Concepts in Calculus, III : Multivariable Calculus

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Concepts in Calculus, III : Multivariable Calculus
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MAC 222
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From the University of Florida Department of Mathematics, this is the third volume in a three volume presentation of calculus from a concepts perspective. The emphasis is on learning the concepts behind the theories, not the rote completion of problems. CLICK THE SHOPPING CART BUTTON TO PURCHASE A LOW COST PRINTED TEXTBOOK ONLINE OR YOU MAY ALSO PHONE THE UNIVERSITY PRESS OF FLORIDA TO ORDER A TEXTBOOK ON THEIR TOLL FREE NUMBER: 800-226-3822.
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MAC2282 - ENGINEERING CALCULUS II, MAC1311 - CALCULUS I, MAC1412 - CALCULUS II, MAC2234 - CALCULUS FOR BUSINESS & SOC. SCIENCE II, MAC2412 - CALCULUS II

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ConceptsinCalculusIIIBetaVersion UNIVERSITYPRESSOFFLORIDAFloridaA&MUniversity,Tallahassee FloridaAtlanticUniversity,BocaRaton FloridaGulfCoastUniversity,Ft.Myers FloridaInternationalUniversity,Miami FloridaStateUniversity,Tallahassee NewCollegeofFlorida,Sarasota UniversityofCentralFlorida,Orlando UniversityofFlorida,Gainesville UniversityofNorthFlorida,Jacksonville UniversityofSouthFlorida,Tampa UniversityofWestFlorida,Pensacola OrangeGroveTexts Plus

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ConceptsinCalculusIIIMultivariableCalculus,BetaVersion SergeiShabanov UniversityofFloridaDepartmentof MathematicsU niversity P ressof F lorida Gainesville € Tallahassee € Tampa € BocaRaton Pensacola € Orlando € Miami € Jacksonville € Ft.Myers € Sarasota

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Copyright2012bytheUniversityofFloridaBoardofTrusteesonbehalfoftheUniversityof FloridaDepartmentofMathematics Thisworkislicensedunderamodi“edCreativeCommonsAttribution-Noncommercial-No DerivativeWorks3.0UnportedLicense.Toviewacopyofthislicense,visithttp:// creativecommons.org/licenses/by-nc-nd/3.0/.Youarefreetoelectronicallycopy,distribute,and transmitthisworkifyouattributeauthorship. However,allprintingrightsarereservedbythe UniversityPressofFlorida(http://www.upf.com).PleasecontactUPFforinformationabout howtoobtaincopiesoftheworkforprintdistribution .Youmustattributetheworkinthe mannerspeci“edbytheauthororlicensor(butnotinanywaythatsuggeststhattheyendorse youoryouruseofthework).Foranyreuseordistribution,youmustmakecleartoothersthe licensetermsofthiswork.Anyoftheaboveconditionscanbewaivedifyougetpermissionfrom theUniversityPressofFlorida.Nothinginthislicenseimpairsorrestrictstheauthorsmoral rights. ISBN978-1-61610-157-2 OrangeGroveTexts Plus isanimprintoftheUniversityPressofFlorida,whichisthescholarly publishingagencyfortheStateUniversitySystemofFlorida,comprisingFloridaA&M University,FloridaAtlanticUniversity,FloridaGulfCoastUniversity,FloridaInternational University,FloridaStateUniversity,NewCollegeofFlorida,UniversityofCentralFlorida, UniversityofFlorida,UniversityofNorthFlorida,UniversityofSouthFlorida,andUniversityof WestFlorida. UniversityPressofFlorida 15Northwest15thStreet Gainesville,FL32611-2079 http://www.upf.com

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Contents Chapter11.VectorsandtheSpaceGeometry171.RectangularCoordinatesinSpace172.VectorsinSpace1173.TheDotProduct2174.TheCrossProduct3175.TheTripleProduct4076.PlanesinSpace4877.LinesinSpace5478.QuadricSurfaces62Chapter12.VectorFunctions7379.CurvesinSpaceandVectorFunctions7980.DierentiationofVectorFunctions8481.IntegrationofVectorFunctions8982.ArcLengthofaCurve9683.CurvatureofaSpaceCurve10184.PracticalApplications109Chapter13.DifferentiationofMultivariableFunctions12385.FunctionsofSeveralVariables12386.LimitsandContinuity12987.AGeneralStrategytoStudyLimits13688.PartialDerivatives14589.Higher-OrderPartialDerivatives14990.ChainRulesandImplicitDierentiation15891.LinearizationofMultivariableFunctions16292.TheDierentialandTaylorPolynomials16793.DirectionalDerivativeandtheGradient17494.MaximumandMinimumValues18095.MaximumandMinimumValues(Continued)18696.LagrangeMultipliers191 Chapterandsectionnumberingcontinuesfromthepreviousvolumeintheseries, ConceptsinCalculusII .

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viCONTENTS Chapter14.MultipleIntegrals20197.DoubleIntegrals20198.PropertiesoftheDoubleIntegral20799.IteratedIntegrals211100.DoubleIntegralsOverGeneralRegions215101.DoubleIntegralsinPolarCoordinates222102.ChangeofVariablesinDoubleIntegrals227103.TripleIntegrals234104.TripleIntegralsinCylindricalandSphericalCoordinates243105.ChangeofVariablesinTripleIntegrals250106.ImproperMultipleIntegrals254107.LineIntegrals261108.SurfaceIntegrals265109.MomentsofInertiaandCenterofMass273Chapter15.VectorCalculus283110.LineIntegralsofaVectorField283111.FundamentalTheoremforLineIntegrals288112.GreensTheorem297113.FluxofaVectorField303114.StokesTheorem310115.Gauss-Ostrogradsky(Divergence)Theorem315

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CHAPTER11 VectorsandtheSpaceGeometry Ourspacemaybeviewedasacollectionofpoints.Everygeometrical“gure,suchasasphere,plane,orline,isaspecialsubsetofpointsin space.Themainpurposeofanalgebraicdescriptionofvariousobjects inspaceistodevelopasystematicrepresentationoftheseobjectsby numbers.Interestinglyenough,ourexperienceshowsthatsofarreal numbersandbasicrulesoftheiralgebraappeartobesucienttodescribeallfundamentallawsofnature,modeleverydayphenomena,and evenpredictmanyofthem.TheevolutionoftheUniverse,forcesbindingparticlesinatomicnuclei,andatomicnucleiandelectronsforming atomsandmolecules,starandplanetformation,chemistry,DNAstructures,andsoon,allcanbeformulatedasrelationsbetweenquantities thataremeasuredandexpressedasrealnumbers.Perhaps,thisis themostintriguingpropertyoftheUniverse,whichmakesmathematicsthemaintoolofourunderstandingoftheUniverse.Thedeeper ourunderstandingofnaturebecomes,themoresophisticatedarethe mathematicalconceptsrequiredtoformulatethelawsofnature.But theyremainbasedonrealnumbers.Inthiscourse,basicmathematical conceptsneededtodescribevariousphenomenainathree-dimensional Euclideanspacearestudied.Theveryfactthatthespaceinwhich weliveisathree-dimensionalEuclideanspaceshouldnotbeviewedas anabsolutetruth.Allonecansayisthatthis mathematicalmodel of thephysicalspaceissucienttodescribearatherlargesetofphysical phenomenaineverydaylife.Asamatteroffact,thismodelfailsto describephenomenaonalargescale(e.g.,ourgalaxy).Itmightalso failattinyscales,butthishasyettobeveri“edbyexperiments.71.RectangularCoordinatesinSpace Theelementaryobjectinspaceisapoint.Sothediscussionshould beginwiththequestion:Howcanonedescribeapointinspacebyreal numbers?Thefollowingprocedurecanbeadopted.Selectaparticular pointinspacecalledthe origin andusuallydenoted O .Setupthree mutuallyperpendicularlinesthroughtheorigin.Arealnumberis associatedwitheverypointoneachlineinthefollowingway.The origincorrespondsto0.Distancestopointsononesideoftheline1

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211.VECTORSANDTHESPACEGEOMETRY fromtheoriginaremarkedbypositiverealnumbers,whiledistances topointsontheotherhalfofthelinearemarkedbynegativenumbers (theabsolutevalueofanegativenumberisthedistance).Thehalf-lines withthegridofpositivenumberswillbeindicatedbyarrowspointing fromtheorigintodistinguishthehalf-lineswiththegridofnegative numbers.Thedescribedsystemoflineswiththegridofrealnumbers onthemiscalleda rectangularcoordinatesystem attheorigin O .The lineswiththeconstructedgridofrealnumbersarecalled coordinate axes .71.1.PointsinSpaceasOrderedTriplesofRealNumbers.Theposition ofanypointinspacecanbe uniquely speci“edasan orderedtripleofreal numbers relativetoagivenrectangularcoordinatesystem.Consider arectanglewhosetwooppositevertices(theendpointsofthelargest diagonal)aretheoriginandapoint P ,whileitssidesthatareadjacent attheoriginlieontheaxesofthecoordinatesystem.Foreverypoint P thereisonlyonesuchrectangle.Therectangleisuniquelydetermined byitsthreesidesadjacentattheorigin.Letthenumber x marksthe positionofonesuchsidethatliesonthe“rstaxis,thenumbers y and z dosoforthesecondandthirdsides,respectively.Notethat,depending onthepositionof P ,thenumbers x y ,and z maybenegative,positive, oreven0.Inotherwords,anypointinspaceisassociatedwitha unique orderedtriple ofrealnumbers( x,y,z )determinedrelativetoa rectangularcoordinatesystem.Thisorderedtripleofnumbersiscalled rectangularcoordinates ofapoint.Tore”ecttheorderin( x,y,z ),the axesofthecoordinatesystemwillbemarkedas x y ,and z axes.Thus, to“ndapointinspacewithrectangularcoordinates(1 2 Š 3),onehas toconstructarectanglewithavertexattheoriginsuchthatitssides adjacentattheoriginoccupytheintervals[0 1],[0 2],and[ Š 3 0]along the x y ,and z axes,respectively.Thepointinquestionisthevertex oppositetotheorigin.71.2.APointasanIntersectionofCoordinatePlanes.Theplanecontainingthe x and y axesiscalledthe xyplane .Forallpointsinthis plane,the z coordinateis0.Theconditionthatapointliesinthe xy planecanthereforebestatedas z =0.The xz and yz planescanbe de“nedsimilarly.Theconditionthatapointliesinthe xz or yz plane reads y =0or x =0,respectively.Theorigin(0 0 0)canbeviewed astheintersectionofthreecoordinateplanes x =0, y =0,and z =0. Considerallpointsinspacewhose z coordinateis“xedtoaparticular value z = z0(e.g., z =1).Theyformaplaneparalleltothe xy plane thatlies | z0| unitsoflengthaboveitif z0> 0orbelowitif z0< 0.

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71.RECTANGULARCOORDINATESINSPACE3 Figure11.1. Left :Anypoint P inspacecanbeviewed astheintersectionofthreecoordinateplanes x = x0, y = y0,and z = z0;hence, P canbegivenanalgebraicdescriptionasanorderedtripleofnumbers P =( x0,y0,z0). Right :Translationofthecoordinatesystem.Theorigin ismovedtoapoint( x0,y0,z0)relativetotheoldcoordinatesystemwhilethecoordinateaxesremainparallel totheaxesoftheoldsystem.Thisisachievedbytranslatingtheorigin“rstalongthe x axisbythedistance x0(asshowninthe“gure),thenalongthe y axisbythe distance y0,and“nallyalongthe z axisbythedistance z0.Asaresult,apoint P thathadcoordinates( x,y,z ) intheoldsystemwillhavethecoordinates x= x Š x0, y= y Š y0,and z= z Š z0inthenewcoordinatesystem. Apoint P withcoordinates( x0,y0,z0)canthereforebeviewedasan intersectionofthree coordinateplanes x = x0, y = y0,and z = z0as showninFigure11.1.Thefacesoftherectangleintroducedtospecify thepositionof P relativetoarectangularcoordinatesystemlieinthe coordinateplanes.Thecoordinateplanesareperpendiculartothecorrespondingcoordinateaxes:theplane x = x0isperpendiculartothe x axis,andsoon.71.3.ChangingtheCoordinateSystem.Sincetheoriginanddirections oftheaxesofacoordinatesystemcanbechosenarbitrarily,thecoordinatesofapointdependonthischoice.Supposeapoint P has

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411.VECTORSANDTHESPACEGEOMETRY coordinates( x,y,z ).Consideranewcoordinatesystemwhoseaxesare paralleltothecorrespondingaxesoftheoldcoordinatesystem,but whoseoriginisshiftedtothepoint Owithcoordinates( x0, 0 0).It isstraightforwardtoseethatthepoint P wouldhavethecoordinates ( x Š x0,y,z )relativetothenewcoordinatesystem(Figure11.1,right panel).Similarly,iftheoriginisshiftedtoapoint Owithcoordinates ( x0,y0,z0),whiletheaxesremainparalleltothecorrespondingaxesof theoldcoordinatesystem,thenthecoordinatesof P aretransformedas (11.1)( x,y,z ) Š ( x Š x0,y Š y0,z Š z0) Onecanchangetheorientationofthecoordinateaxesbyrotating themabouttheorigin.Thecoordinatesofthesamepointinspaceare dierentintheoriginalandrotatedrectangularcoordinatesystems. Algebraicrelationsbetweenoldandnewcoordinates,similarto(11.1), canbeestablished.Asimplecase,whenacoordinatesystemisrotated aboutoneofitsaxes,isdiscussedattheendofthissection. Itisimportanttorealizethatnophysicalorgeometricalquantity shoulddependonthechoiceofacoordinatesystem.Forexample,the lengthofastraightlinesegmentmustbethesameinanycoordinate system,whilethecoordinatesofitsendpointsdependonthechoiceof thecoordinatesystem.Whenstudyingapracticalproblem,acoordinatesystemcanbechoseninanywayconvenienttodescribeobjectsin space.Algebraicrulesforrealnumbers(coordinates)canthenbeused tocomputephysicalandgeometricalcharacteristicsoftheobjects.The numericalvaluesofthesecharacteristicsdonotdependonthechoice ofthecoordinatesystem.71.4.DistanceBetweenTwoPoints.Considertwopointsinspace, P1and P2.Lettheircoordinatesrelativetosomerectangularcoordinate systembe( x1,y1,z1)and( x2,y2,z2),respectively.Howcanonecalculatethedistancebetweenthesepoints,orthelengthofastraightline segmentwithendpoints P1and P2?Thepoint P1istheintersection pointofthreecoordinateplanes x = x1, y = y1,and z = z1.The point P2istheintersectionpointofthreecoordinateplanes x = x2, y = y2,and z = z2.Thesesixplanescontainfacesoftherectangle whoselargestdiagonalisthestraightlinesegmentbetweenthepoints P1and P2.Thequestionthereforeishowto“ndthelengthofthis diagonal. Considerthreesidesofthisrectanglethatareadjacent,say,atthe vertex P1.Thesideparalleltothe x axisliesbetweenthecoordinate planes x = x1and x = x2andisperpendiculartothem.Sothe lengthofthissideis | x2Š x1| .Theabsolutevalueisnecessaryasthe

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71.RECTANGULARCOORDINATESINSPACE5 dierence x2Š x1maybenegative,dependingonthevaluesof x1and x2,whereasthedistancemustbenonnegative.Similarargumentslead totheconclusionthatthelengthsoftheothertwoadjacentsidesare | y2Š y1| and | z2Š z1| .Ifarectanglehasadjacentsidesoflength a b and c ,thenthelength d ofitslargestdiagonalsatis“estheequation d2= a2+ b2+ c2. ItsproofisbasedonthePythagoreantheorem(seeFigure11.2).Considertherectanglefacethatcontainsthesides a and b .Thelength f ofitsdiagonalisdeterminedbythePythagoreantheorem f2= a2+ b2. Considerthecrosssectionoftherectanglebytheplanethatcontains thefacediagonal f andtheside c .Thiscrosssectionisarectangle withtwoadjacentsides c and f andthediagonal d .Theyarerelated as d2= f2+ c2bythePythagoreantheorem,andthedesiredconclusion follows. Figure11.2. Distancebetweentwopointswithcoordinates P1=( x1,y1,z1)and P2=( x2,y2,z2).Theline segment P1P2isviewedasthelargestdiagonalofthe rectanglewhosefacesarethecoordinateplanescorrespondingtothecoordinatesofthepoints.Therefore,the distancesbetweentheoppositefacesare a = | x1Š x2| b = | y1Š y2| ,and c = | z1Š z2| .Thelengthofthediagonal d isobtainedbythedoubleuseofthePythagorean theoremineachoftheindicatedrectangles: d2= c2+ f2(topright)and f2= a2+ b2(bottomright).

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611.VECTORSANDTHESPACEGEOMETRY Put a = | x2Š x1| b = | y2Š y1| ,and c = | z2Š z1| .Then d = | P1P2| is thedistancebetween P1and P2.Thedistanceformulaisimmediately found: (11.2) | P1P2| = ( x2Š x1)2+( y2Š y1)2+( z2Š z1)2. Notethatthenumbers(coordinates)( x1,y1,z1)and( x2,y2,z2)depend onthechoiceofthecoordinatesystem,whereasthenumber | P1P2| remainsthesame inanycoordinatesystem!Forexample,iftheoriginof thecoordinatesystemistranslatedtoapoint( x0,y0,z0)whiletheorientationofthecoordinateaxesremainsunchanged,then,accordingto rule(11.1),thecoordinatesof P1and P2relativetothenewcoordinate become( x1Š x0,y1Š y0,z1Š z0)and( x2Š x0,y2Š y0,z2Š z0),respectively.Thenumericalvalueofthedistancedoesnotchangebecause thecoordinatedierences,( x2Š x0) Š ( x1Š x0)= x2Š x1(similarly forthe y and z coordinates ),donotchange. RotationsinSpace. Theorigincanalwaysbetranslatedto P1so thatinthenewcoordinatesystem P1is(0 0 0)and P2is( x2Š x1,y2Š y1,z2Š z1).Sincethedistanceshouldnotdependontheorientationof thecoordinateaxes,anyrotationcannowbedescribedalgebraically as alineartransformationofanorderedtriple ( x,y,z ) underwhich thecombination x2+ y2+ z2remainsinvariant .Alineartransformationmeansthatthenewcoordinatesarelinearcombinationsofthe oldones.Itshouldbenotedthatre”ectionsofthecoordinateaxes, x Š x (similarlyfor y and z ),arelinearandalsopreservethedistance.However,acoordinatesystemobtainedbyanoddnumberof re”ectionsofthecoordinateaxescannotbeobtainedbyanyrotation oftheoriginalcoordinatesystem.So,intheabovealgebraicde“nition ofarotation,there”ectionsshouldbeexcluded.71.5.SpheresinSpace.Inthiscourse,relationsbetweentwoequivalent descriptionsofobjectsinspace„thegeometricalandthealgebraic„ willalwaysbeemphasized.Oneofthecourseobjectivesistolearn howtointerpretanalgebraicequationbygeometricalmeansandhow todescribegeometricalobjectsinspacealgebraically.Thesimplest exampleofthiskindisasphere. GeometricalDescriptionofaSphere .Asphereisasetof pointsinspacethatareequidistantfroma“xedpoint.The“xedpoint iscalledthe spherecenter .Thedistancefromthespherecentertoany pointofthesphereiscalledthe sphereradius AlgebraicDescriptionofaSphere .Analgebraicdescriptionof asphereimplies“ndinganalgebraicconditiononcoordinates( x,y,z ) ofpointsinspacethatbelongtothesphere.Soletthecenterofthe

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71.RECTANGULARCOORDINATESINSPACE7 Figure11.3. Left :Asphereisde“nedasapointset inspace.Eachpoint P ofthesethasa“xeddistance R froma“xedpoint P0.Thepoint P0isthecenterofthe sphere,and R istheradiusofthesphere. Right :Transformationofcoordinatesunderarotation ofthecoordinatesysteminaplane. spherebeapoint P0withcoordinates( x0,y0,z0)(de“nedrelativeto somerectangularcoordinatesystem).Ifapoint P withcoordinates ( x,y,z )belongstothesphere,thenthenumbers( x,y,z )mustbesuch thatthedistance | PP0| isthesameforanysuch P andequaltothe sphereradius,denoted R ,thatis, | PP0| = R or | PP0|2= R2(see Figure11.3,leftpanel).Usingthedistanceformula,thisconditioncan bewrittenas (11.3)( x Š x0)2+( y Š y0)2+( z Š z0)2= R2. Forexample,thesetofpointswithcoordinates( x,y,z )thatsatisfythe condition x2+ y2+ z2=4isasphereofradius R =2centeredatthe origin x0= y0= z0=0.71.6.AlgebraicDescriptionofPointSetsinSpace.Theideaofanalgebraicdescriptionofaspherecanbeextendedtoothersetsinspace.It isconvenienttointroducesomebriefnotationforanalgebraicdescriptionofsets.Forexample,foraset S ofpointsinspacewithcoordinates ( x,y,z )suchthattheysatisfythealgebraiccondition(11.3),onewrites S = ( x,y,z ) ( x Š x0)2+( y Š y0)2+( z Š z0)2= R2 Thisrelationmeansthattheset S isacollectionofallpoints( x,y,z ) suchthat(theverticalbar)theirrectangularcoordinatessatisfy(11.3).

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811.VECTORSANDTHESPACEGEOMETRY Similarly,the xy planecanbeviewedasasetofpointswhose z coordinatesvanish: P = ( x,y,z ) z =0 Thesolidregioninspacethatconsistsofpointswhosecoordinatesare nonnegativeiscalledthe “rstoctant : O1= ( x,y,z ) x 0 ,y 0 ,z 0 Thespatialregion B = ( x,y,z ) x> 0 ,y> 0 ,z> 0 ,x2+ y2+ z2< 4 isthecollectionofallpointsintheportionofaballofradius2that liesinthe“rstoctant.Thestrictinequalitiesimplythattheboundary ofthisportionoftheballdoesnotbelongtotheset B .71.7.StudyProblems.Problem11.1. Showthatthecoordinatesofthemidpointofastraight linesegmentare x1+ x2 2 y1+ y2 2 z1+ z2 2 ifthecoordinatesofitsendpointsare ( x1,y1,z1) and ( x2,y2,z2) Solution: Let P1and P2betheendpointsandlet M bethemidpoint. Onehastoverifythecondition | P2M | = | MP1| or | P2M |2= | MP1|2bymeansofthedistanceformula.The x -coordinatedierencesfor thesegments P2M and MP1read x2Š ( x1+ x2) / 2=( x2Š x1) / 2and ( x1+ x2) / 2 Š x1=( x2Š x1) / 2,respectively;thatis,theycoincide. Similarly,thedierencesofthecorresponding y and z coordinatesare thesame.Bythedistanceformula,itisthenconcludedthat | P2M |2= | MP1|2. Problem11.2. Let ( x,y,z ) becoordinatesofapoint P .Consider anewcoordinatesystemthatisobtainedbyrotatingthe x and y axes aboutthe z axiscounterclockwiseasviewedfromthetopofthe z axis throughanangle .Let ( x,y,z) becoordinatesof P inthenewcoordinatesystem.Findtherelationsbetweentheoldandnewcoordinates. Solution: Theheightof P relativetothe xy planedoesnotchange uponrotation.So z= z .Itisthereforesucienttoconsiderrotationsinthe xy plane,thatis,forpoints P withcoordinates( x,y, 0). Let r = | OP | (thedistancebetweentheoriginand P )andlet bethe

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71.RECTANGULARCOORDINATESINSPACE9 anglecountedfromthepositive x axistowardtheray OP counterclockwise(seeFigure11.3,rightpanel).Then x = r cos and y = r sin (thepolarcoordinatesof P ).Inthenewcoordinatesystem,theanglebetweenthepositive xaxisandtheray OP becomes = Š Therefore, x= r cos =cos( Š )= r cos cos + r sin sin = x cos + y sin y= r sin = r sin( Š )= r sin cos Š r cos sin = y cos Š x sin Problem11.3. Giveageometricaldescriptionoftheset S = ( x,y,z ) x2+ y2+ z2Š 4 z =0 Solution: Theconditiononthecoordinatesofpointsthatbelong tothesetcontainsthesumofsquaresofthecoordinatesjustlikethe equationofasphere.Thedierenceisthat(11.3)containsthesum ofperfectsquares.Sothesquaresmustbecompletedintheabove equationandtheresultingexpressioncomparedwith(11.3).Onehas z2Š 4 z =( z Š 2)2Š 4sothattheconditionbecomes x2+ y2+( z Š 2)2=4. Itdescribesasphereofradius R =2thatiscenteredatthepoint ( x0,y0,z0)=(0 0 2);thatis,thecenterofthesphereisonthe z axis atadistanceof2unitsabovethe xy plane. Problem11.4. Giveageometricaldescriptionoftheset C = ( x,y,z ) x2+ y2Š 2 x Š 4 y =4 Solution: Asinthepreviousproblem,theconditioncanbewritten asthesumofperfectsquares( x Š 1)2+( y Š 2)2=9bymeansthe ofrelations x2Š 2 x =( x Š 1)2Š 1and y2Š 4 y =( y Š 2)2Š 4.In the xy plane,thisisnothingbuttheequationofacircleofradius3 whosecenteristhepoint(1 2 0).Inanyplane z = z0paralleltothe xy plane,the x and y coordinatessatisfythesameequation,andhence thecorrespondingpointsalsoformacircleofradius3withthecenter (1 2 ,z0).Thus,thesetisacylinderofradius3whoseaxisisparallel tothe z axisandpassesthroughthepoint(1 2 0). Problem11.5. Giveageometricaldescriptionoftheset P = ( x,y,z ) z ( y Š x )=0 .

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1011.VECTORSANDTHESPACEGEOMETRY Solution: Theconditionissatis“edifeither z =0or y = x .The formerequationdescribesthe xy plane,whilethelatterrepresentsa lineinthe xy plane.Sinceitdoesnotimposeanyrestrictiononthe z coordinate,eachpointofthelinecanbemovedupanddownparallel tothe z axis.Theresultingsetisaplanethatcontainstheline y = x inthe xy planeandthe z axis.Thus,theset P istheunionofthis planeandthe xy plane. 71.8.Exercises.(1) Findthedistancebetweenthefollowingspeci“ed points: (i) A (1 2 3)and B ( Š 1 0 2) (ii) A ( Š 1 3 Š 2)and B ( Š 1 2 Š 1) (2) Lettheset S consistofpoints( t, 2 t, 3 t )where Š 0 (iv) x2+ y2Š 4 y< 0, z> 0 (v)4 x2+ y2+ z2 9 (vi) x2+ y2 1, x2+ y2+ z2 4 (vii) x2+ y2+ z2Š 2 z< 0, z> 1 (viii) x2+ y2+ z2Š 2 z =0, z =1 (ix)( x Š a )( y Š b )( z Š c )=0 (4) Sketcheachofthefollowingsetsandgivetheiralgebraicdescription: (i)Aspherewhosediameteristhestraightlinesegment AB where A =(1 2 3)and B =(3 2 1). (ii)Aspherecenteredat(1 2 3)thatliesinthe“rstoctantand touchesoneofthecoordinateplanes. (iii)Thelargestsolidcubethatiscontainedinaballofradius R centeredattheorigin.Solvethesameproblemiftheballis notcenteredattheorigin. (iv)Thesolidregionthatisaballofradius R thathasacylindrical holeofradius R/ 2whoseaxisisatadistanceof R/ 2fromthe centeroftheball.Chooseaconvenientcoordinatesystem.

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72.VECTORSINSPACE11 (v)Theportionofaballofradius R thatliesbetweentwoparallel planeseachofwhichisatsdistanceof a
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1211.VECTORSANDTHESPACEGEOMETRY Figure11.4. Left :Orientedsegmentsobtainedfrom oneanotherbyparalleltransport.Theyallrepresentthe samevector. Right :Avectorasanorderedtripleofnumbers.Anorientedsegmentistransportedparallelsothatitsinitial pointcoincideswiththeoriginofarectangularcoordinatesystem.Thecoordinatesoftheterminalpointof thetransportedsegment,( a1,a2,a3),arecomponentsof thecorrespondingvector.Soavectorcanalwaysbewrittenasanorderedtripleofnumbers: a = a1,a2,a3 .By construction,thecomponentsofavectordependonthe choiceofthecoordinatesystem(theorientationofthe coordinateaxesinspace).72.2.VectorasanOrderedTripleofNumbers.Hereanalgebraicrepresentationofvectorsinspacewillbeintroduced.Consideranoriented segment AB thatrepresentsavector a (i.e., a = AB ).Anorientedsegment ABrepresentsthesamevectorifitisobtainedbytransporting AB paralleltoitself.Inparticular,letustake A= O ,where O isthe originofsomerectangularcoordinatesystem.Then a = AB = OB. Thedirectionandlengthoftheorientedsegment OBisuniquelydeterminedbythecoordinatesofthepoint B.Thus,wehavethefollowing algebraicde“nitionofavector. Definition 11.1 (Vectors) Avectorinspaceisanorderedtripleofrealnumbers: a = a1,a2,a3 Thenumbers a1, a2,and a3arecalled components ofthevector a Notethatthenumericalvaluesofthecomponentsdependonthe choiceofcoordinatesystem.Fromageometricalpointofview,the orderedtriple( a1,a2,a3)isthecoordinatesofthepoint B,thatis,the

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72.VECTORSINSPACE13 endpointoftheorientedsegmentthatrepresents a iftheinitialpoint coincideswiththeorigin. Definition 11.2 (EqualityofTwoVectors) Twovectors a and b areequalorcoincideiftheircorrespondingcomponentsareequal: a = b a1= b1,a2= b2,a3= b3. Thisde“nitionagreeswiththegeometricalde“nitionofavector asaclassofallorientedsegmentsthatareparallelandhavethesame length.Indeed,iftwoorientedsegmentsrepresentthesamevector, then,afterparalleltransportsuchthattheirinitialpointscoincide withtheorigin,their“nalpointscoincidetooandhencehavethesame coordinates. Example 11.1 Findthecomponentsofavector P1P2ifthecoordinatesof P1and P2are ( x1,y1,z1) and ( x2,y2,z2) ,respectively. Solution: Considerarectanglewhoselargestdiagonalcoincideswith thesegment P1P2andwhosesidesareparalleltothecoordinateaxes. Afterparalleltransportofthesegmentsothat P1movestotheorigin, thecoordinatesoftheotherendpointarethecomponentsof P1P2. Alternatively,theoriginofthecoordinatesystemcanbemovedtothe point P1,keepingthedirectionsofthecoordinateaxes.Therefore, P1P2= x2Š x1,y2Š y1,z2Š z1 accordingtothecoordinatetransformationlaw(11.1),where P0= P1. Thus,inorderto“ndthecomponentsofthevector P1P2fromthe coordinatesofitspoints,onehastosubtractthecoordinatesofthe initialpoint P1fromthecorrespondingcomponentsofthe“nalpoint P2. Definition 11.3 (NormofaVector). Thenumber a = a2 1+ a2 2+ a2 3iscalledthe norm ofavector a ByExample11.1andthedistanceformula(11.2),thenormofa vectoristhelengthofanyorientedsegmentrepresentingthevector. Thenormofavectorisalsocalledthe magnitude or length ofavector. Definition 11.4 (ZeroVector) Avectorwithvanishingcomponents, 0 = 0 0 0 ,iscalleda zero vector .

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1411.VECTORSANDTHESPACEGEOMETRY Avector a isazerovectorifandonlyifitsnormvanishes, a =0. Indeed,if a = 0 ,then a1= a2= a3=0andhence a =0.Forthe converse,itfollowsfromthecondition a =0that a2 1+ a2 2+ a2 3=0, whichisonlypossibleif a1= a2= a3=0,or a = 0 .Recallthatanif andonlyifŽstatementactuallyimpliestwostatements.First,if a = 0 then a =0(thedirectstatement).Second,if a =0,then a = 0 (theconversestatement).72.3.VectorAlgebra.Continuingtheanalogybetweenthevectorsand velocitiesofamovingobject,considertwoobjectsmovingparallelbut withdierentrates(speeds).Theirvelocitiesasvectorsareparallel, buttheyhavedierentmagnitudes.Whatistherelationbetweenthe componentsofsuchvectors?Takeavector a = a1,a2,a3 .Itcanbe viewedasthelargestdiagonalofarectanglewithonevertexatthe originandtheoppositevertexatcoordinates( a1,a2,a3).Theadjacent sidesoftherectanglehavelengthsgivenbythecorrespondingcomponentsof a (modulothesignsiftheyhappentobenegative).The directionofthediagonaldoesnotchangeifthesidesoftherectangle arescaledbythesamefactor,whilethelengthofthediagonalisscaled Figure11.5. Left :Multiplicationofavector a bya number s .If s> 0,theresultofthemultiplicationisa vectorparallelto a whoselengthisscaledbythefactor s .If s< 0,then s a isavectorwhosedirectionisthe oppositetothatof a andwhoselengthisscaledby | s | Middle :Constructionofaunitvectorparallelto a .The unitvector a isavectorparallelto a whoselengthis1. Therefore,itisobtainedfrom a bydividingthelatterby itslength a ,i.e., a = s a ,where s =1 / a Right :Aunitvectorinaplanecanalwaysbeviewedas anorientedsegmentwhoseinitialpointisattheorigin ofacoordinatesystemandwhoseterminalpointlieson thecircleofunitradiuscenteredattheorigin.If isthe polarangleintheplane,then a = cos sin 0 .

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72.VECTORSINSPACE15 bythisfactor.Thisgeometricalobservationleadstothefollowingalgebraicrule. Definition 11.5 (MultiplicationofaVectorbyaNumber) Avector a multipliedbyanumber s isavectorwhosecomponentsare multipliedby s : s a = sa1,sa2,sa3 If s> 0,thenthevector s a hasthesamedirectionas a .If s< 0, thenthevector s a hasthedirectionoppositeto a .Forexample,the vector Š a hasthesamemagnitudeas a butpointsinthedirection oppositeto a .Themagnitudeof s a reads: s a = ( sa1)2+( sa2)2+( sa3)2= s2 a2 1+ a2 2+ a2 3= | s | a ; thatis,whenavectorismultipliedbyanumber,itsmagnitudechanges bythefactor | s | .Thegeometricalanalysisofthemultiplicationofa vectorbyanumberleadstothefollowingsimplealgebraiccriterionfor twovectorsbeingparallel. Theorem 11.1 Twononzerovectorsareparalleliftheyareproportional: a b a = s b forsomereal s Ifallthecomponentsofthevectorsinquestiondonotvanish,then thiscriterionmayalsobewrittenas a b s = a1 b1= a2 b2= a3 b3, whichiseasytoverify.If,say, b1=0,then b isparallelto a when a1= b1=0and a2/b2= a3/b3. Definition 11.6 (UnitVector) Avector a iscalleda unitvector ifitsnormequals1, a =1 Anynonzerovector a canbeturnedintoaunitvector a thatis parallelto a .Thenorm(length)ofthevector s a reads s a = | s | a = s a if s> 0.So,bychoosing s =1 / a ,theunitvectorparallelto a isobtained: a = 1 a a = a1 a a2 a a3 a Forexample,owingtothetrigonometricidentity,cos2 +sin2 =1, anyunitvectorinthe xy planecanalwaysbewrittenintheform a = cos sin 0 ,where istheanglecountedfromthepositive x axis towardthevector a counterclockwise.Notethat,inmanypractical

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1611.VECTORSANDTHESPACEGEOMETRY applications,thecomponentsofavectoroftenhavedimensions.For instance,thecomponentsofadisplacementvectoraremeasuredin unitsoflength(meters,inches,etc.),thecomponentsofavelocity vectoraremeasuredin,forexample,meterspersecond,andsoon. Themagnitudeofavector a hasthesamedimensionasitscomponents. Therefore,thecorrespondingunitvector a isdimensionless.Itspeci“es onlythedirectionofavector a .72.3.1.TheParallelogramRule.Supposeapersoniswalkingonthe deckofashipwithspeed v m / s.In1second,thepersongoesadistance v frompoint A to B ofthedeck.Thevelocityvectorrelativetothe deckis v = AB and v = | AB | = v (thespeed).Theshipmoves relativetothewatersothatin1seconditcomestoapoint D froma point C onthesurfaceofthewater.Theshipsvelocityvectorrelative tothewateristhen u = CD withmagnitude u = u = | CD | .What isthevelocityvectorofthepersonrelativetothewater?Suppose thepoint A onthedeckcoincideswiththepoint C onthesurface ofthewater.Thenthevelocityvectoristhedisplacementvectorof thepersonrelativetothewaterin1second.Asthepersonwalkson thedeckalongthesegment AB ,thissegmenttravelsthedistance u paralleltoitselfalongthevector u relativetothewater.In1second, thepoint B ofthedeckismovedtoapoint Bonthesurfaceofthe watersothatthedisplacementvectorofthepersonrelativetothe waterwillbe AB.Apparently,thedisplacementvector BBcoincides withtheshipsvelocity u because B travelsthedistance u parallelto u .Thissuggestsasimplegeometricalrulefor“nding ABasshownin Figure11.6.Takethevector AB = v ,placethevector u sothatits initialpointcoincideswith B ,andmaketheorientedsegmentwiththe initialpointof v andthe“nalpointof u inthisdiagram.Theresulting vectoristhedisplacementvectorofthepersonrelativetothesurface ofthewaterin1secondandhencede“nesthevelocityoftheperson relativetothewater.Thisgeometricalprocedureiscalled additionof vectors Consideraparallelogramwhoseadjacentsides,thevectors a and b ,extendfromthevertexoftheparallelogram.Thesumofthevectors a and b isavector,denoted a + b ,thatisthediagonalofthe parallelogramextendedfromthesamevertex.Notethattheparallel sidesoftheparallelogramrepresentthesamevector(theyareparallel andhavethesamelength).Thisgeometricalruleforaddingvectors iscalledthe parallelogramrule .Itfollowsfromtheparallelogramrule thattheadditionofvectorsis commutative : a + b = b + a ;

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72.VECTORSINSPACE17 Figure11.6. Left :Parallelogramruleforaddingtwo vectors.Iftwovectorsformadjacentsidesofaparallelogramatavertex A ,thenthesumofthevectorsisa vectorthatcoincideswiththediagonaloftheparallelogramandoriginatesatthevertex A Right :Addingseveralvectorsbyusingtheparallelogramrule.Giventhe“rstvectorinthesum,allother vectorsaretransportedparallelsothattheinitialpoint ofthenextvectorinthesumcoincideswiththeterminalpointofthepreviousone.Thesumisthevector thatoriginatesfromtheinitialpointofthe“rstvectorandterminatesattheterminalpointofthelast vector.Itdoesnotdependontheorderofvectorsin thesum. thatis,theorderinwhichthevectorsareaddeddoesnotmatter.To addseveralvectors(e.g., a + b + c ),onecan“rst“nd a + b bythe parallelogramruleandthenadd c tothevector a + b .Alternatively, thevectors b and c canbeadded“rst,andthenthevector a canbe addedto b + c .Accordingtotheparallelogramrule,theresulting vectoristhesame: ( a + b )+ c = a +( b + c ) Thismeansthattheadditionofvectorsis associative .Soseveralvectorscanbeaddedinanyorder.Takethe“rstvector,thenmovethe secondvectorparalleltoitselfsothatitsinitialpointcoincideswith the“nalpointofthe“rstvector.Thethirdvectorismovedparallelso thatitsinitialpointcoincideswiththe“nalpointofthesecondvector, andsoon.Finally,makeavectorwhoseinitialpointcoincideswith theinitialpointofthe“rstvectorandwhose“nalpointcoincideswith the“nalpointofthelastvectorinthesum.Tovisualizethisprocess, imagineamanwalkingalongthe“rstvector,thengoingparallelto thesecondvector,thenparalleltothethirdvector,andsoon.The endpointofhiswalkisindependentoftheorderinwhichhechooses thevectors.

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1811.VECTORSANDTHESPACEGEOMETRY 72.3.2.AlgebraicAdditionofVectors.Definition 11.7 Thesumoftwovectors a = a1,a2,a3 and b = b1,b2,b3 isavectorwhosecomponentsarethesumsofthecorrespondingcomponentsof a and b : a + b = a1+ b1,a2+ b2,a3+ b3 Thisde“nitionisequivalenttothegeometricalde“nitionofadding vectors,thatis,theparallelogramrulethathasbeenmotivatedby studyingthevelocityofacombinedmotion.Indeed,put a = OA wheretheendpoint A hasthecoordinates( a1,a2,a3).Avector b representsallparallelsegmentsofthesamelength b .Inparticular, b is onesuchorientedsegmentwhoseinitialpointcoincideswith A .Supposethat a + b = OC = c1,c2,c3 ,where C hascoordinates( c1,c2,c3). Bytheparallelogramrule, b = AC = c1Š a1,c2Š a2,c3Š a3 ,where therelationbetweenthecomponentsofavectorandthecoordinates ofitsendpointshasbeenused.Theequalityoftwovectorsmeans theequalityofthecorrespondingcomponents,thatis, b1= c1Š a1, b2= c2Š a2,and b3= c3Š a3,or c1= a1+ b1, c2= a2+ b2,and c3= a3+ b3asrequiredbythealgebraicadditionofvectors.72.3.3.RulesofVectorAlgebra.Combiningadditionofvectorswith multiplicationbyrealnumbers,thefollowingsimplerulecanbeestablishedbyeithergeometricaloralgebraicmeans: s ( a + b )= s a + s b ( s + t ) a = s a + t a Thedierenceoftwovectorscanbede“nedas a Š b = a +( Š 1) b Intheparallelogramwithadjacentsides a and b ,thesumofvectors a and( Š 1) b representsthevectorthatoriginatesfromtheendpoint of b andendsattheendpointof a because b +[ a +( Š 1) b ]= a in accordancewiththegeometricalruleforaddingvectors;thatis a b aretwodiagonalsoftheparallelogram.Theprocedureisillustratedin Figure11.7(leftpanel).72.4.StudyProblems.Problem11.6. Considertwononparallelvectors a and b inaplane. Showthatanyvector c inthisplanecanbewrittenasalinearcombination c = t a + s b forsomereal t and s Solution: Byparalleltransport,thevectors a b ,and c canbemoved sothattheirinitialpointscoincide.Thevectors t a and s b areparallel to a and b ,respectively,forallvaluesof s and t .Considerthelines Laand Lbthatcontainthevectors a and b ,respectively.Construct

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72.VECTORSINSPACE19 Figure11.7. Left :Subtractionoftwovectors.The dierence a Š b isviewedasthesumof a and Š b ,the vectorthathasthedirectionoppositeto b andthesame lengthas b .Theparallelogramruleforadding a and Š b showsthatthedierence a Š b = a +( Š b )isthe vectorthatoriginatesfromtheterminalpointof b and endsattheterminalof a if a and b areadjacentsidesof aparallelogram;thatis,thesum a + b andthedierence a Š b arethetwodiagonalsoftheparallelogram. Right :IllustrationtoStudyProblem11.6.Anyvector inaplanecanalwaysberepresentedasalinearcombinationoftwononparallelvectors. twolinesthroughtheendpointof c ;oneisparallelto Laandthe otherto LbasshowninFigure11.7(rightpanel).Theintersection pointsoftheselineswith Laand Lbandtheinitialand“nalpointsof c formtheverticesoftheparallelogramwhosediagonalis c andwhose adjacentsidesareparallelto a and b .Therefore, a and b canalwaysbe scaledsothat t a and s b becometheadjacentsidesoftheconstructed parallelogram.Foragiven c ,thereals t and s areuniquelyde“ned bytheproposedgeometricalconstruction.Bytheparallelogramrule, c = t a + s b Problem11.7. Findthecoordinatesofapoint B thatisatadistance of 6 unitsoflengthfromthepoint A (1 Š 1 2) inthedirectionofthe vector v = 2 1 Š 2 Solution: Thepositionvectorofthepoint A is a = OA = 1 Š 1 2 Thepositionvectorofthepoint B is b = a + s v ,where s isapositive numbertobechosensuchthatthelength | AB | = s v equals6.Since v =3,one“nds s =2.Therefore, b = 1 Š 1 2 +2 2 1 Š 2 = 5 1 Š 2 Problem11.8. Considerastraightlinesegmentwiththeendpoints A (1 2 3) and B ( Š 2 Š 1 0) .Findthecoordinatesofthepoint C onthe segmentsuchthatitistwiceasfarfrom A asitisfrom B .

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2011.VECTORSANDTHESPACEGEOMETRY Solution: Let a = 1 2 3 b = Š 1 0 1 ,and c bepositionvectors of A B ,and C ,respectively.Thequestionistoexpress c via a and b Onehas c = a + AC .Thevector AC isparallelto AB = Š 3 Š 3 Š 3 andhence AC = s AB .Since | AC | =2 | CB | | AC | =2 3| AB | and therefore s =2 3.Thus, c = a +2 3 AB = a +2 3( b Š a )= Š 1 0 1 Problem11.9. InStudyProblem11.6,let a =1 b =2 ,and theanglebetween a and b be 2 / 3 .Findthecoecients s and t ifthe vector c hasanormof 6 andbisectstheanglebetween a and b Solution: ItfollowsfromthesolutionofStudyProblem11.6that thenumbers s and t donotdependonthecoordinatesystemrelativetowhichthecomponentsofallthevectorsarede“ned.So choosethecoordinatesystemsothat a isparalleltothe x axisand b liesinthe xy plane.Withthischoice, a = 1 0 0 and b = b cos(2 / 3) b sin(2 / 3) 0 = Š 1 3 0 .Similarly, c isthevectoroflength c =6thatmakestheangle / 3withthe x axis,and therefore c = 3 3 3 0 .Equatingthecorrespondingcomponentsin therelation c = t a + s b ,one“nds3= t Š s and3 3= s 3,or s =3 and t =6.Hence, c =6 a +3 b Problem11.10. Supposethethreecoordinateplanesareallmirrored. Alightraystrikesthemirrors.Determinethedirectioninwhichthe re”ectedraywillgo. Solution: Let u beavectorparalleltotheincidentray.Under are”ectionfromaplanemirror,thecomponentof u perpendicularto theplanechangesitssign.Therefore,afterthreeconsecutivere”ections fromeachcoordinateplane,allthreecomponentsof u changetheir signs,andthere”ectedraywillgoparalleltotheincidentraybutin theexactoppositedirection.Forexample,supposetherayisre”ected “rstbythe xz plane,thenbythe yz plane,and“nallybythe xy plane. Inthiscase, u = u1,u2,u3 u1, Š u2,u3Š u1, Š u2,u3 Š u1, Š u2, Š u3 = Š u Remark. Thisprincipleisusedtodesignre”ectorslikethecatseyesonbicyclesandthosethatmarktheborderlinesofaroad.No matterfromwhichdirectionsuchare”ectorisilluminated(e.g.,bythe headlightsofacar),itre”ectsthelightintheoppositedirection(so thatitwillalwaysbeseenbythedriver).72.5.Exercises.(1) Findthecomponentsofeachofthefollowing vectorsandtheirnorms: (i)Thevectorhasendpoints A (1 2 3)and B ( Š 1 5 1)andisdirectedfrom A to B .

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73.THEDOTPRODUCT21 (ii)Thevectorhasendpoints A (1 2 3)and B ( Š 1 5 1)andisdirectedfrom B to A (iii)Thevectorhastheinitialpoint A (1 2 3)andthe“nalpoint C thatisthemidpointofthelinesegment AB ,where B = ( Š 1 5 1). (iv)Thepositionvectorisofapoint P obtainedfromthepoint A ( Š 1 2 Š 1)bytransportingthelatteralongthevector u = 2 2 1 3unitsoflengthandthenalongthevector w = Š 3 0 Š 4 10unitsoflength. (v)Thepositionvectorofthevertex C ofatriangle ABC inthe xy planeif A isattheorigin, B =( a, 0 0),theangleatthe vertex B is / 3,and | BC | =3 a (2) ConsideratriangleABC .Let a beavectorfromthevertex A tothemidpointoftheside BC ,let b beavectorfrom B tothe midpointof AC ,andlet c beavectorfrom C tothemidpointof AB Usevectoralgebrato“nd a + b + c (3) Let uk, k =1 2 ,...,n ,beunitvectorsintheplanesuchthat thesmallestanglebetweenthetwovectors ukand uk +1is2 /n .What canbesaidaboutthesum u1+ u2+ + un?Whathappenswhen n ? (4) Aplane”iesataspeedof v mi/hrelativetotheair.Thereis awindblowingataspeedof u mi/hinthedirectionthatmakesthe angle withthedirectioninwhichtheplanemoves.Whatisthespeed oftheplanerelativetotheground? (5) Letpointlikemassiveobjectsbepositionedat Pi, i =1 2 ,...,n andlet mibethemassat Pi.Thepoint P0iscalledthe centerof mass if m1r1+ m2r2+ + mnrn= 0 where riisthevectorfrom P0to Pi.Expressthepositionvectorofthe centerofmassviathepositionvectorsofthepointmasses.Inparticular,“ndthecenterofmassofthreepointmasses, m1= m2= m3= m locatedattheverticesofatriangle ABC for A (1 2 3), B ( Š 1 0 1),and C (1 1 Š 1).73.TheDotProduct Definition 11.8 (DotProduct) The dotproduct a b oftwovectors a = a1,a2,a3 and b = b1,b2,b3 isanumber: a b = a1b1+ a2b2+ a3b3.

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2211.VECTORSANDTHESPACEGEOMETRY Itfollowsfromthisde“nitionthatthedotproducthasthefollowing properties: a b = b a ( s a ) b = s ( a b ) a ( b + c )= a b + a c whichholdforanyvectors a b ,and c andanumber s .The“rst propertystatesthattheorderinwhichtwovectorsaremultipliedinthe dotproductdoesnotmatter;thatis,thedotproductis commutative Thesecondpropertymeansthattheresultofthedotproductdoesnot dependonwhetherthevector a isscaled“rstandthenmultipliedby b orthedotproduct a b iscomputed“rstandtheresultmultiplied by s .Thethirdrelationshowsthatthedotproductis distributive .73.1.GeometricalSigni“canceoftheDotProduct.Asitstands,thedot productisanalgebraicruleforcalculatinganumberoutofsixgiven numbersthatarecomponentsofthetwovectorsinvolved.Thecomponentsofavectordependonthechoiceofthecoordinatesystem. Naturally,oneshouldaskwhetherthenumericalvalueofthedotproductdependsonthecoordinatesystemrelativetowhichthecomponents ofthevectorsaredetermined.Itturnsoutthatitdoesnot.Therefore, itrepresentsanintrinsicgeometricalquantityassociatedwithtwovectorsinvolvedintheproduct.Toelucidatethegeometricalsigni“cance ofthedotproduct,note“rsttherelationbetweenthedotproductand thenorm(length)ofavector: a a = a2 1+ a2 2+ a2 3= a 2or a = a a Thus,if a = b inthedotproduct,thenthelatterdoesnotdepend onthecoordinatesystemwithrespecttowhichthecomponentsof a arede“ned.Next,considerthetrianglewhoseadjacentsidesarethe vectors a and b asdepictedinFigure11.8(leftpanel). Thentheothersideofthetrianglecanberepresentedbythedifference c = b Š a .Thesquaredlengthofthislattersideis (11.4) c c =( b Š a ) ( b Š a )= b b + a a Š 2 a b wherethealgebraicpropertiesofthedotproducthavebeenused. Therefore,thedotproductcanbeexpressedviathegeometricalinvariants,namely,thelengthsofthesidesofthetriangle: (11.5) a b = 1 2 c 2Š b 2Š a 2 .

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73.THEDOTPRODUCT23 Figure11.8. Left :Independenceofthedotproduct fromthechoiceofacoordinatesystem.Thedotproduct oftwovectorsthatareadjacentsidesofatrianglecanbe expressedviathelengthsofthetrianglesidesasshown in(11.5). Right :Geometricalsigni“canceofthedotproduct. Itdeterminestheanglebetweentwovectorsasstated in(11.6).Twononzerovectorsareperpendicularifand onlyiftheirdotproductvanishes.Thisfollowsfrom (11.5)andthePythagoreantheorem: a 2+ b 2= c 2foraright-angledtriangle. Thismeansthatthenumericalvalueofthedotproductisindependent ofthechoiceofcoordinatesystem.Thus,itcanbecomputedinany coordinatesystem.Inparticular,letustakethecoordinatesystem inwhichthevector a isparalleltothe x axisandthevector b lies inthe xy planeasshowninFigure11.8(rightpanel).Lettheangle between a and b be .Byde“nition,thisangleliesintheinterval [0 ].When =0,thevectors a and b pointinthesamedirection. When = / 2,theyareperpendicular,andtheypointintheopposite directionsif = .Inthechosencoordinatesystem, a = a 0 0 and b = b cos b sin 0 .Hence, (11.6) a b = a b cos orcos = a b a b Equation(11.6)revealsthegeometricalsigni“canceofthedotproduct. Itdeterminestheanglebetweentwoorientedsegmentsinspace.It providesasimplealgebraicmethodtoestablishamutualorientation oftwostraightlinesegmentsinspace.Thefollowingtheoremisuseful inpracticalapplications.

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2411.VECTORSANDTHESPACEGEOMETRY Theorem 11.2 (GeometricalSigni“canceoftheDotProduct) Twononzerovectorsareperpendicularifandonlyiftheirdotproduct vanishes: a b a b =0 Inparticular,foratrianglewithsides a b ,and c andanangle betweensides a and b ,itfollowsfromtherelation(11.4)that c2= a2+ b2Š 2 ab cos Foraright-angledtriangle,thePythagoreantheoremisrecovered: c2= a2+ b2. Example 11.2 Consideratrianglewhoseverticesare A (1 1 1) B ( Š 1 2 3) ,and C (1 4 Š 3) .Findalltheanglesofthetriangle. Solution: Lettheanglesatthevertices A B ,and C be ,and respectively.Then + + =180.Soitissucientto“ndanytwo angles.To“ndtheangle ,de“nethevectors a = AB = Š 2 1 2 and b = AC = 0 3 Š 4 .Theinitialpointofthesevectorsis A ,and hencetheanglebetweenthevectorscoincideswith .Since a =3 and b =5,bythegeometricalpropertyofthedotproduct, cos = a b a b = 0+3 Š 8 15 = Š 1 3 = =cosŠ 1( Š 1 / 3) 109 5. To“ndtheangle ,de“nethevectors a = BA = 2 Š 1 Š 2 and b = BC = 2 2 Š 6 withtheinitialpointatthevertex B .Thenthe anglebetweenthesevectorscoincideswith .Since a =3, b = 2 11,and a b =4 Š 2+12=14,one“ndscos =14 / (6 11)and =cosŠ 1(7 / (3 11)) 45 3.Therefore, 180Š 109 5Š 45 3= 25 2.NotethattherangeofthefunctioncosŠ 1mustbetakenfrom 0to180inaccordancewiththede“nitionoftheanglebetweentwo vectors. Theorem 11.3 (Cauchy-SchwarzInequality) Foranytwovectors a and b | a b | a b wheretheequalityisreachedonlyifthevectorsareparallel. Thisinequalityisadirectconsequenceofthe“rstrelationin(11.6) andtheinequality | cos | 1.Theequalityisreachedonlywhen =0 or = ,thatis,when a and b areparallel.

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73.THEDOTPRODUCT25 Theorem 11.4 (TriangleInequality) Foranytwovectors a and b a + b a + b Proof. Put a = a and b = b sothat a a = a 2= a2and similarly b b = b2.Usingthealgebraicrulesforthedotproduct, a + b 2=( a + b ) ( a + b )= a2+ b2+2 a b a2+ b2+2 ab =( a + b )2, wheretheCauchy-S chwarzinequalityhasbeenused.Bytakingthe squarerootofbothsides,thetriangleinequalityisobtained. Thetriangleinequalityhasasimplegeometricalmeaning.Consider atrianglewithsides a b ,and c .Thedirectionsofthevectorsare chosensothat c = a + b .Thetriangleinequalitystatesthatthelength c cannotexceedthetotallengthoftheothertwosides.Itisalso clearthatthemaximallength c = a + b isattainedonlyif a and b areparallelandpointinthesamedirection.Iftheyareparallelbut pointintheoppositedirection,thenthelength c becomesminimal andcoincideswiththedierenceof a and b .Thisobservationcan bestatedinthefollowingalgebraicform: (11.7) a Š b a + b a + b .73.2.DirectionAngles.Considerthreeunitvectors e1= 1 0 0 e2= 0 1 0 ,and e3= 0 0 1 thatareparalleltothecoordinateaxes x y and z ,respectively.Bytherulesofvectoralgebra,anyvectorcanbe writtenasthesumofthreemutuallyperpendicularvectors: a = a1,a2,a3 = a1 e1+ a2 e2+ a3 e3. Thevectors a1 e1, a2 e2,and a3 e3areadjacentsidesoftherectanglewhoselargestdiagonalcoincideswiththevector a asshownin Figure11.9(rightpanel). De“netheangle thatiscountedfromthepositivedirectionofthe x axistowardthevector a .Inotherwords,theangle istheangle between e1and a .Similarly,theangles and are,byde“nition,the anglesbetween a andtheunitvectors e2and e3,respectively.Then cos = e1 a e1 a = a1 a cos = e2 a e2 a = a2 a cos = e3 a e3 a = a3 a .

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2611.VECTORSANDTHESPACEGEOMETRY Figure11.9. Left :Directionanglesofavectorarede“nedastheanglesbetweenthevectorandthreecoordinatesaxes.Eachanglerangesbetween0and and iscountedfromthecorrespondingpositivecoordinate semiaxistowardthevector.Thecosinesofthedirection anglesofavectorarecomponentsoftheunitvectorparalleltothatvector. Right :Decompositionofavectorintothesumofthree mutuallyperpendicularvectorsthatareparalleltothe coordinateaxesofarectangularcoordinatesystem.The vectoristhediagonaloftherectangle,whereasthevectorsinthesumformtheedgesoftherectangle. Thesecosinesarenothingbutthecomponentsoftheunitvectorparallel to a : a = 1 a a = cos cos cos Thus,theangles ,and uniquelydeterminethedirectionofa vector.Forthisreason,theyarecalled directionangles .Notethatthey cannotbesetindependentlybecausetheyalwayssatisfythecondition a =1or cos2 +cos2 +cos2 =1 Inpractice(physics,mechanics,etc.),vectorsareoftenspeci“edby theirmagnitude a = a anddirectionangles.Thecomponentsare thenfoundby a1= a cos a2= a cos ,and a3= a cos .

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73.THEDOTPRODUCT27 73.3.PracticalApplications. 73.3.1.StaticProblems.AccordingtoNewtonsmechanics,apointlike objectthatwasatrestremainsatrestifthevectorsumofallforces appliedtoitvanishes.Thisisthefundamentallawofstatics: F1+ F2+ + Fn= 0 Thisvectorequationimpliesthreescalarequationsthatrequirevanishingeachofthethreecomponentsofthetotalforce.Ifthereisasystem ofpointlikeobjects,thenthesystemisatrestifeachobjectisatrest, andhencethesumofallforcesappliedtoeachobjectvanishes.This givesasystemofvectorequations,eachofwhichistheaboveequilibriumconditionforaparticularobject.Atypicalstaticproblemisto determineeitherthemagnitudesofsomeforcesorthevaluesofsome geometricalparametersatwhichthesysteminquestionisatrest. Example 11.3 Letaballofmass m beattachedtotheceilingby tworopessothatthesmallestanglebetweenthe“rstropeandtheceiling is 1andtheangle 2isde“nedsimilarlyforthesecondrope.Findthe magnitudesofthetensionforcesintheropes. Solution: Setthecoordinatesystemsothatthe x axisishorizontal andorientedfromthe“rstropetothesecondropesasdepictedin Figure11.10(leftpanel).Theropesareinthe xy plane,whilethe gravitationalforceisinthedirectionoppositetothe y axis.Let T1and T2bethemagnitudesofthetensionforces.Theninthiscoordinate systemtheforcesactingontheballare T1= Š T1cos 1,T1sin 1, 0 T2= T2cos 2,T2sin 2, 0 G = 0 Š mg, 0 where G isthegravitationalforceand g istheaccelerationofthe freefall( g 9 8m/s2);thatis, mg istheweightoftheball.The equilibriumcondition T1+ T2+ G = 0 leadstotwoequationsforthecomponents(thethirdcomponentsofall vectorsareidentically0): Š T1cos 1+ T2cos 2=0 ,T1sin 1+ T2sin 2Š mg =0 whichcanbesolvedfor T1and T2.Bymultiplyingthe“rstequation bysin 1andthesecondbycos 1andthenaddingthem,onegets T2= mg cos 1/ sin( 1+ 2).Substituting T2intothe“rstequation, thetension T1isobtained.

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2811.VECTORSANDTHESPACEGEOMETRY Figure11.10. Left :IllustrationtoExample11.3.At equilibrium,thevectorsumofallforcesactingonthe ballvanishes.Thecomponentsoftheforcesareeasyto “ndinthecoordinatesysteminwhichthe x axisishorizontalandthe y axisisvertical. Right :IllustrationtoStudyProblem11.11.Thevector c istheprojectionofavector b onto a .Itisavectorparallelto a .Theinitialpointsof b and c coincide.Theline throughtheterminalpointsof b and c isperpendicular to a .73.3.2.WorkDonebyaForce.Supposethatanobjectofmass m moves withspeed v .Thequantity K = mv2/ 2iscalledthe kineticenergy of theobject.Supposethattheobjecthasmovedalongastraightline segmentfromapoint P1toapoint P2undertheactionofaconstant force F .Alawofphysicsstatesthatachangeinanobjectskinetic energyisequaltothework W donebythisforce: K2Š K1= F P1P2= W, where K1and K2arethekineticenergiesattheinitialand“nalpoints ofthemotion,respectively. Example 11.4 Letanobjectslideonaninclinedplanewithout frictionunderthegravitationalforce.Findthe“nals peed v ofthe objectiftherelativeheightoftheinitialand“nalpointsis h andthe objectwasinitiallyatrest. Solution: Choosethecoordinatesystemsothatthedisplacement vector P1P2andthegravitationalforceareinthe xy plane.Letthe y axisbeverticalsothatthegravitationalforceis F = 0 Š mg, 0 where m isthemassand g istheaccelerationofthefreefall.Theinitial pointischosentohavethecoordinates(0 ,h, 0)whilethe“nalpoint is( L, 0 0),where L isthedistancetheobjecttravelsinthehorizontal

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73.THEDOTPRODUCT29 directionwhilesliding.Thedisplacementvectoris P1P2= L, Š h, 0 Since K1=0,onehas mv2 2 = W = F P1P2= mgh = v = 2 gh. Notethatthespeedisindependentofthemassoftheobjectandthe inclinationangleoftheplane(itstangentis h/L );itisfullydetermined bytherelativeheightonly. 73.4.StudyProblems.Problem11.11.(Projectionof b onto a ) Considertwovectors a and b withacommoninitialpoint O .Consider thelinethroughtheendpointof b thatisperpendicularto a .Let C be thepointintersectionofthislinewiththelinecontainingthevector a Findthevector c = OC .Thisvectoriscalleda projection of b onto a Solution: (SeetherightpanelofFig.11.10).Byconstruction, c isparallelto a andhenceproportionaltoit; c = s a forsomereal s Lettheanglebetween b and a be .Then,byconstruction, s> 0if < 90( c and a pointinthesamedirection)and s< 0if > 90( c and a pointintheoppositedirections).Also,fromtheright-angled triangle, c = b cos if < 90and c = Š b cos if > 90. Therefore, c = s a ,s = b cos a = b a cos a 2= a b a 2. Problem11.12. Findallvaluesof t forwhichthevectors a = 2 t, 3 Š t, Š 1 and b = t,t, 3+ t areorthogonal. Solution: Bythegeometricalpropertyofthedotproduct,twovectorsareorthogonalifandonlyiftheirdotproductvanishes.Therefore, a b =2 t2+ t (3 Š t ) Š (3+ t )=( t +1)2Š 4=0.Thesolutionsofthis equationare t =1and t = Š 3. Problem11.13. Describethesetofpointsinspacewhoseposition vector r satis“esthecondition ( r Š a ) ( r Š b )=0 .Hint: Notethat thepositionvectorsatisfyingthecondition r Š c = R describesa sphereofradius R whosecenterhasthepositionvector c Solution: Theequationofaspherecanalsobewrittenintheform r Š c 2=( r Š c ) ( r Š c )= R2.Theequation( r Š a ) ( r Š b )=0can

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3011.VECTORSANDTHESPACEGEOMETRY betransformedintothesphereequationbycompletingthesquares. Usingthealgebraicpropertiesofthedotproduct, ( r Š a ) ( r Š b )= r r Š r ( a + b )+ a b =( r Š c ) ( r Š c ) Š c c + a b c =1 2( a + b ) c c Š a b = R R R =1 2( a Š b ) Hence,thesetisasphereofradius R = R ,anditscenterispositioned at c 73.5.Exercises.(1) Findthedotproduct a b if (i) a = 1 2 3 and b = Š 1 2 0 (ii) a = e1+3 e2Š e3and b =3 e1Š 2 e2+ e3(2) Forwhatvaluesof b arethevectors Š 6 ,b, 2 and b,b2,b orthogonal? (3) Findtheangleatthevertex A ofatriangle ABC for A (1 0 1), B (1 2 3),and C (0 1 1). (4) Findthecosinesoftheanglesofatriangle ABC for A (0 1 1), B ( Š 2 4 3),and C (1 2 Š 1). (5) Findtheunitvectorparallelto a = 2 Š 1 Š 2 andtheunit vectorwhosedirectionisoppositeto a (6) Consideratrianglewhoseanytwoadjacentsidesareunitvectors.Whatarepossiblevaluesofthedotproductsofanytwosuchunit vectors? (7) Consideracubewhoseedgeshavelength a .Findtheangle betweenitslargestdiagonalandanyedgeadjacenttothediagonal. (8) Avector a makestheangle / 3withthepositive x axis,the angle / 6withthenegative y axis,andtheangle / 4withthepositive z axis.Findthecomponentsof a ifitslengthis6. (9) Findthecomponentsofallunitvectors u thatmaketheangle/ 6withthepositive z axis. Hint: Put u = a v + b e3,where v isaunitvectorinthe xy plane.Find a b ,andall v usingthepolarangleinthe xy plane. (10) If c = a b + b a ,where a and b arenonzerovectors,show that c bisectstheanglebetween a and b (11) Letthevectors a and b havethesamelength.Showthatthe vectors a + b and a Š b areorthogonal. (12) Consideraparallelogramwithadjacentsidesoflength a and b .If d1and d2arethelengthsofthediagonals,provetheparallelogram law: d2 1+ d2 2=2( a2+ b2).

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74.THECROSSPRODUCT31 Hint: Considerthevectors a and b thatareadjacentsidesofthe parallelogramandexpressthediagonalsvia a and b .Usethedot producttoevaluate d2 1+ d2 2. (13) Twoballsofmass m and3 m ,respectively,areconnectedby apieceofropeoflength h .Thentheballsareattachedtodierent pointsonahorizontalceilingbyapieceofropewiththesamelength h sothatthedistance L betweenthepointsisgreaterthan h butless than3 h .Findtheequilibriumpositionsoftheballs.74.TheCrossProduct74.1.DeterminantofaSquareMatrix.Definition 11.9 Thedeterminantofa 2 2 matrixisthenumber computedbythefollowingrule: det a11a12a21a22 = a11a22Š a12a21, thatis,theproductofthediagonalelementsminustheproductofthe o-diagonalelements. Definition 11.10 Thedeterminantofa 3 3 matrix A isthe numberobtainedbythefollowingrule: det a11a12a13a21a22a23a31a32a33 = a11det A11Š a12det A12+ a13det A13=3k =1( Š 1)k +1a1 kdet A1 k, A11= a22a23a32a33 ,A12= a21a23a31a33 ,A13= a21a22a31a32 wherethematrices A1 k, k =1 2 3 ,areobtainedfromtheoriginal matrix A byremovingtherowandcolumncontainingtheelement a1 k. Itisstraightforwardtoverifythatthedeterminantcanbeexpanded overanyroworcolumn: det A =3k =1( Š 1)k + mamkdet Amkforany m =1 2 3 det A =3m =1( Š 1)k + mamkdet Amkforany k =1 2 3 ,

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3211.VECTORSANDTHESPACEGEOMETRY wherethematrix Amkisobtainedfrom A byremovingtherowand columncontaining amk.Thisde“nitionofthedeterminantisextended to N N squarematricesbyletting k and m rangeover1 2 ,...,N Inparticular,thedeterminantofatriangularmatrix(i.e.,thematrixallofwhoseelementseitheraboveorbelowthediagonalvanish)is theproductofitsdiagonalelements: det a1bc 0 a2d 00 a3 =det a100 ba20 cda3 = a1a2a3foranynumbers b c ,and d .Also,itfollowsfromtheexpansionofthe determinantoveranycolumnorrowthat,ifanytworowsoranytwo columnsareswappedinthematrix,itsdeterminantchangessign. Example 11.5 Calculate det A ,where A = 123 013 Š 121 Solution: Expandingthedeterminantoverthe“rstrowyields det A =1(1 1 Š 2 3) Š 2(0 1 Š ( Š 1) 3)+3(0 2 Š ( Š 1) 1)= Š 8 Alternatively,expandingthedeterminantoverthesecondrowyields thesameresult: det A = Š 0(2 1 Š 3 2)+1(1 1 Š ( Š 1) 3) Š 3(1 2 Š ( Š 1) 2)= Š 8 Onecancheckthatthesameresultcanbeobtainedbyexpandingthe determinantoveranyroworcolumn. 74.2.TheCrossProductofTwoVectors.Definition 11.11 (CrossProduct) Thecrossproductoftwovectors a = a1,a2,a3 and b = b1,b2,b3 is avectorthatisthedeterminantoftheformalmatrixexpandedoverthe “rstrow: a b =det e1 e2 e3a1a2a3b1b2b3 = e1det a2a3b2b3 Š e2det a1a3b1b3 + e3det a1a2b1b2 = a2b3Š a3b2,a3b1Š a1b3,a1b2Š a2b1 (11.8)

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74.THECROSSPRODUCT33 Notethatthe“rstrowofthematrixconsistsoftheunitvectors paralleltothecoordinateaxesratherthannumbers.Forthisreason,it isreferredastoa formal matrix.Theuseofthedeterminantismerely acompactwaytowritethealgebraicruletocomputethecomponents ofthecrossproduct. Thecrossproducthasthefollowingpropertiesthatfollowfromits de“nition: a b = Š b a ( a + c ) b = a b + c b ( s a ) b = s ( a b ) The“rstpropertyisobtainedbyswappingthecomponentsof b and a in(11.8).Itstatesthatthecrossproductisskew-symmetric(i.e.,itis notcommutative andtheorderinwhichthevectorsaremultipliedis essential);changingtheorderleadstotheoppositevector.Thecross productis distributive accordingtothesecondproperty.Toproveit, change aito ai+ ci, i =1 2 3,in(11.8).Ifavector a isscaledbya number s andtheresultingvectorismultipliedby b ,theresultisthe sameasthecrossproduct a b computed“rstandthenscaledby s (change aito saiin(11.8)andthenfactorout s ).74.3.GeometricalSigni“canceoftheCrossProduct.Theabovealgebraicde“nitionofthecrossproductusesaparticularcoordinatesystemrelativetowhichthecomponentsofthevectorsarede“ned.Does thecrossproductdependonthechoiceofthecoordinatesystem?To answerthisquestion,oneshouldinvestigatewhetherbothits direction andits magnitude dependonthechoiceofthecoordinatesystem.Let us“rstinvestigatethemutualorientationoftheorientedsegments a b ,and a b .Asimplealgebraiccalculationleadstothefollowing result: a ( a b )= a1( a2b3Š a3b2)+ a2( a3b1Š a1b3)+ a3( a1b2Š a2b1)=0 Bytheskewsymmetryofthecrossproduct,itisalsoconcludedthat b ( a b )= Š b ( b a )=0.Bythegeometricalpropertyofthe dotproduct,thecrossproductmustbeperpendiculartobothvectors a and b : (11.9) a ( a b )= b ( a b )=0 a b a and a b b Thisshowsthatthedirectionofthecrossproductdoesnotdependon thechoiceofthecoordinatesystemmodulothere”ection a b a b So,byasuitablerotation,thecoordinatesystemcanbeorientedsothat thecrossproductisparalleltothe z axis.Thenthevectors a and b

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3411.VECTORSANDTHESPACEGEOMETRY Figure11.11. Left :Geometricalinterpretationofthe crossproductoftwovectors.Thecrossproductisavectorthatisperpendiculartobothvectorsintheproduct. Itslengthequalstheareaoftheparallelogramwhoseadjacentsidesarethevectorsintheproduct.Ifthe“ngers oftherighthandcurlinthedirectionofarotationfrom the“rsttosecondvectorthroughthesmallestanglebetweenthem,thenthethumbpointsinthedirectionof thecrossproductofthevectors. Right :IllustrationtoStudyProblem11.15. areinthe xy plane.Let aand bbeanglescountedfromthepositive x axiscounterclockwisetowardthevectors a and b ,respectively.The componentsofthesevectorsare a = a cos a, a sin a, 0 and b = b cos b, b sin b, 0 (comparewiththepolarcoordinatesoftwo pointsinaplanewiththepositionvectors a and b ).Thenthecross productreads: a b = e3 a b (cos asin bŠ sin acos b)= e3 a b sin( bŠ a) Twoimportantconclusionscanbededucedfromthisexpression.74.3.1.TheRight-HandRule(theCross-ProductDirection).Byde“nition,theangles aand brangeovertheinterval[0 2 ).Let0 bethe(smallest)anglebetweenthevectors a and b (justasde“nedby theirdotproduct).Thelengths a and b andtheangledierence bŠ aareindependentoftheorientationofthecoordinateaxesinthe planeandsomustbethecrossproduct.Inparticular,onecanchoose the x axisparalleltothevector a (or a=0).Then b= if b and b=2 Š if
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74.THECROSSPRODUCT35 knownasthe right-handrule : Ifthe“ngersoftherighthandcurlin thedirectionofarotationfrom a toward b throughthesmallestangle betweenthem,thenthethumbpointsinthedirectionof a b Thus,thecrossproductisalwaysperpendiculartotheplanecontaining a and b andoriented(upŽordownŽrelativetotheplane) accordingtotheright-handrule. Remark. Thetransformationinwhichthecoordinateaxeschange theirdirectiontotheoppositeiscalledthe paritytransformation .Evidently,undertheparitytransformation,coordinatesofeverypoint changetheirsign,andhenceeveryvector(de“nedasanorderedtriple ofnumbers)changesitsdirection, a = a1,a2,a3Š a1, Š a2, Š a3 = Š a .However,thecrossproductoftwovectorsdoesnotchangeunder theparitytransformation: a b ( Š a ) ( Š b )= a b .Forthis reason,thecrossproductissometimesreferredtoasa pseudovector or an axialvector .Thecoordinatesystemsrelatedbytheparitytransformationcannotbeobtainedfromoneanotherbyrotations,justlike leftŽandrightŽareswappedinamirrorre”ection.Thereareforces innaturethatareaxialvectors.Sotheworldanditsmirrorimagecan bedistinguishedbystudyingtheresultsoftheactionsofsuchforces. Physicalexperimentsrevealthattheparitysymmetryisindeedbroken inourUniverse!74.3.2.TheAreaofaParallelogram(theCross-ProductMagnitude).By thede“nitionoftheangle ,sin 0.Therefore,themagnitudeofthe crossproductisexpressedviathegeometricalinvariants„thelength ofthevectorsandtheanglebetweenthem: a b = a b sin Nowconsidertheparallelogramwithadjacentsides a and b .If a isthelengthofitsbase,then h = b sin isitsheight.Thenthe magnitudeofthecrossproduct, a b = a h ,mustbetheareaof theparallelogram.Thiscompletesaproofofthefollowingtheorem. Theorem 11.5 (GeometricalSigni“canceoftheCrossProduct) Thecrossproduct a b ofvectors a and b isthevectorthatisperpendiculartobothvectors, a b a and a b b ,hasamagnitude equaltotheareaoftheparallelogramwithadjacentsides a and b ,and isdirectedaccordingtotheright-handrule. Itshouldbeemphasizedthatnocoordinatesystemisrequiredto determinethecrossproductoftwovectors.Thegeometricalproperties ofthecrossproductcanbeusedtoobtainanotheralgebraiccriterion fortwovectorsthatareparallel.

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3611.VECTORSANDTHESPACEGEOMETRY Corollary 11.1 Twononzerovectorsareparallelifandonlyif theircrossproductvanishes: a b = 0 a b Whentwovectorsareparallel,theareaofthecorrespondingparallelogramvanishes, a b =0.Thelatteristrueifandonlyif a b = 0 .Conversely,fortwoparallelvectors,thereisanumber s suchthat a = s b .Hence, a b =( s a ) b = s ( b b )= 0 Oneofthemostimportantapplicationsofthecrossproductisin calculationsoftheareasofplanar“guresinspace. Corollary 11.2 (AreaofaTriangle) Consideratrianglewithtwoadjacentsidesrepresentedbythevectors a and b suchthatthevectorshavethesameinitialpointatavertexof thetriangle.Thentheareaofthetriangleis Area = 1 2 a b Indeed,bythegeometricalconstruction,theareaofthetriangleis halfoftheareaofaparallelogramwithadjacentsides a and b Example 11.6 Let A =(1 1 1) B =(2 Š 1 3) ,and C =( Š 1 3 1) Findtheareaofthetriangle ABC andavectornormaltotheplane thatcontainsthetriangle. Solution: Accordingtothegeometricalpropertiesofthecrossproduct,inorderto“ndavectornormaltoaplane,oneshouldtake thecrossproductofanytwononparallelvectorsintheplane.For example, a = AB = 1 Š 2 2 and b = AC = Š 2 2 0 .Then a b = Š 4 Š 4 Š 6 isnormaltotheplane.Notethatthecross productofanyotherpairofvectorscorrespondingtothesidesofthe trianglecanonlybeascaledvector s Š 4 Š 4 6 becauseanytwonor-malvectorsofagivenplanemustbeparallelandhenceproportional. Since Š 4 Š 4 Š 6 =2 2 2 3 =2 17,theareaofthetriangle ABC is 17byCorollary11.2.Theunitsherearesquaredunitsof lengthusedtomeasurethecoordinatesofthetrianglevertices(e.g., m2ifthecoordinatesaremeasuredinmeters). 74.4.StudyProblems.Problem11.14. Findthemostgeneralvector r thatsatis“estheequations a r =0 and b r =0 ,where a and b arenonzero,nonparallel vectors.

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74.THECROSSPRODUCT37 Solution: Theconditionsimposedon r holdifandonlyifthevector r isorthogonaltobothvectors a and b .Therefore,itmustbeparallel totheircrossproduct.Thus, r = t ( a b )foranyreal t Problem11.15. Usegeometricalmeansto“ndthecrossproductsof theunitvectorsparalleltothecoordinateaxes. Solution: Consider e1 e2.Since e1 e2and e1 = e2 =1,their crossproductmustbeaunitvectorperpendiculartoboth e1and e2. Thereareonlytwosuchvectors, e3.Bytheright-handrule,itfollows that e1 e2= e3. Similarly,theothercrossproductsareshowntobeobtainedbycyclic permutationsoftheindices1,2,and3intheaboverelation.Apermutationofanytwoindicesleadstoachangeinsign(e.g., e2 e1= Š e3). Sinceacyclicpermutationofthreeindices { ijk }{ kij } (andsoon) consistsoftwopermutationsofanytwoindices,therelationbetween theunitvectorscanbecastintheform ei= ej ek, { ijk } = { 123 } andcyclicpermutations Problem11.16. Provethebac Š cabŽrule: d = a ( b c )= b ( a c ) Š c ( a b ) Solution: If c and b areparallel,then d = 0 .If c and b arenot parallel,then d mustbeperpendiculartoboth a and b c .From thecondition d b c ,itfollowsthat d liesintheplanecontaining b and c andhenceisalinearcombinationofthem, d = s b + t c Fromthecondition d a or a d =0,itfollowsthat s = p ( a c )and t =Š p ( a b )forsomereal p .Sincethemagnitudeofthecrossproduct isindependentofthechoiceofthecoordinatesystem,thenumber p canbe“xedbycomputing d inanyconvenientcoordinatesystem.By rotatingthecoordinatesystem,onecanalwaysdirectthe x axisalong thevector c sothat c = c e1,whilethevector b liesinthe xy plane sothat b = b1 e1+ b2 e2.Then b c = Š e3b2 c andtherefore,fora generic a = a1,a2,a3 a ( b c )= Š e1 c a2b2+ e2b2 c a1= Š c a2b2+( b Š b1 e1) c a1= b c a1Š c ( a1b1+ a2b2)= b ( c a ) Š c ( a b ) thatis, p =1.Ofcourse,thestatementcanalsobeprovedbya directuseofthealgebraicde“nitionofthecrossproduct(abrute-force method).

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3811.VECTORSANDTHESPACEGEOMETRY Problem11.17. ProvetheJacobiidentity a ( b c )+ b ( c a )+ c ( a b )= 0 Solution: Notethatthesecondandthirdtermsontheleftsideare obtainedfromthe“rstbycyclicpermutationsofthevectors.Making useofthe bac…cab ruleforthe“rsttermandthenaddingtoitits twocyclicpermutations,onecanconvinceoneselfthatthecoecients ateachofthevectors a b ,and c areaddeduptomake0. Remark. NotethattheJacobiidentityimpliesinparticularthat a ( b c ) =( a b ) c ; thatis,themultiplicationlawde“nedbythecrossproductdoesnot generallyobeytheassociativelawformultiplicationofnumbers. Problem11.18. Considerallvectorsinaplane.Anysuchvector a canbeuniquelydeterminedbyspecifyingitslength a = a andthe angle athatiscountedfromthepositive x axistowardthevector a (i.e., 0 a< 2 ).Therelation a1,a2 = a cos a,a sin a establishesa one-to-onecorrespondencebetweenorderedpairs ( a1,a2) and ( a,a) De“nethevectorproductoftwovectors a and b asthevector c for which c = ab and c= a+ b.Showthatthisproductisassociative andcommutative,thatis,that c doesnotdependontheorderofvectors intheproduct. Solution: Letusdenotethevectorproductbyasmallcircleto distinguishitfromthedotandcrossproducts, a b = c .Since c = ab cos( a+ b) ,ab sin( a+ b) ,thecommutativityofthevector product a b = b a followsfromthecommutativityoftheproduct andadditionofnumbers: ab = ba and a+ b= b+ a.Similarly, theassociativityofthevectorproduct( a b ) c = a ( b c )follows fromtheassociativityoftheproductandadditionofordinarynumbers: ( ab ) c = a ( bc )and( a+ b)+ c= a+( b+ c). Remark. Thevectorproductintroducedforvectorsinaplaneis knownasthe productofcomplexnumbers .Itisinterestingtonote thatnocommutativeandassociativevectorproduct(i.e.,vectortimes vector=vectorŽ)canbede“nedinaEuclideanspaceofmorethantwo dimensions. Problem11.19. Let u beavectorrotatinginthe xy planeaboutthe z axis.Givenavector v ,“ndthepositionof u suchthatthemagnitude ofthecrossproduct v u ismaximal.

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74.THECROSSPRODUCT39 Solution: Foranytwovectors, v u = v u sin ,where istheanglebetween v and u .Themagnitudeof v is“xed,while themagnitudeof u doesnotchangewhenrotating.Therefore,the absolutemaximumofthecross-productmagnitudeisreachedwhen sin =1orcos =0(i.e.,whenthevectorsareorthogonal).The correspondingalgebraicconditionis v u =0.Since u isrotatinginthe xy plane,itscomponentsare u = u cos u sin 0 ,where0 < 2 istheanglecountedcounterclockwisefromthe x axistoward thecurrentpositionof u .Put v = v1,v2,v3 .Thenthedirectionof u isdeterminedbytheequation v u = u ( v1cos + v2sin )=0, andhencetan = Š v1/v2.Thisequationhastwosolutionsinthe range0 < 2 : = Š tanŠ 1( v1/v2)and = Š tanŠ 1( v1/v2)+ Geometrically,thesesolutionscorrespondtothecasewhen u isparallel totheline y = Š ( v1/v2) x inthe xy plane. 74.5.Exercises.(1) Findthecrossproduct a b if (i) a = 1 2 3 and b = Š 1 0 1 (ii) a = e1+3 e2Š e3and b =3 e1Š 2 e2+ e3(2) Findtheareaofatriangle ABC for A (1 0 1), B (1 2 3),and C (0 1 1)andanonzerovectorperpendiculartotheplanecontaining thetriangle. (3) Suppose a liesinthe xy plane,itsinitialpointisattheorigin, anditsterminalpointisin“rstquadrantofthe xy plane.Let b be parallelto e3.Usetheright-handruletodeterminewhethertheangle between a b andtheunitvectorsparalleltothecoordinateaxeslies intheinterval(0 ,/ 2)or( / 2 )orequals / 2. (4) Ifvectors a b ,and c havetheinitialpointattheoriginand lie,respectively,inthepositivequadrantsofthe xy yz ,and xz planes, “ndtheoctantsinwhichthepairwisecrossproductsofthesevectors lie. (5) Let A =(1 2 1)and B =( Š 1 0 2)beverticesofaparallelogram.Iftheothertwoverticesareobtainedbymoving A and B by 3unitsoflengthalongthevector a = 2 1 Š 2 ,“ndtheareaofthe parallelogram. (6) Considerfourpointsinspace.Supposethatthecoordinates ofthepointsareknown.Describeaprocedurebasedonvectoralgebratodeterminewhetherthepointsareinoneplane.Inparticular,arethepoints(1 2 3),( Š 1 0 1),(1 3 Š 1),and(0 1 2)inone plane? (7) Letthesidesofatrianglehavelengths a b ,and c andletthe anglesattheverticesoppositetothesides a b ,andc be,respectively,

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4011.VECTORSANDTHESPACEGEOMETRY ,and .Provethat sin a = sin b = sin c Hint: De“nethesidesasvectorsandexpresstheareaofthetriangle viathevectorsateachvertexofthetriangle. (8) Considerapolygonwithfourvertices A B C ,and D .Ifthe coordinatesoftheverticesarespeci“ed,describetheprocedurebased onvectoralgebratocalculatetheareaofthepolygon.Inparticular, put A =(0 0) B =( x1,y1), C =( x2,y2),and D =( x3,y3),andexpress theareavia xiand yi, i =1 2 3. (9) Consideraparallelogram.Constructanotherparallelogram whoseadjacentsidesarediagonalsofthe“rstparallelogram.Find therelationbetweentheareasoftheparallelograms. (10) Giventwononparallelvectors a and b ,showthatanyvector r inspacecanbewrittenasalinearcombination r = x a + y b + z a b andthatthenumbers x y ,and z areuniqueforevery r Hint: SeeStudyProblems11.14and11.6. (11) Atetrahedronisasolidwithfourverticesandfourtriangular faces.Let v1, v2, v3,and v4bevectorswithlengthsequaltotheareas ofthefacesanddirectionsperpendiculartothefacesandpointing outward.Showthat v1+ v2+ v3+ v4= 0 (12) If a b = a c and a b = a c ,doesitfollowthat b = c ?75.TheTripleProduct Definition 11.12 Thetripleproductofthreevectors a b ,and c isanumberobtainedbytherule: a ( b c ) Itfollowsfromthealgebraicde“nitionofthecrossproductandthe de“nitionofthedeterminantofa3 3matrixthat a ( b c )= a1det b2b3c2c3 Š a2det b1b3c1c3 + a3det b1b2c1c2 =det a1a2a3b1b2b3c1c2c3 Thisprovidesaconvenientwaytocalculatethenumericalvalueof thetripleproduct.Iftworowsofamatrixareswapped,thenits determinantchangessign.Therefore, a ( b c )= Š b ( a c )= Š c ( b a ) Thismeans,inparticular,thattheabsolutevalueofthetripleproduct isindependentoftheorderofthevectorsinthetripleproduct.

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75.THETRIPLEPRODUCT41 Figure11.12. Left :Geometricalinterpretationofthe tripleproductasthevolumeoftheparallelepipedwhose adjacentsidesarethevectorsintheproduct: h = a cos A = b c V = hA = a b c cos = a ( b c ). Right :Testforthecoplanarityofthreevectors.Three vectorsarecoplanarifandonlyiftheirtripleproduct vanishes: a ( b c )=0.75.1.GeometricalSigni“canceoftheTripleProduct.Supposethat b and c arenotparallel(otherwise, b c = 0 ).Let betheanglebetween a and b c asshowninFigure11.12(leftpanel).If a b c (i.e., = / 2),thenthetripleproductvanishes.Let = / 2.Considerthe parallelepipedwhoseadjacentsidesbeingarethevectors a b ,and c Thefacesoftheparallelepipedaretheparallelogramswhoseadjacent sidesarepairsofthevectors.Inparticular,thecrossproduct b c isperpendiculartothefacecontainingthevectors b and c ,whereas A = b c istheareaofthisfaceoftheparallelepiped(theareaof theparallelogramwithadjacentsides b and c ).Bythegeometrical propertyofthedotproduct, a ( b c )= A a cos .Ontheother hand,thedistancebetweenthetwofacesparalleltoboth b and c (ortheheightoftheparallelepiped)is h = a cos if /2or, h = a | cos | .Thevolumeofthe parallelepipedis V = Ah .Thisleadstothefollowingtheorem. Theorem 11.6 (GeometricalSigni“canceoftheTripleProduct) Thevolume V ofaparallelepipedwhoseadjacentsidesarethevectors a b ,and c istheabsolutevalueoftheirtripleproduct: V = | a ( b c ) | .

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4211.VECTORSANDTHESPACEGEOMETRY Thus,thetripleproductisaconvenientalgebraictoolforcalculatingvolumes.Thereisausefulconsequenceofthistheorem. Definition 11.13 (CoplanarVectors). Vectorsaresaidtobecoplanariftheyareinoneplane. Clearly,anytwovectorsarealwayscoplanar.Whatisanalgebraic conditionforthreevectorsbeingcoplanar? Corollary 11.3 (CriterionforThreeVectorstoBeCoplanar) Threevectorsarecoplanarifandonlyiftheirtripleproductvanishes: a b c arecoplanar a ( b c )=0 Indeed,ifthevectorsarecoplanar(Figure11.12,rightpanel),then thecrossproductofanytwovectorsmustbeperpendiculartothe planewherethevectorsareandthereforethetripleproductvanishes. If,conversely,thetripleproductvanishes,theneither b c = 0 or a b c .Intheformercase, b isparallelto c ,or c = t b ,andhence a alwaysliesinaplanewith b and c .Inthelattercase,allthreevectors a b ,and c areperpendicularto b c andthereforemustbeinone plane(perpendicularto b c ). Example 11.7 Determinewhetherthepoints A (1 1 1) B (2 0 2) C (3 1 Š 1) ,and D(0 2 3) areinthesameplane. Solution: Considerthevectors a = AB = 1 Š 1 1 b = AC = 2 0 2 ,and c = AD = Š 1 1 2 .Thepointsinquestionareinthe sameplaneifandonlyifthevectors a b ,and c arecoplanar,or a ( b c )=0.One“nds. a ( b c )=det 1 Š 11 202 Š 112 =1(0 Š 2)+1(4+2)+1(2 Š 0)=6 =0 Therefore,thepointsarenotinthesameplane. Thetripleproductcanbeusedto“ndthedistancesbetweentwosets ofpointsinspace.Let S1and S2betwosetsofpointsinspace.Leta point A1belongto S1,letapoint A2belongto S2,andlet | A1A2| be thedistancebetweenthem.75.2.DistancesBetweenLinesandPlanes.Ifthelinesorplanesin spacearenotintersecting,thenhowcanone“ndthedistancebetween them?Thisquestioncanbeansweredusingthegeometricalproperties ofthetripleandcrossproducts.

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75.THETRIPLEPRODUCT43 Definition 11.14 (DistanceBetweenSetsinSpace) Thedistance D betweentwosetsofpointsinspace, S1and S2,isthe largestnumberthatislessthanorequaltoallthenumbers | A1A2| when thepoint A1rangesover S1andthepoint A2rangesover S2. Naturally,ifthesetshaveatleastonecommonpoint,thedistance betweenthemvanishes.Thedistancebetweensetsmayvanisheven ifthesetshavenocommonpoints.Forexample,let S1beanopen interval(0 1)on,say,the x axis,while S2istheinterval(1 2)onthe sameaxis.Apparently,thesetshavenocommonpoints(thepoint x =1doesnotbelongstoeitherofthem).Thedistanceisthelargest number D suchthat D | x1Š x2| ,where0 0canbemadesmallerthananypreassigned positivenumberbytaking x1and x2closeenoughto1.Sincethe distance D 0,theonlypossiblevalueis D =0.Intuitively,thesets areseparatedbyasinglepointthatisnotanextendedŽobject,and hencethedistancebetweenthemshouldvanish.Inotherwords,there aresituationsinwhichtheminimumof | A1A2| isnotattainedforsome A1S1,orsome A2S2,orboth.Nevertheless,thedistancebetween thesetsisstillwellde“nedasthelargestnumberthatislessthanor equaltoallnumbers | A1A2| .Suchanumberiscalledthe in“mum of thesetofnumbers | A1A2| anddenotedinf | A1A2| .Thus, D =inf | A1A2| ,A1S1,A2S2. Thenotation A1S1standsforapoint A1belongstotheset S1,Ž orsimply A1isanelementof S1.ŽThede“nitionisillustratedin Figure11.13(leftpanel). Theorem 11.7 (DistanceBetweenParallelPlanes) Thedistancebetweenparallelplanes P1and P2isgivenby D = | AP ( AB AC ) | AB AC where A B ,and C areanythreepointsintheplane P1thatarenot onthesameline,and P isanypointintheplane P2. Proof. Sincethepoints A B ,and C arenotonthesameline,the vectors b = AB and c = AC arenotparallelandtheircrossproduct isavectorperpendiculartotheplanes(seeFigure11.13,rightpanel). Considertheparallelepipedwithadjacentsides a = AP b ,and c .Two ofitsfaceslieintheparallelplanes,onein P1andtheotherin P2(i.e., theparallelogramswithadjacentsides b and c ).Thedistancebetween theplanesis,byconstruction,theparallelepipedheight,whichisequal

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4411.VECTORSANDTHESPACEGEOMETRY Figure11.13. Left :Distancebetweentwopointsets S1and S2de“nedasthelargestdistancethatislessthan orequaltoalldistances | A1A2| ,where A1rangesoverall pointsin S1and A2rangesoverallpointsin S2. Right :Distancebetweentwoparallelplanes(Theorem 11.7).Consideraparallelepipedwhoseoppositefaceslie intheplanes P1and P2.Thenthedistance D between theplanesistheheightoftheparallelepiped,whichcan becomputedastheratio D = V/A ,where V = | a ( b c ) | isthevolumeoftheparallelepipedand A = b c is theareaoftheface. to V/A ,where V and A aretheparallelepipedvolumeandareaofthe faceparallelto b and c .Theconclusionfollowsfromthegeometrical propertiesofthetripleandcrossproducts: V = | a ( b c ) | and A = b c Similarly,thedistancebetweentwoparallellines L1and L2canbe determined.Recallthatlinesareparalleliftheyarenotintersecting andlieinthesameplane.Let A and B beanytwopointsontheline L1andlet C beanypointontheline L2.Considertheparallelogram withadjacentsides a = AB and b = AC asdepictedinFigure11.14 (leftpanel).Thedistancebetweenthelinesistheheightofthisparallelogram,whichis A/ a ,where A = a b ,istheparallelogram areaand a isthelengthofitsbase. Corollary 11.4 (DistanceBetweenParallelLines) Thedistancebetweentwoparallellines L1and L2is D = AB AC AB where A and B areanytwodistinctpointsontheline L1and C isany pointontheline L2. Byconstruction, D istheheightoftheparallelogramwhoseadjacentsidesarethevectors AB and AC .Therefore, D isitsareadivided

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75.THETRIPLEPRODUCT45 Figure11.14. Left :Distancebetweentwoparallel lines.Consideraparallelogramwhosetwoparallelsides lieinthelines.Thenthedistancebetweenthelinesis theheightoftheparallelogram(Corollary11.4). Right :Distancebetweenskewlines.Consideraparallelepipedwhosetwononparalleledges AB and CP in theoppositefaceslieintheskewlines L1and L2,respectively.Thenthedistancebetweenthelinesisthe heightoftheparallelepiped,whichcanbecomputedas theratioofthevolumeandtheareaoftheface(Corollary11.5). bythelengthofthebase AB .Bythegeometricalpropertiesofthe crossproduct, AB AC istheareaoftheparallelogram. Definition 11.15 (SkewLines) Twolinesthatarenotintersectingandnotparallelarecalled skew lines Todeterminethedistancebetweenskewlines L1and L2,consider anytwopoints A and B on L1andanytwopoints C and P on L2. De“nethevectors b = AB and c = CP thatareparalleltolines L1and L2,respectively.Sincethelinesarenotparallel,thecrossproduct b c doesnotvanish.Thelines L1and L2lieintheparallelplanes perpendicularto b c (bythegeometricalpropertiesofthecrossproduct, b c isperpendicularto b and c ).Thedistancebetweenthelines coincideswiththedistancebetweentheseparallelplanes.Considerthe parallelepipedwithadjacentsides a = AC b ,and c asshowninFigure 11.14(rightpanel).Thelineslieintheparallelplanesthatcontainthe facesoftheparallelepipedparalleltothevectors b and c .Thus,the distancebetweenskewlinescanbefoundfromthedistancebetween theparallelplanescontainingthem, D = V/A ,where V and A arethe parallelepipedvolumeandtheareaofthebase A = b c .

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4611.VECTORSANDTHESPACEGEOMETRY Corollary 11.5 (DistanceBetweenSkewLines) Thedistancebetweentwoskewlines L1and L2is D = | AC ( AB CP ) | AB CP where A and B areanytwodistinctpointson L1,while C and P are anytwodistinctpointson L2. Notethat,givenanytwolines,onecancalculate D ,provided,of course,thatthevectors AB and CP arenotparallel;thatis,thelines arenotparallel.If D =0,thenthelinesmustintersect.Thisgivesa simplealgebraiccriterionfortwolinesbeingskeworintersecting.75.3.StudyProblems.Problem11.20. Findthemostgeneralvector r thatsatis“estheequation a ( r b )=0 ,where a and b arenonzero,nonparallelvectors. Solution: Bythealgebraicpropertyofthetripleproduct, a ( r b )= r ( b a )=0.Hence, r a b .Thevector r liesintheplaneparallelto both a and b because a b isorthogonaltothesevectors.Anyvector intheplaneisalinearcombinationofanytwononparallelvectorsin it: r = t a + s b foranyreal t and s (seeStudyProblem11.6). Problem11.21.(VolumeofaTetrahedron). Atetrahedronisasolid withfourverticesandfourtriangularfaces.Itsvolume V =1 3Ah where h isthedistancefromavertextotheoppositefaceand A isthe areaofthatface.Givencoordinatesofthevertices B C D ,and P expressthevolumeofthetetrahedronthroughthem. Solution: Put b = BC c = BD ,and a = AP .Theareaofthe triangle BCD is A =1 2 b c .Thedistancefrom P totheplane P1containingtheface BCD isthedistancebetween P1andtheparallel plane P2throughthevertex P .Hence, V = 1 3 A | a ( b c ) | b c = 1 6 | a ( b c ) | Sothevolumeofatetrahedronwithadjacentsides a b ,and c isonesixththevolumeoftheparallelepipedwiththesameadjacentsides. Notetheresultdoesnotdependonthechoiceofavertex.Anyvertex couldhavebeenchoseninsteadof B intheabovesolution.

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75.THETRIPLEPRODUCT47 75.4.Exercises.(1) Determinewhetherthepoints A =(1 2 3), B = (1 0 1), C =( Š 1 1 2),and D =( Š 2 1 0)areinoneplaneand,ifnot, “ndthevolumeoftheparallelepipedwithadjacentedges AB AC and AD (2) Find (i)allvaluesof s atwhichthepoints A ( s, 0 ,s ), B (1 0 1), C ( s,s, 1), and D (0 1 0)areinthesameplane (ii)allvaluesof s atwhichthevolumeoftheparallelepipedwithadjacentsides AB AC ,and AD is9units (3) Verifywhetherthevectors a = e1+2 e2Š e3, b =2 e1Š e2+ e3, and c =3 e1+ e2Š 2 e3arecoplanar. (4) Letthenumbers u v ,and w besuchthat uvw =1and u3+ v3+ w3=1.Arethevectors a = u e1+ v e2+ w e3, b = v e1+ w e2+ u e3, and c = w e1+ u e2+ v e3coplanar?Ifnot,whatisthevolumeofthe parallelepipedwithadjacentedges a b ,and c ? (5) Provethat ( a b ) ( c d )=det a cb c a db d Hint: UsetheinvarianceofthetripleproductunderacyclicpermutationofvectorsinitandStudyProblem11.16. (6) Letaset S1bethecircle x2+ y2=1andletaset S2bethe linethroughthepoints(0 2)and(2 0).Whatisthedistancebetween thesets S1and S2? (7) Consideraplanethroughthreepoints A =(1 2 3), B = (2 3 1),and C =(3 1 2).Findthedistancebetweentheplaneanda point P obtainedfrom A bymovingthelatter3unitsoflengthalong thevector a = Š 1 2 2 (8) Considertwolines.The“rstlinepassesthroughthepoints (1 2 3)and(2 Š 1 1),whiletheotherpassesthroughthepoints( Š 1 3 1) and(1 1 3).Findthedistancebetweenthelines. (9) Findthedistancebetweenthelinethroughthepoints(1 2 3) and(2 1 4)andtheplanethroughthepoints(1 1 1),(3 1 2),and (1 2 Š 1). Hint: Ifthelineisnotparalleltotheplane,thentheyintersectandthe distanceis0.Socheck“rstwhetherthelineisparalleltotheplane. Howcanthisbedone? (10) Considerthelinethroughthepoints(1 2 3)and(2 1 2).If asecondlinepassesthroughthepoints(1 ,1 ,s )and(2 Š 1 0),“ndall valuesof s ,ifany,atwhichthedistancebetweenthelinesis9 / 2units.

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4811.VECTORSANDTHESPACEGEOMETRY 76.PlanesinSpace76.1.AGeometricalDescriptionofaPlaneinSpace.Letaplane P gothroughapoint P0.Clearly,therearemanyplanesthatcontain aparticularpointinspace.Allsuchplanescanbeobtainedfroma particularplanebyageneralrotationaboutthepoint P0.Toeliminate thisfreedomandde“netheplaneuniquely,onecandemandthatevery lineintheplanebeperpendiculartoagivenvector n .Thisvectoris calleda normal oftheplane P .Thus,thegeometricaldescriptionofa plane P inspaceentailsspecifyingapoint P0thatbelongsto P anda normal n of P .76.2.AnAlgebraicDescriptionofaPlaneinSpace.Letaplane P be de“nedbyapoint P0thatbelongstoitandanormal n .Insome coordinatesystem,thepoint P0hascoordinates( x0,y0,z0)andthe vector n isspeci“edbyitscomponents n = n1,n2,n3 .Ageneric pointinspace P hascoordinates( x,y,z ).Analgebraicdescriptionof aplaneamountstospecifyingconditionsonthevariables( x,y,z )such thatthepoint P ( x,y,z )belongstotheplane P .Let r0= x0,y0,z0 and r = x,y,z bethepositionvectorsofaparticularpoint P0in theplaneandagenericpoint P inspace,respectively.Considerthe vector P0P = r Š r0= x Š x0,y Š y0,z Š z0 .Thisvectorliesinthe plane P ifandonlyifitisorthogonaltothenormal n ,accordingtothe geometricaldescriptionofaplane(seeFigure11.15,leftpanel).The algebraicconditionequivalenttothegeometricalone, n P0P ,reads n P0P =0.Thus,thefollowingtheoremhasjustbeenproved. Theorem 11.8 (EquationofaPlane) Apointwithcoordinates ( x,y,z ) belongstoaplanethroughapoint P0( x0,y0,z0) andnormaltoavector n = n1,n2,n3 ifandonlyif n1( x Š x0)+ n2( y Š y0)+ n3( z Š z0)=0or n r = n r0, where r and r0arepositionvectorsofagenericpointandaparticular point P0intheplane. Soageneralsolutionoftheequation n r = d ,where n isagiven vectorand d isagivennumber,isasetofpositionvectorsofallpoints oftheplanethatisperpendicularto n .Thenumber d determinesthe positionoftheplaneinspaceinthefollowingway.If r0istheposition vectorofaparticularpointintheplane,then d = n r0.Theposition vectorofanotherpointintheverysameplaneis r0+ a ,wherethe vector a isintheplane(aparticularpoint P0hasjustbeendisplaced intheplanealongthevector a ).Thenumber d isindependentofthe

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76.PLANESINSPACE49 Figure11.15. Left :Algebraicdescriptionofaplane. If r0isapositionvectorofaparticularpointintheplane and r isthepositionvectorofagenericpointinthe plane,thenthevector r Š r0liesintheplaneandis perpendiculartoitsnormal,thatis, n ( r Š r0)=0. Right :Equationsofparallelplanesdieronlybytheir constantterms.Thedierenceoftheconstantterms determinesthedistancebetweentheplanesasstatedin (11.12). choiceofaparticularpointintheplanebecause d = n ( r0+ a )= n r0andthevectors n and a areorthogonal, n a .Thenumber d changes ifthepoint P0ismovedalongthenormal n ,buttheresultofsucha displacementof P0isapointthatisnotintheoriginalplane.Thus, theequations n r = d1and n r = d2describetwo parallel planes if d1 = d2;thatis,variationsof d correspondtoshiftsoftheplane paralleltoitselfalongitsnormal(seeFigure11.15,rightpanel). Notealsothatthenormalvectorofagivenplaneisnotuniquely de“nedbecauseitsmagnitudeisirrelevantforthegeometricaldescriptionoftheplane.If n isanormal,then s n isalsoanormalofthesame planeforanynonzeroreal s .Inthealgebraicapproach,thescalingof n doesnotchangetheequationoftheplane,( s n ) r =( s n ) r0,or, bycancellingthescalingfactor s inthisequation, n r = n r0.Thus, twoplanesareparalleliftheirnormalsareparallel.Fromthealgebraic pointofview,twoplanesareparalleliftheirnormalsareproportional: P1P1 n1 n2 n1= s n2forsomereal s .

