unittotheleft.24id17BR}1-12*1.x5.Factoredintheformx=(2+),6.Factoredintheformy=—(x+1)(z—4)/2,theparabolaopenstotherightwith2-interceptsoftheparabolaare—1and4.y-intercepts0and—2.Itsmaximumisatz=3/2.byidEyxa3NxH37.Theparabolaz=1—4?opensleftwith8.Factoredintheformx=—(2y—5)(y+1),x-intercept1andy-intercepis+1.y-interceptsoftheparabolaare—1and5/2.Itsmaximuminthez-directionoccursfory=3/4.yy37]374]Tx3!5x+Studyxy
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Solution Manual for Calculus for Engineers, 4th Edition - Page 13 preview image10EXERCISES1.4A9.Expressedintheformy=4(x?+bz/4)+1010.Thisparabolaopenstotheright.=4x+5/8)*+10—25/16=4(z+5/8)%--135/16,theparabolaopensupwardwithminimumat(—5/8,135/16).rdLyx(-5/8,135/16)toxz11.Expressedintheformy=—{x?—6z+9)12.Thisparabolaopenstotheleft=—(z—3)?,theparabolaopensdownwardandtouchesthey-axisaty=—4.fromthepoint(3,0).y43x-16x4913.(a)They-interceptis—5,andz-interceptsare2AD=1+6.(b)Thea-interceptis4,andfromx=4(y—1)2,they-interceptis1.14.Theequationmustbeoftheformy=az?+1.Since(2,3)isontheparabola,3=da+1=a=1/2.15.Theequationmustbeoftheform==2+ay?.Since(0,4)isontheparabola,0=2+16a=>a=—1/8.16.Sincetheparabolacrossesthez-axisat©=~1andz=3,itmustbeoftheformy=alz+1)(2~3).Since(0,—1)isontheparabola,—1=a(1)(-3)=a=1/3.17.Sincetheparabolatouchesthez-axisat=1,itmustbeoftheformy=a(x—1)%.Since(0,2)isontheparabola,2=a(—1)?=>a=2.18.Wesctz+1=1-2—0=2"+219.Weset—20=1+a?=0=212s+1=a(z+1)=x=0,—1.Pointsof=(z+1)=x=—1.Theonlypointintersectionare(—1,0)and(0,1).ofintersectionis(1,2).y421]2|24i%E;EliK4.220.Weset22—22—6=5+2/5=>21.Wesety(y—1)=y—1/2=>2°—dy+1=0.522%—9+55=0.Withequation1.5,Withequation1.5,y=(4+16—8)/42=(94/81=4(5)(55))/10.Since=(2+v/2)/2.Pointsofintersectionare81—4(5)(55)<0,therearenopoints(1V2)/2,(2£V2)/2).ofintersection.+StudyXy
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Solution Manual for Calculus for Engineers, 4th Edition - Page 14 preview imageEXERCISES1.4A11yyS|3ra)Bi3A10}20}Bi3*Bl22.Weset—y*+1=y*+2y~323.Weset62?—2—=2%+z+1=52—z—3=0.=0=22+2—4=20y+2)(y—1)Equation1.5gives==(1++/T+60)/10==37=1,—2.Pointsofintersectionare=(1+y/B1)/10.Pointsofintersectionare(0,1)and(—3,-2).((1261)/10,(43&3/81)/25).LyyIcallBN‘SellE)E]T*24.TherangeoftheshellisR=(v?/9.81)sin26.Since0<6<7/2,rangeisamaximumwhensin20=1=20=1/2=>=7/4radians.i25.Withcoordinatesasshown,theequationoftheparabolatakestheformy=aa?+10.Sincethepoint(100,50)isontheparabola,50=(10000)+10=>a=1/250.Whenz=70,weobtainy=(70)2/250+10=148/56mfortheoolengthofthesupportingrod.100-10**26.Wesetba=(z—2)"+4—>(z2)"5c+4=0=>2!—82°+2422—372+20=0.Possiblerationalsolutionsare+1,£2,+4,+5,+10,£20.Onesolutionis@=1,sothatat8281240%—87x+20=(2—1)(2®—7a?+172—20)=0.Wefindthat22=4isazeroofthecubic,sothat21—82%+242”—3Tw+20=(2—1)(x—4)(2?—Bx+5).Since#2—8x+5=0hasnorealsolutions,theonlypointsofintersectionare(1,1)and(4,4).27.Withthecoordinatesystemshown,thepdequationoftheparabolatakestheform(0)y=(25/4—2%).Sincethepoint(3/2,4)isontheparabola,4=(25/4—9/4)==>c=1.Thearchistherefore25/4unitshigh.EiERE"28.Iftheparabolaistopassthroughthesepoints,then2=a+b+ec10=9a—3b+c,4=9a+3b+eThesolutionoftheseequationsisa=1/2,b=—1,c=5/2.29.(a)Sinceresistancesattemperatures0°,100°,and700°are10.000,13.946,and24.172,10.000=Ro(1),13.946=Ro(1+100e+10000b),24.172=Ro(1+700a+4900006).ThesolutionoftheseequationsisRg=10.000,+StudyXy