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5011.VECTORSANDTHESPACEGEOMETRY Definition 11.16 (AngleBetweenTwoPlanes) Theanglebetweenthenormalsoftwoplanesiscalledthe anglebetween theplanes If n1and n2arethenormals,thentheangle betweenthemis determinedby cos = n1 n2 n1 n2 = n1 n2. Notethataplaneasageometricalsetofpointsinspaceisnotchanged ifthedirectionofitsnormalisreversed(i.e., n Š n ).Sotherangeof canalwaysberestrictedtotheinterval[0 ,/ 2].Indeed,if happens tobeintheinterval[ / 2 ](i.e.,cos 0),thentheangle Š / 2can alsobeviewedastheanglebetweentheplanesbecauseonecanalways reversethedirectionofoneofthenormals n1Š n1or n2Š n2so thatcos Š cos Theplanesareperpendiculariftheirnormalsareperpendicular. Forexample,theplanes x + y + z =1and x +2 y Š 3 z =4are perpendicularbecausetheirnormals n1= 1 1 1 and n2= 1 2 Š 3 areperpendicular: n1 n2=1+2 Š 3=0(i.e., n1 n2). Example 11.8 Findanequationoftheplanethroughthreegiven points A (1 1 1) B (2 3 0) ,and C ( Š 1 0 3) Solution: Aplaneisspeci“edbyaparticularpoint P0initandby avector n normaltoit.Threepointsontheplanearegiven,soany ofthemcanbetakenas P0,forexample, P0= A or( x0,y0,z0)= (1 1 1).Avectornormaltoaplanecanbefoundasthecrossproduct ofanytwononparallelvectorsinthatplane(seeFigure11.16,left panel).Soput a = AB = 1 2 Š 1 and b = AC = Š 2 Š 1 2 .Then onecantake n = a b = 3 0 Š 3 .Anequationoftheplaneis 3( x Š 1)+0( y Š 1)+( Š 3)( z Š 1)=0,or x Š z =0.Sincetheequation doesnotcontainthevariable y ,theplaneisparalleltothe y axis. Notethatifthe y componentof n vanishes(i.e.,thereisno y inthe equation),then n isorthogonalto e2because n e2=0;thatis,the y axisisperpendicularto n andhenceparalleltotheplane. 76.3.TheDistanceBetweenaPointandaPlane.Considertheplane throughapoint P0andnormaltoavector n .Let P1beapointin space.Whatisthedistancebetween P1andtheplane?Lettheangle between n andthevector P0P1be (seeFigure11.16,rightpanel). Thenthedistanceinquestionis D = P0P1 cos if / 2(the lengthofthestraightlinesegmentconnecting P1andtheplanealong

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76.PLANESINSPACE51 Figure11.16. Left :IllustrationtoExample11.8.The crossproductoftwononparallelvectorsinaplaneisa normaloftheplane. Right :Distancebetweenapoint P1andaplane.An illustrationtothederivationofthedistanceformula (11.10).Thesegment P1B isparalleltothenormal n sothatthetriangle P0P1B isright-angled.Therefore, D = | P1B | = | P0P1| cos thenormal n ).For >/ 2,cos mustbereplacedby Š cos because D 0.So (11.10) D = P0P1| cos | = n P0P1| cos | n = | n P0P1| n Notethatthisdistanceformulacanbeobtainedfromthedistance betweentwoparallelplanes(Corollary11.7).Indeed,thevector AB AC isthecrossproductoftwovectorsintheplaneandhencecanbe usedasthenormal n ,whereasthevector AP canbeusedas P0P1. Let r0and r1bepositionvectorsof P0and P1,respectively.Then P0P1= r1Š r0,and (11.11) D = | n ( r1Š r0) | n = | n r1Š d | n whichisabitmoreconvenientforcalculatingthedistanceiftheplane isde“nedalgebraicallybyanequation n r = d .76.3.1.DistanceBetweenParallelPlanes.Equation(11.11)allowsus toobtainasimpleformulaforthedistancebetweentwoparallelplanes de“nedbytheequations n r = d1and n r = d2(seeFigure11.15, rightpanel): (11.12) D = | d1Š d2| n Indeed,thedistancebetweentwoparallelplanesisthedistancebetween the“rstplaneandanypoint r0inthesecondplane.By(11.11),this

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5211.VECTORSANDTHESPACEGEOMETRY distanceis D = | n r1Š d1| / n = | d2Š d1| / n because n r0= d2foranypointinthesecondplane. Example 11.9 Findanequationofaplanethatisparalleltothe plane 2 x Š y +2 z =2 andatadistanceof3unitsfromit. Solution: Thereareafewwaystosolvethisproblem.Fromthe geometricalpointofview,aplaneisde“nedbyaparticularpointinit anditsnormal.Sincetheplanesareparallel,theymusthavethesame normal n = 2 Š 1 2 .Notethatthecoecientsatthevariablesinthe planeequationde“nethecomponentsofthenormalvector.Therefore, theproblemisreducedto“ndingaparticularpoint.Let P0bea particularpointonthegivenplane.Thenapointonaparallelplane canbeobtainedfromitbyshifting P0byadistanceof3unitsalong thenormal n .If r0isthepositionvectorof P0,thenapointona parallelplanehasapositionvector r0+ s n ,wherethedisplacement vector s n musthavealengthof3,or s n = | s | n =3 | s | =3and therefore s = 1.Naturally,thereshouldbetwoplanesparallelto thegivenoneandatthesamedistancefromit.To“ndaparticular pointonthegivenplane,onecansettwocoordinatesto0and“nd thevalueofthethirdcoordinatefromtheequationoftheplane.Take, forinstance, P0(1 0 0).Particularpointsontheparallelplanesare r0+ n = 1 0 0 + 2 Š 1 2 = 3 Š 1 2 and,similarly, r0Š n = Š 1 1 Š 2 .Usingthesepointsinthestandardequationofaplane, theequationsoftwoparallelplanesareobtained: 2 x Š y +2 z =11and2 x Š y +2 z = Š 7 Analternativealgebraicsolutionisbasedonthedistanceformula (11.12)forparallelplanes.Anequationofaplaneparalleltothegiven oneshouldhavetheform2 x Š y +2 z = d .Thenumber d isdetermined bytheconditionthat | d Š 2 | / n =3or | d Š 2 | =9,or d = 9+2. 76.4.StudyProblems.Problem11.22. Findanequationoftheplanethatisnormalto astraightlinesegment AB andbisectsitif A =(1 1 1) and B = ( Š 1 3 5) Solution: Onehasto“ndaparticularpointintheplaneandits normal.Since AB isperpendiculartotheplane, n = AB = Š 2 2 4 Themidpointofthesegmentliesintheplane.Hence, P0(0 2 3)(the coordinatesofthemidpointsarethehalfsumsofthecorresponding coordinatesoftheendpoints).Theequationreads Š 2 x +2( y Š 2)+ 4( z Š 3)=0or Š x + y +2 z =8.

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76.PLANESINSPACE53 Problem11.23. Findanequationoftheplanethroughthepoint P0(1 2 3) thatisperpendiculartotheplanes x + y + z =1 and x Š y + 2 z =1 Solution: Onehasto“ndaparticularpointintheplaneandany vectorperpendiculartoit.The“rstpartoftheproblemiseasytosolve: P0isgiven.Let n beanormaloftheplaneinquestion.Then,fromthe geometricaldescriptionofaplane,itfollowsthat n n1= 1 1 1 and n n2= 1 Š 1 2 ,where n1and n2arenormalsofthegivenplanes.So n isavectorperpendiculartotwogivenvectors.Bythegeometrical propertyofthecrossproduct,suchavectorcanbeconstructedas n = n1 n2= 3 Š 1 Š 2 .Hence,theequationreads3( x Š 1) Š ( y Š 2) Š 2( z Š 3)=0or3 x Š y Š 2 z = Š 5. Problem11.24. Determinewhethertwoplanes x +2 y Š 2 z =1 and 2 x +4 y +4 z =10 areparalleland,ifnot,“ndtheanglebetweenthem. Solution: Thenormalsare n1= 1 2 Š 2 and n2= 2 4 4 = 2 1 2 2 (i.e.,theyarenotproportional).Hence,theplanesarenot parallel.Since n1 =3, n1 =6,and n1 n2=2,theangleis determinedbycos =2 / 18=1 / 9or =cosŠ 1(1 / 9). 76.5.Exercises.(1) Findanequationoftheplanethroughtheorigin andparalleltotheplane2 x Š 2 y + z =4.Whatisthedistancebetween thetwoplanes? (2) Dotheplanes2 x + y Š z =1and4 x +2 y Š 2 z =10intersect? (3) Consideraparallelepipedwithonevertexattheoriginatwhich theadjacentsidesarethevectors a = 1 2 3 b = 2 1 1 ,and c = Š 1 0 1 .Findequationsoftheplanesthatcontainthefacesofthe parallelepiped. (4) Determinewhethertheplanes2 x + y Š z =3and x + y + z =1 areintersecting.Iftheyare,“ndtheanglebetweenthem. (5) Findanequationoftheplanewith x intercept a y intercept b ,and z intercept c .Whatisthedistancebetweentheoriginandthe plane? (6) Findanequationforthesetofpointsthatareequidistantfrom thepoints(1 2 3)and( Š 1 2, 1).Giveageometricaldescriptionofthe set. (7) Findanequationoftheplanethatisperpendiculartothe plane x + y + z =1andcontainsthelinethroughthepoints(1 2 3) and( Š 1 1 0). (8) Towhichoftheplanes x + y + z =1and x +2 y Š z =2isthe point(1 2 3)theclosest?

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5411.VECTORSANDTHESPACEGEOMETRY (9) Giveageometricaldescriptionofthefollowingfamiliesofplanes: (i) x + y + z = c (ii) x + y + cz =1 (iii) x sin c + y cos c + z =1 where c isaparameter. (10) Considerthreeplaneswithnormals n1, n2,and n3suchthat eachpairoftheplanesisintersecting.Underwhatconditiononthe normalsarethethreelinesofintersectionparallelorevencoincide?77.LinesinSpace77.1.AGeometricalDescriptionofaLineinSpace.Considertwopoints inspace.Theycanbeconnectedbyapath.Amongallthecontinuous pathsthatconnectthetwopoints,thereisadistinctone,namely,the onethathasthesmallestlength.Thispathiscalleda straightline segment Definition 11.17 (GeometricalDescriptionofaLine) Aline L isasetofpointsinspacesuchthattheshortestpathconnecting anypairofpointsof L belongsto L Givenapoint P0,therearein“nitelymanylinesthrough P0,allof whicharerelatedbyrigidrotationsaboutthepoint P0.Therefore,to “xalineuniquely,oneshouldspecifyadirectiontowhichthelineis parallel,inadditiontoitsposition P0.Thedirectioncanbedetermined byavector v .Itfollowsfromthegeometricaldescriptionofalinethat v isavectorconnectinganytwopointsoftheline.Thelengthornorm of v isirrelevantforspecifyingthedirection;thatis,anyparallelvector (ortheorientedsegmentbetweenanotherpairofpointsoftheline)is justasgoodas v .Thus,aline L isuniquelyspeci“edbyaparticular point P0of L andanyvector v towhichthelineisparallel, v L Remark. Theverynotionofaline,de“nedastheshortestpathbetweentwopointsinspace,isdeeplyrootedintheverystructureof spaceitself.Howcanalineberealizedinthespaceinwhichwelive? Onecanuseapieceofrope,asintheancientworld,ortheline ofsightŽ(i.e.,thepathtraveledbylightfromonepointtoanother). EinsteinstheoryofgravitystatesthatstraightlinesŽde“nedastrajectoriestraversedbylightarenotexactlythesameasstraightlinesŽ inaEuclideanspace.SoaEuclideanspacemayonlybeviewedasa mathematicalapproximation(ormodel)ofourspace.Agoodanalogy wouldbetocomparetheshortestpathsinaplaneandonthesurfaceof asphere;theyarenotthesameasthelatterarecurved.ŽTheconcept ofcurvatureofapathisdiscussedinthenextchapter.Theshortest

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77.LINESINSPACE55 pathbetweentwopointsinaspaceiscalleda geodesic (byanalogy withtheshortestpathonthesurfaceoftheEarth).Thegeodesics ofaEuclideanspace(straightlines)donothavecurvature,whereas thegeodesicsofourspace(i.e.,thepathstraversedbylight)dohave curvaturethatisdeterminedbythedistributionofgravitatingmasses (planets,stars,etc.).Adeviationofthegeodesicsfromstraightlines nearthesurfaceoftheEarthisveryhardtonotice.However,adeviationofthetrajectoryoflightfromastraightlinehasbeenobservedfor thelightcomingfromadistantstartotheEarthandpassingnearthe Sun.Einsteinstheoryofgeneralrelativityassertsthatabettermodel ofourspaceisa Riemannspace .Asucientlysmallneighborhoodin aRiemannspacelookslikeaportionofaEuclideanspace.77.2.AnAlgebraicDescriptionofaLine.Insomecoordinatesystem,a particularpointofaline L hascoordinates P0( x0,y0,z0),andavectorparallelto L isde“nedbyitscomponents, v = v1,v2,v3 .Let r = x,y,z beapositionvectorofagenericpointof L andlet r0= x0,y0,z0 bethepositionvectorof P0.Thevector r Š r0must beparallelto L becauseitistheorientedstraightlinesegmentconnectingtwopointsof L (seeFigure11.17,leftpanel).Hence,apoint ( x,y,z )belongsto L ifandonlyif r Š r0 v .Theequivalentalgebraic conditionreads r Š r0= t v forsomereal t .Toobtainallpointsof L oneshouldlet t rangeoverallrealnumbers. Figure11.17. Left :Algebraicdescriptionofaline L through r0andparalleltoavector v .If r0and r are positionvectorsofparticularandgenericpointsofthe line,thenthevector r Š r0isparalleltothelineandhence mustbeproportionaltoavector v ,thatis, r Š r0= t v forsomerealnumber t Right :Distancebetweenapoint P1andaline L through apoint P0andparalleltoavector v .Itistheheightof theparallelogramwhoseadjacentsidesarethevectors P0P1and v .

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5611.VECTORSANDTHESPACEGEOMETRY Theorem 11.9 (EquationsofaLine) Thecoordinatesofthepointsoftheline L throughapoint P0and paralleltoavector v satisfythevectorequation (11.13) r = r0+ t v Š
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77.LINESINSPACE57 77.3.RelativePositionsofLinesinSpace.Twolinesinspacecanbe intersecting,parallel,orskew.Givenanalgebraicdescriptionofthe lines,howcanone“ndoutwhichoftheabovethreecasesoccurs?If thelinesareintersecting,howcanone“ndthecoordinatesofthepoint ofintersection?Supposetheline L1containsapoint P1andisparallel to v1,while L2containsapoint P2andisparallelto v2. Corollary 11.6 (CriterionforTwoLinestoParallel) Twolinesareparallelifandonlyiftheirdirectionvectors v1and v2areparallel: L1L2 v1 v2 v1= s v2forsomereal s. Supposethatthelinesarenotparallel.Thentheyareeitherskew orintersecting.Inthelattercase,thedistancebetweenthelinesis0as theyhaveacommonpoint(seeSection75),whereasintheformercase thedistancecannotbe0.Sincethelinesarenotparallel, v1 v2 = 0 Makinguseofthedistanceformulabetweenskewlines(seeCorollary 11.5),oneprovesthefollowing. Corollary 11.7 (CriteriaforTwoNon-parallelLinestoBeSkewor Intersecting) Let P1beapointof L1andlet P2beapointof L2.Let v1L1and v2L2andletthelines L1and L2benonparallel.Then P1P2 ( v1 v2) =0 L1, L2areskew P1P2 ( v1 v2)=0 L1, L2areintersecting Let L1and L2beintersecting.Howcanone“ndthepointof intersection?Tosolvethisproblem,considerthevectorequationsfor thelines rt= r1+ t v1and rs= r2+ s v2.Whenchangingtheparameter t ,theterminalpointof rtslidesalongtheline L1,whiletheterminal pointof rsslidesalongtheline L2whenchangingtheparameter s as depictedinFigure11.18(leftpanel).Notethattheparametersofboth linesarenotrelatedinanywayaccordingtothegeometricaldescription ofthelines.Iftwolinesareintersecting,thenthereshouldexistapair ofnumbers( t,s )=( t0,s0)atwhichtheterminalpointsofvectors rtand rscoincide, rt= rs.Let vi= ai,bi,ci i =1 2.Writingthis vectorequationincomponentform,thefollowingsystemofequations isobtained: x1+ ta1= x2+ sa2, y1+ tb1= x2+ sb2, z1+ tc1= x2+ sc2.

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5811.VECTORSANDTHESPACEGEOMETRY Figure11.18. Left :Intersectionpointoftwolines L1and L2.Theterminalpointofthevector rttraverses L1as t rangesoverallrealnumbers,whiletheterminal pointofthevector rstraverses L2as s rangesoverallreal numbersindependentlyof t .Ifthelinesareintersecting, thenthereshouldexistapairofnumbers( t,s )=( t0,s0) suchthatthevectors rtand rscoincide,whichmeans thattheircomponentsmustbethesame.Thisde“nes threeequationsontwovariables t and s Right :Intersectionpointofaline L andaplane P .The terminalpointofthevector rttraverses L as t ranges overallrealnumbers.Ifthelineintersectstheplane de“nedbytheequation r n = d ,thenthereshouldexist aparticularvalueof t atwhichthevector rtsatis“esthe equationoftheplane: rt n = d Thesystemhasthreeequationsforonlytwovariables.Itisan overdetermined system,whichmayormaynothaveasolution.Fromthe abovegeometricalanalysis,itfollowsthat,ifthelinesareparallel(i.e., v1 v2= 0 ),thenthesystemhasnosolution(thelinesaredistinct), oritmighthavein“nitelymanysolutions(thelinescoincide).For example,put P1= P2and v1=2 v2.Thenthesystemissatis“edby anypair( t,s =2 t ),where t isanyreal.Thesystemhasnosolutionif thecriterionfortwolinestobeskewissatis“ed.Finally,thesystem musthavetheonlysolutionifthecriterionfortwononparallellinesto beintersectingissatis“ed.Let( t,s )=( t0,s0)beasolution.Thenthe positionvectorofthepointofintersectionis r1+ t0v1or r2+ s0v2.77.4.RelativePositionsofLinesandPlanes.Consideraline L anda plane P .Thequestionofinterestistodeterminewhethertheyare intersectingorparallel.Ifthelinedoesnotintersecttheplane,then theymustbeparallel.Inthelattercase,thelinemustbeperpendicular tothenormaloftheplane.

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77.LINESINSPACE59 Corollary 11.8 (CriterionforaLineandaPlanetoBeParallel) Let v beavectorparalleltoaline L andlet n beanormalofaplane P .Then LP v n v n =0 Ifthelineintersectstheplane,thenthereshouldexistaparticular valueoftheparameter t forwhichthepositionvector rt= r0+ t v ofapointof L alsosatis“estheplaneequation r n = d (seeFigure 11.18,rightpanel).Thevalueoftheparameter t thatcorrespondsto thepointofintersectionisdeterminedbytheequation rt n = d r0 n + t v n = d t = d Š r0 n v n Thepositionvectorofthepointofintersectionisfoundbysubstituting thisvalueof t intothevectorequationoftheline rt= r0+ t v Example 11.11 Findanequationoftheplane P thatisperpendiculartotheplane P1, x + y Š z =1 ,andcontainstheline x Š 1= y/ 2= z +1 Solution: Theplane P mustbeparalleltotheline( P containsit) andthenormal n1= 1 1 Š 1 of P1(as PP1).Sothenormal n of P isperpendiculartoboth n1andthevector v = 1 2 1 thatisparallel totheline.Therefore,onecantake n = n1 v = 3 Š 2 1 .To“nd aparticularpointof P ,notethatthepointofintersectionof P1and thelinebelongstotheplane P .Thelinecontainsthepoint(1 0 Š 1). Put rt= r0+ t v = 1+ t, 2 t, Š 1+ t .Theequation rt n1=1or 2+2 t =1hasthesolution t = Š 1 / 2.Hence,thepositionvectorofa particularpointof P is rt = Š 1 / 2= 1 / 2 Š 1 Š 3 / 2 .Anequationof P reads3( x +1 / 2) Š 2( y +1)+( z +3 / 2)=0or3 x Š 2 y + z = Š 1. 77.5.StudyProblems.Problem11.25. Let L1bethelinethrough P1(1 1 1) andparallelto v1= 1 2 Š 1 andlet L2bethelinethrough P2(4 0 Š 2) andparallel to v2= 2 1 0 .Determinewhetherthelinesareparallel,intersecting, orskewand“ndtheline L thatisperpendiculartoboth L1and L2and intersectsthem. Solution: Thevectors v1and v2arenotproportional,andhence thelinesarenotparallel.Onehas P1P2= 3 Š 1 Š 3 and v1 v2= 1 Š 2 Š 3 .Therefore, P1P2 ( v1 v2)=14 =0,andthelinesare skewbyCorollary11.7.To“ndtheline L ,notethatithastocontain onepointofeachline.Let rt= r1+ t v1beapositionvectorofapoint of L1andlet rs= r2+ s v2beapositionvectorofapointof L2as

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6011.VECTORSANDTHESPACEGEOMETRY Figure11.19. Left :IllustrationtoStudyProblem 11.25.Thevectors rsand rttraceouttwogivenskewed lines L1and L2,respectively.Thereareparticularvalues of t and s atwhichthedistance rtŠ rs becomesminimal.Therefore,theline L throughsuchpoints rtand rsisperpendiculartoboth L1and L2. Right :Intersectionofaline L andasphere S .AnillustrationtoStudyProblem11.26.Theterminalpoint ofthevector rttraversesthelineas t rangesoverallreal numbers.Ifthelineintersectsthesphere,thenthere shouldexistaparticularvalueof t atwhichthecomponentsofthevector rtsatisfytheequationofthesphere. Thisequationisquadraticin t ,andhenceitcanhave twodistinctrealroots,onemultiplerealroot,ornoreal roots.Thesethreecasescorrespondtotwo,one,orno pointsofintersection.Oneintersectionpointmeansthat thelineistangenttothesphere. showninFigure11.19(leftpanel).Astheline L shouldintersectboth L1and L2,thereshouldexistapairofvalues( t,s )oftheparameters atwhichthevector rsŠ rtisparallelto L ;thatis,thevector rsŠ rtbecomesperpendiculartobothvectors v1and v2.Thecorresponding algebraicconditionsare rsŠ rt v1 ( rsŠ rt) v1=4+4 s Š 6 t =0 rsŠ rt v1 ( rsŠ rt) v2=5+5 s Š 4 t =0 Thissystemhasthesolution t =0and s = Š 1.Thus,thepointswith thepositionvectors rt =0= r1and rs = Š 1= r2Š v2= 2 Š 1 Š 2 belong to L .Sothevector v = rs = Š 1Š rt =0= 1 Š 3 Š 1 isparallelto L .

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77.LINESINSPACE61 Takingaparticularpointof L tobe P1,theparametricequationsread x =1+ t y =1 Š 3 t z =1 Š t Problem11.26. Consideralinethroughtheoriginthatisparallelto thevector v = 1 1 1 .Findtheportionofthislinethatliesinsidethe sphere x2+ y2+ z2Š x Š 2 y Š 3 z =9 Solution: Theparametricequationsofthelineare x = t y = t z = t .Ifthelineintersectsthesphere,thenthereshouldexistparticular valuesof t atwhichthecoordinatesofapointofthelinealsosatisfythe sphereequation(seeFigure11.19,rightpanel).Ingeneral,parametric equationsofalinearelinearin t ,whileasphereequationisquadraticin thecoordinates.Therefore,theequationthatdeterminesthevaluesof t correspondingtothepointsofintersectionisquadratic.Aquadratic equationhastwo,one,ornorealsolutions.Accordingly,thesecases correspondtotwo,one,andnopointsofintersection,respectively.In ourcase,3 t2Š 6 t =9or t2Š 2 t =3andhence t = Š 1and t =3.The pointsofintersectionare( Š 1 Š 1 Š 1)and(3 3 3).Thelinesegment connectingthemcanbedescribedbytheparametricequations x = t y = t z = t ,where Š 1 t 3. 77.6.Exercises.(1) Findparametricequationsofthelinethrough thepoint(1 2 3)andperpendiculartotheplane x + y +2 z =1.Find thepointofintersectionofthelineandtheplane. (2) Findparametricandsymmetricequationsofthelineofintersectionoftheplanes x + y + z =1and2 x Š 2 y + z =1. (3) Isthelinethroughthepoints(1 2 3)and(2 Š 1 1)perpendiculartothelinethroughthepoints(0 1 Š 1)and(1 0 2)?Arethelines intersecting?Ifso,“ndthepointofintersection. (4) Determinewhetherthelines x =1+2 t y =3 t z =2 Š t and x +1= y Š 4=( z Š 1) / 3areparallel,skew,orintersecting.Ifthey intersect,“ndthepointofintersection. (5) Findthevectorequationofthestraightlinesegmentfromthe point(1 2 3)tothepoint( Š 1 1 2). (6) Let r1and r2bepositionvectorsoftwopointsinspace.Find thevectorequationofthestraightlinesegmentfrom r1to r2. (7) Considertheplane x + y + z =0andapoint P =(1 2 Š 3)in it.Findparametricequationsofthelinesthroughtheoriginthatare atadistanceof1unitfrom P (8) Findparametric,symmetric,andvectorequationsoftheline through(0 1 2)thatisperpendicularto v = 1 2 1 andparallelto theplane x +2 y + z =3.

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6211.VECTORSANDTHESPACEGEOMETRY (9) Findparametricequationsofthelinethatisparallelto v = 2 Š 1 2 andgoesthroughthecenterofthesphere x2+ y2+ z2= 2 x +6 z Š 6.Restricttherangeoftheparametertodescribetheportion ofthelinethatisinsidethesphere. (10) Lettheline L1passthroughthepoint A (1 1 0)parallelto thevector v = 1 Š 1 2 andlettheline L2passthroughthepoint B (2 0 2)paralleltothevector w = Š 1 1 2 .Showthatthelinesare intersecting.Findthepoint C ofintersectionandparametricequations oftheline L3through C thatisperpendicularto L1and L2.78.QuadricSurfaces Definition 11.18 (QuadricSurface) Thesetofpointswhosecoordinatesinarectangularcoordinatesystem satisfytheequation Ax2+ By2+ Cz2+ pxy + qxz + vyz + x + y + z + D =0 where A B C p q v ,and D arerealnumbers,iscalleda quadricsurface Asphereprovidesasimpleexampleofaquadricsurface: x2+ y2+ z2Š R2=0,thatis, A = B = C =1, p = q = v =0, = = and D = Š R2,where R istheradiusofthesphere.Theequationthat de“nesquadricsurfacesisthemostgeneralequation quadratic inallthe coordinates.Thisiswhysurfacesde“nedbyitarecalled quadric .The taskhereistoclassifyalltheshapesofquadricsurfaces.Theshape doesnotchangeunderitsrigidrotationsandtranslations.Onthe otherhand,theequationthatdescribestheshapewouldchangeunder rotationsandtranslationsofthecoordinatesystem(seeSection71 andExample11.2).Thefreedominchoosingthecoordinatesystem canbeusedtosimplifytheequationforquadricsurfaceandobtaina classi“cationofdierentshapesdescribedbyit.78.1.QuadricCylinders.Consider“rstasimplerprobleminwhichthe equationofaquadricsurfacedoesnotcontainoneofthecoordinates, say, z (i.e., C = q = v = =0).Thentheset S S = ( x,y,z ) Ax2+ By2+ pxy + x + y + D =0 isthesamecurveineveryhorizontalplane z =const.Forexample, if A = B =1, p =0,and D = Š R2,thecrosssectionofthesurface S byanyhorizontalplaneisacircle x2+ y2= R2.Sothesurface S isacylinderofradius R thatissweptbythecirclewhenthelatter isshiftedupanddownparalleltothe z axis.Similarly,ageneral

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78.QUADRICSURFACES63 cylindricalshapeisobtainedbyshiftingacurveinthe xy planeupand downparalleltothe z axis.Thetaskhereistoclassifyallpossible shapesofquadriccylinders. InExample11.2,itwasestablishedthat,underacounterclockwise rotationofthecoordinatesystemthroughanangle x x cos + y sin and y y cos Š x sin .Bysubstitutingthetransformedcoordinatesintotheequationfor S ,oneobtainstheequationforthe very same shapeinthenewrotatedcoordinatesystem.Thefreedomof choosingtherotationangle canbeusedtosimplifytheequation.In particular,itisalwayspossibletoadjust sothatinthenewcoordinatesystemtheequationfor S doesnotcontainthemixedŽterm xy Indeed,afterthesubstitutionofthetransformedcoordinatesintothe equation,thecoecientat xy de“nesanew p : p 2( A Š B )cos sin + a (cos2 Š sin2 )=( A Š B )sin(2 ) + a cos(2 )= p. Therefore,theterm xy disappearsfromtheequationiftheangle satis“esthecondition p=0or (11.15)tan(2 )= q B Š A and = 4 if A = B. Thecoecients A and B (thefactorsat x2and y2)and and (the factorsat x and y )aretransformedas A 1 2[ A + B +( A Š B )cos(2 ) Š a sin(2 )]= A, B 1 2[ A + B Š ( A Š B )cos(2 )+ a sin(2 )]= B, cos Š sin = cos + sin = where satis“es(11.15). Dependingonthevaluesof A B ,and p ,thefollowingthreecases canoccur.First, A= B=0,whichisonlypossibleif A = B = p =0. Inthiscase, S isde“nedbytheequation x + y + D =0,whichis aplaneparalleltothe z axis. Second,onlyoneof Aand Bvanishes,say, B=0(notethatfor establishingtheshapeitisirrelevanthowthehorizontalandvertical coordinatesinthe xy planearecalled).Inthiscase,theequationfor S assumestheform Ax2+ x + y + D =0or,bycompletingthe squares, A x Š x02+( y Š y0)=0 ,x0= 2 A,y0= 1 Ax2 0Š D ;

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6411.VECTORSANDTHESPACEGEOMETRY hereitisassumedthat =0(otherwise,theequationdoesnotde“ne acurveinthe xy planeandhence S isnotasurface).Thisequation de“nesaparabola y = Ax2, A= A/,afterthe translation ofthe coordinatesystem: x x + x0and y y + y0(seeSection71). Third,both Aand Bdonotvanish.Then,afterthecompletion ofsquares,theequationcanbebroughttotheform A( x Š x0)2+ B( y Š y0)2= D, where x0= Š 2 A,y0= Š 2 B,D= Š D + 1 2 ( Ax2 0+ By2 0) Finally,afterthetranslationoftheorigintothepoint( x0,y0),the equationbecomes Ax2+ By2= D. If D=0,thenthisequationde“nestwostraightlines y = mx ,where m =( Š A/B)Š 1 / 2,provided Aand Bhaveoppositesigns(otherwise, theequationhasnosolution).If D =0,thentheequationcanbe writtenas( A/D) x2+( B/D) y2=1.Onecanalwaysassumethat A/D> 0(astheshapeofthecurveisindependentofhowthecoordinateaxesarecalled).Notealsothattherotationofthecoordinate systemthroughtheangle / 2swapstheaxes,( x,y ) ( y, Š x ),which canbeusedtoreversethesignof A/D.Nowput A/D=1 /a2and B/D= 1 /b2(dependingonwhether B/Dispositiveornegative) sothattheequationbecomes x2 a2 y2 b2=1 Whentheplusistaken,thisequationde“nesanellipse.Whenthe minusistaken,thisequationde“nesahyperbola. Theaboveresultsaresummarizedinthefollowingtheorem(see Figure11.20). Theorem 11.10 (Classi“cationofQuadricCylinders) Ageneralequationforquadriccylinders S = ( x,y,z ) Ax2+ By2+ pxy + x + y + D =0 canbebroughttooneofthefollowingstandardformsbyasuitable rotationandtranslationofthecoordinatesystem: y Š ax2=0(paraboliccylinder) x2 a2+ y2 b2=1(ellipticcylinder) ,

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78.QUADRICSURFACES65 Figure11.20. Left :Paraboliccylinder.Thecrosssectionbyanyhorizontalplane z =constisaparabola y = ax2. Middle :Anellipticcylinder.Thecrosssectionbyany horizontalplane z =constisanellipse x2/a2+ y2/b2=1. Right :Ahyperboliccylinder.Thecrosssectionby anyhorizontalplane z =constisahyperbola x2/a2Š y2/b2=1. x2 a2Š y2 b2=1(hyperboliccylinder) provided A B ,and p donotvanishsimultaneously.If A = B = p =0 then S isaplane.78.2.Classi“cationofGeneralQuadricSurfaces.Theclassi“cationof generalquadricsurfacescanbecarriedoutinthesameway(i.e.,by simplifyingthegeneralquadraticequationbymeansofrotationsand translationsofthecoordinatesystem).First,onecanprovethatthere existsarotationofthecoordinatesystemsuchthatinthenewcoordinatesystemthequadraticequationdoesnothavemixedŽterms: p p=0, q q=0,and v v=0.Dependingonhowmanyof thecoecients A, B,and Cdonotvanish,someofthelineartermsor allofthemcanbeeliminatedbytranslationsofthecoordinatesystem. Thecorrespondingtechnicalitiesrequireasubstantialuseoflinearalgebramethods,whichgoesbeyondthescopeofthiscourse.Sothe “nalresultisgivenwithoutaproof. Theorem 11.11 (Classi“cationofQuadricSurfaces) Byrotatingandtranslatingarectangularcoordinatesystem,ageneralequationforquadricsurfacescanbebroughteithertooneofthe

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6611.VECTORSANDTHESPACEGEOMETRY Figure11.21. Left :Anellipsoid.Acrosssectionby anycoordinateplaneisanellipse. Right :Anellipticdoublecone.Acrosssectionbyahorizontalplane z =constisanellipse.Acrosssectionby anyverticalplanethroughthe z axisistwolinesthrough theorigin. standardequationsforquadriccylindersortooneofthefollowingsix forms: x2 a2+ y2 b2+ z2 c2=1(ellipsoid) z2 c2= x2 a2+ y2 b2(ellipticdoublecone) x2 a2+ y2 b2Š z2 c2=1(hyperboloidofonesheet) Š x2 a2Š y2 b2+ z2 c2=1(hyperboloidoftwosheets) z c = x2 a2+ y2 b2(ellipticparaboloid) z c = x2 a2Š y2 b2(hyperbolicparaboloid) Itshouldbenotedagainthattheshapeofthesurfacedoesnot dependonhowthecoordinateaxesarecalled.Sotheshapedoes notchangeunderanypermutationofthecoordinates( x,y,z )inthe standardequations;onlytheorientationoftheshaperelativetothe

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78.QUADRICSURFACES67 Figure11.22. Left :Ahyperboloidofonesheet.A crosssectionbyahorizontalplane z =constisanellipse. Acrosssectionbyaverticalplane x =constor y =const isahyperbola. Right :Ahyperboloidoftwosheets.Anonemptycross sectionbyahorizontalplaneisanellipse.Acrosssection byaverticalplaneisahyperbola. Figure11.23. Left :Anellipticparaboloid.A nonemptycrosssectionbyahorizontalplaneisanellipse.Acrosssectionbyaverticalplaneisaparabola. Right :Ahyperbolicparaboloid(asaddleŽ).Across sectionbyahorizontalplaneisahyperbola.Across sectionbyaverticalplaneisaparabola.

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6811.VECTORSANDTHESPACEGEOMETRY coordinatesystemchanges.Forexample,theequations x2+ y2= R2and y2+ z2= R2describeacylinderofradius R ,butintheformer casethecylinderaxiscoincideswiththe z axis,whilethecylinderaxis isthe x axisinthelattercase.78.3.VisualizationofQuadricSurfaces.Theshapeofaquadricsurface canbeunderstoodbystudyingintersectionsofthesurfacewiththe coordinateplanes x = x0, y = y0,and z = z0.Theseintersectionsare alsocalled traces AnEllipsoid .If a2= b2= c2= R2,thentheellipsoidbecomesa sphereofradius R .So,intuitively,anellipsoidisaspherestretchedŽ alongthecoordinateaxes.Tracesofanellipsoidintheplanes x = x0, | x0| a .Traces intheplanes y = y0and z = z0arealsoellipsesandexistonlyif | y0| b and | z0| c .Thus, thecharacteristicgeometricalpropertyof anellipsoidisthatitstracesareellipses AParaboloid .Suppose c> 0.Thentheparaboloidliesabove the xy planebecauseithasnotraceinallhorizontalplanesbelowthe xy plane, z = z0< 0.Inthe xy plane,itstracecontainsjustthe origin.Similarly,aparaboloidwith c> 0liesbelowthe xy plane. Horizontaltraces(intheplanes z = z0)oftheparaboloidareellipses, x2/a2+ y2/b2= k ,where k = z0/c .Theellipsesbecomewideras z0getslarger( c> 0).Forthesakeofsimplicity,put c =1.Vertical traces(tracesintheplanes x = x0or y = y0)areparabolas z Š k = y2/b2,where k = x2 0/a2,or z Š k = x2/a2,where k = y2 0/b2.So the characteristicgeometricalpropertyofaparaboloidisthatitshorizontal tracesareellipses,whileitsverticalonesareparabolas .If a = b ,the ellipsoidisalsocalleda circularellipsoid becauseitshorizontaltraces arecircles. ADoubleCone .Horizontaltracesareellipses x2/a2+ y2/b2= k where k = z2 0/c2.Theybecomewideras | z0| grows,thatis,asthe horizontalplanemoves away fromthe xy plane( z =0).Inthe xy plane,theconehasatracethatconsistsofasinglepoint(theorigin). Theverticaltracesintheplanes x =0or y =0areapairoflines z = ( c/b ) y or z = ( c/a ) x .Verticaltracesintheplanes x = x0 =0 or y = y0 =0arehyperbolas y2/b2Š z2/c2= k ,where k = Š x2 0/a2,or x2/a2Š z2/c2= k ,where k = Š y2 0/b2.So thecharacteristicgeometrical propertyofaconeisthathorizontaltracesareellipses;itsvertical tracesareeitherapairoflines,iftheplanecontainstheconeaxis,

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78.QUADRICSURFACES69 orhyperbolas .If a = b ,theconeiscalleda circularcone .Inthis case,verticaltracesintheplanescontainingtheconeaxisareapair oflineswiththesameslope c/a = c/b .Theangle betweentheaxis oftheconeandanyoftheselinesde“nestheconeuniquelybecause c/a =cot ,andtheequationoftheconecanbewrittenas z2=cot2( )( x2+ y2) 0 < 0orin x< 0(i.e.,theydonotintersectthe y axis).If z0> 0,thenthehyperbolasaresymmetricaboutthe y axis,andtheir brancheslieeitherin y> 0orin y< 0(i.e.,theydonotintersect the x axis).Verticaltracesintheplanes x = x0areupwardparabolas,whereasintheplanes y = y0theyaredownwardparabolas.The hyperbolicparaboloidhasthecharacteristicshapeofasaddle.Ž AHyperboloidofOneSheet Thecharacteristicgeometrical propertyofahyperboloidofonesheetisthatitshorizontaltracesare ellipsesanditsverticaltracesarehyperbolas .Everyhorizontalplane hasatraceofthehyperboloid,andthesmallestoneisinthe xy plane (anellipsewithsemiaxes a and b ).Thesemiaxesoftheellipsesincrease astheplanemoves away fromthe xy plane. AHyperboloidofTwoSheets .Adistinctivefeatureofthis surfaceisthatitconsistsoftwosheets.Indeed,thehyperboloidhas notraceinthehorizontalplanes z = z0if Š cc or z = z0< Š c areellipses. Theuppersheettouchestheplane z = c atthepoint(0 0 ,c ),whilethe lowersheettouchestheplane z = Š c atthepoint(0 0 Š c ).Vertical tracesarehyperbolas.Sothecharacteristicgeometricalpropertiesof hyperboloidsofonesheetandtwosheetsaresimilar,apartfromthefact thatthelatteroneconsistsoftwosheets.Also,intheasymptoticregion | z | c ,thehyperboloidsapproachthesurfaceofthedoublecone. Indeed,inthiscase, z2/c2 1,andhencetheequations x2/a2+ y2/b2= 1+ z2/c2canbewellapproximatedbythedouble-coneequation( 1 canbeneglectedontherightsideoftheequations).Intheregion z> 0,thehyperboloidofonesheetapproachesthedoubleconefrom below,whilethehyperboloidoftwosheetsapproachesitfromabove.

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7011.VECTORSANDTHESPACEGEOMETRY For z< 0,theconverseholds.Inotherwords,thehyperboloidoftwo sheetsliesinsideŽthecone,whilethehyperboloidofonesheetlies outsideŽit.78.4.StudyProblems.Problem11.27. Classifythequadricsurface 3 x2+3 y3Š 2 xy =4 Solution: Theequationdoesnotcontainonevariable(the z coordinate).Thesurfaceisacylinder.Todeterminethetypeofcylinder,considerarotationofthecoordinatesysteminthe xy planeand choosetherotationanglesothatthecoecientatthemixedŽterm vanishes.Accordingto(11.15), A = B =3andhence = / 4.Then A=( A + B Š a ) / 2=4and B=( A + B + a ) / 2=2.So,inthenew coordinates,theequationbecomes x2+ y2/ 2=1,whichisanellipse withsemiaxes a =1and b = 2.Thesurfaceisanellipticcylinder. Problem11.28. Classifythequadricsurface x2Š 2 x + y + z =0 Solution: Bycompletingthesquares,theequationcanbetransformedintotheform( x Š 1)2+( y Š 1)+ z =0.Aftershiftingtheorigin tothepoint(1 1 0),theequationbecomes x2+ y Š z =0.Considerrotationsofthecoordinatesystemaboutthe x axis: y cos y +sin z z cos z Š sin y .Underthisrotation, y Š z (cos +sin ) y + (sin Š cos ) z .Therefore,for = / 4,theequationassumesoneof thestandardforms x2+ 2 y =0,whichcorrespondstoaparabolic cylinder. Problem11.29. Classifythequadricsurface x2+ z2Š 2 x +2 z Š y =0 Solution: Bycompletingthesquares,theequationcanbetransformedintotheform( x Š 1)2+( z +1)2Š ( y +2)=0.Thelatter canbebroughtintooneofthestandardformsbyshiftingtheoriginto thepoint(1 Š 2 Š 1): x2+ z2= y ,whichisacircularparaboloid.Its symmetryaxisisparalleltothe y axis(thelineofintersectionofthe planes x =1and z = Š 1)anditsvertexis(1 Š 2 Š 1). Problem11.30. Sketchand/ordescribethesetofpointsinspace formedbyafamilyoflinesthroughthepoint (1 2 0) andparallelto v= cos sin 1 ,where [0 2 ] labelslinesinthefamily. Solution: Theparametricequationsofeachlineare x =1+ t cos y =2+ t sin ,and z = t .Therefore,( x Š 1)2+( y Š 2)2= z2forall valuesof t and .Thus,thelinesformadoubleconewhoseaxisis paralleltothe z axisandwhosevertexis(1 2 0).Alternatively,one couldnoticethatthevector vrotatesaboutthe z axisas changes.

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78.QUADRICSURFACES71 Figure11.24. AnillustrationtoStudyProblem11.30. Thevector urotatesabouttheverticallinesothatthe linethrough(1 2 0)andparallelto vsweepsadouble conewiththevertexat(1 2 0). Indeed,put v= u + ez,where u = cos sin 0 istheunitvectorin the xy planeasshowninFigure11.24.Itrotatesas changes,making afullturnas increasesfrom0to2 .Sothesetinquestioncanbe obtainedbyrotatingaparticularline,say,theonecorrespondingto =0,abouttheverticallinethrough(1 2 0).Thelinesweepsthe doublecone. 78.5.Exercises.(1) Usetracestosketchandidentifyeachofthefollowingsurfaces: (i) y2= x2+9 z2(ii) y = x2Š z2(iii)4 x2+2 y2+ z2=4 (iv) x2Š y2+ z2= Š 1 (v). y2+4 z2=16 (vi). x2Š y2+ z2=1 (2) Reduceeachofthefollowingequationstooneofthestandard form,classifythesurface,andsketchit: (i) x2+ y2+4 z2Š 2 x +4 y =0 (ii) x2Š y2+ z2+2 x Š 2 y +4 z +2=0 (iii) x2+4 y2Š 6 x + z =0 (3) Findanequationforthesurfaceobtainedbyrotatingtheline y =2 x aboutthe y axis. (4) Findanequationforthesurfaceconsistingofallpointsthat areequidistantfromthepoint(1 1 1)andtheplane z =2.

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7211.VECTORSANDTHESPACEGEOMETRY (5) Sketchthesolidregionboundedbythesurface z = x2+ y2frombelowandby x2+ y2+ z2Š 2 z =0fromabove. (6) Findanequationforthesurfaceconsistingofallpoints P for whichthedistancefrom P tothe y axisistwicethedistancefrom P tothe zx plane.Identifythesurface. (7) Showthatifthepoint( a,b,c )liesonthehyperbolicparaboloid z = y2Š x2,thenthelinesthrough( a,b,c )andparallelto v = 1 1 2( b Š a ) and u = 1 Š 1 Š 2( b Š a ) bothlieentirelyon thisparaboloid.Deducefromthisresultthatthehyperbolicparaboloidcanbegeneratedbythemotionofastraightline.Showthat hyperboloidsofonesheet,cones,andcylinderscanalsobeobtainedby themotionofastraightline. Remark. Thefactthathyperboloidsofonesheetaregeneratedbythe motionofastraightlineisusedtoproducegeartransmissions.The cogsofthegearsarethegeneratinglinesofthehyperboloids. (8) Findanequationforthecylinderofradius R whoseaxisgoes throughtheoriginandisparalleltoavector v (9) Showthatthecurveofintersectionofthesurfaces x2Š 2 y2+ 3 z2Š 2 x + y Š z =1and2 x2Š 4 y2+6 z2+ x Š y +2 z =4liesina plane.

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CHAPTER12 VectorFunctions 79.CurvesinSpaceandVectorFunctions Todescribethemotionofapointlikeobjectinspace,itsposition vectorsmustbespeci“edateverymomentoftime.Avectorisde“ned bythreecomponentsinacoordinatesystem.Therefore,themotionof theobjectcanbedescribedbyanorderedtripleofreal-valuedfunctionsoftime.Thisobservationleadstotheconceptofvector-valued functionsofarealvariable. Definition 12.1 (VectorFunction) Let D beasetofrealnumbers.Avectorfunction r ( t ) ofarealvariable t isarulethatassignsavectortoeveryvalueof t from D .Theset D iscalledthe domain ofthevectorfunction. Mostcommonlyusedrulestode“neavectorfunctionarealgebraicrulesthatspecifycomponentsofavectorfunctioninacoordinatesystemasfunctionsofarealvariable: r ( t )= x ( t ) ,y ( t ) ,z ( t ) .For example, r ( t )= 1 Š t, ln( t ) ,t2 or x ( t )= 1 Š t,y ( t )=ln( t ) ,z ( t )= t2. Unlessspeci“edotherwise,thedomainofthevectorfunctionistheset D ofallvaluesof t atwhichthealgebraicrulemakessense;thatis, allthreecomponentscanbecomputedforany t from D .Intheabove example,thedomainof x ( t )is Š
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7412.VECTORFUNCTIONS Figure12.1. Left :Theterminalpointofavector r ( t ) whosecomponentsarecontinuousfunctionsof t traces outacurveinspace. Right :Graphingaspacecurve.Drawacurveinthe xy planede“nedbytheparametricequations x = x ( t ), y = y ( t ).Itistracedoutbythevector R ( t )= x ( t ) ,y ( t ) 0 Thisplanarcurvede“nesacylindricalsurfaceinspace inwhichthespacecurveinquestionlies.Thespace curveisobtainedbyraisingorloweringthepointsof theplanarcurvealongthesurfacebytheamount z ( t ), thatis, r ( t )= R ( t )+ e3z ( t ).Inotherwords,thegraph z = z ( t )iswrappedaroundthecylindricalsurface. space,andagraphofavectorfunctionisacurveinspace.Thereis adierencebetweengraphsofordinaryfunctionsandgraphsofvector functions,though.Thefunctionofarealvariableisuniquelyde“ned byitsgraph.Thisisnotsoforvectorfunctions.Supposetheshapeof acurveisdescribedgeometrically,thatis,asapointsetinspace(e.g., alinethroughtwogivenpoints).Onemightaskthequestion:Whatis avectorfunctionthattracesoutagivencurveinspace?Theanswer tothisquestionisnotunique.Forexample,aline L asapointsetin spaceisuniquelyde“nedbyitsparticularpointandavector v parallel toit.If r1and r2arepositionvectorsoftwoparticularpointsof L thenbothvectorfunctions r1( t )= r1+ t v and r2( t )= r2Š 2 t v trace out L becausethevector Š 2 v isalsoparallelto L Thefollowing,moresophisticatedexampleisalsoofinterest.Supposeonewantsto“ndavectorfunctionthattracesoutasemicircle ofradius R .Letthesemicirclebepositionedintheupperpartofthe

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79.CURVESINSPACEANDVECTORFUNCTIONS75 xy plane( y 0).Thefollowingthreevectorfunctionstraceoutthe semicircle: r1( t )= t, R2Š t2, 0 Š R t R, r2( t )= R cos t,R sin t, 0 0 t r3( t )= Š R cos t,R sin t, 0 0 t Thisiseasytoseebycomputingthenormofthesevectorfunctions: ri( t ) 2= R2or x2 i( t )+ y2 i( t )= R2,where i =1 2 3,foranyvalue of t ;thatis,theendpointsofthevectors ri( t )alwaysremainonthe semicircleofradius R as t rangesoverthespeci“edinterval.Itcan thereforebeconcludedthattherearemanyvectorfunctionsthattrace outthesamecurveinspacede“nedasapointsetinspace. Anotherobservationisthattherearevectorfunctionsthattrace outthesamecurveinoppositedirections.Intheaboveexample,the vectorfunction r2( t )tracesoutthesemicirclecounterclockwise,while thefunctions r1( t )and r3( t )dosoclockwise.Soavectorfunction de“nesthe orientation ofaspatialcurve.However,thisnotionofthe orientationofacurveshouldberegardedwithcaution.Forexample, thevectorfunction r ( t )= R cos t,R | sin t | 0 tracesoutthesemicircle twice,backandforth,when t rangesfrom0to2 .Inthiscase,the rangeof r ( t )shouldbeconsideredastwosemicircles(oneisoriented counterclockwiseandtheotherclockwise),andthesesemicirclesare superimposedoneontotheother.79.1.GraphingSpaceCurves.Tovisualizetheshapeofacurve C tracedoutbyavectorfunction,itisconvenienttothinkabout r ( t ) asatrajectoryofmotion.Thepositionofaparticleinspacemaybe determinedbyitspositioninaplaneanditsheightrelativetothat plane.Forexample,thisplanecanbechosentobethe xy plane.Then r ( t )= x ( t ) ,y ( t ) ,z ( t ) = x ( t ) ,y ( t ) 0 + 0 0 ,z ( t ) = R ( t )+ z ( t ) e3. Considerthecurvede“nedbytheparametricequations x = x ( t ), y = y ( t )inthe xy plane.Onecanmarkafewpointsalongthe curvecorrespondingtoparticularvaluesof t ,say, Pnwithcoordinates ( x ( tn) ,y ( tn)), n =1 2 ,...,N .Thenthecorrespondingpointsofthe curve C areobtainedfromthembymovingthepoints Pnalongthe directionnormaltotheplane(i.e.,alongthe z axisinthiscase),by theamount z ( tn);thatis, Pngoesupif z ( tn) > 0ordownif z ( tn) < 0. Inotherwords,asaparticlemovesalongthecurve x = x ( t ), y = y ( t ), itascendsordescendsaccordingtothecorrespondingvalueof z ( t ).

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7612.VECTORFUNCTIONS Thecurvecanalsobevisualizedbyusingapieceofpaper.Consider ageneralcylinderwiththehorizontaltracebeingthecurve x = x ( t ), y = y ( t ),likeawalloftheshapede“nedbythiscurve.Thenmakea graphofthefunction z ( t )onapieceofpaper(wallpaper)andglueitto thewallsothatthe t axisofthegraphisgluedtothecurve x = x ( t ), y = y ( t )(i.e.,eachpoint t onthe t axisisgluedtothecorresponding point( x ( t ) ,y ( t ))ofthecurve).Aftersuchaprocedure,thegraphof z ( t )alongthewallwouldcoincidewiththecurve C tracedoutby r ( t ). TheprocedureisillustratedinFigure12.1(rightpanel). Example 12.1 Graphthevectorfunction r = cos t, sin t,t ,where t rangesovertherealline. Solution: Itisconvenienttorepresent r ( t )asthesumofavector inthe xy planeandavectorparalleltothe z axis.Inthe xy plane, thecurve x =cos t y =sin t isthecircleofunitradiustracedout counterclockwisesothatthepoint(1 0 0)correspondsto t=0.The circularmotionisperiodicwithperiod2 .Theheight z ( t )= t rises linearlyasthepointmovesalongthecircle.Startingfrom(1 0 0), thecurvemakesoneturnonthesurfaceofthecylinderofunitradius climbingupby2 ineachturn.Thinkofapieceofpaperwitha straightlinedepictedonitthatiswrappedaroundthecylinder.Thus, thecurvetracedby r ( t )liesonthesurfaceofacylinderofunitradius andperiodicallywindsaboutitclimbingby2 perturn.Suchacurve iscalleda helix .TheprocedureisshowninFigure12.2. 79.2.LimitsandContinuityofVectorFunctions.Definition 12.2 (LimitofaVectorFunction) Avector r0iscalledthe limit ofavectorfunction r ( t ) as t t0if limt t0 r ( t ) Š r0 =0; thelimitisdenotedas limt t0r ( t )= r0. Theleftandrightlimits,limt tŠ 0r ( t )andlimt t+ 0r ( t ),arede“ned similarly.Thisde“nitionsaysthatthelengthornormofthevector r ( t ) Š r0approaches0as t tendsto t0.Thenormofavectorvanishes ifandonlyifthevectoristhezerovector.Therefore,thefollowing theoremholds. Theorem 12.1 (LimitofaVectorFunction) Let r ( t )= x ( t ) ,y ( t ) ,z ( t ) andlet r0= x0,y0,z0 .Then limt t0r ( t )= r0 limt t0x ( t )= x0, limt t0y ( t )= y0, limt t0z ( t )= z0.

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79.CURVESINSPACEANDVECTORFUNCTIONS77 Figure12.2. Graphingahelix. Right :Thecurve R ( t )= cos t, sin t, 0 isacircleofunitradius,traced outcounterclockwise.Sothehelixliesonthecylinderof unitradiuswhosesymmetryaxisisthe z axis. Middle :Thegraph z = z ( t )= t isastraightlinethat de“nestheheightofhelixpointsrelativetothecircle tracedoutby R ( t ). Right :Thegraphofthehelix r ( t )= R ( t )+ z ( t ) e3. As R ( t )traversesthecircle,theheight z ( t )= t rises linearly.Sothehelixcanbeviewedasastraightline wrappedaroundthecylinder. Thistheoremreducestheproblemof“ndingthelimitofavector functiontotheproblemof“ndinglimitsofthreeordinaryfunctions. Italsosaysthatthelimitofavectorfunctionexistsifandonlyifthe limitsofitsallcomponentsexist. Example 12.2 Findthelimitof r ( t )= sin( t ) /t,t ln t, ( etŠ 1 Š t ) /t2 as t 0+. Solution: BylHospitalsrule, limt 0+sin t t =limt 0+cos t 1 =1 limt 0+t ln t =limt 0+ln t tŠ 1=limt 0+tŠ 1 Š tŠ 2= Š limt 0+t =0 limt 0+etŠ 1 Š t t2=limt 0+etŠ 1 2 t =limt 0+et 2 = 1 2 Therefore,limt 0+r ( t )= 1 0 1 / 2

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7812.VECTORFUNCTIONS Definition 12.3 (ContinuityofaVectorFunction) Avectorfunction r ( t ) t [ a,b ] ,issaidtobecontinuousat t = t0 [ a,b ] if limt t0r ( t )= r ( t0) Avectorfunction r ( t ) iscontinuousintheinterval [ a,b ] ifitiscontinuousateverypointof [ a,b ] ByTheorem12.1, avectorfunctioniscontinuousifandonlyifall itscomponentsarecontinuousfunctions Example 12.3 Determinewhetherthevectorfunction r ( t )= sin (2 t ) /t,t2,et forall t =0 and r (0)= 1 0 1 iscontinuous. Solution: Thecomponents y ( t )= t2and z ( t )= etarecontinuousfor allreal t and y (0)=0and z (0)=1.Thecomponent x ( t )=sin(2 t ) /t iscontinuousforall t =0becausetheratiooftwocontinuousfunctions iscontinuous.BylHospitalsrule, limt 0x ( t )=limt 0sin(2 t ) t =limt 02cos(2 t ) 1 =2 limt 0x ( t ) = x (0)=1; thatis, x ( t )isnotcontinuousat t =0.Thus, r ( t )iscontinuous everywhere,but t =0. 79.3.SpaceCurveasaContinuousVectorFunction.Acurveinspace canbeunderstoodasacontinuoustransformation(oradeformation withoutbreaking)ofastraightlinesegmentinspace.Conversely,every spacecurvecanbecontinuouslydeformedtoastraightlinesegment.So aspacecurveisacontinuousdeformationofastraightlinesegment, andthisdeformationhasacontinuousinverse .Thismotivatesthe following(simpler)de“nitionofaspatialcurvethatissucientforall applicationsdiscussedinthiscourse. Definition 12.4 (CurveinSpace) Acurveinspaceistherangeofacontinuousvectorfunction. Ifacurveinspaceisde“nedasapointsetbygeometricalmeans (e.g.,asanintersectionoftwosurfaces),thenthisde“nitionimplies thatthereisacontinuousvectorfunctionwhoserangecoincideswith thepointset.Itshouldbeunderstoodthattherearedierentcontinuousvectorfunctionswiththesamerange,andacontinuousvector functionmaytraversethesamepointsetseveraltimes.Forexample, thevectorfunction r ( t )=( t2,t2,t2)iscontinuousontheinterval[ Š 1 1] andtracesoutthestraightlinesegment, x = y = z ,betweenthepoints (0 0 0)and(1 1 1)twice.

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79.CURVESINSPACEANDVECTORFUNCTIONS79 Acurveissaidtobe simple ifitdoesnotintersectitselfatany point;thatis,asimplecurveisacontinuousvectorfunction r ( t )for which r ( t1) = r ( t2)forany t1 = t2intheopeninterval( a,b ).Asimple curveisalways oriented becausethefunction r ( t )tracesoutitsrange onlyoncefromtheinitialpoint r ( a )totheterminalpoint r ( b ).Acurve is closed if r ( a )= r ( b ).Thede“nitionofaspacecurveasacontinuous vectorfunctionisratherfruitfulbecauseitallowsustogiveaprecise algebraicdescriptionofthegeometricalpropertiesaspacecurvemay have.79.4.StudyProblems.Problem12.1. Findavectorfunctionthattracesoutahelixofradius R thatclimbsupalongitsaxisby h Solution: Letthehelixaxisbethe z axis.Themotioninthe xy planemustbecircularwithradius R .Suitableparametricequations ofthecircleare x ( t )= R cos t y ( t )= R sin t .Withthisparameterizationofthecircle,themotionhasaperiodof2 .Ontheother hand, z ( t )mustriselinearlyby h as t changesovertheperiod.Therefore, z ( t )= ht/ (2 ).Thevectorfunctionmaybechosenintheform r ( t )= R cos t,R sin t,ht/ (2 ) Problem12.2. Sketchand/ordescribethecurvetracedoutbythe vectorfunction r ( t )= cos t, sin t, sin(4 t ) if t rangesintheinterval [0 2 ] Solution: Inthe xy plane,themotiongoesalongthecircleofunit radius,counterclockwise,startingfromthepoint(1 0 ,0).As t ranges overthespeci“edinterval,thecircleistracedoutonlyonce.The height z ( t )=sin(4 t )hasaperiodof2 / 4= / 2.Therefore,thegraph ofsin(4 t )makesfourupsandfourdownsif0 t 2 .Thecurvelooks likethegraphofsin(4 t )wrappedaroundthecylinderofunitradius. Itmakesoneupandonedownineachquarterofthecylinder.The procedureisshowninFigure12.3. Problem12.3. Sketchand/ordescribethecurvetracedoutbythe vectorfunction r ( t )= t cos t,t sin t,t Solution: Thecomponentsof r ( t )satisfytheequation x2( t )+ y2( t )= z2( t )forallvaluesof t .Therefore,thecurveliesonthedoublecone x2+ y2= z2.Since x2( t )+ y2( t )= t2,themotioninthe xy planeis aspiral(thinkofarotationalmotionabouttheoriginsuchthatthe radiusincreaseslinearlywiththeangleofrotation).If t increasesfrom t =0,thecurveinquestionistracedbyapointthatriseslinearlywith

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8012.VECTORFUNCTIONS Figure12.3. IllustrationtoStudyProblem12.2. Left : Thecurveliesonthecylinderofunitradius.Itmay beviewedasthegraphof z =sin(4 t )ontheinterval 0 t 2 wrappedaroundthecylinder. Right :Thecircletracedoutby R ( t )= cos t, sin t, 0 (top).Itde“nesthecylindricalsurfaceonwhichthe curvelies.Thegraph z = z ( t )=sin(4 t ),whichde“nes theheightofpointsofthecurverelativetothecirclein the xy plane(bottom). thedistancefromtheoriginasittravelsalongthespiral.If t decreases from t =0,insteadofrising,thepointwoulddescend( z ( t )= t< 0). Sothecurvewindsabouttheaxisofthedoubleconewhileremaining onitssurface.TheprocedureisshowninFigure12.4. Problem12.4. Findtheportionoftheelliptichelix r ( t )= 2cos( t ) t, sin( t ) thatliesinsidetheellipsoid x2+ y2+4 z2=13 Solution: Thehelixhereiscalled elliptic becauseitliesonthesurface ofanellipticcylinder.Indeed,inthe xz plane,themotiongoesalong theellipse x2/ 4+ z2=1.Sothecurveremainsonthesurfaceofthe ellipticcylinderparalleltothe y axis.Oneturnaroundtheellipse occursas t changesfrom0to2.Thehelixrisesby2alongthe y axis perturn.Now,tosolvetheproblem,onehasto“ndthevaluesof t atwhichthehelixintersectstheellipsoid.Theintersectionhappens whenthecomponentsof r ( t )satisfytheequationoftheellipsoid,that is,when x2( t )+ y2( t )+4 z2( t )=1or4+ t2=13andhence t = 3.The

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79.CURVESINSPACEANDVECTORFUNCTIONS81 Figure12.4. IllustrationtoStudyProblem12.3. Left : Theheightofthegraphrelativetothe xy plane(top). Thecurve R = t cos t,t sin t, 0 .For t 0,itlookslike anunwindingspiral(bottom). Right :For t> 0,thecurveistraversedbythepoint movingalongthespiralwhilerisinglinearlywiththe distancetraveledalongthespiral.Itcanbeviewedasa straightlinewrappedaroundthecone x2+ y2= z2. positionvectorsofthepointsofintersectionare r ( 3)= Š 2 3 0 Theportionofthehelixthatliesinsidetheellipsoidcorrespondstothe range Š 3 t 3. Problem12.5. Considertwocurves C1and C2tracedoutbythe vectorfunctions r1( t )= t2,t,t2+2 t Š 8 and r2( s )= 8 Š 4 s, 2 s,s2+ s Š 2 ,respectively.Dothecurvesintersect?Ifso,“ndthepointsof intersection.Supposetwoparticleshavethetrajectories r1( t ) and r2( t ) where t istime.Dotheparticlescollide? Solution: Thecurvesintersectiftherearevaluesofthepair( t,s ) suchthat r1( t )= r2( s ).Thisvectorequationisequivalentthesystem ofthreeequations x1( t )= x2( s ), y1( t )= y2( s ), z1( t )= z2( s ).Itfollows fromthesecondequationthat t =2 s .Substitutingthisintothe“rst equation,one“ndsthat(2 s )2=8 Š 4 s whosesolutionsare s = Š 2and s =1.Onehasyettoverifythatthethirdequation t2+2 t Š 8= s2+ s Š 2 holdsforthepairs( t,s )=( Š 4 Š 2)and( t,s )=(2 1)(otherwise,the z componentsdonotmatch).Asimplecalculationshowsthatindeed bothpairssatisfytheequation.Sothepositionvectorsofthepoints

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8212.VECTORFUNCTIONS ofintersectionare r1( Š 4)= r2( Š 2)= 16 Š 4 0 and r1(2)= r2(1)= 4 2 0 .Althoughthecurvesalongwhichtheparticlestravelintersect, thisdoesnotmeanthattheparticleswouldnecessarilycollidebecause theymaynotarriveatapointofintersectionatthesamemomentof time,justliketwocarstravelingalongintersectingstreetsmayormay notcollideatthestreetintersection.Thecollisionconditionismore restrictive, r1( t )= r2( t )(i.e.,thetime t mustsatisfythreeconditions). Fortheproblemathand,theseconditionscannotbeful“lledforany t because,amongallthesolutionsof r1( t )= r2( s ),thereisnosolution forwhich t = s .Thus,theparticlesdonotcollide. Problem12.6. Findavectorfunctionthattracesoutthecurveof intersectionoftheparaboloid z = x2+ y2andtheplane 2 x +2 y + z =2 counterclockwiseasviewedfromthetopofthe z axis. Solution: Onehasto“ndthecomponents x ( t ), y ( t ),and z ( t )such thattheysatisfytheequationsoftheparaboloidandplanesimultaneouslyforallvaluesof t .Thisensuresthattheendpointofthevector r ( t ) remainsonbothsurfaces,thatis,tracesouttheircurveofintersection (seeFigure12.5).Consider“rstthemotioninthe xy plane.Solving theplaneequationfor z z =2 Š 2 x Š 2 y ,andsubstitutingthesolution intotheparaboloidequation,one“nds2 Š 2 x Š 2 y = x2+ y2.After Figure12.5. IllustrationtoStudyProblem12.6.The curveisanintersectionoftheparaboloidandtheplane P .Itistraversedbythepointmovingcounterclockwise aboutthecircleinthe xy plane(indicatedby P0)and risingsothatitremainsontheparaboloid.

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79.CURVESINSPACEANDVECTORFUNCTIONS83 completingthesquares,thisequationbecomes4=( x +1)2+( y +1)2, whichdescribesacircleofradius2centeredat( Š 1 Š 1).Thecircleliesintheplane P0inFigure12.5.Itsparametricequationsread x = x ( t )= Š 1+2cos t y = y ( t )= Š 1+2sin t .As t increases from0to2 ,thecircleistracedoutcounterclockwiseasrequired. Thus, r ( t )= Š 1+2cos t, Š 1+2sin t, 6 Š 2cos t Š 2sin t ,where t [0 2 ]. 79.5.Exercises.(1) Findthedomainofeachofthefollowingvector functions: (i) r ( t )= 9 Š t2, ln t, cos t (ii) r ( t )= ln(9 Š t2) ln | t | (1+ t ) / (2+ t ) (2) Findeachofthefollowinglimitsorshowthatitdoesnotexist: (i)limt 0+ ( e2 tŠ 1) /t, ( 1+ t Š 1) /t,t ln t (ii)limt 0 sin2(2 t ) /t2,t2+2 (cos t Š 1) /t2 (iii)limt 0 ( e2 tŠ t ) /t,t cot t, 1+ t (3) Sketcheachofthefollowingcurvesandidentifythedirection inwhichthecurveistracedoutastheparameter t increases: (i) r ( t )= t, cos(3 t ) sin(3 t ) (ii) r ( t )= 2sin(5 t ) 4 3cos(5 t ) (iii) r ( t )= 2 t sin t, 3 t cos t,t (iv) r ( t )= sin t, cos t, ln t (4) Twoobjectsaresaidtocollideiftheyareatthesameposition atthesametime .Twotrajectoriesaresaidtointersectiftheyhave commonpoints.Let t bethephysicaltime.Lettwoobjectstravelalong thespacecurves r1( t )= t,t2,t3 and r2( t )= 1+2 t, 1+6 t, 1+14 t Dotheobjectscollide?Dotheirtrajectoriesintersect?Ifso,“ndthe collisionandintersectionpoints. (5) Findavectorfunctionthattracesoutthecurveofintersection oftwosurfaces: (i) x2+ y2=9and z = xy (ii) x2+ y2= z2and x + y + z =1 (iii) z = x2+ y2and y = x2(6) Supposethatthelimitslimt av ( t )andlimt au ( t )exist.Prove thebasiclawsoflimitsforthefollowingvectorfunctions: limt a( v ( t )+ u ( t ))=limt av ( t )+limt au ( t ) limt a( s v ( t ))= s limt av ( t ) limt a( v ( t ) u ( t ))=limt av ( t ) limt au ( t ) limt a( v ( t ) u ( t ))=limt av ( t ) limt au ( t ) .

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8412.VECTORFUNCTIONS (7) Toappreciatethebasiclawsoflimitsestablishedinexercise6, put v ( t )= ( e2 tŠ 1) /t, ( 1+ t Š 1) /t,t ln t and u ( t )= sin2(2 t ) /t2, t2+2 (cos t Š 1) /t2 (seeexercise2)and“nd: (i)limt 0+( v ( t )+ u ( t )) (ii)limt 0+( v ( t ) u ( t )) (iii)limt 0+( v ( t ) u ( t )) Thinkoftheamountoftechnicalitiesneededtoobtaintheanswers withoutthelawsoflimits(e.g.,calculatingthecrossproduct“rstand then“ndingthelimitvalue). (8) Findthevaluesoftheparameters a and b atwhichthecurve r ( t )= 1+ at2,b Š t,t3 passesthroughthepoint(1 2 8). (9) Let r (0)= a,b,c andlet r ( t )= t cot(2 t ) ,t1 / 3ln | t | ,t2+2 for t =0.Findthevaluesof a b ,and c atwhichthevectorfunctionis continuous. (10) Supposethatthevectorfunction v ( t ) u ( t )iscontinuous. Doesthisimplythatbothvectorfunctions v ( t )and u ( t )arecontinuous?Supportyourargumentsbyexamples.80.DifferentiationofVectorFunctions Definition 12.5 (DerivativeofaVectorFunction) Supposeavectorfunction r ( t ) isde“nedonaninterval [ a,b ] and t0 [ a,b ] .Ifthelimit limh 0r ( t0+ h ) Š r ( t0) h = r( t0)= d r dt ( t0) exists,thenitiscalledthe derivativeofavectorfunction r ( t ) at t = t0, and r ( t ) issaidtobedierentiableat t0.For t0= a or t0= b ,thelimit isunderstoodastheright( h> 0 )orleft( h< 0 )limit,respectively.If thederivativeexistsforallpointsin [ a,b ] ,thenthevectorfunction r ( t ) issaidtobedierentiableon [ a,b ] Itfollowsfromthede“nitionofthelimitthatavectorfunctionis dierentiableifandonlyifallitscomponentsaredierentiable: r( t )=limh 0 x ( t + h ) Š x ( t ) h y ( t + h ) Š y ( t ) h z ( t + h ) Š z ( t ) h = x( t ) ,y( t ) ,z( t ) (12.1) Forexample, r ( t )= sin(2 t ) ,t2Š t,eŠ 3 t r( t )= 2cos(2 t ) 2 t Š 1 Š 3 eŠ 3 t Definition 12.6 (ContinuouslyDierentiableVectorFunction) Ifthederivative r( t ) isacontinuousvectorfunctiononaninterval

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80.DIFFERENTIATIONOFVECTORFUNCTIONS85 [ a,b ] ,thenthevectorfunction r ( t ) issaidtobecontinuouslydierentiableon [ a,b ] Higher-orderderivativesarede“nedsimilarly:thesecondderivative isthederivativeof r( t ), r( t )=( r( t )),thethirdderivativeisthe derivativeof r( t ), r( t )=( r( t )),and r( n )( t )=( r( n Š 1)( t )),provided theyexist.80.1.DifferentiationRules.Thefollowingrulesofdierentiationofvectorfunctionscandeducedfrom(12.1). Theorem 12.2 (DierentiationRules) Suppose u ( t ) and v ( t ) aredierentiablevectorfunctionsand f ( t ) isa real-valueddierentiablefunction.Then d dt v ( t )+ u ( t ) = v( t )+ u( t ) d dt f ( t ) v ( t ) = f( t ) v ( t )+ f ( t ) v( t ) d dt v ( t ) u ( t ) = v( t ) u ( t )+ v ( t ) u( t ) d dt v ( t ) u ( t ) = v( t ) u ( t )+ v ( t ) u( t ) d dt v ( f ( t )) = f( t ) v( f ( t )) Theproofisbasedonastraightforwarduseoftherule(12.1)and basicrulesofdierentiationforordinaryfunctionsandleftasanexercisetothereader.80.2.DifferentialofaVectorFunction.If r ( t )isdierentiable,then (12.2) r ( t )= r ( t + t ) Š r ( t )= r( t ) t + u ( t ) t, where u ( t ) 0 as t 0.Indeed,bythede“nitionofthederivative, u ( t )= r / t Š r( t ) 0 as t 0.Therefore,the componentsofthedierence r Š r t convergeto0fasterthan t Asforordinaryfunctions,situationsinwhichallsuchtermscanbe neglectedisdescribedbytheconceptofthedierential. Definition 12.7 (DierentialofaVectorFunction) Let r ( t ) beadierentiablevectorfunction.Thenthevector d r ( t )= r( t ) dt iscalledthe dierential of r ( t ) .

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8612.VECTORFUNCTIONS Inparticular,thederivativeistheratioofthedierentials, r( t )= d r /dt .Inasmallenoughneighborhoodofanyparticular t = t0,a dierentiablevectorfunctioncanbewellapproximatedbyalinear vectorfunctionbecause r d r ( t0)orwith dt = t : r ( t0+ t ) L ( t )= r ( t0)+ r( t0) t, t = t Š t0. Thelinearvectorfunction L ( t )de“nesalinepassingthroughthepoint r ( t0).Thislineiscalledthe tangentline tothecurvetracedoutby r ( t ). Thus, thedierential d r ( t ) atapointofthecurve r ( t ) istheincrement ofthepositionvectoralongthelinetangenttothecurveatthatpoint .80.3.GeometricalSigni“canceoftheDerivative.Consideravectorfunctionthattracesoutalineparalleltoavector v r ( t )= r0+ t v .Then r( t )= v ;thatis,thederivativeisavectorparallelortangenttothe line.Thisobservationisofageneralnature;thatis, thevector r( t0) is tangenttothecurvetracedoutby r ( t ) atthepointwhosepositionvector is r ( t0).Let P0and Phhavepositionvectors r ( t0)and r ( t0+ h ).Then P0Ph= r ( t0+ h ) Š r ( t0)isasecantvector.As h 0, P0Phapproaches avectorthatliesonthetangentlineasdepictedinFigure12.6.On theotherhand,itfollowsfrom(12.2)that,forsmallenough h = dt P0Ph= d r ( t0)= r( t0) h ,andthereforethetangentlineisparallelto r( t0).Thedirectionofthetangentvectoralsode“nestheorientation Figure12.6. Left :Asecantlinethroughtwopoints ofthecurve, P0and Ph.As h getssmaller,thedirection ofthevector P0Ph= r ( t0+ h ) Š r ( t0)becomescloserto thetangenttothecurveat P0. Right :Thederivative r( t )de“nesatangentvectorto thecurveatthepointwiththepositionvector r ( t ).It alsospeci“esthedirectioninwhich r ( t )traversesthe curvewithincreasing t T ( t )istheunittangentvector.

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80.DIFFERENTIATIONOFVECTORFUNCTIONS87 ofthecurve,thatis,thedirectioninwhichthecurveistracedout by r ( t ). Example 12.4 Findthelinetangenttothecurve r ( t )= 2 t,t2Š 1 ,t3+2 t atthepoint P0(2 0 3) Solution: Bythegeometricalpropertyofthederivative,avector paralleltothelineis v = r( t0),where t0isthevalueoftheparameter t atwhich r ( t0)= 2 0 3 isthepositionvectorof P0.Therefore, t0=1.Then v = r(1)= 2 2 t, 3 t +2 |t =1= 2 2 5 .Parametric equationsofthelinethrough P0andparallelto v are x =2+2 t y =2 t z =3+5 t Ifthederivative r( t )existsanddoesnotvanish,then,atanypoint ofthecurvetracedoutby r ( t ),a unittangentvector canbede“nedby T ( t )= r( t ) r( t ) InSection79.3,spatialcurveswereidenti“edwithcontinuousvector functions.Intuitively,asmoothcurveasapointsetinspaceshould haveaunittangentvectorthatiscontinuousalongthecurve.Recall alsothat,foranycurveasapointsetinspace,therearemanyvector functionswhoserangecoincideswiththecurve. Definition 12.8 (SmoothCurve) Apointsetinspaceiscalleda smoothcurve ifthereisacontinuously dierentiablevectorfunctionwhoserangecoincideswiththepointset andwhosederivativedoesnotvanish. Asmoothcurve r ( t )is oriented bythedirectionoftheunittangent vector T ( t ). Considertheplanarcurve r = t3,t2, 0 .Thevectorfunctionisdifferentiableeverywhere, r( t )= 2 t, 3 t2, 0 ,andthederivativevanishes attheorigin, r(0)= 0 .Theunittangentvector T ( t )isnotde“nedat t =0.Inthe xy plane,thecurvetracesoutthegraph y = x2 / 3,which hasa cusp at x =0.Thegraphisnotsmoothattheorigin.Thetangentlineistheverticalline x =0because y( x )=(2 / 3) xŠ 1 / 3 as x 0.Thegraphapproachesitfromthepositivehalf-plane y> 0, formingahornlikeshapeattheorigin.Acuspdoesnotnecessarily occuratapointwherethederivative r( t )vanishes.Forexample,consider r ( t )= t3,t5, 0 suchthat r(0)= 0 .Thecurvetracesoutthe graph y = x5 / 3,whichhasnocuspat x =0(ithasanin”ectionpoint at x =0).Thereisanothervectorfunction R ( s )= s,s5 / 3, 0 that tracesoutthesamegraph,but R(0)= 1 0 0 = 0 ,andthecurveis smooth.Sothevanishingofthederivativeismerelyassociatedwitha

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8812.VECTORFUNCTIONS poorchoiceofthevectorfunction.Notethat r ( t )= R ( s )identically if s = t3.Bythechainrule,d dtr ( t )=d dtR ( s )= R( s )( ds/dt ).This showsthat,evenif R( s )nevervanishes,thederivative r( t )canvanish,provided ds/dt vanishesatsomepoint,whichisindeedthecasein theconsideredexampleas ds/dt =3 t2vanishesat t =0.80.4.StudyProblems.Problem12.7. Provethat,foranysmoothcurveonasphere,atangentvectoratanypoint P isperpendiculartothevectorfromthesphere centerto P Solution: Let r0bethepositionvectorofthecenterofasphereof radius R .Thepositionvector r ofanypointofthespheresatis“esthe equation r Š r0 = R or( r Š r0) ( r Š r0)= R2(because a 2= a a foranyvector a ).Let r ( t )beavectorfunctionthattracesoutacurve onthesphere.Then,forallvaluesof t ,( r ( t ) Š r0) ( r ( t ) Š r0)= R2. Dierentiatingbothsidesofthelatterrelation,oneinfers r( t ) ( r ( t ) Š r0)=0.Thisalgebraicconditionisequivalenttothegeometricalone: r( t ) r ( t ) Š r0.If r ( t )isthepositionvectorof P and O isthecenter ofthesphere,then OP = r ( t ) Š r0,andhencethetangentvector r( t )at P isperpendicularto OP forany t oratanypoint P ofthe curve. 80.5.Exercises.(1) Findthederivativesanddierentialsofeachof thefollowingcurves: (i) r ( t )= cos t, sin2( t ) ,t2 (ii) r ( t )= ln( t ) ,e2 t,teŠ t (iii) r ( t )= 3 t Š 2 t2Š 4 ,t (iv) r ( t )= a + b t2Š c et(v) r ( t )= t a ( b Š c et) (2) Determineifthecurvetracedoutbyeachofthefollowingvector functionsissmoothforaspeci“edintervaloftheparameter.Ifthe curveisnotsmoothataparticularpoint,graphitnearthatpoint. (i) r ( t )= t,t2,t3 ,0 t 1 (ii) r ( t )= t2,t3, 2 Š 1 t 1 (iii) r ( t )= t1 / 3,t,t3 Š 1 t 1 (vi) r ( t )= t5,t3,t4 Š 1 t 1 (3) Findtheparametricequationsofthetangentlinetoeachofthe followingcurvesataspeci“edpoint: (i) r ( t )= t2Š t,t3/ 3 2 t P0=(6 9 6) (ii) r ( t )= ln t, 2 t,t2 P0=(0 2 1)

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81.INTEGRATIONOFVECTORFUNCTIONS89 (4) Isthereapointonthecurve r ( t )= t2Š t,t3/ 3 2 t atwhich thetangentlineisparalleltothevector v = Š 5 / 2 2 1 ?Ifso,“nd thepoint. (5) Let r ( t )= et, 2cos t, sin(2 t ) .Usethetangentlineapproximationto“nd r (0 2).Useacalculatortoassesstheaccuracyofthe approximation. (6) Supposeasmoothcurve r ( t )doesnotintersectaplanethrough apoint P0andperpendiculartoavector n .Whatistheanglebetween n and r( t )atthepointofthecurvethatistheclosesttotheplane? (7) Doesthecurve r ( t )= 2 t2, 2 t, 2 Š t2 intersecttheplane x + y + z = Š 3?Ifnot,“ndapointonthecurvethatisclosesttotheplane. Whatisthedistancebetweenthecurveandtheplane. (8) Findthepointintersectionoftwocurves r1( t )= 1 1 Š t, 3+ t2 and r1( s )= 3 Š s,s Š 2 ,s2 .Iftheangleatwhichtwocurves intersectisde“nedastheanglebetweentheirtangentlinesatthepoint ofintersection,“ndtheangleatwhichtheabovetwocurvesintersect. (9) Suppose r ( t )istwicedierentiable.Showthat( r ( t ) r( t ))= r ( t ) r( t ). (10) Supposethat r ( t )isdierentiablethreetimes.Put v = r ( r r).Showthat v= r ( r r).81.IntegrationofVectorFunctions Definition 12.9 (De“niteIntegralofaVectorFunction) Let r ( t ) bede“nedontheinterval [ a,b ] .Thevectorwhosecomponents arethede“niteintegralsofthecorrespondingcomponentsof r ( t )= x ( t ) ,y ( t ) ,z ( t ) iscalledthe de“niteintegral of r ( t ) overtheinterval [ a,b ] anddenotedas (12.3) b ar ( t ) dt = b ax ( t ) dt, b ay ( t ) dt, b ax ( t ) dt Iftheintegral(12.3)exists,then r ( t ) issaidtobeintegrableon [ a,b ] Bythisde“nition,avectorfunctionisintegrableifandonlyifall itscomponentsareintegrablefunctions.Recallthatacontinuousrealvaluedfunctionisintegrable.Therefore,thefollowingtheoremholds. Theorem 12.3 Ifavectorfunctioniscontinuousontheinterval [ a,b ] ,thenitisintegrableon [ a,b ] Definition 12.10 (Inde“niteIntegralofaVectorFunction) Avectorfunction R ( t ) iscalledan inde“niteintegral of r ( t ) if R( t )= r ( t ) .

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9012.VECTORFUNCTIONS If R ( t )= X ( t ) ,Y ( t ) ,Z ( t ) and r ( t )= x ( t ) ,y ( t ) ,z ( t ) .Then,accordingto(12.1),thefunctions X ( t ), Y ( t ),and Z ( t )areantiderivatives of x ( t ), y ( t ),and z ( t ),respectively, X ( t )= x ( t ) dt + c1,Y ( t )= y ( t ) dt + c2,Z ( t )= z ( t ) dt + c3, where c1, c2,and c3areconstants.Thelatterrelationscanbecombined intoasinglevectorrelation: R ( t )= r ( t ) dt + c where c isanarbitraryconstantvector.Applyingthefundamental theoremofcalculustoeverycomponentontherightsideof(12.3),the fundamentaltheoremofcalculuscanbeextendedtovectorfunctions. Theorem 12.4 (FundamentalTheoremofCalculusforVector Functions) If R ( t ) isaninde“niteintegralof r ( t ) ,then b ar ( t ) dt = R ( b ) Š R ( a ) Example 12.5 Find r ( t ) if r( t )= 2 t, 1 6 t2 and r (1)= 2 1 0 Solution: Takingtheantiderivativeof r( t ),one“nds r ( t )= 2 t, 1 6 t2 dt + c = t2,t, 3 t3 + c .Theconstantvector c isdeterminedbythe condition r (1)= 2 1 0 ,whichgives 1 1 3 + c = 2 1 0 .Hence, c = 1 0 Š 3 and r ( t )= t2+1 ,t, 3 t3Š 3 Ingeneral,thesolutionoftheequation r( t )= v ( t )satisfyingthe condition r ( t0)= r0canbewrittenintheform r( t )= v ( t )and r ( t0)= r0 r ( t )= r0+ t t0v ( s ) ds if v ( t )isacontinuousvectorfunction.Recallthatiftheintegrandisa continuousfunction,thenthederivativeoftheintegralwithrespectto itsupperlimitisthevalueoftheintegrandatthatlimit.Therefore, ( d/dt ) t t0v ( s ) ds = v ( t ),andhence r ( t )isanantiderivativeof v ( t ). When t = t0,theintegralvanishesand r ( t0)= r0asrequired.81.1.ApplicationstoMechanics.Let r ( t )bethepositionvectorofa particleasafunctionoftime t .The“rstderivative r( t )= v ( t )is calledthe velocity oftheparticle.Themagnitudeofthevelocityvector v ( t )= v ( t ) iscalledthe speed .Thespeedofacarisanumbershown onthespeedometer.Thevelocityde“nesthedirectioninwhichthe

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81.INTEGRATIONOFVECTORFUNCTIONS91 particletravelsandtheinstantaneousrateatwhichitmovesinthat direction.Thesecondderivative r( t )= v( t )= a ( t )iscalledthe acceleration .If m isthemassofaparticleand F istheforceactingon theparticle,accordingtoNewtonssecondlaw,theaccelerationand forcearerelatedas F = m a Iftheforceisknownasavectorfunctionoftime,thenthisrelation isanequationofmotionthatdeterminesaparticlestrajectory.The problemof“ndingthetrajectoryamountstoreconstructingthevector function r ( t )ifitssecondderivative r( t )=(1 /m ) F ( t )isknown;that is, r ( t )isgivenbythesecondantiderivativeof(1 /m ) F ( t ).Indeed,the velocity v ( t )isanantiderivativeof(1 /m ) F ( t ),andthepositionvector r ( t )isanantiderivativeofthevelocity v ( t ).Asshownintheprevious section,anantiderivativeisnotunique,unlessitsvalueataparticular pointisspeci“ed.So thetrajectoryofmotionisuniquelydetermined byNewtonsequation,providedthepositionandvelocityvectorsare speci“edataparticularmomentoftime ,forexample, r ( t0)= r0and v ( t0)= v0.Thelatterconditionsarecalled initialconditions .Given theinitialconditions,thetrajectoryofmotionisuniquelyde“nedby therelations: v ( t )= v0+ 1 m t t0F ( s ) ds, r ( t )= r0+ t t0v ( s ) ds iftheforceisacontinuousvectorfunctionoftime. Remark. Iftheforceisafunctionofaparticlesposition,thenthe Newtonsequationbecomesasystemof ordinarydierentialequations thatisasetofsomerelationsbetweencomponentsofthevectorfunctions,itsderivatives,andtime. Example 12.6 (MotionUnderaConstantForce) Provethatthetrajectoryofmotionunderaconstantforceisaparabola. Solution: Let F beaconstantforce.Withoutlossofgenerality,the initialconditionscanbesetat t =0, r (0)= r0,and v (0)= v0.Then v ( t )= v0+ t m F r ( t )= r0+ t v0+ t2 2 m F Thevector r ( t ) Š r0isalinearcombinationof v0and F andhencemust beperpendicularto n = v0 F bythegeometricalpropertyofthecross product.Therefore,theparticleremainsintheplanethrough r0thatis parallelto F and v0orperpendicularto n ,thatis,( r ( t ) Š r0) n =0(see Figure12.7,leftpanel).Theshapeofaspacecurvedoesnotdepend onthechoiceofthecoordinatesystem.Letuschoosethecoordinate systemsuchthattheoriginisattheinitialposition r0andtheplane

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9212.VECTORFUNCTIONS Figure12.7. Left :Motionunderaconstantforce F0. Thetrajectoryisaparabolathatliesintheplanethrough theinitialpointofthemotion r0andperpendicularto thevector n = v0 F ,where v0istheinitialvelocity. Right :Motionofaprojectilethrownatanangle and aninitialheight h .Thetrajectoryisaparabola.The pointofimpactde“nestherange L ( ). inwhichthetrajectoryliescoincideswiththe zy planesothat F is paralleltothe z axis.Inthiscoordinatesystem, r0= 0 F = 0 0 Š F and v0= 0 ,v0 y,v0 z .Theparametricequationsofthetrajectoryof motionassumetheform x =0, y = v0 yt ,and z = v0 zt Š t2F/ (2 m ).The substitutionof t = y/v0 yintothelatterequationyields z = ay2+ by where a = Š Fv2 0 y/ (2 m )and b = v0 z/v0 y,whichde“nesaparabolain the zy plane.Thus,thetrajectoryofmotionunderaconstantforceis aparabolathroughthepoint r0thatliesintheplanecontainingthe forceandinitialvelocityvectors F and v0. 81.2.MotionUnderaConstantGravitationalForce.Themagnitudeof thegravitationalforcethatactsonanobjectofmass m nearthesurface oftheEarthis mg ,where g 9 8m / s2isauniversalconstantcalled the accelerationofafreefall .Accordingtotheprevioussection,any projectile“redfromsomepointfollowsaparabolictrajectory.This factallowsonetopredicttheexactpositionsoftheprojectileand, inparticular,thepointatwhichitimpactstheground.Inpractice, theinitialspeed v0oftheprojectileandangleofelevation atwhich theprojectileis“redareknown(seeFigure12.7,rightpanel).Some practicalquestionsare:Atwhatelevationangleisthemaximalrange reached?Atwhatelevationangledoestherangeattainaspeci“ed value(e.g.,tohitatarget)?

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81.INTEGRATIONOFVECTORFUNCTIONS93 Toanswertheseandrelatedquestions,choosethecoordinatesystemsuchthatthe z axisisdirectedupwardfromthegroundandthe parabolictrajectoryliesinthe zy plane.Theprojectileis“redfrom thepoint(0 0 ,h ),where h istheinitialelevationoftheprojectileabove theground(“ringfromahill).Inthenotationoftheprevioussection, F = Š mg ( F isnegativebecausethegravitationalforceisdirected towardtheground,whilethe z axispointsupward), v0 y= v0cos ,and v0 z= v0sin .Thetrajectoryis y = tv0cos ,z = h + tv0sin Š 1 2 gt2,t 0 Itisinterestingtonotethatthetrajectoryisindependentofthemass oftheprojectile.Lightandheavyprojectileswouldfollowthesame parabolictrajectory,providedtheyare“redfromthesameposition, atthesamespeed,andatthesameangleofelevation.Theheightof theprojectilerelativetothegroundisgivenby z ( t ).Thehorizontal displacementis y ( t ).Let tL> 0bethemomentoftimewhenthe projectilelands;thatis,when t = tL,theheightvanishes, z ( tL)=0. Apositivesolutionofthisequationis tL= v0sin + v2 0sin2 +2 gh g Thedistance L traveledbytheprojectileinthehorizontaldirection untilitlandsisthe range : L = y ( tL)= tLv0cos Forexample,iftheprojectileis“redfromtheground, h =0,then tL=2 v0sin /g andtherangeis L = v2 0sin(2 ) /g .Therangeattainsits maximalvalue v2 0/g whentheprojectileis“redatanangleofelevation = / 4.Theangleofelevationatwhichtheprojectilehitsatargetata givenrange L = L0is =(1 / 2)sinŠ 1( L0g/v2 0).For h =0,theangleat which L = L ( )attainsitsmaximalvaluescanbefoundbysolvingthe equation L( )=0,whichde“nescriticalpointsofthefunction L ( ). Theangleofelevationatwhichtheprojectilehitsatargetatagiven rangeisfoundbysolvingtheequation L ( )= L0.Thetechnicalities arelefttothereader. Remark. Inreality,thetrajectoryofaprojectiledeviatesfroma parabolabecausethereisanadditionalforceactingonaprojectile movingintheatmosphere,thefrictionforce.Thefrictionforcedependsonthevelocityoftheprojectile.Soamoreaccurateanalysisof theprojectilemotionintheatmosphererequiresmethodsofordinary dierentialequations.

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9412.VECTORFUNCTIONS 81.3.StudyProblems.Problem12.8. Theaccelerationofaparticleis a = 2 6 t, 0 .Find thepositionvectoroftheparticleanditsvelocityintwounitsoftime t iftheparticlewasinitiallyatthepoint ( Š 1 Š 4 1) andhadthevelocity 0 2 1 Solution: Thevelocityvectoris v ( t )= a ( t ) dt + c = 2 t, 3 t2, 0 + c Theconstantvector c is“xedbytheinitialcondition v (0)= 0 2 1 whichyields c = 0 2 1 .Thus, v ( t )= 2 t, 3 t2+2 1 and v (2)= 2 8 1 .Thepositionvectoris r ( t )= v ( t ) dt + c = t2,t3+2 t,t + c Heretheconstantvector c isdeterminedbytheinitialcondition r (0)= 0 2 1 ,whichyields c = Š 1 Š 4 1 .Thus, r ( t )= t2Š 1 ,t3+2 t Š 4 ,t +1 and r (2)= 3 4 3 Problem12.9. Showthatifthevelocityandpositionvectorsofa particleremainsorthogonalduringthemotion,thenthetrajectorylies onasphere. Solution: If v ( t )= r( t )and r ( t )areorthogonal,then r( t ) r ( t )=0 forall t .Since( r r )= r r + r r=2 r r =0,oneconcludesthat r ( t ) r ( t )= R2=constor r ( t ) = R forall t ;thatis,theparticle remainsata“xeddistance R fromtheoriginallthetime. Problem12.10. A.chargedparticlemovinginamagnetic“eld B is subjecttotheLorentzforce F =( e/c ) v B ,where e istheelectric chargeoftheparticleand c isthespeedof lightinvacuum.Assume thatthemagnetic“eldisaconstantvectorparalleltothe z axisand theinitialvelocityis v (0)= v, 0 ,v .Showthatthetrajectoryisa helix: r ( t )= R sin( t ) ,R cos( t ) ,vt = eB mc ,R = v where B = B isthemagnetic“eldmagnitudeand m istheparticle mass. Solution: Newtonssecondlawreads m v= e c v B Put B = 0 0 ,B .Then v = r= R cos( t ) Š R sin( t ) ,v v B = Š B cos( t ) Š R sin( t ) 0 v= Š 2R cos( t ) Š 2R sin( t ) 0 .

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81.INTEGRATIONOFVECTORFUNCTIONS95 ThesubstitutionoftheserelationsintoNewtonssecondlawyields m2R = eBR/c andhence =( eB ) / ( mc ).Since v (0)= R, 0 ,v = v, 0 ,v ,itfollowsthat R = v/ Remark. Notethattheequationsofmotioninvolveonlythevelocity v .Forthisreason,thevelocityvectorisuniquelydeterminedbythe initialcondition v (0)= v0,whiletheinitialconditionfortheposition vectorisnotneeded(thevectorfunction r ( t )+ r0isalsoasolution foranarbitraryconstantvector r0).Therateatwhichthehelixrises alongthemagnetic“eldisdeterminedbythemagnitude(speed)ofthe initialvelocitycomponent vparalleltothemagnetic“eld,whereasthe radiusofthehelixisdeterminedbythemagnitudeoftheinitialvelocity component vperpendiculartothemagnetic“eld.Aparticlemakes onefullturnaboutthemagnetic“eldintime T =2 / =2 mc/ ( eB ), thatis,thelargerthemagnetic“eld,thefastertheparticlerotates aboutit.81.4.Exercises.(1) Findtheinde“niteandde“niteintegralsovera speci“edintervalforeachofthefollowingfunctions: (i) r ( t )= sin t,t3, cos t Š t (ii) r ( t )= t2,t t Š 1 t ,0 t 1 (iii) r ( t )= t ln t,t2,e2 t ,0 t 0 (2) Find r ( t )ifthederivatives r( t )and r ( t0)aregiven: (i) r( t )= t Š 1 ,t2, t r (1)= 1 0 1 (ii) r( t )= sin(2 t ) 2cos t, sin2t r ( )= 1 2 3 (3) Ifaparticlewasinitiallyatpoint(1 2 1)andhadvelocity v = 0 1 Š 1 .Findthepositionvectoroftheparticleafterithas beenmovingwithacceleration a ( t )= 1 0 ,t for2unitsoftime. (4) Aparticleofunitmassmovesunderaconstantforce F .Ifa particlewasinitiallyatthepoint r0andpassedthroughthepoint r1after2unitsoftime,“ndtheinitialvelocityoftheparticle.Whatwas thevelocityoftheparticlewhenitpassedthrough r1? (5) Thepositionvectorofaparticleis r ( t )= t2, 5 t,t2Š 16 t .Find r ( t )whenthespeedoftheparticleismaximal. (6) Aprojectileis“redataninitialspeedof400m/sandatan angleofelevationof30.Findtherangeoftheprojectile,themaximum heightreached,andthespeedatimpact. (7) Aballofmass m isthrownsouthwardintotheairataninitial speedof v0atanangleof totheground.Aneastwindappliesa steadyforceofmagnitude F totheballinawesterlydirection.Find thetrajectoryoftheball.Wheredoestheballlandandatwhatspeed? Findthedeviationoftheimpactpointfromtheimpactpoint A when

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9612.VECTORFUNCTIONS nowindispresent.Isthereanywaytocorrectthedirectioninwhich theballisthrownsothattheballstillhits A ? (8) Arocketburnsitsonboardfuelwhilemovingthroughspace. Let v ( t )and m ( t )bethevelocityandmassoftherocketattime t Itcanbeshownthattheforceexertedbytherocketjetenginesis m( t ) vg,where vgisthevelocityoftheexhaustgasesrelativetothe rocket.Showthat v ( t )= v (0) Š ln( m (0) /m ( t )) vg.Therocketisto accelerateinastraightlinefromresttotwicethespeedofitsown exhaustgases.Whatfractionofitsinitialmasswouldtherockethave toburnasfuel? (9) Theaccelerationofaprojectileis a ( t )= 0 2 6 t .Theprojectileisshotfrom(0 0 0)withaninitialvelocity v (0)= 1 Š 2 Š 10 .It issupposedtodestroyatargetlocatedat(2 0 Š 12).Thetargetcan bedestroyediftheprojectilesspeedisatleast3.1atimpact.Willthe targetbedestroyed?82.ArcLengthofaCurve Considerapartitionofacurve C ,thatis,acollectionofpointsof C Pk, k =0 1 ,...,N ,where P0and PNaretheendpointsofthecurve.A partitionissaidtobe re“ned ifthenumberofpartitionpointsincreases sothatmaxk| Pk Š 1Pk| 0as N ;here | Pk Š 1Pk| isthedistance betweenthepoints Pk Š 1and Pk. Definition 12.11 (ArcLengthofaCurve) Thearclengthofacurve C isthelimit L =limN Nk =1| Pk Š 1Pk| provideditexistsandisindependentofthechoiceofpartition.If L< ,thecurveiscalled measurable or recti“able Thegeometricalmeaningofthisde“nitionisrathersimple.Here thesumof | Pk Š 1Pk| isthelengthofapolygonalpathwithverticesat P0, P1,..., PNinthisorder.Asthepartitionbecomes“nerand“ner, thispolygonalpathapproachesthecurvemoreandmoreclosely(see Figure12.8,leftpanel).Incertaincases,thearclengthisgivenbythe Riemannintegral. Theorem 12.5 (ArcLengthofaCurve) Let C beacurvetracedoutbyacontinuouslydierentiablevectorfunction r ( t ) ,whichde“nesaone-to-onecorrespondencebetweenpointsof C andtheinterval t [ a,b ] .Then

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82.ARCLENGTHOFACURVE97 Figure12.8. Left :Thearclengthofacurveisde“ned asthelimitofthesequenceoflengthsofpolygonalpaths throughpartitionpointsofthecurve. Right :Naturalparameterizationofacurve.Givena point A ofthecurve,thearclength s iscountedfromit toanypoint P ofthecurve.Thepositionvectorof P isa vector R ( s ).Ifthecurveistracedoutbyanothervector function r ( t ),thenthereisarelation s = s ( t )suchthat r ( t )= R ( s ( t )). L = b a r( t ) dt. Proof .Foranypartition Pkof C ,thereisapartition tkof[ a,b ] suchthat t0= a 0, k =1 2 ,...,N .Inthelimit N tk 0because rkŠ rk Š 1 0 forall k .Let r k Š 1= r( tk Š 1).Thedierentiabilityof r ( t )impliesthat rkŠ rk Š 1= r k Š 1 tk+ uk tk,where uk 0 as tk 0forevery k (cf.(12.2)).Then,bythetriangleinequality(11.7), r k Š 1 tkŠ uk tk rkŠ rk Š 1 r k Š 1 tk+ uk tk. Bythecontinuityofthederivative,thefunction r( t ) iscontinuous andhenceintegrable.Therefore,itsRiemannsumconverges:Nk =1 r k Š 1 tk b a r( t ) dt as N .

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9812.VECTORFUNCTIONS Putmaxk uk = MN(thelargest uk foragivenpartitionsize N ). ThenNk =1 uk tk MN Nk =1 tk= MN( b Š a ) 0as N because uk 0as tk 0forall k ,andhence MN 0as N Itfollowsfromthesqueezeprinciplethatthelimitof N k =1 rkŠ rk Š 1 as N existsandequals b a r( t ) dt Remark. Let r ( t )becontinuouslydierentiableon[ a,b ]butdoes notnecessarilyde“neaone-to-onecorrespondencewithitsrange C Thentheintegral b a r( t ) dt isnotthelengthofthecurve C asa pointsetinspacebecause r ( t )maytraverseapartof C severaltimes. However,itisalsousefulinpracticalapplications.Suppose r ( t )isa trajectoryofaparticle.Thenitsvelocityis v ( t )= r( t )anditsspeed is v ( t )= v ( t ) .Thedistancetraveledbytheparticleinthetime interval[ a,b ]isgivenby D = b av ( t ) dt = b a r( t ) dt. Ifaparticletravelsalongthesamespacecurve(orsomeofitsparts) severaltimes,thenthedistancetraveleddoesnotcoincidewiththearc lengthofthecurve. Example 12.7 Findthearclengthofoneturnofahelixofradius R thatrisesby h pereachturn. Solution: Letthehelixaxisbethe z axis.Thehelixistracedout bythevectorfunction r ( t )= R cos t,R sin t,th/ (2 ) .Oneturncorrespondstotheinterval t [0 2 ].Therefore, r( t ) = Š R sin t,R cos t,h/ (2 ) = R2+( h/ (2 ))2. Sothenormofthederivativeturnsoutbeconstant.Thearclengthis L = 2 0 r( t ) dt = R2+( h/ (2 ))22 0dt = (2 R )2+ h2. Thisresultisrathereasytoobtainwithoutcalculus.Thehelixlieson acylinderofradius R .Ifthecylinderiscutparalleltoitsaxisand unfoldedintoastrip,thenoneturnofthehelixbecomesthehypotenuse oftheright-angledtrianglewithcatheti2 R and h .Theresultfollows fromthePythagoreantheorem.

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82.ARCLENGTHOFACURVE99 82.1.ReparameterizationofaCurve.InSection79,itwasshownthat aspacecurvede“nedasapointsetinspacecanbetracedbydierent vectorfunctions.Forexample,asemicircleofradius R istracedout bythevectorfunctions r ( t )= R cos t,R sin t, 0 ,t [0 ] R ( u )= u, R2Š u2, 0 ,u [ Š R,R ] Thesevectorfunctionsarerelatedtooneanotherbythecomposition rule: R ( u )= r ( t ( u )) ,t ( u )=cosŠ 1( u/R )or r ( t )= R ( u ( t )) ,u ( t )= R cos t. Thisexampleillustratestheconceptofareparameterizationofacurve. Areparameterizationofacurveisachangeoftheparameterthatlabels pointsofthecurve. Definition 12.12 (ReparameterizationofaCurve) Let r ( t ) traceoutacurve C if t [ a,b ] .Consideraone-to-onemapping [ a,b ] [ a,b] ,thatis,afunction u = u ( t ) withthedomain [ a,b ] and therange [ a,b] thathastheinverse t = t ( u ) .Thevectorfunction R ( u )= r ( t ( u )) iscalleda reparameterization of C Itshouldbeemphasizedthatthegeometricalpropertiesofthecurve (e.g.,itsshapeorlength)donotdependonaparameterizationofthe curvebecausethevectorfunctions r ( t )and R ( u )havethe same range. Areparameterizationofacurveisatechnicaltoolto“ndanalgebraic descriptionofthecurveconvenientforparticularapplications.82.2.ANaturalParameterizationofaSmoothCurve.Supposeoneis travelingalongahig hway fromtown A totown B andcomesupon anaccident.Howcanthelocationoftheaccidentbereportedtothe police?IfonehasaGPSnavigator,onecanreportcoordinatesonthe surfaceoftheEarth.Thisimpliesthatthepoliceshoulduseaspeci“c(GPS)coordinatesystemtolocatetheaccident.Isitpossibleto avoidanyreferencetoacoordinatesystem?Asimplerwaytode“ne thepositionoftheaccidentistoreportthedistancetraveledfrom A alongthehig hwaytothepoint wheretheaccidenthappened(byusing, e.g.,milemarkers).Nocoordinatesystemisneededtouniquelylabel allpointsofthehighwaybysp ecifyingthedistancefromaparticular point A tothepointofinterestalongthehig hway. Thisobservation canbeextendedtoallsmoothcurves(seeFigure12.8,rightpanel).

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10012.VECTORFUNCTIONS Definition 12.13 (NaturalorArcLengthParameterization) Let C beasmoothcurveoflength L betweenpoints A and B .Let r ( t ) t [ a,b ] ,beaone-to-onevectorfunctionthattracesout C sothat r ( a ) and r ( b ) arepositionvectorsof A and B ,respectively.Thenthearc length s = s ( t ) oftheportionofthecurvebetween r ( a ) and r ( t ) isa functionoftheparameter t : s = s ( t )= t a r( u ) du,s [0 ,L ] Thevectorfunction R ( s )= r ( t ( s )) iscalleda naturalorarclength parameterization of C ,where t ( s ) istheinversefunctionof s ( t ) Example 12.8 Findthecoordinatesofapoint P thatis 5 / 3 unitsoflengthawayfromthepoint (4 0 0) alongthehelix r ( t )= 4cos( t ) 4sin( t ) 3 t Solution: Theinitialpointofthehelixcorrespondsto t =0.Sothe arclengthcountedfrom(4 0 0)asafunctionof t is s ( t )= t 0 r( u ) du = t 05 du =5 t because r( u )= Š 4 sin( u ) 4 cos( u ) 3 andtherefore r( u ) = 5 .Hence,theinverseis t = s/ (5 ),andthenaturalparameterizationreads R ( s )= r ( t ( s ))= 4cos( s/ 5) 4sin( s/ 5) 3 s/ 5 .Theposition vectorof P is R (5 / 3)= 2 2 3 .Notethattherearetwopoints ofthehelixatthespeci“eddistancefrom(4 0 0).Onesuchpoint isupwardalongthehelix,andtheotherisdownwardalongit.The downwardpointcorrespondsto t< 0.Hence,itspositionvectoris R ( Š 5 / 3)= 2 Š 2 3 Š Byde“nition,thearclengthisindependentofaparameterizationof aspacecurve.Forsmoothcurves,thiscanalsobeestablishedthrough thechangeofvariablesintheintegralthatdeterminesthearclength. Indeed,let r ( t ), t [ a,b ],beaone-to-onecontinuouslydierentiable vectorfunctionthattracesoutacurve C oflength L .Considerthe changeoftheintegrationvariable t = t ( s ), s [0 ,L ].Then ds = s( t ) dt = r( t ) dt (bydierentiatingtheintegralfor s ( t )withrespect totheupperlimit)and L = b a r( t ) dt = L 0ds foranyparameterizationofthecurve C .

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83.CURVATUREOFASPACECURVE101 Figure12.9. Left :Astraightlinedoesnotbend.The unittangentvectorhaszerorateofchangerelativeto thearclengthparameter s Right :Curvatureofasmoothcurve.Themorea smoothcurvebends,thelargertherateofchangeofthe unittangentvectorrelativetothearclengthparameter becomes.Sothemagnitudeofthederivative(curvature) T( s ) = ( s )canbetakenasageometricalmeasureof bending.82.3.Exercises.(1) Findthearclengthofeachofthefollowingcurves: (i) r ( t )= 3cos t, 2 t, 3sin t Š 2 t 2 (ii) r ( t )= 2 t,t3/ 3 ,t2 ,0 t 1 (2) Findthearclengthoftheportionofthehelix r ( t )= cos t, sin t,t thatliesinsidethesphere x2+ y2+ z2=2. (3) Findthearclengthoftheportionofthecurve r ( t )= 2 t, 3 t2, 3 t3 thatliesbetweentheplanes z =3and z =24. (4) Let C bethecurveofintersectionofthesurfaces z2=2 y and 3 x = yz .Findthelengthof C fromtheorigintothepoint(36 18 6). (5) Reparameterizeeachofthefollowingcurveswithrespectto thearclengthmeasurefromthepointwhere t =0inthedirectionof increasing t : (i) r = t, 1 Š 2 t, 5+3 t (ii) r =2 t t2+1 e1+(2 t2+1Š 1) e3(6) Aparticletravelsalongahelixofradius R thatrises h units oflengthperturn.Letthe z axisbethesymmetryaxisofthehelix. Ifaparticletravelsthedistance4 R fromthepoint( R, 0 0),“ndthe positionvectoroftheparticle.83.CurvatureofaSpaceCurve Considertwocurvespassingthroughapoint P .Bothcurvesbend at P .Whichonebendsmorethantheotherandhowmuchmore? Theanswertothisquestionrequiresanumericalcharacterizationof bending,thatis,anumbercomputedat P foreachcurvewiththe

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10212.VECTORFUNCTIONS propertythatitbecomeslargerasthecurvebendsmore.Naturally,this numbershouldnotdependonaparameterizationofacurve.Suppose thatacurveissmoothsothataunittangentvectorcanbeattached toeverypointofthecurve.Astraightlinedoesnotbend(doesnot curveŽ)soithasthesameunittangentvectoratallitspoints.If acurvebends,thenitsunittangentvectorbecomesafunctionofits positiononthecurve.Thepositiononthecurvecanbespeci“edina coordinate-andparameterization-independentwaybythearclength s countedfromaparticularpointofthecurve.If T ( s )istheunit tangentvectorasafunctionof s ,thenitsderivative T( s )vanishesfor astraightline(seeFigure12.9),whilethiswouldnotbethecasefora generalcurve.Fromthede“nitionofthederivative T( s0)=lims s0 T ( s ) Š T ( s0) s Š s0, itfollowsthatthemagnitude T( s0) becomeslargerwhenthecurve bendsmore.ŽFora“xeddistance s Š s0betweentwoneighboring pointsofthecurve,themagnitude T ( s ) Š T ( s0) becomeslarger whenthecurvebendsmoreatthepointcorrespondingto s0.Sothe number T( s0) canbeusedasanumericalmeasureofthebending or curvature ofacurve. Definition 12.14 (CurvatureofaSmoothCurve) Let C beasmoothcurveandlet T ( s ) betheunittangentvectorasa dierentiablefunctionofthearclengthcountedfromaparticularpoint of C .Thenumber ( s )= d ds T ( s ) iscalledthe curvature of C atthepointcorrespondingtothevalue s ofthearclength. Inpractice,acurvemaynotbegiveninthenaturalparameterization.Therefore,aquestionofinterestisto“ndamethodtocalculate thecurvatureinanyparameterization. Let r ( t )beavectorfunctionin[ a,b ]thattracesoutacurve C such thatthearclengthparametercanbede“nedasafunctionof t s = s ( t ), andithastheinversefunction t = t ( s ).Theunittangentvectoras afunctionoftheparameter t hastheform T ( t )= r( t ) / r( t ) .So, tocalculatethecurvatureasafunctionof t ,therelationbetweenthe derivatives d/ds and d/dt hastobefound.Thegraphsof s ( t )andits inverseareobtainedfromoneanotherbythere”ectionabouttheline s = t .Let( t1,s1)and( t2,s2)bepointsonthegraphof s ( t ),thatis,

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83.CURVATUREOFASPACECURVE103 s1 2= s ( t1 2).Thenthepoints( s1,t1)and( s2,t2)lieonthegraphof theinverseof s ( t ),where t1 2= t ( s1 2).Considertheidentity t ( s1) Š t ( s2) s1Š s2= t1Š t2 s1Š s2= 1 s1Š s2 t1Š t2= 1 s ( t1) Š s ( t2) t1Š t2. When t2 t1,therightsidetendsto1 /s( t1)because s ( t )isdierentiableand,moreover, s( t )= r( t ) > 0for t>a .Hence,thelimit oftheleftsideas s2 s1existstooand,bythede“nitionofthederivative,mustbeequalto t( s1).Thisisknownastheinversefunction theoremforareal-valuedfunctionofonerealargument. Theorem 12.6 (InverseFunctionTheorem) Let s ( t ) haveacontinuousderivativesuchthat s( t ) > 0 .Thenthere existsaninversedierentiablefunction t = t ( s ) and t( s )=1 /s( t ) where t = t ( s ) ontherightside. Thecondition s( t ) > 0guaranteestheexistenceofaone-to-one correspondencebetweenthevariables s and t andhencetheexistence oftheinversefunction(seeCalculusI).Recallthatthederivativecan bewrittenastheratioofdierentials s( t )= ds/dt .Theadvantage ofthisrepresentationisthatthedierentialscanbemanipulatedas numbers.Sothetheoremcanbestatedinthecompactform ds ( t ) dt = 1 dt ( s ) ds,s = s ( t ) Makinguseofthisrelation,one“nds d ds = 1 s( t ) d dt = 1 r( t ) d dt andtherefore (12.4) ( t )= T( t ) r( t ) Notethattheexistenceofthecurvaturerequiresthat r ( t )betwice dierentiablebecause T ( t )isproportionalto r( t ).Dierentiationof theunitvector T cansometimesbeatedioustechnicaltask.The followingtheoremprovidesamoreconvenientwaytocalculatethe curvature. Theorem 12.7 (CurvatureofaCurve) Letacurvebetracedoutbyatwice-dierentiablevectorfunction r ( t ) Thenthecurvatureis (12.5) ( t )= r( t ) r( t ) r( t ) 3.

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10412.VECTORFUNCTIONS Proof. Put v ( t )= r( t ) .Withthisnotation, r( t )= v ( t ) T ( t ) Dierentiatingbothsidesofthisrelation,oneinfers (12.6) r( t )= v( t ) T ( t )+ v ( t ) T( t )= v( t ) v ( t ) r( t )+ v ( t ) T( t ) Sincethecrossproductoftwoparallelvectorsvanishes,itfollowsfrom (12.6)that (12.7) r( t ) r( t )= v ( t ) r( t ) T( t ) Therefore, (12.8) r( t ) r( t ) = v ( t ) r( t ) T( t ) = r( t ) 2 T( t ) sin where istheanglebetween T( t )andthetangentvector r( t ).By construction, T ( t )isaunitvector, T ( t ) 2= T ( t ) T ( t )=1.By takingthederivativeofbothsidesofthelatterrelation,oneobtains T( t ) T ( t )=0,whichmeansthatthederivativeofaunittangentvector isalwaysperpendiculartotheunittangentvector.Since v isparallel to T ,thevector Tisperpendicularto v aswell.Hence, = / 2and sin =1.Substitutingthelatterrelationand T( t ) = ( t ) r( t ) (see(12.4))into(12.8),theexpression(12.5)isderived. Example 12.9 Findthecurvatureofthecurve r ( t )= ln t,t2, 2 t atthepoint P0(0 1 2) Solution: Thepoint P0correspondsto t =1because r (1)= 0 1 2 coincideswiththepositionvectorof P0.Hence,onehastocalculate (1): r(1)= tŠ 1, 2 t, 2 t =1= 1 2 2 r(1) =3 r(1)= Š tŠ 2, 2 0 t =1= Š 1 2 0 r(1) r(1) = Š 2 2 1 Š 2 (1)= r(1) r(1) r(1) 3= 2 2 1 Š 2 33= 6 27 = 2 9 Equation(12.5)canbesimpli“edintwoparticularlyinteresting cases.Ifacurveisplanar(i.e.,itliesinaplane),then,bychoosing thecoordinatesystemsothatthe xy planecoincideswiththeplane inwhichthecurvelies,onehas r ( t )= x ( t ) ,y ( t ) 0 .Since rand

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83.CURVATUREOFASPACECURVE105 rareinthe xy plane,theircrossproductisparalleltothe z axis: r r= 0 0 ,xyŠ xy .Thenwehavethefollowingresult. Corollary 12.1 (CurvatureofaPlanarCurve) Foraplanarcurve r ( t )= x ( t ) ,y ( t ) 0 ,thecurvatureisgivenby = | xyŠ xy| [( x)2+( y)2]3 / 2. Afurthersimpli“cationoccurswhentheplanarcurveisagraph y = f ( x ).Thegraphistracedoutbythevectorfunction r ( t )= t,f ( t ) 0 .Then,intheabovecorollary, x( t )=1, x( t )=0,and y( t )= f( t )= f( x ),whichleadstothefollowingresult. Corollary 12.2 (CurvatureofaGraph) Thecurvatureofthegraph y = f ( x ) isgivenby ( x )= | f( x ) | [1+( f( x ))2]3 / 2.83.1.GeometricalSigni“canceoftheCurvature.Letuscalculatethe curvatureofacircleofradius R .Put r ( t )= R cos t R sin t ,0 .Then r( t ) = Š R sin t,R cos t, 0 = R and | xyŠ xy| = R2.Therefore, thecurvatureisconstantalongthecircleandequalsareciprocalofits radius, =1 /R .Thefactthatthecurvatureisindependentofits positiononthecirclecanbeanticipatedfromtherotationalsymmetry ofthecircle(itbendsuniformly).Naturally,iftwocirclesofdierent radiipassthroughthesamepoint,thenthecircleofsmallerradius bendsmore.Notealsothatthecurvaturehasthedimensionofthe inverselength.Thismotivatesthefollowingde“nition. Definition 12.15 (CurvatureRadius) Thereciprocalofthe curvatureofacurveiscalledthe curvatureradius ( t )=1 / ( t ) Letaplanarcurvehaveacurvature atapoint P .Considera circleofradius =1 / throughthesamepoint P .Thecurveand thecirclehavethesamecurvatureat P ;thatis,inasucientlysmall neighborhoodof P ,thecircleapproximateswellthecurveastheyare equallybentat P .So,ifonesaysthatthecurvatureofacurveata point P is inversemeters,thenthecurvelookslikeacircleofradius 1 / metersnear P Forageneralspatialcurve,noteverycircleofradius =1 / that passesthrough P wouldapproximatewellthecurvenear P .Thebest approximationisattainedwhenthecircleandthecurvebendŽinthe

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10612.VECTORFUNCTIONS Figure12.10. Left :Curvatureradius.Asmooth curvenearapoint P canbeapproximatedbyaportionofacircleofradius =1 / .Thecurvebendsinthe samewayasacircleofradiusthatisthereciprocalof thecurvature.Alargecurvatureatapointcorresponds toasmallcurvatureradius. Middle :Osculatingplaneandosculatingcircle.Theosculatingplaneatapoint P containsthetangentvector T anditsderivative Tat P andhenceisperpendicular to n = T T.Theosculatingcircleliesintheosculating plane,ithasradius =1 / ,anditscenterisadistance from P inthedirectionof T.Onesaysthatthecurve bendsŽintheosculatingplane. Right :Foracurvetracedoutbyavectorfunction r ( t ), thederivatives rand ratanypoint P0lieintheosculatingplanethrough P0.Sothenormaltotheosculating planecanalsobecomputedas n = r ( t0) r( t0),where r ( t0)isthepositionvectorof P0. sameplane.Fromthediscussionatthebeginningofthissection,it mightbeconcludedthatacurvebendsintheplanethatcontainsthe unittangentvector T anditsderivative T. Definition 12.16 (OsculatingPlaneandCircle) Theplanethroughapoint P ofacurvethatisparalleltotheunit tangentvector T anditsderivative T = 0 at P iscalledthe osculating plane at P .Thecircleofradius =1 / ,where isthecurvatureat P ,through P thatliesintheosculatingplaneandwhosecenterisin thedirectionof Tfrom P iscalledthe osculatingcircle at P Recallthat T T (seetheproofofTheorem12.7).Bythegeometricalinterpretationofthederivative, Tshouldpointinthedirection inwhichthecurvebends.Hence,theosculatingcirclemusthavethe same Tatacommonpoint P inordertomakethebestapproximation

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83.CURVATUREOFASPACECURVE107 tothecurvenear P .Therefore,itscentermustbeinthedirectionof Tfrom P ,notintheoppositeone. Theorem 12.8 (EquationoftheOsculatingPlane) Letacurve C betracedoutbyatwice-dierentiablevectorfunction r ( t ) Let P0beapointof C suchthatitspositionvectoris r ( t0)= x0,y0,z0 atwhichthevector n = r( t0) r( t0) doesnotvanish.Anequationof theosculatingplanethrough P0is n1( x Š x0)+ n2( y Š y0)+ n3( z Š z0)=0 n = n1,n2,n3 Proof. Itfollowsfrom(12.6)thatthesecondderivative r( t0)lies intheosculatingplanebecauseitisalinearcombinationof T ( t0)and T( t0).Hence,theosculatingplanecontainsthe“rstandsecondderivatives r( t0)and r( t0).Therefore,theircrossproduct n = r( t0) r( t0) isperpendiculartotheosculatingplane,andtheconclusionofthetheoremfollows. Example 12.10 Forthecurve r ( t )= t,t2,t3 ,“ndtheosculating planethroughthepoint (1 1 1) Solution: Thepointinquestioncorrespondsto t =1.Therefore,the normaloftheosculatingplaneis n = r(1) r(1)= 1 2 3 0 2 6 = 6 Š 6 2 .Theosculatingplaneis6( x Š 1) Š 6( y Š 1)+2( z Š 1)=0 or3 x Š 3 y + z =1. 83.2.StudyProblems.Problem12.11. Findthemaximalcurvatureofthegraphoftheexponential, y = ex,andthepoint(s)atwhichitoccurs. Solution: Thecurvatureofthegraphisgivenby ( x )= ex/ (1+ e2 x)3 / 2.Criticalpointsaredeterminedby ( x )=0or ( x )= ex(1+ e2 x)1 / 2[2 e2 xŠ 1] (1+ e2 x)3=0 2 e2 xŠ 1=0 x = Š ln2 2 Fromtheshapeofthegraphoftheexponential,itisclearthat ( x )attainsitsabsolutemaximum(maximalbending)and max= ( Š ln(2) / 2)=2 / 33 / 2. Problem12.12.(EquationoftheOsculatingCircle) Findavectorfunctionthattracesouttheosculatingcircleofacurve r ( t ) atapoint r ( t0) .

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10812.VECTORFUNCTIONS Solution: Put r0= r ( t0)and T0= T ( t0)(theunittangentvector tothecurveatthepointwiththepositionvector r0).Put N0= T( t0) / T( t0) ;itisaunitvectorinthedirectionof T( t0).Let 0= 1 / ( t0)bethecurvatureradiusatthepoint r0.Thecenterofthe osculatingcirclemustlie 0unitsoflengthfromthepoint r0inthe directionof N0.Thus,itspositionvectoris R0= r0+ 0 N0.Let R ( t ) bethepositionvectorofagenericpointoftheosculatingcircle.Then thevector R ( t ) Š R0liesintheosculatingplaneandhencemustbea linearcombinationof T0and N0,thatis, R ( t ) Š R0= a ( t ) N0+ b ( t ) T0. To“ndthefunctions a ( t )and b ( t ),notethatthevector R ( t ) Š R0traces outacircleofradius 0.Inacoordinatesysteminwhich N0coincides with exand T0with ey(suchacoordinatesystemalwaysexistsbecause N0and T0areunitorthogonalvectors),thevector Š 0cos( t ) ex+ 0sin( t ) eytracesoutacircleofradius 0inthe xy plane.Thus,one canalwaysput a ( t )= Š 0cos t and b ( t )= 0sin t ,andthevector functionthattracesouttheosculatingcircleis R ( t )= r0+ 0 1 Š cos t N0+ 0sin t T0, where t [0 2 ]. Problem12.13. Considerahelix r ( t )= R cos( t ) ,R sin( t ) ,ht where and h arenumericalparameters.Thearclengthofoneturnof thehelixisafunctionoftheparameter L = L ( ) ,andthecurvature atany“xedpointofthehelixisalsoafunctionof = ( ) .Use onlygeometricalarguments(nocalculus)to“ndthelimitsof L ( ) and ( ) as Solution: Thevectorfunction r ( t )tracesoutoneturnofthehelix when t rangesovertheperiodofcos( t )orsin( t )(i.e.,overtheinterval oflength2 / ).Thus,thehelixrisesby2 h/ = H ( )alongthe z axispereachturn.When ,theheight H ( )tendsto0so thateachturnofthehelixbecomescloserandclosertoacircleofradius R .Therefore, L( ) 2 R (thecircumference)and ( ) 1 /R (the curvatureofthecircle)as Acalculusapproachrequiresalotmoreworktoestablishthisresult: L ( )= 2 / 0 r( t ) dt = 2 ( R )2+ h2=2 R2+( h/ )2 2 R,

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84.PRACTICALAPPLICATIONS109 ( )= r( t ) r( t ) r( t ) 3= R2[( R )2+ h2]1 / 2 [( R )2+ h2]3 / 2= R R2+( h/ )2 1 R as 83.3.Exercises.(1) Findthecurvatureof r ( t )= t,t2/ 2 ,t3/ 3 atthe pointofitsintersectionwiththeplane z =2 xy +1 / 3. (2) Findthemaximalandminimalcurvaturesofthegraph y = cos( ax )andthepointsatwhichtheyoccur. (3) Useageometricalinterpretationofthecurvaturetoguessthe pointonthegraphs y = ax2and y = ax4wherethemaximalcurvature occurs.Thenverifyyourguessbycalculations. (4) Let r ( t )= t3,t2, 0 .Thiscurvehasacuspat t =0.Findthe curvaturefor t =0andinvestigateitslimitas t 0. (5) Findanequationfortheosculatingandnormalplanesforthe curve r ( t )= ln( t ) 2 t,t2 atthepoint P0ofitsintersectionwiththe plane y Š z =1.Aplaneisnormaltoacurveatapointifthetangent tothecurveatthatpointisnormaltotheplane. (6) Provethatthetrajectoryofaparticleisplanarifitsvelocity v ( t )remainsperpendiculartoaconstantvector n anditsacceleration is a ( t )= n v ( t )+ ( t ) v ( t ),where ( t )isafunctionoftime.Find anequationoftheplaneinwhichthetrajectoryliesiftheparticleis knowntopassapoint r0.84.PracticalApplications84.1.TangentialandNormalAccelerations.Let r ( t )bethetrajectory ofaparticle( t istime).Then v ( t )= r( t )and a ( t )= v( t )arethe velocityandaccelerationoftheparticle.Themagnitudeofthevelocity vectoristhespeed, v ( t )= v ( t ) .If T ( t )istheunittangentvector tothetrajectory,then T( t )isperpendiculartoit.Theunitvector N ( t )= T( t ) / T( t ) iscalledaunit normal tothetrajectory.In particular,theosculatingplaneatanypointofthetrajectorycontains T ( t )and N ( t ).Itfollowsfrom(12.6)thattheaccelerationalwayslies intheosculatingplane: a = v T + v T= vT + v T N .

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11012.VECTORFUNCTIONS Figure12.11. Left :Decompositionoftheacceleration a ofaparticleintonormalandtangentialcomponents. Thetangentialcomponent aTisthescalarprojectionof a ontotheunittangentvector T .Thenormalcomponent isthescalarprojectionof a ontotheunitnormalvector N .Thevectors r and v arethepositionandvelocity vectorsoftheparticle. Right :Thetangent,normal,andbinormalvectorsassociatedwithasmoothcurve.Thesevectorsaremutually orthogonalandhaveunitlength.Thebinormalisde“nedby B = T N .Theshapeofthecurveisuniquely determinedbytheorientationofthetripleofvectors T N ,and B asfunctionsofthearclengthparameterupto generalrigidrotationsandtranslationsofthecurveas thewhole. Furthermore,substitutingtherelations = T /v and =1 / intothelatterequation,one“nds(seeFigure12.11,leftpanel)that a = aT T + aN N aT= v= T a = v a v aN= v2= v2 = v a v Definition 12.17 (TangentialandNormalAccelerations) Scalarprojections aTand aNoftheaccelerationvectorontotheunit tangentandnormalvectorsatanypointofthetrajectoryofmotionare called tangentialandnormalaccelerations ,respectively. Thetangentialacceleration aTdeterminestherateofchangeof aparticlesspeed,whilethenormalaccelerationappearsonlywhen theparticlemakesaturn.ŽInparticular,acircularmotionwitha

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84.PRACTICALAPPLICATIONS111 constantspeed, v = v0,hasnotangentialacceleration, aT=0,and thenormalaccelerationisconstant, aN= v2 0/R ,where R isthecircle radius. Togainanintuitiveunderstandingofthetangentialandnormal accelerations,consideracarmovingalongaroad.Thespeedofthecar canbechangedbypressingthegasorbrakepedals.Whenoneofthese pedalsissuddenlypressed,onecanfeelaforcealongthedirectionof motionofthecar(thetangentialdirection).Thecarspeedometeralso showsthatthespeedchanges,indicatingthatthisforceisduetotheaccelerationalongtheroad(i.e.,thetangentialacceleration aT= v =0). Whenthecarmovesalongastraightroadwithaconstantspeed,its accelerationis0.Whentheroadtakesaturn,thesteeringwheelmust beturnedinordertokeepthecarontheroad,whilethecarmaintainsaconstantspeed.Inthiscase,onecanfeelaforcenormalto theroad.Itislargerforsharperturns(largercurvatureorsmaller curvatureradius)andalsogrowswhenthesameturnispassedwitha greaterspeed.Thisforceisduetothenormalacceleration, aN= v2/ andiscalleda centrifugalforce .Whenmakingaturn,thecardoesnot slideotheroadaslongasthefrictionforcebetweenthetiresandthe roadcompensatesforthecentrifugalforce.Themaximalfrictionforce dependsontheroadandtireconditions(e.g.,awetroadandworn tiresreducesubstantiallythemaximalfrictionforce).Thecentrifugal forceisdeterminedbythespeed(thecurvatureoftheroadis“xedby theroadshape).So,forahighenoughspeed,thecentrifugalforcecan nolongerbecompensatedforbythefrictionforceandthecarwould skidotheroad.Forthisreason,suggestedspeedlimitsignsareoften placedathig hway exits.Ifonedrivesacaronahig hway exitwith aspeedtwiceashighasthesuggestedspeed, theriskofskiddingo theroadisquadrupled,notdoubled ,becausethenormalacceleration aN= v2/ quadrupleswhenthespeed v isdoubled. Example 12.11 Aroadhasaparabolicshape, y = x2/ (2 R ) ,where ( x,y ) arecoordinatesofpointsoftheroadand R isaconstant(all measuredinunitsoflength,e.g.,meters).Asafetyassessmentrequires thatthenormalaccelerationontheroadshouldnotex ceedathreshold value am(e.g.,meterspersecondsquared)toavoidskiddingothe road.Ifacarmoveswithaconstants peed v0alongtheroad,“ndthe portionoftheroadwherethecarmightskidotheroad. Solution: Thenormalaccelerationofthecarasafunctionof position (nottime!)is aN( x )= ( x ) v2 0.Thecurvatureofthegraph y = x2/ (2 R ) is ( x )=(1 /R )[1+( x/R )2]Š 3 / 2.Themaximalcurvatureandhencethe maximalnormalaccelerationareattainedat x =0.So,ifthespeed

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11212.VECTORFUNCTIONS issuchthat aN(0)= v2 0/Ram.Thecar canskidotheroadwhenmovingonitsportioncorrespondingtothe interval Š R ( Š 1)1 / 2 x R ( Š 1)1 / 2. 84.2.Frenet-SerretFormulas.Theshapeofaspacecurveasapointset isindependentofaparameterizationofthecurve.Anaturalquestion arises:Whatparametersofthecurvedetermineitsshape?Suppose thecurveissmoothenoughsothattheunittangentvector T ( s )andits derivative T( s )canbede“nedasfunctionsofthearclength s counted fromanendpointofthecurve.Let N ( s )betheunitnormalvectorof thecurve. Definition 12.18 (BinormalVector) Let T and N betheunittangentandnormalvectorsatapointofa curve.Theunitvector B = T N iscalledthe binormal(unit)vector So,witheverypointofasmoothcurve,onecanassociateatriple unitofmutuallyorthogonalvectorssothatoneofthemistangentto thecurvewhiletheothertwospantheplanenormaltothetangent vector(normaltothecurve).Byasuitablerotation,thetripleof vectors T N ,and B canbeorientedparalleltotheaxesofanygiven coordinatesystem,thatis,parallelto ex, ey,and ez,respectively(note that ex ey= ez;thisiswhythebinormalisde“nedas T N ,notas N T = Š T N ).Theorientationoftheunittangent,normal,and binormalvectorsrelativetosomecoordinatesystemdependsonthe pointofthecurve.Thetripleofthesevectorscanonlyrotateasthe pointslidesalongthecurve(thevectorsaremutuallyorthogonaland unitatanypoint).Therefore,therateswithrespecttothearclength atwhichthesevectorschangemustbecharacteristicfortheshapeof thecurve(seeFigure12.11,rightpanel). Bythede“nitionofthecurvature, T( s )= ( s ) N ( s ).Next,considertherate: B=( T N )= T N + T N= T Nbecause T( s )isparallelto N ( s ).Itfollowsfromthisequationthat Bisperpendicularto T ,and,since B isaunitvector,itsderivative mustalsobeperpendicularto B .Thus, Bmustbeparallelto N .

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84.PRACTICALAPPLICATIONS113 Thisconclusionestablishestheexistenceofanotherscalarquantity thatcharacterizesthecurveshape. Definition 12.19 (TorsionofaCurve) Let N ( s ) and B ( s ) beunitnormalandbinormalvectorsofthecurve asfunctionsofthearclength s .Then d B ( s ) ds = Š ( s ) N ( s ) andthenumber ( s ) iscalledthe torsionofthecurve Byde“nition,thetorsionismeasuredinunitsofareciprocallength, justlikethecurvature,becausetheunitvectors T N ,and B are dimensionless. Atanypointofacurve,thebinormal B isperpendiculartothe osculatingplane.So,ifthecurveisplanar,then B doesnotchange alongthecurve, B( s )= 0 ,becausetheosculatingplaneatanypoint coincideswiththeplaneinwhichthecurvelies.Thus,thetorsionisa localnumericalcharacteristicthatdetermineshowfastthecurvedeviatesfromtheosculatingplanewhilebendinginitwithsomecurvature radius. Itfollowsfromtherelation N = B T (compare ey= ez ex)that N=( B T )= B T + B T= Š N T + B N = B Š T wherethede“nitionsofthetorsionandcurvaturehavebeenused.The obtainedratesoftheunitvectorsareknownasthe Frenet-Serretformulas : T( s )= ( s ) N ( s ) (12.9) N( s )= Š ( s ) T ( s )+ ( s ) B ( s ) (12.10) B( s )= Š ( s ) N ( s ) (12.11) Supposethatthevectors T (0), N (0),and B (0)aregivenataninitial pointofthecurve.Then T ( s ), N ( s ),and B ( s )areuniquelydeterminedbysolvingtheFrenet-Serretequations,providedthecurvature andtorsionaregivenasfunctionsofthearclength.Thisestablishesa fundamentaltheoremabouttheshapeofaspacecurve. Theorem 12.9 (ShapeofaSmoothCurveinSpace) Acurveinspaceisdeterminedbyitscurvatureandtorsionasfunctions ofthearclengthuptorigidrotationsandtranslationsofthecurveas awhole.

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11412.VECTORFUNCTIONS Acurvewithzerocurvatureandtorsionisastraightline.Indeed, inthiscase,thetangent,normal,andbinormalvectorsremainconstant alongthecurve, T ( s )= T (0), N ( s )= N (0),and B ( s )= B (0);that is,ithasaconstantunittangentvector,whichisthecharacteristic propertyofastraightline. Example 12.12 Provethatacurvewithaconstantcurvature ( s )= 0 =0 andzerotorsion ( s )=0 isacircle(oritsportion)ofradius R =1 /0. Solution: Letthecoordinatesystembesetsothat T (0)= ex, N (0)= ey,and B (0)= ez.Sincethetorsionis0,thebinormal doesnotchangealongthecurve, B ( s )= ez.Anyunitvector T in the xy planecanbewrittenas T = cos sin 0 ,where = ( s ) suchthat (0)=0.Thenaunitvector N perpendicularto T such that T N = B = ezmusthavetheform N = Š sin cos 0 Equation(12.9)gives T= N = 0 N andtherefore ( s )= 0or ( s )= 0s .Let r ( s )= x ( s ) ,y ( s ) 0 beanaturalparameterizationof thecurve.Itfollowsfromthede“nitionofthearclengthparameter s that r( s )= T ( s )(seetherelationabove(12.4)).Hence, r( s )= cos( 0s ) sin( 0s ) 0 r ( s )= r0+ Š 1 0sin( 0s ) Š Š 1 0cos( 0s ) 0 where r0= x0,y0,z0 .Thus,thecurveliesintheplane z = z0and ( x ( s ) Š x0)2+( y ( s ) Š y0)2= R2,where R =1 /0,forallvaluesof s ; thatis,thecurveisacircle(oritsportion)ofradius R Theorem 12.10 (TorsionofaCurve) Thetorsionofacurvetracedoutby r ( t ) isgivenby ( t )= ( r( t ) r( t )) r( t ) r( t ) r( t ) 2. Proof. Put r( t ) = v ( t )(if s = s ( t )isthearclengthasafunction of t ,then s= v ).By(12.6)andthede“nitionofthecurvature, (12.12) r= v T + v2 N andby(12.7)andthede“nitionofthebinormal, (12.13) r r= v T r= v3 B Dierentiationofbothsidesof(12.12)gives r= v T + v T+( v2+2 vv) N + v2 N.

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84.PRACTICALAPPLICATIONS115 Thederivatives T( t )and N( t )arefoundbymakinguseofthedierentiationrule d/ds =(1 /s( t ))( d/dt )=(1 /v )( d/dt )intheFrenet-Serret equations(12.9)and(12.10): T= v N N= Š v T + v B Therefore, (12.14) r=( vŠ 2v3) T +(3 vv+ v2) N + v3 B Sincethetangent,normal,andbinormalvectorsareunitandorthogonaltoeachother,( r r) r= v3( r r) B = 2v6 .Therefore, = ( r r) r 2v6andtheconclusionofthetheoremfollowsfromTheorem12.7, = r r /v3. Remark. Relation(12.13)isoftenmoreconvenientforcalculating theunitbinormalvectorratherthanitsde“nition.Theunittangent, normal,andbinormalvectorsataparticularpoint r ( t0)ofthecurve r ( t )are T ( t0)= r( t0) r( t0) B ( t0)= r( t0) r( t0) r( t0) r( t0) N ( t0)= B ( t0) T ( t0) .84.3.StudyProblems.Problem12.14. Findthepositionvector r ( t ) ofaparticleasafunctionoftime t iftheparticlemovesclockwisealongacircularpathof radius R inthe xy planethrough r (0)= R, 0 0 withaconstants peed v0. Solution: Foracircleofradius R inthe xy planethroughthepoint ( R, 0 0), r ( t )= R cos ,R sin 0 ,where = ( t )suchthat (0)= 0.Thenthevelocityis v ( t )= r( t )= Š R sin ,R cos 0 .Hence, thecondition v ( t ) = v0yields R | ( t ) | = v0or ( t )= ( v0/R ) t and r ( t )= R cos( t ) R sin( t ) 0 where = v0/R istheangularvelocity.Thesecondcomponentmustbe takenwiththeminussignbecausetheparticlerevolvesclockwise(the secondcomponentshouldbecomenegativeimmediatelyafter t =0).

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11612.VECTORFUNCTIONS Problem12.15. Lettheparticlepositionvectorasafunctionoftime t be r ( t )= ln( t ) ,t2, 2 t t> 0 .Findthes peed, tangentialandnormal accelerations,theunittangent,normal,andbinormalvectors,andthe torsionofthetrajectoryatthepoint P0(0 1 2) Solution: ByExample12.9,thevelocityandaccelerationvectorsat P0are v = 1 2 2 and a = Š 1 2 0 .Sothespeedis v = v =3. Thetangentialaccelerationis aT= v a /v =1.As v a =2 Š 2 Š 1 2 thenormalaccelerationis aN= v a /v =6 / 3=2.Theunittangent vectoris T = v /v =(1 / 3) 1 2 2 andtheunitbinormalvectoris B = v a / v a =(1 / 3) Š 2 Š 1 2 astheunitvectoralong v a Therefore,theunitnormalvectoris N = T B =(1 / 9) v ( v a )= (1 / 3) Š 2 2 Š 1 .To“ndthetorsionat P0,thethirdderivativeat t =0 hastobecalculated, r(1)= 2 /t2, 0 0 |t =1= 2 0 0 = b .Therefore, (1)=( v a ) b / v a 2= Š 8 / 36= Š 2 / 9. Problem12.16.(CurveswithConstantCurvatureandTorsion) Findtheshapeofacurvethathasconstant,nonzerocurvatureand torsion. Solution: Put ( s )= 0 =0and ( s )= 0 =0.Itfollowsfrom (12.9)and(12.11)thatthevector w = T + B doesnotchangealong thecurve, w( s )=0.Indeed,because ( s )= ( s )=0,onehas w= T+ B=( Š ) N = 0 .Itisthereforeconvenientto introducenewunitvectorsorthogonalto N : u =cos( ) T Š sin( ) B w =sin( ) T +cos( ) B wherecos = 0/ ,sin = 0/ ,and =( 2 0+ 2 0)1 / 2.Byconstruction,theunitvectors u w ,and N aremutuallyorthogonalunit vectors,whichiseasytoverifybycalculatingthecorrespondingdot products, u u = w w =1and u w =0.Also, u w = N Ithasbeenestablishedthatthevector w ( s )isconstantalongthecurve andsoisitsunitvector w ( s )= w ( s ) / .Therefore,onecanalways choosethecoordinatesystemsothat w ( s )= w (0)= ez. Bydierentiatingthevector u andusingtheFrenet-Serretequations, u= N Put u = cos sin 0 ,where = ( s ).Thentheunitnormalvector is N = u w = Š sin cos 0 and u= N .Hence, ( s )=

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84.PRACTICALAPPLICATIONS117 or ( s )= s (theintegrationconstantissetto0;seeExample12.12). Expressingthevector T via u and w T =cos( ) u +sin( ) w oneinfers(compareExample12.12) r( s )= T ( s )= 0 cos( s ) 0 sin( s ) 0 where r ( s )isanaturalparameterizationofthecurve.Theintegration ofthisequationgives r ( s )= r0+ R sin( s ) Š R cos( s ) ,hs ,R = 0 2,h = 0 where =( 2 0+ 2 0)1 / 2.Thisisahelixofradius R whoseaxisgoes throughthepoint r0paralleltothe z axis;thehelixclimbsalongits axisby2 h/ pereachturn. Remark. Acurveinaneighborhoodofitsparticularpoint P0canbe wellapproximatedbyahelixthroughthatpointwhosecurvatureand torsioncoincidewiththecurvatureandtorsionofthecurveat P0.Such anapproximationisbetterthantheapproximationbyanosculating circlebecausethelatterdoesnottakeintoaccounttherateatwhich thecurvedeviatesfromtheosculatingplane(whichisdeterminedby thetorsion). Problem12.17.(MotioninaConstantMagneticField,Revisited) Theforceactingonachargedparticlemovinginthemagnetic“eld B is givenby F =( e/c ) v B ,where e istheelectricchargeoftheparticle, c isthespeedof light,and v isitsvelocity.Showthatthetrajectoryof theparticleinaconstantmagnetic“eldisahelixwhoseaxisisparallel tothemagnetic“eld. Solution: IncontrasttoStudyProblem12.10,heretheshapeofthe trajectoryistobeobtaineddirectlyfromNewtonssecondlawwith arbitraryinitialconditions.Choosethecoordinatesystemsothatthe magnetic“eldisparalleltothe z axis, B = B ez,where B isthe magnitudeofthemagnetic“eld.Newtonslawofmotion, m a = F where m isthemassoftheparticle,determinestheacceleration, a = v B = B v ez,where = e/ ( mc ).First,notethat a3= ez a =0. Hence, v3= v=const.Second,theacceleration,andvelocityremain orthogonalduringthemotion,andthereforethetangentialacceleration vanishes, aT= v a =0.Hence,thespeedoftheparticleisaconstantof motion, v = v0(because v= aT=0).Put v = v+ v ez,where vis theprojectionof v ontothe xy plane.Since v = v0,themagnitudeof

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11812.VECTORFUNCTIONS visalsoconstant, v = v=( v2 0Š v2 )1 / 2.Thevelocityvectormay thereforebewrittenintheform v = vcos ,vsin ,vz 0 ,where = ( t ).Takingthederivative a = vandsubstitutingitintoNewtons equation,one“nds ( t )= B or ( t )= Bt + 0.Integrationofthe equation r= v yieldsthetrajectoryofmotion: r ( t )= r0+ R sin( t + 0) Š R cos( t + 0) ,vt where = eB/ ( mc )istheso-calledcyclotronfrequencyand R = v/ .Thisequationdescribesahelixofradius R whoseaxisgoes through r0paralleltothe z axis.Soachargedparticlemovesalonga helixthatwindsaboutforcelinesofthemagnetic“eld.Theparticle revolvesintheplaneperpendiculartothemagnetic“eldwithfrequency = eB/ ( mc ).Ineachturn,theparticlemovesalongthemagnetic “eldadistance h =2 v/ .Inparticular,iftheinitialvelocityis perpendiculartothemagnetic“eld(i.e., v=0),thenthetrajectory isacircleofradius R Problem12.18. Supposethattheforceactingonaparticleofmass m isproportionaltothepositionvectoroftheparticle(suchforces arecalled central ).Provethattheangularmomentumoftheparticle, L = m r v ,isaconstantofmotion(i.e., d L /dt =0 ). Solution: Sinceacentralforce F isparalleltothepositionvector r ,theircrossproductvanishes, r F = 0 .ByNewtonssecondlaw, m a = F andhence m r a = 0 .Therefore, d L dt = m ( r v )= m ( r v + r v)= m r a = 0 where r= v v = a ,and v v = 0 havebeenused. Problem12.19.(KeplersLawsofPlanetaryMotion). Newtonslawof gravitystatesthattwomasses m and M atadistance r areattracted byaforceofmagnitude GmM/r2,where G istheuniversalconstant (called Newtonsconstant ).ProveKeplerslawsofplanetarymotion: 1.AplanetrevolvesaroundtheSuninanellipticalorbitwiththeSun atonefocus. 2.ThelinejoiningtheSuntoaplanetsweepsoutequalareasinequal times. 3.Thesquareoftheperiodofrevolutionofaplanetisproportionalto thecubeofthelengthofthemajoraxisofitsorbit.

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84.PRACTICALAPPLICATIONS119 Solution: LettheSunbeattheoriginofacoordinatesystemand let r bethepositionvectorofaplanet.Let r = r /r betheunitvector parallelto r .Thenthegravitationalforceis F = Š GMm r2 r = Š GMm r3r where M isthemassoftheSunand m isthemassofaplanet.The minussignisnecessarybecauseanattractiveforcemustbeopposite tothepositionvector.ByNewtonssecondlaw,thetrajectoryofa planetsatis“estheequation m a = F andhence a = Š GM r3r Thegravitiesforceisacentralforce,and,byStudyProblem12.18,the vector r v = l isaconstantofmotion.Onehas v = r=( r r )= r r + r r.Usingthisidentity,theconstantofmotioncanalsobewritten as l = r v = r r v = r ( r r r + r r r)= r2( r r) Usingtheruleforthedoublecrossproduct(seeStudyProblem11.17), oneinfersthat a l = Š GM r2 r l = Š GM r ( r r)= GM r, where r r =1hasbeenused.Ontheotherhand, ( v l )= v l + v l= a l because l= 0 .Itfollowsfromthesetwoequationsthat ( v l )= GM r= v l = GM r + c where c isaconstantvector.Themotionischaracterizedbytwo constantvectors l and c .Itoccursintheplanethroughtheorigin thatisperpendiculartotheconstantvector l because l = r v must beorthogonalto r .Itisthereforeconvenienttochoosethecoordinate systemsothat l isparalleltothe z axisand c tothe x axisasshown inFigure12.12(leftpanel). Thevector r liesinthe xy plane.Let bethepolarangleof r (i.e., r c = rc cos ,where c = c isthelengthof c ).Then r ( v l )= r ( GM r + c )= GMr + rc cos Ontheotherhand,usingacyclicpermutationinthetripleproduct, r ( v l )= l ( r v )= l l = l2,

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12012.VECTORFUNCTIONS Figure12.12. Left :ThesetupofthecoordinatesystemforthederivationofKeplers“rstlaw. Right :AnillustrationtothederivationofKeplerssecondlaw. where l = l isthelengthof l .Thecomparisonofthelasttwo equationsyieldstheequationforthetrajectory: l2= r ( GM + b cos )= r = ed 1+ e cos where d = l2/c and e = c/ ( GM ).Thisisthepolarequationofa conicsectionwithfocusattheoriginandeccentricity e (seeCalculus II).Thus, allpossibletrajectoriesofanymassivebodyinasolarsystemareconicsections !Thisisaquiteremarkableresult.Parabolas andhyperbolasdonotcorrespondtoaperiodicmotion.Soaplanet mustfollowanelliptictrajectorywiththeSunatonefocus.Allobjectscomingtothesolarsystemfromouterspace(i.e.,thosethatare notcon“nedbythegravitationalpulloftheSun)shouldfolloweither parabolicorhyperbolictrajectories. ToproveKeplerssecondlaw,put r = cos sin 0 andhence r= Š sin ,cos 0 .Therefore, l = r2( r r)= 0 0 ,r2 = l = r2. Theareaofasectorwithangle d sweptby r is dA =1 2r2d (see CalculusII;theareaboundedbyapolargraph r = r ( )).Hence, dA dt = 1 2 r2d dt = l 2 .

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84.PRACTICALAPPLICATIONS121 Foranymomentsoftime t1and t2,theareaofthesectorbetween r ( t1) and r ( t2)is A12= t2t1dA dt dt = t2t1l 2 dt = l 2 ( t2Š t1) Thus,thepositionvector r sweepsoutequalareasinequaltimes(see Figure12.12,rightpanel). Keplersthirdlawfollowsfromthelastequation.Indeed,theentire areaoftheellipse A issweptwhen t2Š t1= T istheperiodofthe motion.Ifthemajorandminoraxesoftheellipseare2 a and2 b respectively, a>b ,then A = ab = lT/ 2and T =2 ab/l .Now recallthat ed = b2/a foranellipticconicsection(seeCalculusII)or b2= eda = l2a/ ( GM ).Hence, T2= 4 2a2b2 l2= 4 2 GM a3. Notethattheproportionalityconstant4 2/ ( GM )isindependentofthe massofaplanet;therefore,Keplerslawsare universal forallmassive objectstrappedbytheSun(planets,asteroids,andcomets). 84.4.Exercises.(1) Findthenormalandtangentialaccelerationsof aparticlewiththepositionvector r ( t )= t2+1 ,t3,t2Š 1 whenthe particleisattheleastdistancefromtheorigin. (2) Findthetangentialandnormalaccelerationsofaparticlewith thepositionvector r ( t )= R sin( t + 0) Š R cos( t + 0) ,v0t ,where R 0,and v0areconstants(seeStudyProblem12.17). (3) Theshapeofawindingroadcanbeapproximatedbythegraph y = L cos( x/L ),wherethecoordinatesareinmilesand L =1mile. Theconditionoftheroadissuchthatifthenormalaccelerationofa caronitexceeds10m / s2,thecarmayskidotheroad.Recommend aspeedlimitforthisportionoftheroad. (4) Supposeaparticlemovessothatitstangentialacceleration isconstant,whilethenormalaccelerationremains0.Whatisthe trajectoryoftheparticle? (5) Supposeaparticlemovessothatitstangentaccelerationremains0,whilethenormalaccelerationisconstant.Whatisthetrajectoryoftheparticle? Hint: Prove“rstthattheaccelerationoftheparticlehastheform a = b v ,where v isthevelocityand b isavectorthatcandepend ontime.Usethisfacttoprovethatthetorsionofthetrajectoryis constant.ThenseeStudyProblem12.16.

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CHAPTER13 DifferentiationofMultivariable Functions 85.FunctionsofSeveralVariables Theconceptofafunctionofseveralvariablescanbequalitatively understoodfromsimpleexamplesineverydaylife.Thetemperaturein aroommayvaryfrompointtopoint.Apointinspacecanbede“ned byanorderedtripleofnumbersthatarecoordinatesofthepointin somecoordinatesystem,say,( x,y,z ).Measurementsofthetemperatureateverypointfromaset D inspaceassignarealnumber T (the temperature)toeverypointof D .Thedependenceof T oncoordinatesofthepointisindicatedbywriting T = T ( x,y,z ).Similarly, theconcentrationofachemicalcandependonapointinspace.In addition,ifthechemicalreactswithotherchemicals,itsconcentration atapointmayalsochangewithtime.Inthiscase,theconcentration C dependsonfourvariables„threespatialcoordinatesandthetime t „ C = C ( x,y,z,t ).Ingeneral,ifthevalueofaquantity f depends onvaluesofseveralotherquantities,say, x1, x2,..., xn,thisdependenceisindicatedbywriting f = f ( x1,x2,...,xn).Inotherwords, f = f ( x1,x2,...,xn)indicatesarulethatassignsanumber f toeach ordered n -tupleofrealnumbers( x1,x2,...,xn).Eachnumberinthe n -tuplemaybeofadierentnature.Intheaboveexample,theconcentrationdependsonorderedquadruples( x,y,z,t ),where x y ,and z arethecoordinatesofapointinspaceand t istime. Definition 13.1 (Real-ValuedFunctionofSeveralVariables) Let D beasetofordered n -tuplesofrealnumbers ( x1,x2,...,xn) .A function f of n variablesisarulethatassignstoeach n -tupleinthe set D auniquerealnumberdenotedby f ( x1,x2,...,xn) .Theset D is thedomainof f ,anditsrangeisthesetofvaluesthat f takesonit, thatis, { f ( x1,x2,...,xn) | ( x1,x2,...,xn) D } Therulemaybede“nedbydierentmeans.If D isa“niteset, afunction f canbede“nedbyatable( Pi,f ( Pi)),where Pi D i = 1 2 ,...,N ,areelements(ordered n -tuples)of D ,and f ( Pi)isthevalue of f at Pi.Afunction f canbede“nedgeometrically.Forexample,the123

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12413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS heightofamountainrelativetosealevelisafunctionofitspositionon theglobe.Sotheheightisafunctionoftwovariables,thelongitudeand latitude.Afunctioncanbede“nedbyanalgebraicrulethatprescribes algebraicoperationstobecarriedoutwithrealnumbersinany n -tuple toobtainthevalueofthefunction.Forexample, f ( x,y,z )= x2Š y + z3. Thevalueofthisfunctionat(1 2 3)is f (1 2 3)=12Š 2+33=28. Unlessspeci“edotherwise,thedomain D ofafunctionde“nedbyan algebraicruleisthesetof n -tuplesforwhichtherulemakessense. Example 13.1 Findthedomainandtherangeofthefunctionof twovariables f ( x,y )=ln(1 Š x2Š y2) Solution: Thelogarithmisde“nedforanystrictlypositivenumber. Therefore,thedoublets( x,y )mustbesuchthat1 Š x2Š y2> 0or x2+ y2< 1.Hence, D = { ( x,y ) | x2+ y2< 1 } .Sinceanydoublet( x,y ) canbeuniquelyassociatedwithapointonaplane,theset D canbe givenageometricaldescriptionasadiskofradius1whoseboundary, thecircle x2+ y2=1,isnotincludedin D .Foranypointinthe interiorofthedisk,theargumentofthelogarithmliesintheinterval 0 1 Š x2Š y2< 1.Sotherangeof f isthesetofvaluesofthe logarithmintheinterval(0 1],whichis Š
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85.FUNCTIONSOFSEVERALVARIABLES125 Definition 13.2 (GraphofaFunctionofTwoVariables) Thegraphofafunction f ( x,y ) withdomain D isthepointsetinspace { ( x,y,z ) | z = f ( x,y ) ( x,y ) D } Thedomain D isasetofpointsinthe xy plane.Thegraphis thenobtainedbymovingeachpointof D paralleltothe z axisbyan amountequaltothecorrespondingvalueofthefunction z = f ( x,y ).If D isaportionoftheplane,thenthegraphof f isgenerallyasurface. OnecanthinkofthegraphasmountainsŽofheight f ( x,y )onthe xy plane. Example 13.3 Sketchthegraphofthefunction f ( x,y )= 1 Š ( x/ 2)2Š ( y/ 3)2. Solution: Thedomainistheportionofthe xy plane( x/ 2)2+( y/ 3)2 1;thatis,itisboundedbytheellipsewithsemiaxes2and3.Thegraph isthesurfacede“nedbytheequation z = 1 Š ( x/ 2)2Š ( y/ 3)2.By squaringbothsidesofthisequation,one“nds( x/ 2)2+( y/ 3)2+ z2= 1,whichde“nesanellipsoid.Thegraphisitsupperportionwith z 0. Theconceptofthegraphisobviouslyhardtoextendtofunctions ofmorethantwovariables.Thegraphofafunctionofthreevariables wouldbeathree-dimensionalsurfaceinfour-dimensionalspace.Sothe qualitativebehaviorofafunctionofthreevariablesshouldbestudied bydierentgraphicalmeans.85.2.LevelCurves.Whenvisualizingtheshapeofquadricsurfaces, themethodofcrosssectionsbycoordinateplaneshasbeenhelpful.It canalsobeappliedtovisualizetheshapeofthegraph z = f ( x,y ). Inparticular,considerthecrosssectionsofthegraphwithhorizontal planes z = k .Thecurveofintersectionisde“nedbytheequation f ( x,y )= k .Continuingtheanalogythat f ( x,y )de“nestheheightof amountain,ahikertravelingalongthepath f ( x,y )= k doesnothave toclimbordescendastheheightalongthepathremainsconstant. Definition 13.3 (LevelCurves) Thelevelcurvesofafunction f oftwovariablesarethecurvesalong whichthefunctionremainsconstant;thatis,theyaredeterminedby theequation f ( x,y )= k ,where k isanumberfromtherangeof f Definition 13.4 (ContourMap) Acollectionoflevelcurvesiscalleda contourmap ofthefunction f ThecontourmapofthefunctioninExample13.3consistsofellipses. Indeed,therangeistheinterval[0 1].Forany k [0 1],alevelcurve

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12613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS isanellipse,1 Š ( x/ 2)2+( y/ 3)2= k2or( x/a )2+( y/b )2=1,where a =2 1 Š k2and b =3 1 Š k2. Acontourmapisausefultoolforstudyingthequalitativebehavior ofafunction.Considerthecontourmapthatconsistsoflevelcurves Ci, i =1 2 ,... f ( x,y )= ki,where ki +1Š ki= k is“xed.The valuesofthefunctionalongtheneighboringcurves Ciand Ci +1dier by k .So,intheregionwherethelevelcurvesaredense(closetoone another),thefunction f ( x,y )changesrapidly.Indeed,let P beapoint of Ciandlet s bethedistancefrom P to Ci +1alongthenormalto Ci.Thentheslopeofthegraphof f ortherateofchangeof f at P inthatdirectionis k/ s .Thus,thecloserthecurves Ciaretoone another,thefasterthefunctionchanges.Suchcontourmapsareused intopographytoindicatethesteepnessofmountainsonmaps.85.3.LevelSurfaces.Incontrasttothegraphofafunction,themethod oflevelcurvesdoesnotrequireahigher-dimensionalspacetostudythe behaviorofafunctionoftwovariables.Sotheconceptcanbeextended tofunctionsofthreevariables. Definition 13.5 (LevelSurface) Thelevelsurfacesofafunction f ofthreevariablesarethesurfaces alongwhichthefunctionremainsconstant;thatis,theyaredetermined bytheequation f ( x,y,z )= k ,where k isanumberfromtherangeof f Theshapeofthelevelsurfacesmaybestudied,forexample,bythe methodofcrosssectionswithcoordinateplanes.Acollectionoflevel surfaces Si, f ( x,y,z )= ki, ki +1Š ki= k i =1 2 ,... ,canbedepicted inthedomainof f .Thecloserthelevelsurfaces Siaretooneanother, thefasterthefunctionchanges. Example 13.4 Sketchand/ordescribethelevelsurfacesofthe function f ( x,y,z )= z/ (1+ x2+ y2) Solution: Thedomainistheentirespace,andtherangecontains allrealnumbers.Theequation f ( x,y,z )= k canbewritteninthe form z Š k = k ( x2+ y2),whichde“nesacircularparaboloidwhose symmetryaxisisthe z axisandwhosevertexisat(0 0 ,k ).Forlarger k ,theparaboloidrisesfaster.For k =0,thelevelsurfaceisthe xy plane.For k> 0,thelevelsurfacesareparaboloidsabovethe xy plane, (i.e.,theyareconcavedownward).For k< 0,theparaboloidsarebelow the xy plane(i.e.,theyareconcaveupward).

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85.FUNCTIONSOFSEVERALVARIABLES127 85.4.EuclideanSpaces.Whenthenumberofvariablesisgreaterthan 3,thegeometricalvisualizationofthedomainisnotsosimple.The goalisachievedwiththehelpoftheconceptofahigher-dimensional Euclideanspace.Theplaneandspaceareparticularcasesoftwo-and three-dimensionalEuclideanspaces. Witheveryorderedpairofnumbers( x,y ),onecanassociateapoint inaplaneanditspositionvectorrelativetoa“xedpoint(0 0)(the origin), r = x,y .Witheveryorderedtripleofnumbers( x,y,z ),one canassociateapointinspaceanditspositionvector(againrelative totheorigin(0 0 0)), r = x,y,z .Sotheplanecanbeviewedas thesetofalltwo-componentvectors;similarly,spaceisthesetofall three-componentvectors.Fromthispointofview,theplaneandspace havecharacteristiccommonfeatures.First,theirelementsarevectors. Second,theyareclosedrelativetoadditionofvectorsandmultiplicationofvectorsbyarealnumber;thatis,if a and b areelementsof spaceoraplaneand c isarealnumber,then a + b and c a arealso elementsofspace(orderedtriplesofnumbers)oraplane(orderedpairs ofnumbers).Third,thenormorlengthofavector r vanishesifand onlyifthevectorhaszerocomponents.Consequently,twoelements ofspaceoraplanecoincideifandonlyifthenormoftheirdierence vanishes,thatis, a = b a Š b =0.Fromthispointofview, thereisnodierencebetweenavector a1,a2,a3 andanorderedtriple ( a1,a2,a3)astheyrepresenttheverysamepointinspace;thatis,there isnoconfusioninwriting a =( a1,a2,a3).Finally,thedotproduct a b oftwoelementsisde“nedinthesamewayfortwo-orthree-component vectors(planeorspace)sothat a 2= a a .Theseobservationscan beextendedtoordered n -tuplesforany n andleadtothenotionofa Euclideanspace Definition 13.6 (EuclideanSpace) Foreachpositiveinteger n ,considerthesetofallordered n -tuples ofrealnumbers.Foranytwoelements a =( a1,a2,...,an) and b = ( b1,b2,...,bn) andanumber c ,put a + b =( a1+ b1,a2+ b2,...,an+ bn) c a =( ca1,ca2,...,can) a b = a1b1+ a2b2+ + anbn, a = a a = a2 1+ a2 2+ + a2 n. Thesetofallordered n -tuplesinwhichtheaddition,themultiplication byanumber,thedotproduct,andthenormarede“nedbytheserules iscalledan n dimensionalEuclideanspace .

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12813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS TwopointsofaEuclideanspacearesaidtocoincide, a = b ,ifthe correspondingcomponentsareequal,thatis, ai= bifor i =1 2 ,...,n Itfollowsthat a = b ifandonlyif a Š b =0.Indeed,bythe de“nitionofthenorm, c =0ifandonlyif c =(0 0 ,..., 0).Put c = a Š b .Then a Š b =0ifandonlyif a = b .Thenumber a Š b iscalledthe distance betweenpoints a and b ofaEuclideanspace. ThedotproductinaEuclideanspacehasthesamegeometrical propertiesasintwoandthreedimensions.TheCauchy-S chwarzinequalitycanbeextendedtoanyEuclideanspace(cf.Theorem11.3). Theorem 13.1 (Cauchy-SchwarzInequality) | a b | a b foranyvectors a and b inaEuclideanspace,andtheequalityisreached ifandonlyif a = t b forsomenumber t Proof. Put a = a and b = b ,thatis, a2= a a andsimilarly for b .If b =0,then b = 0 ,andtheconclusionofthetheoremholds. For b =0andanyrealvariable t a Š t b 2=( a Š t b ) ( a Š t b ) 0. Therefore, a2Š 2 tc + t2b2 0,where c = a b .Therightsideofthis inequalityisadownwardconcaveparabolawithrespectto t ,which attainsitsabsoluteminimumat t = c/b2.Sincetheinequalityisvalid forany t ,itissatis“edfor t = b2/c ,thatis, a2Š c2/b2 0or c2 a2b2or | c | ab ,whichistheconclusionofthetheorem.Theinequality becomesanequalityifandonlyif a Š t b 2=0andhenceifandonly if a = t b ItfollowsfromtheS chwarzinequalitythat a b = s a b ,where s isanumbersuchthat | s | 1.Soonecanalwaysput s =cos where [0 ].If =0,then a = t b forsomepositive t> 0 (i.e.,thevectorsareparallel),and a = t b t< 0,when = (i.e.,the vectorsareantiparallel).Thedotproductvanisheswhen = / 2.This allowsonetode“ne astheanglebetweenvectorsinanyEuclidean space:cos = a b / ( a b )muchlikeintwoandthreedimensions. Consequently,thetriangleinequality(11.7)holdsinaEuclideanspace ofanydimension. Inwhatfollows,thedomainofafunctionof n variablesisviewed asasubsetinan n -dimensionalEuclideanspace.Itisalsoconvenient toadoptthevectornotationoftheargument: f ( x1,x2,...,xn)= f ( r ) r =( x1,x2,...,xn) Forexample,thedomainofthefunction f ( r )=(1 Š x2 1Š x2 2ŠŠ x2 n)1 / 2=(1 Š r 2)1 / 2isthesetofpointsinthe n -dimensionalEuclidean

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86.LIMITSANDCONTINUITY129 spacewhosedistancefromtheorigin(thezerovector)doesnotexceed 1, D = { r | r 1 } ;thatis,itisan n -dimensionalballofradius1. Sothedomainofamultivariablefunctionde“nedbyanalgebraicrule canbedescribedbyconditionsonthecomponents(coordinates)ofthe ordered n -tuple r underwhichtherulemakessense.85.5.Exercises.(1) Findandsketchthedomainofeachofthefollowingfunctions: (i) f ( x,y )=ln(9 Š x2Š ( y/ 2)2) (ii) f ( x,y )= 1 Š ( x/ 2)2Š ( y/ 3)2(iii) f ( x,y,z )=ln(1 Š z + x2+ y2) (iv) f ( x,y )=ln(9 Š x2Š ( y/ 2)2) (v) f ( x,y,z )= x2Š y2Š z2(vi) f ( t, x )=( t2Š x 2)Š 1, x =( x1,x2,...,xn) (2) Foreachofthefollowingfunctions,sketchthegraphanda contourmap: (i) f ( x,y )= x2+4 y2(ii) f ( x,y )= xy (iii) f ( x,y )= x2Š y2(iv) f ( x,y )= x2+9 y2(v) f ( x,y )=sin x (3) Describeandsketchthelevelsurfacesofeachofthefollowing functions: (i) f ( x,y,z )= x +2 y +3 z (ii) f ( x,y,z )= x2+4 y2+9 z2(iii) f ( x,y,z )= z + x2+ y2(iv) f ( x,y,z )= x2+ y2Š z2(4) Explainhowthegraph z = g ( x,y )canbeobtainedfromthe graphof f ( x,y )if (i) g ( x,y )= k + f ( x,y ),where k isaconstant (ii) g ( x,y )= mf ( x,y ),where m isanonzeroconstant (iii) g ( x,y )= f ( x Š a,y Š b ),where a and b areconstants (iv) g ( x,y )= f ( px,qy ),where p and q arenonzeroconstants Let f ( x,y )= x2+ y2.Sketchthegraphsof g ( x,y )de“nedabove. Analyzecarefullyvariouscasesforvaluesoftheconstants,forexample, m> 0, m< 0, p> 1,0
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13013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS x 0.Itisknown(seeCalculusI)thatsin( x ) /x 1as x 0.A similarquestioncanbeaskedforfunctionsofseveralvariables.For example,thedomainofthefunction f ( x,y )=sin( x2+ y2) / ( x2+ y2)is theentireplaneexceptthepoint( x,y )=(0 0).If( x,y ) =(0 0),then, incontrasttotheone-dimensionalcase,thepoint( x,y )mayapproach (0 0)alongvariouspaths.Sotheverynotionthat( x,y )approaches (0 0)needstobeaccuratelyde“ned. Asnotedbefore,thedomainofafunction f ofseveralvariablesis asetin n -dimensionalEuclideanspace.Twopoints x =( x1,x2,...,xn) and y =( y1,y2,...,yn)coincideifandonlyifthedistance x Š y = ( x1Š y1)2+( x2Š y2)2+ +( xnŠ yn)2vanishes. Definition 13.7 Apoint r issaidtoapproacha“xedpoint r0if thedistance r Š r0 tendsto0.Thelimit r Š r0 0 isalsodenoted by r r0. Intheaboveexample,thelimit( x,y ) (0 0)meansthat x2+ y2 0or x2+ y2 0.Therefore, sin( x2+ y2) x2+ y2= sin u u 1as x2+ y2= u 0 Notethatherethelimitpoint(0 0)canbeapproachedfromanydirectionintheplane.Thisisnotalwaysso.Forexample,thedomainofthefunction f ( x,y )=sin( xy ) / ( x + y )isthe“rstquadrant,includingitsboundariesexceptthepoint(0 0).Thepoints(0 0) and( Š 1 Š 1)arenotinthedomainofthefunction.However,the limitof f as( x,y ) (0 0)canbede“ned,whereasthelimitof f as ( x,y ) ( Š 1 Š 1)doesnotmakeanysense.Thedierencebetween thesetwopointsisthatanyneighborhoodof(0 0)containspointsof thedomain,whilethisisnotsofor( Š 1 Š 1).Sothelimitcanbe de“nedonlyforsomespecialclassofpointscalled limitpoints ofa set D Definition 13.8 (LimitPointofaSet) Apoint r0issaidtobealimitpointofaset D ifanyopenball B= { r | r Š r0 < } centeredat r0containsapointof D thatdoesnot coincidewith r0if r0 D Inotherwords,alimitpoint r0of D mayormaynotbein D butitcanalwaysbeapproachedfromwithintheset D inthesense that r r0and r D because,nomatterhowsmall is,onecan always“ndapoint r D whosedistancefrom r0islessthan .Inthe

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86.LIMITSANDCONTINUITY131 aboveexampleof D beingthe“rstquadrant,thelimit( x,y ) (0 0) isunderstoodas x2+ y2 0while( x,y ) =(0 0)and x 0, y 0.86.1.LimitsofFunctionsofSeveralVariables.Definition 13.9 (LimitofaFunctionofSeveralVariables) Let f beafunctionofseveralvariableswhosedomainisaset D ina Euclideanspace.Let r0bealimitpointof D .Thenthelimitof f ( r ) as r r0issaidtobeanumber f0if,foreverynumber > 0 ,thereexists acorrespondingnumber > 0 suchthatif r D and 0 < r Š r0 < then | f ( r ) Š f0| < .Inthiscase,onewrites limr r0f ( r )= f0. Thenumber | f ( r ) Š f0| determinesadeviationofthevalueof f from thenumber f0.Theexistenceofthelimitmeansthatnomatterhow small is,thereisaneighborhood N( r0)of r0in D ,whichcontainsall pointsof D whosedistancefrom r0doesnotexceedanumber ,such thatthevaluesofthefunction f in N( r0)deviatefromthelimitvalue f0nomorethan ,thatis, f0Š 0.Toestablishtheexistenceof > 0,notethat theinequality8 R3< or R<3 / 2guaranteesthat | f ( r ) Š f0| < Therefore, =3 / 2.Forexample,put =10Š 6.Then,intheinterior ofaballofradius =0 005,thevaluesofthefunctioncandeviate from f0=0nomorethan10Š 6. Remark. Notethat dependson and,ingeneral,onthelimit point r0. Remark. Thede“nitionofthelimitguaranteesthatifthelimitexists,then itdoesdependonapathalongwhichthelimitpointmaybe approached .Indeed,takeanycurvethatendsatthelimitpoint r0and

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13213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS “x > 0.Then,bytheexistenceofthelimit f0,thereisaballof radius = ( r0) > 0centeredat r0suchthatthevaluesof f liein theinterval f0Š 0,thereisaninterval0
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86.LIMITSANDCONTINUITY133 Theorem 13.3 (SqueezePrinciple) Letthefunctionsofseveralvariables g f ,and h haveacommondomain D suchthat g ( r ) f ( r ) h ( r ) forany r D .Ifthelimitsof g ( r ) and h ( r ) as r r0existandequalanumber f0,thenthelimitof f ( r ) as r r0existsandequals f0,thatis, g ( r ) f ( r ) h ( r )andlimr r0g ( r )=limr r0h ( r )= f0 limr r0f ( r )= f0. Proof. Fromtheconditionofthetheorem,itfollowsthat0 f ( r ) Š g ( r ) h ( r ) Š g ( r ).Put F ( r )= f ( r ) Š g ( r )and H ( r )= h ( r ) Š g ( r ). Then0 F ( r ) H ( r )implies | F ( r ) || H ( r ) | (thepositivityof F isessentialforthisconclusion).Bythebasicpropertiesofthelimit, H ( r ) 0as r r0.Hence,forany > 0,thereisacorresponding number suchthat0 | F ( r ) || H ( r ) | < whenever0 < r Š r0 < .Thisinequalityalsoimpliesthatlimr r0F ( r )=0.Bythebasic propertiesofthelimit,itisthenconcludedthat f ( r )= F ( r )+ g ( r ) 0+ f0= f0as r r0. Thesimpli“edsqueezeprincipleisaparticularcaseofthistheorem becausethecondition | f ( r ) Š f0| h ( R )isequivalentto f0Š h ( R ) f ( r ) f0+ h ( R ). Example 13.6 Showthat lim( x,y ) (0 0)f ( x,y )=0 where f ( x,y )= x3y Š 3 x2y2 x2+ y2+ x4. Solution: Let R = x2+ y2(thedistancefromthelimitpoint (0 0)).Then | x | R and | y | R .Therefore, | x3y Š 3 x2y2| x2+ y2+ x4 | x |3| y | +3 x2y2 x2+ y2+ x4 4 R4 R2+ x4 4 R21 1+( x4/R2) 4 R2. Itfollowsfromthisinequalitythat Š 4( x2+ y2) f ( x,y ) 4( x2+ y2), and,bythesqueezeprinciple, f ( x,y )musttendto0because 4( x2+ y2)= 4 R2 0as R 0.Inthede“nitionofthelimit,forany > 0, thecorrespondingnumber is = / 2. 86.2.ContinuityofFunctionsofSeveralVariables.Definition 13.10 (Continuity) Afunction f ofseveralvariableswithdomain D issaidtobecontinuous atapoint r0 D if limr r0f ( r )= f ( r0) Thefunction f issaidtobecontinuouson D ifitiscontinuousat everypointof D .

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13413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Let f ( x,y )=1if y x andlet f ( x,y )=0if yx0,then f ( x0,y0)=1.Ontheotherhand,foreverysuchpointonecan“nda neighborhood( x Š x0)+( y Š y0)2<2(adiskofradius > 0centeredat ( x0,y0))thatliesintheregion y>x .Therefore, | f ( r ) Š f ( r0) | =1 Š 1= 0 < forany > 0inthisdisk,thatis,limr r0f ( r )= f ( r0)=1.The samelineofargumentsappliestoestablishthecontinuityof f atany point( x0,y0),where y0 0inwhich | f ( r ) Š f ( r0) | = | f ( r ) Š 1 | < because | f ( r ) Š 1 | =1for y 0

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86.LIMITSANDCONTINUITY135 andorigin r0, f ( r0)=0,onehas | f ( r ) Š f ( r0) | = | f ( r ) | RN 0 as R 0because | xi| R = x2 1+ x2 2+ x2 nforanyelement ofthe n -tuple.Bythesqueezeprinciple, f ( r ) 0= f ( r0).The rationalfunction f ( r ) /g ( r )iscontinuousastheratiooftwocontinuous functions. Theorem 13.6 (ContinuityofaComposition) Let g ( u ) becontinuousontheinterval u [ a,b ] andlet h beafunction ofseveralvariablesthatiscontinuouson D andhastherange [ a,b ] Thecomposition f ( r )= g ( h ( r )) iscontinuouson D Theprooffollowsthesamelineofreasoningasinthecaseofthe compositionoftwofunctionsofonevariableinCalculusIandisleft tothereaderasanexercise. Inparticular,somebasicfunctionsstudiedinCalculusI,sin u ,cos u eu,ln u ,andsoon,arecontinuousfunctionsontheirdomains.If f ( r ) isacontinuousfunctionofseveralvariables,theelementaryfunctions whoseargumentisreplacedby f ( r )arecontinuousfunctions.Incombinationwiththepropertiesofcontinuousfunctions,thecomposition rulede“nesalargeclassofcontinuousfunctionsofseveralvariables, whichissucientformanypracticalapplications.86.3.Exercises.(1) Usethede“nitionofthelimittoverifyeachof thefollowinglimits(i.e.,given > 0,“ndthecorresponding ( )): (i)limr 0x3Š 4 y2x +5 y3 x2+ y2=0 (ii)limr 0x3Š 4 y2x +5 y3 3 x2+4 y2=0 (iii)limr 0x3Š 4 y4+5 y3x2 3 x2+4 y2=0 (iv)limr 0x3Š 4 y2x +5 y3 3 x2+4 y2+ y4=0 (2) Verifywhetherthegivenfunctioniscontinuousonitsdomain: (i) f ( x,y )= yx/ ( x2+ y2)if( x,y ) =(0 0)and f (0 0)=1 (ii) f ( x,y,z )= yxz/ ( x2+ y2+ z2)if( x,y,z ) =(0 0 0)and f (0 0 0)=0 (iii) f ( x,y )=sin( xy ) (iv) f ( x,y )=cos( xyz ) / ( x2y2+1) (v) f ( x,y )=( x2+ y2)ln( x2+ y2)if( x,y ) =(0 0)and f (0 0)=0

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13613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 87.AGeneralStrategytoStudyLimits Thede“nitionofthelimitgivesonlythecriterionforwhethera number f0isthelimitof f ( r )as r r0.Inpractice,however,a possiblevalueofthelimitistypicallyunknown.Somestudiesare neededtomakeaneducatedŽguessforapossiblevalueofthelimit. Hereaproceduretostudylimitsisoutlinedthatmightbehelpful.In whatfollows,thelimitpointisoftensettotheorigin r0=(0 0 ,..., 0). Thisisnotalimitationbecauseonecanalwaystranslatetheoriginof thecoordinatesystemtoanyparticularpointbyshiftingthevaluesof theargument,forexample, lim( x,y ) ( x0,y0)f ( x,y )=lim( x,y ) (0 0)f ( x + x0,y + y0) .87.1.Step1:ContinuityArgument.Thesimplestscenarioinstudying thelimithappenswhenthefunction f inquestioniscontinuousatthe limitpoint: limr r0f ( r )= f ( r0) Forexample, lim( x,y ) (1 2)xy x3Š y2= Š 2 3 becausethefunctioninquestionisarationalfunctionthatiscontinuous if x3Š y2 =0.Thelatterisindeedthecaseforthelimitpoint(1 2). Ifthecontinuityargumentdoesnotapply,thenitishelpfultocheck thefollowing.87.2.Step2:CompositionRule.Theorem 13.7 (CompositionRuleforLimits) Let g ( t ) beafunctionofonevariableandlet h beacontinuousfunction ofseveralvariablessuchthat h ( r ) t0= h ( r0) as r r0.Suppose thatthefunction f isthecomposition f ( r )= g ( h ( r )) .Then limr r0f ( r )=limt t0g ( t ) Theproofisomittedasitissimilartothecasewhen h isacontinuousfunctionofonevariable,whichwasprovedinCalculusI.The signi“canceofthistheoremisthat,undertheconditionsofthetheorem,atoughproblemofstudyingamultivariablelimitisreducedto theproblemofthelimitofafunctionofasingleargument.Thelatter problemcanbestudied,by,forexample,lHospitalsrule.Itmustbe emphasizedthat thereisnoanalogoflHospitalsruleformultivariable functions .

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87.AGENERALSTRATEGYTOSTUDYLIMITS137 Example 13.7 Find lim( x,y ) (0 0)cos( xy ) Š 1 x2y2. Solution: Thefunctioninquestionis(cos t Š 1) /t2,wheretheargument t isreplacedbythefunction h ( x,y )= xy .Thefunction h isa polynomialandhencecontinuous.Inparticular, h ( x,y ) h (0 0)=0 as( x,y ) (0 0).Thus,allthehypothesesofthecompositionrule theoremareful“lled: lim( x,y ) (0 0)cos( xy ) Š 1 x2y2=limt 0cos t Š 1 t2=limt 0Š sin t 2 t =limt 0Š cos t 2 = Š 1 2 wherelHospitalsrulehasbeenusedtwicetoevaluatethesinglevariablelimit. Itmustbestressedthatthehypothesisofthecontinuityof h ( r )in thecompositionruletheoremiscrucial.If,forinstance,intheabove exampletheargumentof(cos t Š 1) /t2isreplacedby h ( x,y )= y/x thelimitoftheresultingfunctiondoesnotexistas( x,y ) (0 0).The reasonisthat y/x isnotcontinuousattheoriginanditslimitdoesnot existas( x,y ) (0 0).Simplemeanstoestablishthelatterfactare providedinthenextstep.87.3.Step3:LimitsAlongCurves.Recallthefollowingresultabout thelimitofafunctionofonevariable.Thelimitof f ( x )as x x0existsandequals f0ifandonlyifthecorrespondingrightandleftlimits of f ( x )existandequal f0: limx x+ 0f ( x )=limx xŠ 0f ( x )= f0 limx x0f ( x )= f0. Inotherwords,ifthelimitexists,itdoesnotdependonthedirection fromwhichthelimitpointisapproached.Iftheleftandrightlimits existbutdonotcoincide,thenthelimitdoesnotexist. Forfunctionsofseveralvariables,therearein“nitelymanypaths alongwhichthelimitpointcanbeapproached.Theyincludestraight linesandpathsofanyothershape,incontrasttotheone-variablecase. Nevertheless,asimilarresultholdsformultivariablelimits(seethesecondRemarkattheendofSection86.1),thatis, ifthelimitexists,then itshouldnotdependonthecurvealongwhichthelimitpointmaybe approached.Iftherearetwocurvesalongwhichthelimitsdonotcoincide,thenthemultivariablelimitdoesnotexist .Thisresultprovides apowerfulmethodtoinvestigatetheexistenceofamultivariablelimit andtomakeaneducatedŽguessaboutitspossiblevalue.

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13813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Definition 13.11 (CurveinaEuclideanSpace) AcurveinaEuclideanspaceisasetofpoints r ( t )=( x1( t ) ,x2( t ) ,..., xn( t )) ,where xi( t ) i =1 2 ,...,n ,arecontinuousfunctionsofavariable t [ a,b ] Thisisanaturalgeneralizationoftheconceptofacurveinaplane orspaceasavectorfunctionde“nedbytheparametricequations xi= xi( t ), i =1 2 ,...,n Definition 13.12 (LimitAlongaCurve) Let r0bealimitpointofthedomain D ofafunction f .Let r ( t )= ( x1( t ) ,x2( t ) ,...,xn( t )) beacurve C in D suchthat r ( t ) r0as t t+ 0. Thefunction F ( t )= f ( r ( t )) de“nesthevaluesof f onthecurve C Thelimit limt t+ 0F ( t )=limt t+ 0f ( x1( t ) ,x2( t ) ,...,xn( t )) iscalledthelimitof f alongthecurve C ifitexists. Supposethatthelimitof f ( r )as r r0existsandequals f0.Let C beacurvesuchthat r ( t ) r0as t t+ 0.Fix > 0.Bytheexistenceof thelimit,thereisaneighborhood N( r0)= { r | r D, 0 < r Š r0 < } inwhichthevaluesof f deviatefrom f0nomorethan | f ( r ) Š f0| < Sincethecurve C iscontinuousandpassesthrough r0,thereshould beaportionofitthatliesin N( r0);thatis,thereisanumber such that r ( t ) Š r0 < forall t ( t0,t0+ ).Hence,forany > 0, thedeviationofvaluesof f alongthecurve, F ( t )= f ( r ( t )),doesnot exceed | F ( t ) Š f0| < whenever0 < | t Š t0| <.Bythede“nition oftheone-variablelimit,thisimpliesthat F ( t ) f0as t t0forany curve C through r0.Thisprovesthefollowing. Theorem 13.8 (IndependenceoftheLimitfromaCurveThrough theLimitPoint) Ifthelimitof f ( r ) existsas r r0,thenthelimitof f alonganycurve leadingto r0fromwithinthedomainof f existsanddoesnotdepend onthecurve. Animmediateconsequenceofthistheoremisthefollowingresult, whichisveryusefulinmanypracticalapplications Corollary 13.2 (CriterionforNonexistenceoftheLimit) Let f beafunctionofseveralvariableson D .Ifthereisacurve r ( t ) in D suchthat r ( t ) r0as t t+ 0andthelimit limt t+ 0f ( r ( t )) doesnot exist,thenthemultivariablelimit limr r0f ( r ) doesnotexisteither.If therearetwocurvesin D leadingto r0suchthatthelimitsof f along

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87.AGENERALSTRATEGYTOSTUDYLIMITS139 themexistbutdonotcoincide,thenthemultivariablelimit limr r0f ( r ) doesnotexist.87.3.1.LimitsAlongStraightLines.Letthelimitpointbetheorigin r0=(0 0 ,..., 0).Thesimplestcurveleadingto r0isastraightline xi= vit ,where t 0+forsomenumbers vi, i =1 2 ,...,n .The limitofafunctionofseveralvariables f alongastraightlineisthen limt 0+f ( v1t,v2t,...,vnt ),shouldexistandbethesameforanychoice ofnumbers vi.Forcomparison,recallthevectorequationofastraight lineinspacethroughtheorigin: r = t v ,where v isavectorparallelto theline. Example 13.8 Investigatethetwo-variablelimit lim( x,y ) (0 0)xy3 x4+2 y4. Solution: Considerthelimitsalongstraightlines x = t y = at (or y = ax ,where a istheslope)as t 0+: limt 0+f ( t,at )=limt 0+a3t4 t4(1+2 a4) = a3 1+2 a4. Sothelimitalongastraightlinedependsontheslopeoftheline. Therefore,thetwo-variabledoesnotexist. Example 13.9 Investigatethelimit lim( x,y ) (0 0)sin( xy ) x + y Solution: Thedomainofthefunctionconsistsofthe“rstandthird quadrantsas xy 0excepttheorigin.Linesapproaching(0 0)from withinthedomainare x = t y = at a 0and t 0.Notetheline x =0, y = t alsoliesinthedomain(thelinewithanin“niteslope). Thelimitalongastraightlineapproachingtheoriginfromwithinthe “rstquadrantis limt 0+f ( t,at )=limt 0+sin( t a ) t (1+ a ) =limt 0+ a cos( t a ) 1+ a = a 1+ a wherelHospitalsrulehasbeenusedtocalculatethelimit.Thelimit dependsontheslopeoftheline,andhencethetwo-variablelimitdoes notexist.

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14013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 87.3.2.LimitsAlongPowerCurves(Optional).Ifthelimitalongstraight linesexistsandisindependentofthechoiceoftheline,thenumerical valueofthislimitprovidesadesirededucatedŽguessfortheactual multivariablelimit.However,thishasyettobeprovedbymeansof eitherthede“nitionofthemultivariablelimitor,forexample,the squeezeprinciple.Thiscomprisesthelaststepoftheanalysisoflimits (Step4;seebelow). Thefollowingshouldbestressed. Ifthelimitsalongallstraightlines happentobethesamenumber,thisdoesnotmeanthatthemultivariablelimitexistsandequalsthatnumber becausetheremightexistother curvesthroughthelimitpointalongwhichthelimitattainsadierent valueordoesnotevenexist Example 13.10 Investigatethelimit lim( x,y ) (0 0)y3 x Solution: Thedomainofthefunctionisthewholeplanewiththe y axisremoved( x =0).Thelimitalongastraightline limt 0+f ( t,at )=limt 0+a3t3 t = a3limt 0+t2=0 vanishesforanyslope;thatis,itisindependentofthechoiceofthe line.However,thetwo-variabledoesnotexist!Considerthepower curve x = t y = at1 / 3approachingtheoriginas t 0+.Thelimit alongthiscurvecanattainanyvaluebyvaryingtheparameter a : limt 0+f ( t,at1 / 3)=limt 0+a3t t = a3. Thus,themultivariablelimitdoesnotexist. Ingeneral,limitsalongpowercurvesareconvenientforstudying limitsofrationalfunctionsbecausethevaluesofarationalfunction ofseveralvariablesonapowercurvearegivenbyarationalfunction ofthecurveparameter t .Onecanthenadjust,ifpossible,thepower parameterofthecurvesothattheleadingtermsofthetopandbottom powerfunctionsmatchinthelimit t 0+.Forinstance,intheexample considered,put x = t and y = atn.Then f ( t,atn)=( a3t3 n) /t .The powersofthetopandbottomfunctionsinthisratiomatchif3 n =1; hence,for n =1 / 3,thelimitalongthepowerpathdependsonthe parameter a andcanbeanynumber.87.4.Step4:UsingtheSqueezePrinciple.IfSteps1and2donotapply tothemultivariablelimitinquestion,thenaneducatedŽguessfora

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87.AGENERALSTRATEGYTOSTUDYLIMITS141 possiblevalueofthelimitishelpful.ThisistheoutcomeofStep3.If limitsalongafamilyofcurves(e.g.,straightlines)happentobethe samenumber f0,thenthisnumberisthesought-foreducatedŽguess. Thede“nitionofthemultivariablelimitorthesqueezeprinciplecan beusedtoproveordisprovethat f0isthemultivariablelimit. Example 13.11 Findthelimitorprovethatitdoesnotexist: lim( x,y ) (0 0)sin( xy2) x2+ y2. Solution: Step1 .Thefunctionisnotde“nedattheorigin.Thecontinuity argumentdoesnotapply. Step2 .Nosubstitutionexiststotransformthetwo-variablelimittoa one-variablelimit. Step3 .Put( x,y )=( t,at ),where t 0+.Thelimitalongstraight lines limt 0+f ( t,at )=limt 0+sin( a2t3) t2=limu 0+sin( a2u3 / 2) u =limu 0+(3 / 2) a2u1 / 2cos( a2u3 / 2) 1 =0 vanishes(herethesubstitution u = t2andlHospitalsrulehavebeen usedtocalculatethelimit). Step4 .Ifthetwo-variablelimitexists,thenitmustbeequalto0. Thiscanbeveri“edbymeansofthesimpli“edsqueezeprinciple;that is,onehastoverifythatthereexists h ( R )suchthat | f ( x,y ) Š f0| = | f ( x,y ) | h ( R ) 0as R = x2+ y2 0.Akeytechnicaltrickhere istheinequality | sin u || u | ,whichholdsforanyreal u .Onehas | f ( x,y ) Š 0 | = | sin( xy2) | x2+ y2 | xy2| x2+ y2 R3 R2= R 0 wheretheinequalities | x | R and | y | R havebeenused.Thus,the two-variablelimitexistsandequals0. Fortwo-variablelimits,itissometimesconvenienttousepolarcoordinatescenteredatthelimitpoint x Š x0= R cos y Š y0= R sin Theideaisto“ndoutwhetherthedeviationofthefunction f ( x,y ) from f0(theeducatedŽguessfromStep3)canbeboundedby h ( R ) uniformlyforall [0 2 ]: | f ( x,y ) Š f0| = | f ( x0+ R cos ,y0+ R sin ) Š f0| h ( R ) 0 R 0 .

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14213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Thistechnicaltaskcanbeaccomplishedwiththehelpofthebasic propertiesoftrigonometricfunctions,forexample, | sin | 1, | cos | 1,andsoon. InExample13.10,Step3gives f0=0ifonlythelimitsalong straightlineshavebeenstudied.Then | f ( R cos ,R sin ) | = R2sin2( ) | tan | .Despitethatthedeviationisproportionalto R2 0as R 0, itcannotbemadeassmallasdesiredbydecreasing R becausetan is notaboundedfunction.Thereisasectorintheplanecorresponding toanglesnear = / 2wheretan canbeanylargenumberwhereas sin2 is strictly positiveinitsothatthedeviationof f from0can beaslargeasdesirednomatterhowsmall R is.So,forany > 0, theinequality | f ( r ) Š f0| < isviolatedinthatsectorofanydisk r Š r0 < ,andhencethelimitdoesnotexist. Remark. Formultivariablelimitswith n> 2,asimilarapproach exists.If,forsimplicity, r0=(0 0 ,..., 0).Thenput xi= Rui,where thevariables uisatisfythecondition u2 1+ u2 2+ + u2 n=1.For n =2, u1=cos and u2=sin .For n 3,thevariables uicanbeviewedas thedirectionalcosines,thatis,thecosinesoftheanglesbetween r and ei, ui= r ei/ r .Thenonehastoinvestigatewhetherthereis h ( R ) suchthat | f ( Ru1,Ru2,...,Run) Š f0| h ( R ) 0 ,R 0 Thistechnical,oftenratherdicult,taskmaybeaccomplishedusing theinequalities | ui| 1andsomespeci“cpropertiesofthefunction f Asnoted,thevariables uiarethedirectionalcosines.Theycanalsobe trigonometricfunctionsoftheanglesinthesphericalcoordinatesystem inan n -dimensionalEuclideanspace.Theproblemoftheexistenceor non-existenceofthelimitamountstostudyingthebehaviorofsome trigonometricfunctions.87.5.StudyProblems.Problem13.1. Findthelimit limr r0f ( r ) orshowthatitdoesnot exist,where f ( r )= f ( x,y,z )=( x2+2 y2+4 z2)ln( x4+ y4+ z4) r0=(0 0 0) Solution: Step1 .Thecontinuityargumentdoesnotapplybecause f isnot de“nedat r0. Step2 .Nosubstitutionispossibletotransformthelimittoaonevariablelimit.

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87.AGENERALSTRATEGYTOSTUDYLIMITS143 Step3 .Put r ( t )=( t,at,bt )forsomeconstants a and b thatde“ne thedirectionoftheline.Then f ( r ( t ))= At2ln( Bt4)=4 At2ln( t )+ A ln( B ) t2 0as t 0+,where A =1+2 a2+4 b2and B =1+ a4+ b4(recallthatbylHospitalsrule t ln( t )=ln( t ) /tŠ 1 0as t 0+).So, ifthelimitexists,thenitmustbeequalto0. Step4 .Put R2= x2+ y2+ z2.Bymakinguseoftheinequalities | x | R | y | R | z | R ,onehas x2+2 y2+4 z2 7 R2and x4+ y4+ z4 3 R4. Hence,bythe monotonicity ofthelogarithmfunction, | f ( r ) Š 0 | 7 R2ln(3 R4)=7 R2(4ln( R )+ln(3)) 0as R 0+. Bythesqueezeprinciple,thelimitexistsandequals0. Problem13.2. Provethatthelimit limr r0f ( r ) exists,where f ( r )= f ( x,y )= 1 Š cos( x2y ) x2+2 y2, r0=(0 0) and“ndadiskcenteredat r0inwhichvaluesof f deviatefromthe limitnomorethan =0 5 10Š 4. Solution: Step1 .Thecontinuityargumentdoesnotapplybecause f isnot de“nedat r0. Step2 .Nosubstitutionispossibletotransformthelimittoaonevariablelimit. Step3 .Put r ( t )=( t,at ).Then limt 0+f ( r ( t ))=limt 0+1 Š cos( at3) t2(1+2 a2) = 1 1+2 a2limu 0+1 Š cos( au3 / 2) u = 1 1+2 a2limu 0+au1 / 2sin( au3 / 2) 1 =0 wherethesubstitution u = t2andlHospitalsrulehavebeenusedto evaluatethelimit.Therefore,ifthelimitexists,itmustbeequalto0. Step4 .Note“rstthat1 Š cos u =2sin2( u/ 2) u2/ 2,wherethe inequality | sin x || x | hasbeenused.Put R2= x2+ y2.Then,by makinguseoftheaboveinequalitywith u = x2y togetherwith | x | R and | y | R ,thefollowingchainofinequalitiesisobtained: | f ( r ) Š 0 | ( x2y )2/ 2 x2+2 y2= ( x2y )2/ 2 R2+ y2 ( x2y )2/ 2 R2 1 2 R6 R2= R4 2 0 as R 0+.Bythesqueezeprinciple,thelimitexistsandequals0. Fromtheaboveinequality,itfollowsthat | f ( r ) | < if R4/ 2 < and hence r Š r0 = R< ( )=(2 )1 / 4=0 1.

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14413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Problem13.3. Findthelimit limr r0f ( r ) orshowthatitdoesnot exist,where f ( r )= f ( x,y )= x2y x2Š y2, r0=(0 0) Solution: Step1 .Thecontinuityargumentdoesnotapplybecause f isnot de“nedat r0. Step2 .Nosubstitutionispossibletotransformthelimittoaonevariablelimit. Step3 .Thedomain D ofthefunctionisthewholeplanewiththe lines y = x excluded.Soput r ( t )=( t,at ),where a = 1.Then f ( r ( t ))= at3/t2(1 Š a2)= a (1 Š a2)Š 1t 0as t 0+.So,ifthelimit exists,thenitmustbeequalto0. Step4 .Inpolarcoordinates, x = R cos and y = R sin ,where r Š r0 = R f ( r )= R3cos2 sin R2(cos2 Š sin2 ) = 1 2 R cos sin(2 ) cos(2 ) = R cos 2 tan(2 ) Therefore,inanydisk0 < r Š r0 0small enoughbecausetan(2 )isnotboundedinthisinterval.Hence,forany > 0,thereisno > 0suchthat | f ( r ) | < whenever r D liesin thedisk r Š r0 < .Thus,thelimitdoesnotexist. Step3(Optional) .ThenonexistenceofthelimitestablishedinStep 4impliesthatthereshouldexistcurvesalongwhichthelimitdiers from0.Itisinstructivetodemonstratethisexplicitly.Anysuchcurve shouldapproachtheoriginfromwithinoneofthenarrowsectorscontainingthelines y = x (wheretan(2 )takeslargevalues).Soput, forexample, r ( t )=( t,t Š atn),where n> 1and a =0isanumber. Then f ( r ( t ))=( t3+ atn +2) / (2 atn +1Š a2t2 n).Thisfunctiontendsto anumberas t 0+if n ischosentomatchtheleading(smallest) powersofthetopandbottomoftheratiointhislimit(i.e.,3= n +1 or n =2).Thus,for n =2, f ( r ( t )) 1 / (2 a )as t 0+and f ( r ( t )) divergesfor n> 2inthislimit.

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88.PARTIALDERIVATIVES145 87.6.Exercises.(1) Findeachofthefollowinglimitsorshowthatit doesnotexist: (i)cos( xy + z ) x4+ y2z2+4(ii)limr 0 cos2( xy ) Š 1 xy(iii)limr 0 xy2+1 Š 1 xy2(iv)limr 0 sin( xy3) x2(v)limr 0 x3+ y5 x2+2 y2(vi)limr 0 e r Š 1 Š r r 2(vii)limr 0 x2+sin2y x2+2 y2(viii)limr 0 xy2+ x sin( xy ) x2+2 y288.PartialDerivatives Thederivative f( x0)ofafunction f ( x )at x = x0containsimportantinformationaboutthelocalbehaviorofthefunctionnear x = x0. Itde“nestheslopeofthetangentline L ( x )= f ( x0)+ f( x0)( x Š x0), and,for x closeenoughto x0,valuesof f canbewellapproximatedby thelinearization L ( x ),thatis, f ( x ) L ( x ).Inparticular,if f( x0) > 0, f increasesnear x0,and,if f( x0) < 0, f decreasesnear x0.Furthermore,thesecondderivative f( x0)suppliesmoreinformationabout f near x0,namely,itsconcavity. Itisthereforeimportanttodevelopasimilarconceptforfunctions ofseveralvariablesinordertostudytheirlocalbehavior.Asigni“cant dierenceisthat,givenapointinthedomain,therateofchangeis goingtodependonthedirectioninwhichitismeasured.Forexample, if f ( r )istheheightofahillasafunctionofposition r ,thentheslopes fromwesttoeastandfromsouthtonorthmaybedierent.This observationleadstotheconceptofpartialderivatives.If x and y are thecoordinatesfromwesttoeastandfromsouthtonorth,respectively, thenthegraphof f isthesurface z = f ( x,y ).Ata“xedpoint r0= ( x0,y0),theheightchangesas h ( x )= f ( x,y0)alongthewest…east direction,andas g ( y )= f ( x0,y )alongthesouth…northdirection.Their graphsareintersectionsofthesurface z = f ( x,y )withthecoordinate planes x = x0and y = y0,thatis, z = f ( x0,y )= g ( y )and z = f ( x,y0)= h ( x ).Theslopealongthewest…eastdirectionisthen h( x0), andalongthesouth…northdirection,is g( y0).Theseslopesarecalled partialderivatives of f anddenotedas f x ( x0,y0)= d dx f ( x,y0) x = x0, f y ( x0,y0)= d dy f ( x0,y ) y = y0.

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14613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Thepartialderivativesareoftendenotedas f x ( x0,y0)= f x( x0,y0) f y ( x0,y0)= f y( x0,y0) Thesubscriptof findicatesthevariablewithrespecttowhichthe derivativeiscalculated.Theconceptofpartialderivativescaneasily beextendedtofunctionsofmorethantwovariables.88.1.PartialDerivativesofaFunctionofSeveralVariables.Let D bea subsetofan n -dimensionalEuclideanspace. Definition 13.13 (InteriorPointofaSet) Apoint r0issaidtobeaninteriorpointof D ifthereisanopenball B( r0)= { r | r Š r0 < } ofradius thatliesin D (i.e., B( r ) D ). Inotherwords, r0isaninteriorpointof D ifthereisapositive number > 0suchthatallpointswhosedistancefrom r0islessthan alsoliein D .Forexample,if D isasetpointsinaplanewhose coordinatesareintegers,then D hasnointeriorpointsatallbecause thepointsofadiskofradius0
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88.PARTIALDERIVATIVES147 domainof f andlet r0beaninteriorpointof D .Ifthelimit f xi( r0)=limh 0f ( r0+ h ei) Š f ( r0) h exists,thenitiscalledthepartialderivativeof f withrespectto xiat r0. Thereasonthepoint r0needstobeaninteriorpointissimple.By thede“nitionoftheone-variablelimit, h canbenegativeorpositive. Sothepoints r0+ h ei, i =1 2 ,...,n ,mustbeinthedomainofthe functionbecauseotherwise f ( r0+ h ei)isnotevende“ned.Thisis guaranteedif r0isaninteriorpointbecauseallpoints r intheball Ba( r0)ofsucientlysmallradius a = | h | arein D Let r0=( a1,a2,...,an),where aiare“xednumbers.Considerthe function F ( xi)ofonevariable xi( i is“xed),whichisobtainedfrom f ( r )by“xingallthevariables xj= ajexceptthe i thone(i.e., xj= ajforall j = i ).Bythede“nitionoftheordinaryderivative,the partialderivative f xi( r0)existsifandonlyifthederivative F( ai)exists because (13.1) f xi( r0)=limh 0F ( ai+ h ) Š F ( ai) h = dF ( xi) dxi xi= aijustlikeinthecaseoftwovariablesdiscussedatthebeginningofthis section.Thisruleispracticalforcalculatingpartialderivativesasit reducestheproblemtocomputingordinaryderivatives. Example 13.12 Findthepartialderivativesof f ( x,y,z )= x3Š y2z atthepoint (1 2 3) Solution: Bytherule(13.1), f x(1 2 3)= d dx f ( x, 2 3) x =1= d dx ( x3Š 12) x =1=3 f y(1 2 3)= d dy f (1 ,y, 3) y =2= d dy (1 Š 3 y2) y =2= Š 12 f z(1 2 3)= d dz f (1 2 ,z ) z =3= d dz (1 Š 4 z ) z =3= Š 4 88.1.1.GeometricalSigni“canceofPartialDerivatives.Fromtherule (13.1),itfollowsthat thepartialderivative f xi( r0) de“nestherateof changeofthefunction f whenonlythevariable xichangeswhilethe othervariablesarekept“xed .If,forinstance,thefunction f inExample13.12de“nesthetemperatureindegreesCelsiusasafunction ofpositionwhosecoordinatesaregiveninmeters,then,atthepoint

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14813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (1 2 3),thetemperatureincreasesattherate4degreesCelsiusper meterinthedirectionofthe x axis,anditdecreasesattherates Š 12 and Š 4degreesCelsiuspermeterinthedirectionofthe y and z axes, respectively.88.2.PartialDerivativesasFunctions.Supposethatthepartialderivativesof f existatallpointsofaset D (whichisasubsetofthedomain of f ).Theneachpartialderivativecanbeviewedasafunctionof severalvariableson D .Thesefunctionsaredenotedas f xi( r ),where r D .Theycanbefoundbythesamerule(13.1)if,whendierentiatingwithrespectto xi,allothervariablesarenotsettoanyspeci“c valuesbutratherviewedasindependentof xi(i.e., dxj/dxi=0forall j = i ).Thisagreementisre”ectedbythenotation f xi( x1,x2,...,xn)= xif ( x1,x2,...,xn); thatis,thesymbol /ximeansdierentiationwithrespectto xiwhile regardingallothervariablesasnumericalparametersindependentof xi. Example 13.13 Find f x( x,y ) and f y( x,y ) if f ( x,y )= x sin( xy ) Solution: Assuming“rstthat y isanumericalparameterindependentof x ,oneobtains f x( x,y )= x f ( x,y )= x x sin( xy )+ x x sin( xy ) =sin( xy )+ xy cos( xy ) bytheproductruleforthederivative.Ifnowthevariable x isviewed asanumericalparameterindependentof y ,oneobtains f y( x,y )= y f ( x,y )= x y sin( xy )= x2cos( xy ) 88.3.BasicRulesofDifferentiation.Sinceapartialderivativeisjust anordinaryderivativewithoneadditionalagreementthatallother variablesareviewedasnumericalparameters,thebasicrulesofdifferentiationapplytopartialderivatives.Let f and g befunctionsof severalvariablesandlet c beanumber.Then xi( cf )= c f xi, xi( f + g )= f xi+ g xi, xi( fg )= f xig + f g xi, xi f g =f xig Š fg xi g2.

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89.HIGHER-ORDERPARTIALDERIVATIVES149 Let h ( u )beadierentiablefunctionofonevariableandlet g ( r )bea functionofseveralvariableswhoserangeliesinthedomainof f .Then onecande“nethecomposition f ( r )= h ( g ( r )).Assumingthatthe partialderivativesof g exist,thechainruleholds (13.2) f xi= h( g ) g xi. Example 13.14 Findthepartialderivativesofthefunction f ( r )= r Š 1,where r =( x1,x2,...,xn) Solution: Put h ( u )= uŠ 1 / 2and g ( r )= x2 1+ x2 2+ + x2 n= r 2. Then f ( r )= h ( g ( r )).Since h( u )=( Š 1 / 2) uŠ 3 / 2and g/xi=2 xi, thechainrulegives xi r Š 1= Š xi r 3. 88.4.Exercises.(1) Findthespeci“edpartialderivativesofeachof thefollowingfunctions: (i) f ( x,y )=( x Š y ) / ( x + y ), f x(1 2), f y(1 2) (ii) f ( x,y,z )=( xy + z ) / ( z + y ), f x(1 2 3), f y(1 2 3), f z(1 2 3) (iii) f ( r )=( x1+2 x2+ + nxn) / (1+ r 2), f xi( 0 ), i =1 2 ,...,n (iv) f ( x,y,z )= x sin( yz ), f x(1 2 ,/ 2), f y(1 2 ,/ 2), f z(1 2 ,/ 2) (2) Findthespeci“edpartialderivativesofeachofthefollowing functions: (i) f ( x,y )=( x + y2)n, f x( x,y ), f y( x,y ) (ii) f ( x,y )= xy, f x( x,y ), f y( x,y ) (iii) f ( x,y )= xe( x +2 y )2, f x( x,y ), f y( x,y ) (iv) f ( x,y )=sin( xy )cos( x2+ y2), f x( x,y ), f y( x,y ) (v) f ( x,y,z )=ln( x + y2+ z3), f x( x,y,z ), f y( x,y,z ), f z( x,y,z ) (v) f ( x,y,z )= xy2cos( z2x ), f x( x,y,z ), f y( x,y,z ), f z( x,y,z ) (vi) f ( r )=( a1x1+ a2x2+ + anxn)m=( a r )m, f xi( r ), i =1 2 ,...,n (3) Determinewhetherthefunction f ( x,y )increasesordecreases when x increases,while y is“xed,andwhen y increases,while x is “xedataspeci“edpoint P0: (i) f ( x,y )= xy/ ( x + y ), P0(1 2) (ii) f ( x,y )=( x2Š 2 y2)1 / 3, P0(1 1) (iii) f ( x,y )= x2sin( xy ), P0( Š 1 )89.Higher-OrderPartialDerivatives Sincepartialderivativesofafunctionarealsofunctionsofseveral variables,theycanbedierentiatedwithrespecttoanyvariable.For

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15013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS example,forafunctionoftwovariables,allpossiblesecondderivatives are f x Š x f x = 2f x2, y f x = 2f yx f y Š x f y = 2f xy y f y = 2f y2. Throughoutthetext,briefnotationsforhigher-orderderivativeswill alsobeused.Forexample, 2f x2=( f x) x= f xx, 2f xy =( f y) x= f yxandsimilarlyfor f yyand f xy.Derivativesofthethirdorderarede“ned asderivativesofsecond-orderderivatives,andsoon. Example 13.15 Forthefunction f ( x,y )= x4Š x2y + y2,“ndall second-andthird-orderderivatives. Solution: The“rstderivativesare f x=4 x3Š 2 xy and f y= Š x2+2 y Thenthesecondderivativesare f xx=(4 x3Š 2 xy ) x=12 x2Š 2 y,f yy=( Š x2+2 y ) y=2 f xy=(4 x3Š 2 xy ) y= Š 2 x,f yx=( Š x2+2 y ) x= Š 2 x. Thethirdderivativesarefoundsimilarly: fxxx=(12 x2Š 2 y ) x=24 x,fyyy=(2) y=0 fxxy=(12 x2Š 2 y ) y= Š 2 ,fxyx= fyxx=( Š 2 x ) x= Š 2 fyyx=(2) x=0 ,fyxy= fxyy=( Š 2 x ) y=0 Incontrasttotheone-variablecase,therearehigher-orderderivativesofanewtypethatareobtainedbydierentiatingwithrespect todierentvariablesindierentorders,like f xyand f yx.Intheabove example,ithasbeenfoundthat f xy= f yx, fxxy= fxyx= fyxx, fxyy= fyyx= fyxy; thatis,theresultofdierentiationis independentoftheorderinwhich thederivativeshavebeentaken .Isthisapeculiarityofthefunctionconsideredorageneralfeatureofhigher-orderderivatives?Thefollowing theoremanswersthisquestion.

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89.HIGHER-ORDERPARTIALDERIVATIVES151 Theorem 13.9 (ClairautsorSchwarzsTheorem) Let f beafunctionofseveralvariables ( x1,x2,...,xn) thatisde“nedon anopenball D inaEuclideanspace.Ifthesecondderivatives f xixjand f xjxi,where j = i ,arecontinuousfunctionson D ,then f xixj= f xjxiat anypointof D AconsequenceofClairautstheoremcanbeproved.Itassertsthat, ifhigher-orderderivativesarecontinuousfunctions,thentheresultof dierentiationisindependentoftheorderinwhichthederivativeshave beentaken.Inmanypracticalapplications,itisnotnecessarytocalculatehigher-orderderivativesinallpossibleorderstoverifythehypothesisofClairautstheorem(i.e.,thecontinuityofthederivatives).Derivativesofpolynomialsarepolynomialsandhencecontinuous.Derivatives ofbasicelementaryfunctionslikethesineandcosineandexponential functionsarecontinuous.Socompositionsofthesefunctionswithmultivariablepolynomialshavecontinuousderivativesofanyorder.In otherwords,thecontinuityofthederivativescanoftenbeestablished bydierent,simplermeans. Example 13.16 Findthethirdderivatives fxyz, fyzx, fzxy,andso on,forallpermutationsof x y ,and z ,if f ( x,y,z )=sin( x2+ yz ) Solution: Thesineandcosinefunctionsarecontinuouslydierentiableasmanytimesasdesired.Theargumentofthesinefunctionis amultivariablepolynomial.Bythecompositionrule(sin g ) x= g xcos g andsimilarlyfortheotherderivatives,partialderivativesofanyordermustbeproductsofpolynomialsandthesineandcosinefunctions whoseargumentisapolynomial.Therefore,theyarecontinuousin theentirespace.ThehypothesisofClairautstheoremissatis“ed,and henceallthederivativesinquestioncoincideandareequalto fxyz=( f x) yz=(2 x cos( x2+ yz )) yz=( Š 2 xz sin( x2+ yz )) z= Š 2 x sin( x2+ yz ) Š 2 xyz cos( x2+ yz ) 89.1.ReconstructionofaFunctionfromItsDerivatives.Oneofthestandardproblemsincalculusis“ndingafunction f ( x )ifitsderivative f( x )= F ( x )isknown.Asucientconditionfortheexistenceofa solutionisthecontinuityof F ( x ).Inthiscase, f( x )= F ( x )= f ( x )= F ( x ) dx + c,

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15213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS where c isaconstant.Asimilarproblemcanbeposedforafunction ofseveralvariables.Giventhe“rstpartialderivatives (13.3) f xi( r )= Fi( r ) ,i =1 2 ,...,n, “nd f ( r )ifitexists.Theexistenceofsuch f isamoresubtlequestion inthecaseofseveralvariables.Supposepartialderivatives Fi/xjare continuousfunctionsinanopenball.Thentakingthederivative /xjofbothsidesof(13.3)andapplyingClairautstheorem,oneinfersthat (13.4) f xixj= f xjxi= Fi xj= Fj xi. Thus,theconditions(13.4)onthefunctions Fimustbeful“lled;otherwise, f satisfying(13.3)doesnotexist.Theconditions(13.4)are called integrabilityconditions forthesystemofequations(13.3). Example 13.17 Supposethat f x( x,y )=2 x + y and f y( x,y )= 2 y Š x .Doessuchafunction f exist? Solution: The“rstpartialderivativesof f F1( x,y )=2 x + y and F2( x,y )=2 y Š x ,arepolynomials,andhencetheirderivativesare continuousintheentireplane.Inorderfor f toexist,theintegrability condition F1/y = F2/x mustholdintheentireplane.Thisis notsobecause F1/y =1,whereas F2/x = Š 1.Thus,nosuch f exists. Supposenowthattheintegrabilityconditions(13.4)aresatis“ed. Howisasolution f to(13.3)tobefound?Evidently,onehasto calculateanantiderivativeofthepartialderivative.Intheone-variable case,anantiderivativeisde“neduptoanadditiveconstant.Thisis notsointhemultivariablecase.Forexample,let f x( x,y )=3 x2y .An antiderivativeof f xisafunctionwhose partial derivativewithrespect to x is3 x2y .Itiseasytoverifythat x3y satisfythisrequirement.It isobtainedbyintegrating3 x2y withrespectto x whileviewing y as anumericalparameterindependentof x .Justlikeintheone-variable case,onecanalwaysaddaconstanttotheintegral, x3y + c andobtain anothersolution.Thekeypointtoobserveisthattheintegration constantmaybeafunctionof y !Indeed,( x3y + g ( y )) x=3 x2y .Thus, thegeneralsolutionof f x( x,y )=3 x2y is f ( x,y )= x3y + g ( y ),where g ( y )isarbitrary. If,inaddition,theotherpartialderivative f yisgiven,thenan explicitformof g ( y )canbefound.Put,forexample, f y( x,y )= x3+ 2 y .Theintegrabilityconditionsareful“lled:( f x)y =(3 x2y ) y=3 x2and( f y) x=( x3+2 y ) x=3 x2.Soafunctionwiththesaidpartial derivativesdoesexist.Thesubstitutionof f ( x,y )= x3y + g ( y )into

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89.HIGHER-ORDERPARTIALDERIVATIVES153 theequation f y= x3+2 y yields x3+ g( y )= x3+2 y or g( y )=2 y andhence g ( y )= y2+ c .Notethecancellationofthe x3term.This isadirectconsequenceoftheful“lledintegrabilitycondition.Had onetriedtoapplythisprocedurewithoutcheckingtheintegrability conditions,onecouldhavefoundthat,ingeneral,nosuch g ( y )exists. InExample13.17,theequation f x=2 x + y hasageneralsolution f ( x,y )= x2+ yx + g ( y ).Itssubstitutionintothesecondequation f y=2 y Š x yields x + g( y )=2 y Š x or g( y )=2 y Š 2 x .Thederivative of g ( y )cannotdependon x andhencenosuch g ( y )exists. Example 13.18 Find f ( x,y,z ) if f x= yz +2 x = F1, f y= xz + 3 y2= F2,and f z= xy +4 z3= F3orshowthatitdoesnotexist. Solution: Theintegrabilityconditions( F1) y=( F2) x,( F1) z=( F3) x, and( F2) z=( F3) zaresatis“ed(theirveri“cationislefttothereader). So f exists.Takingtheantiderivativewithrespectto x inthe“rst equation,one“nds f x= yz +2 x = f ( x,y,z )= ( yz +2 x ) dx = xyz + x2+ g ( y,z ) where g ( y,z )isarbitrary.Thesubstitutionof f intothesecondequationsyields f y= xz +3 y2= xz + g y( y,z )= xz +3 y2= g y( y,z )=3 y2= g ( y,z )= 3 y2dy = y3+ h ( z ) = f ( x,y,z )= xyz + x2+ y3+ h ( z ) where h ( z )isarbitrary.Thesubstitutionof f intothethirdequation yields f z= xy +4 z3= xy + h( z )= xy +4 z3= h( z )=4 z3= h ( z )= z4+ c = f ( x,y,z )= xyz + x2+ y3+ z4+ c, where c isaconstant. Theprocedureofreconstructing f fromits“rstpartialderivatives aswellastheintegrabilityconditions(13.4)willbeimportantwhen discussing conservativevector“elds andthe potential ofaconservative vector“eld.

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15413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 89.2.PartialDifferentialEquations.Therelationbetweenafunctionof severalvariablesanditspartialderivatives(ofanyorder)iscalleda partialdierentialequation .Partialdierentialequationsareakey tooltostudyvariousphenomenainnature.Manyfundamentallawsof naturecanbestateintheformofpartialdierentialequations.89.2.1.DiffusionEquation.Let n ( r ,t ),where r =( x,y,z )isthepositionvectorinspaceand t istime,beaconcentrationofasubstance, say,inairorwateroreveninasolid.Evenifthereisnomacroscopic motioninthemedium,theconcentrationchangeswithtimedueto thermalmotionofthemolecules.Thisprocessisknownas diusion Insomesimplesituations,therateatwhichtheconcentrationchanges withtimeatapointis n t= D ( n xx+ n yy+ n zz) wheretheparameter D isthediusionconstant.Sotheconcentration asafunctionofthespatialpositionandtimemustsatisfytheabove partialdierentialequation.89.2.2.WaveEquation.Soundinairispropagatingdisturbancesofthe airdensity.If u ( r ,t )isthedeviationoftheairdensityfromitsconstant (nondisturbed)value u0atthespatialpoint r =( x,y,z )andattime t ,thenitcanbeshownthatsmalldisturbances u/u0 1satisfythe waveequation : u tt= c2( u xx+ u yy+ u zz) where c isthespeedofsoundintheair.Lightisanelectromagnetic wave.Its propagationisalsodescribedbythe wave equation,where c isthespeedoflightinvacuum(orinamedium,iflightgoesthrough amedium)and u istheamplitudeofelectricormagnetic“elds.89.2.3.LaplaceandPoissonEquations.Theequation uxx+ uyy+ uzz= f, where f isagivennon-zerofunctionofposition r =( x,y,z )inspace, iscalledthe Poissonequation .Inthespecialcasewhen f =0,this equationisknownasthe Laplaceequation .ThePoissonandLaplace equationsareusedtodeterminestaticelectromagnetic“eldscreated bystaticelectricchargesandcurrents. Example 13.19 Let h ( q ) beatwice-dierentiablefunctionofa variable q .Showthat u ( r ,t )= h ( ct Š n r ) isasolutionofthewave equationforany“xedunitvector n .