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Solution Manual for Calculus for Engineers, 4th Edition - Page 15 preview image12EXERCISES1.4Ba~0.0042662,andb=—3.2024x1075."(b)Theparabolaisplottedtotheright.(c)Theresistanceis20ohmswhen220=10(1--aT"+bT2)=»BT?+al~1=0.N—fazSolutionsofthisequationareT'=a5Sinceonlythepositivesolutionisacceptable,T=(—a+Va?+46)/(2b)~304.Woe780.IfP(a,b)isthepointatwhichtheropemeetstheparabola,theequationofthelinePQisy—4=[(b—4)/(a—8)](z—3).Thez-coordinatesyofpointsofintersectionofthislinewithA06Htheparabolaaregivenbytheequation21==)(z—3)+4a—3»=,(bdb—4E!*2(27227%)5=Bn»(9)=+3(5)5-0,Sincethelineistomeettheparabolainonlyonepoint,thediscriminantofthisquadraticmustbezerob—4\?b—4y2Jimi228)gl=Z4)?—12(b—4)(a—3)+20(a—3)=0.(=)als(3)3]0==(b—4)?—12(b—4)(a—3)+20(a)SinceP(a,b)isontheparabola,b=a®—1,andwhenwesubstitutethisintotheaboveequation,0=(a2~5)?—12(a®—5)(a—3)+20(a—3)=o*—120°+46a”—60a+25.Possiblerationalsolutionsofthisequationare£1,£5,£25.Wefindthata=1isasolution.Whenitisfactoredfromthequartic,0=a1243+4642--60a+25==(a—1)(a®—11a®+35a—25).Onceagaina=1isazeroofthecubic,sothat0=at—120°+460°—60a+25=(a—1)(a-1)(a®—10a+25)=(a—1)*(a—5)*.Clearly,¢=5isinadmissible,andtherequiredpointis(1,0).EXERCISES1.4B1.Thecircleiscentredattheorigin2.Thecentreofthecircleis(—5,2)withradius5v/2.anditsradiusisvB.JY.yah:xS——%3.Whenwecompletethesquareonthea-terms,4.Whenwecompletethesquareonthey-terms,(z+1)?+y2=16.Thecentreofthecirclea?+(y—2)?=3.Thecentreofthecircleis(—1,0)anditsradiusis4.is(0,2)anditsradiusisv3.+StudyXy
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Solution Manual for Calculus for Engineers, 4th Edition - Page 16 preview imageEXERCISES1.4B13odyJ%p—%5.Whenwecompletesquareson2-andy-terms,6.Whenwecompletethesquareonthez-terms,(z—1)%+(y—1)?=1.Thecentreofthecircle(2+3/2)%+y*=59/4.Thecentreofthecircleis(1,1)anditsradiusis1.is(—3/2,0)anditsradiusisv/59/2.214INNETx=7.Whenwecompletesquaresonz-andy-terms,8.Whenwecompletesquareson@-andy-terms,(x+2/3)%+(y—1/3)=23/9.Thecentreof(z+2)%+(y—1)®=10.Thecentreofthethecircleis(—2/3,1/3)anditsradiuscircleis(—2,1)anditsradiusisvI0.isV23/3.y2d9.Whenwecompletesquareson«-andy-terms,(z—1)*+(y—2)?=0.Theonlypointsatisfyingthisequationis(1,2).10.Whenwecompletesquareson@-andy-terms,(23)+(y+3/2)?=—35/4.Nopointcansatisfythisequation.11.Withcentre(0,0)andradius2,theequationofthecircleis2?fy?=4.12.Sincethecentreis(1,0)anditsradiusis1,theequationofthecircleis(z—1)%+yr=113.Sincethecentreis(3,4)anditsradiusis2,theequationofthecircleis(z—H+(y—4)?=414.Thefigureindicatesthatthecentreofthecircleis(3/2,—3/2).Itsradiusisthen1/(3/2)%+(3/22=3/2.Theequationofthecircleistherefore(z—3/2)?+(y+3/2)=9/2.15.Ifwetaketheequationofthecircleinform1.22,andsubstituteeachofthepoints(3,0),(2,7),and(5,6),9+04+3f+e=0,3f+e=-9,4449127+Tgte=0,=>2+7g+e=53,25436—5f+6g+e=0,—5f+6g+e=—61.Thesolutionoftheseequationsisf=2,g=—6,ande=—15.Theequationofthecircleisx2+y?+226y—15=0.+StudyXxy
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Calculus For Engineers by Donald Tr...

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