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89.HIGHER-ORDERPARTIALDERIVATIVES155 Solution: Let n =( n1,n2,n3),where n2 1+ n2 2+ n2 3=1as n isthe unitvector.Put q = ct Š n r = ct Š n1x Š n2y Š n3z .Bythechainrule (13.2), u t= q th( q )andsimilarlyfortheotherderivatives,one“nds u t= ch( q ), u tt= c2h( q ), u x= Š n1h( q ), u xx= n2 1h( q ),and,inthe samefashion, u yy= n2 2h( q ), u zz= n2 3h( q ).Then u xx+ u yy+ u zz= ( n2 1+ n2 2+ n2 3) h( q )= h( q ),whichcoincideswith u tt/c2,meaningthat thewave equationissatis“edforany h Considerthelevelsurfacesofthesolutionofthe wave equation discussedinthisexample.Theycorrespondtoa“xedvalueof q = q0. So,foreachmomentoftime t ,thedisturbanceoftheairdensity u ( r ,t ) hasaconstantvalue h ( q0)intheplane n r = ct Š q0= d ( t ).Allplanes withdierentvaluesoftheparameter d areparallelastheyhavethe samenormalvector n .Sincehere d ( t )isafunctionoftime,theplaneon whichtheairdensityhasa“xedvaluemovesalongthevector n atthe rate d( t )= c .Thus,adisturbanceoftheairdensitypropagateswith speed c .Thisisthereasonthattheconstant c inthewave equation iscalledthe speedof sound .Evidently,thesamelineofarguments appliestoelectromagnetic waves;thatis,theymovethroughspaceat thespeedoflight.Thespeedofsoundintheairisabout342meters persecond,orabout768mph.Thespeedoflightis3 108meters persecond,or186milespersecond.Ifalightningstrikeoccursamile away duringathunderstorm,itcanbeseenalmostinstantaneously, whilethethunderwillbeheardinabout5secondslater.89.3.StudyProblems.Problem13.4. Findthevalueofaconstant k forwhichthefunction u ( r ,t )= tŠ 3 / 2eŠ kr2/t,r = r satis“esthediusionequationforall t> 0 Solution: Notethat u dependsonthecombination r2= x2+ y2+ z2. To“ndthepartialderivativesof u ,itisconvenienttousethechain rule: u x = u r2r2 x =2 x u r2= Š 2 kx t u, u xx= x u x = Š 2 k t u Š 2 kx t u x = Š 2 k t + 4 k2x2 t2 u. Toobtain u yyand u zz,notethat r2issymmetricwithrespecttopermutationsof x y ,and z .Therefore, u yyand u zzareobtainedfrom u xxbyreplacing,inthelatter, x by y and x by z ,respectively.Hence,the

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15613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS rightsideofthediusionequationreads D ( u xx+ u yy+ u zz)= Š 6 Dk t + 4 Dk2r2 t2 u. Usingtheproductruletocalculatethetimederivative,one“ndsfor theleftside u t= Š 3 2 tŠ 5 / 2eŠ kr2/t+ tŠ 3 / 2eŠ kr2/tkr2 t2= Š 3 2 t + kr2 t2 u. Sincebothsidesmustbeequalfor all valuesof t> 0and r2,the comparisonofthelasttwoexpressionsyields two conditions:6 Dk = 3 / 2(astheequalityofthecoecientsat1 /t )and k =4 Dk2(asthe equalityofthecoecientsat r2/t2).Theonlycommonsolutionof theseconditionsis k =1 / (4 D ). Problem13.5. Considerthefunction f ( x,y )= x3y Š xy3 x2+ y2if( x,y ) =(0 0)and f (0 0)=0 Find f x( x,y ) and f y( x,y ) for ( x,y ) =(0 0) .Usetherule(13.1)to “nd f x(0 0) and f y(0 0) and,thereby,toestablishthat f xand f yexist everywhere.Usetherule(13.1)againtoshowthat f xy(0 0)= Š 1 and f yx(0 0)=1 ,thatis, f xy(0 0) = f yx(0 0) .Doesthisresultcontradict Clairautstheorem? Solution: Usingtheratiorulefordierentiation,one“nds f x( x,y )= x4y +4 x2y3Š y5 ( x2+ y2)2,f y( x,y )= x5Š 4 x3y2Š xy4 ( x2+ y2)2if( x,y ) =(0 0).Notethat,owingtothesymmetry f ( x,y )= Š f ( y,x ), thederivative f yisobtainedfrom f xbychangingthesignofthelatter andswapping x and y .Thederivativesat(0 0)arefoundbytherule (13.1): f x(0 0)= d dx f ( x, 0) x =0=0 ,f y(0 0)= d dy f (0 ,y ) y =0=0 Thederivativesarecontinuousfunctions(theproofislefttothereader asanexercise).Next,onehas f xy(0 0)= d dy f x(0 ,y ) y =0=limh 0f x(0 ,h ) Š f x(0 0) h =limh 0Š h Š 0 h = Š 1 ,

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89.HIGHER-ORDERPARTIALDERIVATIVES157 f yx(0 0)= d dx f y( x, 0) x =0=limh 0f y( h, 0) Š f y(0 0) h =limh 0h Š 0 h =1 TheresultdoesnotcontradictClairautstheorembecause f xy( x,y ) and f yx( x,y )arenotcontinuousat(0 0).Byusingtheratioruleto dierentiate f x( x,y )withrespectto y ,anexplicitformof f xy( x,y )for ( x,y ) =(0 0)canbeobtained.Bytakingthelimitof f xy( x,y )as ( x,y ) (0 0)alongthestraightline( x,y )=( t,at ), t 0,oneinfers thatthelimitdependsontheslope a andhencethetwo-dimensional limitdoesnotexist,thatis,lim( x,y ) (0 0)f xy( x,y ) = f xy(0 0)= Š 1and f xyisnotcontinuousat(0 0).Thetechnicaldetailsarelefttothe reader. 89.4.Exercises.(1) Findallsecondpartialderivativesofeachofthe followingfunctionsandverifyClairautstheorem: (i) f ( x,y )=tanŠ 1xy (ii) f ( x,y,z )= x sin( zy2) (iii) f ( x,y,z )= x3+ zy + z2(iv) f ( x,y,z )=( x + y ) / ( x +2 z ) (2) Findtheindicatedpartialderivativesofeachofthefollowing functions: (i) f ( x,y )= xn+ xy + ym, f xxy, f xyx, f yyx, f xyy(ii) f ( x,y,z )= x cos( yx )+ z3, f xyz, f xxz, f yyz(iii) f ( x,y,z )=sin( xy ) ez, f5/z5, f(4) xyzz, f(4) zyxz, f(4) zxzy(3) Givenpartialderivatives,“ndthefunctionorshowthatitdoes notexist: (i) f x=3 x2y f y= x3+3 y2(ii) f x= yz +3 x2, f y= xz +4 y f z= xy +1 (iii) f xk= kxk, k =1 2 ,...,n (iv) f x= xy + z f y= x2/ 2, f z= x + y (v) f x=sin( xy )+ xy cos( xy ), f y= x2cos( xy )+1 (4) Verifythatagivenfunctionisasolutionoftheindicateddifferentialequation: (i) f ( t,x )= A sin( ct Š x )+ B cos( ct + x ), cŠ 2f ttŠ f xx=0 (ii) f ( x,y )=ln( x2+ y2), f xx+ f yy=0 (iii) f ( x,y )=ln( ex+ ey), f x+ f y=1and f xxf yyŠ ( f xy)2=0 (iv) f ( r )=exp( a r ),where a a =1, f x1x1+ f x2x2+ + f xnxn= f

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15813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 90.ChainRulesandImplicitDifferentiation90.1.ChainRules.Considerthefunction f ( x,y )= x3+ xy2whose domainistheentireplane.Pointsoftheplanecanbelabeledina dierentway.Forexample,thepolarcoordinates x = r cos y = r sin maybeviewedasarulethatassignsanorderedpair( x,y )to anorderedpair( r, ).Usingthisrule,thefunctioncanbeexpressed inthenewvariablesas f ( r cos ,r sin )= r3sin = F ( r, ).Onecan computetheratesofchangeof f withrespecttothenewvariables: f r = F r =3 r2sin f = F = r3cos Alternatively,theseratescanbecomputedas f r = f x x r + f y y r =(3 x2+ y2)cos +2 xy sin =3 r2sin f = f x x + f y y = Š (3 x2+ y2) r sin +2 xyr cos = r3cos where x and y havebeenexpressedinthepolarcoordinatestoobtain the“nalexpressions.Thedierencebetweenthetwoapproachesisthat inthesecondone anexplicitformofthefunctioninthenewvariables isnotrequiredto“nditsrateswithrespecttothenewvariables Furthermore,considerthevaluesofthisfunctionalongthecurve x = t y = t2, f ( t,t2)= t3+ t5= F ( t ).Therateofchangeof f with respecttothecurveparameteris df dt = dF dt =3 t2+5 t4. Itcanalsobeobtainedwithoutcalculating“rsttheexplicitformof thefunction f asafunctionofthecurveparameterinmuchthesame fashionasinthecaseofpolarcoordinates: df dt = f x dx dt + f y dy dt =(3 x2+ y2)+(2 xy )(2 t )=3 t2+5 t4. Qualitatively,thisexpressionoftherate df/dt israthernatural.Ifonly x dependson t while y doesnot(i.e., y =const),thentherate df/dt isdeterminedbyanordinarychainrule df/dt = f xx( t );the y variable ismerelyanumericalparameter.If y = y ( t )and x =const,thechain ruleforone-variablefunctionsgives df/dt = f yy( t ).Whenboth x and y dependon t ,then df/dt becomesthesumofthesetwotermsasboth rates x( t )and y( t )shouldcontributeto df/dt Theexamplesconsideredaboveillustrateageneralruleofdierentiationcalledthe chainrule .

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90.CHAINRULESANDIMPLICITDIFFERENTIATION159 Theorem 13.10 (ChainRule) Let f beafunctionof n variables r =( x1,x2,...,xn) suchthatallits partialderivativesexist.Supposethateachvariable xiis,inturn,a functionof m variables u =( u1,u2,...,um) suchthatallitspartial derivativesexist.Thecompositionof xi= xi( u ) with f ( r ) de“nes f asafunctionof u .Thenitsrateofchangewithrespectto uj, j = 1 2 ,...,m ,reads f uj= f x1x1 uj+ f x2x2 uj+ + f xnxn uj=ni =1f xixi uj. Theproofofthistheoremisrathertechnicalandisomitted. For n = m =1,thisisthefamiliarchainruleforfunctionsofone variable df/du = f( x ) x( u ).If n =1and m> 1,itisthechain rule(13.2)establishedearlier.Theexampleofpolarcoordinatescorrespondstothecase n = m =2,where r =( x,y )and u =( r, ).The rateofchangealongacurveinaplaneisthecase n =2, m =1,where r =( x,y )and u = t .Notealsothatsomeofthefunctions xi( u )may notdependonallvariables uj,andthecorrespondingpartialderivativesinthechainrulevanish. Example 13.20 Afunction f ( x,y,z ) hasthefollowingratesof changeatthepoint r0=(1 2 3) f x( r0)=1 f y( r0)=2 ,and f z( r0)= Š 2 .Supposethat x = x ( t,s )= t2s y = y ( t,s )= s + t ,and z = z ( t,s )=3 s .Findtheratesofchange f withrespectto t and s atthe point r0. Solution: Inthechainrule,put r =( x,y,z )and u =( t,s ).The point r0=(1 2 3)correspondstothepoint u0=(1 1)inthenew variables.Notethat z =3gives3 s =3andhence s =1.Then,from y =2,itfollowsthat s + t =2or1+ t =2or t =1.Also, x (1 1)=1 asrequired.Thepartialderivativesoftheoldvariableswithrespectto thenewonesare x t=2 ts y t=1, z t=0, x s= t2, y s=1,and z s=3. Bythechainrule, f t( r0)= f x( r0) x t( u0)+ f y( r0) y t( u0)+ f z( r0) z t( u0) =2+2+0=4 f s( r0)= f x( r0) x s( u0)+ f y( r0) y s( u0)+ f z( r0) z s( u0) =1+2 Š 6= Š 3 90.2.ImplicitDifferentiation.Considerthefunctionofthreevariables, F ( x,y,z )= x2+ y4Š z .Theequation F ( x,y,z )=0canbesolvedfor

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16013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS oneofthevariables,say, z toobtain z asafunctionoftwovariables: F ( x,y,z )=0= z = z ( x,y )= x2+ y4; thatis,thefunction z ( x,y )isde“nedasarootof F ( x,y,z )andhas thecharacteristicpropertythat (13.5) F ( x,y,z ( x,y ))=0forall( x,y ) Intheexampleconsidered,theequation F ( x,y,z )=0canbesolved analytically,andan explicit formofitsrootasafunctionof( x,y )can befound. Ingeneral,givenafunction F ( x,y,z ),anexplicitformofasolutiontotheequation F ( x,y,z )=0isnotalwayspossibleto“nd. Puttingasidethequestionabouttheveryexistenceofasolutionand itsuniqueness,supposethatthisequationisprovedtohaveaunique solutionwhen( x,y ) D .Inthiscase,thefunction z ( x,y )withthe property(13.5)forall( x,y ) D issaidtobede“ned implicitly on D Althoughananalyticformofanimplicitlyde“nedfunctionisunknown,itsratesofchangecanbefoundandprovideimportantinformationaboutitslocalbehavior.Supposethatpartialderivativesof F exist.Furthermore,therates z x( x,y )and z y( x,y )arealsoassumedto existonanopendisk D intheplane.Sincerelation(13.5)holdsfor all( x,y ) D ,thepartialderivativesofitsleftsidemustalsovanishin D .Thederivativescanbecomputedbythechainrule, n =3, m =2, r =( x,y,z ),and u =( u,v ),wheretherelationsbetweenoldandnew variablesare x = u y = v ,and z = z ( u,v ).Onehas u F ( x,y,z ( x,y ))= F x + F z z x =0= z x= Š F x F z, v F ( x,y,z ( x,y ))= F y + F z z y =0= z y= Š F y F z, where z u( u,v )= z x( x,y )and z v( u,v )= z y( x,y )because x = u and y = v .Theseequationsdeterminetheratesofchangeofanimplicitlyde“nedfunctionoftwovariables.Notethatinorderforthese equationstomakesense,thecondition F z =0mustbeimposed.Severalquestionsabouttheveryexistenceanduniquenessof z ( x,y )for agiven F ( x,y,z )andtheexistenceofderivativesof z ( x,y )havebeen leftunansweredintheaboveanalysis.Thefollowingtheoremaddresses themall. Theorem 13.11 (ImplicitFunctionTheorem) Let F beafunctionof n +1 variables, F ( r ,z ) ,where r =( x1,x2,...,xn) and z isreal,suchthatallitspartialderivativesarecontinuousinan

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90.CHAINRULESANDIMPLICITDIFFERENTIATION161 openball B .Supposethatthereexistsapoint ( r0,z0) B suchthat F ( r0,z0)=0 and F z( r0,z0) =0 .Thenthereexistaneighborhood D of r0andauniquefunction z = z ( r ) withcontinuousderivativeson D suchthat F ( r ,z ( r ))=0and z xi( r )= Š F xi( r ,z ( r )) F z( r ,z ( r )) forall r in D Theproofofthistheoremgoesbeyondthescopeofthiscourse.It includesproofsoftheexistenceanduniquenessof z ( r )andtheexistence ofitsderivatives.Oncethesefactsareestablished,aderivationofthe implicitdierentiationformulafollowsthesamewayasinthe n =2 case: F xi+ F z z xi=0= z xi( r )= Š F xi( r ,z ( r )) F z( r ,z ( r )) Example 13.21 Showthattheequation z (3 x Š y )= sin( xyz ) hasauniquesolution z = z ( x,y ) inaneighborhoodof (1 1) suchthat z (1 1)= / 2 and“ndtheratesofchange z x(1 1) and z y(1 1) Solution: Put F ( x,y,z )= sin( xyz ) Š z (3 x Š y ).Thentheexistenceanduniquenessofthesolutioncanbeestablishedbyverifying thehypothesesoftheimplicitfunctiontheoreminwhich r =( x,y ), r0=(1 1),and z0= / 2.First,notethatthefunction F isthesumof apolynomialandthesinefunctionofapolynomial.Soitsderivatives F x= yz cos( xyz ) Š 3 x,F y= xz cos( xyz )+ z, F z= xy cos( xyz ) Š 3 x + y arecontinuousforall( x,y,z ).Next, F (1 1 ,/ 2)=0asrequired. Finally, F z(1 1 ,/ 2)= Š 2 =0.Therefore,bytheimplicitfunction theorem,thereisanopendiskinthe xy planecontainingthepoint (1 1)inwhichtheequationhasauniquesolution z = z ( x,y ).Bythe implicitdierentiationformulas, z x(1 1)= Š F x(1 1 ,/ 2) F z(1 1 ,/ 2) = Š 3 4 ,z y(1 1)= Š F y(1 1 ,/ 2) F z(1 1 ,/ 2) = 4 Inparticular,thisresultimpliesthat,nearthepoint(1 1),theroot z ( x,y )decreasesinthedirectionofthe x axisandincreasesinthe directionofthe y axis.Itshouldbenotedthatthenumericalvaluesofthederivativescanbeusedtoaccuratelyapproximatetheroot z ( x,y )ofanonlinearequationinaneighborhoodof(1 1)byinvoking theconceptof linearization discussedbelow(seealsoStudy Problem13.6).

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16213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 90.3.Exercises.(1) Usethechainruleto“nd dz/dt if z = 1+ x2+2 y2and x =2 t3, y =ln t (2) Usethechainruleto“nd z/s and z/t if z = eŠ xsin( xy ) and x = ts y = s2+ t2. (3) Usethechainruletowritethepartialderivativesof F with respecttothenewvariables: (i) F = f ( x,y ), x = x ( u,v,w ), y = y ( u,v,w ) (ii) F = f ( x,y,z,t ), x = x ( u,v ), y = y ( u,v ), z = z ( w,s ), t = t ( w,s ) (4) Findtheratesofchange z/u z/v z/w when( u,v,w )= (2 1 1)if z = x2+ yx + y3and x = uv2+ w3, y = u + v ln w (5) Findtheratesofchange f/u f/v f/w when( x,y,z )= (1 / 3 2 0)if x =2 /u Š v + w y = vuw z = ew. (6) Findthepartialderivativesof z = f ( x,y )de“nedimplicitlyby theequation x Š z =tan1( yz ). (7) Letthetemperatureoftheairatapoint( x,y,z )be T ( x,y,z ) degreesCelsius.Aninsect”iesthroughtheairsothatitspositionasa functionoftime t insecondsisgivenby x = 1+ t y =2 t z = t2Š 1. If T x(2 6 8)=2, T y(2 6 8)= Š 1,and T x(2 6 8)=1,howfastis thetemperaturerising(ordecreasing)ontheinsectspathasit”ies throughthepoint(2 6 8)? (8) Letarectangularboxhavethedimensions x y ,and z that changewithtime.Supposethatatacertaininstantthedimensions are x =1m, y = z =2m,and x and y areincreasingattherate2 m/sand z isdecreasingattherate3m/s.Atthatinstance,“ndthe ratesatwhichthevolume,thesurfacearea,andthelargestdiagonal arechanging. (9) Afunctionissaidtobehomogeneousofdegree n if,forany number t ,ithastheproperty f ( tx,ty )= tnf ( x,y ).Giveanexample ofapolynomialfunctionthatishomogeneousofdegree n .Showthata homogeneousdierentiablefunctionsatis“estheequation xf x+ yf y= nf .Showalsothat f x( tx,ty )= tn Š 1f ( x,y ). (10) Supposethattheequation F ( x,y,z )=0de“nesimplicitly z = f ( x,y ),or y = g ( x,z ),or x = h ( y,z ).Assumingthatthederivatives F x, F y,and F zdonotvanish,provethat( z/x )( x/y )( y/z )= Š 1.91.LinearizationofMultivariableFunctions Adierentiableone-variablefunction f ( x )canbeapproximated near x = x0byitslinearization L ( x )= f ( x0)+ f( x0)( x Š x0)orthe tangentline.If x = x0+ x ,then f ( x ) Š L ( x ) x = f ( x0+ x ) Š f ( x0) x Š f( x0) 0as x 0

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91.LINEARIZATIONOFMULTIVARIABLEFUNCTIONS163 bythede“nitionofthederivative f( x0).Thisrelationimpliesthatthe errorofthelinearapproximationgoesto0fasterthanthedeviation x = x Š x0of x from x0,thatis, f ( x )= L ( x )+ ( x ) x, where ( x ) 0as x 0 Forexample,if f ( x )= x2,thenitslinearizationat x =1is L ( x )= 1+2( x Š 1).Itfollowsthat f (1+ x ) Š L (1+ x )= x2or ( x )= x .Intheinterval x (0 9 1 1),theabsoluteerrorofthelinear approximationislessthan0 01.Sothelinearizationof f at x0provides agoodapproximationof f inasucientlysmallneighborhoodof x0. Naturally,suchausefultoolneedstobeextendedtomultivariable functions.91.1.TangentPlaneApproximation.Consider“rstthecaseoftwovariablefunctions.Thegraphof f ( x,y )isthesurface z = f ( x,y ). Considerthecurveofintersectionofthissurfacewiththecoordinate plane x = x0.Itsequationis z = f ( x0,y ).Thevectorfunction r ( t )=( x0,t,f ( x0,t ))tracesoutthecurveofintersection.Thecurve goesthroughthepoint r0=( x0,y0,z0),where z0= f ( x0,y0),because r ( y0)= r0.Itstangentvectoratthepoint r0is v1= r( y0)= (0 1 ,f y( x0,y0)).Thelineparallelto v1throughthepoint r0liesin theplane x = x0andistangenttotheintersectioncurve z = f ( x0,y ). Similarly,thegraph z = f ( x,y )intersectsthecoordinateplane y = y0alongthecurve z = f ( x,y0)whoseparametricequationsare r ( t )= ( t,y0,f ( t,y0)).Thetangentvectortothiscurveatthepoint r0is v2= r( x0)=(1 0 ,f x( x0,y0)).Thelineparallelto v2through r0lies intheplane y = y0andistangenttothecurve z = f ( x,y0). Nowonecande“neaplanethroughthepoint r0ofthegraphthat containsthetwotangentlines.Thisplaneiscalledthe tangentplane to thegraph.Itsnormalmustbeperpendiculartobothvectors v1and v2and,bythegeometricalpropertiesofthecrossproduct,maybetaken as n = v1 v2=( f x( x0,y0) ,f y( x0,y0) Š 1).Thestandardequationof theplane n ( r Š r0)=0canthenbewrittenintheform z = z0+ n1( x Š x0)+ n2( y Š y0) ,n1= f x( x0,y0) ,n2= f y( x0,y0) Thegraphgoesthroughthepoint r0andsodoesthetangentplane.So thetangentplaneisexpectedtobeclosetothegraphinaneighborhood of r0.TheexactmeaningofcloseŽwillbeclari“edinthefollowing section. Example 13.22 Findthetangentplanetotheparaboloid z = x2+ 3 y2atthepoint (2 1 7) .

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16413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Solution: Theparaboloidisthegraphofthefunction f ( x,y )= x2+3 y2.Thecomponentsofthenormalare n1= f x(2 1)=2 x |(2 1)=4, n2= f y(2 1)=6 y |(2 1)=6,and n3= Š 1.Anequationofthetangent planeis4( x Š 2)+6( y Š 1) Š ( z Š 7)=0or4 x +6 y Š z =7. Byanalogywiththeone-variablecase,onecande“nethe linearization of f ( x,y )atapoint( x0,y0)inthedomainof f bythelinear function L ( x,y )= f ( x0,y0)+ f x( x0,y0)( x Š x0)+ f y( x0,y0)( y Š y0) Theapproximation f ( x,y ) L ( x,y )iscalleda linearapproximation of f near( x0,y0).Byanalogy,theconceptsoflinearizationandthelinear approximationareextendedtofunctionsofmorethantwovariables. Definition 13.16 (LinearizationofaMultivariableFunction) Let f beafunctionof m variables r =( x1,x2,...,xm) on D suchthat itsderivativesexistataninteriorpoint r0=( a1,a2,...,am) of D .Put ni= f xi( r0) i =1 2 ,...,m .Thefunction L ( r )= f ( r0)+ n1( x1Š a1)+ n2( x2Š a2)+ + nm( xmŠ am) iscalledthe linearization of f at r0,andtheapproximation f ( r ) L ( r ) iscalledthe linearapproximation of f near r0. Itisconvenienttowritethelinearizationinamorecompactform (13.6) L ( r )= f ( r0)+ n1 x1+ n2 x2+ + nm xm,ni= f xi( r0) where xiisthedeviationof xifrom ai. Example 13.23 Usethelinearapproximationtoestimatethe number [(2 03)2+(1 97)2+(0 94)2]1 / 2. Solution: Considerthefunctionofthreevariables f ( x,y,z )=[ x2+ y2+ z2]1 / 2.Thenumberinquestionisthevalueofthisfunctionat ( x,y,z )=(2 03 1 97 0 94).Thispointiscloseto r0=(2 2 1)at which f ( r0)=3.Thedeviationsare x = x Š 2=0 03, y = y Š 2= Š 0 03,and z =0 94 Š 1= Š 0 06.Thepartialderivatives are f x= x/ ( x2+ y2+ z2)1 / 2, f y= y/ ( x2+ y2+ z2)1 / 2,and f z= z/ ( x2+ y2+ z2)1 / 2.Therefore, n1=2 / 3, n2=2 / 3,and n3=1 / 3.The linearapproximationgives f ( x,y,z ) L ( x,y,z )=3+(2 / 3) x +(2 / 3) y +(1 / 3) z =2 98

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91.LINEARIZATIONOFMULTIVARIABLEFUNCTIONS165 91.2.DifferentiabilityofMultivariableFunctions.Theconceptsoflinearizationandthelinearapproximationhavebeenformallyextended fromtheone-variablecaseinwhichtheveryexistenceofthederivativeatapointissucientforthelinearapproximationtobegood, asarguedatthebeginningofthissection.Inthecaseoftwo-variable functions,theexistenceofpartialderivativesatapointallowsoneto de“nethetangentplanetothegraph.Basedonthisgeometricalobservationthattheplane z = L ( x,y )istangenttothegraph z = f ( x,y ),it hasbeenassumedthatthedierence f ( x,y ) Š L ( x,y )shoulddecrease withthedecreasingdistance( x2+ y2)1 / 2betweenthepoints( x,y ) and( x0,y0).Thekeydierencebetweenone-variableandmultivariable casesisthatthemereexistenceofpartialderivativesis notsucient tomakethelinearapproximationagoodone. Thiscanbeillustratedbythefollowingexample.Put (13.7) f ( x,y )= xy x2+ y2if( x,y ) =(0 0) 0if( x,y )=(0 0) Thisfunctionhasthepropertythat f ( x, 0)=0forall x ,whichimplies thatitsratealongthe x axisvanishes, f x( x, 0)=0.Similarly,the functionvanishesonthe y axis, f (0 ,y )=0,andhencehasnorate ofchangealongit, f y(0 ,y )=0.Inparticular,thepartialderivatives existattheorigin, f x(0 0)= fy(0 0)=0.Sothetangentplaneshould coincidewiththe xy plane, z =0,andthelinearizationisaconstant (zero)function, L ( x,y )=0.Canonesaythat f ( x,y ) Š L ( x,y )= f ( x,y ) 0as( x,y ) (0 0),thatis,thatthelinearapproximation isagoodone?Toanswerthisquestion,thetwo-variablelimitmust bestudied.Considerthislimitalongthestraightline( x,y )=( t,at ), t 0+.Onehas f ( t,at )= a/ (1+ a2) =0.Thus,thelimitdoesnot exist.Inparticular,thedierence f Š L remainsanonzeroconstantas theargumentapproachestheoriginalongastraightline.Furthermore, thefunctionisnotevencontinuousat (0 0) despitethatthepartial derivativesexistat (0 0)!Thisisquiteadeparturefromtheonevariablecasewheretheexistenceofthederivativeimpliesthatthe functionisnecessarilycontinuous.Whathastobechangedinthe multivariablecaseinordertoachieveasimilaritywiththeone-variable case?Thequestionisansweredwiththeconceptof dierentiability of afunctionofseveralvariables. Definition 13.17 (DierentiableFunctions) Thefunction f ofseveralvariables r =( x1,x2,...,xm) onanopenset D issaidtobedierentiableatapoint r0 D ifthereexistsalinear

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16613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS function L ( r ) suchthat (13.8)limr r0f ( r ) Š L ( r ) r Š r0 =0 If f isdierentiableatallpointsof D ,then f issaidtobedierentiable on D Thisde“nitiondemandsthat f ( r ) Š L ( r )= ( r ) r Š r0 and ( r ) 0as r r0;thatis,theerrorofthelinearapproximationdecreases fasterthanthedistance r Š r0 between r and r0justlikeintheonevariablecase.Thefunction(13.7)isnotdierentiableat(0 0)despite theexistenceofitspartialderivativesbecauseitdoesnothaveagood linearapproximationat(0 0)inthesenseof(13.8).Ingeneral,one canprovethefollowingresult. Theorem 13.12 (ExistenceofPartialDerivatives) If f isdierentiableatapoint r0,thenitspartialderivativesexist at r0. Theconverseisnottrue!Thisisthereasonthelinearapproximationmaybebaddespitetheexistenceofpartialderivatives.With someadditionalassumptionsaboutthepartialderivatives,theconversestatementcanbeestablished.Itprovidesausefulcriterionfor dierentiability. Theorem 13.13 (DierentiabilityandPartialDerivatives) Let f beafunctiononanopenset D ofaEuclideanspace.Then f isdierentiableon D ifandonlyifitspartialderivativesexistandare continuousfunctionson D Thus,if,inadditiontotheirexistence,thepartialderivativeshappentobecontinuousfunctions,thenthelinearapproximationisalways agoodoneinthesenseof(13.8).Conversely,thepartialderivatives ofadierentiablefunctionarecontinuousfunctions.Itisstraightforwardtoverifythat f x( x,y )and f y( x,y )forthefunction(13.7)arenot continuousat(0 0).91.3.StudyProblems.Problem13.6. Showthattheequation z (3 x Š y )= sin( xyz ) has auniquesolution z = z ( x,y ) inaneighborhoodof (1 1) suchthat z (1 1)= / 2 .Estimate z (1 04 0 96) Solution: InExample13.21,theexistenceanduniquenessof z ( x,y ) hasbeenestablishedbytheimplicitfunctiontheorem.Thepartial

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92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS167 derivativeshavealsobeenevaluated, z x(1 1)= Š 3 / 4and z y(1 1)= / 4.Thelinearizationof z ( x,y )near(1 1)is z (1+ x, 1+ y ) z (1 1)+ z x(1 1) x + z y(1 1) y = 2 1 Š 3 x 2 + y 2 Putting x =0 04and y = Š 0 04,thisequationyieldstheestimate z (1 04 0 96) 0 45 .Notethatthecombinationoftheimplicit dierentiationandlinearizationallowsonetoapproximatetherootof anonlinearequationinasmallneighborhoodofthepointwherethe valueoftherootisknown.Thisisanextremelyusefulconceptinmany practicalapplications. 92.TheDifferentialandTaylorPolynomials Justlikeintheone-variablecase,givenvariables r =( x1,x2,...,xm), onecanintroduceindependentvariables d r =( dx1,dx2,...,dxm)that arein“nitesimalvariationsof r andalsocalled dierentials of r .The wordin“nitesimalŽmeansherethatpowers( dxi)kcanalwaysbeneglectedfor k> 1. Definition 13.18 (Dierential) Let f ( r ) beadierentiablefunction.Thefunction df ( r )= f x1( r ) dx1+ f x2( r ) dx2+ + f xm( r ) dxmiscalledthe dierential of f Notethatthedierentialisafunctionof2 m independent variables r and d r .Thegeometricalsigni“canceofthedierentialfollowsfrom itsrelationwiththelinearizationof f atapoint r0: L ( r )= f ( r0)+ df ( r0) ,dxi= xi,i =1 2 ,...,m ; thatis,ifthein“nitesimalvariations(ordierentials) d r aresettobe thedeviations r = r Š r0ofthevariables r from r0,then thedifferential df atthepoint r0de“nesthelinearizationof f at r0.This linearizationisagoodoneinthesense(13.8).Fromanalgebraicpoint ofview,thedierentialdeterminesvariationsofvaluesof f underin“nitesimalindependentvariationsofitsargumentssuchthatcontributions ofpowersofthevariations( dxi)k, k> 1,canbeneglected.92.1.ErrorAnalysis.Thevolumeofarectanglewithdimensions x y and z isthefunctionofthreevariables V ( x,y,z )= xyz .Inpractice, measurementsofthedimensionsalwayscontainerrors;thatis,repetitivemeasurementsgivethevaluesof x y ,and z fromtheintervals

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16813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS x [ x0Š x,x0+ x ], y [ y0Š y,y0+ y ],and z [ z0Š z,z0+ z ], where r0=( x0,y0,z0)arethemeanvaluesofthedimensions,while r =( x, y, z )aretheabsoluteerrorsofthemeasurements.Differentmethodsofthelengthmeasurementwouldhavedierentabsoluteerrors.Inotherwords,thedimensions x y ,and z andthe errors x y ,and z areallindependentvariables.Sincetheerrors shouldbesmall(atleast,onewishesso),thevaluesofthedimensions obtainedineachmeasurementare x = x0+ dx y = y0+ dy ,and z = z0+ dz ,wherethedierentialsorin“nitesimalvariationscantake theirvaluesintheintervals dx [ Š x, x ]= I xandsimilarlyfor dy and dz .Thequestionarises:Giventhemeanvalues r0=( x0,y0,z0) andtheabsoluteerrors r ,whatistheabsoluteerrorofthevolume valuecalculatedat r0?Foreachparticularmeasurement,theerroris V ( r0+ d r ) Š V ( r0)= dV ( r0)ashigherpowersofthedierentials d r areirrelevant(small).Thecomponentsof d r areindependentvariables takingtheirvaluesinthespeci“edintervals.Allsuchtriples d r correspondtopointsoftheerrorrectangle R= I x I y I z.Then themaximalorabsoluteerroris V = | max dV ( r0) | ,wherethemaximumistakenoverall d r R.Forexample,if r0=(1 2 3)isin centimetersand r =(1 1 1)isinmillimeters,thentheabsoluteerrorofthevolumeis V = | max dV ( r0) | =max( y0z0dx + x0z0dy + x0y0dz )=0 6+0 3+0 2=1 1cm3,and V =6 1 1cm3.Herethe maximumisreachedat dx = dy = dz =0 1cm.Thisconceptcanbe generalized. Definition 13.19 (AbsoluteandRelativeErrors) Let f beaquantitythatdependsonotherquantities r =( x1,x2,...,xm) ; thatis, f = f ( r ) isafunctionof r .Supposethatthevalues xi= aiare knownwiththeabsoluteerrors xi.Put r0=( a1,a2,...,am) and f = | max df ( r0) | wherethemaximumistakenoverall dxi [ Š xi, xi] Thenumbers f and f/ | f ( r0) | arecalled,respectively,the absolute andrelativeerrors ofthevalueof f at r = r0. Intheaboveexample,therelativeerrorofthevolumemeasurements is1 1 / 6 0 18;thatis,theaccuracyofthemeasurementsisabout18%. Ingeneral,since df ( r0)=mi =1f xi( r0) dxiislinearin dxi,themaximumisattainedat dxi= xiifthecoecient f xi( r0)ispositive,andat dxi= Š xiifthecoecient f xi( r0)is

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92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS169 negative.Sotheabsoluteerrorcanbewrittenintheform f =mi =1| f xi( r0) | xi.92.2.AccuracyoftheLinearApproximation.Intheone-variablecase, theTaylortheoremassertsthatif f ( x )isan n timescontinuously dierentiablefunction,thenthereisaconstant Mn> 0suchthat f ( x )= Tn( x )+ n +1( x ) Tn( x )= f ( x0)+ f( x0) 1! + f( x0) 2! 2+ + f( n )( x0) n n, | n +1( x ) | Mn +1 ( n +1)! | x Š x0|n +1, (13.9) where= x Š x0.Thepolynomial Tn( x )iscalledthe Taylorpolynomial ofdegree n .Theremainder n +1( x )determinestheaccuracyofthe approximation f ( x ) Tn( x ).If,inaddition,thederivative f( n +1)exists,theremainderisprovedtohavetheform n +1( x )= f( n +1)( ) ( n +1)! ( x Š x0)n +1. Inthiscase, Mn +1isanupperboundof | f( n +1)( ) | overtheinterval between x and x0. The“rst-orderTaylorpolynomialisthelinearizationof f at x = x0, T1( x )= L ( x ),andtheremainder 2( x )determinestheaccuracyofthe linearapproximation.Supposethat f( x )isboundedonaninterval ( a,b ),thatis, | f( x ) | M2forall x ( a,b ).Then,for x0 ( a,b ),the accuracyofthelinearapproximationisgivenby (13.10) | 2( x ) | = | f ( x ) Š L ( x ) | M2 2 ( b Š a )2,x ( a,b ) .92.3.TwoVariableTaylorPolynomials.ThereisamultivariableextensionoftheTaylortheoremthatcanbeusedtoassesstheaccuracyof thelinearapproximationaswellastoobtaintheTaylorpolynomial approximationwhenthelinearoneisnotsucientlyaccurate.The two-variablecaseisdiscussed“rst.Itisassumedinwhatfollowsthat functionshavecontinuouspartialderivativesofneededorders.The followingnotationisadopted.Let f ( x,y )beafunctionoftwovariables.Thesymbols xand ydenotetheoperationoftakingpartial derivativeswithrespectto x and y ,thatis, ( ax+ by)2f =( a22 x+2 abxy+ b22 y) f = a2f xx+2 abf xy+ b2f yy

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17013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS foranynumbers a and b .Byanearlierassumption,allthederivativesarecontinuous,andthereforeClairautstheoremapplies, xyf = yxf ;thatis,theorderofdierentiationisirrelevant. Let r0=( x0,y0)bea“xedpointandletafunction f ( r ), r = ( x,y ),havecontinuouspartialderivativesuptoorder n inanopen disk D containing r0.Let =(1, 2)beanorderedpairofnumbers. Considerapolynomialinthevariables ofdegree n de“nedby Pn( )= f ( r0)+ 1 1! (1x+2y) f ( r0) + 1 2! (1x+2y)2f ( r0)+ + 1 n (1x+2y)nf ( r0) =nk =01 k (1x+2y)kf ( r0) wherethepartialderivativesarecomputedatthepoint r0.Byconstruction,thesepolynomialssatisfytherecurrencerelation: (13.11) Pn( )= Pn Š 1( )+ 1 n (1x+2y)nf ( r0) Herethelasttermiscomputedbymeansofbinomialcoecients( a + b )n= n k =0Bk nan Š kbk,where Bn k= n / ( k !( n Š k )!). Definition 13.20 (TaylorPolynomials) Thepolynomial Tn( r )= Pn( ) ,where = r Š r0=( x Š x0,y Š y0) iscalledthe Taylorpolynomial forafunction f nearthepoint r0.Theapproximation f ( r ) Tn( r ) iscalledthe Taylorpolynomial approximationofdegree n near r0. Forexample,with1= x Š x0and2= y Š y0,the“rstfour Taylorpolynomialsare T0( r )= f ( r0) T1( r )= f ( r0)+ f x( r0)1+ f y( r0)2= L ( r ) T2( r )= T1( r )+ f xx( r0) 2 2 1+ f xy( r0)12+ f yy( r0) 2 2 2, T3( r )= T2( r )+ f xxx( r0) 6 3 1+ f xxy( r0) 2 2 12+ f xyy( r0) 2 12 2+ f yyy( r0) 6 3 2.

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92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS171 Thelinearortangentplaneapproximation f ( r ) L ( r )= T1( r )is aparticularcaseoftheTaylorpolynomialapproximationofthe“rst degree. ThefollowingtheoremassessestheaccuracyoftheTaylorpolynomialapproximation. Theorem 13.14 (AccuracyoftheTaylorPolynomialApproximation) Let D beanopendiskcenteredat r0andletthepartialderivativesofa function f becontinuousuptoorder n on D .Supposethatthepartial derivativesoforder n arealsoboundedon D ;thatis,therearenumbers Mnk, k =1 2 ,...,n ,suchthat | n Š k xk yf ( r ) | Mnkforall r D .Then f ( r )= Tn Š 1( r )+ n( r ) wheretheremainder nsatis“es | n( r ) |nk =0Bn kMnk n | x Š x0|n Š k| y Š y0|kforall ( x,y ) D ,where Bn k= n / ( k !( n Š k )!) arebinomialcoecients. Remark. Infact,thecontinuityofthe n th-orderpartialderivatives isnotnecessary;theirexistenceandboundednessaresucientforthe accuracyassessment.Thediscussionofthisremarkaswellastheproof ofthetheoremgoesbeyondthescopeofthecourse. Tomaketheanalogyofthistheoremwiththeone-variablecase, notethat | x Š x0| r Š r0 and | y Š y0| r Š r0 andhence | x Š x0|n Š k| y Š y0|k r Š r0n.Makinguseofthisinequality,one infersthat (13.12) | n( r ) | Mn n r Š r0n, wheretheconstant Mn= n k =0Bn kMnk.Inparticular,forthelinear approximation n =2, | f ( r ) Š L ( r ) | M20 2 ( x Š x0)2+ M11| ( x Š x0)( y Š y0) | + M02 2 ( y Š y0)2 M2 2 r Š r02 M2 2 R2, (13.13) where M2= M20+2 M11+ M02and R istheradiusofthedisk D Theresults(13.12)and(13.13)aretobecomparedwiththesimilar results(13.9)and(13.10)intheone-variablecase.So,ifthesecond derivativesexistandarebounded,theerrorofthelinearortangent planeapproximationdecreasesasthesquareddistance r Š r02. Example 13.24 Usethelinearapproximationorthedierential toestimatetheamountofaluminuminaclosedaluminumcanwith

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17213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS diameter10cmandheight10cmifthealuminumis0.05cmthick. Assesstheaccuracyoftheestimate. Solution: Thevolumeofacylinderofradius r andheight h is f ( h,r )= hr2.Thevolumeofaclosedcylindricalshell(orthecan) ofthickness istherefore V = f ( h +2 ,r + ) Š f ( h,r ),where h and r aretheinternalheightandradiusoftheshell.Sincethevariations h =2 =0 1and r = =0 05aresmall,thisdierence canbeestimatedbylinearizingthefunction f at( h,r )=(10 5). Onehas f h= r2and f r=2 hr ;hence, V Va= df (10 5)= f h(10 5) dh + f r(10 5) dr =25 h +100 r =7 5 cm3,where dh = h and dr = r Toassesstheaccuracy,notethattheapproximation V Vais basedonthelinearapproximationof f ( h,r ),nearthepoint(10 5). Put f ( h,r )= L ( h,r )+ 2( h,r ),where L ( h,r )isthelinearizationof f at(10 5)and 2( h,r )istheremainder.Then,for h =10+2 and r =5+ ,onehas V = f ( h,r ) Š f (10 5)= L ( h,r ) Š f (10 5)+ 2( h,r )= Va+ 2( h,r ).Toestimatetheremainder,onehasto“ndthe upperboundsonthesecond-orderderivativesof f when( h,r )liesin thediskofradius R =2 centeredat(10 5)(see(13.13)).Onehas f hh=0= M20, f hr=2 r 10 2 = M11,and f rr=2 h 20 2 = M02.Therefore, M2=40 4 ,andtheabsoluteerroroftheestimate is | V Š Va| = | 2| ( M2/ 2) R2=80 8 2=0 202 cm3.Therelative errorreads | V Š Va| /Va 0 026;thatis,itisabout2 6%. 92.4.MultivariableTaylorPolynomials.Formorethantwovariables, Taylorpolynomialsarede“nedsimilarly.Let r =( x1,x2,...,xm)andlet =(1, 2,..., m).Considerpolynomialsde“nedbytherecurrence relation(13.11),where1x+2yisreplacedby11+22+ + mm= Dand imeanstheoperationoftakingthederivativewith respectto xi,thatis, if = f/xi.TheTaylorpolynomial Tn( r )of degree n nearthepoint r0=( a1,a2,...,am)isthenobtainedbysetting i= xiŠ aiinthepolynomial Pn( );thatis,inthecompactform therulereads Pn( )=nk =11 k Dk f ( r0) ,Tn( r )= Pn( r Š r0) Theaccuracyoftheapproximation f ( r ) Tn Š 1( r )canbeassessedby puttingupperboundsonthe n th-orderderivativesof f inaballcenteredat r0.Duetoexcessivetechnicalities,thedetailsareomitted.It

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92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS173 isworthnotingthatinequality(13.12)holdsforanynumberofvariables;thatisthedierence | f ( r ) Š Tn Š 1( r ) | tendsto0noslowerthan r Š r0nas r r0. Forpracticalpurposes,ifthedierence | Tn( r ) Š Tn Š 1( r ) | issmallas comparedto | Tn Š 1( r ) | inaball r Š r0 R ,thentheapproximation f ( r ) Tn Š 1( r )isaccurate;thatis,theapproximationbyahigher-order polynomial Tninsteadof Tn Š 1isnotgoingtosigni“cantlyimprovethe accuracy.92.5.StudyProblems.Problem13.7. Calculationofhigher-orderderivativesto“ndTaylor polynomialsmightbeatechnicallytediousproblem.Insomespecial cases,however,itcanbeavoided.Suppose f isacompositionoftwo functions: f ( r )= g ( u ) ,where u = u ( r ) .Supposethat u ( r0)=0 ,that is, f ( r0)= g (0) .Let Tn( r )= Pn( ) beaTaylorpolynomialforthe function u at r0.Ithasthepropertythatithasnoconstanttermas u ( r0)=0 .Therefore,its k thpower, ( Pn( ))k,isapolynomialofdegree nk whoseconstantterm,linearterms,quadraticterms,andsoonup tothe ( k Š 1) th-degreetermsvanish.ConsidertheTaylorpolynomial forthefunction g : Tg n( u )= g (0)+ g(0) u + g(0) 2! u2+ + g( n )(0) n un. If u isapproximatedbyitsTaylorpolynomial Pn( ) inthisexpression, thentheresultingexpressionwillbeapolynomialin whosedegreeis n2andwhosetermsuptodegree n coincidewiththeTaylorpolynomial Tf nfor f at r0.Indeed,anypolynomialin isuniquelydeterminedby itscoecients,including Tf n.Ontheotherhand,anapproximationof g byahigher-degreepolynomial Tg n +1willaddtheterm un +1,whichcan onlycontaintermsin ofdegree n +1 orhigher,andhencewillnot changethetermsofdegreelessthanorequalto n .Usethisobservation to“nd T3forthefunction f ( x,y,z )=sin( xy + z ) attheorigin. Solution: Put u = xy + z .Thethird-degreeTaylorpolynomialfor g ( u )=sin u at u =0is Tg 3( u )= u Š1 6u3.As u isapolynomialof degree2,itsTaylorpolynomialofdegree3coincideswithit, T3( r )= u ( r ).Hence, Tf 3isobtainedfrom Tg 3byomittingalltermsofdegree higherthan3: Tg 3=( xy + z ) Š 1 6 ( xy + z )3= z + xy Š 1 6 ( z3+3( xy ) z2+3( xy )2z +( xy )3) Therefore, Tf 3( r )= z + xy Š1 6z3.Evidently,theprocedureisfarsimpler thancalculating19partialderivatives(uptothethirdorder)!

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17413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 93.DirectionalDerivativeandtheGradient93.1.DirectionalDerivative.Let f beafunctionofseveralvariables r =( x1,x2,...,xm).Thepartialderivative f xi( r0)istherateofchange inthedirectionofthe i thcoordinateaxis.Thisdirectionisde“nedby theunitvector eiparalleltothecorrespondingcoordinateaxis.Let u beaunitvectorthatdoesnotcoincidewithanyofthevectors ei.What istherateofchangeof f at r0inthedirectionof u ?Forexample,if f ( x,y )istheheightofamountain,wherethe x and y axesareoriented alongthewest…eastandsouth…northdirections,respectively,thenitis reasonabletoaskabouttheslopes,forexample,inthesouth…eastor north…westdirections.Naturally,theseslopesgenerallydierfromthe slopes f xand f y. Toanswerthequestionabouttheslopeinthedirectionofaunit vector u ,considerastraightlinethrough r0parallelto u .Itsvector equationis r ( h )= r0+ h u ,where h isaparameterthatlabelspoints oftheline.Thevaluesof f alongthelinearegivenbythecomposition g ( h )= f ( r ( h )).Thenumbers g (0)and g ( h )arethevaluesof f ata givenpoint r0andthepoint r ( h ), h> 0,thatisatthedistance h from r0inthedirectionof u .Sotheslopeisgivenbythederivative g(0). Therefore,thefollowingde“nitionisnatural. Definition 13.21 (DirectionalDerivative) Let f beafunctiononanopenset D .Thedirectionalderivativeof f at r0 D inthedirectionofaunitvector u isthelimit Duf ( r0)=limh 0f ( r0+ h u ) Š f ( r0) h ifthelimitexists. Thenumber Duf ( r0)istherateofchangeof f at r0inthedirection of u .Byde“nition, Duf ( r0)= df ( r ( h )) /dh takenat h =0,where r ( h )= r0+ h u .So,bythechainrule,thedirectionalderivativeexists ifthepartialderivativesof f at r0exist: df ( r ( h )) dh = f x1( r ( h )) x 1( h )+ f x2( r ( h )) x 2( h )+ + f xm( r ( h )) x m( h ) Setting h =0inthisrelationandtakingintoaccountthat r( h )= u or x i( h )= ui,where u =( u1,u2,...,um),oneinfersthat (13.14) Duf ( r0)= f x1( r0) u1+ f x2( r0) u2+ + f xm( r0) um.

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93.DIRECTIONALDERIVATIVEANDTHEGRADIENT175 Equation(13.14)providesaconvenientwaytocomputethedirectional derivative.Recallalsothatifthedirectionisspeci“edbyanonunitvector u ,thenthecorrespondingunitvectorcanbeobtainedbydividing itbyitslength u ,thatis, u = u / u Example 13.25 Theheightofahillis f ( x,y )=(9 Š 3 x2Š y2)1 / 2, wherethe x and y axesaredirectedfromwesttoeastandfromsouth tonorth,respectively.Ahikerisatthepoint r0=(1 2) .Supposethe hikerisfacinginthenorth-westdirection.Whatistheslopethehiker sees? Solution: Aunitvectorintheplanecanalwaysbewrittenintheform u =(cos sin ),wheretheangle iscountedcounterclockwisefrom thepositive x axis;thatis, =0correspondstotheeastdirection, = / 2tothenorthdirection, = tothewestdirection,andsoon. Soforthenorth…westdirection =3 / 2and u =( Š 1 / 2 1 / 2)= ( u1,u2).Thepartialderivativesare f x= Š 3 x/ (9 Š 3 x2Š y2)1 / 2and f y= Š y/ (9 Š 3 x2Š y2)1 / 2.Theirvaluesat r0=(1 2)are f x(1 2)= Š 3 / 2and f y(1 2)= Š 2 / 2.By(13.14),theslopeis Duf ( r0)= f x( r0) u1+ f y( r0) u2=3 / 2 Š 1 / 2=1 Ifthehikergoesnorth…west,hehastoclimbatanangleof45relative tothehorizon. Example 13.26 Findthedirectionalderivativeof f ( x,y,z )= x2+3 xz + z2y atthepoint (1 1 Š 1) inthedirectiontowardthepoint (3 Š 1 0) .Doesthefunctionincreaseordecreaseinthisdirection? Solution: Put r0=(1 1 Š 1)and r1=(3 Š 1 0).Thenthevector u = r1Š r0=(2 Š 2 1)pointsfromthepoint r0towardthepoint r1accordingtotherulesofvectoralgebra.Butitisnotaunitvector becauseitslengthis u =3.Sotheunitvectorinthesamedirection is u = u / 3=(2 / 3 Š 2 / 3 1 / 3)=( u1,u2,u3).Thepartialderivatives are f x=2 x +3 z f y= z2,and f z=3 x +2 zy .Theirvaluesat r0read f x( r0)= Š 1, f y( r0)=1,and f z( r0)=1.By(13.14),thedirectional derivativeis Duf ( r0)= f x( r0) u1+ f y( r0) u2+ f z( r0) u3= Š 2 / 3 Š 2 / 3+1 / 3= Š 1 Sincethedirectionalderivativeisnegative,thefunctiondecreasesat r0inthedirectiontoward r1(therateofchangeisnegativeinthat direction).

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17613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 93.2.TheGradientandItsGeometricalSigni“cance.Definition 13.22 (TheGradient) Let f beafunctionofseveralvariables r =( x1,x2,...,xm) onanopen set D andlet r0 D .Thevectorwhosecomponentsarepartialderivativesof f at r0, f ( r0)=( f x1( r0) ,f x2( r0) ,...,f xm( r0)) isthegradientof f atthepoint r0. So,fortwo-variablefunctions f ( x,y ),thegradientis f =( f x,f y); forthree-variablefunctions f ( x,y,z ),thegradientis f =( f x,f y,f z); andsoon.Comparing(13.14)withthede“nitionofthegradientand recallingthede“nitionofthedotproduct,thedirectionalderivative cannowbewritteninthecompactform (13.15) Duf ( r0)= f ( r0) u Thisequationisthemostsuitableforanalyzingthesigni“canceofthe gradient. Consider“rstthecasesoftwo-andthree-variablefunctions.The gradientiseitheravectorinaplaneorspace.InExample13.25, thegradientat(1,2)is f (1 2)=( Š 3 / 2 Š 2 / 2).InExample 13.26,thegradientat(1,1, Š 1)is f (1 1 Š 1)=( Š 1 1 1).Recallthe geometricalpropertyofthedotproduct a b = a b cos ,where [0 ]istheanglebetweenthevectors a and b .Thevalue =0 correspondstoparallelvectors a and b .When = / 2,thevectors areorthogonal.Thevectorspointintheoppositedirectionsif = Let betheanglebetweenthegradient f ( r0)andtheunitvector u Then (13.16) Duf ( r0)= f ( r0) u = f ( r0) u cos = f ( r0) cos because u =1(theunitvector).Asthecomponentsofthegradient are“xednumbers(thevaluesofthepartialderivativesataparticular point r0),thedirectionalderivativeat r0variesonlyifthevector u changes.Thus,theratesofchangeof f inalldirectionsthathave thesameangle withthegradientarethesame.Inthetwo-variable case,onlytwosuchdirectionsarepossibleif u isnotparalleltothe gradient,whileinthethree-variablecasetheraysfrom r0inallsuch directionsformaconewhoseaxisisalongthegradient.Itisthen concludedthattherateofchangeof f attainsitsabsolutemaximum orminimumwhencos does.Therefore, themaximalrateisattained inthedirectionofthegradient( =0 )andisequaltothemagnitudeof thegradient f ( r0) ,whereastheminimalrateofchange Š f ( r0)

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93.DIRECTIONALDERIVATIVEANDTHEGRADIENT177 occursinthedirectionof Š f ( r0) ,thatis,oppositetothegradient ( = ). Thegraphofafunctionoftwovariables z = f ( x,y )maybeviewed astheshapeofahill.Thenthegradientataparticularpointshows thedirectionofthe steepestascent ,whileitsoppositepointsinthe directionofthe steepestdescent .InExample13.25,themaximalslope atthepoint(1,2)is f ( r0) =(1 / 2) ( Š 3 2) = 13 / 2.Itoccurs inthedirectionof( Š 3 / 2 2 / 2)or( Š 3 2)(themultiplicationofa vectorbyapositiveconstantdoesnotchangeitsdirection).If isthe anglebetweenthepositive x axis(orthevector e1)andthegradient, thentan = Š 2 / 3or 146.Ifthehikergoesinthisdirection,he hastoclimbupatanangleoftanŠ 1( 13 / 2) 69withthehorizon. Also,notethehikersoriginaldirectionwas =135,whichmakes theangle11withthedirectionofthesteepestascent.Sotheslopein thedirection =146+11=157hasthesameslopeasthehikers originalone.Ashasbeenargued,inthetwo-variablecase,therecan onlybetwodirectionswiththesameslope. Next,consideralevelcurve f ( x,y )= k ofafunctionoftwovariables.Thereisavectorfunction r ( t )=( x ( t ) ,y ( t ))thattracesout thelevelcurve.Thisvectorfunctionisde“nedbytheconditionthat f ( x ( t ) ,y ( t ))= k forallvaluesoftheparameter t .Bythede“nitionof levelcurves,thefunction f hasaconstantvalue k alongitslevelcurve. Therefore,bythechainrule, d dt f ( x ( t ) ,y ( t ))=0= df dt = f xx( t )+ f yy( t )= f ( r ( t )) r( t )=0 foranyvalueof t .Foranyparticularvalue t = t0,thepoint r0= r ( t0) liesonthelevelcurve,whilethederivative r( t0)isatangentvectorto thecurveatthepoint r0.Thus, thegradient f ( r0) isorthogonaltoa tangentvectoratthepoint r0tothelevelcurveof f throughthatpoint. Onecanalsosaythegradientof f isalwaysnormaltothelevelcurves of f Recallthatafunction f ( x,y )canbedescribedbyacontourmap, whichisacollectionoflevelcurves.Iflevelcurvesaresmoothenoughto havetangentvectorseverywhere,thenonecande“neacurvethrougha particularpointthatisnormaltoalllevelcurvesinsomeneighborhood ofthatpoint.Thiscurveiscalledthe curveofsteepestdescentor ascent .Thetangentvectorofthiscurveatanypointisparalleltothe gradientatthatpoint.Thevaluesofthefunctionincrease(ordecrease) mostrapidlyalongthiscurve.Ifahikerfollowsthedirectionofthe

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17813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS gradientoftheheight,hewouldgoalongthepathofsteepestascent ordescent. Consideralevelsurface f ( x,y,z )= k .Let r ( t )=( x ( t ) ,y ( t ) ,z ( t )) beasmoothcurveonthelevelsurface,thatis, f ( r ( t ))= k forallvalues of t .Sincethevaluesof f donotchangealongthecurve, df/dt =0. Makinguseofthechainrule,itisconcludedthat df/dt = f xx+ f yy+ f zz= f ( r ( t )) r( t )=0 Therearemanycurvesthroughapoint r0= r ( t0)thatlieinthelevel surface.Thegradient f ( r0)isorthogonaltoatangentvectorto any suchcurve,thatis,to any linethatistangenttothelevelsurfaceat r0. Thus, thegradient f ( r0) isnormaltothetangentplanetothelevel surfacethroughthepoint r0. Allthe“ndingsaresummarizedinthefollowingtheorem,which hasbeenprovedabove. Theorem 13.15 (GeometricalPropertiesoftheGradient) Let f bedierentiableonanopenset D andlet r0 D .Let S bethe levelsurface(orcurve)throughthepoint r0.Then (1) Themaximalrateofchangeof f at r0occursinthedirection ofthegradient f ( r0) andisequaltoitsmagnitude f ( r0) (2) Theminimalrateofchangeof f at r0occursinthedirection oppositetothegradient Š f ( r0) andequals Š f ( r0) (3) Thegradient f ( r0) isnormalto S at r0. Example 13.27 Findanequationofthetangentplanetotheellipsoid x2+2 y2+3 z2=11 atthepoint (2 1 1) Solution: Theequationoftheellipsoidcanbeviewedasthelevel surface f ( x,y,z )=11ofthefunction f ( x,y,z )= x2+2 y2+3 z2through thepoint r0=(2 1 1)because f (2 1 1)=11.Bythegeometrical propertyofthegradient,thevector n = f ( r0)isnormaltotheplane inquestion.Since f =(2 x, 4 y, 6 z ),onehas n =(4 4 6).Anequation oftheplanethroughthepoint(2,1,1)andnormalto n is4( x Š 2)+ 4( y Š 1)+6( z Š 1)=0or2 x +2 y +3 z =9. Theorem13.15holdsforfunctionsofmorethanthreevariables aswell.Equation(13.15)wasobtainedforanynumberofvariables, andtherepresentationofthedotproduct(13.16)holdsinanyEuclideanspace.Thus,the“rsttwopropertiesofthegradientarevalid inanymultivariablecase.Thethirdpropertyishardertovisualizeasthelevelsurfaceofafunctionof m variablesisan( m Š 1)dimensionalsurfaceembeddedinan m -dimensionalEuclideanspace. Suchsurfacesarecalled hypersurfaces toemphasizethefactthatthey

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93.DIRECTIONALDERIVATIVEANDTHEGRADIENT179 arenottwo-dimensionalsurfacesembeddedinathree-dimensionalEuclideanspace.However,if r ( t )isacurveinthelevelhypersurface f ( r )= k ,thenthemultivariablefunction f hasaconstantvaluealong anysuchcurve,and,bythechainrule,itimmediatelyfollowsthat df ( r ( t )) /dt = f ( r ( t )) r( t )=0forany t .Hence, thegradientis normaltothelevelhypersurfaceinthesensethatitisnormaltoany curveinit Remark. Itisinterestingtonotecompactformulasforthelinearization L ( r )of f ( r )at r0andthedierential df : L ( r )= f ( r0)+ f ( r0) ( r Š r0) ,df = f d r thatarevalidforanynumberofvariables.93.3.StudyProblems.Problem13.8. Supposethatthreelevelsurfaces f ( x,y,z )=1 g ( x, y,z )=2 ,and h ( x,y,z )=3 ofdierentiablefunctionsareintersecting alongasmoothcurve C .Let P beapointon C .Find f ( g h ) at P Solution: Let v beatangentvectorto C atthepoint P (itexists becausethecurveissmooth).Since C liesinthesurface f ( x,y,z )=1, thegradient f ( P )isorthogonalto v .Similarly,thegradients g ( P ) and h ( P )mustbeorthogonalto v .Therefore,allthegradientsmust beinaplaneperpendiculartothevector v .Thetripleproductfor anythreecoplanarvectorsvanishes,andhence f ( g h )=0 at P Problem13.9. ConsiderNewtonssecondlaw m a = F .Suppose thattheforceisthegradient F = ŠU ,where U = U ( r ) .Let r = r ( t ) bethetrajectorysatisfyingNewtonslaw.Provethatthequantity E = mv2/ 2+ U ( r ) ,where v = r( t ) isthespeed,isac onstantof motion,thatis, dE/dt =0 .Thisconstantiscalledthetotalenergyofa particle. Solution: First,notethat v2= v v .Hence,( v2)=2 v v= 2 v a .Usingthechainrule, dU/dt = U xx( t )+ U yy( t )+ U zz( t )= r U = v U .Itfollowsfromthesetworelationsthat dE dt = m 2 ( v2)+ dU dt = m v a + v U = v ( m a Š F )=0 Sothetotalenergyisconservedforthetrajectoryofthemotion.

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18013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 94.MaximumandMinimumValues94.1.CriticalPointsofMultivariableFunctions.Thepositionsofthelocalmaximaandminimaofaone-variablefunctionplayanimportant rolewhenanalyzingitsoverallbehavior.InCalculusI,itwasshown howthederivativescanbeusedto“ndlocalmaximaandminima.Here thisanalysisisextendedtomultivariablefunctions. Thefollowingnotationwillbeused.Anopenballofradius centeredatapoint r0isdenoted B= { r | r Š r0 < } ;thatis,itisaset ofpointswhosedistancefrom r0islessthan > 0.Aneighborhood Nofapoint r0inaset D isasetofcommonpointsof D and B;that is, N= D Bcontainsallpointsin D whosedistancefrom r0isless than Definition 13.23 (AbsoluteandLocalMaximaorMinima) Afunction f onaset D issaidtohavealocalmaximumat r0 D ifthereisaneighborhood Nof r0suchthat f ( r0) f ( r ) forall r N.Thenumber f ( r0) iscalleda localmaximumvalue .Ifthere isaneighborhood Nof r0suchthat f ( r0) f ( r ) forall r N, then f issaidtohavealocalminimumat r0andthenumber f ( r0) iscalleda localminimumvalue .Iftheinequality f ( r0) f ( r ) or f ( r0) f ( r ) holdsforallpoints r inthedomainof f ,then f hasan absolutemaximumorabsoluteminimumat r0,respectively. ThereisanextensionofFermatstheoremtomultivariablefunctionsthatishelpfulin“ndingthepossiblepositionsofthelocalmaxima andminima,providedthefunctionhas“rst-orderpartial derivatives. Theorem 13.16 If f hasalocalmaximumorminimumatan interiorpoint r0ofitsdomain D andthe“rst-orderpartialderivatives existat r0,thentheyvanishat r0, f xi( r0)=0 i =1 2 ,...,m Proof. Consideraline r ( t )= r0+ t u through r0andparalleltoa unitvector u .Thefunction F ( t )= f ( r ( t ))de“nesthevaluesof f alongtheline.Therefore, F ( t )musthavealocalmaximumorminimumat t =0.Thederivative Fexistsat t =0because,bythe de“nitionofthedirectionalderivative, F(0)= Duf ( r0)= f ( r0) u andthegradient f ( r0)existsbythehypothesis.ByFermatstheorem, F(0)=0.Sotherateofchangeof f vanishesinanydirectionand, inparticular,alongthecoordinateaxes,thatis, u = eiand Duf ( r0)= f xi( r0)=0. Theconverseofthistheoremisnottrue.Thisisillustratedby theexampleofthefunction f ( x,y )= xy Š 2 y Š x .Itisdierentiable

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94.MAXIMUMANDMINIMUMVALUES181 everywhere.Inparticular,thesystemofequations f x= y Š 1=and f y= x Š 2=0hasthesolution( x,y )=(2 1).However,thefunction hasneitheralocalmaximumnoralocalminimum.Indeed,considera straightlinethrough(2 1), x =2+ u1t y =1+ u2t ,thatisparalleltoa unitvector u =( u1,u2).Thenthevaluesof f alongthelineare F ( t )= f ( x ( t ) ,y ( t ))= Š 1+ at2,where a = u1u2.So F ( t )hasaminimumat t =0if a> 0oramaximumif a< 0.However,thecoecient a may beeitherpositiveornegative,dependingonthechoiceoftheline(or thecomponentsof u ).Thus,thevalue F (0)= f (2 1)= Š 1cannotbe alocalmaximumoralocalminimumvalue.Thegraphof f lookslike asaddleintheneighborhoodof(2 1). Definition 13.24 (SaddlePoint) Ifthenumber f ( r0) isthemaximalvalue f alongsomelinesthrough r0inasmallopenball(disk)centeredat r0,whereas f ( r0) istheminimal valueof f alongalltheothersuchlines,then r0iscalleda saddlepoint of f Remark. Consideralllinesthroughapoint r0inthedomainofa function f .Supposethat f attainsalocalmaximumalongeveryline through r0;thatis,thefunction g ( t )= f ( r0+ u t )hasalocalmaximum at t =0foranychoiceofthevector u .Onemighttendtoconclude thatinthiscasethefunction f shouldhavealocalmaximumat r0. Thisconclusionis wrong !Anexampleisgivenattheendofthissection (seeStudyProblems13.9).Theremarkalsoappliestothecaseofa localminimum. Alocalmaximumorminimummayoccuratapointwheresomeof thepartialderivativesdonotexist.Forexample, f ( x,y )= | x | + | y | is de“nedeverywhereandhasanabsoluteminimumat(0 0).However, thepartialderivatives f x(0 0)and f y(0 0)donotexist. Finally,alocalminimumormaximummayoccuratapointofthe domainthatisnotaninteriorpoint,andhencethepartialderivatives arenotde“nedatthatpoint.Forexample,thedomainofthefunction f ( x,y )= 1 Š x2Š y2isthedisk x2+ y2 1.Itsboundarypoints x2+ y2=1arenotinteriorpoints.Butthisfunctionattainsitsabsolute minimumonthecircle x2+ y2=1. Definition 13.25 (CriticalPoints) Aninteriorpoint r0ofthedomainofafunction f issaidtobeacritical pointof f ifeither f ( r0)= 0 orthegradientdoesnotexistat r0.

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18213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Thus, if f hasalocalmaximumorminimumat r0,then r0isa criticalpointof f .However,notallcriticalpointscorrespondtoeither alocalmaximumoralocalminimum .94.2.Second-DerivativeTest.Supposethatafunction f ( x,y )hascontinuoussecondderivativesinanopenballcenteredat r0.Thesecondderivatives a = f xx( r0), b = f yy( r0),and c = f xy( r0)= f yx( r0) (Clairautstheorem)canbearrangedintoa2 2symmetricmatrix whosediagonalelementsare a and b andwhoseo-diagonalelements c Thequadraticpolynomialofavariable P2( )=det a Š c cb Š =( a Š )( b Š ) Š c2, iscalledthe characteristicpolynomial ofthematrixofsecondpartial derivativesof f at r0. Theorem 13.17 (Second-DerivativeTest) Let r0beacriticalpointofafunction f .Supposethatthesecondorderpartialderivativesof f arecontinuousinanopenball(disk) Bcenteredat r0.Let P2( ) bethecharacteristicpolynomialofthematrix ofsecondderivativesat r0.Let i, i =1 2 ,betherootsof P2( ) .Then € Iftherootsarestrictlypositive, i> 0 ,then f hasalocal minimumat r0. € Iftherootsarestrictlynegative, i< 0 ,then f hasalocal maximumat r0. € Iftherootsdonotvanishbuthavedierentsigns,then r0isa saddlepointof f € Ifatleastoneoftherootsvanishes,then f mayhavealocal maximum,alocalminimum,asaddle,ornoneoftheabove (thesecond-derivativetestisinconclusive). Proof. Considertheseconddirectionalderivativeof f atanypoint r inaneighborhood Nof r0: D2 uf = Du( f xu1+ f yu2)= f xxu2 1+2 f xyu1u2+ f yyu2 2, whichisaquadraticfunctionincomponentsofthevector u .Itdeterminestheconcavityofthecurveobtainedbytheintersectionofthe graphof f withaplaneparalleltoboththe z axisand u andgoing throughthepoint r =( x,y ).Hence,if D2 uf> 0forall r in N,then thegraphisconcavedownwardand f musthavealocalminimum at r0.Similarly, f hasalocalmaximumat r0if D2 uf< 0forall r in N.

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94.MAXIMUMANDMINIMUMVALUES183 Foranyunitvector, u1=cos and u2=sin .Makinguseofthe double-angleformulascos2 =(1+cos(2 ) / 2,sin2 =(1 Š cos(2 ) / 2, and2sin cos =sin(2 ),theseconddirectionalderivativecanbe writtenintheform 2 D2 uf =( f xx+ f yy)+( f xxŠ f yy)cos(2 )+2 f xysin(2 ) Put A2=( f xxŠ f yy)2+4( f xy)2andde“neanangle bycos = ( f xxŠ f yy) /A andsin = Š 2 f xy/A .Sincecos(2 + )=cos cos(2 ) Š sin sin(2 ),oneinfersthat 2 D2 uf =( f xx+ f yy)+ A cos(2 + )=( f xx+ f yy)+ A cos where =2 + takesvaluesin[0 2 ]forall u .De“ne 1and 2by therelations 1+ 2= f xx+ f yy,12= f xxf yyŠ ( f xy)2. Notethatif r = r0,then 1and 2arerootsofthecharacteristic polynomial P2( )=0astheysatisfytheconditions 1+ 2= a + b and 12= ab Š c2.Bythecontinuityofthesecondderivatives, 1and 2arecontinuousaswell.Hence,iftherootsof P2( )donotvanish,then 1and 2donotvanishinsomeneighborhood N.Thenitfollowsthat ( 1Š 2)2=( 1+ 2)2Š 4 12= A2or A = | 1Š 2| ,andthesecond derivativebecomes D2 uf = 1+ 2 2 + | 1Š 2| 2 cos = 1cos2( / 2)+ 2sin2( / 2) ,1 2, 1sin2( / 2)+ 2cos2( / 2) ,1<2, where(1+cos )=2cos2 and(1 Š cos )=2sin2 havebeenused. Supposethat 1 2> 0atthecriticalpoint r0.Then,bythecontinuity ofthesecondderivatives, 1 2> 0insomeneighborhood Nof r0. Thus,inthiscase, D2 uf> 0in Nand f hasalocalminimumat r0. Similarly,iftherootsarestrictlynegative,then D2 uf< 0in Nand f hasalocalmaximumat r0.If 1 2 =0buthavedierentsigns,then D2 uf changesitssignin N.Sincethesignof D2 udoesnotchangewhen u Š u ,onanystraightlinethrough r0andparallelto u f hasa “xedconcavityalongeachline,whichmeansthat f hasasaddlepoint at r0. Theinconclusivenessofthesecond-derivativetestwhenatleastone oftherootsvanishesiseasilyestablishedbyspeci“cexamples. Considerthefunction f ( x,y )= x2+ sy4,where s isanumber. Ithasacriticalpoint(0 0)because f x(0 0)= f y(0 0)=0and a = f xx(0 0)=2, b = f yy(0 0)=0,and c = f xy(0 0)=0.Therefore,

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18413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS P2( )= Š (2 Š ) hastheroots 1=2and 2=0.If s> 0,then f ( x,y ) 0forall( x,y )and f hasaminimumat(0 0).If s< 0, thefunction f hasaminimumalongtheline x = t y =0( F ( t )= t2), whileithasamaximumalongtheline x =0, y = t ( F ( t )= st4, s< 0); thatis,(0 0)isasaddlepoint.Thefunction f ( x,y )= Š ( x2+ sy4) hasamaximumat(0 0)if s> 0,andif s< 0,thecriticalpoint (0 0)isasaddlepoint.So,ifoneoftherootsvanishes,then f may havealocalmaximumoralocalminimum,orasaddle.Thesame conclusionisreachedwhen 1= 2=0bystudyingthefunctions f ( x,y )= ( x4+ sy4)alongthesimilarlinesofarguments. Furthermore,considerthefunction f ( x,y )= xy2.Italsohasa criticalpointattheorigin,andallitssecondderivativesvanishat(0 0), thatis, P2( )= 2and 1= 2=0.Thevaluesof f alonganyline throughtheorigin x = u1t y = u2t are F ( t )= st3,where s = u1u2 2. Forany s =0, F ( t )hasan in”ection pointat t =0.Therefore, f cannothaveamaximumorminimumorsaddleat(0 0)becausein anyofthesesituations f shouldhaveeitheraminimumoramaximum alonganystraightline.Acriticalpointmaybeageneralin”ection pointifonetheofrootsdoesnotvanish.Anexampleisprovidedby f ( x,y )= x2+ y3,whichhasacriticalpoint(0 0)and a =2, b =0,and c =0,thatis 1=2and 2=0.Therearelinesthroughtheorigin alongwhich f haseitheraminimumoranin”ectionat t =0.For example,alongtheline x =0, y = t ,thefunction f hasanin”ection point( F ( t )= t3)at t =0,whereas,alongtheline x = t y =0,ithas aminimum( F ( t )= t2).Theveryexistenceoflinesalongwhich f has anin”ectionprecludesusfromconcludingthat f canhaveamaximum oraminimumorasaddleat(0 0). Thisconcludestheproofofthesecond-derivativetestinthecaseof two-variablefunctions. Example 13.28 Supposethat 1=0 and 2< 0 for f ( x,y ) atits criticalpoint.Findthedirectionsatwhich D2 uf vanishesatthecritical point. Solution: Put u =(cos sin ).Then D2 uf = 2sin2( + / 2)=0 or = Š / 2and = Š / 2+ ,where = Š sinŠ 1(2 c/A ), A = ( a Š b )2+4 c2= | 1Š 2| = | 2| (seetheproofofthesecondderivativetest).Therefore,thedirectionsare u = (cos( / 2) Š sin( / 2)). Example 13.29 Findallcriticalpointsofthefunction f ( x,y )=1 3x3+ xy2Š x2Š y2anddeterminewhether f hasalocalmaximum, minimum,orsaddleatthem.

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94.MAXIMUMANDMINIMUMVALUES185 Solution: Criticalpoints .Thefunctionisapolynomial,andtherefore ithaspartialderivativeseverywhere.Soitscriticalpointsaresolutions ofthesystemofequations f x= x2+ y2Š 2 x =0 f y=2 xy Š 2 y =0 Itisimportantnottolosesolutionswhentransformingthesystemof equations f ( r )= 0 forthecriticalpoints .Itfollowsfromthesecondequationthat y =0or x =2.Therefore,theoriginalsystemof equationsis equivalent totwosystemsofequations: f x= x2+ y2Š 2 x =0 x =1 or f x= x2+ y2Š 2 x =0 y =0 Solutionsofthe“rstsystemare(1 1)and(1 Š 1).Solutionsofthe secondsystemare(0 0)and(2 0).Thus,thefunctionhasfourcritical points. Itisadvisabletocheckifallpointsfounddosatisfytheoriginal system because,whentransformingasystemofnonlinearequations, onemightgetpointsthatdonotsatisfytheoriginalsystemorone mightsimplymakeanerror Second-derivativetest .Thesecondderivativesare f xx=2 x Š 2 ,f yy=2 x Š 2 ,f xy=2 y. Forthepoints(1 1), a = b =0and c = 2.Thecharacteristic polynomialis P2( )= 2Š 4.Itsroots = 2donotvanishand haveoppositesigns.Therefore,thefunctionhasasaddleatthepoints (1 1).Forthepoint(0 0), a = b = Š 2and c =0.Thecharacteristic polynomialis P2( )=( Š 2 Š )2.Ithasonerootofmultiplicity2,that is, 1= 2= Š 2 < 0,and f hasalocalmaximumat(0 0).Finally,for thepoint(2 0), a = b =2and c =0.Thecharacteristicpolynomial P2( )=(2 Š )2hasonerootofmultiplicity2, 1= 2=2 > 0;that is,thefunctionhasalocalminimumat(2 0). 94.3.StudyProblems.Problem13.10. De“ne f (0 0)=0 and f ( x,y )= x2+ y2Š 2 x2y Š 4 x6y2 ( x4+ y2)2if ( x,y ) =(0 0) .Showthat,forall ( x,y ) ,thefollowinginequality holds: 4 x4y2 ( x4+ y2)2.Useitandthesqueezeprincipletoconclude that f iscontinuous.Next,consideralinethrough (0 0) andparallel to u =(cos sin ) andthevaluesof f onit: F( t )= f ( t cos ,t sin ) .

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18613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Showthat F(0)=0 F (0)=0 ,and F (0)=2 forall 0 2 Thus, f hasaminimumat (0 0) alonganystraightlinethrough (0 0) Showthatnevertheless f hasnominimumat (0 0) bystudyingitsvalue alongtheparaboliccurve ( x,y )=( t,t2) Solution: Onehas0 ( a Š b )2= a2Š 2 ab + b2andhence2 ab a2+ b2foranynumbers a and b .Therefore,4 ab =2 ab +2 ab 2 ab + a2+ b2= ( a + b )2.Bysetting a = x4and b = y2,thesaidinequalityisestablished. Thecontinuityofthelasttermin f at(0 0)hastobeveri“ed.Bythe foundinequality, 4 x6y2 ( x4+ y2)2 4 x6y2 4 x4y2= x2 0as( x,y ) (0 0) Thus, f ( x,y ) f (0 0)=0as( x,y ) (0 0),and f iscontinuouseverywhere.If = / 2,thatis,thelinecoincideswiththe x axis,( x,y )=( t, 0),onehas F( t )= t2,fromwhichitfollowsthat F(0)= F (0)=0and F (0)=2.When = / 2sothatsin =0, onehas F( t )= t2+ at3+ bt4 (1+ ct2)2, a = Š 2cos2 sin ,b = Š 4cos6 sin2 ,c = cos4 sin2 Astraightforwarddierentiationshowsthat F(0)= F (0)=0and F (0)=2asstated,and F( t )hasanabsoluteminimumat t =0,or f attainsanabsoluteminimumat(0 0)alonganystraightlinethrough (0 0).Nevertheless,thelatter doesnotimplythat f hasaminimum at (0 0)!Indeed,alongtheparabola( x,y )=( t,t2),thefunction f behavesas f ( t,t2)= Š t4, whichattainsan absolutemaximum at t =0.Thus,alongtheparabola, f hasamaximumvalueattheoriginandhencecannothavealocal minimumthere.Theproblemillustratestheremarkgivenearlierin thissection. 95.MaximumandMinimumValues(Continued)95.1.Second-DerivativeTestforMultivariableFunctions.Intheproofof thesecond-derivativetest,ithasbeenestablishedthat (13.17) Df u= f xxu2 1+2 f xyu1u2+ f yyu2 2= 1v2 1+ 2v2 2,

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95.MAXIMUMANDMINIMUMVALUES(CONTINUED)187 where u =( u1,u2)=(cos sin )and v =( v1,v2)=(cos sin )are unitvectorssuchthat = / 2= + / 2.Thisfacthasaremarkable geometricalinterpretation.Notethat D2 uf isaquadraticfunctionin componentsof u .Whendiscussingtheshapesofquadricsurfaces,in particular,quadriccylinders,ithasbeenshownthat,byasuitable rotationofthecoordinateaxes,themixedŽterm2 f xyu1u2canbe eliminated.Whenthecoordinatesystemisrotated,theangle is simplyshiftedbytherotationangle.Sothevector v isobtainedfrom thevector u byrotatingthelatterthroughtheangle / 2,whichis determinedbythesecond-orderpartialderivativesof f .Indoingso, the quadraticfunction D2 uf isbroughtintothestandardforminwhichthe coecientsaredeterminedbytherootsofthecharacteristicpolynomial Thisresultholdsforanynumberofvariables. First,notethat Du( Duf )= u1Duf x1+ u2Duf x2+ + umDuf xm. Put Dij= f xixj= Dji(byClairautstheorem).Then D2 uf =mi =1 mj =1Dijuiuj, whichisaquadraticfunction.Thenumbers Dijcanbearrangedinto asquare m m matrix.Thepolynomialofdegree m Pm( )=det D11Š D12D13 D1 mD21D22Š D23 D2 mD31D32D33Š D3 m. Dm 1Dm 2Dm 3 DmmŠ iscalledthecharacteristicpolynomialofthematrixofsecondderivatives. Foranysymmetricrealmatrix ( Dij= Dji) ,therootsofitscharacteristicpolynomialareprovedtobereal .Itcanfurtherbeproved thatthereexistsarotationofthecoordinatesystemunderwhich u goesinto v suchthat (13.18) D2 uf =mi =1 mj =1Dijuiuj= 1v2 1+ 2v2 2+ + mv2 m, where iarerootsof Pm.Thisequationshowsthatthesecond-derivative testformulatedintheprecedingsectionholdsforanynumberofvariables. Theorem 13.18 (Second-DerivativeTestfor m Variables) Let r0beacriticalpointof f andsupposethat f hascontinuoussecondorderpartialderivatives Dijinsomeopenballcenteredat r0.Thenthe

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18813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS characteristicpolynomial Pm( ) ofthematrix Dij( r0) hasrealroots i, i =1 2 ,...,m ,and € Ifalltherootsarestrictlypositive, i> 0 ,then f hasalocal minimum. € Ifalltherootsarestrictlynegative, i< 0 ,then f hasalocal maximum. € Ifalltherootsdonotvanishbuthavedierentsigns,then f hasan m -dimensionalsaddlepoint., € Ifsomeoftherootsvanish,then f mayhavealocalmaximum, alocalminimum,asaddle,ornoneoftheabove(thetestis inconclusive). Inthetwo-variablecase,theproofusesspecialpropertiesofthe rootsofquadraticpolynomials.Inthegeneralcase,thegoalisachieved bymeansof linearalgebra methods. Remark. Ifatleastoneoftherootsofthecharacteristicpolynomialsvanishes,thesecond-derivativetestisinconclusive.Howcan thelocalbehaviorofafunctionbeanalyzednearitscriticalpoint? Ifthefunctioninquestionisdierentiablesucientlymanytimes nearacriticalpoint,thentheTaylorpolynomialapproximationat thecriticalpointprovidesausefultechniqueforansweringthisquestionbecauseitiseasiertostudyapolynomialratherthanageneral function. Example 13.30 Investigatethelocalbehaviorofthefunctionde“nedby f (0 ,y )= f ( x, 0)=1 and f ( x,y )=sin( xy ) / ( xy ) if x =0 and y =0 Solution: Since u = xy issmallneartheorigin,sin u canbeapproximatedbyitsTaylorpolynomial T3( u )= u Š u3/ 6.Hence,the correspondingTaylorpolynomial Tf n( x,y )forthefunction f at(0 0) reads: Tf n( x,y )= T3( u ) u =1 Š u2 6 =1 Š x2y2 6 Sothefunctionattainsalocalmaximumat(0 0)because x2y2 0 forall( x,y ).Itisworthnotingthatthe“rstnonconstantpolynomial hasdegree n =4andtherefore Tf 2=1,whichmeansthatallthe “rstandsecondderivativesof f vanishat(0 0);thatis,thecharacteristicpolynomialforthematrixofsecondderivativesis P2( )= 2hasroot =0ofmultiplicity2.Thesecond-derivativetestwouldbe inconclusive.

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95.MAXIMUMANDMINIMUMVALUES(CONTINUED)189 95.2.AbsoluteMaximalandMinimalValues.Forafunction f ofone variable,theextremevaluetheoremsaysthatif f iscontinuousona closedinterval[ a,b ],then f hasanabsoluteminimumvalueandanabsolutemaximumvalue.Forexample,thefunction f ( x )= x2on[ Š 1 2] attainsanabsoluteminimumvalueat x =0andanabsolutemaximum valueat x =2.Notethat f isde“nedforall x ,andthereforeitscritical pointsaredeterminedby f( x )=0.Sotheabsoluteminimumvalue occursatthecriticalpoint x =0insidetheinterval,whiletheabsolute maximumvalueoccursontheboundaryoftheintervalthatisnota criticalpointof f .Thus,to“ndtheabsolutemaximumandminimum valuesofafunction f inaclosedintervalinthedomainof f ,thevalues of f mustbeevaluatednotonlyatthecriticalpointsbutalsoatthe boundariesoftheinterval. Thesituationformultivariablefunctionsissimilar. Definition 13.26 (ClosedSet) Aset D inaEuclideanspaceissaidtobeclosedifitcontainsallits limitpoints. Recallthatanyneighborhoodofalimitpointof D containspoints of D .Ifalimitpointof D isnotaninteriorpointof D ,thenitliesona boundaryof D .Soaclosedsetcontainsitsboundaries.Allpointsofan openinterval( a,b )areitslimitpoints,but,inaddition,theboundaries a and b arealsoitslimitpoints,sowhentheyareadded,aclosedset [ a,b ]isobtained.Similarly,thesetintheplane D { ( x,y ) | x2+ y2< 1 } haslimitpointsonthecircle x2+ y2=1(theboundaryof D ), whichisnotin D .Byaddingthesepoints,aclosedsetisobtained, x2+ y2 1. Definition 13.27 (BoundedSet) Aset D inaEuclideanspaceissaidtobeboundedifitiscontainedin someball. Inotherwords,foranytwopointsinaboundedset,thedistance betweenthemcannotexceedsomevalue(thediameteroftheballthat containstheset). Theorem 13.19 (ExtremeValueTheorem) If f iscontinuousonaclosed,boundedset D inaEuclideanspace,then f attainsanabsolutemaximumvalue f ( r1) andanabsoluteminimum value f ( r2) atsomepoints r1 D and r2 D Bythistheorem,itfollowsthatthepoints r1and r2areeithercriticalpointsof f (becausealocalmaximumorminimumalwaysoccurs

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19013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS atacriticalpoint)orlieontheboundaryof D .So,to“ndtheabsolute minimumandmaximumvaluesofacontinuousfunction f onaclosed, boundedset D ,onehasto (1)Findthevaluesof f atthecriticalpointsof f in D (2)Findtheextremevaluesof f ontheboundaryof D (3)ThelargestofthevaluesobtainedinSteps1and2isthe absolutemaximumvalue,andthesmallestofthesevaluesis theabsoluteminimumvalue. Example 13.31 Findtheabsolutemaximumandminimumvalues of f ( x,y )= x2+ y2+ xy onthedisk x2+ y2 4 andthepointsat whichtheyoccur. Solution: Thefunction f isapolynomial.Itiscontinuousanddifferentiableonthewholeplane. Step1 .Criticalpointsof f satisfythesystemofequations f x=2 x + y = 0and f y=2 y + x =0;thatis,(0 0)istheonlycriticalpointof f andithappenstobeinthedisk.Thevalueof f atthecriticalpoint is f (0 0)=0. Step2 .Theboundaryofthediskisthecircle x2+ y2=4.To “ndtheextremevaluesof f onit,taketheparametricequationsof thecircle x ( t )=2cos t y ( t )=2sin t ,where t [0 2 ].Onehas F ( t )= f ( x ( t ) ,y ( t ))=4+4cos t sin t =4+2sin(2 t ).Thefunction F ( t )attainsitsmaximalvalue6on[0 2 ]whensin(2 t )=1or t = / 4 and t = / 4+ .Thesevaluesof t correspondtothepoints( 2 2) and( Š 2 Š 2).Similarly, F ( t )attainsitsminimalvalue2on[0 2 ] whensin(2 t )= Š 1or t =3 / 4and t =3 / 4+ .Thesevaluesof t correspondtothepoints( Š 2 2)and( 2 Š 2). Step3 .Thelargestnumberof0,2,and6is6.Sotheabsoluteminimumvalueof f is6;itoccursatthepoints( 2 2)and( Š 2 Š 2). Thesmallestnumberof0,2,and6is0.Sotheabsoluteminimum valueof f is0;itoccursatthepoint(0 0). Example 13.32 Findtheabsolutemaximumandminimumvalues of f ( x,y,z )= x2+ y2Š z2+2 z ontheclosedset D = { ( x,y,z ) | x2+ y2 z 4 } Solution: Theset D isthesolidboundedfrombelowbytheparaboloid z = x2+ y2andfromthetopbytheplane z =4.Itisabounded set,and f iscontinuousonitasanypolynomialfunction.

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96.LAGRANGEMULTIPLIERS191 Step1 .Since f isdierentiableeverywhere,itscriticalpointssatisfy theequations f x=2 x =0, f y=2 y =0,and f z= Š 2 z +2=0.There isonlyonecriticalpoint(0 0 1),andithappenstobein D .Thevalue of f atitis f (0 0 1)=1. Step2 .Theboundaryconsistsoftwosurfaces,thedisk S1= { ( x,y,z ) | z =4 ,x2+ y2 4 } intheplane z =4andtheportionofthe paraboloid S2= { ( x,y,z ) | z = x2+ y2,x2+ y2 4 } .Thevaluesof f on S1are F1( x,y )= f ( x,y, 4),wherethepoints( x,y )lie inthediskofradius2, x2+ y2 4.Theproblemnowisto“nd themaximalandminimalvaluesofatwo-variablefunction F1on thedisk.Inprinciple,atthispoint,Steps1,2,and3havetobe appliedto F1.Thesetechnicalitiescanbeavoidedinthisparticularcasebynotingthat F1( x,y )= x2+ y2Š 8= r2Š 8,where r2= x2+ y2 4.Therefore,themaximalvalueof F1isreached when r2=4,anditsminimalvalueisreachedwhen r2=0.Sothe maximalandminimalvaluesof f on S1are Š 4and Š 8.Thevalues of f on S2are F2( x,y )= f ( x,y,x2+ y2)=3 r2Š r4= g ( r ),where r2= x2+ y2 4or r [0 2].Thecriticalpointsof g ( r )satisfythe equation g( r )=6 r Š 4 r3=0whosesolutionsare r =0, r = 3 / 2. Therefore,themaximalvalueof f on S2is9 / 4,whichisthelargestof g (0)=0, g ( 3 / 2)=9 / 4,and g (2)= Š 4,andtheminimalvalueis Š 4 asthesmallestofthesenumbers. Step3 .Theabsolutemaximumvalueof f on D ismax { 1 Š 8 Š 4 9 / 4 } = 9 / 4,andtheabsoluteminimumvalueof f on D ismin { 1 Š 8 Š 4 9 / 4 } = Š 8.Bothvaluesoccurontheboundaryof D : f (0 0 4)= Š 8andthe absolutemaximalvalueisattainedalongthecircleofintersectionof theplane z =3 / 2withtheparaboloid z = x2+ y2. 96.LagrangeMultipliers Let f ( x,y )betheheightofahillasafunctionofposition.Ahiker walksalongapath r ( t )=( x ( t ) ,y ( t )).Whatarethelocalmaximaand minimaalongthepath?Whatarethemaximumandminimumheights alongthepath?Thesequestionsareeasytoansweriftheparametric equationsofthepathareexplicitlyknown.Indeed,theheightalongthe pathisthesingle-variablefunction F ( t )= f ( r ( t ))andtheproblemis reducedtothestandardextremevalueproblemfor F ( t )onaninterval t [ a,b ]. Example 13.33 Theheightasafunctionofpositionis f ( x,y )= xy .Findthelocalmaximaandminimaoftheheightalongthecircular path x2+ y2=4 .

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19213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Solution: Theparametricequationofthecirclecanbetakeninthe form r ( t )=(2cos t, 2sin t ),where t [0 2 ].Theheightalongthe pathis F ( t )=4cos t sin t =2sin(2 t ).Ontheinterval[0 2 ],the functionsin(2 t )attainsitsabsolutemaximumvalueat t = / 4and t = / 4+ anditsabsoluteminimumvalueat t =3 / 4and t =3 / 4+ So,alongthepath,thefunction f attainstheabsolutemaximumvalue 2at( 2 2)and( Š 2 Š 2)andtheabsoluteminimumvalue Š 2 at( Š 2 2)and( 2 Š 2). However,inmanysimilarquestions,anexplicitformof r ( t )isnot knownornoteasyto“nd.Analgebraiccondition g ( x,y )=0isamore generalwaytodescribeacurve.Itsimplysaysthatonlythepoints ( x,y )thatsatisfythisconditionarepermittedintheargumentof f ; thatis,thevariables x and y arenolongerindependent.Thecondition g ( x,y )=0iscalleda constraint Problemsofthistypeoccurforfunctionsofmorethantwovariables.Forexample,let f ( x,y,z )bethetemperatureasafunctionof position.Areasonablequestiontoaskis:Whatarethemaximum andminimumtemperaturesonasurface?Asurfacemaybedescribed byimposingoneconstraint g ( x,y,z )=0onthevariables x y ,and z .Nothingprecludesusfromaskingaboutthemaximumandminimumtemperaturesalongacurvede“nedasanintersectionoftwo surfaces g1( x,y,z )=0and g2( x,y,z )=0.Sothevariables x y and z arenowsubjecttotwoconstraints.Ingeneral,whatarethe extremevaluesofamultivariablefunction f ( r )whoseargumentsare subjecttoseveralconstraints ga( r )=0, a =1 2 ,...,M ?Naturally,the numberofindependentconstraintsshouldnotexceedthenumberof variables. Definition 13.28 (LocalMaximaandMinimaSubjectto Constraints) .Afunction f ( r ) hasalocalmaximum(orminimum)at r0onthesetde“nedbytheconstraints ga( r )=0 if f ( r ) f ( r0) (or f ( r ) f ( r0) )forall r insomeneighborhoodof r0thatsatisfythe constraints,thatis, ga( r )=0 Notethatafunction f maynothavelocalmaximaorminimainits domain.However,whenitsargumentsbecomesubjecttoconstraints, itmaywellhavelocalmaximaandminimaonthesetde“nedbythe constraints.Intheexampleconsidered, f ( x,y )= xy hasnolocal maximaorminima,but,whenitisrestrictedonthecirclebyimposing theconstraint g ( x,y )= x2+ y2Š 4=0,ithappenstohavetwolocal minimaandmaxima.

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96.LAGRANGEMULTIPLIERS193 96.1.CriticalPointsofaFunctionSubjecttoaConstraint.Theextreme valueproblemwithconstraintsamountsto“ndingthecriticalpoints ofafunctionwhoseargumentsaresubjecttoconstraints.Theexample discussedaboveshowsthattheequation f = 0 nolongerdetermines thecriticalpointsfordierentiablefunctionsifitsargumentsareconstrained. Consider“rstthecaseofasingleconstraintfortwovariables r = ( x,y ).Supposethefunction f andthefunction g thatde“netheconstraintaredierentiable.Let r0beapointatwhich f ( r )hasalocal maximumorminimumontheset S de“nedbytheconstraint g ( r )=0, whichisacurveinthetwo-variablecase.Let r ( t )beparametricequationsofthiscurveinaneighborhoodof r0,thatis,forsome t = t0, r ( t0)= r0.Assumingthat r ( t )isdierentiable,itisconcludedthat F( t0)=0,where F ( t )= f ( r ( t ))arevaluesof f alongthecurve.The chainruleyields F( t0)= f x( r0) x( t0)+ f y( r0) y( t0)= f ( r0) r( t0)=0= f ( r0) r( t0) Thegradient f ( r0)isorthogonaltoatangentvectortothecurve at thepointwhere f hasalocalmaximumorminimumonthecurve Ontheotherhand,thegradient g ( r )at any pointisnormaltothe levelcurve g ( r )=0,thatis, g ( r ( t )) r( t )forany t .Therefore,the gradients f ( r0)and g ( r0)mustbeparallelat r0.Thisgeometrical statementcanbetranslatedintoanalgebraicone:thereshouldexist anumber suchthat f ( r0)= g ( r0).Thisprovesthefollowing theorem. Theorem 13.20 (CriticalPointsSubjecttoaConstraint) Supposethat f and g aredierentiableat r0and f hasalocalmaximum orminimumat r0inthesetde“nedbytheconstraint g ( r )=0 .Then thereexistsanumber suchthat f ( r0)= g ( r0) Thetheoremholdsforthree-variablefunctionsaswell.Indeed,if r ( t )isacurvethrough r0inthelevelsurface g ( x,y,z )=0.Thenthe derivative F( t )=( d/dt ) f ( r ( t ))= f xx+ f yy+ f zz= f rmust vanishat t0,thatis, F( t0)= f ( r0) r( t0)=0.Therefore, f ( r0)is orthogonaltoatangentvectorof any curveinthesurface S at r0,and hence f ( r0)isnormaltothetangentplaneto S through r0.Onthe otherhand,thegradient g isnormaltothetangentplanetoalevel surfaceof g atanypoint(seethepropertiesofthegradient).Therefore,

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19413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS atthepoint r0,thegradientsof f and g mustbeparallel.Asimilar lineofreasoningprovesthetheoremforanynumberofvariables. Thistheoremprovidesapowerfulmethodto“ndthecriticalpoints of f subjecttoaconstraint g =0if f and g aredierentiable.Itis calledthe methodofLagrangemultipliers .To“ndthecriticalpoints of f ,thefollowingsystemofequationsmustbesolved: (13.19) f ( r )= g ( r ) ,g ( r )=0 If r =( x,y ),thisisasystemofthreeequations, f x= g x, f y= g y, and g =0forthreevariables( x,y, ).Foreachsolution( x0,y0,0), thecorrespondingcriticalpointof f is( x0,y0).Thenumericalvalue of isnotrelevant;onlyitsexistencemustbeestablishedbysolving thesystem.Inthethree-variablecase,thesystemcontainsfourequationsforfourvariables( x,y,z, ).Foreachsolution( x0,y0,z0,0),the correspondingcriticalpointof f is( x0,y0,z0). Example 13.34 UsethemethodofLagrangemultiplierstosolve theprobleminExample13.33. Solution: Put g ( x,y )= x2+ y2Š 4.Then f x= g xf y= g yg =0 = y =2 x x =2 y x2+ y2=4 Thesubstitutionofthe“rstequationintothesecondonegives x = 4 2x .Thismeansthateither x =0or = 1 / 2.If x =0,then y =0 bythe“rstequation,whichcontradictstheconstraint.For =1 / 2, x = y andtheconstraintgives2 x2=4or x = 2.Thecriticalpoints correspondingto =1 / 2are( 2 2)and( Š 2 Š 2).If = Š 1 / 2, x = Š y andtheconstraintgives2 x2=4or x = 2.Thecritical pointscorrespondingto = Š 1 / 2are( 2 Š 2)and( Š 2 2).So f ( 2 2)=2isthemaximalvalueand f ( 2 2)= Š 2is theminimalone. Example 13.35 Arectangularboxwithoutalidistobemadefrom cardboard.Findthedimensionsoftheboxofagivenvolume V such thatthecostofmaterialisminimal. Solution: Letthedimensionsbe x y ,and z ,where z istheheight. Theamountofcardboardneededisdeterminedbythesurfacearea f ( x,y,z )= xy +2 xz +2 yz .Thequestionisto“ndtheminimalvalue of f subjecttoconstraint g ( x,y,z )= xyz Š V =0.TheLagrange

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96.LAGRANGEMULTIPLIERS195 multipliermethodgives f x= g xf y= g yf z= g zg =0 = y +2 z = yz x +2 z = xz 2 x +2 y = xy xyz = V = xy +2 xz = V xy +2 zy = V 2 xz +2 yz = V xyz = V wherethelastsystemhasbeenobtainedbymultiplyingthe“rstequationby x ,thesecondoneby y ,andthethirdoneby z withthesubsequentuseoftheconstraint.Combiningthe“rsttwoequations,one infers2 z ( y Š x )=0.Since z =0( V =0),onehas y = x .Combining the“rstandthirdequations,oneinfers y ( x Š 2 z )=0andhence x =2 z Thesubstitution y = x =2 z intotheconstraintyields4 z3= V .Hence, theoptimaldimensionsare x = y =(2 V )1 / 3and z =(2 V )1 / 3/ 2.The amountofcardboardminimizingthecostis3(2 V )2 / 3(thevalueof f at thecriticalpoint).Fromthegeometryoftheproblem,itisclearthat f attainsitsminimumvalueattheonlycriticalpoint. 96.2.TheCaseofTwoorMoreConstraints.Let f beafunctionofthree variablessubjecttotwoconstraints g1( r )=0and g2( r )=0.Each constraintde“nesasurfaceinthedomainof f (levelsurfacesof g1and g2).Sothesetde“nedbytheconstraintsisthecurveofintersectionof thelevelsurfaces g1=0and g2=0.Let r0beapointofthecurveat which f hasalocalmaximumorminimum.Let v beatangentvector tothecurveat r0.Sincethecurveliesinthelevelsurface g1=0,by theearlierarguments, f ( r0) v and g1( r0) v .Ontheotherhand, thecurvealsoliesinthelevelsurface g2=0andhence g2( r0) v .It followsthatthegradients f g1,and g2become coplanar atthe point r0astheylieintheplanenormalto v .Therefore,thereexist numbers 1and 2suchthat f ( r )= 1 g1( r )+ 2 g2( r ) ,g1( r )= g2( r )=0 when r = r0(seeStudyProblem11.6).Thisisasystemof“veequationsfor“vevariables( x,y,z,1,2).Foranysolution( x0,y0,z0,10, 20),thepoint( x0,y0,z0)isacriticalpointof f onthesetde“nedby theconstraints.Ingeneral,thefollowingresultcanbeprovedbya similarlineofreasoning. Theorem 13.21 (CriticalPointsSubjecttoConstraints) Supposethat f and ga, a =1 2 ,...,M ,arefunctionsof m variables, m>M ,whicharedierentiableat r0,and f hasalocalmaximumor minimumat r0inthesetde“nedbytheconstraints ga( r )=0 .Then

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19613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS thereexistnumbers asuchthat f ( r0)= 1 g1( r0)+ 2 g2( r0)+ + M gM( r0) Let f ( r )beafunctionsubjecttoaconstraint g ( r ).De“nethe function F ( r )= f ( r ) Š g ( r ) where isviewedasanadditionalindependentvariable.Thencriticalpointsof F aredeterminedby(13.19).Indeed,thecondition F/ =0yieldstheconstraint g ( r )=0,whilethedierentiation withrespecttothevariables r gives F = f Š g =0,whichcoincideswiththe“rstequationin(13.19).Similarly,ifthereareseveral constraints,criticalpointsofthefunctionwithadditionalvariables a, a =1 2 ,...,M (13.20) F ( r ,1,2,...,n)= f ( r ) Š 1g1( r ) Š 2g2( r ) ŠŠ MgM( r ) coincidewiththecriticalpointsof f subjecttotheconstraints ga=0 asstatedinTheorem13.21.Thefunctions F and f havethesame valuesonthesetde“nedbytheconstraints ga=0becausetheydier byalinearcombinationofconstraintfunctionswiththecoecients beingthe Lagrangemultipliers .96.3.FindingLocalMaximaandMinima.Inthesimplestcaseofatwovariablefunction f subjecttoaconstraint,thenatureofcriticalpoints (localmaximumorminimum)canbesolvedbygeometricalmeans. Supposethatthelevelcurve g ( x,y )=0isclosed.Then,bytheextreme valuetheorem, f attainsitsmaximumandminimumvaluesonitat someofthecriticalpoints.Suppose f attainsitsabsolutemaximum atacriticalpoint r1.Then f shouldhaveeitheralocalminimum oranin”ectionattheneighboringcriticalpoint r2alongthecurve. Let r3bethecriticalpointnextto r2alongthecurve.Then f hasa localminimumat r2if f ( r2) f ( r3). Thisproceduremaybecontinueduntilallcriticalpointsareexhausted. Comparethispatternofcriticalpointswiththebehaviorofaheight alongaclosedhikingpath. Remark. Iftheconstraintscanbesolved,thenanexplicitformof f onthesetde“nedbytheconstraintscanbefound,andthestandardsecond-derivativetestapplies!Forinstance,inExample13.35, theconstraintcanbesolved z = V/ ( xy ).Thevaluesofthefunction f ontheconstraintsurfaceare F ( x,y )= f ( x,y,V/ ( xy ))= xy + 2 V ( x + y ) / ( xy ).Theequations F x=0and F y=0determinethe

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96.LAGRANGEMULTIPLIERS197 criticalpoint x = y =(2 V )1 / 3(and z = V/ ( xy )=(2 V )1 / 3/ 2).So thesecond-derivativetestcanbeappliedtothefunction F ( x,y )atthe criticalpoint x = y =(2 V )1 / 3toshowthatindeed F hasaminimum andhence f hasaminimumontheconstraintsurface. Thereisananalogofthesecond-derivativetestforcriticalpointsof functionssubjecttoconstraints.Itsgeneralformulationisnotsimple. Sothediscussionislimitedtothesimplestcaseofafunctionoftwo variablessubjecttoaconstraint. Supposethat g ( r0) = 0 .Then g xand g ycannotsimultaneously vanishatthecriticalpoint.Withoutlossofgenerality,assumethat g y =0at r0=( x0,y0).Bytheimplicitfunctiontheorem(Theorem13.11),thereisaneighborhoodof r0inwhichtheequation g ( x,y )=0hasauniquesolution y = h ( x ).Thevaluesof f onthe levelcurve g =0nearthecriticalpointare F ( x )= f ( x,h ( x )).Bythe chainrule,oneinfersthat F= f x+ f yhand (13.21) F=( d/dx )( f x+ f yh)= f xx+2 f xyh+ f yy( h)2+ f yh. So,inorderto“nd F( x0),onehastocalculate h( x0)and h( x0).This taskisaccomplishedbytheimplicitdierentiation.Bythede“nition of h ( x ), G ( x )= g ( x,h ( x ))=0forall x inanopenintervalcontaining x0.Therefore, G( x )=0,whichde“nes hbecause G= g x+ g yh=0 and h= Š g x/g y.Similarly, G( x )=0yields (13.22) G= g xx+2 g xyh+ g yy( h)2+ g yh=0 whichcanbesolvedfor h,where h= Š g x/g y.Thesubstitutionof h( x0), h( x0),andallthevaluesofallthepartialderivativesof f atthe criticalpoint( x0,y0)into(13.21)givesthevalue F( x0).If F( x0) > 0 (or F( x0) < 0),then f hasalocalminimum(ormaximum)at( x0,y0) alongthecurve g =0.Notealsothat F( x0)=0asrequiredowingto theconditions f x= g xand f y= g ysatis“edatthecriticalpoint. If g y( r0)=0,then g x( r0) =0,andthereisafunction x = h ( y )that solvestheequation g ( x,y )=0.So,byswapping x and y intheabove arguments,thesameconclusionisprovedtohold. Example 13.36 Showthatthepoint r0=(0 0) isacriticalpointof thefunction f ( x,y )= x2y + y + x subjecttotheconstraint exy= x + y +1 anddeterminewhether f hasalocalminimumormaximumatit. Solution: Criticalpoint .Put g ( x,y )= exyŠ x Š y Š 1.Then g (0 0)=0;thatis, thepoint(0 0)satis“estheconstraint.The“rstpartialderivativesof f and g are f x=2 xy +1, f y= x2+1, g x= yexyŠ 1,and g y= xexyŠ 1.

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19813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Therefore,bothequations f x(0 0)= g x(0 0)and f y(0 0)= g y(0 0) aresatis“edat = Š 1.Thus,thepoint(0 0)isacriticalpointof f subjecttotheconstraint g =0. Second-derivativetest .Since g y(0 0)= Š 1 =0,thereisafunction y = h ( x )near x =0suchthat G ( x )= g ( x,h ( x ))=0.Bytheimplicit dierentiation, h(0)= Š g x(0 0) /g y(0 0)= Š 1 Thesecondpartialderivativesof g are g xx= y2exy,g yy= x2exy,g xy= exy+ xyexy. Thederivative h(0)isfoundfrom(13.22),where g xx(0 0)= g yy(0 0)= 0, g xy(0 0)=1, h(0)= Š 1,and g y(0 0)= Š 1: h(0)= Š [ g xx(0 0)+2 g xy(0 0) h(0)+ g yy(0 0)( h(0))2] /g y(0 0)= Š 2 Thesecondpartialderivativesof f are f xx=2 y,f yy=0 ,f xy=2 x. Thesubstitutionof f xx(0 0)= f yy(0 0)= f xy(0 0)=0, h(0)= Š 1, f y(0 0)=1,and h(0)= Š 2into(13.21)gives F(0)= Š 2 < 0. Therefore, f attainsalocalmaximumat(0 0)alongthecurve g = 0.Notealsothat F(0)= f x(0 0)+ f y(0 0) h(0)=1 Š 1=0as required. Theimplicitdierentiationandtheimplicitfunctiontheoremcan beusedtoestablishthesecond-derivativetestforthemultivariable casewithconstraints(seeanotherexampleinStudyProblem13.11).96.4.StudyProblems.Problem13.11. Let f beatwicecontinuouslydierentiablefunction of r =( x,y,z ) subjecttoaconstraint g ( r )=0 .Assumethat g isatwice continuouslydierentiablefunction.Usetheimplicitdierentiationto establishthesecond-derivativetestforcriticalpointsof f onthesurface g =0 Solution: Supposethat g ( r0) = 0 atacriticalpoint r0.Withoutlossofgenerality,onecanassumethat g z( r0) =0.Bytheimplicitfunctiontheorem,thereexistsafunction z = h ( x,y )suchthat G ( x,y )= g ( x,y,h ( x,y ))=0insomeneighborhoodofthecritical point.Thentheequations G x( x,y )=0and G y( x,y )=0determine the“rstpartialderivativesof h : g x+ g zh x=0= h x= Š g x/g z; g y+ g zh y=0= h y= Š g y/g z.

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96.LAGRANGEMULTIPLIERS199 Thesecondpartialderivativesof h arefoundfromtheequations G xx=0= g xx+2 g xzh x+ g zz( h x)2+ g zh xx=0 G yy=0= g yy+2 g yzh y+ g zz( h y)2+ g zh yy=0 G xy=0= g xy+ g xzh x+ g yzh y+ g zzh xh y+ g zh xy=0 Thevaluesofthefunction f ( x,y,z )ofthelevelsurface g ( x,y,z )=0 nearthecriticalpointsare F ( x,y )= f ( x,y,h ( x,y )).Toapplythe second-derivativetesttothefunction F ,itssecondpartialderivatives havetobecomputedatthecriticalpoint.Theimplicitdierentiation gives F xx=( f x+ f zh x) x= f xx+2 f xzh x+ f zz( h x)2+ f zh xx, F yy=( f y+ f zh y) y= f yy+2 f yzh y+ f zz( h y)2+ f zh yy, F xy=( f x+ f zh x) y= f xy+ f xzh x+ f yzh y+ f zzh xh y+ f zh xy, wherethepartialderivativesof h aredeterminedbythepartialderivativesoftheconstraintfunction g asspeci“ed.If( x0,y0,z0)isthecritical pointfoundbytheLagrangemultipliermethod,then a = F xx( x0,y0), b = F yy( x0,y0),and c = F xy( x0,y0)inthesecond-derivativetestfor thetwo-variablefunction F

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CHAPTER14 MultipleIntegrals 97.DoubleIntegrals97.1.TheVolumeProblem.Supposeoneneedstodeterminethevolumeofahillwhoseheight f ( r )asafunctionofposition r =( x,y )is known.Forexample,thehillmustbeleveledtoconstructahig hway. Itsvolumeisrequiredtoestimatethenumberoftruckloadsneededto movethesoil away.The followingprocedurecanbeusedtoestimate thevolume.Thebase D ofthehillis“rstpartitionedintosmallpieces Dpofarea Ap,where p =1 2 ,...,N enumeratesthepieces;thatis, theunionofallthepieces Dpistheregion D .Thepartitionelements shouldbesmallenoughsothattheheight f ( r )hasnosigni“cantvariationwhen r isin Dp.Thevolumeoftheportionofthehillaboveeach partitionelement Dpisapproximately Vp f ( rp) Ap,where rpisa pointin Dp.Theapproximationbecomesbetterforsmaller Dp.The volumeofthehillcanthereforebeestimatedas V Np =1f ( rp) Ap. Forpracticalpurposes,thevalues f ( rp)canbefound,forexample, fromadetailedcontourmapof f Theapproximationisexpectedtobecomebetterandbetterasthe sizeofthepartitionelementsgetssmaller(naturally,theirnumber N hastoincrease).If Rpisthesmallestradiusofadiskthatcontains Dp, thenput RN=maxpRp,whichdeterminesthesizeofthelargestpartitionelement.Whenalargernumber N ofpartitionelementsistaken toimprovetheaccuracyoftheapproximation,onehastoreduce RNatthesametimetomakevariationsof f withineachpartitionelement smaller.Notethatthereductionofthemaximalareamaxp Apversus themaximalsize RNmaynotbegoodenoughtoimprovetheaccuracy oftheestimate.If Dplookslikeanarrowstrip,itsareaissmall,but thevariationoftheheight f alongthestripmaybesigni“cantand theaccuracyoftheapproximation Vp f ( rp) Apispoor.Onecan201

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20214.MULTIPLEINTEGRALS thereforeexpectthattheexactvalueofthevolumeisobtainedinthe limit (14.1) V =limN ( RN 0) Np =1f ( rp) Ap. Thevolume V maybeviewedasthevolumeofasolidboundedfrom abovebythesurface z = f ( x,y ),whichisthegraphof f ,andbythe portion D ofthe xy plane.Naturally,itisnotexpectedtodependon thewaytheregion D ispartitioned,neithershoulditdependonthe choiceofsamplepoints rpineachpartitionelement. Thelimit(14.1)resemblesthelimitofaRiemannsumforasinglevariablefunction f ( x )onaninterval[ a,b ]usedtodeterminethearea underthegraphof f .Indeed,if xk, k =0 1 ,...,N x0= a
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97.DOUBLEINTEGRALS203 the supremum of f on D anddenotedby supDf .Thelargestlower boundof f on D iscalledthe in“mum of f on D anddenotedby infDf Asaboundedregion, D canalwaysbeembeddedinarectangle RD= { ( x,y ) | x [ a,b ] ,y [ c,d ] } (i.e., D isasubsetof RD). Thefunction f isthen extended totherectangle RDbysettingits valuesto0forallpointsoutside D ,thatis, f ( r )=0if r RDand r / D .Considerapartition xk, k =0 1 ,...,N1,oftheinterval[ a,b ],where xk= a + k x x =( b Š a ) /N1,andapartition yj, j =0 1 ,...,N2,oftheinterval[ c,d ],where yj= c + j y and y =( d Š c ) /N2.Thesepartitionsinduceapartitionoftherectangle RDbyrectangles Rkj= { ( x,y ) | x [ xk Š 1,xk] ,y [ yj Š 1,yj] } where k =1 2 ,...,N1and j =1 2 ,...,N2.Theareaofeachpartition rectangle Rkjis A = x y .Thispartitioniscalleda rectangular partition of RD.Foreverypartitionrectangle Rkj,therearenumbers Mjk=sup f ( r )and mjk=inf f ( r ),thesupremumandin“mumof f on Rkj. Definition 14.2 (UpperandLowersums) Let f beaboundedfunctiononaclosedboundedregion D .Let RDbea rectanglethatcontains D andletthefunction f bede“nedtohavezero valueforallpointsof RDthatdonotbelongto D .Givenarectangular partition Rkjof RD,let Mjk=sup f and mjk=inf f bethesupremum andin“mumof f on Rjk.Thesums U ( f,N1,N2)=N1j =1 N2k =1Mkj A,L ( f,N1,N2)=N1j =1 N2k =1mkj A arecalledthe upperandlowersums Thesequencesofupperandlowersumshavethefollowingimportantproperty. Corollary 14.1 (PropertyofUpperandLowerSums) Thesequenceofuppersumsisdecreasing,whilethesequenceoflower sumsisincreasing,thatis, U ( f,N1,N2) U ( f,N 1,N 2) and L ( f,N1, N2) L ( f,N 1,N 2) if N 1 N1and N 2 N2. Proof. Consideranypartitionrectangle R (theindices jk areomitted)ofarea A .Put M =supRf and m =infRf .Whenthepartition isre“ned, R becomesaunionofseveralrectangles Rpofarea Ap, p = 1 2 ,...,q ,sothat A = p Ap.Put Mp=supRpf and mp=infRpf Since Rpiscontainedin R ,onehas Mp M and mp m .Therefore, ifthepartitionisre“ned,theterm M A intheuppersumisreplaced

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20414.MULTIPLEINTEGRALS by pMp Ap M p Ap= M A ,andtheterm m A inthe lowersumisreplacedby pmp Ap m p Ap= m A ;thatis, theuppersumeitherdecreasesordoesnotchange,whilethelowersum eitherincreasesordoesnotchange. Continuingtheanalogywiththevolumeproblem,theupperand lowersumsrepresentupperandlowerestimatesofthevolume.They shouldbecomecloserandclosertothevolumeasthepartitionbecomes “nerand“ner.Thisleadstothefollowingnaturalde“nitionofthe doubleintegral. Definition 14.3 (DoubleIntegral) Ifthelimitsoftheupperandlowersumsexistas N1 2 (or ( x, y ) (0 0) )andcoincide,then f issaidtobeRiemannintegrableon D ,andthelimitoftheupperandlowersums Df ( x,y ) dA =limN1 2U ( f,N1,N2)=limN1 2L ( f,N1,N2) iscalledthe doubleintegral of f overtheregion D Itshouldbeemphasizedthatthedoubleintegralisde“nedasthe two-variablelimit( x, y ) (0 0).Theupperandlowersumsare functionsof x and y because N1=( b Š a ) / x and N2=( d Š c ) / y Theexistenceofthelimitanditsvaluemustbeestablishedaccordingly. Letusdiscussthisde“nitionfromthepointofviewofthevolume problem.First,notethataspeci“cpartitionof D byrectangleshas beenused.Inthisway,thearea Apofthepartitionelementhas beengivenaprecisemeaningastheareaofarectangle.Later,itwill beshownthatifthedoubleintegralexistsinthesenseoftheabove de“nition,thenitexistsiftherectangularpartitionisreplacedbyany partitionof D byelements Dpofanarbitraryshapesubjecttocertain conditionsthatallowforapreciseevaluationoftheirarea.Second,the volume(14.1)isindeedgivenbythedoubleintegralof f ,anditsvalue is independentofthechoiceofsamplepoints rp.Thisisanextremely usefulpropertythatallowsonetoapproximatethedoubleintegralwith anydesiredaccuracybyevaluatingasuitable Riemannsum Definition 14.4 (RiemannSum) Let f beafunctionon D thatiscontainedinarectangle RD.Let f bede“nedbyzerovaluesoutsideof D in RD.Let r jkbeapointina partitionrectangle Rjk,where Rjkformapartitionof RD.Thesum R ( f,N1,N2)=N1j =1 N2k =1f ( r jk) A iscalleda Riemannsum .

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97.DOUBLEINTEGRALS205 Theorem 14.1 (ConvergenceofRiemannSums) Ifafunction f isintegrableon D ,thenitsRiemannsumsforany choiceofsamplepoints r jkconvergetothedoubleintegral: limN1 2R ( f,N1,N2)= DfdA. Proof. Foranypartitionrectangle Rjkandanysamplepoint r jkinit, mjk f ( r jk) Mjk.Itfollowsfromthisinequalitythat L ( f,N1,N2) R ( f,N1,N2) U ( f,N1,N2).Since f isintegrable,thelimitsofthe upperandlowersumsexistandcoincide.Theconclusionofthetheorem followsfromthesqueezeprincipleforlimits. ApproximationofDoubleIntegrals. If f isintegrable,itsdoubleintegralcanbeapproximatedbyasuitableRiemannsum.Acommonlyusedchoiceofsamplepointsistotake r jktobetheintersection ofthediagonalsofpartitionrectangles Rjk,thatis, r jk=( xj, yk),where xjand ykarethemidpointsoftheintervals[ xj Š 1,xj]and[ yk Š 1,yk],respectively.Thisruleiscalledthe midpointrule .Theaccuracyofthe midpointruleapproximationcanbeassessedby“ndingtheupperand lowersums;theirdierencegivestheupperboundontheabsolute erroroftheapproximation.Alternatively,iftheintegralistobeevaluateduptosomesigni“cantdecimals,thepartitionintheRiemann sumhastobere“neduntilitsvaluedoesnotchangeinthesigni“cant digits.Theintegrabilityof f guaranteestheconvergenceofRiemann sumsandtheindependenceofthelimitfromthechoiceofsample points.97.3.ContinuityandIntegrability.AnExampleofaNonintegrableFunction. Noteverybounded functionisintegrable.Suppose f isde“nedonthesquare x [0 1]and y [0 1]sothat f ( x,y )=1ifboth x and y arerational, f ( x,y )=2if both x and y areirrational,and f ( x,y )=0otherwise.Thisfunction isnotintegrable.Recallthatanyinterval[ a,b ]containsbothrational andirrationalnumbers.Therefore,anypartitionrectangle Rjkcontains pointswhosecoordinatesarebothrational,orbothirrational,pairsof rationalandirrationalnumbers.Hence, Mjk=2and mjk=0.The lowersumvanishesforanypartitionandthereforeitslimitis0,whereas theuppersumis2 jk A =2 A =2foranypartition,where A is theareaofthesquare.Thelimitsoftheupperandlowersumsdo notcoincide,2 =0,andthedoubleintegralof f doesnotexist.The Riemannsumforthisfunctioncanconvergetoanynumberbetween2 and0,dependingonthechoiceofsamplepoints.Forexample,ifthe

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20614.MULTIPLEINTEGRALS samplepointshaverationalcoordinates,thentheRiemannsumequals 1.Ifthesamplepointshaveirrationalcoordinates,thentheRiemann sumequals2.Ifthesamplepointsaresuchthatonecoordinateis rationalwhiletheotherisirrational,thentheRiemannsumvanishes. Thefollowingtheoremdescribesaclassofintegrablefunctionsthat issucientinmanypracticalapplications. Theorem 14.2 (IntegrabilityofContinuousFunctions) Let D beaclosed,boundedregionwhoseboundariesarepiecewisesmoothcurves.Ifafunction f iscontinuouson D ,thenitisintegrable on D Notethattheconverseisnottrue;thatis,theclassofintegrable functionsiswiderthantheclassofallcontinuousfunctions.Thisis arathernaturalconclusioninviewoftheanalogybetweenthedouble integralandthevolume.Thevolumeofasolidbelowagraph z = f ( x,y ) 0ofacontinuousfunctionon D shouldexist.Ontheother hand,let f ( x,y )bede“nedon D = { ( x,y ) | x [0 2] ,y [0 1] } so that f ( x,y )= m if x 1and f ( x,y )= M if x> 1.Thefunctionis piecewiseconstantandhasajumpdiscontinuityalongtheline x =1in D .Thevolumebelowthegraph z = f ( x,y )andabove D iseasyto“nd; itisthesumofvolumesoftworectangularboxeswiththesamebase area A1= A2=1anddierentheights M and m V = MA1+ mA2= M + m .Thedoubleintegralof f existsandalsoequals M + m .Indeed, foranyrectangularpartition,thenumbers Mjkand mjkdiersonlyfor partitionrectanglesintersectedbythediscontinuityline x =1,thatis, MjkŠ mjk= M Š m forallsuchrectangles.Therefore,thedierence betweentheupperandlowersumsis l x ( M Š m ),where l =1isthe lengthofthediscontinuitycurve.Inthelimit x 0,thedierence vanishes.Asnotedearlier,theupperandlowersumsaretheupperand lowerestimatesofthevolumeandshouldthereforeconvergetoitas theirlimitscoincide.Usingasimilarlineofarguments,onecanprove thefollowing. Corollary 14.2 If f isboundedon D anddiscontinuousonlyon a“nitenumberofsmoothcurves,thenitisintegrableon D .98.PropertiesoftheDoubleIntegral Thepropertiesofthedoubleintegralaresimilartothoseofan ordinaryintegralandcanbeestablisheddirectlyfromthede“nition.

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98.PROPERTIESOFTHEDOUBLEINTEGRAL207 Linearity. Let f and g befunctionsintegrableon D andlet c bea number.Then D( f + g ) dA = DfdA + DgdA, DcfdA = c DfdA. Area. Thefunction iscalledthe characteristicfunction oftheregion D if ( r )=1if r D and ( r )=0otherwise.Since isconstanton D ,itisalsocontinuouson D andhenceintegrable.Iffollowsthat (14.2) DdA = DdA = A ( D ) where A ( D )iscalledthe area of D .Theregion D canalwaysbecovered bytheunionofadjacentrectanglesofarea A = x y .Inthelimit ( x, y ) (0 0),thetotalareaoftheserectanglesconvergestothe areaof D Additivity. Supposethat D istheunionof D1and D2suchthatthe areaoftheirintersectionis0;thatis, D1and D2mayonlyhavecommon pointsattheirboundariesornocommonpointsatall.If f isintegrable on D ,then DfdA = D1fdA + D2fdA. Thispropertyisthemostdiculttoprovedirectlyfromthede“nition. However,itappearsrathernaturalwhenmakingtheanalogyofthe doubleintegralandthevolume.Iftheregion D iscutintotwopieces D1and D2,thenthesolidabove D isalsocutintotwosolids,oneabove D1andtheotherabove D2.Naturally,thevolumeisadditive. Supposethat f isnonnegativeon D1andnonpositiveon D2.The doubleintegralover D1isthevolumeofthesolidabove D1andbelow thegraphof f .Since Š f 0on D2,thedoubleintegralover D2is the negative volumeofthesolid below D2and above thegraphof f When f becomesnegative,itsgraphgoesbelowtheplane z =0(the xy plane).So,ingeneral,thedoubleintegralmayvanishortakenegative values,dependingonwhichvolume(aboveorbelowthe xy planeis larger).Thispropertyisanalogoustothefamiliarrelationbetweenthe ordinaryintegralandtheareaunderthegraph. Positivity. If f ( r ) 0forall r D ,then DfdA 0 ,

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20814.MULTIPLEINTEGRALS and,asaconsequenceofthelinearity, DfdA DgdA if f ( r ) g ( r )forall r D UpperandLowerBounds. Let m =infDf and M =supDf .Then m f ( r ) M forall r D .Fromthepositivityofthedouble integral,itfollowsthat mA ( D ) DfdA MA ( D ) Thisinequalityiseasytovisualize.If f ispositive,thenthedouble integralisthevolumeofthesolidbelowthegraphof f .Thesolidlies inthecylinderwiththecrosssection D .Thegraphof f liesbetween theplanes z = m and z = M .Therefore,thevolumeofthecylinder ofheight m cannotexceedthevolumeofthesolid,whereasthelatter cannotexceedthevolumeofthecylinderofheight M Theorem 14.3 (IntegralMeanValueTheorem) If f iscontinuouson D ,thenthereexistsapoint r0 D suchthat DfdA = f ( r0) A ( D ) Proof. Let h beanumber.Put g ( h )= D( f Š h ) dA = DfdA Š hA ( D ).Fromtheupperandlowerboundsforthedoubleintegral,it followsthat g ( M ) 0and g ( m ) 0.Since g ( h )islinearin h ,there exists h = h0 [ m,M ]suchthat g ( h0)=0.Ontheotherhand,a continuousfunctiononaclosed,boundedregion D takesitsmaximal andminimalvaluesaswellasallthevaluesbetweenthem.Therefore, forany m h0 M ,thereis r0 D suchthat f ( r0)= h0. Ageometricalinterpretationoftheintegralmeanvaluetheoremis rathersimple.Imaginethatthesolidbelowthegraphof f ismadeof clay.Theshapeofapieceofclaymaybedeformedwhilethevolume ispreservedunderdeformation.Thenon”attopofthesolidcanbe deformedsothatitbecomes”at,turningthesolidintoacylinderof height h0,which,byvolumepreservation,shouldbebetweenthesmallestandthelargestheightsoftheoriginalsolid.Theintegralmeanvalue theoremmerelystatestheexistenceofsuchan average heightatwhich thevolumeofthecylindercoincideswiththevolumeofthesolidwith anon”attop.Thecontinuityofthefunctionissucient(butnotnecessary)toestablishthatthereisapointatwhichtheaverageheight coincideswiththevalueofthefunction.

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98.PROPERTIESOFTHEDOUBLEINTEGRAL209 IntegrabilityoftheAbsoluteValue. Supposethat f isintegrable onabounded,closedregion D .Thenitsabsolutevalue | f | isalso integrableand DfdA D| f | dA. Aproofoftheintegrabilityof | f | israthertechnical.Oncetheintegrabilityof | f | isestablished,theinequalityisasimpleconsequenceof | a + b || a | + | b | appliedtoaRiemannsumof f .Makingtheanalogy betweenthedoubleintegralandthevolume,supposethat f 0on D1and f 0on D2,where D1 2aretwoportionsof D .If V1and V2stand forthevolumesofthesolidsboundedbythegraphof f and D1and D2,respectively,thenthedoubleintegralof f over D is V1Š V2,while thedoubleintegralof | f | is V1+ V2.Naturally, | V1Š V2| V1+ V2for positive V1 2. IndependenceofPartition. Ithasbeenarguedthatthevolumeof asolidunderthegraphof f andabovearegion D canbecomputed by(14.1)inwhichtheRiemannsumisde“nedforan arbitrary (nonrectangular)partitionof D .Canthedoubleintegralof f over D be computedinthesameway?Theanswerisarmative.However,the proofgoesbeyondthescopeofthiscourse.If f iscontinuous,then theassertionisnotsodiculttoestablish.Let D bepartitionedby piecewisesmoothcurvesintopartitionelements Dp, p =1 2 ,...,N ,so thattheunionof Dpis D and A ( D )= N p =1 Ap,where Apisthe areaof Dpde“nedby(14.2).If Rpisthesmallestradiusofadiskthat contains Dp,put RN=max Rp;thatis, Rpcharacterizesthesizeof thepartitionelement Dpand RNisthesizeofthelargestpartition element.Recallthatthelargestpartitionelementdoesnotnecessarily havethelargestarea.Thepartitionissaidtobere“nedif RN
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21014.MULTIPLEINTEGRALS equalsthedoubleintegral.Indeed,put cp= | f ( r p) Š f ( rp) | and cN=max cp, p =1 2 ,...,N .Since f iscontinuous, cN 0as N becauseanypartitionelement Dpiscontainedinadiskofradius Rp RN 0as N .Therefore,thedeviationoftheRiemann sumfromthedoubleintegralconvergesto0: fdA Š R ( f,N ) = Np =1( f ( rp) Š f ( r p)) Ap Np =1| f ( rp) Š f ( r p) | Ap=Np =1cp Ap cN Np =1 Ap= cNA ( D ) 0 as N .SothedoubleintegralcanbeapproximatedbyRiemann sumsforarbitrarypartitionssubjecttotheconditionsspeci“edabove, thatis, (14.3) DfdA =limN ( RN 0) Np =1f ( r p) Apforanychoiceofsamplepoints r p.Notethattheregion D isnolonger requiredtobeembeddedinarectangleand f doesnothavetobe extendedoutsideof D .Thispropertyisusefulforevaluatingdouble integralsbymeansof changeofvariables discussedlaterinthischapter. ItisalsousefultosimplifycalculationsofRiemannsums. Example 14.1 Findthedoubleintegralof f ( x,y )= x2+ y2over thedisk Dx2+ y2 1 usingthepartitionof D byconcentriccircles andraysfromtheorigin. Solution: Considercircles x2+ y2= r2 p,where rp= p r r =1 /N and p =0 1 2 ,...,N .If isthepolarangleintheplane,thenpoints witha“xedvalueof formarayfromtheorigin.Letthedisk D be partitionedbycirclesofradii rpandrays = k= k =2 /n k =1 2 ,...,n .Eachpartitionelementliesinthesectorofangle andisboundedbytwocircleswhoseradiidierby r .Theareaofa sectorofradius rpis r2 p / 2.Therefore,theareaofapartitionelement betweencirclesofradii rpand rp +1is Ap= r2 p +1 / 2 Š r2 p / 2= ( r2 p +1Š r2 p) / 2 =( rp +1+ rp) r / 2.IntheRiemannsum,usethe midpointrule;thatis,thesamplepointsareintersectionsofthecircles ofradius rp=( rp +1+ rp) / 2andtherayswithangles k=( k +1+ k) / 2.

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99.ITERATEDINTEGRALS211 Thevaluesof f atthesamplepointsare f ( r p)= r2 p,theareaelements are Ap= rp r ,andthecorrespondingRiemannsumreads R ( f,N,n )=nk =1 Np =1 r3 p r =2 Np =1 r3 p r because n k =1 =2 ,thetotalrangeof inthedisk D .Thesum over p istheRiemannsumforthesingle-variablefunction g ( r )= r3ontheinterval r [0 1].Inthelimit N ,thissumconvergesto theintegralof g overtheinterval[0 1],thatis, D( x2+ y2) dA =2 limN Np =1 r3 p r =2 1 0r3dr = / 2 So,bychoosingthepartitionaccordingtotheshapeof D ,thedouble RiemannsumhasbeenreducedtoaRiemannsumforasingle-variable function. Thenumericalvalueofthedoubleintegralinthisexampleisthe volumeofthesolidthatliesbetweentheparaboloid z = x2+ y2and thedisk D ofunitradius.Itcanalsoberepresentedasthevolume ofthecylinderwithheight h =1 / 2, V = hA ( D )= h = / 2.This observationillustratestheintegralmeanvaluetheorem.Thefunction f takesthevalue h =1 / 2onthecircle x2+ y2=1 / 2ofradius1 / 2 in D .99.IteratedIntegrals Hereapracticalmethodforevaluatingdoubleintegralswillbedeveloped.Tosimplifythetechnicalities,thederivationofthemethodis givenforcontinuousfunctions.However,themethodisalsovalidfor boundedfunctionsthatarediscontinuousona“nitenumberofsmooth curves,whichissucientformanypracticalapplications.99.1.RectangularDomains.Letafunction f becontinuouson D .Suppose“rstthat D isarectangle x [ a,b ]and y [ c,d ].Let Rjkbea rectangularpartitionof D asde“nedearlier.Foranychoiceofsample points r jk=( x j,y k),where x j [ xj Š 1,xj]and y k [ yk Š 1,yk],theRiemannsumconvergestothedoubleintegralof f over D .SincethedoublelimitoftheRiemannsum(as( x, y ) (0 0))exists,itshould notdependontheorderinwhichthelimits N1 (or x 0)and

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21214.MULTIPLEINTEGRALS N2 (or y 0)arecomputed.Thisisthekeyobservationfor whatfollows.Supposethelimit y 0istobeevaluated“rst: DfdA =limN1 2R ( f,N1,N2) =limN1 N1j =1# limN2 N2k =1f ( x j,y k) y $ x. TheexpressioninparentheseisnothingbuttheRiemannsumforthe single-variablefunction gj( y )= f ( x j,y )ontheinterval y [ c,d ].So, ifthefunctions gj( y )areintegrableon[ c,d ],thenthelimitoftheir Riemannsumsistheintegralof gjovertheinterval.If f iscontinuous on D ,thenitmustalsobecontinuousalongthelines x = x jin D ;that is, gj( y )= f ( x j,y )iscontinuousandhenceintegrableon[ c,d ].Thus, (14.4)limN2 N2k =1f ( x j,y k) y = d cf ( x j,y ) dy. De“neafunction A ( x )by (14.5) A ( x )= d cf ( x,y ) dy. Thevalueof A at x isgivenbytheintegralof f withrespectto y ; theintegrationwithrespectto y iscarriedoutasif x werea“xed number.Forexample,put f ( x,y )= x2y + exyand[ c,d ]=[0 1]. Thenanantiderivative F ( x,y )of f ( x,y )withrespectto y is F ( x,y )= x2y2/ 2+ exy/x ,whichmeansthat F y( x,y )= f ( x,y ).Therefore, A ( x )= 1 0( x2y + exy) dy = x2y2/ 2+ exy/x 1 0= x2/ 2+ ex/x Š 1 /x. Ageometricalinterpretationof A ( x )issimple.If f 0,then A ( x j)is theareaofthecrosssectionofthesolidbelowthegraph z = f ( x,y )by theplane x = x j,and A ( x j) x isthevolumeofthesliceofthesolid ofwidth x ThesecondsumintheRiemannsumforthedoubleintegralinthe Riemannsumof A ( x )ontheinterval[ a,b ]: DfdA =limN1 N1j =1A ( x j) x = b aA ( x ) dx = b a d cf ( x,y ) dy dx,

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99.ITERATEDINTEGRALS213 wheretheintegralexistsbythecontinuityof A .Theintegralonthe rightsideofthisequalityiscalledthe iteratedintegral .Inwhatfollows, theparentheseintheiteratedintegralwillbeomitted.Theorder inwhichtheintegralsareevaluatedisspeci“edbytheorderofthe dierentialsinit;forexample, dydx meansthattheintegrationwith respectto y istobecarriedout“rst.Inasimilarfashion,bycomputing thelimit x 0“rst,thedoubleintegralcanbeexpressedasan iteratedintegralinwhichtheintegrationiscarriedoutwithrespect to x andthenwithrespectto y .Sothefollowingresulthasbeen established. Theorem 14.4 (FubinisTheorem) If f iscontinuousontherectangle D = { ( x,y ) | x [ a,b ] ,y [ c,d ] } then Df ( x,y ) dA = d cb af ( x,y ) dxdy = b ad cf ( x,y ) dydx. Moregenerally,thisistrueforanybounded f on D thatisdiscontinuousona“nitenumberofsmoothcurves. Thinkofaloafofbreadwitharectangularbaseandwithatop havingtheshapeofthegraph z = f ( x,y ).Itcanbeslicedalongeither ofthetwodirectionsparalleltoadjacentsidesofitsbase.Fubinis theoremsaysthatthevolumeoftheloafisthesumofthevolumesof theslicesandisindependentofhowtheslicingisdone. Example 14.2 Findthevolumeofthesolidboundedfromabove bytheportionoftheparaboloid z =4 Š x2Š 2 y2andfrombelowby theportionoftheparaboloid z = Š 4+ x2+2 y2,where x [0 1] and y [0 1] Solution: Iftheheightofthesolidatany( x,y ) D is h ( x,y )= ztop( x,y ) Š zbot( x,y ),wherethegraphs z = ztop( x,y )and z = zbot( x,y ) arethetopandbottomboundariesofthesolid,thenthevolumeis V = Dh ( x,y ) dA = D[ ztop( x,y ) Š zbot( x,y )] dA = D(8 Š 2 x2Š 4 y2) dA = 1 01 0(8 Š 2 x2Š 4 y2) dydx = 1 0[(8 Š 2 x2) y Š 4 y3/ 3] 1 0dx = 1 0(8 Š 2 x2Š 4 / 3) dx =6

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21414.MULTIPLEINTEGRALS Corollary 14.3 (FactorizationofIteratedIntegrals) Let D bearectangle { ( x,y ) | x [ a,b ] ,y [ c,d ] } .Suppose f ( x,y )= g ( x ) h ( y ) ,wherethefunctions g and h areintegrableon [ a,b ] and [ c,d ] respectively.Then Df ( x,y ) dA = b ag ( x ) dx d ch ( y ) dy. ThissimpleconsequenceofFubinistheoremisquiteuseful. Example 14.3 Evaluatethedoubleintegralof f ( x,y )=sin( x + y ) overtherectangle x [0 ] and y [ Š / 2 ,/ 2] Solution: Onehassin( x + y )=sin x cos y +cos x sin y .Theintegral ofsin y over[ Š / 2 ,/ 2]vanishesbysymmetry.So,bythefactorization propertyoftheiteratedintegral,onlythe“rsttermcontributestothe doubleintegral: Dsin( x + y ) dA = 0sin xdx / 2 Š / 2cos ydy =4 Thefollowingexampleillustratestheuseoftheadditivityofadoubleintegral. Example 14.4 Evaluatethedoubleintegralof f ( x,y )=15 x4y2overtheregion D ,whichistherectangle [ Š 2 2] [ Š 2 2] withtherectangularhole [ Š 1 1] [ Š 1 1] Solution: Let D1=[ Š 2 2] [ Š 2 2]andlet D2=[ Š 1 1] [ Š 1 1]. Therectangle D1istheunionof D and D2suchthattheirintersection hasnoarea.Hence, D2fdA = DfdA + D1fdA DfdA = D2fdA Š D1fdA. Byevaluatingthedoubleintegralsover D1 2, D115 x4y2dA =15 2 Š 2x4dx 2 Š 2y2dy =210, D215 x4y2dA =15 1 Š 1x4dx 1 Š 1y2dy =4 thedoubleintegralover D isobtained,1024 Š 4=1020.

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100.DOUBLEINTEGRALSOVERGENERALREGIONS215 99.2.StudyProblems.Problem14.1. Supposeafunction f hascontinuoussecondderivativesontherectangle R =[0 1] [0 1] .Find Rf xydA if f (0 0)=1 f (0 1)=2 f (1 0)=3 ,and f (1 1)=5 Solution: ByFubinistheorem, Rf xydA = 1 01 0 x f y( x,y ) dxdy = 1 0f y( x,y ) 1 0dy = 1 0[ f y(1 ,y ) Š f y(0 ,y )] dy = 1 0d dy [ f (1 ,y ) Š f (0 ,y )] dy =[ f (1 ,y ) Š f (0 ,y )] 1 0=[ f (1 1) Š f (0 1)] Š [ f (1 0) Š f (0 0)]=1 ByClairautstheorem f xy= f yxandthevalueoftheintegralisindependentoftheorderofintegration. 100.DoubleIntegralsOverGeneralRegions Theconceptoftheiteratedintegralcanbeextendedtogeneral regionssubjecttothefollowingconditions.100.1.SimpleRegions.Definition 14.5 (SimpleandConvexRegions) Aregion D issaidtobe simple inthedirection u ifanylineparallel tothevector u intersects D alongatmostonestraightlinesegment.A region D iscalled convex ifitissimpleinanydirection. Suppose D issimpleinthedirectionofthe y axis.Itwillbereferred toas y simple or verticallysimple .Since D isbounded,thereisan interval[ a,b ]suchthatverticallines x = x0intersect D if x0 [ a,b ]. Inotherwords,theregion D lieswithintheverticalstrip a x b Takeaverticalline x = x0 [ a,b ]andconsiderallpointsof D thatalso belongtotheline,thatis,pairs( x0,y ) D ,wherethe“rstcoordinate is“xed.Sincethelineintersects D alongasegment,thevariable y rangesoveraninterval.Theendpointsofthisintervaldependon thelineorthevalueof x0;thatis,forevery x0 [ a,b ], ybot y ytop,wherethenumbers ybotand ytopdependon x0.Inotherwords, verticallysimpleregionsadmitthefollowingalgebraicdescription.

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21614.MULTIPLEINTEGRALS AlgebraicDescriptionofVerticallySimpleRegions. If D isverticallysimple,thenitliesintheverticalstrip a x b andisbounded frombelowbythegraph y = ybot( x )andfromabovebythegraph y = ytop( x ): (14.6) D = { ( x,y ) | ybot( x ) y ytop( x ) ,x [ a,b ] } Thenumbers a and b are,respectively,thesmallestandthelargest valuesofthe x coordinateofpointsof D .Forexample,thehalf-disk x2+ y2 1, y 0,isaverticallysimpleregion.The x coordinateof anypointinthediskliesintheinterval[ a,b ]=[ Š 1 1].Forevery x inthisinterval,the y coordinateliesintheinterval0 y 1 Š x2; thatis,intheverticaldirection,thetopboundaryofthediskisthe graph y = 1 Š x2= ytop( x )andthebottomboundaryisthegraph y =0= ybot( x ). Suppose D issimpleinthedirectionofthe x axis.Itwillbereferred toas x simple or horizontallysimple .Since D isbounded,thereisan interval[ c,d ]suchthathorizontallines y = y0intersect D if y0 [ c,d ]. Inotherwords,theregion D lieswithinthehorizontalstrip c y d .Takeahorizontalline y = y0 [ c,d ]andconsiderallpointsof D thatalsobelongtotheline,thatis,pairs( x,y0) D ,wherethe secondcoordinateis“xed.Sincethelineintersects D alongasegment, thevariable x rangesoveraninterval.Theendpointsofthisinterval dependonthelineorthevalueof y0;thatis,forevery y0 [ c,d ], xbot x xtop,wherethenumbers xbotand xtopdependon y0.In otherwords,horizontallysimpleregionsadmitthefollowingalgebraic description. AlgebraicDescriptionofHorizontallySimpleRegions.If D ishorizontallysimple,thenitliesinahorizontalstrip c y b andisbounded frombelowbythegraph x = xbot( y )andfromabovebythegraph x = xtop( y ): (14.7) D = { ( x,y ) | xbot( y ) x xtop( y ) ,y [ c,d ] } Thenumbers c and d are,respectively,thesmallestandthelargest valuesofthe y coordinateofpointsof D .ThetermsbelowŽand aboveŽarenowde“nedrelativetothelineofsightinthedirection ofthe x axis.Forexample,thehalf-disk x2+ y2 1, y 0,isalso ahorizontallysimpleregion.The y coordinateofanypointinthe diskliesintheinterval[ c,d ]=[0 1].Forevery y inthisinterval, the x coordinateliesintheinterval Š 1 Š y2 x 1 Š y2;that is,inthehorizontaldirection,thetopboundaryofthediskisthe graph x = 1 Š y2= xtop( y )andthebottomboundaryisthegraph x = Š 1 Š y2= xbot( y ).

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100.DOUBLEINTEGRALSOVERGENERALREGIONS217 100.2.IteratedIntegralsforSimpleRegions.Suppose D isvertically simple.Thenitshouldhaveanalgebraicdescriptionaccordingto (14.6).Fortheembeddingrectangle RD,onecantake[ a,b ] [ c,d ], where c ybot( x ) ytop( x ) d forall x [ a,b ].Thefunction f iscontinuousin D andde“nedbyzerovaluesoutside D ;thatis, f ( x,y )=0if c y
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21814.MULTIPLEINTEGRALS thealgebraicdescription(14.7).Thedoubleintegralof f over D is thengivenbytheiteratedintegral (14.9) Df ( x,y ) dA = d cxtop( y ) xbot( y )f ( x,y ) dxdy. IteratedIntegralsforNonsimpleRegions. Iftheintegrationregion D isnotsimple,howcanoneevaluatethedoubleintegral?Any nonsimpleregioncanbecutintosimpleregions Dp, p =1 2 ,...,n .The doubleintegraloversimpleregionscanthenbeevaluated.Thedouble integralover D isthenthesumofthedoubleintegralsover Dpbythe additivityproperty(seeExample14.4). Example 14.5 Evaluatethedoubleintegralof f ( x,y )=6 yx2over theregion D boundedbytheline y =1 andtheparabola y = x2. Solution: Theregion D isbothhorizontallyandverticallysimple.It isthereforepossibletouseeither(14.8)or(14.9).To“ndanalgebraic descriptionof D asaverticallysimpleregion,onehasto“rstspecify themaximalrangeofthe x coordinatein D .Itisdeterminedby theintersectionoftheline y =1andtheparabola y = x2,thatis, 1= x2,andhence x [ a,b ]=[ Š 1 1]forallpointsof D .Forany x [ Š 1 1],the y coordinateofpointsof D attainsthesmallestvalue ontheparabola(i.e., ybot( x )= x2),andthelargestvalueontheline (i.e., ytop( x )=1).Onehas D6 yx2dA =6 1 Š 1x21 x2ydydx =3 1 Š 1x2(1 Š x4) dx =8 / 7 Itisalsoinstructivetoobtainthisresultusingthereverseorderof integration.To“ndanalgebraicdescriptionof D asahorizontally simpleregion,onehasto“rstspecifythemaximalrangeofthe y coordinatein D .Thesmallestvalueof y is0andthelargestvalueis1; thatis, y [ c,d ]=[0 1]forallpointsof D .Forany“xed y [0 1], the x coordinateofpointsof D attainsthesmallestandlargestvalues when y = x2or x = y ,thatis, xbot( y )= Š y and xtop( y )= y Onehas D6 yx2dA =6 1 0y y Š yx2dxdy =2 1 0y (2 y3 / 2) dy =4 1 0y5 / 2dy =8 / 7

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100.DOUBLEINTEGRALSOVERGENERALREGIONS219 100.3.ReversingtheOrderofIntegration.Byreversingtheorderofintegration,asimpli“cationoftechnicalitiesinvolvedinevaluatingdouble integralscanbeachieved,butnotalways,though. Example 14.6 Evaluatethedoubleintegralof f ( x,y )=2 x over theregion D boundedbytheline x =2 y +2 andtheparabola x = y2Š 1 Solution: Theregion D isbothverticallyandhorizontallysimple. However,theiteratedintegralbasedonthealgebraicdescriptionof D asaverticallysimpleregionismoreinvolved.Indeed,thelargestvalue ofthe x coordinatein D occursatoneofthepointsofintersectionof thelineandtheparabola,2 y +2= y2Š 1or( y Š 1)2=4,andhence, y = Š 1 3.Thelargestvalueof x in D is x =32Š 1=8.Thesmallest valueof x occursatthepointofintersectionoftheparabolawiththe x axis, x = Š 1.So[ a,b ]=[ Š 1 8].However,thealgebraicexpression forthetopboundary ytop( x )isnotthesameforall x [ Š 1 8].For any“xed x [ Š 1 0],therangeofthe y coordinateisdeterminedby theparabola, Š x +1 y x +1,whileforany“xed x [0 8], thetopandbottomboundariesoftherangeof y aredeterminedby thelineandparabola,respectively, Š x +1 y ( x Š 2) / 2.This dictatesthenecessitytosplittheregion D intotworegions D1and D2suchthat x [ Š 1 0]forallpointsin D1and x [0 8]forallpoints in D2.Thecorrespondingiteratedintegralreads D2 xdA = D12 xdA + D22 xdA =20Š 1x x +1Š x +1dydx +280xx/ 2+1Š x +1dydx. Ontheotherhand,iftheiteratedintegralcorrespondingtothealgebraicdescriptionof D asahorizontallysimpleregionisused,thetechnicalitiesaregreatlysimpli“ed.Thesmallestandlargestvaluesof y in D occuratthepointsofintersectionofthelineandtheparabolafound above, y = Š 1 3,thatis,[ c,d ]=[ Š 1 3].Forany“xed y [ Š 1 3], the x coordinaterangesfromitsvalueontheparabolatoitsvalueon theline, xbot( y )= y2Š 1 x 2 y +2= xbot( y ).Thecorresponding iteratedintegralreads D2 xdA =2 3 Š 12 y +2 y2Š 1xdxdy = 3 Š 1( Š y4+6 y2+8 y +3) dy =256 / 5 whichissimplertoevaluatethanthepreviousone.

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22014.MULTIPLEINTEGRALS Sometimestheiteratedintegrationcannotevenbecarriedoutin oneorder,butitcanstillbedoneintheotherorder. Example 14.7 Evaluatethedoubleintegralof f ( x,y )=sin( y2) overtheregion D ,whichistherectangleboundedby x =0 y = x ,and y = Solution: Supposetheiteratedintegralforverticallysimpleregions isused.Therangeofthe x coordinateis x [0 ]=[ a,b ],and,for every“xed x [0 ],therangeofthe y coordinateis ybot( x )= x y = ytop( x )in D .Theiteratedintegralreads Dsin( y2) dA = 0 xsin( y2) dydx. However,theantiderivativeofsin( y2)cannotbeexpressedinelementaryfunctions!Letusreversetheorderofintegration.Themaximal rangeofthe y coordinatein D is[0 ]=[ c,d ].Forevery“xed y [0 ],therangeofthe x coordinateis xbot( y )=0 x y = xtop( y ) in D .Therefore,theiteratedintegralreads Dsin( y2) dA = 0sin( y2)y0dxdy = 0sin( y2) ydy = Š 1 2 cos( y2) 0=1 100.4.TheUseofSymmetry.Thesymmetrypropertyhasbeenestablishedinsingle-variableintegration: f ( Š x )= Š f ( x ) a Š af ( x ) dx =0 whichhasprovedtobequiteuseful.Forexample,theintegralof sin( x2011)overanysymmetricinterval[ Š a,a ]vanishesbecausesin( x2011) isanantisymmetricfunction.Asimilarpropertycanbeestablished fordoubleintegrals.Consideratransformationthatmapseachpoint ( x,y )oftheplanetoanotherpoint( xs,ys).Aregion D issaidto be symmetric underatransformation( x,y ) ( xs,ys)iftheimage Dsof D coincideswith D (i.e., Ds= D ).Forexample,let D be boundedbyanellipse x2/a2+ y2/b2=1.Then D issymmetricunder re”ectionsaboutthe x axis,the y axis,ortheircombination,thatis,

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100.DOUBLEINTEGRALSOVERGENERALREGIONS221 ( x,y ) ( xs,ys)=( Š x,y ),( x,y ) ( xs,ys)=( x, Š y ),or( x,y ) ( xs,ys)=( Š x, Š y ).Atransformationoftheplane( x,y ) ( xs,ys) issaidtobe areapreserving iftheimage Dsofanyregion D under thistransformationhasthesamearea,thatis, A ( D )= A ( Ds).For example,translations,rotations,re”ectionsaboutlines,andtheircombinationsarearea-preservingtransformations. Theorem 14.5 (SymmetryProperty) Letaregion D besymmetricunderanarea-preservingtransformation ( x,y ) ( xs,ys) suchthat f ( xs,ys)= Š f ( x,y ) .Thentheintegralof f over D vanishes: Df ( x,y ) dA =0 Ageneralproofispostponeduntilthechangeofvariablesindouble integralsisdiscussed.Herethesimplestcaseofare”ectionabouta lineisconsidered.If D issymmetricunderthisre”ection,thenthe linecuts D intotwoequal-arearegions D1and D2sothat Ds 1= D2and Ds 2= D1.Thedoubleintegralisindependentofthechoiceof partition(see(14.3)).Considerapartitionof D1byelements D1 p, p =1 2 ,...,N .Bysymmetry,theimages Ds 1 pofthepartitionelements D1 pformapartitionof D2suchthat Ap= A ( D1 p)= A ( Ds 1 p)byarea preservation.Chooseelements D1 pand Ds 1 ptopartitiontheregion D Nowrecallthatthedoubleintegralisalsoindependentofthechoice ofsamplepoints.Suppose( xp,yp)aresamplepointsin D1 p.Choose samplepointsin Ds 1 ptobetheimages( xps,xps)of( xp,yp)underthe re”ection.Withthesechoicesofthepartitionof D andsamplepoints, theRiemannsum(14.3)vanishes: DfdA =limN Np =1 f ( xp,yp) Ap+ f ( xps,yps) Ap =0 wherethetwotermsinthesumcorrespondtopartitionsof D1and D2in D ;bythehypothesis,thefunction f isantisymmetricunder there”ectionandtherefore f ( xps,yps)= Š f ( xp,yp)forall p .Froma geometricalpointofview,theportionofthesolidboundedbythegraph z = f ( x,y )thatliesabovethe xy planehasexactlythesameshapeas thatbelowthe xy plane,andthereforetheirvolumescontributewith oppositesignstothedoubleintegralandcanceleachother. Example 14.8 Evaluatethedoubleintegralof sin[( x Š y )3] over theportion D ofthedisk x2+ y2 1 thatliesinthe“rstquadrant ( x,y 0 ).

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22214.MULTIPLEINTEGRALS Solution: Theregion D issymmetricunderthere”ectionaboutthe line y = x ,thatis,( x,y ) ( xs,ys)=( y,x ),whereasthefunctionis anti-symmetric, f ( xs,ys)= f ( y,x )=sin[( y Š x )3]=sin[ Š ( x Š y )3]= Š sin[( x Š y )3]= Š f ( x,y ).Bythesymmetryproperty,thedouble integralvanishes. Example 14.9 Evaluatethedoubleintegralof f ( x,y )= x2y3over theregion D ,whichisobtainedfromtheellipticregion x2/ 4+ y2/ 9 1 byremovingthesquare [0 1] [0 1] Solution: Let D1and D2betheellipticandsquareregions,respectively.Theellipticregion D1islargeenoughtoincludethesquare D2.Therefore,theadditivityofthedoubleintegralcanbeused(compareExample14.4)totransformthedoubleintegraloveranon-simple region D intotwodoubleintegralsoversimpleregions: Dx2y3dA = D1x2y3dA Š D2x2y3dA = Š D2x2y3dA = Š 1 0x2dx 1 0y3dy = Š 1 / 12; theintegralover D1vanishesbecausetheellipticregion D1issymmetric underthere”ection( x,y ) ( xs,ys)=( x, Š y ),whereastheintegrand isanti-symmetric, f ( x, Š y )= x2( Š y )3= Š x2y3= Š f ( x,y ). 101.DoubleIntegralsinPolarCoordinates Thepolarcoordinatesarede“nedbythefollowingrelations: x = r cos ,y = r sin or r = x2+ y2, =tanŠ 1( y/x ) where r isthedistancefromtheorigintothepoint( x,y )and isthe anglebetweenthepositive x axisandtherayfromtheoriginthrough thepoint( x,y )countedcounterclockwise.ThevalueoftanŠ 1must betakenaccordingtothegeometricalde“nitionof .If( x,y )lies inthe“rstquadrant,thenthevalueoftanŠ 1mustbeintheinterval [0 ,/ 2)andtanŠ 1( )= / 2andsimilarlyfortheotherquadrants. Theseequationsde“nea one-to-one correspondencebetweenallpoints ( x,y )oftheplaneandpointsofthestrip( r, ) (0 ) [0 2 ). Alternatively,onecanalsosettherangeof tobetheinterval[ Š ). Theorderedpair( r, )canbeviewedasapointofanauxiliaryplane or polarplane .Inwhatfollows,the r axisinthisplaneissettobe vertical,andthe axisissettobehorizontal.Foranyregion D ,there isanimage Dof D inthepolarplanede“nedbythetransformation ofanorderedpair( x,y ) D totheorderedpair( r, ) D.Note thattheboundariesof Daremappedontotheboundariesof D by

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101.DOUBLEINTEGRALSINPOLARCOORDINATES223 x = r cos and y = r sin .Forexample,let D betheportionofthe disk x2+ y2 1inthe“rstquadrant.Thentheshapeof Dcanbe foundfromtheimagesofboundariesof D inthepolarplane: boundariesof D boundariesof Dx2+ y2=1 r =1 y =0 ,x 0 =0 x =0 ,y 0 = / 2 Since r 0,theregion Distherectangle( r, ) [0 1] [0 ,/ 2]= D. Let Dbetheimageof D onthepolarplaneandlet R Dbearectanglecontaining Dsothattheimageof R Dcontains D .Asbefore,a function f on D isextendedoutside D bysettingitsvaluesto0.Considerarectangularpartitionof Dsuchthateachpartitionrectangle D jkisboundedbythecoordinatelines r = rj, r = rj +1= rj+ r = k,and = k +1= k+ .Eachpartitionrectanglehasthe area A= r .Theimageofthecoordinateline r = rkinthe xy planeisthecircleofradius rkcenteredattheorigin.Theimageofthe coordinateline = konthe xy planeistherayfromtheoriginthat makestheangle kwiththepositive x axiscountedcounterclockwise. Theraysandcirclesarecalled coordinatecurves ofthepolarcoordinatesystem,thatis,thecurvesalongwhicheitherthecoordinate r or thecoordinate remainsconstant(circlesandrays,respectively).A rectangularpartitionof Dinducesapartitionof D bythecoordinate curves.Eachpartitionelement Djkistheimageoftherectangle D jkandisboundedbytwocirclesandtworays. Let f ( x,y )beanintegrablefunctionon D .Thedoubleintegralof f over D canbecomputedasthelimitoftheRiemannsum.According to(14.3),thelimitdoesnotdependoneitherthechoiceofpartitionor thesamplepoints.Let Ajkbetheareaof Djk.Theareaofthesector ofthediskofradius rjthathastheangle is r2 j / 2.Therefore, Ajk= 1 2 ( r2 j +1Š r2 j) = 1 2 ( rj +1+ rj) r = 1 2 ( rj +1+ rj) A. In(14.3),put Ap= Ajk, rp Djkbeingtheimageofasample point( r j, k) D jksothat f ( rp)= f ( r jcos k,r jsin k).Thelimitin (14.3)isunderstoodasthedoublelimit( r, ) (0 0).Owingto theindependenceofthelimitofthechoiceofsamplepoints,put r j= ( rj +1+ rj) / 2(themidpointrule).Withthischoice,( rj +1+ rj) r/ 2= r j r .BytakingthelimitoftheRiemannsum(14.3) limN ( RN 0) Np =1f ( r p) Ap=limN 1 2 ( r, ) (0 0) N1j =1 N2k =1f ( r jcos k,r jsin k) r j A,

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22414.MULTIPLEINTEGRALS oneobtainsthedoubleintegralofthefunction f ( r cos ,r sin ) J ( r ) overtheregion D(theimageof D ),where J ( r )= r iscalledthe Jacobian ofthepolarcoordinates.TheJacobiande“nestheareaelement transformation dA = JdA= rdA. Definition 14.6 (DoubleIntegralinPolarCoordinates) Let Dbetheimageof D inthepolarplanespannedbyorderedpairs ( r, ) ofpolarcoordinates.Thedoubleintegralof f over D inpolar coordinatesis Df ( x,y ) dA = Df ( r cos ,r sin ) J ( r ) dA,J ( r )= r. Asimilaritybetweenthedoubleintegralinrectangularandpolar coordinatesisthattheybothusepartitionsbycorrespondingcoordinatecurves.Notethathorizontalandverticallinesarecoordinate curvesoftherectangularcoordinates.SotheverytermadoubleintegralinpolarcoordinatesŽreferstoaspeci“cpartitioning D inthe Riemannsum,namely,by coordinatecurvesofpolarcoordinates (by circlesandrays). Thedoubleintegralover Dcanbeevaluatedbythe standardmeans,thatis,byconvertingittoasuitableiteratedintegral withrespectto r and Example 14.10 Usepolarcoordinatestoevaluatethedoubleintegralof f ( x,y )= xy2 x2+ y2over D ,whichistheportionofthedisk x2+ y2 1 thatliesinthe“rstquadrant. Solution: First,theimage Dof D hastobefound.Usingthe boundarytransformation,asexplainedatthebeginningofthissection, Distherectangle r [0 1]and [0 ,/ 2].Second,thefunctionhas tobewritteninpolarcoordinates, f ( r cos ,r sin )= r4cos sin2 Third,thedoubleintegralofthisfunction, multipliedbytheJacobian r hastobeevaluatedover D.As Disarectangle,byFubinistheorem, theorderofintegrationintheiteratedintegralisirrelevant: DfdA = / 2 0sin2 cos d 1 0r5dr = 1 3 sin3 / 2 0 1 6 r6 1 0= 1 18 Thisexampleshowsthatthetechnicalitiesinvolvedinevaluating thedoubleintegralhavebeensubstantiallysimpli“edbypassingto polarcoordinates.Thesimpli“cationistwofold.First,thedomainof integrationhasbeensimpli“ed;thenewdomainisarectangle,which ismuchsimplertohandleintheiteratedintegralthanaportionofa

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101.DOUBLEINTEGRALSINPOLARCOORDINATES225 disk.Second,theevaluationofordinaryintegralswithrespectto r and appearstobesimplerthantheintegrationof f withrespecttoeither x or y neededintheiteratedintegral.However,thesesimpli“cations cannotalwaysbeachievedbyconvertingthedoubleintegraltopolar coordinates.Theregion D andtheintegrand f shouldhavesome particularpropertiesthatguaranteetheobservedsimpli“cationsand therebyjustifytheuseofpolarcoordinates.Herearesomeguiding principlestodecidewhethertheconversionofadoubleintegraltopolar coordinatescouldbehelpful: € Thedomain D isboundedbycircles,linesthroughtheorigin, andpolargraphs. € Thefunction f ( x,y )dependsoneitherthecombination x2+ y2= r2or y/x =tan Indeed,if D isboundedonlybycirclescenteredattheoriginand linesthroughtheorigin,thentheimage Disarectanglebecausethe boundariesof D are coordinatecurves ofpolarcoordinates.Ifthe boundariesof D containcirclesnotcenteredattheoriginor,generally, polargraphs,thatis,curvesde“nedbytherelations r = g ( ),then analgebraicdescriptionoftheboundariesof Dissimplerthanthat oftheboundariesof D .If f ( x,y )= h ( u ),where u = x2+ y2= r2or u = y/x =tan ,thenintheiteratedintegraloneoftheintegrations, eitherwithrespectto or r ,becomestrivial. Example 14.11 Evaluatethedoubleintegralof f ( x,y )= xy over theregion D thatliesinthe“rstquadrantandisboundedbythecircles x2+ y2=4 and x2+ y2=2 x Solution: First,theimage Dof D mustbefound.Usingtheprinciple thattheboundariesof D aremappedontotheboundariesof D,one “ndstheequationsoftheboundariesof Dbyconvertingtheequations fortheboundariesof D intopolarcoordinates.Theboundaryofthe region D consistsofthreecurves: x2+ y2=4 r =2 x2+ y2=2 x r =2cos x =0 ,y 0 = / 2 So,inthepolarplane,theregion Disboundedbythehorizontal line r =2,thegraph r =2cos ,andtheverticalline = / 2.In particular,itisconvenienttouseanalgebraicdescriptionof Das averticallysimpleregion;thatis,( r, ) Dif rbot( )=2cos r 2= rtop( )and [0 ,/ 2]=[ a,b ](because rtop(0)= rbot(0)). Second,thefunctioniswritteninpolarcoordinates, f ( r cos ,r sin )=

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22614.MULTIPLEINTEGRALS r2sin cos .MultiplyingitbytheJacobian J = r ,theintegrandis obtained.Onehas DxydA = Dr3sin cos dA= b asin cos rtop( ) rbot( )r3drd = / 2 0sin cos 2 2cos r3drd =4 / 2 0(1 Š cos )4cos sin d =4 1 0(1 Š u )4udu =4 1 0v4(1 Š v ) dv = 4 15 wheretwochangesofvariableshavebeenusedtosimplifythecalculations, u =cos and v =1 Š u Example 14.12 Findtheareaoftheregion D thatisbounded bytwospirals r = and r =2 ,where [0 2 ] ,andthepositive x axis. Beforesolvingtheproblem,letusmakeafewcommentsaboutthe shapeof D .Theboundaries r = and r =2 areexamplesofpolar graphs, r = g ( ),where g ( )= and g ( )=2 .Theycanbevisualized bymeansofasimplegeometricalprocedure.Takearaycorresponding toa“xedvalueofthepolarangle .Onthisray,markthepointthat isadistance r = g ( )fromtheorigin.Allsuchpointsobtainedforall valuesof formacurve,calledthe polargraph .When g ( )= ,the distance r = increasesastherayrotatesabouttheorigin,andthe polargraphisaspiralwindingabouttheorigin.Theregion D liesbetweentwospirals;itisnotsimpleinanydirection.Anysmoothcurve inthe xyplanecanalwaysbede“nedbyanequation h ( x,y )=0.In thiscase,byconvertingthepolargraphequationintotherectangular coordinates,onehas x2+ y2=tanŠ 1( y/x )or y = x tan( x2+ y2). Thereisnowayto“ndananalyticsolutionofthisequationtoexpress y asafunctionof x orviceversa.Therefore,hadonetriedtoevaluatethedoubleintegralintherectangularcoordinates,onewouldhave facedan unsolvable problemof“ndingtheequationsfortheboundaries of D intheform y = ytop( x )and y = ybot( x )! Solution: Theregion D isboundedbythreecurves,twospirals(polar graphs),andtheline y =0 ,x 0.Theirimagesonthepolarplane arethelines r = r =2 ,andtheverticalline =2 .Theyformthe boundariesof D.Analgebraicdescriptionof Dasaverticallysimple

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102.CHANGEOFVARIABLESINDOUBLEINTEGRALS227 regionisconvenienttouse,( r, ) Dif rbot( )= r 2 = rtop( ) and [0 2 ]=[ a,b ].Hence, A ( D )= DdA = DrdA= 2 02 rdrd = 3 2 2 02d =4 3. Example 14.13 Findthevolumeoftheportionofthesolidbounded bythecone z =2 x2+ y2andtheparaboloid z =2 Š x2Š y2thatlies inthe“rstoctant. Solution: Theintersectionoftheconeandparaboloidisacircleof unitradius.Indeed,put r = x2+ y2.Thenthepointsofintersection satisfythecondition2 r =2 Š r2or r =1.Sotheprojection D ofthe solidontothe xy planeistheportionofthedisk r 1inthe“rst quadrant.Forany( x,y ) D ,theheightis h =2 Š r2Š 2 r (i.e., independentofthepolarangle ).Theimage Dof D inthepolar planeistherectangle( r, ) [0 1] [0 ,/ 2].Thevolumeis Dh ( x,y ) dA = D(2 Š r2Š 2 r ) rdA= / 2 0d 1 0(2 r Š r3Š 2 r2) dr = 24 102.ChangeofVariablesinDoubleIntegrals Withanexampleofpolarcoordinates,itisquiteclearthatasmart choiceofintegrationvariablescansigni“cantlysimplifythetechnicalitiesinvolvedwhenevaluatingdoubleintegrals.Thesimpli“cationis twofold:simplifyingtheshapeoftheintegrationregion(arectangular shapeismostdesirable)and“ndingantiderivativeswhencalculating theiteratedintegral.Itisthereforeofinteresttodevelopatechnique forageneralchangeofvariableinthedoubleintegralsothatonewould beableto design newvariablesspeci“ctothedoubleintegralinquestioninwhichthesought-forsimpli“cationisachieved.102.1.ChangeofVariables.Letthefunctions x ( u,v )and y ( u,v )be de“nedonanopenregion D.Then,foreverypair( u,v ) D,one can“ndapair( x,y ),where x = x ( u,v )and y = y ( u,v ).Allsuch pairsformaregioninthe xy planethatisdenoted D .Inotherwords, thefunctions x ( u,v )and y ( u,v )de“nea transformation ofaregion Dinthe uv planeontoaregion D inthe xy plane.Ifnotwopoints

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22814.MULTIPLEINTEGRALS in Dhavethesameimagepointin D ,thenthetransformationis called one-to-one .Foraone-to-onetransformation,onecande“nethe inversetransformation,thatis,thefunctions u ( x,y )and v ( x,y )that assignapair( u,v ) Dtoapair( x,y ) D ,where u = u ( x,y ) and v = v ( x,y ).Owingtothisone-to-onecorrespondencebetween rectangularcoordinates( x,y )andpairs( u,v ),onecandescribepoints inaplaneby newcoordinates ( u,v ).Forexample,ifpolarcoordinates areintroducedbytherelations x = x ( r, )= r cos and y = y ( r, )= r sin foranyopenset Dofpairs( r, )thatliewithinthehalf-strip [0 ) [0 2 ),thenthereisaone-to-onecorrespondencebetweenthe pairs( x,y ) D and( r, ) D.Inparticular,theinversefunctions are r ( x,y )= x2+ y2and ( x,y )=tanŠ 1( y/x ). Definition 14.7 (ChangeofVariablesinaPlane) Aone-to-onetransformationofanopenregion Dde“nedby x = x ( u,v ) and y = y ( u,v ) iscalleda changeofvariables ifthefunctions x ( u,v ) and y ( u,v ) havecontinuous“rst-orderpartialderivatives on D. Thepairs( u,v )areoftencalled curvilinearcoordinates .Recallthat apointofaplanecanbedescribedasanintersectionpointoftwo coordinatelinesofarectangularcoordinatesystem x = xpand y = yp. Thepoint( xp,yp) D isauniqueimageofapoint( up,vp) D. Considertheinversetransformation u = u ( x,y )and v = v ( x,y ).Since u ( xp,yp)= upand v ( xp,yp)= vp,thepoint( xp,yp) D canbeviewed asthepointofintersectionoftwocurves u ( x,y )= upand v ( x,y )= vp.Thecurves u ( x,y )= upand v ( x,y )= vparecalled coordinate curves ofthenewcoordinates u and v ;thatis,thecoordinate u hasa “xedvaluealongitscoordinatecurve u ( x,y )= up,and,similarly,the coordinate v hasa“xedvaluealongitscoordinatecurve v ( x,y )= vp. Thecoordinatecurvesareimagesofthestraightlines u = upand v = vpin Dundertheinversetransformation.Ifthecoordinatecurves arenotstraightlines(asinarectangularcoordinatesystem),then suchcoordinatesarenaturallycurvilinear.Forexample,thecoordinate curvesofpolarcoordinatesareconcentriccircles(a“xedvalueof r ) andraysfromtheorigin(a“xedvalueof ),andeverypointinaplane canbeviewedasanintersectionofonesuchrayandonesuchcircle.102.2.ChangeofVariablesinaDoubleIntegral.Consideradoubleintegralofafunction f ( x,y )overaregion D .Let x = x ( u,v )and y = y ( u,v )de“neatransformationofaregion Dto D ,where Dis boundedbypiecewise-smoothcurvesinthe uv plane.Supposethat thetransformationisachangeofvariablesonanopenregionthat

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102.CHANGEOFVARIABLESINDOUBLEINTEGRALS229 includes D.Thenthereisaninversetransformation,thatis,atransformationof D to D,whichisde“nedbythefunctions u = u ( x,y ) and v = v ( x,y ).Accordingto(14.3),thedoubleintegralof f over D is thelimitofaRiemannsum.Thelimitdependsneitheronapartition of D byareaelementsnoronsamplepointsinthepartitionelements. Followingtheanalogywithpolarcoordinates,considerapartitionof D bycoordinatecurves u ( x,y )= ui, i =1 2 ,...,N1,and v ( x,y )= vj, j =1 2 ,...,N2,suchthat ui +1Š ui= u and vj +1Š vj= v .This partitionof D isinducedbyarectangularpartitionof Dbyhorizontal lines v = vjandverticallines u = uiinthe uv plane.Eachpartition element D ijof Dhasthearea A= u v .Itsimageisapartition element Dijof D .If( u i,v j) D ijisasamplepoint,thenthecorrespondingsamplepointin Dijis r ij=( x ( u i,v j) ,y ( u i,v j)),and(14.3) becomes DfdA =limN1,N2 N1i =1 N2j =1f ( r ij) Aij, where Aijistheareaofthepartitionelement Dij.Thelimit N1,N2 isunderstoodinthesenseofadoublelimit( u, v ) (0 0).As before,thevaluesof f ( x ( u,v ) ,y ( u,v ))outside Daresetto0when calculatingthevalueof f inapartitionrectanglethatintersectsthe boundaryof D. Asinthecaseofpolarcoordinates,theaimistoconvertthislimit intoadoubleintegralof f ( x ( u,v ) ,y ( u,v ))overtheregion D.This canbeaccomplishedby“ndingarelationbetween Aijand A, thatis,theruleoftheareaelementtransformationunderachange ofvariables.Tosimplifythenotation,lettheindex p labelpartition elements Dij,thatis, Dp= Dij,( up,vp)=( ui,vj),etc.Nowtakea point( up,vp)and“xthenumbers u v .Considerarectangle D pin the uv planeboundedbythelines u = up, u = up+ u v = vp,and v = vp+ v .Let Abethevertex( up,vp), Bbe( up+ u,vp),and Cbe( up,vp+ v ).Theimage Dpof D pinthe xy planeisaregion boundedbythecoordinatecurvesofthevariables u and v .Theimages A and B ofthepoints Aand Blieonthecoordinatecurve v = vp, while A and C (theimageof C)areonthecoordinatecurve u = up. Thenumbers u and v canbeviewedasin“nitesimalvariationsof u and v ortheirdierentials.So,whencalculatingthearea A of Dp, itissucienttokeeponlyterms linear in v and u ;theirhigher powersaretobeneglected(byde“nitionofthedierential),thatis, A = J u v = J A,

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23014.MULTIPLEINTEGRALS wherethecoecient J istobefound.Recallthat J = r forpolar coordinates. Since u and v arein“nitesimallysmall,theareaof Dpcanbe approximatedbytheareaofaparallelogramwithadjacentsides AB = b and AC = c .Thecoordinatesof A are( x ( up,vp) ,y ( up,vp)),while thecoordinatesof B are( x ( up+ u,vp) ,y ( up+ u,vp))becausethey areimagesof Aand B,respectively,undertheinversetransformation x = x ( u,v )and y = y ( u,v ).Therefore, b = x ( up+ u,vp) Š x ( up,vp) ,y ( up+ u,vp) Š y ( up,vp) 0 = x u( up,vp) u,y u( up,vp) u, 0 = u x u( up,vp) ,y u( up,vp) 0 where x ( up+ u,vp)= x ( up,vp)+ x u( up,vp) u hasbeenlinearized, thatis,higherpowersof u areneglected,andsimilarlyfor y ( up+ u,vp);thethirdcomponentof b issetto0asthevectorisplanar. Ananalogouscalculationforthecomponentsof c yields c = v x v( up,vp) ,y v( up,vp) 0 Theareaoftheparallelogramreads (14.10) A = b c = det x ux vy uy v u v = J u v. Notethatthevectors b and c areinthe xy plane.Therefore,their crossproducthasonlyonenonzerocomponent(the z component) givenbythedeterminant.Theabsolutevalueofthedeterminantis neededbecausethe z componentofthecrossproductmaybenegative, (0 0 ,z ) = z2= | z | Definition 14.8 (JacobianofaTransformation) TheJacobianofatransformationde“nedby x = x ( u,v ) and y = y ( u,v ) is ( x,y ) ( u,v ) =det x u x v y u y v = x uy vŠ x vy u. TheJacobiancoincideswiththedeterminantin(14.10).Inthis de“nition,aconvenientnotationhasbeenintroduced.Thematrix whosedeterminantisevaluatedhasthe “rst rowcomposedofthepartialderivativesofthe “rst variablethenumeratorwithrespecttoall variablesinthedenominator,andsimilarlyforthesecondrow.This ruleiseasytoremember.

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102.CHANGEOFVARIABLESINDOUBLEINTEGRALS231 Furthermore,thecoecient J in(14.10)isthe absolutevalue of theJacobian.TheJacobianofachangeofvariablesinthedouble integralshouldnotvanishon Dbecause A =0.Sincethepartial derivativesof x and y withrespectto u and v arecontinuous, J is continuous,too.Therefore,foranypartitionelement Dij,thedierence ( AijŠ J ( u i,v j) A) / Avanishesinthelimit( u, v ) (0 0). So,inthislimit,onecanput Aij= J ( u i,v j) u v intheRiemann sum.ThelimitoftheRiemannsumde“nesthedoubleintegralofthe function f ( x ( u,v ) ,y ( u,v )) J ( u,v )overtheregion D.Theforegoing argumentssuggestthatthefollowingtheoremistrue(afullproofis giveninadvancedcalculuscourses). Theorem 14.6 (ChangeofVariablesinaDoubleIntegral) Supposeatransformation x = x ( u,v ) y = y ( u,v ) hascontinuous“rstorderpartialderivativesandmapsaregion Dboundedbypiecewisesmoothcurvesontoaregion D .Supposethatthistransformationis one-to-oneandhasanonvanishingJacobian,exceptperhapsonthe boundaryof D.Then Df ( x,y ) dA = Df ( x ( u,v ) ,y ( u,v )) J ( u,v ) dA, J ( u,v )= ( x,y ) ( u,v ) Notethatinthecaseofpolarcoordinates,theboundaryof Dmay containtheline r =0onwhichtheJacobian J = r vanishes.This entirelinecollapsesintoasinglepoint,theorigin( x,y )=(0 0)inthe xy plane,uponthetransformation x = r cos and y = r sin ;that is,thistransformationisnotone-to-oneonthisline.Afullproofof thetheoremrequiresananalysisofsuchsubtletiesinageneralchange ofvariables,whichwasexcludedintheabovederivationbyassuming thatthetransformationisa genuine changeofvariablesonaregion thatcontains D. Thechangeofvariablesinadoubleintegralentailsthefollowing steps: 1.Findingtheimage Dof D undertheinversetransformation u = u ( x,y ), v = v ( x,y ).Ausefulruletorememberhereis: boundariesof D boundariesof D; thatis,ifequationsofboundariesof D aregiven,thenequationsofthecorrespondingboundariesof Dcanbeobtainedby

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23214.MULTIPLEINTEGRALS expressingtheformerinthenewvariablesbythesubstitution x = x ( u,v )and y = y ( u,v ). 2.Transformationofthefunctiontonewvariables f ( x,y )= f ( x ( u,v ) ,y ( u,v )) 3.CalculationoftheJacobianthatde“nestheareaelementtransformation: dA = ( x,y ) ( u,v ) dudv = JdA,J = ( x,y ) ( u,v ) 4.Evaluationofthedoubleintegralof fJ over Dbyconverting ittoasuitableiteratedintegral.Thechoiceofnewvariables shouldbemotivatedbysimplifyingtheshape D(arectangular shapeisthemostdesirable). WhencalculatingtheJacobian,thefollowingstatement,givenwithout aproof,mightbeuseful. Corollary 14.4 If u = u ( x,y ) and v = v ( x,y ) istheinverseof thetransformation x = x ( u,v ) and y = y ( u,v ) ,thenitsJacobianis (14.11) ( x,y ) ( u,v ) = 1 ( u,v ) ( x,y )= 1 det u xu yv xv y Equation(14.11)de“nestheJacobianasafunctionof( x,y ).Sometimesitistechnicallysimplertoexpresstheproduct f ( x,y ) J ( x,y )in thenewvariablesratherthandoingsofor f and J separately.Thisis illustratedbythefollowingexample. Example 14.14 Useasuitablechangeofvariablestoevaluatethe doubleintegralof f ( x,y )= xy3overtheregionDthatliesinthe“rst quadrantandisboundedbythelines y = x and y =3 x andbythe hyperbolas yx =1 and yx =2 Solution: Thelineequationscanbewrittenintheform y/x =1and y/x =3because y,x> 0in D .Notethattheequationsofboundaries of D dependonjusttwoparticularcombinations y/x and yx that takeconstantvaluesontheboundariesof D .So,ifthenewvariables de“nedbytherelations u = u ( x,y )= y/x and v = v ( x,y ),thenthe imageregion Dinthe uv planeisarectangle u [1 3]and v [1 2]. Indeed,theboundaries y/x =1and y/x =3aremappedontothe verticallines u =1and u =3,whilethehyperbolas yx =1and yx =2 aremappedontothehorizontallines v =1and v =2.Letusputaside foramomenttheproblemofexpressing x and y asfunctionsofnew

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102.CHANGEOFVARIABLESINDOUBLEINTEGRALS233 variables,whichisneededtoexpress f and J asfunctionsof u and v and“nd“rsttheJacobianasafunctionof x and y bymeansof(14.11): J = det u xu yv xv y Š 1= det Š y/x21 /x yx Š 1= |Š 2 y/x |Š 1= x 2 y Notethat x and y arestrictlypositivein D .Theintegrandbecomes fJ = x2y2/ 2= v2/ 2.So“ndingthefunctions x = x ( u,x )and y = y ( u,v )happenstobeunnecessaryinthisexample!Hence, Dxy3dA = 1 2 Dv2dA= 1 2 3 1du 2 1v2dv =7 / 3 Thereaderisadvisedtoevaluatethedoubleintegralintheoriginal rectangularcoordinatestocomparetheamountofworkneededwith thissolution. Thefollowingexampleillustrateshowachangeofvariablescanbe usedtosimplifytheintegrandofadoubleintegral. Example 14.15 Evaluatethedoubleintegralofthefunction f ( x,y ) =cos[( y Š x ) / ( y + x )] overthetrapezoidalregionwithvertices (1 0) (2 0) (0 1) ,and (0 2) Solution: Aniteratedintegralintherectangularcoordinateswould containtheintegralofthecosinefunctionofarationalargument(either withrespectto x or y ),whichisdiculttoevaluate.Soachangeof variablesshouldbeusedtosimplifytheargumentofthecosinefunction. Theregion D isboundedbythelines x + y =1, x + y =2, x =0,and y =0.Put u = x + y and v = y Š x sothatthefunctioninthenew variablesbecomes f =cos( v/u ).Thelines x + y =1and x + y =2are mappedontotheverticallines u =1and u =2.Since y =( u+ v ) / 2 and x =( u Š v ) / 2,theline x =0ismappedontotheline v = u whiletheline y =0ismappedontotheline v = Š u .Thus,theregion D= { ( u,v ) |Š u v u,u [1 2] } .TheJacobianofthechangeof variablesis J =1 / 2.Hence, Dcos y Š x y + x dA = 1 2 Dcos v u dA= 1 2 2 1u Š ucos v u dvdu = 1 2 2 1u sin v u u Š udu =sin(1) 2 1udu =3sin(1) / 2

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23414.MULTIPLEINTEGRALS Example 14.16 (AreaofanEllipse) Findtheareaoftheregion D boundedbytheellipse x2/a2+ y2/b2=1 Solution: Underthechangeofvariables u = x/a v = y/b ,the ellipseistransformedintothecircle u2+ v2=1ofunitradius.Since theJacobianofthetransformationis J = ab A ( D )= DdA = DJdA= ab DdA= abA ( D)= ab. When a = b ,theellipsebecomesacircleofradius R = a = b ,andthe areaoftheellipsebecomestheareaofthedisk, A = R2.102.3.SymmetriesandaChangeofVariables.Asnotedearlier,the symmetrypropertiesofdoubleintegralsarequitehelpfulfortheirevaluation.Atransformation x = x ( u,v ), y = y ( u,v )thatmaps Donto D issaidtobeareapreservingiftheabsolutevalueofitsJacobianis 1,thatis, dA = dA,fromwhichitimmediatelyfollowsthattheareas of D and Dcoincide, A ( D )= A ( D).Forexample,rotations,translations,andre”ectionsarearea-preservingtransformationsforobvious geometricalreasons.Thefollowingtheoremholds. Theorem 14.7 Supposethatanarea-preservingtransformation x = x ( u,v ) y = y ( u,v ) mapsaregion D ontoitself.Supposethat afunction f isskew-symmetricunderthistransformation,thatis, f ( x ( u,v ) ,y ( u,v ))= Š f ( u,v ) .Thenthedoubleintegralof f over D vanishes. Proof. Since D= D and dA = dA(i.e., dxdy = dudv ),thechange ofvariablesyields I = Df ( x,y ) dA = Df ( x ( u,v ) ,y ( u,v )) dA= Š Df ( u,v ) dA= Š I, thatis, I = Š I ,or I =0. 103.TripleIntegrals Supposeasolidregion E is“lledwithaninhomogeneousmaterial. Thelattermeansthat,ifasmallvolume V ofthematerialistaken attwodistinctpointsof E ,thenthemassesofthesetwopiecesare dierent,despitetheequalityoftheirvolumes.Theinhomogeneityof thematerialcanbecharacterizedbythe massdensity asafunctionof

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103.TRIPLEINTEGRALS235 position.Let m ( r )bethemassofasmallpieceofmaterialofvolume V cutoutaroundapoint r .Thenthemassdensityisde“nedby ( r )=lim V 0 m ( r ) V Thelimitisunderstoodinthefollowingsense.If R istheradiusof thesmallestballthatcontainstheregionofvolume V ,thenthe limitmeansthat R 0(i.e.,roughlyspeaking,allthedimensionsof thepiecedecreasesimultaneouslyinthelimit).Themassdensityis measuredinunitsofmassperunitvolume.Forexample,thevalue ( r )=5g / cm3meansthatapieceofmaterialofvolume1cm3cut outaroundthepoint r hasamassof5gr. Supposethatthemassdensityofthematerialinaregion E is known.Thequestionis:Whatisthetotalmassofthematerialin E ? Apracticalanswertothisquestionistopartitiontheregion E sothat eachpartitionelement Ep, p =1 2 ,...,N ,hasamass mp.Thetotal massis M = p mp.Ifapartitionelement Ephasavolume Vp, then mp ( rp) Vpforsome rp Ep.If Rpistheradiusofthe smallestballthatcontains Ep,put RN=max Rp.Then,byincreasing thenumber N ofpartitionelementssothat Rp RN 0as N theapproximation mp ( rp) Vpbecomesmoreandmoreaccurate bythede“nitionofthemassdensitybecause Vp 0forall p .Sothe totalmassis (14.12) M =limN ( RN 0) Np =1 ( rp) Vp, whichistobecomparedwith(14.1).Incontrastto(14.1),thesummationoverthepartitionshouldincludeatriplesum,onesumpereach directioninspace.Thisgivesanintuitiveideaofatripleintegral.Its abstractmathematicalconstructionfollowsexactlythefootstepsofthe double-integralconstruction.103.1.De“nitionofaTripleIntegral.RectangularPartition. Aregion E inspaceisassumedtobebounded; thatis,itiscontainedinaballofsome(“nite)radius.Theboundariesof E areassumedtobepiecewise-smoothsurfaces.Asmooth surfacecanbeviewedasalevelsurfaceofadierentiablefunction ofthreevariables.Theregion E isthenembeddedinarectangle RE=[ a,b ] [ c,d ] [ s,q ],thatis, x [ a,b ], y [ c,d ],and z [ s,q ].If f ( r )isaboundedfunctionon E ,thenitisextendedto REbysetting itsvaluesto0outside E .Therectangle REispartitionedbythecoordinateplanes x = xi= a + i x i =0 1 ,...,N1,where x =( b Š a ) /N1;

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23614.MULTIPLEINTEGRALS y = yj= c + j y i =0 1 ,...,N2,where y =( d Š c ) /N2;and z = zi= s + k z k =0 1 ,...,N3,where z =( q Š s ) /N3.Thevolumeof eachpartitionelementisarectangle Rijkofvolume V = x y z Thetotalnumberofrectanglesis N = N1N2N3. UpperandLowerSums .ByanalogywithDe“nition14.2,the loweranduppersumsarede“ned.Put Mijk=sup f ( r ),and mijk= inf f ( r )wherethesupremumandin“mumaretakenoverthepartition rectangle Rijk.Thentheupperandlowersumsare U ( f, N )=N1i =1 N2j =1 N3k =1Mijk V,L ( f, N )=N1i =1 N2j =1 N3k =1mijk V, where N =( N1,N2,N3). Definition 14.9 (TripleIntegral) Ifthelimitsoftheupperandlowersumsexistas N1 2 3 (or ( x, y, z ) (0 0 0) )andcoincide,then f issaidtobeRiemann integrableon E ,andthelimitoftheupperandlowersums Ef ( x,y,z ) dV =limN U ( f, N )=limN L ( f, N ) iscalledthe tripleintegral of f overtheregion E Thelimitisunderstoodasathree-variablelimit( x, y, z ) (0 0 0).103.2.PropertiesofTripleIntegrals.Thepropertiesoftripleintegrals arethesameasthoseofthedoubleintegraldiscussedinSection98; thatis,thelinearity,additivity,positivity,integrabilityoftheabsolute value | f | ,andupperandlowerboundsholdsfortripleintegrals. ContinuityandIntegrability .Therelationbetweencontinuity andintegrabilityisprettymuchthesameasinthecaseofdouble integrals. Theorem 14.8 (IntegrabilityofContinuousFunctions) Let E beaclosed,boundedspatialregionwhoseboundariesarepiecewisesmoothsurfaces.Ifafunction f iscontinuouson E ,thenitisintegrable on E .Furthermore,if f hasboundeddiscontinuitiesonlyona“nite numberofsmoothsurfacesin E ,thenitisalsointegrableon E Inparticular,aconstantfunctionisintegrable,andthevolumeof aregion E isgivenbythetripleintegral V ( E )= EdV.

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103.TRIPLEINTEGRALS237 TheIntegralMean-ValueTheorem. Theintegralmeanvalue theorem(Theorem14.3)isextendedtotripleintegrals,wherethedoubleintegralisreplacedbythetripleintegraland A ( D )bythevolume V ( E ).Itsprooffollowsthesamelinesasinthecaseofdoubleintegrals. RiemannSums .Ifafunction f isintegrable,thenitstripleintegralisthelimitofaRiemannsum,anditsvalueisindependentofthe partitionof E andachoiceofsamplepointsinthepartitionelements, thatis,(14.12)holds: (14.13) Ef ( r ) dV =limN ( RN 0) Np =1f ( rp) Vp. Thisequationcanbeusedforapproximationsoftripleintegrals,when evaluatingthelatternumericallyjustlikeinthecaseofdoubleintegrals. Symmetry .Ifatransformationinspacepreservesthevolumeof anyregion,thenitiscalled volumepreserving .Obviously,rotations, re”ections,andtranslationsinspacearevolume-preservingtransformations.Supposethat,underavolume-preservingtransformation,a region E ismappedontoitself;thatis, E is symmetric relativetothis transformation.If rs E istheimageof r E underthistransformationandtheintegrandisskew-symmetric, f ( rs)= Š f ( r ),thenthe tripleintegralof f over E vanishes. Example 14.17 Evaluatethetripleintegralof f ( x,y,z )= x2sin( y4z )+2 overaballcenteredattheoriginofradius R Solution: Put g ( x,y,z )= x2sin( y4z )sothat f = g + h ,where h =2isaconstantfunction.Bythelinearityproperty,thetriple integralof f isthesumoftripleintegralsof g and h overtheball.The ballissymmetricrelativetothere”ectiontransformation( x,y,z ) ( x,y, Š z ),whereasthefunction g isskew-symmetric, g ( x,y, Š z )= Š g ( x,y,z ).Therefore,itstripleintegralvanishes,and EfdV = EgdV + EhdV =0+2 EdV =2 V ( E )=8 R3/ 3 Onecanthinkofthenumericalvalueofatripleintegralof f over E asthetotalamountofaquantitydistributedintheregion E withthe density f (theamountofthequantityperunitvolume).Forexample, f canbeviewedasthedensityofelectricchargedistributedinadielectric occupyingaregion E .Thetotalelectricchargestoredintheregion E is

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23814.MULTIPLEINTEGRALS thengivenbytripleintegralofthedensityover E .Theelectriccharge canbepositiveandnegative.So,ifthetotalpositivechargein E is exactlythesameasthenegativecharge,thetripleintegralvanishes.103.3.IteratedTripleIntegrals.Similartoadoubleintegral,atriple integralcanbeconvertedtoatripleiteratedintegral,whichcanthen beevaluatedbymeansofordinarysingle-variableintegration. Definition 14.10 (SimpleRegion) Aspatialregion E issaidtobesimpleinthedirectionofavector v if anystraightlineparallelto v intersects E alongatmostonestraight linesegment. Atripleintegralcanbeconvertedtoaniteratedintegralif E is simpleinaparticulardirection.Ifthereisnosuchdirection,then E shouldbesplitintoaunionofsimpleregionswiththeconsequentuse oftheadditivitypropertyoftripleintegrals.Supposethat v = e3;that is, E issimplealongthe z axis.Thentheregion E admitsthefollowing description: E = { ( x,y,z ) | zbot( x,y ) z ztop( x,y ) ( x,y ) Dxy} Indeed,consideralllinesparalleltothe z axisthatintersect E .These linesalsointersectthe xy plane.Theregion Dxyinthe xy planeis thesetofallsuchpointsofintersection.Onemightthinkof Dxyas ashadowmadebythesolid E whenitisilluminatedbyraysoflight paralleltothe z axis.Takeanylinethrough( x,y ) Dxyparalleltothe z axis.Bythesimplicityof E ,anysuchlineintersects E alongasingle segment.If zbotand ztoparetheminimalandmaximalvaluesofthe z coordinatealongtheintersectionsegment,then,forany( x,y,z ) E zbot z ztopandany( x,y ) Dxy.Naturally,thevalues zbotand ztopmaydependon( x,y ) Dxy.Thus,theregion E isboundedfrom thetopbythegraph z = ztop( x,y )andfromthebottombythegraph z = zbot( x,y ).If E issimplealongthe y or x axis,then E admits similardescriptions: E = { ( x,y,z ) | ybot( x,z ) y ytop( x,z ) ( x,z ) Dxz} (14.14) E = { ( x,y,z ) | xbot( y,z ) x xtop( y,z ) ( y,z ) Dyz} (14.15) where Dxzand Dyzareprojectionsof E intothe xz and yz planes, respectively;theyarede“nedanalogouslyto Dxy. Accordingto(14.13),thelimitoftheRiemannsumisindependent ofpartitioning E andchoosingsamplepoints.Let Dp, p =1 2 ,...,N partitiontheregion Dxy.Consideraportion Epof E thatisprojected onthepartitionelement Dp; Episacolumnwith Dpitscrosssection

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103.TRIPLEINTEGRALS239 byahorizontalplane.Since E isbounded,therearenumbers s and q suchthat s zbot( x,y ) ztop( x,y ) q forall( x,y ) Dxy;that is, E alwaysliesbetweentwohorizontalplanes z = s and z = q Considerslicingthesolid E byequispacedhorizontalplanes z = s + k z k =0 1 ,...,N3, z =( q Š s ) /N3.Theneachcolumn Epis partitionedbytheseplanesintosmallregions Epk.Theunionofall Epkformsapartitionof E ,whichwillbeusedintheRiemannsum (14.13).Thevolumeof Epkis Vpk= z Ap,where Apisthearea of Dp.Assuming,asusual,that f isde“nedbyzerovaluesoutside E ,samplepointsmaybeselectedsothat,if( xp,yp, 0) Dp,then ( xp,yp,z k) Epk,thatis, zk Š 1 z k zkfor k =1 2 ,...,N3.Thethreevariablelimit(14.13)existsandhencecanbetakeninanyparticular order.Take“rstthelimit N3 or z 0.Thedoublelimitof thesumoverthepartitionof Dxyisunderstoodasbefore;thatis,as N 0,theradii Rpofsmallestdiskscontaining Dpgoto0uniformly, Rp RN 0.Therefore, EfdV =limN ( RN 0) Np =1# limN3 N3k =1f ( xp,yp,z k) z $ Ap=limN ( RN 0) Np =1# ztop( xp,yp) zbot( xp,yp)f ( xp,yp,z ) dz $ Apbecause,forevery( xp,yp) Dxy,thefunction f vanishesoutsidethe interval z [ zbot( xp,yp) ,ztop( xp,yp)].Theintegrationof f withrespect to z overtheinterval[ zbot( x,y ) ,ztop( x,y )]de“nesafunction F ( x,y ) whosevalues F ( xp,yp)atsamplepointsinthepartitionelements Dpappearintheparenthese.Acomparisonoftheresultingexpression with(14.3)leadstotheconclusionthat,aftertakingthesecondlimit, oneobtainsthedoubleintegralof F ( x,y )over Dxy. Theorem 14.9 (IteratedTripleIntegral) Let f beintegrableonasolidregion E .Supposethat E issimplein the z directionsothatitisboundedbythegraphs z = zbot( x,y ) and z = ztop( x,y ) for ( x,y ) Dxy.Then Ef ( x,y,z ) dV = Dxyztop( x,y ) zbot( x,y )f ( x,y,z ) dzdA = DxyF ( x,y ) dA.103.3.1.EvaluationofTripleIntegrals.Inpracticalterms,anevaluation ofatripleintegraloveraregion E iscarriedoutbythefollowingsteps:

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24014.MULTIPLEINTEGRALS Step1 .Determinethedirectionalongwhich E issimple.Ifnosuch directionexists,split E intoaunionofsimpleregionsandusethe additivityproperty.Forde“nitiveness,supposethat E happenstobe z simple. Step2 .Findtheprojection Dxyof E intothe xy plane. Step3 .Findthebottomandtopboundariesof E asthegraphsof somefunctions z = zbot( x,y )and z = ztop( x,y ). Step4 .Evaluatetheintegralof f withrespectto z toobtain F ( x,y ). Step5 .Evaluatethedoubleintegralof F ( x,y )over Dxybyconverting ittoasuitableiteratedintegral. Similariteratedintegralscanbewrittenwhen E issimpleinthe y or x direction.Accordingto(14.14)(or(14.15)),the“rstintegrationis carriedoutwithrespectto y (or x ),andthedoubleintegralisevaluated over Dxz(or Dyz).If E issimpleinanydirection,thenanyofthe iteratedintegralscanbeused.Inparticular,justlikeinthecaseof doubleintegrals,thechoiceofaniteratedintegralforasimpleregion E shouldbemotivatedbythesimplicityofanalgebraicdescriptionof thetopandbottomboundariesorbythesimplicityoftheintegrations involved.Technicaldicultiesmaystronglydependontheorderin whichtheiteratedintegralisevaluated. Fubinistheoremcanbeextendedtotripleintegrals. Theorem 14.10 (FubinisTheorem) Let f beintegrableonarectangularregion E =[ a,b ] [ c,d ] [ s,q ] Then EfdV = b ad cq sf ( x,y,z ) dzdydx andtheiteratedintegralcanbeevaluatedinanyorder. Here Dxy=[ a,b ] [ c,d ],andthetopandbottomboundariesarethe planes z = q and z = s .Alternatively,onecantake Dyz=[ c,d ] [ s,q ], xbot( y,z )= a ,and xtop( y,z )= b toobtainaniteratedintegralina dierentorder(wherethe x integrationiscarriedout“rst). Example 14.18 Evaluatethetripleintegralof f ( x,y,z )= xy2z3overtherectangle E =[0 2] [1 2] [0 3] Solution: ByFubinistheorem, Exy2z3dV = 2 0xdx 2 1y2dy 3 0z3dz =2 (7 / 3) 9=52 Thisexampleshowsthatthefactorizationpropertyalsoholdsfortriple integrals(seeCorollary14.3).

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103.TRIPLEINTEGRALS241 Example 14.19 Evaluatethetripleintegralof f ( x,y,z )=( x2+ y2) z overtheportionofthesolidboundedbythecone z = x2+ y2andparaboloid z =2 Š x2Š y2inthe“rstoctant. Solution: Followingthestep-by-stepprocedureoutlinedabove,the integrationregionis z simple.Thetopboundaryisthegraphof ztop( x,y )=2 Š x2Š y2,andthegraphof zbot( x,y )= x2+ y2is thebottomboundary.Todeterminetheregion Dxy,notethatithas tobeboundedbytheprojectionofthecurveoftheintersectionofthe coneandparaboloidontothe xy plane.Theintersectioncurveisde“nedby zbot= ztopor r =2 Š r2,where r = x2+ y2,andhence r =1, whichisthecircleofunitradius.Since E isinthe“rstoctant, Dxyis thequarterofthediskofunitradiusinthe“rstquadrant.Onehas E( x2+ y2) zdV = Dxy( x2+ y2) 2 Š x2Š y2 x2+ y2zdzdA = 1 2 Dxy( x2+ y2)[(2 Š x2Š y2)2Š ( x2+ y2)] dA = 1 2 / 2 0d 1 0r2[(2 Š r2)2Š r2] rdr = 8 1 0u [(2 Š u )2Š u ] du = 7 96 wherethedoubleintegralhasbetransformedintopolarcoordinates because Dxybecomestherectangle D xy=[0 1] [0 ,/ 2]inthepolar plane.Theintegrationwithrespectto r iscarriedoutbythechange ofvariable u = r2. Example 14.20 Evaluatethetripleintegralof f ( x,y,z )= y2+ z2overtheregion E boundedbytheparaboloid x = y2+ z2andtheplane x =4 Solution: Itisconvenienttochooseaniteratedintegralfor E describedasan x simpleregion(see(14.15)).Therearetworeasons fordoingso.First,theintegrand f isindependentof x ,andhence the“rstintegrationwithrespectto x istrivial.Second,theboundariesof E arealreadygivenintheformrequiredby(14.15),thatis, xbot( y,z )= y2+ z2and xtop( y,z )=4.Theregion Dyzisdetermined bythecurveofintersectionoftheboundariesof E xtop= xbotor y2+ z2=4.Therefore, Dyzisthediskorradius2.Onehas E y2+ z2dV = Dyz y2+ z24 y2+ z2dxdA

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24214.MULTIPLEINTEGRALS = Dyz y2+ z2[4 Š ( y2+ z2)] dA = 2 0d 2 0r [4 Š r2] rdr = 128 15 wherethedoubleintegralover Dyzhasbeenconvertedtopolarcoordinatesinthe yz plane. 103.4.StudyProblems.Problem14.2. Evaluatethetripleintegralof f ( x,y,z )= z overthe region E boundedbythecylinder y2+ z2=1 andtheplanes z =0 y =1 ,and y = x inthe“rstoctant. Solution: Theregionis z simpleandboundedbythe xy planefrom thebottom(i.e., zbot( x,y )=0),andbythecylinderfromthetop(i.e., ztop( x,y )= 1 Š y2)(bytakingthepositivesolutionof y2+ z2=1). Theregion Dxyisboundedbythelinesofintersectionoftheplanes x =0and y = x andthecylinder y2+ z2=1withthe xy plane(orthe plane z =0).Thecylinderintersectsthisplanealongtheline y =1 inthe“rstquadrant.Thus, Dxyisthetriangleboundedbythelines x =0, y =0,and y = x .Onehas EzdV = Dxy 1 Š y20zdzdA = 1 2 Dxy(1 Š y2) dA = 1 2 1 0(1 Š y2) y 0dxdy = 1 8 wherethedoubleintegralhasbeenevaluatedbyusingthedescription of Dxyasahorizontallysimpleregion, xbot=0 x y = xtopforall y [0 1]=[ c,d ]. Problem14.3. Evaluatethetripleintegralofthefunction f ( x,y,z )= xy2z3overtheregion E thatisaballofradius 3 centeredattheorigin withacubiccavity [0 1] [0 1] [0 1] Solution: Theregion E isnotsimpleinanydirection.Theadditivity propertymustbeused.Let E1betheballandlet E2bethecavity. Bytheadditivityproperty, Exy2z3dV = E1xy2z3dV Š E2xy2z3dV =0 Š 1 0xdx 1 0y2dy 1 0z3dz = Š 1 12 .

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104.TRIPLEINTEGRALSINCYLINDRICALCOORDINATES243 Thetripleintegralover E1vanishesbythesymmetryargument(the ballissymmetricunderthere”ection( Š x,y,z ) ( Š x,y,z )whereas f ( Š x,y,z )= Š f ( x,y,z )).ThesecondintegralisevaluatedbyFubinis theorem. 104.TripleIntegralsinCylindricalandSphericalCoordinates Achangeofvariableshasbeenprovedtobequiteusefulinsimplifyingthetechnicalitiesinvolvedinevaluatingdoubleintegrals.An essentialadvantageisasimpli“cationoftheintegrationregion.The conceptofchangingvariablescanbeextendedtotripleintegrals.104.1.CylindricalCoordinates.Oneofthesimplestexamplesofcurvilinearcoordinatesinspaceiscylindricalcoordinates.Theyarede“nedby (14.16) x = r cos ,y = r cos ,z = z. Inanyplaneparalleltothe xy plane,thepointsarelabeledbypolarcoordinates,whilethe z coordinateisnottransformed.Equations (14.16)de“neatransformationofanorderedtripleofnumbers( r,,z ) toanotherorderedtriple( x,y,z ).Asetoftriples( r,,z )canbeviewed asasetofpoints EinaEuclideanspaceinwhichthecoordinateaxes arespannedby r ,and z .Then,underthetransformation(14.16), theregion Eismappedontoan image region E .Fromthestudyofpolarcoordinates,thetransformation(14.16)isone-to-oneif r (0 ), [0 2 ),and z ( Š ).Theinversetransformationisgivenby r = x2+ y2, =tanŠ 1( y/x ) ,z = z, wherethevalueoftanŠ 1istakenaccordingtothequadrantinwhich thepair( x,y )belong(seethediscussionofpolarcoordinates).Itmaps anyregion E intheEuclideanspacespannedby( x,y,z )ontotheimage region E.To“ndtheshapeof Easwellasitsalgebraicdescription, thesamestrategyasinthetwo-variablecaseshouldbeused: boundariesof E boundariesof Eunderthetransformation(14.16)anditsinverse.Aparticularlyimportantquestionishowtoinvestigatetheshapeof coordinatesurfaces ofcylindricalcoordinates,thatis,surfacesonwhicheachofthecylindricalcoordinateshasaconstantvalue.If E isboundedbycoordinate surfacesonly,thenitsimage Eisarectangle,whichisthesimplest, mostdesirable,shapewhenevaluatingamultipleintegral. Thecoordinatesurfacesof r arecylinders, r = x2+ y2= r0or x2+ y2= r2 0.Inthe xy plane,theequation = 0de“nesarayfromthe

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24414.MULTIPLEINTEGRALS originattheangle 0tothepositive x axiscountedcounterclockwise. Since dependsonlyon x and y ,thecoordinatesurfaceof isthe half-planeboundedbythe z axisthatmakesanangle 0withthe xz plane(itissweptbytheraywhenthelatterismovedparallelupand downalongthe z axis).Sincethe z coordinateisnotchanged,neither changesitscoordinatesurfaces;theyareplanesparalleltothe xy plane. Sothecoordinatesurfacesofcylindricalcoordinatesare r = r0 x2+ y2= r2 0(cylinder) = 0 y cos 0= x sin 0(half-plane) z = z0 z = z0(plane) Apointinspacecorrespondingtoanorderedtriple( r0,0,z0)isan intersectionpointofacylinder,half-planeboundedbythecylinder axis,andaplaneperpendiculartothecylinderaxis. Example 14.21 Findtheimage Eofthesolidregion E thatis boundedbytheparaboloid z = x2+ y2andtheplanes z =4 y = x ,and y =0 inthe“rstoctant. Solution: Incylindricalcoordinates,theequationsofboundariesbecome,respectively, z = r2, z =4, = / 4,and =0.Since E lies belowtheplane z =4andabovetheparaboloid z = r2,therange of r isdeterminedbytheirintersection4= r2or r =2as r 0. Thus, E= ( r,,z ) | r2 z 4 ( r, ) [0 2] [0 ,/ 4] 104.1.1.TripleIntegralsinCylindricalCoordinates.Tochangevariables inatripleintegraltocylindricalcoordinates,onehastoconsidera partitionoftheintegrationregion E by coordinatesurfaces ,thatis,by cylinders,half-planes,andhorizontalplanes,whichcorrespondstoa rectangularpartitionof E(theimageof E underthetransformation fromrectangulartocylindricalcoordinates).Thenthelimitofthe correspondingRiemannsum(14.13)hastobeevaluated.Inthecase ofcylindricalcoordinates,thistaskcanbeaccomplishedbysimpler means. Suppose E is z simplesothatbyTheorem14.9thetripleintegral canbewrittenasaniteratedintegralconsistingofadoubleintegral over Dxyandanordinaryintegralwithrespectto z .Thetransformation(14.16)merelyde“nespolarcoordinatesintheregion Dxy.So, if D xyistheimageof Dxyinthepolarplanespannedbypairs( r, ),

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104.TRIPLEINTEGRALSINCYLINDRICALCOORDINATES245 then,byconvertingthedoubleintegraltopolarcoordinate,soneinfers that Ef ( x,y,z ) dV = D xyztop( r, ) zbot( r, )f ( r cos ,r sin ,z ) rdzdA= Ef ( r cos ,r sin ,z ) rdV, (14.17) wheretheregion Eistheimageof E underthetransformationfrom rectangulartocylindricalcoordinates, E= { ( r,,z ) | zbot( r, ) z ztop( r, ) ( r, ) D xy} and z = zbot( r, ), z = ztop( r, )areequationsofthebottomandtop boundariesof E writteninpolarcoordinatesbysubstituting(14.16) intotheequationsforboundarieswritteninrectangularcoordinates. Notethat dV= dzdrd = dzdAisthevolumeofanin“nitesimal rectangleinthespacespannedbythetriples( r,,z ).Itsimageinthe spacespannedby( x,y,z )liesbetweentwocylinderswhoseradiidier by dr ,betweentwohalf-planeswiththeangle d betweenthem,and betweentwohorizontalplanesseparatedbythedistance dz .Soits volumeistheproductofthearea dA ofthebaseandtheheight dz dV = dzdA = rdzdAaccordingtotheareatransformationlawfor polarcoordinates, dA = rdA.Sothevolumetransformationlawfor cylindricalcoordinatesreads dV = JdV,J = r, where J = r istheJacobianoftransformationtocylindricalcoordinates. Cylindricalcoordinatesareadvantageouswhentheboundariesof E containcylinders,half-planes,horizontalplanes,oranysurfaceswith axialsymmetry .Asetinspaceissaidtobe axiallysymmetric ifthere isanaxissuchthatanyrotationaboutitmapsthesetontoitself.For example,circularcones,circularparaboloids,andspheresareaxially symmetric.Notealsothattheaxisofcylindricalcoordinatesmaybe chosentobethe x or y axis,whichwouldcorrespondtopolarcoordinatesinthe yz or xz plane. Example 14.22 Evaluatethetripleintegralof f ( x,y,z )= x2z overtheregion E boundedbythecylinder x2+ y2=1 ,theparaboloid z = x2+ y2,andtheplane z =0 Solution: Thesolid E isaxiallysymmetricbecauseitisboundedfrom thebottombytheplane z =0,bythecircularparaboloidfromthe top,andthesideboundaryisthecylinder.Hence, Dxyisadiskof

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24614.MULTIPLEINTEGRALS unitradius,and D xyisarectangle,( r, ) [0 1] [0 2 ].Thetop andbottomboundariesare z = ztop( r, )= r2and z = zbot( r, )=0. Hence, Ex2zdV = 2 01 0r20r2cos2zrdzdrd = 1 2 2 0cos2d 1 0r7dr = 16 wherethedouble-angleformula,cos2 =(1+cos(2 )) / 2,hasbeenused toevaluatetheintegral. 104.2.SphericalCoordinates.Sphericalcoordinatesareintroducedby thefollowinggeometricalprocedure.Let( x,y,z )beapointinspace. Considerarayfromtheoriginthroughthispoint.Anysuchraylies inthehalf-planecorrespondingtoa“xedvalueofthepolarangle Therefore,therayisuniquelydeterminedbythepolarangle andthe angle betweentherayandthepositive z axis.If isthedistance fromtheorigintothepoint( x,y,z ),thentheorderedtripleofnumbers ( ,, )de“nesuniquelyanypointinspace.Thetriples( ,, )are called sphericalcoordinates inspace. To“ndthetransformationlawfromsphericaltorectangularcoordinates,considertheplanethatcontainsthe z axisandtherayfromthe originthrough( x,y,z )andtherectanglewithvertices(0 0 0),(0 0 ,z ), ( x,y, 0),and( x,y,z )inthisplane.Thediagonalofthisrectanglehas length (thedistancebetween(0 0 0)and( x,y,z )).Therefore,its verticalsidehaslength z = cos becausetheanglebetweenthisside andthediagonalis .Itshorizontalsidehaslength sin .Onthe otherhand,itisalsothedistancebetween(0 0 0)and( x,y, 0),that is, r = sin ,where r = x2+ y2.Since x = r cos and y = r sin itisconcludedthat (14.18) x = sin cos ,y = sin sin ,z = cos Theinversetransformationfollowsfromthegeometricalinterpretation ofthesphericalcoordinates: (14.19) = x2+ y2+ z2, cot = z r = z x2+ y2, tan = y x If( x,y,z )spantheentirespace,themaximalrangeofthevariable is thehalf-axis [0 ).Thevariable rangesovertheinterval[0 2 ) asitcoincideswiththepolarangle.Todeterminetherangeofthe azimuthal angle ,notethatananglebetweenthepositive z axisand anyrayfromtheoriginmustbeintheinterval[0 ].If =0,theray

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104.TRIPLEINTEGRALSINSPHERICALCOORDINATES247 coincideswiththepositive z axis.If = ,therayisthenegative z axis.Anyraywith = / 2liesinthe xy plane.104.2.1.CoordinateSurfacesofSphericalCoordinates.Allpointsthat havethesamevalueof = 0formasphereofradius 0centeredat theoriginbecausetheyareatthesamedistance 0fromtheorigin. Naturally,thecoordinatesurfacesof arethehalf-planesdescribed earlierwhendiscussingcylindricalcoordinates.Considerarayfrom theoriginthathastheangle = 0withthepositive z axis.By rotatingthisrayaboutthe z axis,allrayswiththe“xedvalueof areobtained.Therefore,thecoordinatesurface = 0isacircular conewhoseaxisisthe z axis.Forsmallvaluesof ,theconeisa narrowconeaboutthepositive z axis.Theconebecomeswideras increasessothatitcoincideswiththe xy planewhen = / 2.For >/ 2,theconeliesbelowthe xy plane,anditeventuallycollapses intothenegative z axisassoonas reachesthevalue .Thealgebraic equationsofthecoordinatesurfacesfollowfrom(14.19): = 0 x2+ y2+ z2= 2 0(sphere) = 0 z =cot( 0) x2+ y2(cone) = 0 y cos 0= x sin 0(half-plane) Soanypointinspacecanbeviewedasthepointofintersectionof threecoordinatesurfaces:thesphere,cone,andhalf-plane.Underthe transformation(14.19),anyregion E ismappedontoaregion Ein thespacespannedbytheorderedtriples( ,, ).If E isboundedby spheres,cones,andhalf-planesonly,thenitsimage Eisarectangle. Thus,achangeofvariablesinatripleintegraltosphericalcoordinates isadvantageouswhen E isboundedbyspheres,cones,andhalf-planes. Example 14.23 Let E betheportionofthesolidboundedbythe sphere x2+ y2+ z2=4 andthecone z2=3( x2+ y2) thatliesin the“rstoctant.Finditsimage Eundertransformationtospherical coordinates. Solution: Theregion E hasfourboundaries:thesphere,thecone z = 3 x2+ y2,the xz plane( x 0),andthe yz plane( y 0). Theseboundariesaremappedonto,respectively, =2,cot = 3 or = / 3, =0,and = / 2.So Eistherectangle[0 2] [0 ,/ 3] [0 ,/ 2].Theregion E isintersectedbyallsphereswithradii 0 2,allconeswithangles0 / 3,andallhalf-planeswith angles0 / 2.

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24814.MULTIPLEINTEGRALS 104.2.2.VolumeTransformationLaw.Let Ebetheimageofaregion E underthetransformationtosphericalcoordinates(14.19).Consider arectangularpartitionof Ebyequispacedplanes = i, = j,and = ksuchthat i +1Š i= j +1Š j= ,and k +1Š k= where ,and aresmallnumbersthatcanberegardedas dierentials(orin“nitesimalvariations)ofthesphericalcoordinates. Eachpartitionrectanglehasvolume V= .Therectangularpartitionof Einducesapartitionof E byspheres,cones,and half-planes.Eachpartitionelementisboundedbytwosphereswhose radiidierby ,bytwoconeswhoseanglesdierby ,andbytwo half-planestheanglebetweenwhichis .Thevolumeofanysuch partitionelementcanbewrittenas V = J Vbecauseonlytermslinearinthevariations = d = d ,and = d havetoberetained.Thevalueof J dependsonapartitionelement(e.g.,partitionelementsclosertotheoriginshouldhave smallervolumesbythegeometryofthepartition).Thefunction J is the Jacobian forsphericalcoordinates. Bymeansof(14.18),anintegrablefunction f ( x,y,z )canbewritteninsphericalcoordinates.Accordingto(14.13),inthethree-variable limit( ) (0 0 0),theRiemannsumfor f forthepartition constructedconvergestoatripleintegralof fJ expressedinthevariables( ,, )overtheregion Eandtherebyde“nesthetripleintegral of f over E insphericalcoordinates. To“nd J ,considertheimageoftherectangle [ 0,0+ ], [ 0,0+ ], [ 0,0+ ]underthetransformation(14.18). Sinceitliesbetweentwospheresofradii 0and 0+ ,itsvolume canbewrittenas V = A ,where A istheareaoftheportion ofthesphereofradius 0thatliesbetweentwoconesandtwohalfplanes.Anyhalf-plane = 0intersectsthesphere = 0alonga half-circleofradius 0.Thearclengthoftheportionofthiscircle thatliesbetweenthetwocones = 0and = 0+ istherefore a = 0 .Thecone = 0intersectsthesphere = 0alonga circleofradius r0= 0sin 0(seethetextabove(14.18)).Hence,the arclengthoftheportionofthiscircleofintersectionthatliesbetween thehalf-planes = 0and = 0+ is b = r0 = 0sin Thearea A canbeapproximatedbytheareaofarectanglewith adjacentsides a and b .Sinceonlytermslinearin and are toberetained,onecanwrite A = a b = 2 0sin 0 .Thus, thevolumetransformationlawreads dV = JdV,J = 2sin .

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104.TRIPLEINTEGRALSINSPHERICALCOORDINATES249 BythecontinuityoftheJacobian,thedierenceofthevaluesof J atanytwosamplepointsinapartitionrectanglein Evanishesin thelimit( ) (0 0 0);thatis,thevalueoftheJacobianin V = J Vcanbetakenatanypointwithinthepartitionelement whenevaluatingthelimit.Therefore,foranychoiceofsamplepoints, thelimitoftheRiemannsum(14.13)fortheconstructedpartitionis Ef ( x,y,z ) dV = Ef sin cos sin sin cos 2sin dV. Thisrelationde“nesthetripleintegralof f over E insphericalcoordinates.Thetripleintegralover Ehastobeevaluatedbyconverting ittoasuitableiteratedintegral. Example 14.24 Findthevolumeofthesolid E boundedbythe sphere x2+ y2+ z2=2 z andthecone z = x2+ y2. Solution: Bycompletingthesquares,theequation x2+ y2+ z2=2 z iswritteninthestandardform x2+ y2+( z Š 1)2=1,whichdescribes asphereofunitradiuscenteredat(0 0 1).So E isboundedfromthe topbythissphere,whilethebottomboundaryof E isthecone,and E hasnootherboundaries.Insphericalcoordinates,thetopboundary becomes 2=2 cos or =2cos .Thebottomboundaryis = / 4. Theboundariesof E imposenorestrictionon ,whichcantherefore betakenoveritsfullrange.Hence,theimage Eadmitsthefollowing algebraicdescription: E= ( ,, ) | 0 2cos ( ) [0 ,/ 4] [0 2 ] Sincetherangeof dependsontheothervariables,theintegrationwith respecttoitmustbecarriedout“rstwhenconvertingthetripleintegral over Eintoaniteratedintegral( Eis simple,andtheprojectionof Eontothe planeistherectangle[0 ,/ 4] [0 2 ]).Theorderin whichtheintegrationwithrespectto and iscarriedoutisirrelevant becausetheangularvariablesrangeoverarectangle.Onehas V ( E )= EdV = E2sin dV= 2 0/ 4 0sin 2cos 02ddd = 8 3 2 0d / 4 0cos3 sin d = 16 3 1 1 / 2u3du = wherethechangeofvariables u =cos hasbeencarriedoutinthelast integral.

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25014.MULTIPLEINTEGRALS 105.ChangeofVariablesinTripleIntegrals Considerthetransformationofanopenregion Einspaceinto anopenregion E de“nedby x = x ( u,v,w ), y = y ( u,v,z ),and z = z ( u,v,w );thatis,foreverypoint( u,v,w ) E,thesefunctionsde“ne animagepoint( x,y,z ) E .Ifnotwopointsin Ehavethesame imagepoint,thetransformationis one-to-one ,andthereisa one-to-one correspondence betweenpointsof E and E.Theinversetransformation existsandisde“nedbythefunctions u = u ( x,y,z ), v = v ( x,y,z ),and w = w ( x,y,z ).Apoint( x0,y0,z0)= r0istheintersectionpointof threecoordinateplanes x = x0, y = y0,and z = z0.Alternatively, itcanalsobeviewedasthepointofintersectionofthree coordinate surfaces u ( x,y,z )= u0, v ( x,y,z )= v0,and w ( x,y,z )= w0,wherethe imageof( u0,v0,w0)undertheone-to-onetransformationis r0. Definition 14.11 (JacobianofaMapping) Supposethataone-to-onemappingofanopenset Eonto E hascontinuous“rst-orderpartialderivatives.Thequantity ( x,y,z ) ( u,v,w ) =det x uy uz ux vy vz vx wy wz w iscalledtheJacobianofthemapping. Definition 14.12 (ChangeofVariables) Acontinuouslydierentiableone-to-onemappingofanopenset Eonto E iscalleda changeofvariables (orachangeofcoordinates)if theJacobianofthemappingdoesnotvanishin E. Asinthecaseofdoubleintegrals,achangeofvariablesinspace canbeusedtosimplifytheevaluationoftripleintegrals.Forexample,ifthereisachangeofvariableswhosecoordinatesurfacesforma boundaryoftheintegrationregion E ,thenthenewintegrationregion Eisarectangle,andthelimitsinthecorrespondingiteratedintegral aregreatlysimpli“edinaccordancewithFubinistheorem.105.1.TheVolumeTransformationLaw.Itisconvenienttointroduce thefollowingnotations:( u,v,w )= rand( x,y,z )= r ;thatis,under thechangeofvariables, (14.20) r = x ( r) ,y ( r) ,z ( r) or r= u ( r ) ,v ( r ) ,w ( r ) Let E 0beanin“nitesimalrectanglein E, u [ u0,u0+ u ], v [ v0,v0+ v ],and w [ w0,w0+ w ],where u v ,and w are in“nitesimalvariationsthatcanbeviewedasdierentialsofthenew

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105.CHANGEOFVARIABLESINTRIPLEINTEGRALS251 variables.Thismeansthatallalgebraicexpressionsinvolving u v and w aretobelinearizedwithrespecttothem,andtheirhigher powersneglected.If E0istheimageof E 0,thevolumesof E0and E 0areproportional: V = J V, V= u v w. Theobjectiveistocalculate J .Bytheexamplesofcylindricaland sphericalcoordinates, J isafunctionofthepoint( u0,v0,w0)atwhich therectangle E 0istaken.Thederivationof J isfullyanalogoustothe two-variablecase. Let O, A, B,and Chavethecoordinates,respectively, r 0=( u0,v0,w0) r a=( u0+ u,v0,w0)= r 0+ e1 u, r b=( u0,v0+ v,w0)= r 0+ e2 v, r c=( u0,v0,w0+ w )= r 0+ e3 w, where e1 2 3areunitvectorsalongthe“rst,second,andthirdcoordinateaxes.Inotherwords,thesegments OA, OB,and OCarethe adjacentsidesoftherectangle E 0.Let O A B ,and C betheimages of O, A, B,and Cintheregion E .Thevolume V of E0canbe approximatedbythevolumeoftheparallelepipedwithadjacentsides a = OA b = OB ,and c = OC .Then a = x ( r a) Š x ( r 0) ,y ( r a) Š y ( r 0) ,z ( r a) Š z ( r 0) =( x u,y u,z u) u, b = x ( r b) Š x ( r 0) ,y ( r b) Š y ( r 0) ,z ( r b) Š z ( r 0) =( x v,y v,z v) v, c = x ( r c) Š x ( r 0) ,y ( r c) Š y ( r 0) ,z ( r c) Š z ( r 0) =( x w,y w,z w) w, whereallthedierenceshavebeenlinearized,forinstance, x ( r a) Š x ( r 0)= x ( r 0+ e1 u ) Š x ( r 0)= x u( r 0) u ,bythede“nitionofthe partialderivativeof x ( u,v,w )withrespecttothe“rstvariable u .The volumeoftheparallelepipedisgivenbytheabsolutevalueofthetriple product: (14.21) V = | a ( b c ) | = det x uy uz ux vy vz vx wy wz w u v w = J V, wherethederivativesareevaluatedat( u0,v0,w0).Thefunction J in (14.21)isthe absolutevalue oftheJacobian.The“rst-orderpartial derivativesarecontinuousforachangeofvariablesandsoarethe

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25214.MULTIPLEINTEGRALS Jacobiananditsabsolutevalue.Similarlytothetwo-dimensionalcase, itcanalsobeprovedthat (14.22) J = ( x,y,z ) ( u,v,w ) = 1 ( u,v,w ) ( x,y,z ) = det u xu yu zv xv yv zw xw yw z Š 1. Thisexpressionde“nes J asafunctionoftheoldvariables( x,y,z ).105.2.TripleIntegralinCurvilinearCoordinates.Considerapartitionof Ebyequispacedplanes u = ui, v = vj,and w = wkui +1Š ui= u vj +1Š vj= v ,and wk +1Š wk= w .Theindices( i,j,k )enumerate planesthatintersect E.Thisrectangularpartitionof Ecorresponds toapartitionof E bythecoordinatesurfaces u ( r )= ui, v ( r )= vj, and w ( r )= wk.If E ijkistherectangle u [ ui,ui +1], vj [ vj,vj +1], and w [ wk,wk +1],thenitsimage,beingthecorrespondingpartition elementof E ,isdenotedby Eijk.ARiemannsumcanbeconstructed forthispartitionof E (assumingasbeforethat f isde“nedbyzeros outside E ).Thetripleintegralof f over E isthelimit(14.13)whichis understoodasthethree-variablelimit( u, v, w ) (0 0 0).The volume Vijkof Eijkisrelatedtothevolumeoftherectangle E ijkby (14.21).Bycontinuityof J ,itsvaluein(14.21)canbetakenatany samplepointin E ijk.Accordingtothede“nitionofthetripleintegral, thelimitoftheRiemannsumisthetripleintegralof fJ overthe region E. Theorem 14.11 (ChangeofVariablesinaTripleIntegral) Letacontinuouslydierentiablemapping E E haveanon-vanishing Jacobian,exceptperhapsontheboundaryof E.Supposethat f is continuouson E and E isboundedbypiecewise-smoothsurfaces.Then Ef ( r ) dV = Ef ( x ( r) ,y ( r) ,z ( r)) J ( r) dV, J ( r)= ( x,y,z ) ( u,v,w ) Evaluationofatripleintegralincurvilinearcoordinatesfollowsthe samestepsasforadoubleintegralincurvilinearcoordinates. Example 14.25 (VolumeofanEllipsoid) Findthevolumeofasolidregion E boundedbyanellipsoid x2/a2+ y2/b2+ z2/c2=1 Solution: Theintegrationdomaincanbesimpli“edbyascaling transformation x = au y = bv ,and z = cw underwhichtheellipsoid

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105.CHANGEOFVARIABLESINTRIPLEINTEGRALS253 ismappedontoasphereofunitradius u2+ v2+ w2=1.Theimage Eof E isaballofunitradius.TheJacobianofthistransformationis J = det a 00 0 b 0 00 c = abc. Therefore, V ( E )= EdV = EJdV= abc EdV= abcV ( E)= 4 3 abc. When a = b = c = R ,theellipsoidbecomesaballofradius R ,anda familiarexpressionforthevolumeisrecovered: V =(4 / 3) R3.105.3.StudyProblems.Problem14.4.(VolumeofaTetrahedron) Atetrahedronisasolidwithfourverticesandfourtriangularfaces. Letthevectors a b ,and c bethreeadjacentsidesofthetetrahedron. Finditsvolume. Solution: Consider“rstatetrahedronwhoseadjacentsidesarealong thecoordinateaxesandhavethesamelength q .Fromthegeometry,it isclearthatsixsuchtetrahedronsformacubeofvolume q3.Therefore, thevolumeofeachtetrahedronis q3/ 6(ifsodesiredthiscanalsobe establishedbyevaluatingthecorrespondingtripleintegral;thisisleft tothereader).Theideaistomakeachangeofvariablessuchthata generictetrahedronismappedontoatetrahedronwhoseadjacentfaces lieinthethreecoordinateplanes.Theadjacentfacesareportionsof theplanesthroughtheorigin.Thefacecontainingvectors a and b is perpendiculartovector n = a b sotheequationofthisboundaryis n r =0.Theotheradjacentfacesaresimilar: n r =0or n1x + n2y + n3z =0 n = a b l r =0or l1x + l2y + l3z =0 l = c a m r =0or m1x + m2y + m3z =0 m = b c where r =( x,y,z ).So,byputting u = m r v = l r ,and w = n r theimagesoftheseplanesbecomethecoordinateplanes, w =0, v =0, and u =0.Alinearequationintheoldvariablesbecomesalinear equationinthenewvariablesunderalineartransformation.Therefore, animageofaplaneisaplane.Sothefourthboundaryof Eisaplane

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25414.MULTIPLEINTEGRALS throughthepoints a, b,and c,whicharetheimagesof r = a r = b and r = c ,respectively.Onehas a=( u ( a ) ,v ( a ) ,w ( a ))=( q, 0 0), where q = a m = a ( b c )because a n =0and a l =0bythe geometricalpropertiesofthecrossproduct.Similarly, b=(0 ,q, 0) and c=(0 0 ,q ).Thus,thevolumeoftheimageregion Eis V ( E)= | q |3/ 6(theabsolutevalueisneededbecausethetripleproductcanbe negative).To“ndthevolume V ( E ),theJacobianofthetransformation hastobefound.Itisconvenienttousetherepresentation(14.22): J = det m1m2m3l1l2l3n1n2n3 Š 1= 1 | m ( n l ) | Therefore, V ( E )= EdV = EJdV= J EdV= JV ( E)= | q |3J 6 Thevolume V ( E )isindependentoftheorientationofthecoordinate axes.Itisconvenienttodirectthe x axisalongthevector a .The y axis isdirectedsothat b isinthe xy plane.Withthischoice, a =( a1, 0 0), b =( b1,b2, 0),and c =( c1,c2,c3).Astraightforwardcalculationshows that q = a1b2c3and J =( a2 1b2 2c2 3)Š 1.Hence, V ( E )= | a1b2c3| / 6.Finally, notethat | c3| = h istheheightofthetetrahedron,thatis,thedistance fromavertex c totheoppositeface(tothe xy plane).Theareaofthat faceis A = a b / 2= | a1b2| / 2.Thus, V ( E )= 1 3 hA ; thatis,thevolumeofatetrahedronisone-thirdthedistancefroma vertextotheoppositeface,timestheareaofthatface. 106.ImproperMultipleIntegrals Inthecaseofone-variableintegration,improperintegralsoccur whentheintegrandisnotde“nedataboundarypointoftheintegration intervalortheintegrationintervalisnotbounded.Forexample, (14.23) 1 0dx x=lima 01 adx x=lima 01 Š a1 Š 1 Š = 1 1 Š ,< 1 or 01 1+ x2dx =lima a 01 1+ x2dx =lima tanŠ 1a = 2 Impropermultipleintegralsarequitecommoninmanypracticalapplications.

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106.IMPROPERMULTIPLEINTEGRALS255 106.1.MultipleIntegralsofUnboundedFunctions.Supposeafunction f ( r )isnotde“nedatapoint r0thatisalimitpointofthedomainof f (anyneighborhoodof r0containspointsofthedomainof f ).Here r =( x,y,z ) E or r =( x,y ) D .Ifinanysmallball(ordisk) Bof radius centeredat r0thevaluesof | f ( r ) | arenotbounded,thenthe function f issaidtobe singular at r0.Inthiscase,theupperandlower sumscannotbede“nedbecauseforsomepartitionrectanglessup f or inf f orbothdonotexist,andneitherisde“nedamultipleintegralof f .Ifaregion E (or D )containssuchapoint,de“netheregion E(or D)byremovingallpointsof E (or D )thatalsolieintheball(disk) B.Supposethat f isintegrableon E(or D)forany > 0(e.g.,itis continuous).Then,byanalogywiththeone-variablecase,amultiple integralof f over E (or D )is de“ned asthelimit (14.24) EfdV =lim 0EfdV or DfdA =lim 0DfdA, provided,ofcourse,thelimitexists.If f issingularinapointset S thenonecanconstructaset Sthatistheunionofballsofradius centeredateachpointof S .Then D(or E)isobtainedbyremoving Sfrom D (or E ). Althoughthisde“nitionseemsarathernaturalgeneralizationofthe one-variablecase,therearesubtletiesthatarespeci“ctomultivariable integrals.Thisisillustratedbythefollowingexample.Supposethat (14.25) f ( x,y )= y2Š x2 ( x2+ y2)2istobeintegratedoverthesector0 0ofadisk x2+ y2 1, where isthepolarangle.Ifthede“nition(14.24)isapplied,then Distheportionofthering 2 x2+ y2 1correspondingto0 0. Then,byevaluatingtheintegralinpolarcoordinates,one“ndsthat Dy2Š x2 ( x2+ y2)2dA = Š 00cos(2 ) d 1 dr r = 1 2 sin(2 0)ln Thelimit 0doesnotexistforall 0suchthatsin(2 0) =0,whereas theintegralvanishesif 0= k/ 2, k =1 2 3 4,forany > 0.Let 0= / 2.Theintegralvanishesbecauseofsymmetry,( x,y ) ( y,x ), f ( y,x )= Š f ( x,y ),whiletheintegrationregionisinvariantunderthis transformation.Theintegrandispositiveinthepartofthedomain where x2x2,andthereisamutualcancellationofcontributionsfromtheseregions.Iftheimproperintegralof

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25614.MULTIPLEINTEGRALS theabsolutevalue | f ( x,y ) | isconsidered,thennosuchcancellationcan occur,andtheimproperintegralalwaysdiverges. Definition 14.13 (ConditionalConvergence) Theimproperintegralissaidtoconvergeconditionallyifthelimit (14.24)exists. Definition 14.14 (AbsoluteConvergenceandIntegrability) Theimproperintegralofafunction f issaidtoconvergeabsolutelyif theimproperintegraloftheabsolutevalue | f | converges,andinthis case,thefunction f issaidtobeabsolutelyintegrable. ConsideraRiemannsumforanimproperintegraloverabounded regionwherenosamplepointscoincidewith r0.Theabsoluteintegrabilityof f guaranteesthatallRiemannsumsremainbounded,and hencetheycannotdiverge.Indeed, | R ( f,N ) | = pf ( rp) Vp p| f ( rp) | Vp< andthesumontherightsideconvergesinthelimit N bythe convergenceoftheintegralof | f | .Thisconclusionholdsindependently ofthechoiceofapartitionoftheintegrationregionandthechoiceof samplepoints(withtheaforementionedrestriction). Ifafunctionisnotabsolutelyintegrable,itsiteratedintegralsmay stillbewellde“ned.However,thevalueoftheiteratedintegraldepends ontheorderofintegration(e.g.,Fubinistheoremmaynothold).For example,considerthefunction(14.25)overtherectangle D =[0 1] [0 1].Itisnotabsolutelyintegrableon D astheimproperintegral of | f | divergesasargued.Ontheotherhand,considerarectangular partitionof D whereeachpartitionrectanglehasthearea x y and thesamplepointinthepartitionrectangle[0 x ] [0 y ]doesnot coincidewiththeorigin(where f isnotde“ned).Thelimitofthe Riemannsumisthe two-variable limit( x, y ) (0 0).Bytaking “rst y 0andthen x 0,oneobtainsaniteratedintegralin whichtheintegrationwithrespectto y iscarriedout“rst: lima 01 alimb 01 bx2Š y2 ( x2+ y2)2dydx =lima 01 alimb 01 b y y x2+ y2dydx =lima 01 alimb 0 1 1+ x2Š b x2+ b2 dx =lima 01 adx 1+ x2= 1 0dx 1+ x2= 4 .

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106.IMPROPERMULTIPLEINTEGRALS257 Here( a,b )=( x, y ).Alternatively,thelimit x 0canbetaken “rstandthen y 0,whichresultsintheiteratedintegralinthe reverseorder: limb 01 blima 01 ax2Š y2 ( x2+ y2)2dxdy = Š limb 01 blima 01 a x x x2+ y2dydx = Š limb 01 blima 0 1 1+ y2Š a y2+ a2 dy = Š limb 01 bdy 1+ y2= Š 1 0dy 1+ y2= Š 4 ThisshowsthatthelimitoftheRiemannsumasafunctionof two variables x and y doesnotexistbecauseitdependsonapathalong whichthelimitpointisapproached. Thus,whendealingwithanimproperintegral,theabsoluteconvergence(absoluteintegrability)mustbeestablished“rst.Thentheimproperintegralcanbeevaluatedbymeansofthelimitp rocedure (14.24). AnanalogycanbemadewiththeconditionallyandabsolutelyconvergentseriesstudiedinCalculusII.Ifaseriesconvergesabsolutely,then thesumdoesnotdependontheorderofsummationorrearrangement oftheseriesterms.Ifaseriesconvergesconditionally,butnotabsolutely,then,byrearrangingtheterms,thesumcantakeanyvalue orevendiverge.Riemannsumsofconditionallyconvergentintegrals behaveprettymuchasconditionallyconvergentseries.Thefollowing theoremisusefultoassesstheintegrability. Theorem 14.12 (AbsoluteIntegrabilityTest) If | f ( r ) | g ( r ) forall r in D and g ( r ) isintegrableon D ,then f is absolutelyintegrableon D Example 14.26 Evaluatethetripleintegralof f ( x,y,z )=( x2+ y2+ z2)Š 1overaballofradius R centeredattheoriginifitexists. Solution: Thefunctionissingularonlyattheorigin.Sotherestrictedregion Eliesbetweentwospheres: 2 x2+ y2+ z2 R2. Since | f | = f> 0in D ,theconvergenceoftheintegralover Ealso impliestheabsoluteintegrabilityof f .Bymakinguseofthespherical coordinates,oneobtains. EdV x2+ y2+ z2= 2 0 0R 2sin 2ddd =4 ( R Š ) 4 R as 0.Sotheimproperintegralexistsandequals4 R

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25814.MULTIPLEINTEGRALS Example 14.27 Investigatetheabsoluteintegrabilityof f ( x,y )= x/ ( x2+ y2)/ 2, > 0 ,onaboundedregion D .Findtheintegral,ifit exists,over D ,thepartofthediskofunitradiusinthe“rstquadrant. Solution: Thefunctionissingularattheorigin.Since f iscontinuous everywhereexcepttheorigin,itissucienttoinvestigatetheintegrabilityonadiskcenteredattheorigin.Put r = x2+ y2(thepolar radialcoordinate).Then | x | r andhence | f | r/r= r1 Š = g .In thepolarcoordiantes,theimproperintegral(14.24)of g overadiskof unitradiusis 2 0d 1 g ( r ) rdr =2 1 r2 Š dr =2 Š ln =3 1 Š3 Š 3 Š =3 Thelimit 0is“niteif < 3.Bytheintegrabilitytest(Theorem14.12),thefunction f isabsolutelyintegrableif < 3.For < 3 and D ,thepartoftheunitdiskinthe“rstquadrant,oneinfersthat lim 0DfdA =lim 0 / 20 1r cos rrdrd =lim 0 1r2 Š dr =1 Thetwoexamplesstudiedexhibitacommonfeatureofhowthefunction shouldchangewiththedistancefromthepointofsingularityinorder tobeintegrable. Theorem 14.13 Letafunction f becontinuousonaboundedregion D ofaEuclideanspaceandlet f besingularatalimitpoint r0of D .Supposethat | f ( r ) | M r Š r0Š forall r in D suchthat r Š r0 0 and M> 0 .Then f isabsolutely integrableon D if
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106.IMPROPERMULTIPLEINTEGRALS259 2sin ddd where r = .Soasimilarestimateoftheimproper tripleintegralof f over BD Ryieldsanupperbound4 M R 02 Š d whichis“niteif < 3. 106.2.MultipleIntegralsOverUnboundedRegions.Suppose f ( r )isa continuousfunctiononanunboundedplanar(orspatial)region D (or E ).Let DRbetheintersectionof D withadiskofradius R centered attheoriginandlet ERbetheintersectionof E withaballofradius R centeredattheorigin.Theintegralof f over D (or E )is de“ned by Df ( r ) dA =limR DRf ( r ) dA or Ef ( r ) dV =limR ERf ( r ) dV. Theimproperintegralissaidtoconvergeabsolutelyifthelimitof integralsoftheabsolutevalue | f | existsas R andthefunction f iscalledabsolutelyintegrableon D .Theimproperintegraliscalled conditionallyconvergentifthelimitofintegralsof f exists,while f isnotabsolutelyintegrable.Since f | f | ,theabsoluteintegrability impliestheexistenceoftheimproperintegral. Example 14.28 Evaluatethedoubleintegralof f ( x,y )=exp( Š x2Š y2) overtheentireplane. Solution: Thefunctionispositiveand | f | = f .Soitissucientto investigatetheconvergenceoftheintegralover DR,whichisthedisk ofradius R as R 0.Bymakinguseofthepolarcoordinates, DeŠ x2Š y2dA =limR 2 0R 0eŠ r2rdrd = limR R20eŠ udu = limR (1 Š eŠ R2)= wherethesubstitution u = r2hasbeenmade. Itisinterestingtoobservethefollowing.Sincethedoubleintegral exists,itcanalsoberepresentedasaniteratedintegralinrectangular coordinates,whichistheproductoftwoimproperordinaryintegrals: DeŠ x2Š y2dA = ŠeŠ x2dx ŠeŠ y2dy = I2, I = ŠeŠ x2dx =

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26014.MULTIPLEINTEGRALS because I2= bythevalueofthedoubleintegral.Adirectevaluation of I bymeansofthefundamentaltheoremofcalculusisproblematicas anantiderivativeof eŠ x2cannotbeexpressedinelementaryfunction. Theintegrabilitytest(Theorem14.12)holdsforthecaseofunboundedregions.Theasymptoticbehaviorofafunctionsucientfor integrabilityonanunboundedregionisstatedinthefollowingtheorem, whichisananalogofTheorem14.13. Theorem 14.14 Suppose f isacontinuousfunctiononanunboundedregion D ofaEuclideanspacesuchthat | f ( r ) | M r Š for all r R in D andsome R> 0 and M 0 .Then f isabsolutely integrableon D if >n ,where n isthedimensionofthespace. Proof. Let R> 0.Considerthefollowingone-dimensionalimproper integral: Rdx x=lima a Rdx x=lima x1 Š 1 Š a R= Š R1 Š 1 Š +lima a1 Š 1 Š if =1.Thelimitis“niteif > 1.When =1,theintegraldiverges asln a .Let D Rbethepartof D thatliesoutsidetheball BRofradius R andlet B Bbethepartofthespaceoutside BR.Notethat B Rincludes D R.Inthetwo-variablecase,theuseofthepolarcoordinatesgives D R| f | dA B R| f | dA B RMdA r = M2 0dRrdr r=2 MRdr r Š 1, whichis“nite,provided Š 1 > Š 1or > 2.Thecaseoftripleintegrals isprovedsimilarlybymeansofthesphericalcoordinates.Thevolume elementis dV = 2sin ddd .Theintegrationoverthespherical anglesyieldsthefactor4 as0 and0 2 fortheregion B Rsothat D R| f | dV B R| f | dV B RMdV r =4 MR2d =4 MRd Š 2, whichconvergesif > 3.

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107.LINEINTEGRALS261 106.3.StudyProblems.Problem14.5. Evaluatethetripleintegralof f ( x,y,z )=( x2+ y2)Š 1 / 2( x2+ y2+ z2)Š 1 / 2over E ,whichisboundedbythecone z = x2+ y2andthesphere x2+ y2+ z2=1 ifitexists. Solution: Thefunctionissingularatallpointsonthe z axis.Consider Eobtainedfrom E byeliminatingfromthelatteracone andaball ,where and aresphericalcoordinates.Toinvestigatetheintegrability,consider | f | dV = fdV inthesphericalcoordinates: fdV =( 2sin )Š 12sin ddd = ddd whichisregular. Sothefunction f isintegrableastheimage Eof E inthespherical coordinateisarectangle(i.e.,itisbounded).Hence, lim 0EfdV =lim 0E ddd = 2 0d / 4 0d 1 0d = 2 2 SotheJacobiancancelsoutallthesingularitiesofthefunction. 107.LineIntegrals Considerawiremadeofanonhomogeneousmaterial.Theinhomogeneitymeansthat,ifonetakesasmallpieceofthewireoflength s atapoint r ,thenitsmass m dependsonthepoint r .Itcantherefore becharacterizedbya linear massdensity(themassperunitlengthat apoint r ): ( r )=lim s 0 m ( r ) s Supposethatthelinearmassdensityisknownasafunctionof r .What isthetotalmassofthewirethatoccupiesaspacecurve C ?Ifthecurve C hasalength L ,thenitcanbepartitionedinto N smallsegmentsof length s = L/N .If r pisasamplepointinthe p thsegment,thenthe totalmassreads M =limN Np =1 ( r p) s, wherethemassofthe p thsegmentisapproximatedby mp ( r p) s andthelimitisrequiredbecausethisapproximationbecomesexact onlyinthelimit s 0.Theexpressionfor M resemblesthelimitof aRiemannsumandleadstotheconceptofa lineintegral of alonga curve C .

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26214.MULTIPLEINTEGRALS 107.1.LineIntegralofaFunction.Let f beaboundedfunctionin E andlet C beasmooth(orpiecewise-smooth)curvein E .Suppose C hasa“nitearclength.Considerapartitionof C byits N pieces Cpoflength sp, p =1 2 ,...,N ,whichisthearclengthof Cp(itexists forasmoothcurve!).Put mp=infCpf and Mp=supCPf ;thatis, mpisthelargestlowerboundofvaluesof f forall r Cp,and Mpisthesmallestupperboundonthevaluesof f forall r Cp.The upperandlowersumsarede“nedby U ( f,N )= N p =1Mp spand L ( f,N )= N p =1mp sp. Definition 14.15 (LineIntegralofaFunction) Thelineintegralofafunction f alongapiecewise-smoothcurve C is Cf ( r ) ds =limN U ( f,N )=limN L ( f,N ) providedthelimitsoftheupperandlowersumsexistandcoincide. Thelimitisunderstoodinthesensethat max sp 0 as N (thepartitionelementofthemaximallength becomessmalleras N increases). ThelineintegralcanalsoberepresentedbythelimitofaRiemann sum: Cf ( r ) ds =limN Np =1f ( r p) sp=limN R ( f,N ) Ifthelineintegralexists,itfollowsfromtheinequality mp f ( r ) Mpforall r Cpthat L ( f,N ) R ( f,N ) U ( f,N ),andthelimitofthe Riemannsumis independent ofthechoiceofsamplepoints r p. Itisalsointerestingtoestablisharelationofthelineintegralwith atriple(ordouble)integral.Supposethat f isintegrableonaregion E thatlookslikeawireoftheshape C withacrosssectionofasmall area A atanypoint;thatis, E isacylinderŽwhoseaxisisthecurve C .Then,inthelimit A 0(inthesensethatthediameterofthe areaelementgoesto0), (14.26) 1 A Ef ( r ) dV Cf ( r ) ds. Inotherwords,lineintegralscanbeviewedasthelimitingcaseoftriple (ordouble)integralswhentwo(orone)dimensionsoftheintegration regionbecomein“nitesimallysmall.Thisfollowsimmediatelyfrom consideringapartitionof E byvolumeelements Vp= A spin (14.13).Inparticular,itcanbeconcludedthat thelineintegralexists forany f thatiscontinuousorhasonlya“nitenumberofbounded

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107.LINEINTEGRALS263 jumpdiscontinuitiesalong C .Also, thelineintegralinheritsallthe propertiesofmultipleintegrals Theevaluationofalineintegralisbasedonthefollowingtheorem. Theorem 14.15 (EvaluationofaLineIntegral) Supposethat f iscontinuousinaregionthatcontainsasmoothcurve C .Letavectorfunction r ( t ) t [ a,b ] ,traceoutthecurve C justonce. Then (14.27) Cf ( r ) ds = b af ( r ( t )) r( t ) dt. Proof. Considerapartitionof[ a,b ], tp= a + p t p =0 1 2 ,...,N where t =( b Š a ) /N .Itinducesapartitionof C bypieces Cpso that r ( t )tracesout Cpwhen t [ tp Š 1,tp], p =1 2 ,...,N .Thearc lengthof Cpis tptp Š 1 r( t ) dt = sp.Since C issmooth,thetangent vector r( t )isacontinuousfunctionandsoisitslength r( t ) .Bythe integralmeanvaluetheorem,thereis t p [ tp Š 1,tp]suchthat sp= r( t p) t .Since f isintegrablealong C ,thelimitofitsRiemann sumisindependentofthechoiceofsamplepointsandapartitionof C .Choosethesamplepointstobe r p= r ( t p).Therefore, Cfds =limN Np =1f ( r ( t p)) r( t p) t = b af ( r ) r( t ) dt. NotethattheRiemannsumforthelineintegralbecomesaRiemann sumofthefunction F ( t )= f ( r ( t )) r( t ) overaninterval t [ a,b ]. Itslimitexistsbythecontinuityof F andequalstheintegralof F over[ a,b ]. Theconclusionofthetheoremstillholdsif f hasa“nitenumberof boundedjumpdiscontinuitiesand C ispiecewisesmooth.Thelatter impliesthatthetangentvectormayonlyhavea“nitenumberofdiscontinuitiesandsodoes r .Therefore, F ( t )hasonlya“nitenumber ofboundedjumpdiscontinuitiesandhenceisintegrable.107.2.EvaluationofaLineIntegral.Step1 .Findtheparametricequationofacurve C r ( t )=( x ( t ) y ( t ) ,z ( t )). Step2 .Restricttherangeoftheparameter t toaninterval[ a,b ]so that r ( t )tracesout C onlyoncewhen t [ a,b ]. Step3 .Calculatethederivative r( t )anditsnorm r( t ) .

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26414.MULTIPLEINTEGRALS Step4 .Substitute x = x ( t ), y = y ( t ),and z = z ( t )into f ( x,y,z )and evaluatetheintegral(14.27). Remark. Acurve C maybetracedoutbydierentvectorfunctions. Thevalueofthelineintegralis independent ofthechoiceofparametricequationsbecauseitsde“nitionisgivenonlyinparameterizationinvariantterms(thearclengthandvaluesofthefunctiononthecurve). Theintegrals(14.27)writtenfortwodierentparameterizationsof C arerelatedbyachangeoftheintegrationvariable(recalltheconcept ofreparameterizationofaspatialcurve). Example 14.29 Evaluatethelineintegralof f ( x,y )= x2y overa circleofradius R centeredatthepoint (0 ,a ) Solution: Theequationofacircleofradius R centeredattheorigin is x2+ y2= R2.Ithasfamiliarparametricequations x = R cos t and y = R sin t ,where t istheanglebetween r ( t )andthepositive x axiscountedcounterclockwise.Theequationofthecircleinquestionis x2+( y Š a )2= R2.So,byanalogy,onecanput. x = R cos t and y Š a = R sin t (byshiftingtheorigintothepoint(0 ,a )).Parametricequation ofthecirclecanbetakenintheform r ( t )=( R cos t,a + R sin t ). Therangeof t mustberestrictedtotheinterval t [0 2 ]sothat r ( t )tracesthecircleonlyonce.Then r( t )=( Š R sin t,R cos t )and r( t ) = R2sin2t + R2cos2t = R .Therefore, Cx2yds = 2 0( R cos t )2( a + R sin t ) Rdt = R2a 2 0cos2tdt = R2a, wheretheintegralofcos2t sin t over[0 2 ]vanishesbytheperiodicity ofthecosinefunction.Thelastintegralisevaluatedwiththehelpof thedouble-angleformulacos2t =(1+cos(2 t )) / 2. Example 14.30 Evaluatethelineintegralof f ( x,y,z )= 3 x2+3 y2Š z2overthecurveofintersectionofthecylinder x2+ y2= 1 andtheplane x + y + z =1 Solution: Sincethecurveliesonthecylinder,onecanalwaysput x =cos t y =sin t ,and z = z ( t ),where z ( t ),istobefoundfromthe conditionthatthecurvealsoliesintheplane: x ( t )+ y ( t )+ z ( t )=0 or z ( t )= Š cos t Š sin t .Theintervalof t is[0 2 ]asthecurvewinds aboutthecylinder.Therefore, r( t )=( Š sin t, cos t, sin t Š cos t )and r( t ) = 2 Š 2sin t cos t = 2 Š sin(2 t ).Thevaluesofthefunction alongthecurveare f = 3 Š (cos t +sin t )2= 2 Š sin(2 t ).Note thatthefunctionisde“nedonlyintheregion3( x2+ y2) z2(outside thedoublecone).Ithappensthatthecurve C liesinthedomainof f

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108.SURFACEINTEGRALS265 becauseitsvaluesalong C arewellde“nedas2 > sin(2 t )forany t Hence, Cfds = 2 0 2 Š sin(2 t ) 2 Š sin(2 t ) dt = 2 0(2 Š sin(2 t )) dt =4 108.SurfaceIntegrals108.1.SurfaceArea.Let S beasurfaceinspace.Supposethatit admitsanalgebraicdescriptionasagraphofafunctionoftwovariables, z = f ( x,y ),where( x,y ) D ,or,atleast,itcanbeviewedasaunion ofafewgraphs.Forexample,asphere x2+ y2+ z2=1istheunionof twographs, z = 1 Š x2Š y2and z = Š 1 Š x2Š y2,where( x,y ) areinthedisk D ofunitradius, x2+ y2 1.Whatistheareaofthe surface? Thequestioncanbeansweredbythestandardtrickofintegral calculus.Considerarectangularpartitionof D .Let Sijbethearea ofthepartofthegraphthatliesabovetherectangle( x,y ) [ xi,xi+ x ] [ yj,yj+ y ]= Rij.Thetotalsurfaceareaisthesumofall Sij. Ifthegraphisasmoothsurface(i.e.,thefunction f isdierentiableon D )then Sijcanbeapproximatedbytheareaoftheparallelogram thatliesabove Rijinthetangentplanetothegraphthroughthepoint ( x i,y j,z ij),where z ij= f ( x i,y j)and( x i,y j) Rijisanysample point.Recallthatthedierentiabilityof f meansthatthelinearization of f (orthetangentplaneapproximation)becomesmoreandmore accurateas( x, y ) (0 0).Therefore,inthislimit, x and y canbeviewedasthedierentials dx and dy ,andtheareas Sijand A = x y mustbeproportional: Sij= Jij A. Thecoecient Jjiisfoundbycomparingtheareaoftheparallelogram inthetangentplaneabove Rijwiththearea A of Rij.Thinkofthe roofofabuildingofshape z = f ( x,y )coveredbyshinglesofarea Sij. Theequationofthetangentplaneis z = z ij+ f x( x i,y j)( x Š x i)+ f y( x i,y j)( y Š y j)= L ( x,y ) Let O, A,and Bbe,respectively,thevertices( xi,yj, 0),( xi+ x,yj, 0), and( xi,yj+ y, 0)oftherectangle Rij;thatis,thesegments OAand OBaretheadjacentsidesof Rij.If O A ,and B arethepointsin thetangentplaneabove O, A,and B,respectively,thentheadjacent sidesoftheparallelograminquestionare a = OA and b = OB and Sij= a b .

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26614.MULTIPLEINTEGRALS Bysubstituting Ointothetangentplaneequation,thecoordinates ofthepoint O arefound,( xi,yj,L ( xi,yj)).Bysubstituting Ainto thetangentplaneequation,thecoordinatesofthepoint A arefound, ( xi+ x,yj,L ( xi+ x,yj)).Bythelinearityofthefunction L L ( xi+ x,yj) Š L ( xi,yj)= f x( x i,y j) x and a =( x, 0 ,f x x ).Similarly, b =(0 y,f y y ).Hence, a b =( Š f x, Š f y, 1) x y, Sij= a b = 1+( f x)2+( f y)2 A = J ( x i,y j) A, where J ( x,y )= 1+( f x)2+( f y)2.Thus,thesurfaceareaisgivenby A ( S )=lim( x, y ) (0 0)ijJ ( x i,y j) A. Sincethederivativesof f arecontinuous,thefunction J ( x,y )iscontinuouson D ,andtheRiemannsumconvergestothedoubleintegral of J over D Theorem 14.16 (SurfaceArea) Supposethat f ( x,y ) hascontinuous“rst-orderpartialderivativeson D .Thenthesurfaceareaofthegraph z = f ( x,y ) isgivenby A ( S )= D 1+( f x)2+( f y)2dA. If z =const,then f x= f y=0and A ( S )= A ( D )asrequired because S is D movedparallelintotheplane z =const. Example 14.31 Provethatthesurfaceareaofasphereofradius R is 4 R2. Solution: Thehemisphereisthegraph z = f ( x,y )= R2Š x2Š y2onthedisk x2+ y2 R2ofradius R .Theareaofthesphereistwice theareaofthisgraph.Onehas f x= Š x/f and f y= Š y/f .Therefore, J =(1+ x2/f2+ y2/f2)1 / 2=( f2Š x2Š y2)1 / 2/f = R/f .Hence, A ( S )=2 R DdA R2Š x2Š y2=2 R 2 0d R 0rdr R2Š r2=4 R R 0rdr R2Š r2=2 R R20du u =4 R2, wherethedoubleintegralhasbeenconvertedtopolarcoordinates andthesubstitution u = R2Š r2hasbeenusedtoevaluatethelast integral.

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108.SURFACEINTEGRALS267 Example 14.32 Findtheareaofthepartoftheparaboloid z = x2+ y2inthe“rstoctantandbelowtheplane z =4 Solution: Thesurfaceinquestionisthegraph z = f ( x,y )= x2+ y2. Next,theregion D mustbespeci“ed(itdeterminesthepartofthe graphwhoseareaistobefound).Onecanview D asthevertical projectionofthesurfaceontothe xy plane.Theplane z =4intersects theparaboloidalongthecircle4= x2+ y2ofradius2.Sincethesurface alsoliesinthe“rstoctant, D isthepartofthedisk x2+ y2 4inthe “rstquadrant.Then f x=2 x f y=2 y ,and J =(1+4 x2+4 y2)1 / 2. Thesurfaceareais A ( S )= D 1+4 x2+4 y2dA = / 2 0d 2 0 1+4 r2rdr = 2 2 0 1+4 r2rdr = 16 17 1 udu = 24 (173 / 2Š 1) wherethedoubleintegralhasbeenconvertedtopolarcoordinates andthesubstitution u =1+4 r2hasbeenusedtoevaluatethelast integral. 108.2.SurfaceIntegralofaFunction.Anintuitiveideaoftheconcept ofthesurfaceintegralofafunctioncanbeunderstoodfromthefollowingexample.Supposeonewantsto“ndthetotalhumanpopulation ontheglobe.Thedataaboutthepopulationisusuallysuppliedas thepopulation density (i.e.,thenumberofpeopleperunitarea).The populationdensityisnotaconstantfunctionontheglobe.Itishigh incitiesandlowindesertsandjungles.Therefore,thesurfaceofthe globemustbepartitionedbysurfaceelementsofarea Sp.If ( r )is thepopulationdensityasafunctionofposition r ontheglobe,then thepopulationoneachpartitionelementisapproximately ( r p) Sp, where r pisasamplepointinthepartitionelement.Theapproximationneglectsvariationsof withineachpartitionelement.Thetotal populationisapproximatelytheRiemannsum p ( r p) Sp.Toget anexactvalue,thepartitionhastobere“nedsothatthesizeofeach partitionelementbecomessmaller.Thelimitisthesurfaceintegral of overthesurfaceoftheglobe,whichisthetotalpopulation.In general,onecanthinkofsomequantitydistributedoverasurfacewith somedensity(theamountofthisquantityperunitareaasafunction ofpositiononthesurface).Thetotalamountisthesurfaceintegralof thedensityoverthesurface. Let f beaboundedfunctionin E andlet S beasurfacein E that hasa“nitesurfacearea.Considerapartitionof S by N pieces Sp,

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26814.MULTIPLEINTEGRALS p =1 2 ,...,N ,whichhavesurfacearea Sp.Put mp=infSpf and Mp=supSPf ;thatis, mpisthelargestlowerboundofvaluesof f forall r Spand Mpisthesmallestupperboundonthevaluesof f forall r Sp.Theupperandlowersumsarede“nedby U ( f,N )= N p =1Mp Spand L ( f,N )= N p =1mp Sp.Let Rpbetheradiusof thesmallestballthatcontains SpandmaxpRp= RN.Apartition of S issaidtobere“nedif RNN .Inotherwords, underthere“nement,thesizes Rpofeachpartitionelementbecome uniformlysmaller. Definition 14.16 (SurfaceIntegralofaFunction) Thesurfaceintegralofafunction f overasurface S is Sf ( r ) dS =limN U ( f,N )=limN L ( f,N ) providedthelimitsoftheupperandlowersumsexistandcoincide.The limitisunderstoodinthesense RN 0 as N ThesurfaceintegralcanalsoberepresentedbythelimitofaRiemannsum: (14.28) Sf ( r ) dS =limN Np =1f ( r p) Sp=limN R ( f,N ) Ifthesurfaceintegralexists,itfollowsfromtheinequality mp f ( r ) Mpforall r Spthat L ( f,N ) R ( f,N ) U ( f,N ),andthelimit oftheRiemannsumis independent ofthechoiceofsamplepoints r p. Riemannsumscanbeusedinnumericalapproximationsofthesurface integral. Similartolineintegrals,surfaceintegralsarerelatedtotripleintegrals.Supposethat f isintegrableonaregion E thatlookslikeashell oftheshape S withaconstantsmallthickness h .Then,inthelimit h 0, (14.29) 1 h Ef ( r ) dV Sf ( r ) dS. Thisfollowsimmediatelybyconsideringapartitionof E byvolume elements Vp= h Spin(14.13).Hence, thesurfaceintegralexists forany f thatiscontinuousorhasboundedjumpdiscontinuitiesalong a“nitenumberofsmoothcurveson S ,anditinheritsalltheproperties ofmultipleintegrals .

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108.SURFACEINTEGRALS269 108.3.EvaluationofaSurfaceIntegral.Theorem 14.17 (EvaluationofaSurfaceIntegral) Supposethat f iscontinuousinaregionthatcontainsasurface S de“nedbythegraph z = g ( x,y ) on D .Supposethat g hascontinuous “rst-orderpartialderivativeson D .Then (14.30) Sf ( x,y,z ) dS = Df ( x,y,g ( x,y )) 1+( g x)2+( g y)2dA. Considerapartitionof D byelements Dpofarea Ap, p =1 2 ,..., N .Let J ( x,y )= 1+( g x)2+( g y)2.Bythecontinuityof g xand g y, J iscontinuouson D .Bytheintegralmeanvaluetheorem,theareaof thepartofthegraph z = g ( x,y )over Dpisgivenby Sp= DpJ ( x,y ) dA = J ( x p,y p) Apforsome( x p,y p) Dp.IntheRiemannsumforthesurfaceintegral (14.28),takethesamplepointstobe r p=( x p,y p,g ( x p,y p)) Sp. TheRiemannsumbecomestheRiemannsum(14.3)ofthefunction F ( x,y )= f ( x,y,g ( x,y )) J ( x,y )on D .Bythecontinuityof F ,itconvergestothedoubleintegralof F over D .Theargumentgivenhereis basedonatacitassumptionthatthesurfaceintegralexistsaccording toDe“nition14.16,andhencethelimitoftheRiemannsumexistsand isindependentofthechoiceofsamplepoints.Itcanbeprovedthat underthehypothesisofthetheoremthesurfaceintegralexists. Theevaluationofthesurfaceintegralinvolvesthefollowingsteps: Step1 .Represent S asagraph z = g ( x,y )(i.e.,“ndthefunction g usingageometricaldescriptionof S ). Step2 .Findtheregion D thatde“nesthepartofthegraphthat coincideswith S (if S isnottheentiregraph). Step3 .Calculatethederivatives g xand g yandtheareatransformationfunction J dS = JdA Step4 .Evaluatethedoubleintegral(14.30). Example 14.33 Evaluatetheintegralof f ( x,y,z )= z overthe partofthesaddlesurface z = xy thatliesinsidethecylinder x2+ y2=1 inthe“rstoctant. Solution: Thesurfaceisapartofthegraph z = g ( x,y )= xy .Since itlieswithinthecylinder,itsprojectionontothe xy planeisbounded bythecircleofunitradius, x2+ y2=1.Thus, D isthequarterofthe

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27014.MULTIPLEINTEGRALS disk x2+ y2 1inthe“rstquadrant.Onehas g x= y g y= x ,and J ( x,y )=(1+ x2+ y2)1 / 2.Thesurfaceintegralis SzdS = Dxy 1+ x2+ y2dA = / 2 0cos sin d 1 0r2 1+ r2rdr = sin2 2 / 2 01 2 2 1( u Š 1) udu = 1 2 u5 / 2 5 Š u3 / 2 3 2 1= 2(4 2+1) 15 wherethedoubleintegralhasbeenconvertedtopolarcoordinatesand thelastintegralisevaluatedbythesubstitution u =1+ r2. 108.4.ParametricEquationsofaSurface.Thegraph z = g ( x,y ),where ( x,y ) D de“nesasurface S inspace.Considerthevectors r ( u,v )= ( u,v,g ( u,v ))wherethepairofparameters( u,v )spanstheregion D Thevectorfunction r ( u,v )oftwovariablesde“nesaone-to-onemappingoftheregion D intospacesothattheimageof D isthesurface S Considernowaregion D spannedbytheorderedpairs( u,v ).Three functions x ( u,v ), y ( u,v ),and z ( u,v )on D de“neamappingof D into space r ( u,v )=( x ( u,v ) ,y ( u,v ) ,z ( u,v )).Therangeofthismappingis calleda surfaceinspace ,andtheequations x = x ( u,v ), y = y ( u,v ), and z = z ( u,v )arecalledparametricequations ofthissurface. Forexample,theequations (14.31) x = R cos v sin u,y = R sin v sin u,z = R cos u areparametricequationsofasphereofradius R .Indeed,bycomparing theseequationswiththesphericalcoordinates,one“ndsthat( ,, )= ( R,u,v );thatis,when( u,v )rangeovertherectangle[0 ] [0 2 ], thevector( x,y,z )= r ( u,v )tracesoutthesphere = R .Anapparent advantageofusingparametricequationsofasurfaceisthatthesurface nolongerneedsberepresentedasagraph.Hereasphereisdescribed byonevector-valuedfunctionoftwovariables. Definition 14.17 Let r ( u,v ) beavectorfunctiononanopenregion D thathascontinuouspartialderivatives r uand r von D .The range S ofthevectorfunctioniscalledasmoothsurfaceif S iscovered justonceas ( u,v ) rangesthroughout D andthevector r u r visnot zero.

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108.SURFACEINTEGRALS271 Ananalogycanbemadewithparametricequationsofacurvein space.Acurveinspaceisamappingofan interval [ a,b ]intospace de“nedbyavectorfunctionof onevariable r ( t ).If r( t )iscontinuous and r( t ) = 0 ,thenthecurvehasacontinuoustangentvectorand thecurveissmooth.Similarly,thecondition r u r v = 0 ensures thatthesurfacehasacontinuousnormalvectorjustlikeagraphof acontinuouslydierentiablefunctionoftwovariables.Thiswillbe explainedshortlyafterthediscussionofafewexamples. Example 14.34 Findtheparametricequationsofthedoublecone z2= x2+ y2. Solution: Suppose z =0.Then( x/z )2+( y/z )2=1.Thesolution ofthisequationis x/z =cos u and y/z =sin u ,where u [0 2 ). Therefore,theparametricequationsare x = v cos u,y = v sin u,z = v, where( u,v ) [0 2 ) ( Š )forthewholedoublecone.Ofcourse, therearemanydierentparameterizationsofthesamesurface.They arerelatedbyachangeofvariables( u,v ) D ( s,t ) D,where s = s ( u,v )and t = t ( u,v ). Example 14.35 Atorusisasurfaceobtainedbyrotatingacircle aboutanaxisoutsidethecircleandparalleltoitsdiameter.Findthe parametricequationsofatorus. Solution: Lettherotationaxisbethe z axis.Let R bethedistance fromthe z axistothecenteroftherotatedcircleandlet a betheradius ofthelatter, a R .Inthe xz plane,therotatedcircleis z2+( x Š a )2= R2.Let( x0, 0 ,z0)beasolutiontothisequation.Thepoint( x0, 0 ,z0) tracesoutthecircleofradius x0upontherotationaboutthe z axis.All suchpointsare( x0cos v,x0sin v,z0),where v [0 2 ).Allthatisleft istoparameterizeallsolutions( x0, 0 ,z0),whichissimply z0= R sin u and x0Š a = R cos u .Thus,theparametricequationsofatorusare (14.32) x =( R + a cos u )cos v,y =( R + a cos u )sin v,z = R sin u, where( u,v ) [0 2 ) [0 2 ) Theparametricequationsofasurfaceareconvenientforevaluating thesurfaceintegrals.Iftheregion D spannedbytheparameters( u,v ) ispartitionedbyrectanglesofarea A = u v ,thenthemapping r ( u,v )de“nesapartitionofthesurface.Sothesurfaceintegralcan beconvertedintoadoubleintegralover D ifoneestablishesthearea transformationlaw S = J A .Considerarectangle( u,v ) [ u0,u0+

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27214.MULTIPLEINTEGRALS u ] [ v0,v0+ v ]= R0.Letitsvertices O, A,and Bhavethe coordinates( u0,v0),( u0+ u,v0),and( u0,v0+ v ),respectively.The segments OAand OBaretheadjacentsidesoftherectangle R0.Let O A ,and B betheimagesofthesepointsinthesurface.Theirposition vectorsare r0= r ( u0,v0), ra= r ( u0+ u,v0),and ra= r ( u0,v0+ v ), respectively.Thearea S oftheimageoftherectangle R0canbe approximatedbytheareaoftheparallelogramwithadjacentsides: a = OA = raŠ r0= r ( u0+ u,v0) Š r ( u0,v0)= r u( u0,v0) u, b = OB = rbŠ r0= r ( u0,v0+ v ) Š r ( u0,v0)= r v( u0,v0) v, whichholdinthelimit( u, v ) (0 0)(when du = u and dv = v )undertheassumptionthatthecomponentsof r ( u,v )are continuouslydierentiable .Notethatifthesurfaceisagraph z = g ( x,y ), then r ( u,v )=( u,v,g ( u,v )),andthevectors a and b aregivenbythe familiarexpressions a =(1 0 ,g u) u and b =(0 1 ,g v) v .Thevector r ( u,v0)(oneargumentis“xed, v = v0)tracesoutacurveinthe surface.Thederivative r u( u,v0)istangenttothecurveandhenceto thesurface.Asimilarargumentappliesto r v.Thus,thederivatives r uand r varetangenttothesurfaceand,hence,theircrossproductmust benormaltoit. Corollary 14.5 (NormaltoaParametricSurface) Letasmoothsurfacebedescribedbytheparametricequations r = r ( u,v ) .Thenthevector n = r u r visnormaltothesurface. Theareatransformationlawisnoweasyto“nd: S = a b = r u r v A .Thesurfaceintegralcanbewrittenasthedoubleintegral Sf ( r ) dS = Df ( r ( u,v )) r u r v dA and,inparticular, A ( S )= D r u r v dA. Example 14.36 Findthesurfaceareaofthetorus(14.32). Solution: Tosimplifythenotation,put w = R + a cos u .Onehas r u=( Š a sin u cos v, Š a sin u sin v,R cos u ) r v=( Š ( R + a cos u )sin v, ( R + a cos u )cos v, 0) = w ( Š sin v, cos v, 0) n = r u r v= w ( Š a cos v cos u, Š a cos v cos u, Š a sin u ) J = r u r v = aw = a ( R + a cos u ) .

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109.MOMENTSOFINERTIAANDCENTEROFMASS273 Thesurfaceareais A ( S )= DJ ( u,v ) dA = 2 02 0a ( R + a cos u ) dvdu =4 2Ra Example 14.37 Evaluatethesurfaceintegralof f ( x,y,z )= z2( x2+ y2) overasphereofradius R centeredattheorigin. Solution: Usingtheparametricequations(14.31),one“nds r u=( R cos v cos u,R sin v cos u, Š R sin u ) r v=( Š R sin v sin u,R cos v sin u, 0) = R sin u ( Š sin v, cos v, 0) n = r u r v= R sin u ( R sin u cos v,R sin u sin v,R cos u ) = R sin u r ( u,v ) J = r u r v = R2sin u, f ( r ( u,v ))=( R cos u )2R2sin2u = R4cos2u (1 Š cos2u ) Notethatsin u 0because u [0 ]( u = and v = ).Therefore, thenormalvector n isoutward(paralleltothepositionvector;the inwardnormalwouldbeoppositetothepositionvector.)Thesurface integralis SfdS = Df ( r ( u,v )) J ( u,v ) dA = R62 0dv 0cos2u (1 Š cos2u )sin udu =2 R61 Š 1w2(1 Š w2) dw = 8 15 R6, wherethesubstitution w =cos u hasbeenmadetoevaluatethelast integral. 109.MomentsofInertiaandCenterofMass Animportantapplicationofmultipleintegralsis“ndingthe center ofmass and momentsofinertia ofanextendedobject.Thelawsof mechanicssaythatthecenterofmassofanextendedobjectonwhich noexternalforceactsmovesalongastraightlinewithaconstantspeed. Inotherwords,thecenterofmassisaparticularpointofanextended objectthatde“nesthetrajectoryoftheobjectasawhole.Themotion ofanextendedobjectcanbeviewedasacombinationofthemotionof

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27414.MULTIPLEINTEGRALS itscenterofmassandrotationaboutitscenterofmass.Thekinetic energyoftheobjectis K = Mv2 2 + Krot, where M isthetotalmassoftheobject, v isthespeedofitscenterof mass,and Krotisthekineticenergyofrotationoftheobjectaboutits centerofmass;thelatterquantityisdeterminedbymomentsofinertia. Forexample,whendockingaspacecrafttoaspacestation,oneneeds toknowexactlyhowlongtheengineshouldbe“redtoachievethe requiredpositionofitscenterofmassandtheorientationofthecraft relativetoit,thatis,howexactlyitskineticenergyhastobechanged by“ringtheengines.Soitscenterofmassandmomentsofinertia mustbeknowntoaccomplishthetask.109.1.CenterofMass.Considerapointmass m “xedatanendpoint ofarodthatcanrotateaboutitsotherend.Iftherodhaslength L andthegravitationalforceisnormaltotherod,thenthequantity gmL iscalledthe rotationalmoment ofthegravitationalforce mg ,where g isthefreeacceleration.Iftherotationisclockwise(themassisat therightendpoint),themomentisassumedtobepositive,anditis negative, Š gmL ,foracounterclockwiserotation(themassisattheleft endpoint).Moregenerally,ifthemasshasaposition x onthe x axis, thenitsrotationmomentaboutapoint xcis M =( x Š xc) m (omitting theconstant g ).Itisnegativeif xxc. Thecenterofmassisunderstoodthroughtheconceptofrotational moments. Thesimplestextendedobjectconsistsoftwopointmasses m1and m2connectedbyamasslessrod.Supposethatonepointoftherod is“xedsothatitcanonlyrotateaboutthatpoint.Thecenterof massisthepointontherodsuchthattheobjectwouldnotrotate aboutitunderauniformgravitationalforceappliedalongthedirection perpendiculartotherod.Evidently,thepositionofthecenterofmass isdeterminedbytheconditionthatthetotalrotationalmomentabout itvanishes.Supposethattherodliesonthe x axissothatthemasses havethecoordinates x1and x2.Thetotalrotationalmomentofthe objectaboutthepoint xcis M = M1+ M2=( x1Š xc) m1+( x2Š xc) m2. If xcissuchthat M =0,then m1( x1Š xc)+ m2( x2Š xc)=0= xc= m1x1+ m2x2 m1+ m2.

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109.MOMENTSOFINERTIAANDCENTEROFMASS275 Thecenterofmass( xc,yc)ofpointmasses mi, i =1 2 ,...,N ,positioned onaplaneat( xi,yi)canbeunderstoodasfollows.Thinkoftheplane asaplateonwhichthemassesarepositioned.Thegravitationalforce isnormaltotheplane.Ifarodisputunderneaththeplane,then dueanevendistributionofmassestheplanecanrotateaboutthe rod.Whentherodisalignedalongeithertheline x = xcortheline y = yc,theplanewithdistributedmassesonitdoesnotrotateunder thegravitationalpull.Inotherwords,therotationalmomentsabout thelines x = xcand y = ycvanish.Therotationalmomentaboutthe line x = xcor y = ycisdeterminedbythedistancesofthemassesfrom thisline:Ni =1( xiŠ xc) mi=0= xc= 1 mNi =1mixi= My m ,m =Ni =1mi,Ni =1( yiŠ yc) mi=0= yc= 1 mNi =1miyi= Mx m ,m =Ni =1mi, where m isthetotalmass.Thequantity Myisthemomentaboutthe y axis(theline x =0),whereas Mxisthemomentaboutthe x axis (theline y =0). Thecenterofmassofageneralextendedobjectisde“nedsimilarly bydemandingthatthetotalmomentsabouteitheroftheplanes x = xc, or y = yc,or z = zcvanish.Thus,if rcisthepositionvectorofthe centerofmass,itsatis“esthecondition: imi( riŠ rc)= 0 wherethevectors riŠ rcarepositionvectorsofmasses relativetothe centerofmass Definition 14.18 (CenterofMass) Supposethatanextendedobjectconsistsof N pointmasses mi, i = 1 2 ,...,N ,whosepositionvectorsare ri.Thenitscenterofmassisa pointwiththepositionvector (14.33) rc= 1 mNi =1miri,m =Ni =1mi, where m isthetotalmassoftheobject. Ifanextendedobjectcontainscontinuouslydistributedmasses, thentheobjectcanbepartitionedinto N smallpieces.Let Bibe thesmallestballofradius Riwithinwhichthe i thpartitionpiecelies.

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27614.MULTIPLEINTEGRALS Althoughallthepartitionpiecesaresmall,theystillhave“nitesizes Ri,andthede“nition(14.33)cannotbeusedbecausethepoint ricould beanypointin Bi.Bymakingtheusualtrickofintegralcalculus,this uncertaintycanbeeliminatedbytakingthelimit N inthesense thatallthepartitionsizestendto0uniformly, Ri maxiRi= RN 0 as N .Inthislimit,thepositionofeachpartitionpiececanbe describedbyanysamplepoint r i Bi.ThelimitoftheRiemannsum isgivenbytheintegralovertheregion E inspaceoccupiedbytheobject.If ( r )isthemassdensityoftheobject,then mi= ( r i) Vi, where Viisthevolumeofthe i thpartitionelementand rc= 1 m limN Ni =1r i mi= 1 m Er ( r ) dV,m = E ( r ) dV. Inpracticalapplications,oneoftenencountersextendedobjectswhose oneortwodimensionsaresmallrelativetotheother(e.g.,shell-like objectsorwirelikeobjects).Inthiscase,thetripleintegralissimpli“ed toeitherasurface(ordouble)integralforshell-like E ,accordingto (14.29),ortoalineintegral,accordingto(14.26).Fortwo-andonedimensionalextendedobjects,thecenterofmasscanbewrittenas, respectively, rc= 1 m Sr ( r ) dS,m = S ( r ) dS, rc= 1 m Cr ( r ) ds,m = C ( r ) ds, where,accordingly, isthesurfacemassdensityorthelinemassdensityfortwo-orone-dimensionalobjects.Inparticular,when S isa planar,”atsurface,thesurfaceintegralturnsintoadoubleintegral. Theconceptofrotationalmomentsisalsousefulfor“ndingthe centerofmassusingthesymmetriesofthemassdistributionofan extendedobject.Forexample,thecenterofmassofadiskwitha uniformmassdistributionapparentlycoincideswiththediskcenter (thediskwouldnotrotateaboutitsdiameterunderthegravitational pull). Example 14.38 Findthecenterofmassofthehalf-disk x2+ y2 R2, y 0 ,ifthemassdensityatanypointisproportionaltothe distanceofthatpointfromthe x axis. Solution: Themassisdistributedevenlytotheleftandrightfrom the y axisbecausethemassdensityisindependentof x ( x,y )= ky

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109.MOMENTSOFINERTIAANDCENTEROFMASS277 ( k isaconstant).So,therotationalmomentaboutthe y axisvanishes; My=0bysymmetryandhence xc= My/m =0.Thetotalmassis m = DdA = k DydA = k 0R 0r sin rdrd =2 k R 0r2dr = 2 kR3 3 wheretheintegralhasbeenconvertedtopolarcoordinates.Themomentaboutthe x axis(abouttheline y =0)is Mx= DydA = 0R 0k ( r sin )2rdrd = k 2 R 0r3dr = kR4 8 So yc= Mx/m =3 R/ 16. Example 14.39 Findthecenterofmassofthesolidthatliesbetweenspheresofradii a
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27814.MULTIPLEINTEGRALS theradiusofthecirculartrajectoryis R ,thenthelinearvelocityofthe objectis v = R .Theobjecthasthekineticenergy Krot= mv2 2 = mR22 2 = I2 2 Theconstant Iiscalledthe momentofinertia ofthepointmass m abouttheaxis .Similarly,consideranextendedobjectconsistingof N pointmasses.Therelativepositionsofthemassesdonotchange whentheobjectmoves.So,iftheobjectrotatesaboutanaxis at aconstantrate ,theneachpointmassrotatesatthesamerateand hencehaskineticenergy miR2 i2/ 2,where Riisthedistancefromthe mass mitotheaxis .Thetotalkineticenergyis Krot= I2/ 2,where theconstant I=Ni =1miR2 iiscalledthe momentofinertiaoftheobjectabouttheaxis .Itis independentofthemotionitselfanddeterminedsolelybythemass distributionanddistancesofthemassesfromtherotationaxis. Supposethatthemassiscontinuouslydistributedinaregion E with themassdensity ( r ).Let R( r )bethedistancefromapoint r E toanaxis(line) .Considerapartitionof E bysmallelements Eiof volume Vi.Themassofeachpartitionelementis mi= ( r i) V forsomesamplepoint r i Eiinthelimitwhenallthesizesofpartition elementstendto0uniformly.Themomentofinertiaabouttheaxis is I=limN Ni =1R2 ( ri) ( r i) Vi= ER2 ( r ) ( r ) dV inaccordancewiththeRiemannsumfortripleintegrals(14.13).In particular,thedistanceofapoint( x,y,z )fromthe x -, y -,and z axes is,respectively, Rx= y2+ z2, Ry= x2+ z2,and Rz= x2+ y2. Sothemomentsofinertiaaboutthecoordinateaxesare Ix= E( y2+ z2) dV,Iy= E( x2+ z2) dV, Iz= E( x2+ y2) dV. Ingeneral,iftheaxis goesthroughtheoriginparalleltoaunitvector u ,thenbythedistanceformulabetweenapoint r andtheline, (14.34) R2 ( r )= u r 2=( u r ) ( u r )= u ( r ( u r ))= r2Š ( u r )2,

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109.MOMENTSOFINERTIAANDCENTEROFMASS279 wherethe bac Š cab rule(seeStudyProblem11.16)hasbeenusedto transformthedoublecrossproduct. Ifoneortwodimensionsoftheobjectaresmallrelativetothe other,thetripleintegralisreducedtoeitherasurfaceintegralora lineintegral,respectively,inaccordancewith(14.29)or(14.26);that is,fortwo-orone-dimensionalobjects,themomentofinertiabecomes, respectively, I= SR2 ( r ) ( r ) dS,I= CR2 ( r ) ( r ) ds, where iseitherthesurfaceorlinearmassdensity. Example 14.40 Arockettipismadeofthinplateswithaconstant surfacemassdensity = k .Ithasacircularconicshapewithbase diameter 2 a anddistance h fromthetiptothebase.Findthemoment ofinertiaofthetipaboutitsaxisofsymmetry. Solution: Setupthecoordinatesystemsothatthetipisatthe originandthebaseliesintheplane z = h ;thatisthesymmetry axiscoincideswiththe z axis.If istheanglebetweenthe z axis andthesurfaceofthecone,thencot = h/a andtheequationofthe coneis z =cot x2+ y2.Thus,theobjectinquestionisthesurface (graph) z = g ( x,y )=( h/a ) x2+ y2overtheregion D : x2+ y2 a2. Toevaluatetheneededsurfaceintegral,theareatransformationlaw dS = JdA shouldbeestablished.Onehas g x=( hx/a )( x2+ y2)Š 1 / 2and g x=( hy/a )( x2+ y2)Š 1 / 2sothat J = 1+( g x)2+( g y)2= 1+( h/a )2= h2+ a2 a Themomentofinertiaaboutthe z axisis Iz= S( x2+ y2) dS = k D( x2+ y2) JdA = kJ 2 0d a 0r3dr = k 2 a3 h2+ a2 Example 14.41 Findthemomentofinertiaofahomogeneousball ofradius a andmass m aboutitsdiameter. Solution: Setupthecoordinatesystemsothattheoriginisatthe centeroftheball.Thenthemomentofinertiaaboutthe z axishas tobeevaluated.Sincetheballishomogeneous,itsmassdensityis

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28014.MULTIPLEINTEGRALS constant, = m/V ,where V =4 a3/ 3isthevolumeoftheball.One has Iz= E( x2+ y2) dV = 3 m 4 a32 0 0a 0( sin )22sin ddd = 3 10 ma2 0sin3d = 3 10 ma21 Š 1(1 Š u2) du = 2 5 ma2, wherethesubstitution u =cos hasbeenmadetoevaluatetheintegral.Itisnoteworthythattheproblemadmitsasmartersolutionby notingthat Iz= Ix= Iyowingtotherotationalsymmetryofthemass distribution.Bytheidentity Iz=( Ix+ Iy+ Iz) / 3,thetripleintegral canbesimpli“ed: Iz= 1 3 E2( x2+ y2+ z2) dV = 1 3 8 a 04d = 2 5 ma2109.3.StudyProblems.Problem14.6. FindthecenterofmassoftheshelldescribedinExample14.40. Solution: Bythesymmetryofthemassdistributionabouttheaxis oftheconicshell,thecenterofmassmustbeonthataxis.Usingthe algebraicdescriptionofashellgiveninExample14.41,thetotalmass oftheshellis m = SdS = k SdS = kJ DdA = kJA ( D )= ka h2+ a2. Themomentaboutthe xy planeis Mxy= SzdS = k D( h/a ) x2+ y2JdA = kJh a D x2+ y2dA = kJh a 2 0a 0r2drd = 2 kha 3 h2+ a2. Thus,thecenterofmassisatthedistance zc= Mxy/m =2 h/ 3from thetipofthecone. Problem14.7.(ParallelAxisTheorem) Let Ibethemomentofinertiaofanextendedobjectaboutanaxis andlet cbeaparallelaxisthroughthecenterofmassoftheobject. Provethat I= Ic+ mR2 c,

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109.MOMENTSOFINERTIAANDCENTEROFMASS281 where Rcisthedistancebetweentheaxis andthecenterofmass,and m isthetotalmass. Solution: Choosethecoordinatesystemsothattheaxis goes throughtheorigin.Letitbeparalleltoaunitvector u .Thedifference IŠ Icistobeinvestigated.If rcisthepositionvectorofthe centerofmass,thentheaxis cisobtainedfrom byparalleltransportofthelatteralongthevector rc.Therefore,thedistance R2 c( r )is obtainedfrom R2 ( r )(see(14.34))bychangingthepositionvector r in thelattertothepositionvectorrelativetothecenterofmass, r Š rc. Inparticular, R2 ( rc)= R2 cbythede“nitionof R.Hence, R2 ( r ) Š R2 c( r )= R2 ( r ) Š R2 ( r Š rc) =2 rc r Š r2 cŠ ( u rc)(2 u r Š u rc) = r2 cŠ ( u rc)2+2 rc ( r Š rc) Š 2( u rc) u ( r Š rc) = R2 cŠ 2 a ( r Š rc) where a = rcŠ ( u rc) u .Therefore, IŠ Ic= E R2 ( r ) Š R2 c( r ) ( r ) dV = R2 cE ( r ) dV Š 2 a E( r Š rc) ( r ) dV = R2 cm, wherethesecondintegralvanishesbythede“nitionofthecenterof mass. Problem14.8. Findthemomentofinertiaofahomogeneousballof radius a andmass m aboutanaxisthatisatadistance R fromtheball center. Solution: Thecenterofmassoftheballcoincideswithitscenter becausethemassdistributionisinvariantunderrotationsaboutthe center.Themomentofinertiaoftheballaboutitsdiameteris Ic= (2 / 5) ma2byExample14.41.Bytheparallelaxistheorem,forany axis atadistance R fromthecenterofmass, I= Ic+ mR2= m ( R2+2 a2/ 5).

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CHAPTER15 VectorCalculus 110.LineIntegralsofaVectorField110.1.VectorFields.Consideranair”owintheatmosphere.Theair velocityvariesfrompointtopoint.Inordertodescribethemotionof theair,theairvelocitymustbede“nedasafunctionofposition,which meansthatavelocity vector hastobeassignedtoeverypointinspace. Inotherwords,incontrasttoordinaryfunctions,theairvelocityisa vector-valued functionofthepositionvectorinspace. Definition 15.1 (VectorField) Let E beasubsetinspace.Avector“eldon E isafunction F that assignstoeachpoint r =( x,y,z ) avector F ( r )=( F1( r ) ,F2( r ) ,F3( r )) Thefunctions F1, F2,and F3arecalledthecomponentsofthevector “eld F Avector“eldis continuous ifitscomponentsarecontinuous.A vector“eldis dierentiable ifitscomponentsaredierentiable. Asimpleexampleofavector“eldisthegradientofafunction, F ( r )= f ( r ).Thecomponentsofthisvector“eldarethe“rst-order partialderivatives: F ( r )= f ( r ) F1( r )= f x( r ) ,F2( r )= f y( r ) ,F3( r )= f z( r ) Manyphysicalquantitiesaredescribedbyvector“elds.Electricand magnetic“eldsarevector“elds.Light waves,radio waves,TVwaves, andwaves usedincellphonecommunicationsareallelectromagnetic waves thatarealternatingelectromagnetic“elds.Thepropagationof electromagnetic wavesin spaceisdescribedbydierentialequations thatrelateelectromagnetic“eldsateachpointinspaceandeachmomentoftimetoadistributionofelectricchargesandcurrents(e.g., antennas).Thegravitationalforcelooksconstantnearthesurfaceof theEarth,butonthescaleofthesolarsystemthisisnotso.The gravitationalforceexertedbyaplanetofmass M onaspacecraftof mass m dependsonthepositionofthespacecraftrelativetotheplanet centeraccordingtoNewtonslawofgravity: F ( r )= Š GMm r3r = Š GMm x r3, Š GMm y r3, Š GMm z r3 ,283

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28415.VECTORCALCULUS where G isNewtonsgravitationalconstant, r isthepositionvector relativetotheplanetcenter,and r = r isitslength(thedistance betweentheplanetcenterandthespacecraft).Theforceisproportionaltothepositionvectorandhenceparalleltoitforeachpointin space.Theminussignindicatesthat F isdirectedoppositeto r ,that is,theforceis attractive ;thegravitationalforcepullstowarditssource (theplanet).Themagnitude F = GMmrŠ 2decreaseswithincreasingdistance r .Sothegravitationalvector“eldcanbevisualizedby plottingvectorsoflength F ateachpointinspacepointingtoward theorigin.Themagnitudesofthesevectorsbecomesmallerforpoints farther away fromtheorigin.Thisobservationleadstotheconceptof ”owlines ofavector“eld.110.2.FlowLinesofaVectorField.Definition 15.2 (FlowLinesofaVectorField) The”owlineofavector“eld F isacurveinspacesuchthat,atany point r ,thevector“eld F ( r ) istangenttoit. Thedirectionof F de“nesthe orientation of”owlines;thatis, thedirectionofatangentvector F isshownbyarrowsonthe”ow lines.Forexample,the”owlinesoftheplanetsgravitational“eldare straightlinesorientedtowardthecenteroftheplanet.Flowlinesof agradientvector“eld F = f arenormaltolevelsurfacesof f and orientedinthedirectioninwhich f increases(mostrapidly).Theyare thecurvesofsteepestascentofthefunction f .Flowlinesoftheair velocityvector“eldareoftenshowninweatherforecaststoindicate thewinddirectionoverlargeareas.Forexample,”owlinesoftheair velocityofahurricanewouldlooklikeclosedloopsaroundtheeyeof thehurricane. Thequalitativebehaviorof”owlinesmaybeunderstoodbyplotting vectors F atseveralpoints riandsketchingcurvesthroughthemso thatthevectors Fi= F ( ri)aretangenttothecurves.Findingthe exactshapeofthe”owlinesrequiressolvingdierentialequations.If r = r ( t )isaparametricequationofa”owline,then r( t )isparallel to F ( r ( t )).Sothederivative r( t )mustbeproportionalto F ( r ( t )), whichde“nesasystemofdierentialequationsforthecomponentsof thevectorfunction r ( t ),forexample, r( t )= F ( r ( t )). Example 15.1 Analyze”owlinesoftheplanarvector“eld F = ( Š y,x, 0) Solution: Bynotingthat F r =0,itisconcludedthatatany point F isperpendiculartothepositionvector r =( x,y, 0)inthe

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110.LINEINTEGRALSOFAVECTORFIELD285 plane.So”owlinesarecurveswhosetangentvectorisperpendicular tothepositionvector.If r = r ( t )isaparametricequationofsucha curve,then r ( t ) r( t )=0or( d/dt ) r2( t )=0andhence r2( t )=const, whichisacirclecenteredattheorigin.So”owlinesareconcentric circles.Atthepoint(1 0 0),thevector“eldisdirectedalongthe y axis: F (1 0 0)=(0 1 0)= e2.Therefore,the”owlinesareoriented counterclockwise.Themagnitude F = x2+ y2remainsconstant oneachcircleandincreaseswithincreasingcircleradius. 110.3.LineIntegralofaVectorField.Theworkdonebyaconstant force F inmovinganobjectalongastraightlineisgivenby W = F d where d isthedisplacementvector.Supposethattheforcevariesin spaceandthedisplacementtrajectoryisnolongerastraightline.What istheworkdonebytheforce?Thisquestionleadstotheconceptof thelineintegralofavector“eld. Let C beasmoothcurvethatgoesfromapoint ratoapoint rbandhasalength L .Considerapartitionof C bysegments Ci, i =1 2 ,...,N ,oflength s = L/N .Sincethecurveissmooth,each segmentcanbeapproximatedbyastraightlinesegmentoflength s orientedalongtheunittangentvector T ( r i)atasamplepoint r i Ci.Theworkalongthesegment Cicanthereforebeapproximatedby Wi= F ( r i) T ( r i) s sothatthetotalworkisapproximatelythesum W = W1+ W2+ + WN.Theactualworkshouldnotdependon thechoiceofsamplepoints.Thisproblemisresolvedbytheusualtrick ofintegralcalculus,thatis,byre“ningapartition,“ndingthelowand uppersums,andtakingtheirlimits.Iftheselimitsexistandcoincide, thelimitingvalueisthesought-forwork.Thetechnicalitiesinvolved maybesparedbynotingthat Wi= f ( r i) s ,where f ( r )= F ( r ) T ( r ) and T ( r )denotestheunittangentvectoratapoint r C .The approximatetotalworkappearstobeaRiemannsumfortheline integralof f along C .Sothe workisthelineintegralwithrespectto thearclengthofthetangentialcomponent F T oftheforce Definition 15.3 (LineIntegralofaVectorField) Thelineintegralofavector“eld F alongasmoothcurve C is CF d r = CF T ds, where T istheunittangentvectorto C ,providedthetangentialcomponent F T ofthevector“eldisintegrableon C .

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28615.VECTORCALCULUS Theintegrabilityof F T isde“nedinthesenseoflineintegralsfor ordinaryfunctions(seeDe“nition14.15)110.4.EvaluationofLineIntegralsofVectorFields.Thelineintegralof avector“eldisevaluatedinmuchthesamewayasthelineintegralof afunction. Theorem 15.1 (EvaluationofLineIntegrals) Let F =( F1,F2,F3) beacontinuousvector“eldon E andlet C bea smoothcurve C in E thatoriginatesfromapoint raandterminatesat apoint rb.Supposethat r ( t )=( x ( t ) ,y ( t ) ,z ( t )) t [ a,b ] ,isavector functionthattracesoutthecurve C sothat r ( a )= raand r ( b )= rb. Then CF ( r ) d r = CF T ds = b aF ( r ( t )) r( t ) dt = b a F1( r ( t )) x( t )+ F2( r ( t )) y( t )+ F3( r ( t )) z( t ) dt. (15.35) Proof. Theunittangentvectorreads T = r/ r and ds = r dt Therefore, T ds = r( t ) dt .Thefunction f ( t )= F ( r ( t )) r( t )iscontinuouson[ a,b ],andtheconclusionofthetheoremfollowsfromTheorem 14.15. Equation(15.35)alsoholdsif C ispiecewisesmoothand F hasa “nitenumberofboundedjumpdiscontinuitiesalong C muchlikein thecaseofthelineintegralofordinaryfunctions. Incontrasttothelineintegralofordinaryfunctions,thelineintegral ofavector“elddependsontheorientationof C .Theorientationof C is“xedbytheconditions r ( a )= raand r ( b )= rbforavectorfunction r ( t ),where a t b ,providedthevectorfunctiontracesoutthecurve onlyonce.If r ( t )tracesout C from rbto ra,thentheorientationis reversed,andsuchacurveisdenotedby Š C .Thelineintegralchanges itssignwhentheorientationofthecurveisreversed: (15.36) Š CF d r = Š CF d r becausethedirectionofthederivative r( t )isreversedforall t Theevaluationofalineintegralincludesthefollowingsteps: Step1 .Ifthecurve C isde“nedasapointsetinspacebysome geometricalmeans,then“nditsparametricequations r = r ( t )that agreewiththeorientationof C .Hereitisusefultorememberthat,if r ( t )correspondstotheoppositeorientation,thenitcanstillbeused accordingto(15.36). Step2 .Restricttherangeof t toaninterval[ a,b ]sothat C istraced

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110.LINEINTEGRALSOFAVECTORFIELD287 outonlyonceby r ( t ). Step3 .Substitute r = r ( t )intotheargumentsof F toobtainthe valuesof F on C andcalculatethederivative r( t )andthedotproduct F ( r ( t )) r( t ). Step4 .Evaluatethe(ordinary)integral(15.35). Remark. If C ispiecewisesmooth(e.g.,theunionofsmoothcurves C1and C2),thentheadditivityoftheintegralshouldbeused: CF d r = C1F d r + C2F d r Remark. Ifacurveis de“ned asavectorfunctionon[ a,b ](seeSection79.3),then r ( t )maytraceitsrange(asapointsetinspace)or somepartsofitseveraltimesas t changesfrom a to b .Iftwovector functions r1( t )and r2( t )on[ a,b ]havethesamerangebut r1( t ) = r2( t ) forsomevaluesof t [ a,b ],theyareconsidered dierent curves.For example, r1=(cos t, sin t, 0)and r1( t )=(cos(2 t ) sin(2 t ) 0)havethe samerangeon[0 2 ],whichisthecircleofunitradius,but r2( t )traces outthecircletwice.Notethatthesecurveshavedierentlengths, L1=2 and L2=4 .Sothelineintegral(15.35)maybedierentfor twocurves de“ned astwovectorfunctions,eventhoughtherangesof thesefunctionscoincideaspointsetsinspace.Thecurvede“nedbya vectorismuchlikethetrajectoryofaparticlethatcanpassthrough thesamepointsmultipletimes. Example 15.2 Evaluatethelineintegralof F =( Š y,x,z2) alonga closedcontour C thatconsistsofoneturnofahelixofradius R ,which beginsatthepoint ra=( R, 0 0) andendsatthepoint rb=( R, 0 2 h ) andastraightlinesegmentfrom rbto ra. Solution: Let C1beoneturnofthehelixandlet C2bethestraight linesegment.Twolineintegralshavetobeevaluated.Theparametricequationsofthehelixare r ( t )=( R cos t,R sin t,ht )sothat r (0)=( R, 0 0)and r (2 )=( R, 0 ,h )asrequiredbytheorientationof C1.Therangeof t hastoberestrictedto[0 2 ].Then r( t )=( Š R sin t,R cos t,h ).Therefore, F ( r ( t )) r( t )=( Š R sin t,R cos t,h2t2) ( Š R sin t,R cos t,h ) = R2+ h3t2, C1F d r = 2 0F ( r ( t )) r( t ) dt = 2 0( R2+ h3t2) dt =2 R2+ (2 h )3 3 .

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28815.VECTORCALCULUS Theparametricequationsofthelinethroughtwopoints raand rbare r ( t )= ra+ v t ,where v = rbŠ raisthevectorparalleltotheline, orinthecomponents r =( R, 0 0)+ t (0 0 2 h )=( R, 0 2 ht ).Then r (0)= raand r (1)= rbsothattheorientationisreversedif t [0 1]. Thefoundparametricequationsdescribethecurve Š C2.Onehas r( t )=(0 0 2 h )andhence F ( r ( t )) r( t )=(0 ,R, (2 h )2t2) (0 0 2 h )=(2 h )3t2, C2F d r = Š Š C2F d r = Š (2 h )31 0t2dt = Š (2 h )3 3 Thelineintegralalong C isthesumoftheseintegrals,whichisequal to2 R2. 111.FundamentalTheoremforLineIntegrals Recallthefundamentaltheoremofcalculus,whichassertsthat,if thederivative f( x )iscontinuousonaninterval[ a,b ],then b af( x ) dx = f ( b ) Š f ( a ) Itappearsthatthereisananalogofthistheoremforlineintegrals.111.1.ConservativeVectorFields.Definition 15.4 (ConservativeVectorFieldandItsPotential) Avector“eld F inaregion E issaidtobeconservativeifthereisa function f ,calleda potential of F ,suchthat F = f in E Conservativevector“eldsplayasigni“cantroleinmanypractical applications.Ithasbeenprovedearlier(seeStudyProblem13.9)thatif aparticlemovesalongatrajectory r = r ( t )undertheforce F = Š U thenitsenergy E = m v2/ 2+ U ( r ),where v = risthevelocity,isconservedalongthetrajectory, dE/dt =0.Inparticular,Newtonsgravitationalforceisconservative, F = Š U ,where U ( r )= Š GMm r Š 1. Astaticelectric“eld(theCoulomb“eld)createdbyadistributionof staticelectricchargesisalsoconservative.Conservativevector“elds havearemarkableproperty. Theorem 15.2 (FundamentalTheoremforLineIntegrals) Let C beasmoothcurveinaregion E withinitialandterminalpoints raand rb,respectively.Let f beafunctionon E whosegradient f is continuouson C .Then (15.37) C f d r = f ( rb) Š f ( ra) .

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111.FUNDAMENTALTHEOREMFORLINEINTEGRALS289 Proof. Let r = r ( t ), t [ a,b ],betheparametricequationsof C such that r ( a )= raand r ( b )= rb.Then,by(15.35)andthechainrule, C f d r = b a( f xx+ f yy+ f zz) dt = b ad dt f ( r ( t )) dt = f ( rb) Š f ( ra) Thelatterequalityholdsbythefundamentaltheoremofcalculusand thecontinuityofthe“rst-orderderivativesof f and r( t )forasmooth curve. 111.2.PathIndependenceofLineIntegrals.Definition 15.5 (PathIndependenceofLineIntegrals) Acontinuousvector“eld F haspath-independentlineintegralsif C1F d r = C2F d r foranytwosimple,piecewise-smoothcurvesinthedomainof F with thesameendpoints. Recallthatacurveissimpleifitdoesnotintersectitself(seeSection79.3).Animportantconsequenceofthefundamentaltheoremfor lineintegralsisthattheworkdonebyaconservativeforce, F = f is path-independent .Soacriterionforavector“eldtobeconservative wouldbeadvantageousforevaluatinglineintegralsbecauseforaconservativevector“eldacurvemaybedeformedatconveniencewithout changingthevalueoftheintegral. Theorem 15.3 (Path-IndependentProperty) Let F beacontinuousvector“eldonanopenregion E .Then F has path-independentlineintegralsifandonlyifitslineintegralvanishes alongeverypiecewise-smooth,simple,closedcurve C in E .Inthat case,thereexistsafunction f suchthat F = f : F = f %CF d r =0 Thesymbol &Cisoftenusedtodenotelineintegralsalongaclosed path. Proof. Pickapoint r0in E andconsideranysmoothcurve C from r0toapoint r =( x,y,z ) E .Theideaistoprovethatthefunction (15.38) f ( r )= CF d r isapotentialof F ,thatis,toprovethat f = F underthecondition thatthelineintegralof F vanishesforeveryclosedcurvein E .This

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29015.VECTORCALCULUS guessŽfor f ismotivatedbythefundamentaltheoremforlineintegrals (15.37),where rbisreplacedbyagenericpoint r E .Thepotential isde“neduptoanadditiveconstant( ( f +const)= f )sothe choiceofa“xedpoint r0isirrelevant.First,notethatthevalueof f isindependentofthechoiceof C .Considertwosuchcurves C1and C2.Thentheunionof C1and Š C2(thecurve C2whoseorientation isreversed)isaclosedcurve,andthelineintegralalongitvanishes bythehypothesis.Ontheotherhand,thislineintegralisthesum oflineintegralsalong C1and Š C2.Bytheproperty(15.36),theline integralsalong C1and C2coincide.Tocalculatethederivative f x( r )= limh 0( f ( r + h e1) Š f ( r )) /h ,where e1=(1 0 0),letusexpressthe dierence f ( r + h e1) Š f ( r )viaalineintegral.Notethat E isopen, whichmeansthataballofsucientlysmallradiuscenteredatany pointin E iscontainedin E (i.e., r + h e1 E forasucientlysmall h ).Sincethevalueof f ispath-independent,forthepoint r + h e1, thecurvecanbechosensothatitgoesfrom r0to r andthenfrom r to r + h e1alongthestraightlinesegment.Denotethelatterby C Therefore, f ( r + h e1) Š f ( r )= CF d r becausethelineintegralof F from r0to r ispath-independent.A vectorfunctionthattracesout C is r ( t )=( t,y,z )if x t x + h Therefore, r( t )= e1and F ( r ( t )) r( t )= F1( t,y,z ).Thus, f x( r )=limh 01 h x + h xF1( t,y,z ) dt =limh 01 h x + h aŠ x a F1( t,y,z ) dt = x x aF1( t,y,z ) dt = F1( x,y,z )= F1( r ) bythecontinuityof F1.Theequalities f y= F2and f z= F3are establishedsimilarly.Thedetailsareomitted. Althoughthepathindependencepropertydoesprovideanecessaryandsucientconditionforavector“eldtobeconservative,itis ratherimpracticaltoverify(onecannotevaluatelineintegralsalong everyclosedcurve!).Amorefeasibleandpracticalcriterionisneeded, whichisestablishednext.Itisworthnotingthat(15.38)givesapracticalmethodof“ndingapotentialifthevector“eldisfoundtobe conservative(seethestudyproblemsattheendofthissection).111.3.TheCurlofaVectorField.Accordingtotherulesofvectoralgebra,theproductofavector a =( a1,a2,a3)andanumber s isde“ned by s a =( sa1,sa2,sa3).Byanalogy,thegradient f canbeviewedas

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111.FUNDAMENTALTHEOREMFORLINEINTEGRALS291 theproductofthevector =( /x,/y,/z )andascalar f : f = x y z f = f x f y f z Thecomponentsof arenotordinarynumbers,butrathertheyare operators (i.e.,symbolsstandingforaspeci“edoperationthathasto becarriedout).Forexample,( /x ) f meansthattheoperator /x isappliedtoafunction f andtheresultofitsactionon f isthepartial derivativeof f withrespectto x .Thedirectionalderivative Duf can beviewedastheresultoftheactionoftheoperator Du= u = u1( /x )+ u2( /y )+ u3( /z )onafunction f .Inwhatfollows,the formalvector isviewedasanoperatorwhoseactionobeystherules ofvectoralgebra. Definition 15.6 (CurlofaVectorField) Thecurlofadierentiablevector“eld F is curl F = F Thecurlofavector“eldisalsoavector“eldwhosecomponents canbecomputedaccordingtothede“nitionofthecrossproduct: F =det e1 e2 e3 x y zF1F2F3 = F3 y Š F2 z e1+ F1 z Š F3 x e2+ F2 x Š F1 y e3. Whencalculatingthecomponentsofthecurl,theproductofacomponentof andacomponentof F meansthatthecomponentof operatesonthecomponentof F ,producingthecorrespondingpartial derivative.Ofcourse,itisassumedthatpartialderivativesofcomponentsof F existinorderforthecurltoexist. Example 15.3 Findthecurlofthevector“eld F =( yz,xyz,x2) Solution: F =det e1 e2 e3 x y zyzxyzx2 = ( x2) yŠ ( xyz ) z, Š ( x2) x+( yz ) z, ( xyz ) xŠ ( yz ) y =( Š xy,y Š 2 x,yz Š z )

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29215.VECTORCALCULUS Thegeometricalsigni“canceofthecurlofavector“eldwillbediscussed later.Herethecurlisusedtoformulatesucientconditionsforavector “eldtobeconservative.111.4.OntheUseoftheOperator.Therulesofvectoralgebraare usefultosimplifyalgebraicoperationsinvolvingtheoperator .For example, curl f = ( f )=( ) f = 0 becausethecrossproductofavectorwithitselfvanishes.However, thisformalalgebraicmanipulationshouldbeadoptedwithprecaution becauseitcontainsatacitassumptionthattheactionofthecomponentsof on f vanishes.Thelatterimposesconditionsonthe classoffunctionsforwhichsuchformalalgebraicmanipulationsare justi“ed.Indeed,accordingtothede“nition, f =det e1 e2 e3 x y zf xf yf z =( f zyŠ f yz,f zxŠ f xz,f xyŠ f yx) Thisvectorvanishes,providedtheorderofdierentiationdoesnot matter(i.e.,Clairautstheoremholdsfor f ).Thus, therulesofvector algebracanbeusedtosimplifytheactionofanoperatorinvolving ifthepartialderivativesofafunctiononwhichthisoperatoractsare continuousuptotheorderdeterminedbythataction .111.5.TestforaVectorFieldtoBeConservative.Aconservative,continuouslydierentiablevector“eldinanregion E hasbeenshownto havethevanishingcurl: F = f = curl F = 0 Unfortunately,theconverseis not trueingeneral.Inotherwords,the vanishingofthecurlofavector“elddoes not guaranteethatthevector “eldisconservative.Theconverseistrueonlyiftheregioninwhich thecurlvanishesbelongstoaspecialclass.Aregion E issaidtobe connected ifanytwopointsinitcanbeconnectedbyapaththatlies in E .Inotherwords,aconnectedregioncannotberepresentedasthe unionoftwoormorenon-intersecting(disjoint)regions. Definition 15.7 (SimplyConnectedRegion) Aconnectedregion E issimplyconnectedifeverysimpleclosedcurve in E canbecontinuouslyshrunktoapointin E whileremainingin E throughoutthedeformation.

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111.FUNDAMENTALTHEOREMFORLINEINTEGRALS293 Naturally,theentireEuclideanspaceissimplyconnected.Aballin spaceisalsosimplyconnected.If E istheregionoutsideaball,thenit isalsosimplyconnected.However,if E isobtainedbyremovingaline (oracylinder)fromtheentirespace,then E isnotsimplyconnected. Indeed,takeacirclesuchthatthelinepiercesthroughthediskbounded bythecircle.Thereisnowaythiscirclecanbecontinuouslycontracted toapointof E withoutcrossingtheline.Asolidtorusisnotsimply connected.(Explainwhy!)Asimplyconnectedregion D inaplane cannothaveholesŽinit. Theorem 15.4 (TestforaVectorFieldtoBeConservative) Supposethatavector“eld F iscontinuouslydierentiableonasimply connectedopenregion E .Then F isconservativein E ifandonlyif itscurlvanishesforallpointsof E : curl F = 0 onsimplyconnected E F = f on E. ThistheoremfollowsfromStokestheoremdiscussedlaterandhas twousefulconsequences.First,thetestforthepathindependenceof lineintegrals: curl F = 0 onsimplyconnected E C1F d r = C2F d r foranytwopaths C1and C2in E originatingfromapoint ra E and terminatingatanotherpoint rb E .Second,thetestforvanishing lineintegralsalongclosedpaths: curl F = 0 onsimplyconnected E %CF d r =0 where C isaclosedcurvein E .Theconditionthat E issimplyconnectediscrucialhere.Evenifcurl F = 0 ,but E isnotsimplyconnected,thelineintegralof F maystilldependonthepathandtheline integralalongaclosedpathmaynotvanish!Anexampleisgivenin oneofthestudyproblemsattheendofthissection. Newtonsgravitationalforcecanbewrittenasthegradient F = Š U ,where U ( r )= Š GMm r Š 1everywhereexcepttheorigin.Therefore,itscurlvanishesin E thatistheentirespacewithonepoint removed;itissimplyconnected.Hence,theworkdonebythegravitationalforceis independent ofthepathtraveledbytheobjectand determinedbythedierenceofvaluesofitspotential U attheinitial andterminalpointsofthepath. Example 15.4 Evaluatethelineintegralofthevector“eld F = ( F1,F2,F3)=( yz,xz + z +2 y,xy + y +2 z ) alongthepath C thatconsists

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29415.VECTORCALCULUS ofstraightlinesegments AB1, B1B2,and B2D ,wheretheinitialpoint is A =(0 0 0) B1=(2010 2011 2012) B2=(102 1102 2102) ,and theterminalpointis D =(1 1 1) Solution: Thepathlookscomplicatedenoughtocheckwhether F isconservativebeforeevaluatingthelineintegralusingtheparametric equationsof C .First,notethatthecomponentsof F arepolynomials andhencecontinuouslydierentiableintheentirespace.Therefore,if itscurlvanishes,then F isconservativeintheentirespaceastheentire spaceissimplyconnected: F =det e1 e2 e3 x y zF1F2F3 =det e1 e2 e3 x y zyzxz + z +2 yxy + y +2 z = ( F3) yŠ ( F2) z, Š ( F3) x+( F1) z, ( F2) xŠ ( F1) y =( x +1 Š ( x +1) Š y + y,z Š z )=(0 0 0) Thus, F isconservative.Nowtherearetwooptionsto“nishtheproblem. Option1 .Onecanusethepathindependenceofthelineintegral, whichmeansthatonecanpickanyotherpath C1connectingtheinitialpoint A andtheterminalpoint D toevaluatethelineintegralin question.Forexample,astraightlinesegmentconnecting A and D is asimpleenoughtoevaluatethelineintegral.Itsparametricequations are r = r ( t )=( t,t,t ),where t [0 1].Therefore, F ( r ( t )) r( t )=( t2,t2+3 t,t2+3 t ) (1 1 1)=3 t2+6 t andhence CF d r = C1F d r = 1 0(3 t2+6 t ) dt =4 Option2 .TheprocedureofSection89.1maybeusedto“ndapotential f of F (seealsothestudyproblemsattheendofthissection foranalternativeprocedure).Thelineintegralisthenfoundbythe fundamentaltheoremforlineintegrals.Put f = F .Thentheproblemisreducedto“nding f fromits“rst-orderpartialderivatives(the existenceof f hasalreadybeenestablished).Followingtheprocedure ofSection89.1, f x= F1= yz = f ( x,y,z )= xyz + g ( y,z ) ,

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111.FUNDAMENTALTHEOREMFORLINEINTEGRALS295 where g ( y,z )isarbitrary.Thesubstitutionof f intothesecondequation f y= F2yields xz + g y( y,z )= xz + z +2 y = g ( y,z )= y2+ zy + h ( z ) where h ( z )isarbitrary.Thesubstitutionof f = xyz + y2+ zy + h ( z ) intothethirdequation f z= F3yields xy + y + h( z )= xy + y +2 z = h ( z )= z2+ c, where c isaconstant.Thus, f ( x,y,z )= xyz + yz + z2+ y2+ c and CF d r = f (1 1 1) Š f (0 0 0)=4 bythefundamentaltheoremforlineintegrals. 111.6.StudyProblems.Problem15.1. Verifythat F = f = Š y x2+ y2, x x2+ y2, 2 z ,f ( x,y,z )=tanŠ 1( y/x )+ z2and curl F = 0 inthedomainof F .Evaluatethelineintegralof F alongthecircularpath C : x2+ y2= R2intheplane z = a .The pathisorientedcounterclockwiseasviewedfromthetopofthe z axis. Doestheresultcontradicttothefundamentaltheoremforlineintegrals? Explain. Solution: Astraightforwarddierentiationof f showsthatindeed f = F andthereforecurl F = 0 everywhereexcepttheline x = y =0where F isnotde“ned.Thepath C istracedoutby r ( t )= ( R cos t,R sin t,a ),where t [0 2 ].Then F ( r ( t ))=( Š RŠ 1sin t, RŠ 1cos t, 2 a )and r( t )=( Š R sin t,R cos t, 0).Therefore, F ( r ( t )) r( t )=1and %CF d r = 2 0dt =2 Sotheintegralovertheclosedcontourdoesnotvanishdespitethe factthat F = f ,whichseemstobeincon”ictwiththefundamental theoremforlineintegralsasbythelattertheintegralshouldhave vanished. Considerthevaluesof f alongthecircle.Byconstruction, f ( x,y,a ) = ( x,y )+ a2,where ( x,y )isthepolarangleinanyplane z = a .Itis0 onthepositive x axisandincreasesasthepointmovesabouttheorigin. Asthepointarrivesbacktothepositive x axis,theanglereachesthe value2 ;thatis, f isnotreallyafunctionontheclosedcontourbecause ittakes two values,0and2 ,atthesamepointonthepositive x axis.

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29615.VECTORCALCULUS Theonlywaytomake f afunctionistoremovethehalf-plane =0 fromthedomainof f .Thinkofacutinspacealongthehalf-plane. Butinthiscase,anyclosedpaththatintersectsthehalf-planebecomes nonclosedasithastwo distinct endpointsontheoppositeedgesofthe cut.Ifthefundamentaltheoremforlineintegralsisappliedtosucha path,thennocontradictionarisesbecausethevaluesof f ontheedges ofthecutdierexactlyby2 infullaccordancewiththeconclusionof thetheorem. Alternatively,theissuecanbeanalyzedbystudyingwhether F is conservativeinitsdomain E .Thevector“eldisde“nedeverywherein spaceexcepttheline x = y =0(the z axis).So E isnotsimplyconnected.Therefore,theconditioncurl F = 0 isnotsucienttoclaim thatthevector“eldisconservativeonitsdomain.Indeed,theevaluatedlineintegralalongtheclosedpath(whichcannotbecontinuously contracted,stayingwithin E ,toapointin E )showsthatthevector “eldcannotbeconservativeon E .Ifthehalf-plane =0isremoved from E ,then F isconservativeonthisreducedŽregionbecausethe latterissimplyconnected.Naturally,thelineintegralalonganyclosed paththatdoesnotcrossthehalf-plane =0(i.e.,itlieswithinthe reduceddomain)vanishes. Problem15.2. Provethatif F =( F1,F2,F3) isconservative,then itspotentialis f ( x,y,z )= x x0F1( t,y0,z0) dt + y y0F2( x,t,z0) dt + z z0F3( x,y,t ) dt, where ( x0,y0,z0) isanypointinthedomainof F .Usethisequationto “ndapotentialof F fromExample15.4. Solution: In(15.38),take C thatconsistsofthreestraightlinesegments,( x0,y0,z0) ( x,y0,z0) ( x,y,z0) ( x,y,z ).Theparametric equationofthe“rstline C1is r ( t )=( t,y0,z0),where x0 t x Therefore, r( t )=(1 0 0)and F ( r ( t )) r( t )= F1( t,y0,z0).Sotheline integralof F along C1givesthe“rsttermintheaboveexpressionfor f Similarly,thesecondtermisthelineintegralof F alongthesecondline r ( t )=( x,t,z0),where y0 t y ,sothat r( t )=(0 1 0).Thethird termisthelineintegralof F alongthethirdline r ( t )=( x,y,t ),where z0 t z .InExample15.4,itwasestablishedthat F =( F1,F2,F3)= ( yz,xz + z +2 y,xy + y +2 z )isconservative.Forsimplicity,choose

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112.GREENSTHEOREM297 ( x0,y0,z0)=(0 0 0).Then f ( x,y,z )= x 0F1( t, 0 0) dt + y 0F2( x,t, 0) dt + z 0F3( x,y,t ) dt =0+ y2+( xyz + yz + z2)= xyz + yz + z2+ y2, whichnaturallycoincideswith f foundbyadierent(longer) method. 112.GreensTheorem Greenstheoremshouldberegardedasthecounterpartofthefundamentaltheoremofcalculusforthedoubleintegral. Definition 15.8 (OrientationofPlanarClosedCurves). Asimple closedcurve C inaplanewhosesingletraversaliscounterclockwise (clockwise)issaidtobepositively(negatively)oriented. Asimpleclosedcurvedividestheplaneintotwoconnectedregions. Ifaplanarregion D isboundedbyasimpleclosedcurve,thenthe positivelyorientedboundaryof D isdenotedbythesymbol D Recallthatasimpleclosedcurvecanberegardedasacontinuous vectorfunction r ( t )=( x ( t ) ,y ( t ))on[ a,b ]suchthat r ( a )= r ( b )and, forany t1 = t2intheopeninterval( a,b ), r ( t1) = r ( t2);thatis, r ( t ) tracesout C onlyoncewithoutself-intersection.Apositiveorientation meansthat r ( t )tracesoutitsrangecounterclockwise.Forexample, thevectorfunctions r ( t )=(cos t, sin t )and r ( t )=(cos t, Š sin t )onthe interval[0 2 ]de“nethepositivelyandnegativelyorientedcirclesof unitradius,respectively. Theorem 15.5 (GreensTheorem) Let C beapositivelyoriented,piecewise-smooth,simple,closedcurvein theplaneandlet D betheregionboundedby C = D .Ifthefunctions F1and F2havecontinuouspartialderivativesonanopenregionthat contains D ,then D F2 x Š F1 y dA = %DF1dx + F2dy. Justlikethefundamentaltheoremofcalculus,Greenstheoremrelatesthederivativesof F1and F2intheintegrandtothevaluesof F1and F2ontheboundaryoftheintegrationregion.AproofofGreens theoremisratherinvolved.Hereitislimitedtothecasewhenthe region D issimple.

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29815.VECTORCALCULUS Proof(forsimpleregions) .Asimpleregion D admitstwoequivalentalgebraicdescriptions: D = { ( x,y ) | ybot( x ) y ytop( x ) ,x [ a,b ] } (15.39) D = { ( x,y ) | xbot( y ) x xtop( y ) ,y [ c,d ] } (15.40) Theideaoftheproofistoestablishtheequalities (15.41) %DF1dx = Š DF1 y dA, %DF2dy = DF2 x dA using,respectively,(15.39)and(15.40).Theconclusionofthetheorem isthenobtainedbyaddingtheseequations. Thelineintegralistransformedintoanordinaryintegral“rst.The boundary D containsfourcurves,denoted C1, C2, C3,and C4.The curve C1isthegraph y = ybot( x )whoseparametricequationsare r =( t,ybot( t )),where t [ a,b ].So C1istracedoutfromleftto rightasrequiredbythepositiveorientationof D .Thecurve C3is thetopboundary y = ytop( x ),and,similarly,itsparametricequations r ( t )=( t,ytop( t )),where t [ a,b ].Since C3istracedoutfromleft toright,theorientationof C3mustbereversed;thatis, D contains thecurve Š C3.Theboundarycurves C2and C4(thesidesof D ) aresegmentsoftheverticallines x = b (orientedupward)and x = a (orienteddownward),whichmaycollapsetoasinglepointifthegraphs y = ybot( x )and y = ytop( x )intersectat x = a or x = b orboth.The lineintegralsalong C2and C4donotcontributetothelineintegral withrespectto x along D because dx =0along C2and C4.By construction, x = t and dx = dt forthecurves C1and C2.Hence, %DF1dx = C1F1dx + Š C2F1dx = b a F ( x,ybot( x )) Š F ( x,ytop( x )) dx, wheretheproperty(15.36)hasbeenused.Next,thedoubleintegral istransformedintoanordinaryintegralbyconvertingittoaniterated integral: DF1 y dA = b aytop( x ) ybot( x )F1 y dydx = b a F ( x,ytop( x )) Š F ( x,ybot( x )) dx, wherethelatterequalityfollowsfromthefundamentaltheoremof calculusandthecontinuityof F1onanopenintervalthatcontains

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112.GREENSTHEOREM299 [ ybot( x ) ,ytop( x )]forany x [ a,b ](thehypothesisofGreenstheorem). Comparingtheexpressionofthelineanddoubleintegralsviaordinary integrals,thevalidityofthe“rstrelationin(15.41)isestablished.The secondequalityin(15.41)isprovedanalogouslybyusing(15.40).The detailsareomitted. Remark. Supposethatasmooth,orientedcurve C dividesaregion D intotwo simple regions D1and D2.Iftheboundary D1contains C (i.e.,theorientationof C coincideswiththepositiveorientationof D1),then D2mustcontainthecurve Š C andviceversa.Usingthe conventionalnotation F1dx + F2dy = F d r ,where F =( F1,F2),one infersthat %DF d r = %D1F d r + %D2F d r = D1 F2 x Š F1 y dA + D2 F2 x Š F1 y dA = D F2 x Š F1 y dA. The“rstequalityholdsbecauseofthecancellationofthelineintegralsalong C and Š C accordingto(15.36).Thevalidityofthesecond equalityfollowsfromtheproofofGreenstheoremforsimpleregions. Finally,theequalityisestablishedbytheadditivitypropertyofdoubleintegrals.Bymakinguseofsimilararguments,theproofcanbe extendedtoaregion D thatcanberepresentedastheunionofa“nite numberofsimpleregions. Remark. Lettheregions D1and D2beboundedbysimple,piecewisesmooth,closedcurvesandlet D2lieintheinteriorof D1.Considerthe region D thatwasobtainedfrom D1byremoving D2(theregion D has aholeoftheshape D2).MakinguseofGreenstheorem,one“nds D F2 x Š F1 y dA = D1 F2 x Š F1 y dA Š D2 F2 x Š F1 y dA = %D1F d r Š %D2F d r = %D1F d r + %Š D2F d r = %DF d r (15.42) ThisestablishesthevalidityofGreenstheoremfornotsimplyconnectedregions.Theboundary D consistsof D1and Š D2;thatis, theouterboundaryhasapositiveorientation,whiletheinnerboundary

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30015.VECTORCALCULUS isnegativelyoriented.Asimilarlineofreasoningleadstotheconclusionthatthisholdsforanynumberofholesin D :allinnerboundaries of D mustbenegativelyoriented.Suchorientationoftheboundaries canalsobeunderstoodasfollows.Letacurve C connectsapointof theouterboundarywithapointoftheinnerboundary.Letusmake acutoftheregion D along C .Thentheregion D becomessimply connectedand D consistsofa continuous curve(theinnerandouter boundaries,andthecurves C and Š C ).Theboundary D canalways bepositivelyoriented.Thelatterrequiresthattheouterboundary betracedcounterclockwise,whiletheinnerboundaryistracedclockwise(theorientationof C and Š C ischosenaccordingly).Byapplying Greenstheoremto D ,onecanseethatthelineintegralsover C and Š C arecancelledand(15.42)followsfromtheadditivityofthedoubleintegral.Evidently,thesameargumentcanbeusedtoestablish Greenstheoremforaregionwithmultipleholes(allinnerboundaries mustbeorientedclockwise).112.1.EvaluatingLineIntegralsviaDoubleIntegrals.Greenstheorem providesatechnicallyconvenienttooltoevaluatelineintegralsalong planarclosedcurves.Itisespeciallybene“cialwhenthecurveconsists ofseveralsmoothpiecesthatarede“nedbydierentvectorfunctions; thatis,thelineintegralmustbesplitintoasumoflineintegralstobe convertedintoordinaryintegrals.Sometimes,thelineintegralturns outtobemuchmorediculttoevaluatethanthedoubleintegral. Example 15.5 Evaluatethelineintegralof F =( y2+ ecos x, 3 xy Š sin( y4)) alongthecurve C thatistheboundaryofthehalfofthering: 1 x2+ y2 4 and y 0 ; C isorientedclockwise. Thecurve C consistsoffoursmoothpieces,thehalf-circlesofradii 1and2andtwostraightlinesegmentsofthe x axis,[ Š 2 Š 1]and[1 2]. Eachcurvecanbeeasilyparameterizedandthelineintegralinquestion canbetransformedintothesumoffourordinaryintegralswhichare thenevaluated.Thereaderisadvisedtopursuethisavenueofactions toappreciatethefollowingalternativewaybasedonGreenstheorem (thisisnotimpossibletoaccomplishifone“guresouthowtohandle theintegrationofthefunctions ecos xandsin( y4)whoseanti-derivatives arenotexpressibleinelementaryfunctions). Solution: Thecurve C isasimple,piecewise-smooth,closedcurve andthevector“eld F iscontinuouslydierentiable.Thus,Greens theoremappliesif D = Š C (becausetheorientationof C isnegative) and D isthehalf-ring.Onehas F1/y =2 y and F2/x =3 y .By

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112.GREENSTHEOREM301 Greenstheorem, %CF d r = Š %DF d r = Š D F2 x Š F1 y dA = Š DydA = Š 02 1r sin rdrd = Š 0sin d 2 1r2dr = Š 14 3 wherethedoubleintegralhasbeentransformedintopolar coordinates. 112.2.AreaofaPlanarRegionasaLineIntegral.Put F2= x and F1= 0.Then D F2 x Š F1 y dA = DdA = A ( D ) Thearea A ( D )canalsobeobtainedif F =( Š y, 0)or F =( Š y/ 2 ,x/ 2). ByGreenstheorem,theareaof D canbeexpressedbylineintegrals: (15.43) A ( D )= %Dxdy = Š %Dydx = 1 2 %Dxdy Š ydx, assuming,ofcourse,thattheboundaryof D isasimple,piecewisesmooth,closedcurve(orseveralsuchcurvesif D hasholes).Thereason thevaluesoftheselineintegralscoincideissimple.Thedierenceof anytwovector“eldsinvolvedisaconservativevector“eldwhoseline integralalongaclosedcurvevanishes.Forexample,for F =(0 ,x )and G ( Š y, 0),thedierenceis F Š G =( y,x )= f ,where f ( x,y )= xy sothat %DF d r Š %DG d r = %D( F Š G ) d r = %D f d r =0 Therepresentation(15.43)oftheareaofaplanarregionastheline integralalongitsboundaryisquiteusefulwhentheshapeof D istoo complicatedtobecomputedusingadoubleintegral(e.g.,when D isnot simpleand/orarepresentationofboundariesof D bygraphsbecomes technicallydicult). Example 15.6 Consideranarbitrarypolygonwhoseverticesin counterclockwiseorderare ( x1,y1) ( x2,y2) ,..., ( xn,yn) .Finditsarea. Solution: Evidently,agenericpolygonisnotasimpleregion(e.g.,it mayhaveastarlikeshape).Sothedoubleintegralisnotatallsuitable for“ndingthearea.Incontrast,thelineintegralapproachseemsfar morefeasibleastheboundaryofthepolygonconsistsof n straight linesegmentsconnectingneighboringvertices.If Ciissuchasegment orientedfrom( xi,yi)to( xi +1,yi +1)for i =1 2 ,...,n Š 1,then Cngoes from( xn,yn)to( x1,y1).Avectorfunctionthattracesoutastraight

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30215.VECTORCALCULUS linesegmentfromapoint ratoapoint rbis r ( t )= ra+( rbŠ ra) t ,where 0 t 1.Forthesegment Ci,take ra=( xi,yi)and rb=( xi +1,yi +1). Hence, x ( t )= xiŠ ( xi +1Š xi) t = xi+ xit and y ( t )= yi+( yi +1Š yi) t = yi+ yit .Forthevector“eld F =( Š y,x )on Ci,onehas F ( r ( t )) r( t )=( Š y ( t ) ,x ( t )) ( xi, yi)= xi yiŠ yi xi= xiyi +1Š yixi +1; thatis,the t dependencecancelsout.Therefore,takingintoaccount that Cngoesfrom( xn,yn)to( x1,y1),theareais A = 1 2 %Dxdy Š ydx = 1 2ni =1Cixdy Š ydx = 1 2n Š 1i =11 0( xiyi +1Š yixi +1) dt + 1 2 1 0( xny1Š ynx1) dt = 1 2 n Š 1i =1( xiyi +1Š yixi +1)+( xny1Š ynx1) SoGreenstheoremoersanelegantwayto“ndtheareaofageneralpolygonifthecoordinatesofitsverticesareknown.Asimple, piecewise-smooth,closedcurve C inaplanecanalwaysbeapproximatedbyapolygon.Theareaoftheregionenclosedby C cantherefore beapproximatedbytheareaofapolygonwithalargeenoughnumber ofvertices,whichisoftenusedinmanypracticalapplications.112.3.TheTestforPlanarVectorFieldstoBeConservative.GreenstheoremcanbeusedtoproveTheorem15.4forplanarvector“elds.Consideraplanarvector“eld F =( F1( x,y ) ,F2( x,y ) 0).Itscurlhasonly onecomponent: F =det e1 e2 e3 x y zF1( x,y ) F2( x,y )0 = e3 F2 x Š F1 y Supposethatthecurlof F vanishesthroughoutasimplyconnected openregion D F = 0 .Byde“nition,anysimpleclosedcurve C inasimplyconnectedregion D canbeshrunktoapointof D while remainingin D throughoutthedeformation(i.e.,anysuch C bounds asubregion Dsof D ).ByGreenstheorem,where C = Ds, %CF d r = Ds F2 x Š F1 y dA = Ds0 dA =0

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113.FLUXOFAVECTORFIELD303 foranyclosedsimplecurve C in D .Bythepath-independenceproperty, thevector“eld F isconservativein D .112.4.StudyProblems.Problem15.3. Evaluatethelineintegralof F =( y + ex2, 3 x Š sin( y2)) alongthecounterclockwise-orientedboundaryof D thatisenclosedby theparabolas y = x2and x = y2. Solution: Onehas F1/y =1and F2/x =3.ByGreens theorem, %DF d r = D2 dA =2 1 0 x x2dydx =2 1 0( x Š x2) dx = 1 3 Problem15.4. Provethatthelineintegraloftheplanarvector“eld F = Š y x2+ y2, x x2+ y2 alonganypositivelyoriented,simple,smooth,closedcurve C thatencirclestheoriginis 2 andthatitvanishesforanysuchcurvethatdoes notencircletheorigin. Solution: Ithasbeenestablished(seeStudyProblem15.1)that thecurlofthisvector“eldvanishesinthedomainthatistheentire planewiththeoriginremoved.If C doesnotencircletheorigin,then F2/x Š F1/y =0throughouttheregionencircledby C ,andthe lineintegralalong C vanishesbyGreenstheorem.Givenaclosedcurve C thatencirclestheorigin,butdoesnotgothroughit,onecanalways “ndadiskofasmallenoughradius a suchthatthecurve C doesnot intersectit.Let Dabetheregionboundedbythecircle Caofradius a andthecurve C .Then F2/x Š F1/y =0throughout Da.Let C beorientedcounterclockwise,while Caisorientedclockwise.Then Daistheunionof C and Ca.ByGreenstheorem, %DF d r =0 %CF d r = Š %CaF d r = %Š CaF d r =2 because Š Caisthecircleorientedcounterclockwiseandforsuchacircle thelineintegralhasbeenfoundtobe2 (seeStudyProblem15.1). 113.FluxofaVectorField Theideaofa”uxofavector“eldstemsfromanengineeringproblemofmasstransferacrossasurface.Supposethereisa”owofa”uid

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30415.VECTORCALCULUS orgaswithaconstantvelocity v andaconstantmassdensity (mass perunitvolume).Let A beaplanarareaelementplacedintothe ”ow.Atwhatrateisthe”uidorgascarriedbythe”owacrossthearea A ?Inotherwords,whatisthemassof”uidtransferredacross A perunittime?Thisquantityiscalleda ”ux ofthemass”owacross thearea A Suppose“rstthatthemass”owisnormaltotheareaelement. Considerthecylinderwithanaxisparallelto v withcrosssectionarea A andheight h = v t ,where v = v isthe”owspeedand t isa timeinterval.Thevolumeofthecylinderis V = h A = v t A Intime t ,allthemassstoredinthiscylinderistransferredbythe ”owacross A .Thismassis m = V = v t A ,andthe”ux is = m t = v A. The”uxdependsontheorientationofanareaelementrelativeto the”ow.Ifthe”owisparalleltotheareaelement,thennomassis transferredacrossit.Thevelocityvectorcanbeviewedasthesumof avectornormaltotheareaelementandavectortangentialtoit.Only thenormalcomponentofthe”owcontributestothe”ux.If n isthe unitnormalvectortotheareaelementand istheanglebetween v and n ,thenthenormalcomponentofthevelocityis vn= v cos = v n and (15.44)= vn A = v n A = F n A = Fn A, wherethevector F = v characterizesthemass”ow(howmuchŽ ( )andhowfastŽ( v ))and Fnisitscomponentnormaltothearea element. Ifnowthemass”owisnotconstant(i.e., F becomesavector“eld), thenits”uxacrossasurface S canbede“nedbypartitioning S into smallsurfaceareaelements Si, i =1 2 ,...,N ,whosesurfaceareasare Si.Let r ibeasamplepointin Siandlet nibetheunitvectornormal to Siat r i.Ifthesize(theradiusofthesmallestballcontaining Si)is small,then,byneglectingvariationsin F andthenormal n within Si, the”uxacross Sicanbeapproximatedby(15.44),i F ( r i) ni Si. Theapproximationbecomesbetterwhen N sothatthesizesof Sidecreaseto0uniformlyandhencethetotal”uxis =limN Ni =1i=limN Ni =1F ( r i) ni Si=limN Ni =1Fn( r i) Si. ThesuminthisequationisnothingbuttheRiemannsumofthefunction Fn( r )overapartitionofthesurface S .Naturally,itslimitisthe

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113.FLUXOFAVECTORFIELD305 surfaceintegralof Fn( r )over S .Thus, the”uxofavector“eldacross asurfaceisthesurfaceintegralofthenormalcomponentofthevector “eld .113.1.OrientableSurfaces.Theabovede“nitionofthe”uxsounds ratherplausible.However,itcontainsatacitassumptionthatthe normalcomponentofavector“eldcanalwaysbe uniquely de“nedas acontinuousfunctiononasmoothsurface.Itappearsthatthereare smoothsurfacesforwhichthiscannotbedone! Thenormal n = n ( r )dependsonthepointofasurface.Soitisa vector“eldon S .Inorderforthenormalcomponent Fntobeuniquely de“ned,therule n = n ( r )shouldassignjustone n foreverypointof S Furthermore, n ( r )shouldbecontinuouson S andhencealongevery closedcurve C inasmoothsurface S .Inotherwords,if n istransported alongaclosedcurve C in S ,theinitial n mustcoincidewiththe“nal n .Since,ateverypointof S ,thereareonlytwopossibilitiestodirect theunitnormalvector,bycontinuitythedirectionof n ( r )de“nesone sideof S ,whilethedirectionof Š n ( r )de“nestheotherside.Thus, thenormalcomponentofavector“eldiswellde“nedfortwo-sided surfaces.Forexample,theoutwardnormalofasphereiscontinuous alonganyclosedcurveonthesphere(itremainsoutwardalongany closedcurve)andhencede“nestheoutersideofthesphere.Ifthe normalonthesphereischosentobeinward,thenitisalsocontinuous andde“nestheinnersideofthesphere. Arethereone-sidedsurfaces?Ifsuchasurfaceexists,itshouldhave quiteremarkableproperties.Takeapointonit.Inaneighborhoodof thispoint,onealwaysthinksabouttwosides(asurfaceissmooth). Onesideisde“nedbyanormal n (face-uppatch),whiletheotherhas thesameshapebutitsnormalis Š n (face-downpatch).Foraonesidedsurface,theface-upandface-downpatchesmustbeonthesame sideofthesurface.Thisimpliesthatthereshouldexistacurveonthe surfacethatstartsatapointononesideandcanreachtheverysame pointbutfromtheotherside without crossingthesurfaceboundaries (ifany)orpiercingthesurface.Bymovingtheface-uppatchalongsuch acurve,itbecomestheface-downpatch.Thus,thenormalcannotbe uniquelyde“nedonaone-sided S .113.1.1.ExamplesofOne-SidedSurfaces.One-sidedsurfacesdoexist. Toconstructanexample,takearectangularpieceofpaper.Putupward arrowsonitsverticalsidesandgluethesesidessothatthearrows remainparallel.Indoingso,acylinderisobtained,whichisatwosidedsurface(thereisnocurvethattraversesfromonesidetothe

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30615.VECTORCALCULUS otherwithoutcrossingtheboundarycirclesformedbythehorizontal sidesoftherectangle).Thegluingcanbedonedierently.Before gluingtheverticalsides,twisttherectanglesothatthearrowson thembecomeoppositeandthengluethem.Theresultingsurfaceisthe famous M¨ obiusstrip (namedaftertheGermanmathematicianAugust M¨ obius).Itisone-sided.Allcurveswindingaboutittraversebothsides ofthegluedrectanglewithoutcrossingitsboundaries(thehorizontal edges). Thereareone-sidedsurfaceswithoutboundaries(likeasphere). Themostfamousoneisa Kleinbottle .Takeabottle.Drillaholeon thesidesurfaceandinthebottomofthebottle.Supposetheneck ofthebottleis”exible(arubberŽbottle).Benditsneckandpullit throughtheholeonthebottlessidesurface(sothatneck“tstightly intothehole).Finally,attachtheedgeofthebottlesnecktothe edgeoftheholeinthebottlebottom.Theresultisasurfacewithout boundariesanditisone-sided.Abugcancrawlalongthissurfaceand getinandoutofthebottle.113.1.2.FluxandOne-SidedSurfaces.The”uxmakessenseonlyfor two-sidedsurfaces.Indeed,the”uxmeansthatsomethingisbeing transferredfromonesidetotheothersideofthesurface(i.e., across it)atacertainrate.Ifthesurfaceisone-sided,thenonecangetacross itŽbymerelyslidingalongit!Forexample,amass”ow tangential to aone-sidedsurfacecantransfermassacrossthesurface. Definition 15.9 (OrientableSurface) Asmoothsurfaceiscalled orientable ifthereisnoclosedcurveinit suchthatthenormalvectorisreversedwhenmovedaroundthiscurve. Soorientablesurfacesaretwo-sidedsurfaces.The”uxofavector “eldcanonlybede“nedacrossanorientablesurface.113.2.FluxasaSurfaceIntegral.Definition 15.10 (FluxofaVectorField) Let S beanorientablesmoothsurfaceandlet n betheunitnormal vectoron S .The”uxofavector“eld F across S isthesurfaceintegral = SF n dS, providedthenormalcomponent F n ofthevector“eldisintegrable on S .

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113.FLUXOFAVECTORFIELD307 Theintegrabilityofthenormalcomponent Fn( r )= F n isde“ned inthesenseofsurfaceintegralsofordinaryfunctions(seeDe“nition 14.16).113.3.EvaluationoftheFluxofaVectorField.Supposethatasurface S isagraph z = g ( x,y )overaregion( x,y ) D .Therearetwo possibleorientationsof S .Thenormalvectortothetangentplaneat apointof S is n =( Š g x, Š g y, 1)(seeSection91.1).Its z component ispositive.Forthisreason,thegraphissaidtobe orientedupward Alternatively,onecantakethenormalvectorintheoppositedirection, n =( g x,g y, Š 1).Inthiscase,thegraphissaidtobe orienteddownward Accordingly,the upward ( downward )”ux,denoted(),ofavector “eldisassociatedwiththeupward(downward)orientationofthegraph. Whentheorientationofasurfaceisreversed,the”uxchangesitssign: = Š Considertheupward-orientedgraph z = g ( x,y ).Theunitnormal vectorreads n = 1 n n = 1 J ( Š g x,g y, 1) ,J = 1+( g x)2+( g y)2. Recallthattheareatransformationlawforagraphis dS = JdA Therefore,inthein“nitesimal”uxacrossthesurfacearea, dS canbe writtenintheform F n dS = F n 1 J JdA = F n dA, wherethevector“eldmustbeevaluatedon S ,thatis, F = F ( x,y, g ( x,y ))(thevariable z isreplacedby g ( x,y )because z = g ( x,y )for anypoint( x,y,z ) S ).Ifthedotproduct F n isanintegrable functionon D ,the”uxexistsandisgivenbythedoubleintegralover D .Thefollowingtheoremhasbeenproved. Theorem 15.6 (EvaluationoftheFluxAcrossaGraph) Supposethat S isagraph z = g ( x,y ) ofafunction g whose“rst-order partialderivativesarecontinuouson D .Let S beorientedupwardby thenormalvector n =( Š g x, Š g y, 1) andlet F beacontinuousvector “eldon S .Then = SF n dS = DFn( x,y ) dA, Fn( x,y )= F n z = g ( x,y )= Š g xF1( x,y,g ) Š g yF2( x,y,g )+ F3( x,y,g ) .

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30815.VECTORCALCULUS Theevaluationofthesurfaceintegralinvolvesthefollowing steps: Step1 .Represent S asagraph z = g ( x,y )(i.e.“ndthefunction g usingageometricaldescriptionof S ).If S cannotberepresentedas graphofasinglefunction,thenithastobesplitintopiecessothat eachpiececanbedescribedasagraph.Bytheadditivityproperty,the surfaceintegralover S isthesumofintegralsovereachpiece. Step2 .Findtheregion D thatde“nesthepartofthegraphthat coincideswith S (if S isnotthegraphonthewholedomainof g ). Step3 .Determinetheorientationof S (upwardordownward)from theproblemdescription.Thesignofthe”uxisdeterminedbytheorientation.Calculatethenormalcomponent Fn( x,y )ofthevector“eld asafunctionon D Step4 .Evaluatethedoubleintegralof Fnover D Example 15.7 Evaluatethedownward”uxofthevector“eld F = ( xz,yz,z ) acrossthepartoftheparaboloid z =1 Š x2Š y2inthe“rst octant. Solution: Thesurfaceisthepartofthegraph z = g ( x,y )=1 Š x2Š y2inthe“rstoctant.Theparaboloidintersectsthe xy plane( z =0)along thecircle x2+ y2=1.Therefore,theregion D isthequarterofthe diskboundedbythiscircleinthe“rstquadrant( x,y 0).Since S is orienteddownward, n =( g x,g y, Š 1)=( Š 2 x, Š 2 y, Š 1)andthenormal componentof F is Fn( x,y )=( xg,yg,g ) ( Š 2 x, Š 2 y, Š 1)= Š (1 Š x2Š y2)(1+2 x2+2 y2) Convertingthedoubleintegralof Fntopolarcoordinates, = DFn( x,y ) dA = Š / 2 01 0(1+ r2)(1+2 r2) rdrd = Š 19 24 Thenegativevalueofthedownward”uxmeansthattheactualtransfer ofaquantity(likeamass),whose”owisdescribedbythevector“eld F ,occursintheupwarddirectionacross S 113.4.ParametricSurfaces.Ifthesurface S inthe”uxintegralisde“nedbytheparametricequations r = r ( u,v ),where( u,v ) D ,then, byCorollary14.5,thenormalvectorto S is n = r u r v(or Š n ;thesign ischosenaccordingtothegeometricaldescriptionoftheorientationof S ).Since n = J ,where J determinestheareatransformationlaw dS = JdA ( dA = dudv ),the”uxofavector“eld F acrossthesurface

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113.FLUXOFAVECTORFIELD309 area dS reads: F ( r ( u,v )) n dS = F ( r ( u,v )) n dA = F ( r ( u,v )) ( r u r v) dA = Fn( u,v ) dA andthe”uxisgivenbythedoubleintegral = FF n dS = DF ( r ( u,v )) ( r u r v) dA = DFn( u,v ) dA. Naturally,agraph z = g ( x,y )isdescribedbytheparametricequations r ( u,v )=( u,v,g ( u,v )),whichisaparticularcaseoftheaboveexpression;itcoincideswiththatgiveninTheorem15.6( x = u and y = v ). Adescriptionofsurfacesbyparametricequationsisespeciallyconvenientforclosedsurfaces(i.e.,whenthesurfacecannotberepresented asagraphofasinglefunction). Example 15.8 Evaluatetheoutward”uxofthevector“eld F = ( z2x,z2y,z3) acrossthesphereofunitradiuscenteredattheorigin. Solution: Theparametricequationsofthesphereofradius R =1are givenin(14.31),andthenormalvectoriscomputedinExample14.38: n =sin( u ) r ( u,v ),where r ( u,v )=(cos v sin u, sin v sin u, cos u )and ( u,v ) D =[0 ] [0 2 ];itisanoutwardnormalbecausesin u 0. Itisconvenienttorepresent F = z2r sothat Fn( u,v )= F ( r ( u,v )) n =cos2u sin u r ( u,v ) r ( u,v ) =cos2u sin u r ( u,v ) 2=cos2u sin u because r ( u,v ) 2= R2=1.Theoutward”uxreads = SF n dS = Dcos2u sin udA = 2 0dv 0cos2u sin udu = 4 3 NonorientableSurfaces.Nonorientablesurfacescanbedescribedby theparametricequations r = r ( u,v )orbyanalgebraicequation F ( x,y,z )=0(asalevelsurfaceofafunction).Forexample,aM¨ obius stripofwidth2 h withmidcircleofradius R andheight z =0isde“ned bytheparametricequations (15.45) r ( u,v )= [ R + u cos( v/ 2)]cos v, [ R + u cos( v/ 2)]sin v,u sin( v/ 2) ,

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31015.VECTORCALCULUS where( u,v ) D =[ Š h,h ] [0 2 ].ItalsofollowsfromtheseparametricequationsthattheM¨ obiusstripisde“nedbya cubic surface: Š R2y + x2y + y3Š 2 Rxz Š 2 x2z Š 2 y2z + yz3=0 Thisisveri“edbysubstitutingtheparametricequationsintothisalgebraicequationandshowingthattheleftsidevanishesforall( u,v ) D Letusprovethatthesurfacede“nedbytheparametricequations (15.45)isnotorientable.Todoso,oneshouldanalyzethebehaviorof anormalvectorwhenthelatterismovedaroundaclosedcurveinthe surface.Considerthecircleinthe xy planede“nedbythecondition u =0: r (0 ,v )=( R cos v,R sin v, 0).Itiseasytoshowthat r u(0 ,v )=(cos( v/ 2)cos v, cos( v/ 2)sin v, sin( v/ 2)) r v(0 ,v )=( Š R sin v,R cos v, 0) PWhen r (0 ,v )returnstotheinitialpoint,thatis, r (0 ,v +2 )= r (0 ,v ), thenormalvectorisreversed.Indeed, r u(0 ,v +2 )= Š r u(0 ,v )and r v(0 ,v +2 )= r v(0 ,v ).Hence, n (0 ,v +2 )= r u(0 ,v +2 ) r v(0 ,v +2 )= Š r u(0 ,v ) r v(0 ,v ) = Š n (0 ,v ); thatis,thesurfacede“nedbytheseparametricequationsis not orientablebecausethenormalvectorisreversedwhenmovedarounda closedcurve. So,ifasurface S isde“nedbyparametricoralgebraicequations, onestillhastoverifythatitisorientable(i.e.,itistwo-sided!),when evaluatingthe”uxacrossit;otherwise,the”uxmakesnosense.114.StokesTheorem114.1.VectorFormofGreensTheorem.ItwasshowninSection112.3 thatthecurlofaplanarvector“eld F ( x,y )=( F1( x,y ) ,F2( x,y ) 0)is paralleltothe z axis, F =( F2/ Š F1/x ) e3.Thisobservation allowsustoreformulateGreenstheoreminthefollowingvectorform: %DF d r = D(curl F ) e3dA. Thus, thelineintegralofavector“eldalongaclosedsimplecurveis determinedbythe”uxofthecurlofthevector“eldacrossthesurface boundedbythiscurve .Itturnsoutthatthisstatementholdsnotonly inaplane,butalsoinspace.Itisknownas Stokestheorem .114.2.StokesTheorem.

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114.STOKESTHEOREM311 114.2.1.Positive(Induced)OrientationofaClosedCurve.Suppose S is asmoothsurfaceorientedbyitsnormalvector n andboundedbya closedsimplecurve C .Consideratangentplaneatapoint r0of S Anycircleinthetangentplanecenteredat r0canalwaysbeoriented counterclockwiseasviewedfromthetopofthenormalvector n = n0at r0.Thiscircleissaidtobe positivelyorientedrelativetotheorientation of S .Sincethesurfaceissmooth,acircleofasucientlysmallradius canalwaysbeprojectedontoaclosedsimplecurvein S bymovingeach pointofthecircleparallelto n0.Thiscurveisalso positivelyoriented relativeto n0.Itcanthenbecontinuously(i.e.,withoutbreaking) deformedalong S sothatitspartliesontheboundary C afterthe deformation.Theorientationispreservedthroughoutthedeformation, andhenceitinducesa positiveorientation oftheboundary C .The positivelyorientedboundaryof S isdenotedby S Inotherwords,thepositive(orinduced)orientationof C means thatifonewalksinthepositivedirectionalong C withoneshead pointinginthedirectionof n ,thenthesurfacewillalwaysbeonones left.Let S beagraph z = g ( x,y )over D orientedupward.Then S isobtainedfrom D (apositivelyorientedboundaryof D )bylifting pointsof D to S paralleltothe z axis. Theorem 15.7 (StokesTheorem) Let S beanoriented,piecewise-smoothsurfacethatisboundedbya simple,closed,piecewise-smoothcurve C withpositiveorientation C = S .Let F beacontinuouslydierentiablevector“eldonanopenspatial regionthatcontains S .Then %SF d r = Scurl F n dS, where n istheunitnormalvectoron S Stokestheoremisdiculttoproveingeneral.Hereitisproved foraparticularcasewhen S isagraphofafunction. Proof(for S beingagraph) .Let S betheupward-orientedgraph z = g ( x,y ),( x,y ) D ,where g istwicecontinuouslydierentiableon D and D isasimpleplanarregionwhoseboundary D corresponds totheboundary S .Inthiscase,thenormalvector n =( Š g x, Š g y, 1) andtheupward”uxofcurl F across S canbeevaluatedaccordingto

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31215.VECTORCALCULUS Theorem15.6,where F isreplacedby F : Scurl F n dS = D(curl F )ndA, (curl F )n= Š F3 y Š F2 z z x Š F1 z Š F3 z z y + + F2 x Š F1 y where z/x = g xand z/y = g y.Let x = x ( t )and y = y ( t ), t [ a,b ],beparametricequationsof D sothat x ( a )= x ( b )and y ( a )= y ( b )( D isaclosedcurve).Thenthevectorfunction r ( t )=( x ( t ) ,y ( t ) ,g ( x ( t ) ,y ( t )) ,t [ a,b ] tracesouttheboundary S r ( a )= r ( b ).MakinguseofTheorem15.1, thelineintegralof F along S canbeevaluated.Onehas r= ( x,y,g xx+ g yy).Therefore, F r=( F1+ F3g x) x+( F2+ F3g y) yandhence %SF d r = b a[( F1+ F3g x) x+( F2+ F3g y) y] dt = %D F1+ F3z x dx + F2+ F3z y dy because xdt = dx and ydt = dy along D ,where z = g ( x,y )inall componentsof F .Thelatterlineintegralcanbetransformedintothe doubleintegralover D byGreenstheorem: %SF d r = D x F2+ F3z y Š y F1+ F3z x dA = D(curl F )ndA = Scurl F n dS, wherethemiddleequalityisveri“edbythedirectevaluationofthe partialderivativesusingthechainrule.Forexample,( /x ) F2( x,y, g ( x,y ))= F2/x +( F2/z )( z/x ).Thetermscontainingthemixed derivatives 2z/xy = g xy= g yxarecancelledoutbyClairautstheorem,whiletheothertermscanbearrangedtocoincidewiththeexpressionforthenormalcomponent(curl F )nfoundabove.Thelast equalityholdsbyTheorem14.17( dS = JdA and n = J n ). 114.3.UseofStokesTheorem.Stokestheoremisveryhelpfulforevaluatinglineintegralsalongclosedcurvesofcomplicatedshapeswhena directuseofTheorem15.1istechnicallytooinvolved.Theprocedure includesafewbasicsteps.

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114.STOKESTHEOREM313 Step1 .Givenaclosedsimplecurve C ,choose any smoothorientable surface S whoseboundaryis C .Notethat,accordingtoStokestheorem,thevalueofthelineintegralisindependentofthechoiceof S Thisfreedomshouldbeusedtomake S assimpleaspossible. Step2 .Findtheorientationof S (thedirectionofthenormalvector) sothattheorientationof C ispositiverelativetothenormalof S ,that is, C = S Step3 .Evaluate B =curl F andcalculatethe”uxof B across S Example 15.9 Evaluatethelineintegralof F =( xy,yz,xz ) along thecurveofintersectionofthecylinder x2+ y2=1 andtheplane x + y + z =1 .Thecurveisorientedclockwiseasviewedfromabove. Solution: Thecurve C liesintheplane x + y + z =1.Therefore, thesimplestchoiceof S istheportionofthisplanethatlieswithinthe cylinder: z = g ( x,y )=1 Š x Š y ,where( x,y ) D and D isthedisk x2+ y2 1.Since C isorientedclockwiseasviewedfromabove,the orientationof S mustbedownwardtomaketheorientationpositive relativetothenormalon S ,thatis, n =( g x,g y, Š 1)=( Š 1 Š 1 Š 1). Next, B = F =det e1 e2 e3 x y zxyyzxz =( Š y, Š z, Š x ) Therefore, Bn( x,y )= B n =( Š y, Š g, Š x ) ( Š 1 Š 1 Š 1)= g ( x,y )+ y + x =1,andhence CF d r = SF d r = SB n dS = DBn( x,y ) dA = DdA = A ( D )= Example 15.10 Evaluatethelineintegralof F =( z2y, Š z2x,z ) alongthecurve C thatistheboundaryofthepartoftheparaboloid z = 1 Š x2Š y2inthe“rstoctant.Thecurve C isorientedcounterclockwise asviewedfromabove. Solution: Choose S tobethespeci“edpartoftheparaboloid z = g ( x,y )=1 Š x2Š y2,where( x,y ) D and D isthepartofthedisk x2+ y2 1inthe“rstquadrant.Theparaboloidmustbeoriented upwardsothatthegivenorientationof C ispositiverelativetothe normalon S .Therefore,thenormalvectoris n =( Š g x, Š g y, 1)=

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31415.VECTORCALCULUS (2 x, 2 y, 1).Next, B = F =det e1 e2 e3 x y zz2y Š z2xz =(2 zx, 2 zy, Š 2 z2) sothat Bn( x,y )= B n =(2 gx, 2 gy, Š 2 g2) (2 x, 2 y, 1)=4 g ( x2+ y2) Š 2 g2=4 g (1 Š g ) Š 2 g2=4 g Š 6 g2.Thus, CF d r = SB n dS = DBn( x,y ) dA = / 2 01 0[4(1 Š r2) Š 6(1 Š r2)2] rdrd = 7 15 wherethedoubleintegralhasbeenconvertedtopolarcoordinates, g ( x,y )=1 Š r2. 114.4.GeometricalSigni“canceoftheCurl.Stokestheoremrevealsthe geometricalsigni“canceofthecurlofavector“eld.Thelineintegralof avector“eldalongaclosedcurve C isoftencalledthe circulation ofa vector“eldalong C .Let B =curl F andlet B0= B ( r0)atsomepoint r0.Consideraplanethrough r0normaltoaunitvector n .Let Cabeapositivelyorientedsimple,closed,smoothcurveintheplanethat encirclesaregion Saoftheplaneand r0 Sa.Let a betheradiusofthe smallestdiskcenteredat r0thatcontains Sa.Considerthecirculation ofavector“eld perunitarea atapoint r0de“nedby lima 01 S %CaF d r =lima 01 S SaB n dS = B0 n =(curl F )0 n Thisfollowsfromtheintegralmeanvaluetheorem.Sincethefunction f ( r )= B n iscontinuouson Sa,thereisapoint ra Sasuchthat thesurfaceintegralof f equals Sf ( ra).As a 0, ra r0and, bythecontinuityof f f ( ra) f ( r0).Thisrelationhasthefollowing mechanicalinterpretation.Let F describea”uid”ow F = v ,where v isthe”uidvelocityvector“eld.Imagineatinypaddlewheelinthe ”uidatapoint r0whoseaxisisdirectedalong n .The”uidexerts pressureonthepaddles,causingthepaddlewheeltorotate.Themore workdonebythepressureforcealongtheloop Ca,thefasterthewheel rotates.Thewheelrotatesfastest(maximalwork)whenitsaxis n is paralleltocurl v because,inthiscase,thenormalcomponentofthe curlcurl v n = curl v ismaximal.Forthisreason,thecurlisoften calledthe rotation ofavector“eld.

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115.GAUSS-OSTROGRADSKY(DIVERGENCE)THEOREM315 Definition 15.11 (RotationalVectorField) Avector“eld F thatcanberepresentedasthecurlofanothervector “eld A ,thatis, F = A ,iscalleda rotationalvector“eld Thefollowingtheoremholds(theproofisomitted). Theorem 15.8 (HelmholtzsTheorem) Let F beavector“eldonaboundeddomain E ,whichistwicecontinuouslydierentiable.Then F canbedecomposedintothesumof conservativeandrotationalvector“eld;thatis,thereisafunction f andavector“eld A suchthat F = f + A Forexample,electromagnetic wavesare rotationalcomponentsof electromagnetic“elds,whiletheCoulomb“eldcreatedbystaticcharges isconservative.114.5.TestforaVectorFieldtoBeConservative.Thetestforavector “eldtobeconservative(Theorem15.4)followsfromStokestheorem. Indeed,inasimplyconnectedregion E ,anysimple,closedcurvecan beshrunktoapointwhileremainingin E throughthedeformation. Therefore,foranysuchcurve C ,onecanalways“ndasurface S in E suchthat S = C (e.g., C canbeshrunktoapointalongsuch S ).If curl F = 0 throughout E ,then,byStokestheorem, %CF d r = Scurl F n dS =0 foranysimpleclosedcurve C in E .Bythepathindependenceproperty, F isconservative.Theassumptionthat E issimplyconnectediscrucial. Forexample,if E istheentirespacewiththe z axisremoved(seeStudy Problem15.1),thenthe z axisalwayspiercesthroughanysurface S boundedbyaclosedsimplecurveencirclingthe z axis,andonecannot claimthatthecurlvanisheseverywhereon S .115.Gauss-Ostrogradsky(Divergence)Theorem115.1.DivergenceofaVectorField.Definition 15.12 (DivergenceofaVectorField) Supposethatavector“eld F =( F1,F2,F3) isdierentiable.Thenthe scalarfunction div F = F = F1 x + F2 y + F3 z iscalledthe divergenceofavector“eld .

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31615.VECTORCALCULUS Example 15.11 Findthedivergenceofthevector“eld F =( x3+ cos( yz ) ,y +sin( x2z ) ,xyz ) Solution: Onehas div F =( x3+cos( yz )) x+( y +sin( x2z )) y+( xyz ) z=3 x2+1+ yx. Corollary 15.1 Acontinuouslydierentiablerotationalvector “eldisdivergencefree, divcurl A =0 Proof. Byde“nition,arotationalvector“eldhastheform F = curl A = A ,where A istwicecontinuouslydierentiablebecause, bythehypothesis, F iscontinuouslydierentiable.Therefore, div F =divcurl A = curl A = ( A )=0 bytherulesofvectoralgebra(thetripleproductvanishesifanytwovectorsinitcoincide).Theserulesareapplicablebecause A istwicecontinuouslydierentiable(Clairautstheoremholdsforitscomponents; seeSection111.4). 115.2.AnotherVectorFormofGreensTheorem.Greenstheoremrelatesalineintegralalongaclosedcurveofthe tangential componentof aplanarvector“eldtothe”uxofthecurlacrosstheregionbounded bythecurve.Letusinvestigatethelineintegralofthe normal component.Ifthevectorfunction r ( t )=( x ( t ) ,y ( t )), a t b ,tracesout theboundary C of D inthepositive(counterclockwise)direction,then T ( t )= 1 r( t ) x( t ) ,y( t ) n ( t )= 1 r( t ) y( t ) Š x( t ) T n =0 aretheunittangentvectorandtheoutwardunitnormalvectortothe curve C ,respectively.Considerthelineintegral &CF n ds ofthenormal componentofaplanarvector“eldalong C .Onehas ds = r( t ) dt andhence F n ds = F1ydt Š F2xdt = F1dy Š F2dx = G d r where G =( Š F2,F1).ByGreenstheoremappliedtothelineintegral ofthevector“eld G %CF n ds = %CG d r = D G2 x Š G1 y dA = D F1 x + F2 y dA.

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115.GAUSS-OSTROGRADSKY(DIVERGENCE)THEOREM317 Theintegrandinthedoubleintegralisthedivergenceof F .Thus, anothervectorformofGreenstheoremhasbeenobtained: %DF n ds = Ddiv F dA. Foraplanarvector“eld(thinkofamass”owonaplane),theline integralontheleftsidecanbeviewedastheoutward”uxof F across theboundaryofaregion D .AnextensionofthisformofGreens theoremtothree-dimensionalvector“eldsisknownasthe divergence orGauss-Ostrogradskytheorem .115.3.TheDivergenceTheorem.Letasolidregion E beboundedbya closedsurface S .Ifthesurfaceisorientedoutward(thenormalvector pointsoutsideof E ),thenitisdenoted S = E Theorem 15.9 (Gauss-Ostrogradsky(Divergence)Theorem) Suppose E isabounded,closedregioninspacethathasapiecewisesmoothboundary S = E orientedoutward.If F isacontinuously dierentiablevector“eldonanopenregionthatcontains E ,then EF n dS = Ediv F dV. Thedivergencetheoremstatesthattheoutward”uxofavector “eldacrossaclosedsurface S isgivenbythetripleintegralofthe divergenceofthevector“eldoverthesolidregionboundedby S .It providesaconvenienttechnicaltooltoevaluatethe”uxofavector “eldacrossaclosedsurface. Remark. Itshouldbenotedthattheboundary E maycontain severaldisjointpieces.Forexample,let E beasolidregionwitha cavity.Then E consistsoftwopieces,theouterboundaryandthe cavityboundary.Bothpiecesareorientedoutwardinthedivergence theorem. Example 15.12 Evaluatethe”uxofthevector“eld F =(4 xy2z + ez, 4 yx2z,z4+sin( xy )) acrosstheclosedsurfaceorientedoutwardthat istheboundaryofthepartoftheball x2+ y2+ z2 R2inthe“rst octant( x,y,z 0 ). Solution: Thedivergenceofthevector“eldis div F =(4 xy2z + ez) x+(4 yx2z ) y+( z4+sin( xy )) z=4 z ( x2+ y2+ z2) .

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31815.VECTORCALCULUS Bythedivergencetheorem, SF n dS = E4 z ( x2+ y2+ z2) dV = / 2 0/ 2 0R 04 3cos 2sin ddd = R6 24 wherethetripleintegralhasbeenconvertedtosphericalcoordinates. Thereaderisadvisedtoevaluatethe”ux without usingthedivergence theoremtoappreciatethepowerofthelatter! Thedivergencetheoremcanbeusedtochange(simplify)thesurface inthe”uxintegral. Corollary 15.2 Lettheboundary E ofasolidregion E bethe unionoftwosurfaces S1and S2.Supposethatallthehypothesesofthe divergencetheoremhold.Then S2F n dS = Ediv F dV Š S1F n dS. Thisestablishesarelationbetweenthe”uxacross S1andthe”ux across S2withacommonboundarycurve.Indeed,since E isthe unionoftwodisjointpieces S1and S2,thesurfaceintegralover E isthesumoftheintegralsover S1and S2.Ontheotherhand,the integralover E canbeexpressedasatripleintegralbythedivergence theorem,whichestablishesthestatedrelationbetweenthe”uxesacross S1and S2. Example 15.13 Evaluatetheupward”uxofthevector“eld F = ( z2tanŠ 1( y2+1) z4ln( x2+1) ,z ) acrossthepartoftheparaboloid z =2 Š x2Š y2thatliesabovetheplane z =1 Solution: Considerasolid E boundedbytheparaboloidandthe plane z =1.Let S2bethepartoftheparaboloidthatbounds E andlet S1bethepartoftheplane z =1thatbounds E .If S2is orientedupwardand S1isorienteddownward,thentheboundaryof E isorientedoutward,andCorollary15.2applies.Thesurface S1is thepartoftheplane z =1boundedbytheintersectioncurveofthe paraboloidandtheplane:1=2 Š x2Š y2or x2+ y2=1.So S2is thegraph z = g ( x,y )=1over D ,whichisthedisk x2+ y2 1.The downwardnormalvectorto S1is n =( g x,g y, Š 1)=(0 0 Š 1),and hence Fn= F n = Š F3( x,y,g )= Š 1on S1and S1F n dS = DFn( x,y ) dA = Š DdA = Š A ( D )= Š .

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115.GAUSS-OSTROGRADSKY(DIVERGENCE)THEOREM319 Next,thedivergenceof F is div F =( z2tanŠ 1( y2+1)) x+( z4ln( x2+1)) y+( z ) z=0+0+1=1 Hence, Ediv F dV = EdV = 2 01 02 Š r21rdzdrd =2 1 0(1 Š r2) rdr = 2 wherethetripleintegralhasbeentransformedintocylindricalcoordinatesfor E = { ( x,y,z ) | zbot=1 z 2 Š x2Š y2= ztop, ( x,y ) D } Theupward”uxof F acrosstheparaboloidisnoweasyto“ndby Corollary15.2: S2F n dS = Ediv F dV Š S1F n dS = 2 + = 3 2 Thereaderisagainadvisedtotrytoevaluatethisdirectlytoappreciate thepowerofthedivergencetheorem! Corollary 15.3 The”uxofacontinuouslydierentiablerotationalvector“eldacrossanorientable,closed,piecewise-smoothsurface S vanishes: Scurl A n dS =0 Proof. Thehypothesesofthedivergencetheoremaresatis“ed.Therefore, Scurl A n dS = Edivcurl A dV =0 byCorollary15.1. ByHelmholtzstheorem,avector“eldcanalwaysbedecomposed intothesumofconservativeandrotationalvector“elds.Itfollowsthen thatonlytheconservativecomponentofthevector“eldcontributesto the”uxacrossaclosedsurface.Thisobservationisfurtherelucidated withthehelpoftheconceptofvector“eldsources.115.4.SourcesofaVectorField.Considerasimpleregion Eaofvolume V .Let a betheradiusofthesmallestballthatcontains Eaandis centeredatapoint r0.Letuscalculatetheoutward”ux perunitvolume ofacontinuouslydierentiablevector“eld F acrosstheboundary Ea,

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32015.VECTORCALCULUS whichisde“nedby lima 01 V EaF n dS =lima 01 V Eadiv F dV =div F ( r0) Thelatterequalityfollowsfromtheintegralmeanvaluetheorem.By thecontinuityofdiv F ,andtheintegralmeanvaluetheorem,thereisa point ra Easuchthatthetripleintegralequals V div F ( ra).Inthe limit a 0, ra r0.Thus,ifthedivergenceispositivediv F ( r0) > 0, the”uxofthevector“eldacrossanysmallsurfacearound r0ispositive. This,inturn,meansthatthe”owlinesof F areoutgoingfrom r0asif thereisa source creatinga”owat r0.Followingtheanalogywithwater ”ow,suchasourceiscalleda faucet .Ifdiv F ( r0) < 0,the”owlines disappearat r0(theinward”owispositive).Suchasourceiscalleda sink .Thus,thedivergenceofavector“elddeterminesthedensityof thesourcesofavector“eld.Forexample,”owlinesofastaticelectric “eldoriginatefrompositiveelectricchargesandendonnegativeelectric charges.Sothedivergenceoftheelectric“elddeterminestheelectric chargedensityinspace. Thedivergencetheoremstatesthattheoutward”uxofavector “eldacrossaclosedsurfaceisdeterminedbythetotalsourceofthe vector“eldintheregionboundedbythesurface.



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ConceptsinCalculusIIIUNIVERSITYPRESSOFFLORIDAFloridaA&MUniversity,Tallahassee FloridaAtlanticUniversity,BocaRaton FloridaGulfCoastUniversity,Ft.Myers FloridaInternationalUniversity,Miami FloridaStateUniversity,Tallahassee NewCollegeofFlorida,Sarasota UniversityofCentralFlorida,Orlando UniversityofFlorida,Gainesville UniversityofNorthFlorida,Jacksonville UniversityofSouthFlorida,Tampa UniversityofWestFlorida,Pensacola OrangeGroveTexts Plus

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ConceptsinCalculusIIIMultivariableCalculus SergeiShabanov UniversityofFloridaDepartmentof MathematicsUniversityPressofFlorida Gainesville  Tallahassee  Tampa  BocaRaton Pensacola  Orlando  Miami  Jacksonville  Ft.Myers  Sarasota

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Copyright2012bytheUniversityofFloridaBoardofTrusteesonbehalfoftheUniversityof FloridaDepartmentofMathematics ThisworkislicensedunderamodiedCreativeCommonsAttribution-Noncommercial-No DerivativeWorks3.0UnportedLicense.Toviewacopyofthislicense,visithttp:// creativecommons.org/licenses/by-nc-nd/3.0/.Youarefreetoelectronicallycopy,distribute,and transmitthisworkifyouattributeauthorship. However,allprintingrightsarereservedbythe UniversityPressofFlorida(http://www.upf.com).PleasecontactUPFforinformationabout howtoobtaincopiesoftheworkforprintdistribution .Youmustattributetheworkinthe mannerspeciedbytheauthororlicensor(butnotinanywaythatsuggeststhattheyendorse youoryouruseofthework).Foranyreuseordistribution,youmustmakecleartoothersthe licensetermsofthiswork.Anyoftheaboveconditionscanbewaivedifyougetpermissionfrom theUniversityPressofFlorida.Nothinginthislicenseimpairsorrestrictstheauthor'smoral rights. ISBN978-1-61610-162-6 OrangeGroveTexts Plus isanimprintoftheUniversityPressofFlorida,whichisthescholarly publishingagencyfortheStateUniversitySystemofFlorida,comprisingFloridaA&M University,FloridaAtlanticUniversity,FloridaGulfCoastUniversity,FloridaInternational University,FloridaStateUniversity,NewCollegeofFlorida,UniversityofCentralFlorida, UniversityofFlorida,UniversityofNorthFlorida,UniversityofSouthFlorida,andUniversityof WestFlorida. UniversityPressofFlorida 15Northwest15thStreet Gainesville,FL32611-2079 http://www.upf.com

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Contents Chapter11.VectorsandtheSpaceGeometry1 71.RectangularCoordinatesinSpace172.VectorsinSpace1273.TheDotProduct2574.TheCrossProduct3875.TheTripleProduct5176.PlanesinSpace6577.LinesinSpace7378.QuadricSurfaces82Chapter12.VectorFunctions97 79.CurvesinSpaceandVectorFunctions9780.DierentiationofVectorFunctions11181.IntegrationofVectorFunctions12082.ArcLengthofaCurve12883.CurvatureofaSpaceCurve13684.ApplicationstoMechanicsandGeometry147Chapter13.DifferentiationofMultivariableFunctions163 85.FunctionsofSeveralVariables16386.LimitsandContinuity17387.AGeneralStrategyforStudyingLimits18388.PartialDerivatives19689.Higher-OrderPartialDerivatives20290.LinearizationofMultivariableFunctions21191.ChainRulesandImplicitDierentiation22192.TheDierentialandTaylorPolynomials23193.DirectionalDerivativeandtheGradient24594.MaximumandMinimumValues25795.MaximumandMinimumValues(Continued)26896.LagrangeMultipliers278 Chapterandsectionnumberingcontinuesfromthepreviousvolumeintheseries, ConceptsinCalculusII .

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viCONTENTS Chapter14.MultipleIntegrals293 97.DoubleIntegrals29398.PropertiesoftheDoubleIntegral30199.IteratedIntegrals310100.DoubleIntegralsOverGeneralRegions315101.DoubleIntegralsinPolarCoordinates330102.ChangeofVariablesinDoubleIntegrals341103.TripleIntegrals356104.TripleIntegralsinCylindricalandSphericalCoordinates369105.ChangeofVariablesinTripleIntegrals382106.ImproperMultipleIntegrals392107.LineIntegrals403108.SurfaceIntegrals408109.MomentsofInertiaandCenterofMass423Chapter15.VectorCalculus437 110.LineIntegralsofaVectorField437111.FundamentalTheoremforLineIntegrals446112.GreensTheorem458113.FluxofaVectorField470114.StokesTheorem481115.Gauss-Ostrogradsky(Divergence)Theorem490Acknowledgments501

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CHAPTER11 VectorsandtheSpaceGeometry Ourspacemaybeviewedasacollectionofpoints.Everygeometrical“gure,suchasasphere,plane,orline,isaspecialsubsetofpointsin space.Themainpurposeofanalgebraicdescriptionofvariousobjects inspaceistodevelopasystematicrepresentationoftheseobjectsby numbers.Interestinglyenough,ourexperienceshowsthatsofarreal numbersandbasicrulesoftheiralgebraappeartobesucienttodescribeallfundamentallawsofnature,modeleverydayphenomena,and evenpredictmanyofthem.TheevolutionoftheUniverse,forcesbindingparticlesinatomicnuclei,andatomicnucleiandelectronsforming atomsandmolecules,starandplanetformation,chemistry,DNAstructures,andsoon,allcanbeformulatedasrelationsbetweenquantities thataremeasuredandexpressedasrealnumbers.Perhaps,thisis themostintriguingpropertyoftheUniverse,whichmakesmathematicsthemaintoolofourunderstandingoftheUniverse.Thedeeper ourunderstandingofnaturebecomes,themoresophisticatedarethe mathematicalconceptsrequiredtoformulatethelawsofnature.But theyremainbasedonrealnumbers.Inthiscourse,basicmathematical conceptsneededtodescribevariousphenomenainathree-dimensional Euclideanspacearestudied.Theveryfactthatthespaceinwhich weliveisathree-dimensionalEuclideanspaceshouldnotbeviewedas anabsolutetruth.Allonecansayisthatthis mathematicalmodel of thephysicalspaceissucienttodescribearatherlargesetofphysical phenomenaineverydaylife.Asamatteroffact,thismodelfailsto describephenomenaonalargescale(e.g.,ourgalaxy).Itmightalso failattinyscales,butthishasyettobeveri“edbyexperiments. 71.RectangularCoordinatesinSpace Theelementaryobjectinspaceisapoint.Sothediscussionshould beginwiththequestion:Howcanonedescribeapointinspacebyreal numbers?Thefollowingprocedurecanbeadopted.Selectaparticular pointinspacecalledthe origin andusuallydenoted O .Setupthree mutuallyperpendicularlinesthroughtheorigin.Arealnumberis associatedwitheverypointoneachlineinthefollowingway.The origincorrespondsto0.Distancestopointsononesideoftheline1

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211.VECTORSANDTHESPACEGEOMETRY fromtheoriginaredenotedbypositiverealnumbers,whiledistances topointsontheotherhalfofthelinearedenotedbynegativenumbers (theabsolutevalueofanegativenumberisthedistance).Thehalf-lines withthegridofpositivenumberswillbeindicatedbyarrowspointing fromtheorigintodistinguishthehalf-lineswiththegridofnegative numbers.Thedescribedsystemoflineswiththegridofrealnumbers onthemiscalleda rectangularcoordinatesystem attheorigin O .The lineswiththeconstructedgridofrealnumbersarecalled coordinate axes .71.1.PointsinSpaceasOrderedTriplesofRealNumbers.Theposition ofanypointinspacecanbe uniquely speci“edasan orderedtripleofreal numbers relativetoagivenrectangularcoordinatesystem.Consider arectangularboxwhosetwooppositevertices(theendpointsofthe largestdiagonal)aretheoriginandapoint P ,whileitssidesthatare adjacentattheoriginlieontheaxesofthecoordinatesystem.For everypoint P ,thereisonlyonesuchrectangularbox.Itisuniquely determinedbyitsthreesidesadjacentattheorigin.Letthenumber x denotethepositionofonesuchsidethatliesonthe“rstaxis;the numbers y and z dosoforthesecondandthirdsides,respectively. Notethat,dependingonthepositionof P ,thenumbers x y ,and z maybenegative,positive,oreven0.Inotherwords,anypointin spaceisassociatedwithaunique orderedtriple ofrealnumbers( x,y,z ) determinedrelativetoarectangularcoordinatesystem.Thisordered tripleofnumbersiscalled rectangularcoordinates ofapoint.Tore”ect theorderin( x,y,z ),theaxesofthecoordinatesystemwillbedenoted as x y ,and z axes.Thus,to“ndapointinspacewithrectangular coordinates(1 2 Š 3),onehastoconstructarectangularboxwitha vertexattheoriginsuchthatitssidesadjacentattheoriginoccupythe intervals[0 1],[0 2],and[ Š 3 0]alongthe x y ,and z axes,respectively. Thepointinquestionisthevertexoppositetotheorigin.71.2.APointasanIntersectionofCoordinatePlanes.Theplanecontainingthe x and y axesiscalledthe xyplane .Forallpointsinthis plane,the z coordinateis0.Theconditionthatapointliesinthe xy planecanthereforebestatedas z =0.The xz and yz planescanbe de“nedsimilarly.Theconditionthatapointliesinthe xz or yz plane reads y =0or x =0,respectively.Theorigin(0 0 0)canbeviewed astheintersectionofthreecoordinateplanes x =0, y =0,and z =0. Considerallpointsinspacewhose z coordinateis“xedtoaparticular value z = z0(e.g., z =1).Theyformaplaneparalleltothe xy plane thatlies | z0| unitsoflengthaboveitif z0> 0orbelowitif z0< 0.

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71.RECTANGULARCOORDINATESINSPACE3 Figure11.1.Left :Anypoint P inspacecanbeviewed astheintersectionofthreecoordinateplanes x = x0, y = y0, and z = z0;hence, P canbegivenanalgebraicdescription asanorderedtripleofnumbers P =( x0,y0,z0). Right : Translationofthecoordinatesystem.Theoriginismoved toapoint( x0,y0,z0)relativetotheoldcoordinatesystem whilethecoordinateaxesremainparalleltotheaxesofthe oldsystem.Thisisachievedbytranslatingtheorigin“rst alongthe x axisbythedistance x0(asshowninthe“gure), thenalongthe y axisbythedistance y0,and“nallyalong the z axisbythedistance z0.Asaresult,apoint P thathad coordinates( x,y,z )intheoldsystemwillhavethecoordinates x= x Š x0, y= y Š y0,and z= z Š z0inthenew coordinatesystem.Apoint P withcoordinates( x0,y0,z0)canthereforebeviewedasan intersectionofthree coordinateplanes x = x0, y = y0,and z = z0as showninFigure11.1.Thefacesoftherectangleintroducedtospecify thepositionof P relativetoarectangularcoordinatesystemlieinthe coordinateplanes.Thecoordinateplanesareperpendiculartothecorrespondingcoordinateaxes:theplane x = x0isperpendiculartothe x axis,andsoon.71.3.ChangingtheCoordinateSystem.Sincetheoriginanddirections oftheaxesofacoordinatesystemcanbechosenarbitrarily,thecoordinatesofapointdependonthischoice.Supposeapoint P has coordinates( x,y,z ).Consideranewcoordinatesystemwhoseaxesare

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411.VECTORSANDTHESPACEGEOMETRY paralleltothecorrespondingaxesoftheoldcoordinatesystem,but whoseoriginisshiftedtothepoint Owithcoordinates( x0, 0 0).It isstraightforwardtoseethatthepoint P wouldhavethecoordinates ( x Š x0,y,z )relativetothenewcoordinatesystem(Figure11.1,right panel).Similarly,iftheoriginisshiftedtoapoint Owithcoordinates ( x0,y0,z0),whiletheaxesremainparalleltothecorrespondingaxesof theoldcoordinatesystem,thenthecoordinatesof P aretransformedas (11.1)( x,y,z ) Š ( x Š x0,y Š y0,z Š z0) Onecanchangetheorientationofthecoordinateaxesbyrotating themabouttheorigin.Thecoordinatesofthesamepointinspaceare dierentintheoriginalandrotatedrectangularcoordinatesystems. Algebraicrelationsbetweenoldandnewcoordinatescanbeestablished. Asimplecase,whenacoordinatesystemisrotatedaboutoneofits axes,isdiscussedinStudyProblem11.2. Itisimportanttorealizethatnophysicalorgeometricalquantity shoulddependonthechoiceofacoordinatesystem.Forexample,the lengthofastraightlinesegmentmustbethesameinanycoordinate system,whilethecoordinatesofitsendpointsdependonthechoiceof thecoordinatesystem.Whenstudyingapracticalproblem,acoordinatesystemcanbechoseninanywayconvenienttodescribeobjectsin space.Algebraicrulesforrealnumbers(coordinates)canthenbeused tocomputephysicalandgeometricalcharacteristicsoftheobjects.The numericalvaluesofthesecharacteristicsdonotdependonthechoice ofthecoordinatesystem.71.4.DistanceBetweenTwoPoints.Considertwopointsinspace, P1and P2.Lettheircoordinatesrelativetosomerectangularcoordinate systembe( x1,y1,z1)and( x2,y2,z2),respectively.Howcanonecalculatethedistancebetweenthesepoints,orthelengthofastraightline segmentwithendpoints P1and P2?Thepoint P1istheintersection pointofthreecoordinateplanes x = x1, y = y1,and z = z1.Thepoint P2istheintersectionpointofthreecoordinateplanes x = x2, y = y2, and z = z2.Thesesixplanescontainfacesoftherectangularboxwhose largestdiagonalisthestraightlinesegmentbetweenthepoints P1and P2.Thequestionthereforeishowto“ndthelengthofthisdiagonal. Considerthreesidesofthisrectangularboxthatareadjacent,say, atthevertex P1.Thesideparalleltothe x axisliesbetweenthe coordinateplanes x = x1and x = x2andisperpendiculartothem.So thelengthofthissideis | x2Š x1| .Theabsolutevalueisnecessaryas thedierence x2Š x1maybenegative,dependingonthevaluesof x1and x2,whereasthedistancemustbenonnegative.Similararguments

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71.RECTANGULARCOORDINATESINSPACE5 leadtotheconclusionthatthelengthsoftheothertwoadjacentsides are | y2Š y1| and | z2Š z1| .Ifarectangularboxhasadjacentsidesof length a b ,and c ,thenthelength d ofitslargestdiagonalsatis“esthe equation d2= a2+ b2+ c2. ItsproofisbasedonthePythagoreantheorem(seeFigure11.2).Considertherectangularfacethatcontainsthesides a and b .Thelength f ofitsdiagonalisdeterminedbythePythagoreantheorem f2= a2+ b2. Considerthecrosssectionoftherectangularboxbytheplanethat containsthefacediagonal f andtheside c .Thiscrosssectionisa rectanglewithtwoadjacentsides c and f andthediagonal d .Theyare relatedas d2= f2+ c2bythePythagoreantheorem,andthedesired conclusionfollows. Put a = | x2Š x1| b = | y2Š y1| ,and c = | z2Š z1| .Then d = | P1P2| is thedistancebetween P1and P2.Thedistanceformulaisimmediately Figure11.2.Distancebetweentwopointswithcoordinates P1=( x1,y1,z1)and P2=( x2,y2,z2).Thelinesegment P1P2isviewedasthelargestdiagonaloftherectangularboxwhosefacesarethecoordinateplanescorresponding tothecoordinatesofthepoints.Therefore,thedistancesbetweentheoppositefacesare a = | x1Š x2| b = | y1Š y2| ,and c = | z1Š z2| .Thelengthofthediagonal d isobtainedbythe doubleuseofthePythagoreantheoremineachoftheindicatedrectangles: d2= c2+ f2(topright)and f2= a2+ b2(bottomright).

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611.VECTORSANDTHESPACEGEOMETRY found: (11.2) | P1P2| = ( x2Š x1)2+( y2Š y1)2+( z2Š z1)2. Notethatthenumbers(coordinates)( x1,y1,z1)and( x2,y2,z2)depend onthechoiceofthecoordinatesystem,whereasthenumber | P1P2| remainsthesame inanycoordinatesystem!Forexample,iftheoriginof thecoordinatesystemistranslatedtoapoint( x0,y0,z0)whiletheorientationofthecoordinateaxesremainsunchanged,then,accordingto rule(11.1),thecoordinatesof P1and P2relativetothenewcoordinate become( x1Š x0,y1Š y0,z1Š z0)and( x2Š x0,y2Š y0,z2Š z0),respectively.Thenumericalvalueofthedistancedoesnotchangebecause thecoordinatedierences,( x2Š x0) Š ( x1Š x0)= x2Š x1(similarly forthe y and z coordinates),donotchange. Example 11.1 Apointmoves 3 unitsoflengthparalleltoaline, thenitmoves 6 unitsparalleltothesecondlinethatisperpendicularto the“rstline,andthenitmoves 6 unitsparalleltothethirdlinethatis perpendiculartothe“rstandsecondlines.Findthedistancebetween theinitialand“nalpositions. Solution: Thedistancebetweenpointsdoesnotdependonthechoice ofthecoordinatesystem.Lettheoriginbepositionedattheinitialpointofthemotionandletthecoordinateaxesbedirectedalong thethreemutuallyperpendicularlinesparalleltowhichthepointhas moved.Inthiscoordinatesystem,the“nalpointhasthecoordinates (3 6 6).Thedistancebetweenthispointandtheorigin(0 0 0)is D = 32+62+62= 9(1+4+4)=9 RotationsinSpace.Theorigincanalwaysbetranslatedto P1sothat inthenewcoordinatesystem P1is(0 0 0)and P2is( x2Š x1,y2Š y1,z2Š z1).Sincethedistanceshouldnotdependontheorientationof thecoordinateaxes,anyrotationcannowbedescribedalgebraically as alineartransformationofanorderedtriple ( x,y,z ) underwhichthe combination x2+ y2+ z2remainsinvariant .Alineartransformation meansthatthenewcoordinatesarelinearcombinationsoftheoldones. Itshouldbenotedthatre”ectionsofthecoordinateaxes, x Š x (similarlyfor y and z ),arelinearandalsopreservethedistance.However, acoordinatesystemobtainedbyanoddnumberofre”ectionsofthe coordinateaxescannotbeobtainedbyanyrotationoftheoriginalcoordinatesystem.So,intheabovealgebraicde“nitionofarotation,the

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71.RECTANGULARCOORDINATESINSPACE7 re”ectionsshouldbeexcluded.Analgebraicdescriptionofrotationsin aplaneandinspaceisgiveninStudyProblems11.2and11.20.71.5.SpheresinSpace.Inthiscourse,relationsbetweentwoequivalent descriptionsofobjectsinspace„thegeometricalandthealgebraic„ willalwaysbeemphasized.Oneofthecourseobjectivesistolearnhow tointerpretanalgebraicequationbygeometricalmeansandhowtodescribegeometricalobjectsinspacealgebraically.Oneofthesimplest examplesofthiskindisasphere.GeometricalDescriptionofaSphere.Asphereisasetofpointsinspace thatareequidistantfroma“xedpoint.The“xedpointiscalledthe center ofthesphere.Thedistancefromthespherecentertoanypoint ofthesphereiscalledthe radius ofthesphere.AlgebraicDescriptionofaSphere.Analgebraicdescriptionofasphere implies“ndinganalgebraicconditiononcoordinates( x,y,z )ofpoints inspacethatbelongtothesphere.Soletthecenterofthesphere beapoint P0withcoordinates( x0,y0,z0)(de“nedrelativetosome rectangularcoordinatesystem).Ifapoint P withcoordinates( x,y,z ) belongstothesphere,thenthenumbers( x,y,z )mustbesuchthatthe distance | PP0| isthesameforanysuch P andequaltotheradiusofthe sphere,denoted R ,thatis, | PP0| = R or | PP0|2= R2(seeFigure11.3, leftpanel).Usingthedistanceformula,thisconditioncanbewritten as (11.3)( x Š x0)2+( y Š y0)2+( z Š z0)2= R2. Forexample,thesetofpointswithcoordinates( x,y,z )thatsatisfythe condition x2+ y2+ z2=4isasphereofradius R =2centeredatthe origin x0= y0= z0=0. Example 11.2 Findthecenterandtheradiusofthesphere x2+ y2+ z2Š 2 x +4 y Š 6 z +5=0 Solution: Inorderto“ndthecoordinatesofthecenterandtheradius ofthesphere,theequationmustbetransformedtothestandardform (11.3)bycompletingthesquares: x2Š 2 x =( x Š 1)2Š 1, y2+4 y = ( y +2)2Š 4,and z2Š 6 z =( z Š 3)2Š 9.Thentheequationofthe spherebecomes ( x Š 1)2Š 1+( y +2)2Š 4+( z Š 3)2Š 9+5=0 ( x Š 1)2+( y +2)2+( z Š 3)2=9 .

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811.VECTORSANDTHESPACEGEOMETRY Figure11.3.Left :Asphereisde“nedasapointsetin space.Eachpoint P ofthesethasa“xeddistance R from a“xedpoint P0.Thepoint P0isthecenterofthesphere, and R istheradiusofthesphere. Right :Anillustrationto StudyProblem11.2.Transformationofcoordinatesundera rotationofthecoordinatesysteminaplane.Acomparisonwith(11.3)showsthatthecenterisat( x0,y0,z0)= (1 Š 2 3)andtheradiusis R = 9=3. 71.6.AlgebraicDescriptionofPointSetsinSpace.Theideaofanalgebraicdescriptionofaspherecanbeextendedtoothersetsinspace.It isconvenienttointroducesomebriefnotationforanalgebraicdescriptionofsets.Forexample,foraset S ofpointsinspacewithcoordinates ( x,y,z )suchthattheysatisfythealgebraiccondition(11.3),onewrites S = ( x,y,z ) ( x Š x0)2+( y Š y0)2+( z Š z0)2= R2 Thisrelationmeansthattheset S isacollectionofallpoints( x,y,z ) suchthat(theverticalbar)theirrectangularcoordinatessatisfy(11.3). Similarly,the xy planecanbeviewedasasetofpointswhose z coordinatesvanish: P = ( x,y,z ) z =0 Thesolidregioninspacethatconsistsofpointswhosecoordinatesare nonnegativeiscalledthe “rstoctant : O1= ( x,y,z ) x 0 ,y 0 ,z 0 Thespatialregion B = ( x,y,z ) x> 0 ,y> 0 ,z> 0 ,x2+ y2+ z2< 4

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71.RECTANGULARCOORDINATESINSPACE9 isthecollectionofallpointsintheportionofaballofradius2that liesinthe“rstoctant.Thestrictinequalitiesimplythattheboundary ofthisportionoftheballdoesnotbelongtotheset B .71.7.StudyProblems.Problem11.1. Showthatthecoordinatesofthemidpointofastraight linesegmentare x1+ x2 2 y1+ y2 2 z1+ z2 2 ifthecoordinatesofitsendpointsare ( x1,y1,z1) and ( x2,y2,z2) Solution: Let P1and P2betheendpointsandlet M bethepoint withcoordinatesequaltohalf-sumsofthecorrespondingcoordinates of P1and P2.Onehastoprovethat | MP1| = | MP2| =1 2| P1P2| .These twoconditionsde“ne M asthemidpoint.Considerarectangularbox B1whoselargestdiagonalis P1M .Thelengthofitssideparallelto the x axisis | ( x1+ x2) / 2 Š x1| = | x2Š x1| / 2.Similarly,itssides paralleltothe y and z axeshavethelengths | y2Š y1| / 2and | z2Š z1| / 2, respectively.Considerarectangularbox B2whoselargestdiagonalis thesegment MP2.Thenitssideparalleltothe x axishasthelength | x2Š ( x1+ x2) / 2 | = | x2Š x1| / 2.Similarly,thesidesparalleltothe y and z axeshavelengths | y2Š y1| / 2and | z2Š z1| / 2,respectively.Thus, thesidesof B1and B2paralleltoeachcoordinateaxishavethesame length.Bythedistanceformula(11.2),thediagonalsof B1and B2musthavethesamelength | P2M | = | MP1| .Thelengthsofthesidesof arectangularboxwhoselargestdiagonalis P1P2are | x2Š x1| | y2Š y1| and | z2Š z1| .Theyaretwiceaslongasthecorrespondingsidesof B1and B2.Ifthelengthofeachsideofarectangularboxisscaledbya positivefactor q ,thenthelengthofitsdiagonalisalsoscaledby q .In particular,thisimpliesthat | MP2| =1 2| P1P2| Problem11.2. Let ( x,y,z ) becoordinatesofapoint P .Consider anewcoordinatesystemthatisobtainedbyrotatingthe x and y axes aboutthe z axiscounterclockwiseasviewedfromthetopofthe z axis throughanangle .Let ( x,y,z) becoordinatesof P inthenewcoordinatesystem.Findtherelationsbetweentheoldandnewcoordinates. Solution: Theheightof P relativetothe xy planedoesnotchange uponrotation.So z= z .Itisthereforesucienttoconsiderrotationsinthe xy plane,thatis,forpoints P withcoordinates( x,y, 0). Let r = | OP | (thedistancebetweentheoriginand P )andlet bethe

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1011.VECTORSANDTHESPACEGEOMETRY anglecountedfromthepositive x axistowardtheray OP counterclockwise(seeFigure11.3,rightpanel).Then x = r cos and y = r sin (thepolarcoordinatesof P ).Inthenewcoordinatesystem,theanglebetweenthepositive xaxisandtheray OP becomes = Š Therefore, x= r cos = r cos( Š )= r cos cos + r sin sin = x cos + y sin y= r sin = r sin( Š )= r sin cos Š r cos sin = y cos Š x sin Theseequationsde“nethetransformation( x,y ) ( x,y)oftheold coordinatestothenewones.Theinversetransformation( x,y) ( x,y )canbefoundbysolvingtheequationsfor( x,y ).Asimplerway istonotethatif( x,y)areviewedasoldŽcoordinatesand( x,y )as newŽcoordinates,thenthetransformationthatrelatesthemisthe rotationthroughtheangle Š (aclockwiserotation).Therefore,the inverserelationscanbeobtainedbyswappingtheoldŽandnewŽ coordinatesandreplacing by Š inthedirectrelations.Thisyields x = xcos Š ysin ,y = ycos + xsin Problem11.3. Giveageometricaldescriptionoftheset S = ( x,y,z ) x2+ y2+ z2Š 4 z =0 Solution: Theconditiononthecoordinatesofpointsthatbelong tothesetcontainsthesumofsquaresofthecoordinatesjustlikethe equationofasphere.Thedierenceisthat(11.3)containsthesum ofperfectsquares.Sothesquaresmustbecompletedintheabove equationandtheresultingexpressioncomparedwith(11.3).Onehas z2Š 4 z =( z Š 2)2Š 4sothattheconditionbecomes x2+ y2+( z Š 2)2=4. Itdescribesasphereofradius R =2thatiscenteredatthepoint ( x0,y0,z0)=(0 0 2);thatis,thecenterofthesphereisonthe z axis atadistanceof2unitsabovethe xy plane. Problem11.4. Giveageometricaldescriptionoftheset C = ( x,y,z ) x2+ y2Š 2 x Š 4 y 4 Solution: Asinthepreviousproblem,theconditioncanbewritten viathesumofperfectsquares( x Š 1)2+( y Š 2)2 9bymeansof therelations x2Š 2 x =( x Š 1)2Š 1and y2Š 4 y =( y Š 2)2Š 4.In the xy plane,theinequalitydescribesadiskofradius3whosecenter

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71.RECTANGULARCOORDINATESINSPACE11 isthepoint(1 2 0).Asthealgebraicconditionimposesnorestriction onthe z coordinateofpointsintheset,inanyplane z = z0parallelto the xy plane,the x and y coordinatessatisfythesameinequality,and hencethecorrespondingpointsalsoformadiskofradius3withthe center(1 2 ,z0).Thus,thesetisasolidcylinderofradius3whoseaxis isparalleltothe z axisandpassesthroughthepoint(1 2 0). Problem11.5. Giveageometricaldescriptionoftheset P = ( x,y,z ) z ( y Š x )=0 Solution: Theconditionissatis“edifeither z =0or y = x .The formerequationdescribesthe xy plane.Thelatterrepresentsaline inthe xy plane.Sinceitdoesnotimposeanyrestrictiononthe z coordinate,eachpointofthelinecanbemovedupanddownparallel tothe z axis.Theresultingsetisaplanethatcontainstheline y = x inthe xy planeandthe z axis.Thus,theset P istheunionofthis planeandthe xy plane. 71.8.Exercises.(1) Findthedistancebetweenthefollowingspeci“edpoints: (i)(1 2 3)and( Š 1 0 2) (ii)( Š 1 3 Š 2)and( Š 1 2 Š 1) (2) Findthedistancefromthepoint(1 2 3)toeachofthecoordinate planesandtoeachofthecoordinateaxes. (3) Findthelengthofthemediansofthetrianglewithvertices A (1 2 3), B ( Š 3 2 Š 1),and C ( Š 1 Š 4 1). (4) Lettheset S consistofpoints( t, 2 t, 3 t ),where Š 0 (iv) x2+ y2Š 4 y< 0, z> 0 (v)4 x2+ y2+ z2 9 (vi) x2+ y2 1, x2+ y2+ z2 4 (vii) x2+ y2+ z2Š 2 z< 0, z> 1 (viii) x2+ y2+ z2Š 2 z =0, z =1 (ix)( x Š a )( y Š b )( z Š c )=0 (x) | x | 1, | y | 2, | z | 3

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1211.VECTORSANDTHESPACEGEOMETRY (6) Sketcheachofthefollowingsetsandgivetheiralgebraicdescription: (i)Aspherewhosediameteristhestraightlinesegment AB where A =(1 2 3)and B =(3 2 1). (ii)Threespherescenteredat(1 2 3)thattouchthe xy yz ,and xz planes,respectively. (iii)Threespherescenteredat(1 Š 2 3)thattouchthe x y ,and z coordinateaxes,respectively. (iv)Thelargestsolidcubethatiscontainedinaballofradius R centeredattheorigin.Solvethesameproblemiftheballisnot centeredattheorigin.Comparethecaseswhentheboundaries ofthesolidareincludedinthesetorexcludedfromit. (v)Thesolidregionthatisaballofradius R thathasacylindricalholeofradius R/ 2whoseaxisisatadistanceof R/ 2 fromthecenteroftheball.Chooseaconvenientcoordinate system.Comparethecaseswhentheboundariesofthesolid areincludedinthesetorexcludedfromit. (vi)Thepartofaballofradius R thatliesbetweentwoparallel planeseachofwhichisatadistanceof a
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72.VECTORSINSPACE13 thenthedistancetraveledis | P1P2| =5m.Therate(orspeed)5m/s doesnotprovideacompletedescriptionofthemotionoftheobjectin spacebecauseitonlyanswersthequestionHowfastdoestheobject move?ŽbutitdoesnotsayanythingaboutWheretodoestheobject move?ŽSincetheinitialand“nalpositionsoftheobjectareknown, bothquestionscanbeanswered,ifoneassociatesan orientedsegment Š Š P1P2withthemovingobject.Thearrowspeci“esthedirection,from P1to P2,Žandthelength | P1P2| de“nestherate(speed)atwhichthe objectmoves.So,foreverymovingobject,onecanassignanoriented segmentwhoselengthequalsitsspeedandwhosedirectioncoincides withthedirectionofmotion.Thisorientedsegmentiscalleda velocity .Considertwoobjectsmovingparallelwiththesamespeed.The orientedsegmentscorrespondingtothevelocitiesoftheobjectshave thesamelengthandthesamedirection,buttheyarestilldierent becausetheirinitialpointsdonotcoincide.Ontheotherhand,the velocityshoulddescribeaparticularphysicalpropertyofthemotion itself(howfastandwheretoŽ),andthereforethespatialposition wherethemotionoccursshouldnotmatter.Thisobservationleads totheconceptofa vector asanabstractmathematicalobjectthat representsallorientedsegmentsthatareparallelandhavethesame length Vectorswillbedenotedbyboldfaceletters. Twoorientedsegments Š AB and Š Š CD representthesamevector a iftheyareparalleland | AB | = | CD | ;thatis,theycanbeobtainedfromoneanotherbytransporting themparalleltothemselvesinspace .Arepresentationofanabstract vectorbyaparticularorientedsegmentisdenotedbytheequality a = Š AB or a = Š Š CD .Thefactthattheorientedsegments Š AB and Š Š CD representthesamevectorisdenotedbytheequality Š AB = Š Š CD .72.2.VectorasanOrderedTripleofNumbers.Hereanalgebraicrepresentationofvectorsinspacewillbeintroduced.Consideranoriented segment Š AB thatrepresentsavector a (i.e., a = Š AB ).Anorientedsegment Š Š ABrepresentsthesamevectorifitisobtainedbytransporting Š AB paralleltoitself.Inparticular,letustake A= O ,where O isthe originofsomerectangularcoordinatesystem.Then a = Š AB = Š Š OB. Thedirectionandlengthoftheorientedsegment Š Š OBisuniquelydeterminedbythecoordinatesofthepoint B.Thus,thefollowingalgebraic de“nitionofavectorcanbeadopted.

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1411.VECTORSANDTHESPACEGEOMETRY Figure11.4.Left :Orientedsegmentsobtainedfromone anotherbyparalleltransport.Theyallrepresentthesame vector. Right :Avectorasanorderedtripleofnumbers. Anorientedsegmentistransportedparallelsothatitsinitialpointcoincideswiththeoriginofarectangularcoordinatesystem.Thecoordinatesoftheterminalpointofthe transportedsegment,( a1,a2,a3),arecomponentsofthecorrespondingvector.Soavectorcanalwaysbewrittenasan orderedtripleofnumbers: a = a1,a2,a3 .Byconstruction,thecomponentsofavectordependonthechoiceofthe coordinatesystem.Definition 11.1 (Vectors) Avectorinspaceisanorderedtripleofrealnumbers: a = a1,a2,a3 Thenumbers a1, a2,and a3arecalled components ofthevector a Considerapoint A withcoordinates( a1,a2,a3)insomerectangularcoordinatesystem.Thevector a = Š OA = a1,a2,a3 iscalledthe positionvector of A relativetothegivencoordinatesystem.Thisestablishesaone-to-onecorrespondencebetweenvectorsandpointsin space.Inparticular,ifthecoordinatesystemischangedbyarotation ofitsaxesabouttheorigin,thecomponentsofavector a aretransformedinthesamewayasthecoordinatesofapointwhoseposition vectoris a Definition 11.2 (EqualityofTwoVectors) Twovectors a and b areequalorcoincideiftheircorrespondingcomponentsareequal: a = b a1= b1,a2= b2,a3= b3. Thisde“nitionagreeswiththegeometricalde“nitionofavectorasa classofallorientedsegmentsthatareparallelandhavethesamelength. Indeed,iftwoorientedsegmentsrepresentthesamevector,then,after

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72.VECTORSINSPACE15 paralleltransportsuchthattheirinitialpointscoincidewiththeorigin, their“nalpointscoincidetooandhencehavethesamecoordinates.By virtueofthecorrespondencebetweenvectorsandpointsinspace,this de“nitionre”ectsthefactthattwosamepointsshouldhavethesame positionvectors. Example 11.3 Findthecomponentsofavector Š Š P1P2ifthecoordinatesof P1and P2are ( x1,y1,z1) and ( x2,y2,z2) ,respectively. Solution: Considerarectangularboxwhoselargestdiagonalcoincideswiththesegment P1P2andwhosesidesareparalleltothecoordinateaxes.Afterparalleltransportofthesegmentsothat P1moves totheorigin,thecoordinatesoftheotherendpointarethecomponentsof Š Š P1P2.Alternatively,theoriginofthecoordinatesystemcan bemovedtothepoint P1,keepingthedirectionsofthecoordinateaxes. Therefore, Š Š P1P2= x2Š x1,y2Š y1,z2Š z1 accordingtothecoordinatetransformationlaw(11.1),where P0= P1. Thus,inorderto“ndthecomponentsofthevector Š Š P1P2,onehasto subtractthecoordinatesoftheinitialpoint P1fromthecorresponding coordinatesofthe“nalpoint P2. Definition 11.3 (NormofaVector). Thenumber a = a2 1+ a2 2+ a2 3iscalledthe norm ofavector a ByExample11.3andthedistanceformula(11.2),thenormofa vectoristhelengthofanyorientedsegmentrepresentingthevector. Thenormofavectorisalsocalledthe magnitude or length ofavector. Definition 11.4 (ZeroVector) Avectorwithvanishingcomponents, 0 = 0 0 0 ,iscalleda zero vector Avector a isazerovectorifandonlyifitsnormvanishes, a =0. Indeed,if a = 0 ,then a1= a2= a3=0andhence a =0.Forthe converse,itfollowsfromthecondition a =0that a2 1+ a2 2+ a2 3=0, whichisonlypossibleif a1= a2= a3=0,or a = 0 .Recallthatan ifandonlyifŽstatementimpliestwostatements.First,if a = 0 ,then a =0(thedirectstatement).Second,if a =0,then a = 0 (the conversestatement).

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1611.VECTORSANDTHESPACEGEOMETRY 72.3.VectorAlgebra.Continuingtheanalogybetweenthevectorsand velocitiesofamovingobject,considertwoobjectsmovingparallelbut withdierentrates(speeds).Theirvelocitiesasvectorsareparallel, buttheyhavedierentmagnitudes.Whatistherelationbetweenthe componentsofsuchvectors?Takeavector a = a1,a2,a3 .Itcanbe viewedasthelargestdiagonalofarectangularboxwithonevertexat theoriginandtheoppositevertexatthepoint( a1,a2,a3).Theadjacent sidesoftherectangularboxhavelengthsgivenbythecorresponding componentsof a (modulothesignsifthecomponentshappentobe negative).Whenthelengthsofthesidesarescaledbyafactor s> 0, anewrectangularboxisobtainedwhoselargestdiagonalisparallelto a .Thisgeometricalobservationleadstothefollowingalgebraicrule. Definition 11.5 (MultiplicationofaVectorbyaNumber) Avector a multipliedbyanumber s isavectorwhosecomponentsare multipliedby s : s a = sa1,sa2,sa3 If s> 0,thenthevector s a hasthesamedirectionas a .If s< 0, thenthevector s a hasthedirectionoppositeto a .Forexample,the vector Š a hasthesamemagnitudeas a butpointsinthedirection Figure11.5.Left :Multiplicationofavector a byanumber s .If s> 0,theresultofthemultiplicationisavector parallelto a whoselengthisscaledbythefactor s .If s< 0, then s a isavectorwhosedirectionistheoppositetothatof a andwhoselengthisscaledby | s | Middle :Construction ofaunitvectorparallelto a .Theunitvector a isavector parallelto a whoselengthis1.Therefore,itisobtainedfrom a bydividingthelatterbyitslength a ,thatis, a = s a where s =1 / a Right :Aunitvectorinaplanecanalwaysbeviewedasanorientedsegmentwhoseinitialpoint isattheoriginofacoordinatesystemandwhoseterminal pointliesonthecircleofunitradiuscenteredattheorigin. If isthepolarangleintheplane,then a = cos sin 0 .

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72.VECTORSINSPACE17 oppositeto a .Themagnitudeof s a is s a = ( sa1)2+( sa2)2+( sa3)2= s2 a2 1+ a2 2+ a2 3= | s | a ; thatis,whenavectorismultipliedbyanumber,itsmagnitudechanges bythefactor | s | .Thegeometricalanalysisofthemultiplicationofa vectorbyanumberleadstothefollowingsimplealgebraiccriterionfor twovectorsbeingparallel. Twononzerovectorsareparalleliftheyare proportional: a b a = s b forsomereal s .Ifallthecomponentsofthevectorsinquestiondonot vanish,thenthiscriterionmayalsobewrittenas a b s = a1 b1= a2 b2= a3 b3, whichiseasytoverify.If,say, b1=0,then b isparallelto a when a1= b1=0and a2/b2= a3/b3.Owingtothegeometricalinterpretation of s b ,allpointsinspacewhosepositionvectorsareparalleltoagiven nonzerovector b formaline(throughtheorigin)thatisparallelto b Definition 11.6 (UnitVector) Avector a iscalleda unitvector ifitsnormequals1, a =1 Anynonzerovector a canbeturnedintoaunitvector a thatis parallelto a .Thenorm(length)ofthevector s a reads s a = | s | a = s a if s> 0.So,bychoosing s =1 / a ,theunitvectorinthe directionof a isobtained: a = 1 a a = a1 a a2 a a3 a Forexample,owingtothetrigonometricidentitycos2 +sin2 =1, anyunitvectorinthe xy planecanalwaysbewrittenintheform a = cos sin 0 ,where istheanglecountedfromthepositive x axistowardthevector a counterclockwise(seetherightpanelof Figure11.5).Notethat,inmanypracticalapplications,thecomponentsofavectoroftenhavedimensions.Forinstance,thecomponents ofapositionvectoraremeasuredinunitsoflength(meters,inches, etc.),thecomponentsofavelocityvectoraremeasuredin,forexample,meterspersecond,andsoon.Themagnitudeofavector a hasthe samedimensionasitscomponents.Therefore,thecorrespondingunit vector a isdimensionless.Itspeci“esonlythedirectionofavector a Example 11.4 Let A =(1 2 3) and B =(3 1 1) .Find a = Š AB b = Š BA ,theunitvectors a and b ,andthevector c = Š 2 Š AB andits norm.

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1811.VECTORSANDTHESPACEGEOMETRY Solution: ByExample11.3, a = 3 Š 1 2 Š 1 1 Š 3 = 2 Š 1 Š 2 .The normof a is a = 22+( Š 1)2+( Š 2)2= 9=3 Theunitvector inthedirectionof a is a =(1 / 3) a = 2 / 3 Š 1 / 3 Š 2 / 3 .Usingtherule ofmultiplicationofavectorbyanumber, c = Š 2 a = Š 2 2 Š 1 Š 2 = Š 4 2 4 and c = ( Š 2) a = |Š 2 | a =2 a =6.Thedirectionof Š BA istheoppositeto Š AB ,andbothvectorshavethesamelength. Therefore, b = Š 2 1 2 b =3,and b =Š a = Š 2 / 3 1 / 3 2 / 3 TheParallelogramRule.Supposeapersoniswalkingonthedeckofa shipwithspeed v m / s.In1second,thepersongoesadistance v from point A topoint B ofthedeck.Thevelocityvectorrelativetothedeck is v = Š AB and v = | AB | = v (thespeed).Theshipmovesrelative tothewatersothatin1seconditcomestoapoint D fromapoint C onthesurfaceofthewater.Theshipsvelocityvectorrelativeto thewateristhen u = Š Š CD withmagnitude u = u = | CD | .What isthevelocityvectorofthepersonrelativetothewater?Suppose thepoint A onthedeckcoincideswiththepoint C onthesurface ofthewater.Thenthevelocityvectoristhedisplacementvectorof thepersonrelativetothewaterin1second.Asthepersonwalkson thedeckalongthesegment AB ,thissegmenttravelsthedistance u paralleltoitselfalongthevector u relativetothewater.In1second, thepoint B ofthedeckismovedtoapoint Bonthesurfaceofthe watersothatthedisplacementvectorofthepersonrelativetothe waterwillbe Š Š AB.Apparently,thedisplacementvector Š Š BBcoincides withtheshipsvelocity u because B travelsthedistance u parallelto u .Thissuggestsasimplegeometricalrulefor“nding Š Š ABasshownin Figure11.6.Takethevector Š AB = v ,placethevector u sothatits initialpointcoincideswith B ,andmaketheorientedsegmentwiththe initialpointof v andthe“nalpointof u inthisdiagram.Theresulting vectoristhedisplacementvectorofthepersonrelativetothesurface ofthewaterin1secondandhencede“nesthevelocityoftheperson relativetothewater.Thisgeometricalprocedureiscalled additionof vectors Consideraparallelogramwhoseadjacentsides,thevectors a and b ,extendfromthevertexoftheparallelogram.Thesumofthevectors a and b isavector,denoted a + b ,thatisthediagonalofthe parallelogramextendedfromthesamevertex.Notethattheparallel sidesoftheparallelogramrepresentthesamevector(theyareparallel andhavethesamelength).Thisgeometricalruleforaddingvectors iscalledthe parallelogramrule .Itfollowsfromtheparallelogramrule

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72.VECTORSINSPACE19 Figure11.6.Left :Parallelogramruleforaddingtwo vectors.Iftwovectorsformadjacentsidesofaparallelogramatavertex A ,thenthesumofthevectorsisavector thatcoincideswiththediagonaloftheparallelogramand originatesatthevertex A Right :Addingseveralvectorsbyusingtheparallelogramrule.Giventhe“rstvectorinthesum,allothervectorsaretransportedparallel sothattheinitialpointofthenextvectorinthesumcoincideswiththeterminalpointofthepreviousone.The sumisthevectorthatoriginatesfromtheinitialpointof the“rstvectorandterminatesattheterminalpointofthe lastvector.Itdoesnotdependontheorderofvectorsin thesum.thattheadditionofvectorsis commutative : a + b = b + a ; thatis,theorderinwhichthevectorsareaddeddoesnotmatter.To addseveralvectors(e.g., a + b + c ),onecan“rst“nd a + b bythe parallelogramruleandthenadd c tothevector a + b .Alternatively, thevectors b and c canbeadded“rst,andthenthevector a canbe addedto b + c .Accordingtotheparallelogramrule,theresulting vectoristhesame: ( a + b )+ c = a +( b + c ) Thismeansthattheadditionofvectorsis associative .Soseveralvectorscanbeaddedinanyorder.Takethe“rstvector,thenmovethe secondvectorparalleltoitselfsothatitsinitialpointcoincideswiththe terminalpointofthe“rstvector.Thethirdvectorismovedparallel sothatitsinitialpointcoincideswiththeterminalpointofthesecond vector,andsoon.Finally,makeavectorwhoseinitialpointcoincides withtheinitialpointofthe“rstvectorandwhoseterminalpointcoincideswiththeterminalpointofthelastvectorinthesum.Tovisualize thisprocess,imagineamanwalkingalongthe“rstvector,thengoing paralleltothesecondvector,thenparalleltothethirdvector,andso

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2011.VECTORSANDTHESPACEGEOMETRY on.Theendpointofhiswalkisindependentoftheorderinwhichhe choosesthevectors.AlgebraicAdditionofVectors.Definition 11.7 Thesumoftwovectors a = a1,a2,a3 and b = b1,b2,b3 isavectorwhosecomponentsarethesumsofthecorrespondingcomponentsof a and b : a + b = a1+ b1,a2+ b2,a3+ b3 Thisde“nitionisequivalenttothegeometricalde“nitionofadding vectors,thatis,theparallelogramrulethathasbeenmotivatedby studyingthevelocityofacombinedmotion.Indeed,put a = Š OA wheretheendpoint A hasthecoordinates( a1,a2,a3).Avector b representsallparallelsegmentsofthesamelength b .Inparticular, b is onesuchorientedsegmentwhoseinitialpointcoincideswith A .Supposethat a + b = Š OC = c1,c2,c3 ,where C hascoordinates( c1,c2,c3). Bytheparallelogramrule, b = Š AC = c1Š a1,c2Š a2,c3Š a3 ,where therelationbetweenthecomponentsofavectorandthecoordinates ofitsendpointshasbeenused(seeExample11.3).Theequalityoftwo vectorsmeanstheequalityofthecorrespondingcomponents,thatis, b1= c1Š a1, b2= c2Š a2,and b3= c3Š a3,or c1= a1+ b1, c2= a2+ b2, and c3= a3+ b3asrequiredbythealgebraicadditionofvectors.RulesofVectorAlgebra.Combiningadditionofvectorswithmultiplicationbyrealnumbers,thefollowingsimplerulecanbeestablishedby eithergeometricaloralgebraicmeans: s ( a + b )= s a + s b ( s + t ) a = s a + t a Thedierenceoftwovectorscanbede“nedas a Š b = a +( Š 1) b Intheparallelogramwithadjacentsides a and b ,thesumofvectors a and( Š 1) b representsthevectorthatoriginatesfromtheendpoint of b andendsattheendpointof a because b +[ a +( Š 1) b ]= a in accordancewiththegeometricalruleforaddingvectors;that,is a b aretwodiagonalsoftheparallelogram.Theprocedureisillustratedin Figure11.7(leftpanel). Example 11.5 Anobjecttravels 3 secondswithvelocity v = 1 2 4 wherethecomponentsaregiveninmeterspersecond,andthen 2 secondswithvelocity u = 2 4 1 .Findthedistancebetweentheinitial andterminalpointsofthemotion.

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72.VECTORSINSPACE21 Figure11.7.Left :Subtractionoftwovectors.Thedifference a Š b isviewedasthesumof a and Š b ,thevector thathasthedirectionoppositeto b andthesamelengthas b .Theparallelogramruleforadding a and Š b showsthat thedierence a Š b = a +( Š b )isthevectorthatoriginates fromtheterminalpointof b andendsattheterminalof a if a and b areadjacentsidesofaparallelogram;thatis,the sum a + b andthedierence a Š b arethetwodiagonals oftheparallelogram. Right :IllustrationtoStudyProblem 11.6.Anyvectorinaplanecanalwaysberepresentedasa linearcombinationoftwononparallelvectors.Solution: Lettheinitialandterminalpointsbe A and B ,respectively.Let C bethepointatwhichthevelocitywaschanged.Then Š AC =3 v and Š Š CB =2 u .Therefore, Š AB = Š AC + Š Š CB =3 v +2 u =3 1 2 4 +2 2 4 1 = 3 6 12 + 4 8 2 = 7 14 14 =7 1 2 2 Thedistance | AB | isthelength(orthenorm)ofthevector Š AB .So | AB | = 7 1 2 2 =7 1 2 2 =7 1+4+4=21meters. 72.4.StudyProblems.Problem11.6. Considertwononparallelvectors a and b inaplane. Showthatanyvector c inthisplanecanbewrittenasalinearcombination c = t a + s b forsomereal t and s Solution: Byparalleltransport,thevectors a b ,and c canbemoved sothattheirinitialpointscoincide.Thevectors t a and s b areparallel to a and b ,respectively,forallvaluesof s and t .Considerthelines Laand Lbthatcontainthevectors a and b ,respectively.Constructtwo linesthroughtheterminalpointof c ;oneisparallelto Laandtheother to LbasshowninFigure11.7(rightpanel).Thepointsofintersection oftheselineswith Laand Lbandtheinitialandterminalpointsof c formtheverticesoftheparallelogramwhosediagonalis c andwhose adjacentsidesareparallelto a and b .Therefore, a and b canalwaysbe scaledsothat t a and s b becometheadjacentsidesoftheconstructed

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2211.VECTORSANDTHESPACEGEOMETRY parallelogram.Foragiven c ,thereals t and s areuniquelyde“ned bytheproposedgeometricalconstruction.Bytheparallelogramrule, c = t a + s b Problem11.7. Findthecoordinatesofapoint B thatisatadistance of 6 unitsoflengthfromthepoint A (1 Š 1 2) inthedirectionofthe vector v = 2 1 Š 2 Solution: Thepositionvectorofthepoint A is a = Š OA = 1 Š 1 2 Thepositionvectorofthepoint B is b = a + s v ,where s isapositive numbertobechosensuchthatthelength | AB | = s v equals6.Since v =3,one“nds s =2.Therefore, b = 1 Š 1 2 +2 2 1 Š 2 = 5 1 Š 2 Problem11.8. Considerastraightlinesegmentwiththeendpoints A (1 2 3) and B ( Š 2 Š 1 0) .Findthecoordinatesofthepoint C onthe segmentsuchthatitistwiceasfarfrom A asitisfrom B Solution: Let a = 1 2 3 b = Š 1 0 1 ,and c bepositionvectors of A B ,and C ,respectively.Thequestionisnowtoexpress c via a and b .Onehas c = a + Š AC .Thevector Š AC isparallelto Š AB = Š 3, Š 3 Š 3 andhence Š AC = s Š AB .Since | AC | =2 | CB | | AC | =2 3| AB | andtherefore s =2 3.Thus, c = a +2 3Š AB = a +2 3( b Š a )= Š 1 0 1 Problem11.9. InStudyProblem11.6,let a =1 b =2 ,and theanglebetween a and b be 2 / 3 .Findthecoecients s and t ifthe vector c hasanormof 6 andbisectstheanglebetween a and b Solution: ItfollowsfromthesolutionofStudyProblem11.6that thenumbers s and t donotdependonthecoordinatesystemrelativetowhichthecomponentsofallthevectorsarede“ned.So choosethecoordinatesystemsothat a isparalleltothe x axisand b liesinthe xy plane.Withthischoice, a = 1 0 0 and b = b cos(2 / 3) b sin(2 / 3) 0 = Š 1 3 0 .Similarly, c isthevectoroflength c =6thatmakestheangle / 3withthe x axis,and therefore c = 3 3 3 0 .Equatingthecorrespondingcomponentsin therelation c = t a + s b ,one“nds3= t Š s and3 3= s 3,or s =3 and t =6.Hence, c =6 a +3 b Problem11.10. Supposethethreecoordinateplanesareallmirrored. Alightraystrikesthemirrors.Determinethedirectioninwhichthe re”ectedraywillgo. Solution: Let u beavectorparalleltotheincidentray.Under are”ectionfromaplanemirror,thecomponentof u perpendicularto

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72.VECTORSINSPACE23 theplanechangesitssign.Therefore,afterthreeconsecutivere”ections fromeachcoordinateplane,allthreecomponentsof u changetheir signs,andthere”ectedraywillgoparalleltotheincidentraybutin theexactoppositedirection.Forexample,supposetherayisre”ected “rstbythe xz plane,thenbythe yz plane,and“nallybythe xy plane. Inthiscase, u = u1,u2,u3 u1, Š u2,u3Š u1, Š u2,u3 Š u1, Š u2, Š u3 = Š u Remark. Thisprincipleisusedtodesignre”ectorslikethecatseyesonbicyclesandthosethatmarktheborderlinesofaroad.No matterfromwhichdirectionsuchare”ectorisilluminated(e.g.,bythe headlightsofacar),itre”ectsthelightintheoppositedirection(so thatitwillalwaysbeseenbythedriver).72.5.Exercises.(1) Findthecomponentsofeachofthefollowingvectorsandtheir norms: (i)Thevectorthathasendpoints A (1 2 3)and B ( Š 1 5 1)and isdirectedfrom A to B (ii)Thevectorthathasendpoints A (1 2 3)and B ( Š 1 5 1)and isdirectedfrom B to A (iii)Thevectorthathastheinitialpoint A (1 2 3)andthe“nal point C thatisthemidpointofthelinesegment AB ,where B =( Š 1 5 1) (iv)Thepositionvectorofapoint P obtainedfromthepoint A ( Š 1 2 Š 1)bytransportingthelatteralongthevector u = 2 2 1 3unitsoflengthandthenalongthevector w = Š 3 0 Š 4 10unitsoflength (v)Thepositionvectorofthevertex C ofatriangle ABC inthe xy planeifA isattheorigin, B =( a, 0 0),theangleatthe vertex B is / 3,and | BC | =3 a (2) Let a and b betwovectorsthatareneitherparallelnorperpendicular.Sketcheachofthefollowingvectors: a +2 b b Š 2 a a Š1 2b and2 a +3 b (3) Let a b ,and c bethreevectorsinaplaneanyofwhichisnot paralleltotheothers.Sketcheachofthefollowingvectors: a +( b Š c ), ( a + b ) Š c ,2 a Š 3( b + c ),and(2 a Š 3 b ) Š 3 c (4) Let a = 2 Š 1 Š 2 and b = Š 3 0 4 .Findunitvectors a and b Express6 a Š 15 b via a and b (5) Let a and b bevectorsinthe xy planesuchthattheirsum c = a + bmakestheangle / 3with a andhaslengthtwicethelengthof a .Find

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2411.VECTORSANDTHESPACEGEOMETRY b if a liesinthe“rstquadrant,makestheangle / 3withthepositive x axis,andhaslength2. (6) Consideratriangle ABC .Let a beavectorfromthevertex A to themidpointoftheside BC ,let b beavectorfrom B tothemidpoint of AC ,andlet c beavectorfrom C tothemidpointof AB .Usevector algebrato“nd a + b + c (7) Let uk, k =1 2 ,...,n ,beunitvectorsinaplanesuchthatthe smallestanglebetweenthetwovectors ukand uk +1is2 /n .Whatis thesum vn= u1+ u2+ + unforaneven n ?Sketchthesumfor n =1, n =3,and n =5.Comparethenorms vn for n =1 3 5. Investigatethelimitof vnas n bystudyingthelimitof vn as n (8) Let uk, k =1 2 ,...,n ,beunitvectorsasde“nedinexercise(7). Let wk= uk +1Š ukfor k =1 2 ,...,n Š 1and wn= u1Š un.Findthe limitof w1 + w2 + + wn as n (9) Aplane”iesataspeedof v mi/hrelativetotheair.Thereis awindblowingataspeedof u mi/hinthedirectionthatmakesthe angle withthedirectioninwhichtheplanemoves.Whatisthespeed oftheplanerelativetotheground? (10) Usevectoralgebratoshowthatthelinesegmentjoiningthemidpointsoftwosidesofatriangleisparalleltothethirdsideandhalfits length. (11) Let a and b bepositionvectorsinthe xy plane.Describetheset ofallpointsintheplanewhosepositionvectors r satisfythecondition r Š a + r Š b = k ,where k> a Š b (12) Letpointlikemassiveobjectsbepositionedat Pi, i =1 2 ,...,n andlet mibethemassat Pi.Thepoint P0iscalledthe centerof mass if m1Š Š P0P1+ m2Š Š P0P2+ + mnŠ ŠŠ P0Pn= 0 Expressthepositionvector r0of P0viathepositionvectors riof Pi. Inparticular,“ndthecenterofmassofthreepointmasses, m1= m2= m3= m ,locatedattheverticesofatriangle ABC for A (1 2 3), B ( Š 1 0 1),and C (1 1 Š 1). (13) Considerthegraph y = f ( x )ofadierentiablefunctionandthe linetangenttoitatapoint x = a .Expresscomponentsofavector paralleltothelinevia f( a )and“ndavectorperpendiculartotheline. Inparticular,“ndsuchvectorsforthegraph y = x2atthepoint x =1. (14) Letthevectors a b ,and c have“xedlengths a b ,and c ,respectively,whiletheirdirectionsmaybechanged.Isitalwayspossibleto achieve a + b + c = 0 ?Ifnot,formulatetheconditionunderwhichit ispossible.

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73.THEDOTPRODUCT25 (15) Letthevectors a and b have“xedlengths,whiletheirdirections maybechanged.Put c= a b .Isitalwayspossibletoachieve cŠ>c+,or cŠ= c+,or cŠ
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2611.VECTORSANDTHESPACEGEOMETRY 73.1.GeometricalSignicanceoftheDotProduct.Asitstands,thedot productisanalgebraicruleforcalculatinganumberoutofsixgiven numbersthatarecomponentsofthetwovectorsinvolved.Thecomponentsofavectordependonthechoiceofthecoordinatesystem. Naturally,oneshouldaskwhetherthenumericalvalueofthedotproductdependsonthecoordinatesystemrelativetowhichthecomponents ofthevectorsaredetermined.Itturnsoutthatitdoesnot.Therefore, itrepresentsanintrinsicgeometricalquantityassociatedwithtwovectorsinvolvedintheproduct.Toelucidatethegeometricalsigni“cance ofthedotproduct,note“rsttherelationbetweenthedotproductand thenorm(length)ofavector: a a = a2 1+ a2 2+ a2 3= a 2or a = a a Thus,if a = b inthedotproduct,thenthelatterdoesnotdepend onthecoordinatesystemwithrespecttowhichthecomponentsof a arede“ned.Next,considerthetrianglewhoseadjacentsidesarethe vectors a and b asdepictedinFigure11.8(leftpanel).Thentheother sideofthetrianglecanberepresentedbythedierence c = b Š a .The squaredlengthofthislattersideis (11.4) c c =( b Š a ) ( b Š a )= b b + a a Š 2 a b wherethealgebraicpropertiesofthedotproducthavebeenused. Therefore,thedotproductcanbeexpressedviathegeometricalinvariants,namely,thelengthsofthesidesofthetriangle: (11.5) a b = 1 2 a 2+ b 2Š c 2 Thismeansthatthenumericalvalueofthedotproductisindependent ofthechoiceofacoordinatesystem.Inparticular,letustakethe coordinatesysteminwhichthevector a isparalleltothe x axisand thevector b liesinthe xy planeasshowninFigure11.8(rightpanel). Lettheanglebetween a and b be .Byde“nition,thisanglelies intheinterval[0 ].When =0,thevectors a and b pointinthe samedirection.When = / 2,theyaresaidtobe orthogonal ,and theypointintheoppositedirectionif = .Inthechosencoordinate system, a = a 0 0 and b = b cos b sin 0 .Hence, (11.6) a b = a b cos orcos = a b a b Equation(11.6)revealsthegeometricalsigni“canceofthedotproduct.Itdeterminestheanglebetweentwoorientedsegmentsinspace. Itprovidesasimplealgebraicmethodtoestablishamutualorientation oftwostraightlinesegmentsinspace.

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73.THEDOTPRODUCT27 Figure11.8.Left :Independenceofthedotproductfrom thechoiceofacoordinatesystem.Thedotproductof twovectorsthatareadjacentsidesofatrianglecanbeexpressedviathelengthsofthetrianglesidesasshownin (11.5). Right :Geometricalsigni“canceofthedotproduct.Itdeterminestheanglebetweentwovectorsasstated in(11.6).Twononzerovectorsareperpendicularifandonly iftheirdotproductvanishes.Thisfollowsfrom(11.5)and thePythagoreantheorem: a 2+ b 2= c 2forarightangledtriangle.Theorem 11.1 (GeometricalSigni“canceoftheDotProduct) If istheanglebetweenthevectors a and b ,then a b = a b cos Inparticular,twononzerovectorsareorthogonalifandonlyiftheir dotproductvanishes: a b a b =0 Foratrianglewithsides a b ,and c andanangle betweensides a and b ,itfollowsfromtherelation(11.4)that c2= a2+ b2Š 2 ab cos Foraright-angledtriangle, c2= a2+ b2(thePythagoreantheorem). Example 11.7 Consideratrianglewhoseverticesare A (1 1 1) B ( Š 1 2 3) ,and C (1 4 Š 3) .Findalltheanglesofthetriangle. Solution: Lettheanglesatthevertices A B ,and C be ,and respectively.Then + + =180.Soitissucientto“ndanytwo angles.To“ndtheangle ,de“nethevectors a = Š AB = Š 2 1 2 and b = Š AC = 0 3 Š 4 .Theinitialpointofthesevectorsis A ,and hencetheanglebetweenthevectorscoincideswith .Since a =3

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2811.VECTORSANDTHESPACEGEOMETRY and b =5,bythegeometricalpropertyofthedotproduct, cos = a b a b = 0+3 Š 8 15 = Š 1 3 = =cosŠ 1( Š 1 / 3) 109 5. To“ndtheangle ,de“nethevectors a = Š BA = 2 Š 1 Š 2 and b = Š Š BC = 2 2 Š 6 withtheinitialpointatthevertex B .Thenthe anglebetweenthesevectorscoincideswith .Since a =3, b = 2 11,and a b =4 Š 2+12=14,one“ndscos =14 / (6 11)and =cosŠ 1(7 / (3 11)) 45 3.Therefore, 180Š 109 5Š 45 3= 25 2.NotethattherangeofthefunctioncosŠ 1mustbetakenfrom 0to180inaccordancewiththede“nitionoftheanglebetweentwo vectors. 73.2.FurtherGeometricalPropertiesoftheDotProduct.Corollary 11.1 (OrthogonalDecompositionofaVector). Givenanonzerovector a ,anyvector b canbeuniquelydecomposed intothesumoftwoorthogonalvectors,oneofwhichisparallelto a : b = b+ b, b= b Š s a b= s a ,s = b a a 2. Indeed,given a and b ,put b= b Š s a andassumethat bis orthogonalto a ,thatis, a b=0.Thisconditionuniquelydetermines thecoecient s : a b Š s a a =0or s = b a / a 2.Thevectors band barecalledthe orthogonal and parallel componentsof b relative tothevector a .Thevector bisalsocalleda vectorprojection of b onto a .Theorthogonaldecomposition b = b+ bisshownin Figure11.10(rightpanel).If a = a / a istheunitvectoralong a then b= b a ,wherethecoecient b= a b / a = b cos iscalled a scalarprojection of b onto a .Itisalsoeasytoseefromthe“gure that b = b sin Example 11.8 Let a = 1 Š 2 1 and b = 5 1 9 .Findthe orthogonaldecomposition b = b+ brelativetothevector a Solution: Onehas a b =5 Š 2+9=12and a 2= a a =1+( Š 2)2+ 1=6.Therefore, s =12 / 6=2, b= s a =2 1 Š 2 1 = 2 Š 4 2 and b= b Š b= 5 1 9 Š 2 Š 4 2 = 3 5 7 .Theresultcanalso beveri“ed: a b=3 Š 10+7=0;thatis, a isorthogonalto bas required.

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73.THEDOTPRODUCT29 Theorem 11.2 (Cauchy-SchwarzInequality) Foranytwovectors a and b | a b | a b wheretheequalityisreachedonlyifthevectorsareparallel. Thisinequalityisadirectconsequenceofthe“rstrelationin(11.6) andtheinequality | cos | 1.Theequalityisreachedonlywhen =0 or = ,thatis,when a and b areparallelorantiparallel. Theorem 11.3 (TriangleInequality) Foranytwovectors a and b a + b a + b Proof. Put a = a and b = b sothat a a = a 2= a2and similarly b b = b2.Usingthealgebraicrulesforthedotproduct, a + b 2=( a + b ) ( a + b )= a2+ b2+2 a b a2+ b2+2 ab =( a + b )2, wheretheCauchy-Schwarzinequalityhasbeenused.Bytakingthe squarerootofbothsides,thetriangleinequalityisobtained. Thetriangleinequalityhasasimplegeometricalmeaning.Consider atrianglewithsides a b ,and c .Thedirectionsofthevectorsare chosensothat c = a + b .Thetriangleinequalitystatesthatthe length c cannotexceedthetotallengthoftheothertwosides.It isalsoclearthatthemaximallength c = a + b isattained onlyif a and b areparallelandpointinthesamedirection.Ifthey areparallelbutpointintheoppositedirection,thenthelength c becomesminimalandcoincideswith | a Š b | .Theabsolutevalue isnecessaryasthelengthof a maybelessthanthelengthof b .This observationcanbestatedinthefollowingalgebraicform: (11.7) a Š b a + b a + b .73.3.DirectionAngles.Considerthreeunitvectors e1= 1 0 0 e2= 0 1 0 ,and e3= 0 0 1 thatareparalleltothecoordinateaxes x y and z ,respectively.Bytherulesofvectoralgebra,anyvectorcanbe writtenasthesumofthreemutuallyperpendicularvectors: a = a1,a2,a3 = a1 e1+ a2 e2+ a3 e3. Thevectors a1 e1, a2 e2,and a3 e3areadjacentsidesoftherectangular boxwhoselargestdiagonalcoincideswiththevector a asshownin Figure11.9(rightpanel).De“netheangle thatiscountedfromthe positivedirectionofthe x axistowardthevector a .Inotherwords, theangle istheanglebetween e1and a .Similarly,theangles and

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3011.VECTORSANDTHESPACEGEOMETRY Figure11.9.Left :Thedirectionanglesofavectorarede“nedastheanglesbetweenthevectorandthreecoordinate axes.Eachanglerangesbetween0and andiscounted fromthecorrespondingpositivecoordinatesemiaxistoward thevector.Thecosinesofthedirectionanglesofavectorare componentsoftheunitvectorparalleltothatvector. Right : Thedecompositionofavector a intothesumofthreemutuallyperpendicularvectorsthatareparalleltothecoordinate axesofarectangularcoordinatesystem.Thevectoristhe diagonaloftherectangularboxwhoseedgesareformedby thevectorsinthesum. are,byde“nition,theanglesbetween a andtheunitvectors e2and e3,respectively.Then cos = e1 a e1 a = a1 a cos = e2 a e2 a = a2 a cos = e3 a e3 a = a3 a Thesecosinesarenothingbutthecomponentsoftheunitvectorparallel to a : a = 1 a a = cos cos cos Thus,theangles ,and uniquelydeterminethedirectionofa vector.Forthisreason,theyarecalled directionangles .Notethatthey cannotbesetindependentlybecausetheyalwayssatisfythecondition a =1or cos2 +cos2 +cos2 =1 .

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73.THEDOTPRODUCT31 Inpractice(physics,mechanics,etc.),vectorsareoftenspeci“edby theirmagnitude a = a anddirectionangles.Thecomponentsare thenfoundby a1= a cos a2= a cos ,and a3= a cos .73.4.BasisVectorsinSpace.Acollectionofallorderedtriplesofreal numbers a1,a2,a3 inwhichtheaddition,themultiplicationbyanumber,thedotproduct,andthenormarede“nedasinvectoralgebrais alsocalledathree-dimensional Euclideanspace .Similarly,acollection ofordereddoubletsofrealnumbersisatwo-dimensionalEuclidean space.Asnoted,anyelementofathree-dimensionalEuclideanspace canbeuniquelyrepresentedasalinearcombinationofthreeparticularelements e1= 1 0 0 e2= 0 1 0 ,and e3= 0 0 1 .They arecalledthe standardbasis .Thereareothertriplesofvectorswith thecharacteristicpropertythatanyvectorisauniquelinearcombinationofthem.Givenanythreemutuallyorthogonalunitvectors ui, i =1 2 3,anyvectorinspacecanbe uniquely expandedintothesum a = a1 u1+ a2 u2+ a3 u3,wherethenumbers aiarethescalarprojections of a onto ui.Anysuchtripleofvectorsiscalledan orthonormalbasis in space.Sowithanyorthonormalbasisonecanassociatearectangular coordinatesysteminwhichthecoordinatesofapointaregivenbythe scalarprojectionsofitspositionvectorontothebasisvectors. Definition 11.9 (BasisinSpace). Atripleofvectors u1, u2,and u3iscalleda basisinspace ifany vector a canbeuniquelyrepresentedasalinearcombinationofthem: a = a1u1+ a2u2+ a3u3. Abasismaycontainvectorsthatarenotnecessarilyorthogonalor unit.Forexample,avectorinaplaneisa unique linearcombination oftwogivennonparallelvectorsintheplane(StudyProblem11.6).In thissense,anytwononparallelvectorsinaplanede“nea(nonorthogonal)basisinaplane.Considerthreevectorsinspace.Ifoneofthem isalinearcombinationoftheothers,thenthevectorsareinoneplane andcalled coplanar .Supposethatnoneofthevectorsisalinearcombinationoftheothertwo;thatis,theyarenotcoplanar.Suchvectors arecalled linearlyindependent .Thus, threevectors a b ,and c are linearlyindependentifandonlyifthevectorequation x a + y b + z c = 0 hasonlyatrivialsolution x = y = z =0because,otherwise,oneofthe vectorsisalinearcombinationoftheothers.Forexample,if x =0, then a = Š ( y/x ) b Š ( z/x ) c .Itcanbeprovedthat anythreelinearlyindependentvectorsformabasisinspace (StudyProblems11.11 and11.12).

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3211.VECTORSANDTHESPACEGEOMETRY 73.5.ApplicationsoftheDotProduct. StaticProblems.AccordingtoNewtonsmechanics,apointlikeobject thatisatrestremainsatrestifthevectorsumofallforcesappliedto itvanishes.Thisisthefundamentallawofstatics: F1+ F2+ + Fn= 0 Thisvectorequationimpliesthreescalarequationsthatrequirevanishingeachofthethreecomponentsofthetotalforce.Asystemof objectsisatrestifallitselementsareatrest.Thus, foranyelement ofasystematrest,thescalarprojectionofthetotalforceontoanyvectorvanishes .Inparticular,thecomponentsofthetotalforceshould vanishin anyorthonormalbasis or,asapointoffact,theyvanishin anybasisinspace(seeStudyProblem11.11).Thisprincipleisusedto determineeitherthemagnitudesofsomeforcesorthevaluesofsome geometricalparametersatwhichthesysteminquestionisatrest. Example 11.9 Letaballofmass m beattachedtotheceilingby tworopessothatthesmallestanglebetweenthe“rstropeandtheceiling is 1andtheangle 2isde“nedsimilarlyforthesecondrope.Findthe magnitudesofthetensionforcesintheropes. Solution: ThesysteminquestionisshowninFigure11.10(left panel).Theequilibriumconditionis T1+ T2+ G = 0 Let e1beaunitvectorthatishorizontalanddirectedfromlefttoright andlet e2beaunitvectordirectedupward.Theyformanorthonormal basisintheplane.Usingthescalarprojections,theforcescanbe expandedinthisbasisas T1= Š T1cos 1 e1+ T1sin 1 e2, T2= T2cos 2 e1+ T2sin 2 e2, G = Š mg e2, where T1and T2arethemagnitudesofthetensionforces.Thescalar projectionsofthetotalforceontothehorizontalandverticaldirections de“nedby e1and e2shouldvanish: Š T1cos 1+ T2cos 2=0 ,T1sin 1+ T2sin 2Š mg =0 Thissystemisthensolvedfor T1and T2.Bymultiplyingthe“rst equationbysin 1andthesecondbycos 1andthenaddingthem,one gets T2= mg cos 1/ sin( 1+ 2).Substituting T2intothe“rstequation, thetension T1= mg cos 2/ sin( 1+ 2)isobtained.

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73.THEDOTPRODUCT33 Figure11.10.Left :IllustrationtoExample11.9.At equilibrium,thevectorsumofallforcesactingontheball vanishes.Thecomponentsoftheforcesareeasyto“ndinthe coordinatesysteminwhichthe x axisishorizontalandthe y axisisvertical. Right :Thevector c isthevectorprojection ofavector b onto a .Thelinethroughtheterminalpoints of b and c isperpendicularto a .Thescalarprojectionof b onto a is b cos ,where istheanglebetween a and b .It ispositiveif / 2.WorkDonebyaForce.Supposethatanobjectofmass m moveswith speed v .Thequantity K = mv2/ 2iscalledthe kineticenergy of theobject.Supposethattheobjecthasmovedalongastraightline segmentfromapoint P1toapoint P2undertheactionofaconstant force F .Alawofphysicsstatesthatachangeinanobjectskinetic energyisequaltothework W donebythisforce: K2Š K1= F Š Š P1P2= W, where K1and K2arethekineticenergiesattheinitialand“nalpoints ofthemotion,respectively. Example 11.10 Letanobjectslideonaninclinedplanewithout frictionunderthegravitationalforce.Themagnitudeofthegravitationalforceisequalto mg ,where m isthemassoftheobjectand g isauniversalconstantforallobjectsnearthesurfaceoftheEarth, g 9 8 m/s2.Findthe“nalspeed v oftheobjectiftherelativeheight oftheinitialand“nalpointsis h andtheobjectwasinitiallyatrest. Solution: Choosethecoordinatesystemsothatthedisplacement vector Š Š P1P2andthegravitationalforceareinthe xy plane.Letthe y axisbeverticalsothatthegravitationalforceis F = 0 Š mg, 0 where m isthemassand g istheaccelerationofthefreefall.Theinitial pointischosentohavethecoordinates(0 ,h, 0)whilethe“nalpoint

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3411.VECTORSANDTHESPACEGEOMETRY is( L, 0 0),where L isthedistancetheobjecttravelsinthehorizontal directionwhilesliding.Thedisplacementvectoris Š Š P1P2= L, Š h, 0 Since K1=0,onehas mv2 2 = W = F Š Š P1P2= 0 Š mg, 0 L, Š h, 0 = mgh v = 2 gh. Notethatthespeedisindependentofthemassoftheobjectandthe inclinationangleoftheplane(itstangentis h/L );itisfullydetermined bytherelativeheightonly. 73.6.StudyProblems.Problem11.11.(GeneralBasisinSpace) Let ui, i =1 2 3 ,bethreelinearlyindependent(non-coplanar)vectors. Showthattheyformabasisinspace;thatis,anyvector a canbe uniquelyexpandedintothesum a = s1u1+ s2u2+ s3u3. Thenumbers siarecalled componentsof a relativetothebasis ui. Solution: Asolutionemploysthesameapproachasinthesolutionof StudyProblem11.6.Let P1betheparallelogramwithadjacentsides u2and u3,and P2betheparallelogramwithsides u1and u3,and P3be theparallelogramwithsides u1and u2.Consideraboxwhosefacesare theparallelograms P1, P2,and P3.Thisboxiscalleda parallelepiped Bytherulesofvectoralgebra,thelargestdiagonaloftheparallelepiped isthesum u1+ u2+ u3.Letthevectors uiandavector a havecommon initialpoint.Considerthreeplanesthroughtheterminalpointof a that areparalleltotheparallelograms P1, P2, P3andsimilarplanesthrough theinitialpointof a .Thesesixplanesencloseaparallelepipedwhose largestdiagonalisthevector a andwhoseadjacentsidesare parallel tothevectors uiandthereforeareproportionaltothem;thatis,the adjacentedgesarethevectors s1u1, s2u2,and s3u3,wherethenumbers s1, s2,and s3areuniquelydeterminedbytheproposedconstruction oftheparallelepiped.Hence, a = s1u1+ s2u2+ s3u3.Notethatthe samegeometricalconstructionhasbeenusedtoexpandavectorinan orthonormal basis eiasshowninFigure11.9. Problem11.12. Let u1= 1 1 0 u2= 1 0 1 ,and u3= 2 2 1 Showthatthesevectorsarelinearlyindependentandhenceformabasis inspace.Findthecomponentsof a = 1 2 3 inthisbasis.

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73.THEDOTPRODUCT35 Solution: Ifthevectors uiarenotlinearlyindependent,thenthere shouldexistnumbers c1, c2,and c3thatdonotsimultaneouslyvanish suchthat c1u1+ c2u2+ c3u3= 0 Indeed,thisalgebraicconditionmeansthatoneofthevectorsisalinear combinationoftheothertwowhenever cidonotvanishsimultaneously. Thisvectorequationcanbewritteninthecomponents: c1+ c2+2 c3=0 c1+2 c3=0 c2+ c3=0 c1+ c2+3 c3=0 c1= Š 2 c3c2= Š c3. Thesubstitutionofthelasttwoequationsintothe“rstoneyields Š c3Š 2 c3+2 c3=0or c3=0andhence c1= c2=0.Thus,thevectors uiarelinearlyindependentandformabasisinspace.Foranyvector a = s1u1+ s2u2+ s3u3,thenumbers si, i =1 2 3,arecomponents of a inthebasis ui.Bywritingthisvectorequationincomponents relativetothestandardbasisfor a = 1 2 3 ,thesystemofequations isobtained: s1+ s2+2 s3=1 s1+2 s3=2 s2+ s3=3 s1+ s2+2 s3=1 s1=2 Š 2 s3s2=3 Š s3. Thesubstitutionofthelasttwoequationsintothe“rstoneyields s3=4 andhence s1= Š 6and s2= Š 1sothat a = Š 6 u1Š u2+4 u3. Problem11.13. Describethesetofallpointsinspacewhoseposition vectors r satisfythecondition ( r Š a ) ( r Š b )=0 .Hint: Notethatthe positionvectorsatisfyingthecondition r Š c = R describesasphere ofradius R whosecenterhasthepositionvector c Solution: Theequationofaspherecanalsobewrittenintheform r Š c 2=( r Š c ) ( r Š c )= R2.Theequation( r Š a ) ( r Š b )=0can betransformedintothesphereequationbycompletingthesquares. Usingthealgebraicpropertiesofthedotproduct, ( r Š a ) ( r Š b )= r r Š r ( a + b )+ a b =( r Š c ) ( r Š c ) Š c c + a b c =1 2( a + b ) c c Š a b = R R R =1 2( a Š b ) Hence,thesetisasphereofradius R = R ,anditscenterispositioned at c .If a and b arethepositionvectorsofpoints A and B ,then,by theparallelogramrule,thecenterofthesphereisthemidpointofthe

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3611.VECTORSANDTHESPACEGEOMETRY straightlinesegment AB andthesegment AB isadiameterofthe sphere. 73.7.Exercises.(1) Findthedotproduct a b if (i) a = 1 2 3 and b = Š 1 2 0 (ii) a = e1+3 e2Š e3and b =3 e1Š 2 e2+ e3(iii) a =2 c Š 3 d and b = c +2 d if c istheunitvectorthatmakes theangle / 3withthevector d and d =2 (2) Forwhatvaluesof b arethevectors Š 6 ,b, 5 and b,b, 1 orthogonal? (3) Findtheangleatthevertex A ofatriangle ABC for A (1 0 1), B (1 2 3),and C (0 1 1). (4) Findthecosinesoftheanglesofatriangle ABC for A (0 1 1), B ( Š 2 4 3),and C (1 2 Š 1). (5) Consideratrianglewhoseonesideisadiameterofacircleandthe vertexoppositetothissideisonthecircle.Usevectoralgebratoprove thatanysuchtriangleisright-angled. Hint: Considerpositionvectors oftheverticesrelativetothecenterofthecircle. (6) Let a = s u + v and b = u + s v ,wheretheanglebetweenunit vectors u and v is / 3.Findthevaluesof s forwhichthedotproduct a b ismaximal,orminimal,orvanishes.Ifsuchvaluesexist,graph thevectors a and b forthesevaluesof s (7) Consideracubewhoseedgeshavelength a .Findtheanglebetweenitslargestdiagonalandanyedgeadjacenttothediagonal. (8) Consideraparallelepipedwithadjacentsides a = 1 Š 2 2 b = 2 2 Š 1 ,and c = 1 1 1 (seethede“nitionofaparallelepipedin StudyProblem11.11).Findtheanglesbetweenitslargestdiagonal andtheadjacentsides. (9) Let a = 1 2 2 .Forthevector b = Š 2 3 1 ,“ndthescalarand vectorprojectionsof b onto aandconstructtheorthogonaldecomposition b = b+ brelativeto a (10) Findallvectorsthathaveagivenlength a andmaketheangle / 3withthepositive x axisandtheangle / 4withthepositive z axis. (11) Findthecomponentsofallunitvectors u thatmaketheangle / 6withthepositive z axis. Hint: Put u = a v + b e3,where v isa unitvectorinthe xy plane.Find a b ,andall v usingthepolarangle inthe xy plane. (12) If c = a b + b a ,where a and b arenonzerovectors,show that c bisectstheanglebetween a and b .

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73.THEDOTPRODUCT37 (13) Letthevectors a and b havethesamelength.Showthatthe vectors a + b and a Š b areorthogonal. (14) Consideraparallelogramwithadjacentsidesoflength a and b If d1and d2arethelengthsofthediagonals,provetheparallelogram law: d2 1+ d2 2=2( a2+ b2). Hint: Considerthevectors a and b that areadjacentsidesoftheparallelogramandexpressthediagonalsvia a and b .Usethedotproducttoevaluate d2 1+ d2 2. (15) Consideraright-angledtrianglewhoseadjacentsidesattheright anglehavelengths a and b .Let P beapointinspaceatadistance c fromallthreeverticesofthetriangle( c a/ 2and c b/ 2).Findthe anglesbetweenthelinesegmentsconnecting P withtheverticesofthe triangle. Hint: Considervectorswiththeinitialpoint P andterminal pointsattheverticesofthetriangle. (16) Showthatthevectors u1= 1 1 2 u2= 1 Š 1 0 ,and u3= 2 2 Š 2 aremutuallyorthogonal.Foravector a = 4 3 4 ,“ndthe scalarorthogonalprojectionsof a onto ui, i =1 2 3,andthenumbers sisuchthat a = s1u1+ s2u2+ s3u3. (17) Apointobjecttraveled3metersfromapoint A inaparticular direction,thenitchangedthedirectionby60andtraveled4meters, andthenitchangedthedirectionagainsothatitwastravelingat60witheachoftheprevioustwodirections.Ifthelaststretchwas2meterslong,howfarfrom A istheobject? (18) Twoballsofthesamemass m areconnectedbyapieceofropeof length h .Thentheballsareattachedtodierentpointsonahorizontal ceilingbyapieceofropewiththesamelength h sothatthedistance L betweenthepointsisgreaterthan h butlessthan3 h .Findthe equilibriumpositionsoftheballsandthemagnitudeoftensionforces intheropes. (19) Aballofmass m isattachedbythreeropesofthesamelength a toahorizontalceilingsothattheattachmentpointsontheceilingform atrianglewithsidesoflength a .Findthemagnitudeofthetension forceintheropes. (20) Fourdogsareattheverticesofasquare.Eachdogstartsrunning towarditsneighborontheright.Thedogsrunwiththesamespeed v Ateverymomentoftime,eachdogkeepsrunninginthedirectionof itsrightneighbor(itsvelocityvectoralwayspointstotheneighbor). Eventually,thedogsmeetinthecenterofthesquare.Whenwillthis happenifthesidesofthesquarehavelength a ?Whatisthedistance traveledbyeachdog? Hint: Isthereaparticulardirectionrelativeto whichthevelocityvectorofadoghasthesamecomponentateach momentoftime?

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3811.VECTORSANDTHESPACEGEOMETRY 74.TheCrossProduct74.1.DeterminantofaSquareMatrix.Definition 11.10 Thedeterminantofa 2 2 matrixisthenumber computedbythefollowingrule: det a11a12a21a22 = a11a22Š a12a21, thatis,theproductofthediagonalelementsminustheproductofthe o-diagonalelements. Definition 11.11 Thedeterminantofa 3 3 matrix A isthe numberobtainedbythefollowingrule: det a11a12a13a21a22a23a31a32a33 = a11det A11Š a12det A12+ a13det A13=3k =1( Š 1)k +1a1 kdet A1 k, A11= a22a23a32a33 ,A12= a21a23a31a33 ,A13= a21a22a31a32 wherethematrices A1 k, k =1 2 3 ,areobtainedfromtheoriginal matrix A byremovingtherowandcolumncontainingtheelement a1 k. Itisstraightforwardtoverifythatthedeterminantcanbeexpanded overanyroworcolumn: det A =3k =1( Š 1)k + mamkdet Amkforany m =1 2 3 det A =3m =1( Š 1)k + mamkdet Amkforany k =1 2 3 wherethematrix Amkisobtainedfrom A byremovingtherowand columncontaining amk.Thisde“nitionofthedeterminantisextended recursivelyto N N squarematricesbyletting k and m rangeover 1 2 ,...,N Inparticular,thedeterminantofatriangularmatrix(i.e.,thematrixallofwhoseelementseitheraboveorbelowthediagonalvanish)is theproductofitsdiagonalelements: det a1bc 0 a2d 00 a3 =det a100 ba20 cda3 = a1a2a3

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74.THECROSSPRODUCT39 foranynumbers b c ,and d .Also,itfollowsfromtheexpansionofthe determinantoveranycolumnorrowthat, ifanytworowsoranytwo columnsareswappedinthematrix,itsdeterminantchangessign .For 2 2matrices,thisiseasytoseedirectlyfromDe“nition11.10.In general,ifthematrix B isobtainedfrom A byswappingthe“rstand secondrows,thatis, b1 k= a2 kand b2 k= a1 k,thenthematrices B2 kand A1 kcoincideandsodotheirdeterminants.Byexpandingdet B overits second row b2 k= a1 k,oneinfersthat det B =3k =1( Š 1)2+ kb2 kdet B2 k=3k =1( Š 1)2+ ka1 kdet A1 k= Š3k =1( Š 1)1+ ka1 kdet A1 k= Š det A. Thisargumentcanbeappliedtoanytworowsorcolumnsinasquare matrixofanydimension. Example 11.11 Calculate det A ,where A = 123 013 Š 121 Solution: Expandingthedeterminantoverthe“rstrowyields det A =1(1 1 Š 2 3) Š 2(0 1 Š ( Š 1) 3)+3(0 2 Š ( Š 1) 1)= Š 8 Alternatively,expandingthedeterminantoverthesecondrowyields thesameresult: det A = Š 0(2 1 Š 3 2)+1(1 1 Š ( Š 1) 3) Š 3(1 2 Š ( Š 1) 2)= Š 8 Onecancheckthatthesameresultcanbeobtainedbyexpandingthe determinantoveranyroworcolumn. 74.2.TheCrossProductofTwoVectors.Definition 11.12 (CrossProduct) Let e1= 1 0 0 e2= 0 1 0 ,and e3= 0 0 1 .Thecrossproduct oftwovectors a = a1,a2,a3 and b = b1,b2,b3 isavectorthatisthe

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4011.VECTORSANDTHESPACEGEOMETRY determinantoftheformalmatrixexpandedoverthe“rstrow: a b =det e1 e2 e3a1a2a3b1b2b3 =det a2a3b2b3 e1Š det a1a3b1b3 e2+det a1a2b1b2 e3= a2b3Š a3b2,a3b1Š a1b3,a1b2Š a2b1 (11.8) Notethatthe“rstrowofthematrixconsistsoftheunitvectors paralleltothecoordinateaxesratherthannumbers.Forthisreason,it isreferredastoa formal matrix.Theuseofthedeterminantismerely acompactwaytowritethealgebraicruletocomputethecomponents ofthecrossproduct. Example 11.12 Evaluatethecrossproduct a b if a = 1 2 3 and b = 2 0 1 Solution: Byde“nition, 1 2 3 2 0 1 =det e1 e2 e3123 201 =det 23 01 e1Š det 13 21 e2+det 12 20 e3=(2 Š 0) e1Š (1 Š 6) e2+(0 Š 4) e3=2 e1+5 e2Š 4 e3= 2 5 Š 4 Thecrossproducthasthefollowingpropertiesthatfollowfromits de“nition: a b = Š b a ( a + c ) b = a b + c b ( s a ) b = s ( a b ) The“rstpropertyisobtainedbyswappingthecomponentsof b and a in(11.8).Alternatively,recallthatthedeterminantofamatrix changesitssigniftworowsareswappedinthematrix(therows a and b inDe“nition11.12).Sothecrossproductisskew-symmetric; thatis,itis notcommutative ,andtheorderinwhichthevectorsare multipliedisessential.Changingtheorderleadstotheoppositevector. Inparticular,if b = a ,then a a = Š a a or2( a a )= 0 or a a = 0 .

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74.THECROSSPRODUCT41 Thecrossproductis distributive accordingtothesecondproperty.To provethischange aito ai+ ci, i =1 2 3,in(11.8).Ifavector a is scaledbyanumber s andtheresultingvectorismultipliedby b ,the resultisthesameasthecrossproduct a b computed“rstandthen scaledby s (change aito saiin(11.8)andthenfactorout s ).The doublecrossproductsatis“esthesocalled bac Š cab rule (11.9) a ( b c )= b ( a c ) Š c ( a b ) andtheJacobiidentity (11.10) a ( b c )+ b ( c a )+ c ( a b )= 0 Notethatthesecondandthirdtermsontheleftsideof(11.10)are obtainedfromthe“rstbycyclicpermutationsofthevectors.The proofsofthe bac Š cab ruleandtheJacobiidentityaregiveninStudy Problems11.16and11.17.TheJacobiidentityimpliesthat a ( b c ) =( a b ) c Thismeansthatthemultiplicationofvectorsde“nedbythecrossproductis notassociative incontrasttomultiplicationofnumbers.This observationisfurtherdiscussedinStudyProblem11.18.74.3.GeometricalSignicanceoftheCrossProduct.Theabovealgebraicde“nitionofthecrossproductusesaparticularcoordinatesystemrelativetowhichthecomponentsofthevectorsarede“ned.Does thecrossproductdependonthechoiceofthecoordinatesystem?To answerthisquestion,oneshouldinvestigatewhetherbothits direction andits magnitude dependonthechoiceofthecoordinatesystem.Let us“rstinvestigatethemutualorientationoftheorientedsegments a b ,and a b .Asimplealgebraiccalculationleadstothefollowing result: a ( a b )= a1( a2b3Š a3b2)+ a2( a3b1Š a1b3)+ a3( a1b2Š a2b1)=0 Bytheskewsymmetryofthecrossproduct,itisalsoconcludedthat b ( a b )= Š b ( b a )=0.Bythegeometricalpropertyofthe dotproduct,thecrossproductmustbeperpendiculartobothvectors a and b : (11.11) a ( a b )= b ( a b )=0 a b a and a b b .

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4211.VECTORSANDTHESPACEGEOMETRY Letuscalculatethelengthofthecrossproduct.Bythede“nition (11.8), a b 2=( a b ) ( a b ) =( a2b3Š a3b2)2+( a3b1Š a1b3)2+( a1b2Š a2b1)2=( a2 1+ a2 2+ a2 3)( b2 1+ b2 2+ b2 3) Š ( a1b1+ a2b2+ a3b3)2= a 2 b 2Š ( a b )2, wherethethirdequalityisobtainedbycomputingthesquaresofthe componentsofthecrossproductandregroupingtermsintheobtained expression.Thelastequalityusesthede“nitionsofthenormandthe dotproduct.Next,recallthegeometricalpropertyofthedotproduct (11.6).If istheanglebetweenthevectors a and b ,then a b 2= a 2 b 2Š a 2 b 2cos2 = a 2 b 2(1 Š cos2 )= a 2 b 2sin2. Since0 ,sin 0andthesquarerootofbothsidesofthis equationcanbetakenwiththeresultthat a b = a b sin Thisrelationshowsthatthelengthofthecrossproductde“nedby (11.8)doesnotdependonthechoiceofthecoordinatesystemasitis expressedviathegeometricalinvariants,thelengthsof a and b and theanglebetweenthem.Nowconsidertheparallelogramwithadjacent sides a and b .If a isthelengthofitsbase,then h = b sin isits height.Therefore,thenormofthecrossproduct, a b = a h = A istheareaoftheparallelogram. Owingtothemutualorientationofthevectors a b ,and a b = 0 establishedin(11.11)aswellasthattheirlengthsarepreserved underrotationsofthecoordinatesystem,thecoordinatesystemcan beorientedsothat a isalongthe x axis, b isinthe xy plane,while a b isparalleltothe z axis.Inthiscoordinatesystem, a = a 0 0 and b= b1,b2, 0 ,where b1= b cos and b2= b sin if b liesin eitherthe“rstorsecondquadrantofthe xy planeand b2= Š b sin if b liesineitherthethirdorfourthquadrant.Intheformercase, a b = 0 0 ,A ,where A istheareaoftheparallelogram.Inthe lattercase,thede“nition(11.8)yields a b = 0 0 Š A .Itturnsout thatthedirectionofthecrossproductinbothcasescanbedescribed byasimpleruleknownasthe right-handrule : Ifthe“ngersoftheright handcurlinthedirectionofarotationfrom a toward b throughthe smallestanglebetweenthem,thenthethumbpointsinthedirectionof a b .

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74.THECROSSPRODUCT43 Figure11.11.Left :Geometricalinterpretationofthe crossproductoftwovectors.Thecrossproductisavectorthatisperpendiculartobothvectorsintheproduct.Its lengthequalstheareaoftheparallelogramwhoseadjacent sidesarethevectorsintheproduct.Ifthe“ngersoftheright handcurlinthedirectionofarotationfromthe“rstvectorto thesecondvectorthroughthesmallestanglebetweenthem, thenthethumbpointsinthedirectionofthecrossproduct ofthevectors. Right :IllustrationtoStudyProblem11.15.Inparticular,byDe“nition11.12, e1 e2= 1 0 0 0 1 0 = 0 0 1 = e3.If a isorthogonalto b ,thentherelativeorientationof thetripleofvectors a b ,and a b isthesameasthatofthestandard basisvectors e1, e2,and e3. Thestatedgeometricalpropertiesaredepictedintheleftpanelof Figure11.11andsummarizedinthefollowingtheorem. Theorem 11.4 (GeometricalSigni“canceoftheCrossProduct) Thecrossproduct a b ofvectors a and b isthevectorthatisperpendiculartobothvectors, a b a and a b b ,hasamagnitude equaltotheareaoftheparallelogramwithadjacentsides a and b ,and isdirectedaccordingtotheright-handrule. Twousefulconsequencescanbededucedfromthistheorem. Corollary 11.2 Twononzerovectorsareparallelifandonlyif theircrossproductvanishes: a b = 0 a b If a b = 0 ,thentheareaofthecorrespondingparallelogram vanishes, a b =0,whichisonlypossibleiftheadjacentsidesofthe parallelogramareparallel.Conversely,fortwoparallelvectors,thereis anumber s suchthat a = s b .Hence, a b =( s b ) b = s ( b b )= 0 .

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4411.VECTORSANDTHESPACEGEOMETRY Ifinthecrossproduct a b thevector b ischangedbyaddingto itanyvectorparallelto a ,thecrossproductdoesnotchange: a ( b + s a )= a b + s ( a a )= a b Let b = b+ bbetheorthogonaldecompositionof b relativetoa nonzerovector a .ByCorollary11.1, a b= 0 because bisparallel to a .Itisthenconcludedthat thecrossproductdependsonlyonthe component bof b thatisorthogonalto a .Thus, a b = a band a b = a b .AreaofaTriangle.Oneofthemostimportantapplicationsofthecross productisincalculationsoftheareasofplanar“guresinspace. Corollary 11.3 (AreaofaTriangle) Letvectors a and b betwosidesofatrianglethathavethesameinitial pointatavertexofatriangle.Thentheareaofthetriangleis Area = 1 2 a b Indeed,bythegeometricalconstruction,theareaofthetriangleis halfoftheareaofaparallelogramwithadjacentsides a and b Example 11.13 Let A =(1 1 1) B =(2 Š 1 3) ,and C = ( Š 1 3 1) .Findtheareaofthetriangle ABC andavectororthogonaltotheplanethatcontainsthetriangle. Solution: Taketwovectorswiththeinitialpointatanyofthevertices ofthetrianglethatformtheadjacentsidesofthetriangleatthatvertex. Forexample, a = Š AB = 1 Š 2 2 and b = Š AC = Š 2 2 0 .Then a b = Š 4 Š 4 Š 6 .Since Š 4 Š 4 Š 6 =2 2 2 3 =2 17,the areaofthetriangle ABC is 17byCorollary11.3.Theunitshereare squaredunitsoflengthusedtomeasurethecoordinatesofthetriangle vertices(e.g.,m2ifthecoordinatesaremeasuredinmeters).Any vectorintheplanethatcontainsthetriangleisalinearcombinationof a and b .Therefore,thevector a b isorthogonaltoanysuchvector andhencetotheplanebecause a b isorthogonaltoboth a and b Thechoice a = Š Š CB and b = Š CA or a = Š BA and b = Š Š BC wouldgive thesameanswer(modulothesignchangeinthecrossproduct). ApplicationsinPhysics:Torque.Torque, or momentofforce ,isthe tendencyofaforcetorotateanobjectaboutanaxisorapivot.Just asaforceisapushorapull,atorquecanbethoughtofasatwist.

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74.THECROSSPRODUCT45 If r isthevectorfromapivotpointtothepointwhereaforce F is applied,thenthetorqueisde“nedasthecrossproduct = r F Thetorquedependsonlyonthecomponent Foftheforcethatis orthogonalto r ,thatis, = r F.If istheanglebetween r and F ,thenthemagnitudeofthetorqueis = r F = rF sin ;here r = r isthedistancefromthepivotpointtothepointwheretheforce ofmagnitude F isapplied.Onecanthinkof r asaleverattachedtoa pivotpointandtheforce F isappliedtotheotherendoftheleverto rotateitaboutthepivotpoint.Naturally,theleverwouldnotrotate iftheforceisparalleltoit( =0or = ),whereasthemaximal rotationaleectiscreatedwhentheforceisappliedinthedirection perpendiculartothelever( = / 2).Thedirectionof determines theaxisaboutwhichtheleverrotates.Bythepropertyofthecross product,thisaxisisperpendiculartotheplanecontainingtheforceand positionvectors.Accordingtotheright-handrule,therotationoccurs counterclockwisewhenviewedfromthetopofthetorquevector.When drivingacar,atorqueisappliedtothesteeringwheeltochangethe directionofthecar.Whenaboltistightenedbyapplyingaforcetoa wrench,theproducedturningeectisthetorque. Anextendedobjectissaidtobe rigid ifthedistancebetweenany twoofitspointsremainsconstantintimeregardlessoftheexternal forcesexertedonit.Let P bea“xed(pivot)pointaboutwhicharigid objectcanrotate.Supposethattheforces Fi, i =1 2 ,...,n ,areapplied totheobjectatthepointswhosepositionvectorsrelativetothepoint P are ri.The principleofmoments statesthatarigidobjectdoesnot rotateaboutthepoint P ifitwasinitiallyatrestandthetotaltorque vanishes: = 1+ 2+ + n= r1 F1+ r2 F2+ + rn Fn= 0 If,inaddition,thetotalforcevanishes, F = F1+ F2+ + Fn= 0 thenarigidobjectremainsatrestandwillnotrotateaboutanyother pivotpoint.Indeed,supposethatthetorqueabout P vanishesandlet r0beapositionvectorof P relativetoanotherpoint P0.Thenthe positionvectorsofthepointsatwhichtheforcesareappliedrelative tothenewpivotpoint P0are ri+ r0.Thetotaltorque,orthetotal momentoftheforces,about P0alsovanishes: 0=( r1+ r0) F1+ +( rn+ r0) Fn= r1 F1+ + rn Fn+ r0 ( F1+ + Fn) = + r0 F = 0

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4611.VECTORSANDTHESPACEGEOMETRY because,bythehypothesis, = 0 and F = 0 Theconditions = 0 and F = 0 comprisethefundamentallawofstaticsforrigidobjects Example 11.14 Theendsofrigidrodsoflength L1and L2are rigidlyjoinedattheangle / 2 .Aballofmass m1isattachedtothe freeendoftherodoflength L1andaballofmass m2isattachedto thefreeendoftherodoflength L2.Thesystemishungbythejoining pointsothatthesystemcanrotatefreelyaboutitunderthegravitational force.Findtheequilibriumpositionofthesystemifthemassesofthe rodscanbeneglectedascomparedtothemassesoftheballs. Solution: Thegravitationalforceshavemagnitudes F1= m1g and F2= m2g forthe“rstandsecondballs,respectively( g isthefree fallacceleration).Theyaredirecteddownwardandthereforeliein theplanethatcontainsthepositionvectorsoftheballsrelativetothe pivotpoint.Sothetorquesofthegravitationalforcesareorthogonal tothisplane,andtheequilibriumcondition 1+ 2= 0 isequivalent to 1Š 2=0,where 1 2arethemagnitudesofthetorques.Theminus signfollowsfromtheright-handrulebywhichthevectors 1and 2are parallelbuthaveoppositedirections.Inotherwords,thegravitational forcesappliedtotheballsgenerateoppositerotationalmoments.When thelatterareequalinmagnitude,thesystemisatrest.Intheplane thatcontainsthesystem,let 1and 2bethesmallestanglesbetween therodsandahorizontalline.Sincetherodsareperpendicular, 1+ 2= / 2.Theanglebetweenthepositionvectorofthe“rstballand thegravitationalforceactingonitis 1= / 2 Š 1,andsimilarly 2= / 2 Š 2istheanglebetweenthepositionvectorofthesecond ballandthegravitationalforceactingonit.Therefore, 1= L1F1sin 1and 2= L2F2sin 2.Owingtotheidentitysin( / 2 Š )=cos ,it followsthat 1= 2 m1L1cos 1= m2L2cos 2 tan 1= m1L1 m2L2, wheretherelation 2= / 2 Š 1hasbeenused. 74.4.StudyProblems.Problem11.14. Findthemostgeneralvector r thatsatis“estheequations a r =0 and b r =0 ,where a and b arenonzero,nonparallel vectors. Solution: Theconditionsimposedon r holdifandonlyifthevector r isorthogonaltobothvectors a and b .Therefore,itmustbeparallel totheircrossproduct.Thus, r = t ( a b )foranyreal t

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74.THECROSSPRODUCT47 Problem11.15. Usegeometricalmeansto“ndthecrossproductsof theunitvectorsparalleltothecoordinateaxes. Solution: Consider e1 e2.Since e1 e2and e1 = e2 =1,their crossproductmustbeaunitvectorperpendiculartoboth e1and e2. Thereareonlytwosuchvectors, e3.Bytheright-handrule, e1 e2= e3. Similarly,theothercrossproductsareshowntobeobtainedbycyclic permutationsoftheindices1,2,and3intheaboverelation.Apermutationofanytwoindicesleadstoachangeinsign(e.g., e2 e1= Š e3). Sinceacyclicpermutationofthreeindices { ijk }{ kij } (andsoon) consistsoftwopermutationsofanytwoindices,therelationbetween theunitvectorscanbecastintheform ei= ej ek, { ijk } = { 123 } andcyclicpermutations Problem11.16. Provethebac Š cabrule(11.9). Solution: If c and b areparallel, b = s c forsomereal s ,thenthe relationistruebecausebothitssidesvanish.If c and b arenotparallel, then,bytheremarkafterCorollary11.2,thedoublecrossproduct a ( b c )dependsonlyonthecomponentof a thatisorthogonalto b c .Thiscomponentliesintheplanecontaining b and c andhence isalinearcombinationofthem(seeStudyProblem11.6).So,without lossofgenerality, a = t b + p c .Also, b c = b c,where c= c Š s b s = c b / b 2,isthecomponentof c orthogonalto b (note b c=0). Thevectors b c,and b caremutuallyorthogonalandoriented accordingtotheright-handrule.Inparticular, b c = b c Byapplyingtheright-handruletwice,itisconcludedthat b ( b c) hasthedirectionoppositeto c.Since b and b careorthogonal, b ( b c) = b b c = b 2 c .Therefore, b ( b c )= Š c b 2= b ( b c ) Š c ( b b ) Byswappingthevector b and c inthisequation,onealsoobtains c ( b c )= Š c ( c b )= b c 2= Š c ( c b )+ b ( c c ) Itfollowsfromtheserelationsthat a ( b c )= t b ( b c )+ p c ( b c ) = b [( t b + p c ) c ] Š c [( t b + p c ) b ] = b ( a c ) Š c ( a b )

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4811.VECTORSANDTHESPACEGEOMETRY Problem11.17. ProvetheJacobiidentity(11.10). Solution: Bythe bac Š cab rule(11.9)appliedtoeachterm, a ( b c )= b ( a c ) Š c ( a b ) b ( c a )= c ( b a ) Š a ( b c ) c ( a b )= a ( c b ) Š b ( a c ) Byaddingtheseequalities,itiseasytoseethatthecoecientsateach ofthevectors a b ,and c ontherightsideareaddeduptomake0. Problem11.18. Considerallvectorsinaplane.Anysuchvector a canbeuniquelydeterminedbyspecifyingitslength a = a and theangle athatiscountedcounterclockwisefromthepositive x axis towardthevector a (i.e., 0 a< 2 ).Therelation a1,a2 = a cos a,a sin a establishesaone-to-onecorrespondencebetweenorderedpairs ( a1,a2) and ( a,a) .De“nethevectorproductoftwovectors a and b asthevector c forwhich c = ab and c= a+ b.Showthat, incontrasttothecrossproduct,thisproductisassociativeandcommutative,thatis,that c doesnotdependontheorderofvectorsinthe product. Solution: Letusdenotethevectorproductbyasmallcircleto distinguishitfromthedotandcrossproducts, a b = c .Since c = ab cos( a+ b) ,ab sin( a+ b) ,thecommutativityofthevector product a b = b a followsfromthecommutativityoftheproduct andadditionofnumbers: ab = ba and a+ b= b+ a.Similarly, theassociativityofthevectorproduct( a b ) c = a ( b c )follows fromtheassociativityoftheproductandadditionofordinarynumbers: ( ab ) c = a ( bc )and( a+ b)+ c= a+( b+ c). Remark. Thevectorproductintroducedforvectorsinaplaneis knownasthe productofcomplexnumbers ,whichcanbeviewedastwodimensionalvectors.Itisinterestingtonotethatnocommutativeand associativevectorproduct(i.e.,vectortimesvector=vectorŽ)canbe de“nedinaEuclideanspaceofmorethantwodimensions. Problem11.19. Let u beavectorrotatinginthe xy planeaboutthe z axis.Givenavector v ,“ndthepositionof u suchthatthemagnitude ofthecrossproduct v u ismaximal. Solution: Foranytwovectors, v u = v u sin ,where istheanglebetween v and u .Themagnitudeof v is“xed,while themagnitudeof u doesnotchangewhenrotating.Therefore,the absolutemaximumofthecross-productmagnitudeisreachedwhen sin =1orcos =0(i.e.,whenthevectorsareorthogonal).The

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74.THECROSSPRODUCT49 correspondingalgebraicconditionis v u =0.Since u isrotatinginthe xy plane,itscomponentsare u = u cos u sin 0 ,where0 < 2 istheanglecountedcounterclockwisefromthe x axistoward thecurrentpositionof u .Put v = v1,v2,v3 .Thenthedirectionof u isdeterminedbytheequation v u = u ( v1cos + v2sin )=0, andhencetan = Š v1/v2.Thisequationhastwosolutionsinthe range0 < 2 : = Š tanŠ 1( v1/v2)and = Š tanŠ 1( v1/v2)+ Geometrically,thesesolutionscorrespondtothecasewhen u isparallel totheline v2y + v1x =0inthe xy plane. 74.5.Exercises.(1) Findthecrossproduct a b if (i) a = 1 2 3 and b = Š 1 0 1 (ii) a = 1 Š 1 2 and b = 3 Š 2 1 (iii) a = e1+3 e2Š e3and b =3 e1Š 2 e2+ e3(iv) a =2 c Š d and b =3 c +4 d ,where c d = 1 2 3 (2) Let a = 3 2 1 b = Š 2 1 Š 1 ,and c = 1 0 Š 1 .Find a ( b c ), b ( c a ),and c ( a b ). (3) Let a beaunitvectororthogonalto b and c .If c = 1 2 2 ,“nd thelengthofthevector a [(a + b ) ( a + b + c )]. (4) Giventwononparallelvectors a and b ,showthatthevectors a a b ,and a ( a b )aremutuallyorthogonal. (5) Suppose a liesinthe xy plane,itsinitialpointisattheorigin,and itsterminalpointisin“rstquadrantofthe xy plane.Let b beparallel to e3.Usetheright-handruletodeterminewhethertheanglebetween a b andtheunitvectorsparalleltothecoordinateaxesliesinthe interval(0 ,/ 2)or( / 2 )orequals / 2. (6) Ifvectors a b ,and c havetheinitialpointattheoriginandlie, respectively,inthepositivequadrantsofthe xy yz ,and xz planes, “ndtheoctantsinwhichthepairwisecrossproductsofthesevectors lie. (7) Findtheareaofatriangle ABC for A (1 0 1), B (1 2 3),and C (0 1 1)andanonzerovectororthogonaltotheplanecontainingthe triangle. (8) Usethecrossproducttoshowthattheareaofthetrianglewhose verticesaremidpointsofthesidesofatrianglewitharea A is A/ 4. (9) Consideratrianglewhoseverticesaremidpointsofanythreesides ofaparallelogram.Iftheareaoftheparallelogramis A ,“ndthearea ofthetriangle. (10) Let A =(1 2 1)and B =( Š 1 0 2)beverticesofaparallelogram.Iftheothertwoverticesareobtainedbymoving A and B by

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5011.VECTORSANDTHESPACEGEOMETRY 3unitsoflengthalongthevector a = 2 1 Š 2 ,“ndtheareaofthe parallelogram. (11) Considerfourpointsinspace.Supposethatthecoordinatesofthe pointsareknown.Describeaprocedurebasedonthepropertiesofthe crossproducttodeterminewhetherthepointsareinoneplane.Inparticular,arethepoints(1 2 3),( Š 1 0 1),(1 3 Š 1),and(0 1 2)inone plane? (12) Letthesidesofatrianglehavelengths a b ,and c andletthe anglesattheverticesoppositetothesides a b ,and c be,respectively, ,and .Provethat sin a = sin b = sin c Hint: De“nethesidesasvectorsandexpresstheareaofthetriangle viathevectorsateachvertexofthetriangle. (13) Considerapolygonwithfourvertices A B C ,and D .Ifthe coordinatesoftheverticesarespeci“ed,describetheprocedurebased onvectoralgebratocalculatetheareaofthepolygon.Inparticular, put A =(0 0), B =( x1,y1), C =( x2,y2),and D =( x3,y3)andexpress theareavia xiand yi, i =1 2 3. (14) Consideraparallelogram.Constructanotherparallelogramwhose adjacentsidesarediagonalsofthe“rstparallelogram.Findtherelation betweentheareasoftheparallelograms. (15) Giventwononparallelvectors a and b ,showthatanyvector r in spacecanbewrittenasalinearcombination r = x a + y b + z a b and thatthenumbers x y ,and z areuniqueforevery r .Express z via r a ,and b .Inparticular,put a = 1 1 1 and b = 1 1 0 .Findthe coecients x y ,and z for r = 1 2 3 Hint: SeeStudyProblems11.6 and11.14. (16) Atetrahedronisasolidwithfourverticesandfourtriangular faces.Let v1, v2, v3,and v4bevectorswithlengthsequaltotheareas ofthefacesanddirectionsperpendiculartothefacesandpointing outward.Showthat v1+ v2+ v3+ v4= 0 (17) If a b = a c and a b = a c ,doesitfollowthat b = c ? (18) Giventwononparallelvectors a and b ,constructthreemutually orthogonalunitvectors ui, i =1 2 3,oneofwhichisparallelto a Aresuchunitvectorsunique?Inparticular,put a = 1 2 2 and b = 1 0 2 and“nd ui. (19) Let ui, i =1 2 3,beanorthonormalbasisinspacewiththe propertythat u3= u1 u2.If a1, a2,and a3arethecomponentsof vector a relativetothisbasisand b1, b2,and b3arethecomponentsof b ,showthatthecomponentsofthecrossproduct a b canalsobe

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75.THETRIPLEPRODUCT51 computedbythedeterminantrulegiveninDe“nition11.12where eiarereplacedby ui. Hint: Usethe bac Š cab ruleto“ndallpairwise crossproductsofthebasisvectors ui. (20) LettheanglebetweentherigidrodsinExample11.14be0 < < .Findtheequilibriumpositionofthesystem. (21) Tworigidrodsofthesamelengtharerigidlyattachedtoaball ofmass m sothattheanglebetweentherodsis / 2.Aballofmass 2 m isattachedtooneofthefreeendsofthesystem.Theremaining freeendisusedtohangthesystem.Findtheanglebetweentherod connectingthepivotpointandtheballofmass m andtheverticalaxis alongwhichthegravitationalforceisacting.Assumethatthemasses oftherodscanbeneglectedascomparedto m (22) Threerigidrodsofthesamelengtharerigidlyjoinedbyoneend sothattherodslieinaplaneandtheotherendofeachrodisfree. Letthreeballsofmasses m1, m2,and m3beattachedtothefreeends oftherods.Thesystemishungbythejoiningpointandcanrotate freelyaboutit.Assumethatthemassesoftherodscanbeneglected ascomparedtothemassesoftheballs.Findtheanglesbetweenthe rodsatwhichtheballsremaininahorizontalplaneundergravitational forcesactingvertically.Dosuchanglesexistforanymassesoftheballs? 75.TheTripleProduct Definition 11.13 The tripleproduct ofthreevectors a b ,and c isanumberobtainedbytherule: a ( b c ) Itfollowsfromthealgebraicde“nitionofthecrossproductandthe de“nitionofthedeterminantofa3 3matrixthat a ( b c )= a1det b2b3c2c3 Š a2det b1b3c1c3 + a3det b1b2c1c2 =det a1a2a3b1b2b3c1c2c3 Thisprovidesaconvenientwaytocalculatethenumericalvalueof thetripleproduct.Iftworowsofamatrixareswapped,thenits determinantchangessign.Therefore, a ( b c )= Š b ( a c )= Š c ( b a ) Thismeans,inparticular,thattheabsolutevalueofthetripleproduct isindependentoftheorderofthevectorsinthetripleproduct.Also, thevalueofthetripleproductisinvariantundercyclicpermutations ofvectorsinit: a ( b c )= b ( c a )= c ( a b ).

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5211.VECTORSANDTHESPACEGEOMETRY Figure11.12.Left :Geometricalinterpretationofthe tripleproductasthevolumeoftheparallelepipedwhoseadjacentsidesarethevectorsintheproduct: h = a cos A = b c V = hA = a b c cos = a ( b c ). Right :Testforthecoplanarityofthreevectors.Threevectorsarecoplanarifandonlyiftheirtripleproductvanishes: a ( b c )=0.75.1.GeometricalSignicanceoftheTripleProduct.Supposethat b and c arenotparallel(otherwise, b c = 0 ).Let betheangle between a and b c asshowninFigure11.12(leftpanel).If a b c (i.e., = / 2),thenthetripleproductvanishes.Let = / 2.Consider parallelogramswhoseadjacentsidesarepairsofthevectors a b ,and c .Theyencloseanonrectangularboxwhoseedgesarethevectors a b ,and c .Aboxwithparallelogramfacesiscalleda parallelepiped with adjacentsides a b ,and c .Thecrossproduct b c isorthogonalto thefacecontainingthevectors b and c ,whereas A = b c isthe areaofthisfaceoftheparallelepiped(theareaoftheparallelogram withadjacentsides b and c ).Bythegeometricalpropertyofthedot product, a ( b c )= A a cos .Ontheotherhand,thedistance betweenthetwofacesparalleltoboth b and c (ortheheightofthe parallelepiped)is h = a cos if / 2,or h = a | cos | .Thevolumeoftheparallelepipedis V = Ah Thisleadstothefollowingtheorem. Theorem 11.5 (GeometricalSigni“canceoftheTripleProduct) Thevolume V ofaparallelepipedwhoseadjacentsidesarethevectors a b ,and c istheabsolutevalueoftheirtripleproduct: V = | a ( b c ) | Thus,thetripleproductisaconvenientalgebraictoolforcalculatingvolumes.Notealsothatthevectorscanbetakeninanyorderin

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75.THETRIPLEPRODUCT53 thetripleproducttocomputethevolumebecausethetripleproduct onlychangesitssignwhentwovectorsareswappedinit. Example 11.15 Findthevolumeofaparallelepipedwithadjacent sides a = 1 2 3 b = Š 2 0 1 ,and c = 2 1 2 Solution: Theexpansionofthedeterminantoverthe“rstrowyields b ( a c )=det Š 201 123 212 = Š 2(4 Š 3)+1(1 Š 4)= Š 5 Takingtheabsolutevalueofthetripleproduct,thevolumeisobtained, V = |Š 5 | =5.Thecomponentsofthevectorsmustbegiveninthe sameunitsoflength(e.g.,meters).Thenthevolumeis5cubicmeters. Anyvectorinaplaneisalinearcombinationoftwoparticular vectorsintheplane.Vectorsthatlieinaplanearecalled coplanar (see Section73.4).Clearly,anytwovectorsarealwayscoplanar.However, threenonzerovectorsdonotgenerallylieinoneplane.Noneofthree non-coplanarvectorscanbeexpressedasalinearcombinationofthe othertwovectors.Suchvectorsaresaidtobe linearlyindependent .As notedinSection73.4,anythreelinearlyindependentvectorsforma basisinspace.Simplecriteriaforthreevectorstobeeithercoplanar orlinearlyindependentcanbededucedfromTheorem11.5. Corollary 11.4 (CriterionforThreeVectorstoBeCoplanar) Threevectorsarecoplanarifandonlyiftheirtripleproductvanishes: a b c arecoplanar a (b c )=0 Consequently,threenonzerovectorsarelinearlyindependentifandonly iftheirtripleproductdoesnotvanish. Indeed,ifthevectorsarecoplanar(Figure11.12,rightpanel),then thecrossproductofanytwovectorsmustbeperpendiculartothe planewherethevectorsareandthereforethetripleproductvanishes. If,conversely,thetripleproductvanishes,theneither b c = 0 or a b c .Intheformercase, b isparallelto c ,or c = t b ,andhence a alwaysliesinaplanewith b and c .Inthelattercase,allthreevectors a b ,and c areperpendicularto b c andthereforemustbeinone plane(orthogonalto b c ). InSection74.3,itwasstatedthatthreelinearlyindependentvectors formabasisinspace.Thelinearindependencemeansthatnoneof thevectorsisalinearcombinationoftheothertwo.Geometrically,

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5411.VECTORSANDTHESPACEGEOMETRY thismeansthatthevectorsarenotcoplanar.Therefore,thefollowing simplecriterionholdsforthreevectorstoformabasisinspace. Corollary 11.5 (BasisinSpace). Threevectors u1, u2,and u3arelinearlyindependentandhenceform abasisinspaceifandonlyiftheirtripleproductdoesnotvanish. Example 11.16 Determinewhetherthepoints A (1 1 1) B (2 0 2) C (3 1 Š 1) ,and D (0 2 3) areinthesameplane. Solution: Considerthevectors a = Š AB = 1 Š 1 1 b = Š AC = 2 0 2 ,and c = Š Š AD = Š 1 1 2 .Thepointsinquestionareinthe sameplaneifandonlyifthevectors a b ,and c arecoplanar,or a ( b c )=0.Theevaluationofthetripleproductyields a ( b c )=det 1 Š 11 202 Š 112 =1(0 Š 2)+1(4+2)+1(2 Š 0)=6 =0 Therefore,thepointsarenotinthesameplane. Example 11.17 Let u1= 1 2 3 u2= 2 1 Š 6 ,and u3= 1 1 Š 1 .Canthevector a = 1 1 1 berepresentedasalinearcombinationofthevectors u1, u2,and u3? Solution: Anyvectorinspaceisalinearcombinationof u1, u2,and u3iftheyformabasis.Letusverify“rstwhetherornottheyforma basis.ByCorollary11.5, u1 ( u2 u3)=det 123 21 Š 6 11 Š 1 =1( Š 1+6) Š 2( Š 2+6)+3(2 Š 1)=0 Therefore,thesevectorsdonotformabasisandarecoplanar.Note that u3=1 3( u1+ u2).Ifthevector a liesinthesameplaneasthevectors u1, u2,and u3,thenitisalinearcombinationofanytwononparallel vectors,say, u1and u3.Sincethefollowingtripleproductdoesnot vanish, a ( u1 u3)=det 111 123 11 Š 1 =1( Š 2 Š 3) Š 1( Š 1 Š 3)+1(1 Š 2)= Š 2 thevector a isnotintheplaneinwhichthevectors u1, u2,and u3lie, andtherefore a isnotalinearcombinationofthem.

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75.THETRIPLEPRODUCT55 75.2.Right-andLeft-HandedCoordinateSystems .Considerarectangular boxwhosesidesareparalleltothreegivenunitvectors ei, i =1 2 3. Anyvectorinspacecanbeviewedasthediagonalofonesuchbox andthereforeisuniquelyexpandedintothesum r = x e1+ y e2+ z e3, wheretheorderedtripleofnumbers( x,y,z )isdeterminedbyscalar projectionsof r onto ei.Thus, withanytripleofmutuallyorthogonal unitvectorsonecanassociatearectangularcoordinatesystem .The vector e1 e2mustbeparallelto e3becausethelatterisorthogonal toboth e1and e2.Furthermore,owingtotheorthogonalityof e1,and e2, e1 e2 = e1 e2 =1andhence e1 e2= e3.Consequently, e3 ( e1 e2)= e1 ( e2 e3)= 1or,owingtothemutualorthogonalityof thevectors, e2 e3= e1.Acoordinatesystemiscalled right-handed if e1 ( e2 e3)=1,andacoordinatesystemforwhich e1 ( e2 e3)= Š 1 iscalled left-handed .Aright-handedsystemcanbevisualizedasfollows.Withthethumb,index,andmiddle“ngersofthe righthand at rightanglestoeachother(withtheindex“ngerpointedstraight),the middle“ngerpointsinthedirectionof e1= e2 e3whenthethumb represents e2andtheindex“ngerrepresents e3.Aleft-handedsystem isobtainedbythere”ection e1Š e1andthereforeisvisualizedby the“ngersofthe lefthand inthesameway.Sincethedotproductcannotbechangedbyrotationsandtranslationsinspace,thehandedness ofthecoordinatesystemdoesnotchangeundersimultaneousrotations andtranslationsofthetripleofvectors ei(threemutuallyorthogonal “ngersofthelefthandcannotbemadepointinginthesamedirection asthecorresponding“ngersoftherighthandbyanyrotationofthe hand).There”ectionofallthreevectors eiŠ eiorjustoneofthem turnsaright-handedsystemintoaleft-handedoneandviceversa.A mirrorre”ectionofaright-handedsystemistheleft-handedone.The coordinatesystemassociatedwiththestandardbasis e1= 1 0 0 e2= 0 1 0 ,and e3= 0 0 1 is right-handed because e1 ( e2 e3)=1.75.3.DistancesBetweenLinesandPlanes.Ifthelinesorplanesin spacearenotintersecting,thenhowcanone“ndthedistancebetween them?Thisquestioncanbeansweredusingthegeometricalproperties ofthetripleandcrossproducts(Theorems11.4and11.5).Let S1and S2betwosetsofpointsinspace.Letapoint A1belongto S1,leta point A2belongto S2,andlet | A1A2| bethedistancebetweenthem. Definition 11.14 (DistanceBetweenSetsinSpace) Thedistance D betweentwosetsofpointsinspace, S1and S2,isthe largestnumberthatislessthanorequaltoallthenumbers | A1A2| when thepoint A1rangesover S1andthepoint A2rangesover S2.

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5611.VECTORSANDTHESPACEGEOMETRY Naturally,ifthesetshaveatleastonecommonpoint,thedistance betweenthemvanishes.Thedistancebetweensetsmayvanisheven ifthesetshavenocommonpoints.Forexample,let S1beanopen interval(0 1)on,say,the x axis,while S2istheinterval(1 2)onthe sameaxis.Apparently,thesetshavenocommonpoints(thepoint x =1doesnotbelongtoeitherofthem).Thedistanceisthelargest number D suchthat D | x1Š x2| ,where0 0canbemadesmallerthananypreassigned positivenumberbytaking x1and x2closeenoughto1.Sincethe distance D 0,theonlypossiblevalueis D =0.Intuitively,thesets areseparatedbyasinglepointthatisnotanextendedŽobject,and hencethedistancebetweenthemshouldvanish.Inotherwords,there aresituationsinwhichtheminimumof | A1A2| isnotattainedforsome A1S1,orsome A2S2,orboth.Nevertheless,thedistancebetween thesetsisstillwellde“nedasthelargestnumberthatislessthanor equaltoallnumbers | A1A2| .Suchanumberiscalledthe in“mum of thesetofnumbers | A1A2| anddenotedinf | A1A2| .Thus, D =inf | A1A2| ,A1S1,A2S2. Thenotation A1S1standsforapoint A1belongstotheset S1,Ž orsimply A1isanelementof S1.ŽThede“nitionisillustratedin Figure11.13(leftpanel). Figure11.13.Left :Distancebetweentwopointsets S1and S2de“nedasthelargestnumberthatislessthanor equaltoalldistances | A1A2| ,where A1rangesoverallpoints in S1and A2rangesoverallpointsin S2. Right :Distance betweentwoparallelplanes(Corollary11.6).Consideraparallelepipedwhoseoppositefaceslieintheplanes P1and P2. Thenthedistance D betweentheplanesistheheightofthe parallelepiped,whichcanbecomputedastheratio D = V/A where V = | a ( b c ) | isthevolumeoftheparallelepiped and A = b c istheareaoftheface.

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75.THETRIPLEPRODUCT57 Corollary 11.6 (DistanceBetweenParallelPlanes) Thedistancebetweenparallelplanes P1and P2isgivenby D = | Š AP ( Š AB Š AC ) | Š AB Š AC where A B ,and C areanythreepointsintheplane P1thatarenot onthesameline,and P isanypointintheplane P2. Proof. Sincethepoints A B ,and C arenotonthesameline,the vectors b = Š AB and c = Š AC arenotparallel,andtheircrossproduct isavectorperpendiculartotheplanes(seeFigure11.13,rightpanel). Considertheparallelepipedwithadjacentsides a = Š AP b ,and c .Two ofitsfaces,theparallelogramswithadjacentsides b and c ,lieinthe parallelplanes,onein P1andtheotherin P2.Thedistancebetween theplanesis,byconstruction,theheightoftheparallelepiped,whichis equalto V/Ap,where Apistheareaofthefaceparallelto b and c and V isthevolumeoftheparallelepiped.Theconclusionfollowsfromthe geometricalpropertiesofthetripleandcrossproducts: V = | a ( b c ) | and Ap= b c Similarly,thedistancebetweentwoparallellines L1and L2canbe determined.Twolinesareparalleliftheyarenotintersectingandliein thesameplane.Let A and B beanytwopointsontheline L1andlet C beanypointontheline L2.Considertheparallelogramwithadjacent sides a = Š AB and b = Š AC asdepictedinFigure11.14(leftpanel).The distancebetweenthelinesistheheightofthisparallelogram,whichis D = Ap/ a ,where Ap= a b istheareaoftheparallelogramand a isthelengthofitsbase. Corollary 11.7 (DistanceBetweenParallelLines) Thedistancebetweentwoparallellines L1and L2is D = Š AB Š AC Š AB where A and B areanytwodistinctpointsontheline L1and C isany pointontheline L2. Definition 11.15 (SkewLines) Twolinesthatarenotintersectingandnotparallelarecalled skew lines Todeterminethedistancebetweenskewlines L1and L2,consider anytwopoints A and B on L1andanytwopoints C and P on L2.

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5811.VECTORSANDTHESPACEGEOMETRY Figure11.14.Left :Distancebetweentwoparallellines. Consideraparallelogramwhosetwoparallelsideslieinthe lines.Thenthedistancebetweenthelinesistheheightofthe parallelogram(Corollary11.7). Right :Distancebetween skewlines.Consideraparallelepipedwhosetwononparallel edges AB and CP intheoppositefaceslieintheskewlines L1and L2,respectively.Thenthedistancebetweenthelines istheheightoftheparallelepiped,whichcanbecomputed astheratioofthevolumeandtheareaoftheface(Corollary11.8).De“nethevectors b = Š AB and c = Š CP thatareparalleltolines L1and L2,respectively.Sincethelinesarenotparallel,thecrossproduct b c doesnotvanish.Thelines L1and L2lieintheparallelplanes perpendicularto b c (bythegeometricalpropertiesofthecrossproduct, b c isperpendicularto b and c ).Thedistancebetweenthelines coincideswiththedistancebetweentheseparallelplanes.Considerthe parallelepipedwithadjacentsides a = Š AC b ,and c asshowninFigure 11.14(rightpanel).Thelineslieintheparallelplanesthatcontainthe facesoftheparallelepipedparalleltothevectors b and c .Thus,the distancebetweenskewlinesisthedistancebetweentheparallelplanes containingthem.ByCorollary11.6,thisdistanceis D = V/Ap,where V and Ap= b c are,respectively,thevolumeoftheparallelepiped andtheareaofitsbase. Corollary 11.8 (DistanceBetweenSkewLines) Thedistancebetweentwoskewlines L1and L2is D = | Š AC ( Š AB Š CP ) | Š AB Š CP where A and B areanytwodistinctpointson L1,while C and P are anytwodistinctpointson L2.

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75.THETRIPLEPRODUCT59 Asaconsequenceoftheobtaineddistanceformulas,thefollowing criterionformutualorientationoftwolinesinspaceholds. Corollary 11.9 Let L1bealinethrough A and B = A ,andlet L2bealinethrough C and P = C .Then (1) L1and L2areparallelif Š AB Š CP = 0 (2) L1and L2areskewif Š AC ( Š AB Š CP ) =0 (3) L1and L2intersectif Š AC ( Š AB Š CP )=0 and Š AB Š CP = 0 (4) L1and L2coincideif Š AC ( Š AB Š CP )=0 and Š AB Š CP = 0 Forparallellines L1and L2,thevectors Š AB and Š CP areparallel, andhencetheircrossproductvanishes.If Š AC ( Š AB Š CP ) =0,then thelinesarenotparallelandlieintheparallelfacesofaparallelepiped thathasnonzerovolume.Suchlinesmustbeskew.If Š AB Š CP = 0 thelinesarenotparallel.Theadditionalcondition Š AC ( Š AB Š CP )=0 impliesthatthedistancebetweenthemvanishes,andhencethelines mustintersectatapoint.Theconditions Š AC ( Š AB Š CP )=0and Š AB Š CP = 0 implythatthelinesareparallelandintersecting.So theymustcoincide. Example 11.18 Findthedistancebetweenthelinethroughthe points A =(1 1 2) and B =(1 2 3) andthelinethrough C =(1 0 Š 1) and P =( Š 1 1 2) Solution:Let a = Š AB = 0 1 1 and b = Š CP = Š 2 1 3 .Then a b = 3 Š 1 Š (0+2) 0+2 = 2 Š 2 2 = 0 .Sothelinesarenot parallelbyproperty(1)inCorollary11.9.Put c = Š AC = 0 Š 1 Š 3 Then c ( a b )= 0 Š 1 Š 2 2 Š 2 2 =0+2 Š 4= Š 2 =0 Byproperty(2)inCorollary11.9,thelinesareskew.Next, a b = 2 Š 2 2 = 2 1 Š 1 1 =2 1 Š 1 1 =2 3.ByCorollary11.8, thedistancebetweenthelinesis D = | c ( a b ) | a b = |Š 2 | 2 3 = 1 3 75.4.StudyProblems.Problem11.20.(RotationsinSpace). Let a = a1,a2,a3 and a= a 1,a 2,a 3 bepositionvectorsofapoint relativetwocoordinatesystemsrelatedtooneanotherbyarotation.As

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6011.VECTORSANDTHESPACEGEOMETRY notedinSection74.1,thecoordinates a iand aiarerelatedbyalinear transformationthatpreservesthelength, a i= vi 1a1+ vi 2a2+ vi 3a3,i =1 2 3 a = a andexcludesthere”ectionsofallcoordinateaxesorjustoneofthem. Soarotationisdescribedbya 3 3 matrix V withelements vij.The vectors vi= vi 1,vi 2,vi 3 and wi= v1 i,v2 i,v3 i i =1 2 3 ,aretherows andcolumnsof V ,respectively.Showthattherowsof V aremutually orthonormal,thecolumnsof V arealsomutuallyorthonormal,andthe determinantof V isunit: (11.12) vi vj= wi wj= 1if i = j 0if i = j anddet V =1 Inparticular,showthatthedirectandinversetransformationsofthe coordinatesunderarotationinspaceare: (11.13) a i= vi a and ai= wi a. Howmanyindependentparameterscanthematrix V haveforageneric rotationinspace? Hint: If uiand ei, i =1 2 3,areorthonormalunitvectors(bases) associatedwiththerotatedandoriginalcoordinatesystems,showthat therowsof V determinethecomponentsof uirelativetothebasis ei,whereasthecolumnsof V determinethecomponentsof eirelative tothebasis ui.Usethisobservationtoshowthat ui uj= vi vj, ei ej= wi wj,and (11.14) u1 ( u2 u3)=det V e1 ( e2 e3) Solution: Avector a canbeexpandedintothesumofmutually orthogonalvectorsineachofthecoordinatesystems a = a1 e1+ a2 e2+ a3 e3= a 1 u1+ a 2 u2+ a 3 u3. Notethat a = a bytheorthonormalityofthebases eiand ui.Let usmultiplybothsidesofthisequalityby ui.Put vij= ui ej.Then a i= ui a = ui e1a1+ ui e2a2+ ui e3a3= vi 1a1+ vi 2a2+ vi 3a3= vi a Thus,thecomponentsoftherotationmatrix V arethedotproducts vij= ui ej.Fora“xed i ,thenumbers vijarescalarprojectionsof uionto ej, j =1 2 3,andhencearecomponentsof uirelativetothe basis ej,thatis, ui= vi 1 e1+ vi 2 e2+ vi 3 e3.Itisthenconcludedthat

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75.THETRIPLEPRODUCT61 the i throwof V coincideswiththecomponentsof uirelativetothe basis ej.Similarly, aj= ej a = ej u1a 1+ ej u2a 2+ ej u3a 3= v1 ja 1+ v2 ja 2+ v3 ja 3= wj a. Fora“xed j ,thenumbers vijarescalarprojectionsof ejonto ui, i =1 2 3,andhencearecomponentsof ejrelativetothebasis ui,that is, ej= v1 j u1+ vj 2 u2+ vj 3 u3.Thus,the j thcolumnof V coincides withthecomponentsof ejrelativetothebasis ui.Thisproves(11.13). Makinguseoftheexpansionof uiinthe orthonormal basis ejandof theexpansionof eiinthebasis ui,oneobtains ui uj= vi 1vj 1+ vi 2vj 2+ vi 3vj 3= vi vj, ei ej= v1 iv1 j+ v2 iv2 j+ v3 iv3 j= wi wj. The“rstrelationin(11.12)followsfromtheorthonormalityofthebasis vectors uiandthebasisvectors ei.Next,considerthecrossproduct u2 u3=( v21 e1+ v22 e2+ v23 e3) ( v31 e1+ v32 e2+ v33 e3) =det V11( e2 e3) Š det V12( e3 e1)+det V13( e1 e2) V11= v22v23v23v33 ,V12= v21v23v31v33 ,V13= v22v22v31v32 wheretheskewsymmetryofthecrossproduct ei ej= Š ej eiand thede“nitionofthedeterminantofa2 2matrixhavebeenused; thematrices V1 i, i =1 2 3,areobtainedbyremovingfrom V therow andcolumnthatcontain v1 i.Usingthesymmetryofthetripleproduct undercyclicpermutationsofthevectors,onehas u1 ( u2 u3)= v11det V11Š v12det V12+ v13det V13 e1 ( e2 e3) Equation(11.14)followsfromthisrelationandthede“nitionofthe determinantofa3 3matrix.Nowrecallthatthehandednessof acoordinatesystemispreservedunderrotations, u1 ( u2 u3)= e1 ( e2 e3)= 1,andthereforedet V =1.Itisalsoworthnoting thatacombinationofrotationsandre”ectionsisdescribedbymatrices V whoserowsandcolumnsareorthonormal,butdet V = 1.The handednessofacoordinatesystemischangedifdet V = Š 1. Thevector u3isdeterminedbyitsthreedirectionalanglesinthe originalcoordinatesystem.Onlytwooftheseanglesareindependent. Arotationabouttheaxiscontainingthevector u3doesnotaect u3andcanbespeci“edbyarotationangleinaplaneperpendicularto theaxis.Thisangledeterminesthevectors u1and u2relativetothe

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6211.VECTORSANDTHESPACEGEOMETRY originalbasis.Soageneralrotationmatrix V hasthreeindependent parameters. Inparticular,thematrix V forcounterclockwiserotationsabout the z axisthroughanangle (seeStudyProblem11.2)is V = cos sin 0 Š sin cos 0 001 Relations(11.12)for V anditsrowsandcolumns v1= cos sin 0 v2= Š sin cos 0 v3= 0 0 1 w1= cos Š sin 0 w2= sin cos 0 w3= 0 0 1 areeasytoverify.TheresultofStudyProblem11.2canbestatedin theform(11.13),where a = x,y,z and a= x,y,z Problem11.21. Findthemostgeneralvector r thatsatis“estheequation a ( r b )=0 ,where a and b arenonzero,nonparallelvectors. Solution: Bythealgebraicpropertyofthetripleproduct, a ( r b )= r ( b a )=0.Hence, r a b .Thevector r liesintheplaneparallelto both a and b because a b isorthogonaltothesevectors.Anyvector intheplaneisalinearcombinationofanytwononparallelvectorsin it: r = t a + s b foranyreal t and s (seeStudyProblem11.6). Problem11.22.(VolumeofaTetrahedron). Atetrahedronisasolidwithfourverticesandfourtriangularfaces.Its volume V =1 3Ah ,where h isthedistancefromavertextotheopposite faceand A istheareaofthatface.Givencoordinatesofthevertices B C D ,and P ,expressthevolumeofthetetrahedronthroughthem. Solution: Put b = Š Š BC c = Š Š BD ,and a = Š AP .Theareaofthe triangle BCD is A =1 2 b c .Thedistancefrom P totheplane P1containingtheface BCD isthedistancebetween P1andtheparallel plane P2throughthevertex P .Hence, V = 1 3 A | a ( b c ) | b c = 1 6 | a ( b c ) | Sothevolumeofatetrahedronwithadjacentsides a b ,and c isonesixththevolumeoftheparallelepipedwiththesameadjacentsides. Notetheresultdoesnotdependonthechoiceofavertex.Anyvertex couldhavebeenchoseninsteadof B intheabovesolution.

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75.THETRIPLEPRODUCT63 Problem11.23.(SystemsofLinearEquations). Considerasystemoflinearequationsforthevariables x y ,and z : a1x + b1y + c1z = d1a2x + b2y + c2z = d2a3x + b3y + c3z = d3. De“nevectors a = a1,a2,a3 b = b1,b2,b3 c = c1,c2,c3 ,and d = d1,d2,d3 .Showthatthesystemhasauniquesolutionforany d if a ( b c ) =0 .If a ( b c )=0 ,formulateconditionson d under whichthesystemhasasolution. Solution: Thesystemoflinearequationscanbecastinthevector form x a + y b + z c = d Thisequationstatesthatagivenvector d isalinearcombinationof threegivenvectors.InStudyProblem11.11,itwasdemonstratedthat anyvectorinspacecanbeuniquelyrepresentedasalinearcombination ofthreenon-coplanarvectors.So,byCorollary11.4,thenumbers x y and z existandareuniqueif a ( b c ) =0.When a ( b c )=0,the vectors a b ,and c lieinoneplane.If d isnotinthisplane,thesystem cannothaveasolutionbecause d cannotberepresentedasalinear combinationofvectorsinthisplane.Supposethattwoofthevectors a b ,and c arenotparallel.Thentheircrossproductisorthogonalto theplane,and d mustbeorthogonaltothecrossproductinorderto beintheplane.If,say, a b = 0 ,thenthesystemhasasolutionif d ( a b )=0.Finally,itispossiblethatallthevectors a, b ,and c areparallel;thatis,allpairwisecrossproductsvanish.Then d must beparalleltothem.If,say, a = 0 ,thenthesystemhasasolutionif d a = 0 75.5.Exercises.(1) Findthetripleproducts a ( b c ), b ( a c ),and c ( a b )if (i) a = 1 Š 1 2 b = 2 1 2 ,and c = 2 1 3 (ii) a = u1+2 u2, b = u1Š u2+2 u3,and c = u2Š 3 u3if u1 ( u2 u3)=2 (2) Verifywhetherthevectors a = e1+2 e2Š e3, b =2 e1Š e2+ e3, and c =3 e1+ e2Š 2 e3arecoplanar. (3) Findthevalueof s ,ifany,forwhichthevectors a = 1 2 3 b = Š 1 0 1 ,and c = s, 1 2 s arecoplanar. (4) Let a = 1 2 3 b = 2 1 0 ,and c = 3 0 1 .Findthevolume oftheparallelepipedwithadjacentsides s a + b c Š t b ,and a Š p c ,

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6411.VECTORSANDTHESPACEGEOMETRY where s t ,and p arenumberssuchthat stp =1. (5) Letthenumbers u v ,and w besuchthat uvw =1and u3+ v3+ w3=1.Arethevectors a = u e1+ v e2+ w e3, b = v e1+ w e2+ u e3, and c = w e1+ u e2+ v e3coplanar?Ifnot,whatisthevolumeofthe parallelepipedwithadjacentedges a b ,and c ? (6) Determinewhetherthepoints A =(1 2 3), B =(1 0 1), C = ( Š 1 1 2),and D =( Š 2 1 0)areinoneplaneand,ifnot,“ndthe volumeoftheparallelepipedwithadjacentedges AB AC ,and AD (7) Find: (i)Allvaluesof s atwhichthepoints A ( s, 0 ,s ), B (1 0 1), C ( s,s, 1), and D (0 1 0)areinthesameplane (ii)Allvaluesof s atwhichthevolumeoftheparallelepipedwith adjacentedges AB AC ,and AD is9units (8) Provethat ( a b ) ( c d )=det a cb c a db d Hint: Usetheinvarianceofthetripleproductundercyclicpermutationsofvectorsinitandthe bac Š cab rule(11.9). (9) Let P beaparallelepipedofvolume V .Find: (i)Thevolumesofallparallelepipedswhoseadjacentedgesare diagonalsoftheadjacentfacesof P (ii)Thevolumesofallparallelepipedswhosetwoadjacentedges arediagonalsoftwononparallelfacesof P ,whilethethird adjacentedgeisthediagonalof P (10) Giventwononparallelvectors a and b ,“ndthemostgeneral vector r if (i) a ( r b )=0 (ii) a ( r b )=0and b r =0 (11) Letaset S1bethecircle x2+ y2=1andletaset S2betheline throughthepoints(0 2)and(2 0).Whatisthedistancebetweenthe sets S1and S2? (12) Consideraplanethroughthreepoints A =(1 2 3), B =(2 3 1), and C =(3 1 2).Findthedistancebetweentheplaneandapoint P obtainedfrom A bymovingthelatter3unitsoflengthalongthevector a = Š 1 2 2 (13) Considertwolines.The“rstlinepassesthroughthepoints (1 2 3)and(2 Š 1 1),whiletheotherpassesthroughthepoints( Š 1 3 1) and(1 1 3).Findthedistancebetweenthelines. (14) Findthedistancebetweenthelinethroughthepoints(1 2 3) and(2 1 4)andtheplanethroughthepoints(1 1 1),(3 1 2),and

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76.PLANESINSPACE65 (1 2 Š 1). Hint: Ifthelineisnotparalleltotheplane,thentheyintersectandthedistanceis0.Socheck“rstwhetherthelineisparallel totheplane.Howcanthisbedone? (15) Considerthelinethroughthepoints(1 2 3)and(2 1 2).Ifa secondlinepassesthroughthepoints(1 1 ,s )and(2 Š 1 0),“ndall valuesof s ,ifany,atwhichthedistancebetweenthelinesis9 / 2units. (16) Considertwoparallelstraightlinesegmentsinspace.Formulate analgorithmtocomputethedistancebetweenthemifthecoordinates oftheirendpointsaregiven.Inparticular,“ndthedistancebetween AB and CD if (i) A =(1 1 1), B =(4 1 5), C =(2 3 3),and D =(5 3 7) (ii) A =(1 1 1), B =(4 1 5), C =(3 5 5),and D =(6 5 9) Notethatthisdistancedoesnotgenerallycoincidewiththedistance betweentheparallellinescontaining AB and CD (17) Considertheparallelepipedwithadjacentedges AB ,AC ,and AD ,where A =(3 0 1), B =( Š 1 2 5), C =(5 1 Š 1),and D = (0 4 2).Findthedistances (i)Betweentheedge AB andallotheredgesparalleltoit (ii)Betweentheedge AC andallotheredgesparalleltoit (iii)Betweentheedge AD andallotheredgesparalleltoit (iv)Betweenallparallelplanescontainingthefacesoftheparallelepiped 76.PlanesinSpace76.1.AGeometricalDescriptionofaPlaneinSpace.Considerthecoordinateplane z =0.Itcontainstheoriginandallvectorsthatare orthogonaltothe z axis(allvectorsthatareorthogonalto e3).Since thecoordinatesystemcanbearbitrarilychosenbytranslatingtheoriginandrotatingthecoordinateaxes, aplaneinspaceisde“nedasa setofpointswhosepositionvectorsrelativetoaparticularpointinthe setareorthogonaltoagivennonzerovector n .Thevector n iscalled a normal oftheplane.Thus,thegeometricaldescriptionofaplane P inspaceentailsspecifyingapoint P0thatbelongsto P andanormal n of P .76.2.AnAlgebraicDescriptionofaPlaneinSpace.Letaplane P be de“nedbyapoint P0thatbelongstoitandanormal n .Insome coordinatesystem,thepoint P0hascoordinates( x0,y0,z0)andthe vector n isspeci“edbyitscomponents n = n1,n2,n3 .Ageneric pointinspace P hascoordinates( x,y,z ).Analgebraicdescriptionof aplaneamountstospecifyingconditionsonthevariables( x,y,z )such

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6611.VECTORSANDTHESPACEGEOMETRY thatthepoint P ( x,y,z )belongstotheplane P .Let r0= x0,y0,z0 and r = x,y,z bethepositionvectorsofaparticularpoint P0inthe planeandagenericpoint P inspace,respectively.Thentheposition vectorof P relativeto P0is Š Š P0P = r Š r0= x Š x0,y Š y0,z Š z0 .This vectorliesintheplane P ifitisorthogonaltothenormal n ,according tothegeometricaldescriptionofaplane(seeFigure11.15,leftpanel). Thealgebraicconditionequivalenttothegeometricalone, n Š Š P0P reads n Š Š P0P =0.Thus,thefollowingtheoremhasjustbeenproved. Theorem 11.6 (EquationofaPlane) Apointwithcoordinates ( x,y,z ) belongstoaplanethroughapoint P0( x0,y0,z0) andnormaltoavector n = n1,n2,n3 if n1( x Š x0)+ n2( y Š y0)+ n3( z Š z0)=0or n r = n r0, where r and r0arepositionvectorsofagenericpointandaparticular point P0intheplane. Givenanonzerovector n andanumber d ,itisalwayspossibleto “ndaparticularvector r0suchthat n r0= d .Sinceatleastone componentof n doesnotvanish,say, n1 =0,then r0= d/n1, 0 0 Therefore,ageneralsolutionofthelinearequation n r = d isaset ofpositionvectorsofallpointsofaplanethatisorthogonalto n .The number d determinesthepositionoftheplaneinspaceinthefollowing way.Supposethateverypointoftheplaneisdisplacedbyavector a thatis, r r + a .Theequationofthedisplacedplaneis n ( r + a )= d or n r = d Š n a .If n a =0,eachpointoftheplaneistranslatedwithin theplanebecause a isorthogonalto n .Theplaneasapointsetdoes notchangeandneitherdoesthenumber d .Ifthedisplacementvector a isnotorthogonalto n ,then d changesbytheamount Š n a =0. Sinceeverypointoftheoriginalplaneistranslatedbythesamevector, theresultofthistransformationisaparallelplane.Variationsof d correspondtoshiftsoftheplaneparalleltoitselfalongitsnormal(see Figure11.15,rightpanel).Thus,theequations n r = d1and n r = d2describetwo parallel planes.Theplanescoincideifandonlyif d1= d2. Consequently,twoplanes n1 r = d1and n2 r = d2areparallelifand onlyiftheirnormalsareproportionalorifandonlyiftheirnormals areparallel: P1P1 n1 n2 n1= s n2forsomereal s =0.Forexample,theplanes x Š 2 y Š z =5and Š 2 x +4 y +2 z =1areparallelbecausetheirnormals, n1= 1 Š 2 Š 1 and n2= Š 2 4 2 ,areproportional n2= Š 2 n1.

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76.PLANESINSPACE67 Figure11.15.Left :Algebraicdescriptionofaplane.If r0isapositionvectorofaparticularpointintheplaneand r isthepositionvectorofagenericpointintheplane,then thevector r Š r0liesintheplaneandisorthogonaltoits normal,thatis, n ( r Š r0)=0. Right :Equationsofparallel planesdieronlybytheirconstantterms.Thedierence oftheconstanttermsdeterminesthedistancebetweenthe planesasstatedin(11.17).Anormaltoagivenplanecanalwaysbeobtainedbytakingthe crossproductofanytwononparallelvectorsintheplane.Indeed,any vectorinaplaneisalinearcombinationoftwononparallelvectors a and b (StudyProblem11.6).Thevector n = a b isorthogonalto both a and b andhencetoanylinearcombinationofthem. Example 11.19 Findanequationoftheplanethroughthreegiven points A (1 1 1) B (2 3 0) ,and C ( Š 1 0 3) Solution: Aplaneisspeci“edbyaparticularpoint P0initandby avector n normaltoit.Threepointsintheplanearegiven,soany ofthemcanbetakenas P0,forexample, P0= A or( x0,y0,z0)= (1 1 1).Avectornormaltoaplanecanbefoundasthecrossproduct ofanytwononparallelvectorsinthatplane(seeFigure11.16,left panel).Soput a = Š AB = 1 2 Š 1 and b = Š AC = Š 2 Š 1 2 .Then onecantake n = a b = 3 0 Š 3 .Anequationoftheplaneis 3( x Š 1)+0( y Š 1)+( Š 3)( z Š 1)=0,or x Š z =0.Sincetheequation doesnotcontainthevariable y ,theplaneisparalleltothe y axis. Notethatifthe y componentof n vanishes(i.e.,thereisno y inthe equation),then n isorthogonalto e2because n e2=0;thatis,the y axisisorthogonalto n andhenceparalleltotheplane.

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6811.VECTORSANDTHESPACEGEOMETRY Figure11.16.Left :IllustrationtoExample11.19.The crossproductoftwononparallelvectorsinaplaneisanormaloftheplane. Right :Distancebetweenapoint P1and aplane.Anillustrationtothederivationofthedistance formula(11.15).Thesegment P1B isparalleltothenormal n sothatthetriangle P0P1B isright-angled.Therefore, D = | P1B | = | P0P1| cos .Definition 11.16 (AngleBetweenTwoPlanes) Theanglebetweenthenormalsoftwoplanesiscalledthe anglebetween theplanes If n1and n2arethenormals,thentheangle betweenthemis determinedby cos = n1 n2 n1 n2 = n1 n2. Notethataplaneasapointsetinspaceisnotchangedifthedirection ofitsnormalisreversed(i.e., n Š n ).Sotherangeof canalways berestrictedtotheinterval[0 ,/ 2].Indeed,if happenstobeinthe interval[ / 2 ](i.e.,cos 0),thentheangle Š / 2canalsobe viewedastheanglebetweentheplanesbecauseonecanalwaysreverse thedirectionofoneofthenormals n1Š n1or n2Š n2sothat cos Š cos Theanglebetweentheplanesisusefulfordeterminingtheirrelative orientation.Twoplanesintersectiftheanglebetweenthemisnot0. Twoplanesareparalleliftheanglebetweenthemvanishes.Theplanes areperpendiculariftheirnormalsareorthogonal.Forexample,the planes x + y + z =1and x +2 y Š 3 z =4areperpendicularbecause theirnormals n1= 1 1 1 and n2= 1 2 Š 3 areorthogonal: n1 n2= 1+2 Š 3=0(i.e., n1 n2).76.3.TheDistanceBetweenaPointandaPlane.Considertheplane throughapoint P0andnormaltoavector n .Let P1beapointin space.Whatisthedistancebetween P1andtheplane?Lettheangle between n andthevector Š Š P0P1be (seeFigure11.16,rightpanel).

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76.PLANESINSPACE69 Thenthedistanceinquestionis D = Š Š P0P1 cos if / 2(the lengthofthestraightlinesegmentconnecting P1andtheplanealong thenormal n ).For >/ 2,cos mustbereplacedby Š cos because D 0.So (11.15) D = Š Š P0P1| cos | = n Š Š P0P1| cos | n = | n Š Š P0P1| n Let r0and r1bepositionvectorsof P0and P1,respectively.Then Š Š P0P1= r1Š r0,and (11.16) D = | n ( r1Š r0) | n = | n r1Š d | n whichisabitmoreconvenientthan(11.15)iftheplaneisde“nedby anequation n r = d .DistanceBetweenParallelPlanes.Equation(11.16)allowsustoobtain asimpleformulaforthedistancebetweentwoparallelplanesde“ned bytheequations n r = d1and n r = d2(seeFigure11.15,rightpanel): (11.17) D = | d2Š d1| n Indeed,thedistancebetweentwoparallelplanesisthedistancebetween the“rstplaneandapoint r2inthesecondplane.By(11.16),this distanceis D = | n r2Š d1| / n = | d2Š d1| / n because n r2= d2foranypointinthesecondplane. Example 11.20 Findanequationofaplanethatisparalleltothe plane 2 x Š y +2 z =2 andatadistanceof3unitsfromit. Solution: Thereareafewwaystosolvethisproblem.Fromthe geometricalpointofview,aplaneisde“nedbyaparticularpointinit anditsnormal.Sincetheplanesareparallel,theymusthavethesame normal n = 2 Š 1 2 .Notethatthecoecientsatthevariablesinthe planeequationde“nethecomponentsofthenormalvector.Therefore, theproblemisreducedto“ndingaparticularpoint.Let P0bea particularpointonthegivenplane.Thenapointonaparallelplane canbeobtainedfromitbyshifting P0byadistanceof3unitsalong thenormal n .If r0isthepositionvectorof P0,thenapointona parallelplanehasapositionvector r0+ s n ,wherethedisplacement vector s n musthavealengthof3,or s n = | s | n =3 | s | =3and therefore s = 1.Naturally,thereshouldbetwoplanesparallelto thegivenoneandatthesamedistancefromit.To“ndaparticular pointonthegivenplane,onecansettwocoordinatesto0and“nd

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7011.VECTORSANDTHESPACEGEOMETRY thevalueofthethirdcoordinatefromtheequationoftheplane.Take, forinstance, P0(1 0 0).Particularpointsontheparallelplanesare r0+ n = 1 0 0 + 2 Š 1 2 = 3 Š 1 2 and,similarly, r0Š n = Š 1 1 Š 2 .Usingthesepointsinthestandardequationofaplane, theequationsoftwoparallelplanesareobtained: 2 x Š y +2 z =11and2 x Š y +2 z = Š 7 Analternativealgebraicsolutionisbasedonthedistanceformula (11.17)forparallelplanes.Anequationofaplaneparalleltothegiven oneshouldhavetheform2 x Š y +2 z = d .Thenumber d isdetermined bytheconditionthat | d Š 2 | / n =3or | d Š 2 | =9,or d = 9+2. 76.4.StudyProblems.Problem11.24. Findanequationoftheplanethatisnormalto astraightlinesegment AB andbisectsitif A =(1 1 1) and B = ( Š 1 3 5) Solution: Onehasto“ndaparticularpointintheplaneandits normal.Since AB isperpendiculartotheplane, n = Š AB = Š 2 2 4 Themidpointofthesegmentliesintheplane.Hence, P0(0 2 3)(the coordinatesofthemidpointsarethehalf-sumsofthecorresponding coordinatesoftheendpoints).Theequationreads Š 2 x +2( y Š 2)+ 4( z Š 3)=0or Š x + y +2 z =8. Problem11.25. Findanequationoftheplanethroughthepoint P0(1 2 3) thatisperpendiculartotheplanes x + y + z =1 and x Š y + 2 z =1 Solution: Onehasto“ndaparticularpointintheplaneandany vectororthogonaltoit.The“rstpartoftheproblemiseasytosolve: P0isgiven.Let n beanormaloftheplaneinquestion.Then,fromthe geometricaldescriptionofaplane,itfollowsthat n n1= 1 1 1 and n n2= 1 Š 1 2 ,where n1and n2arenormalsofthegivenplanes. So n isavectororthogonaltotwogivenvectors.Bythegeometrical propertyofthecrossproduct,suchavectorcanbeconstructedas n = n1 n2= 3 Š 1 Š 2 .Hence,theequationreads3( x Š 1) Š ( y Š 2) Š 2( z Š 3)=0or3 x Š y Š 2 z = Š 5. Problem11.26. Determinewhethertwoplanes x +2 y Š 2 z =1 and 2 x +4 y +4 z =10 areparalleland,ifnot,“ndtheanglebetweenthem. Solution: Thenormalsare n1= 1 2 Š 2 and n2= 2 4 4 = 2 1 2 2 .Theyarenotproportional.Hence,theplanesarenotparallel. Since n1 =3, n2 =6,and n1 n2=2,theangleisdeterminedby cos =2 / 18=1 / 9or =cosŠ 1(1 / 9).

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76.PLANESINSPACE71 Problem11.27. Findafamilyofallplanesthatcontainsthestraight linesegment AB if A =(1 2 Š 1) and B =(2 4 1) Solution: Alltheplanesinquestioncontainthepoint A .Soitcan bechosenasaparticularpointineveryplane.Sincethesegment AB liesineveryplaneofthefamily,thequestionamountstodescribingall vectorsorthogonalto a = Š AB = 1 2 2 thatdeterminethenormalsof theplanesinthefamily.Itiseasyto“ndaparticularvectororthogonal to a .Forexample, b = 0 1 Š 1 isorthogonalto a because a b =0. Next,thevector a b = Š 4 1 1 isorthogonaltoboth a and b .Any vectororthogonalto a liesinaplaneorthogonalto a andhencemust bealinearcombinationofanytwononparallelvectorsinthisplane.So thesought-afternormalsarealllinearcombinationsof b and c = a b Sincethelengthofeachnormalisirrelevant,thefamilyoftheplanes isdescribedbyallunitvectorsorthogonalto a .Recallthatanyunit vectorinaplanecanbewrittenintheform n=cos u1+sin u2, where u1 2aretwounitorthogonalvectorsintheplaneand0 < 2 (seeFigure11.5,rightpanel).Soput u1= b = b / b and u2= c = c / c ,where b = 2and c =3 2.Thefamilyoftheplanes isdescribedbyequations n ( r Š r0)=0,where0 < 2 and r0= 1 2 Š 1 (thepositionvectorof A ). 76.5.Exercises.(1) Findanequationoftheplanethroughtheoriginandparalleltothe plane2 x Š 2 y + z =4.Whatisthedistancebetweenthetwoplanes? (2) Dotheplanes2 x + y Š z =1and4 x +2 y Š 2 z =10intersect? (3) Determinewhethertheplanes2 x + y Š z =3and x + y + z =1 areintersecting.Iftheyare,“ndtheanglebetweenthem. (4) Consideraparallelepipedwithonevertexattheorigin O atwhich theadjacentsidesarethevectors a = 1 2 3 b = 2 1 1 ,and c = Š 1 0 1 .Let OP bethelargestdiagonaloftheparallelepiped.Find anequationoftheplanesthatcontain: (i)Thefacesoftheparallelepiped (ii)Thelargestdiagonaloftheparallelepipedandthediagonalof eachofthreeofitsfacesadjacentat P (iii)Paralleldiagonalsintheoppositefacesoftheparallelepiped (5) Findanequationoftheplanewith x intercept a y intercept b ,and z intercept c .Whatisthedistancebetweentheoriginandtheplane? (6) Findequationsoftheplanesthatareperpendiculartotheline through(1 ,Š 1 1)and(2 0 1)andthatareatthedistance2fromthe point(1 2 3).

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7211.VECTORSANDTHESPACEGEOMETRY (7) Findanequationforthesetofpointsthatareequidistantfrom thepoints(1 2 3)and( Š 1 2 1).Giveageometricaldescriptionofthe set. (8) Findanequationoftheplanethatisperpendiculartotheplane x + y + z =1andcontainsthelinethroughthepoints(1 2 3)and ( Š 1 1 0). (9) Towhichoftheplanes x + y + z =1and x +2 y Š z =2isthe point(1 2 3)theclosest? (10) Giveageometricaldescriptionofthefollowingfamiliesofplanes: (i) x + y + z = c (ii) x + y + cz =1 (iii) x sin c + y cos c + z =1 where c isaparameter. (11) Findvaluesof c forwhichtheplane x + y + cz =1isclosestto thepoint P (1 2 1)andfarthestfrom P. (12) Considerthreeplaneswithnormals n1, n2,and n3suchthateach pairoftheplanesisintersecting.Underwhatconditiononthenormals arethethreelinesofintersectionparallelorevencoincide? (13) Findequationsofalltheplanesthatareperpendiculartothe plane x + y + z =1,havetheangle / 3withtheplane x + y =1,and passthroughthepoint(1 1 1). (14) Let a = 1 2 3 and b = 1 0 Š 1 .Findanequationoftheplane thatcontainsthepoint(1 2 Š 1),thevector a ,andavectororthogonal toboth a and b (15) Considertheplane P throughthreepoints A (1 1 1), B (2 0 1), and C ( Š 1 3 2).Findalltheplanesthatcontainthesegment AB and havetheangle / 3withtheplane P Hint: SeeStudyProblem11.27. (16) Findanequationoftheplanethatcontainsthelinethrough (1 2 3)and(2 1, 1)andcutsthesphere x2+ y2+ z2Š 2 x +4 y Š 6 z =0 intotwohemispheres. (17) Findanequationoftheplanethatistangenttothesphere x2+ y2+ z2Š 2 x Š 4 y Š 6 z +11=0atthepoint(2 1 2). Hint: Whatisthe anglebetweenalinetangenttoacircleatapoint P andthesegment OP where O isthecenterofthecircle?Extendthisobservationtoa planetangenttoaspheretodetermineanormalofthetangentplane. (18) Considerasphereofradius R centeredattheoriginandtwopoints P1and P2whosepositionvectorsare r1and r2.Supposethat r1 >R and r2 >R (thepointsareoutsidethesphere).Findtheequation n r = d oftheplanethrough P1and P2whosedistancefromthesphere ismaximal.Whatisthedistance? Hint: Show“rstthatanormalof

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77.LINESINSPACE73 theplanecanalwaysbewrittenintheform n = r1+ c ( r2Š r1).Then “ndaconditiontodeterminetheconstant c 77.LinesinSpace77.1.AGeometricalDescriptionofaLineinSpace.Considertheline thatcoincideswithacoordinateaxisofarectangularsystem,say,the x axis.Anypointonithasthecharacteristicpropertythatitsposition vectorisproportionaltothepositionvectorofaparticularpoint(e.g., to e1).Sincethecoordinatesystemcanbearbitrarilychosenbytranslatingtheoriginandrotatingthecoordinateaxes, alineinspaceis de“nedasasetofpointswhosepositionvectorsrelativetoaparticular pointinthesetareparalleltoagivennonzerovector v .Thus,the geometricaldescriptionofaline L inspaceentailsspecifyingapoint P0thatbelongsto L andavector v thatisparallelto L Remark. Considertwopointsinspace.Theycanbeconnectedby apath.Amongallthecontinuouspathsthatconnectthetwopoints, thereisadistinctone,namely,theonethathasthesmallestlength. Thispathiscalleda straightlinesegment .Alineinspacecanalso bede“nedas asetofpointsinspacesuchthattheshortestpathconnectinganypairofpointsofthesetbelongstoit .Thisde“nitionof thelineisdeeplyrootedintheverystructureofspaceitself.How canalineberealizedinthespaceinwhichwelive?Onecanusea pieceofrope,asintheancientworld,orthelineofsightŽ(i.e.,the pathtraveledbylightfromonepointtoanother).Einsteinstheory ofgravitystatesthatstraightlinesŽde“nedastrajectoriestraversed bylightarenotexactlythesameasstraightlinesŽinaEuclidean space.SoaEuclideanspacemayonlybeviewedasamathematical approximation(ormodel)ofourspace.Agoodanalogywouldbeto comparetheshortestpathsinaplaneandonthesurfaceofasphere; theyarenotthesame,asthelatteraresegmentsofcirclesandhence arebentŽorcurved.ŽTheconceptofcurvatureofapathisdiscussed inthenextchapter.Theshortestpathbetweentwopointsinaspace iscalleda geodesic (byanalogywiththeshortestpathonthesurface oftheEarth).ThegeodesicsofaEuclideanspacearestraightlines anddonothavecurvature,whereasthegeodesicsofourspace(i.e.,the pathstraversedbylight)dohavecurvaturethatisdeterminedbythe distributionofgravitatingmasses(planets,stars,etc.).Adeviationof thegeodesicsfromstraightlinesnearthesurfaceoftheEarthisvery hardtonotice.However,adeviationofthetrajectoryoflightfrom astraightlinehasbeenobservedforthelightcomingfromadistant

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7411.VECTORSANDTHESPACEGEOMETRY startotheEarthandpassingneartheSun.Einsteinstheoryofgeneralrelativityassertsthatabettermodelofourspaceisa Riemann space .AsucientlysmallneighborhoodinaRiemannspacelookslike aportionofaEuclideanspace.77.2.AnAlgebraicDescriptionofaLine.Insomecoordinatesystems, aparticularpointofaline L hascoordinates P0( x0,y0,z0),andavectorparallelto L isde“nedbyitscomponents, v = v1,v2,v3 .Let r = x,y,z beapositionvectorofagenericpointof L andlet r0= x0,y0,z0 bethepositionvectorof P0.Thenthevector r Š r0is thepositionvectorof P relativeto P0.Bythegeometricaldescription oftheline,itmustbeparallelto v .Sinceanytwoparallelvectorsare proportional,apoint( x,y,z )belongsto L ifandonlyif r Š r0= t v forsomereal t (seeFigure11.17,leftpanel). Figure11.17.Left :Algebraicdescriptionofaline L through r0andparalleltoavector v .If r0and r arepositionvectorsofparticularandgenericpointsoftheline, thenthevector r Š r0isparalleltothelineandhencemust beproportionaltoavector v ,thatis, r Š r0= t v forsome realnumber t Right :Distancebetweenapoint P1anda line L throughapoint P0andparalleltoavector v .Itis theheightoftheparallelogramwhoseadjacentsidesarethe vectors Š ŠŠ P0P1and v .Theorem 11.7 (EquationsofaLine) Thecoordinatesofthepointsoftheline L throughapoint P0( x0,y0,z0) andparalleltoavector v = v1,v2,v3 satisfythevectorequation (11.18) r = r0+ t v Š
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77.LINESINSPACE75 Theparametricequationsofthelinecanbesolvedfor t .Asaresult, oneinfersequationsforthecoordinates x y ,and z : t = x Š x0 v1= y Š y0 v2= z Š z0 v3. Theseequationsarecalled symmetric equationsofaline.Notethat theseequationsmakesenseonlyifallthecomponentsof v donot vanish.If,say, v1=0,thenthe“rstequationin(11.19)doesnot containtheparameter t atall.Sothesymmetricequationsarewritten intheform x = x0, y Š y0 v2= z Š z0 v3. Example 11.21 Findthevector,parametric,andsymmetricequationsofthelinethroughthepoints A (1 1 1) and B (1 2 3) Solution: Take v = Š AB = 0 1 2 and P0= A .Then r = 1 1 1 + t 0 1 2 x =1 ,y =1+ t,z =1+2 t, x =1 ,y Š 1= z Š 1 2 arethevector,parametric,andsymmetricequationsoftheline, respectively. Thelineinthisexamplecanalsobeviewedasthesetofpointsof intersectionoftheplanes x =1and2 y Š z =1asfollowsfromthe symmetricequations.Clearly,alinecanalwaysbedescribedastheset ofpointsofintersectionoftwononparallelplanes.Sincethelinelies ineachplane,itmustbeorthogonaltothenormalsoftheseplanes. Therefore,avectorparalleltothelinecanalwaysbechosenasthe crossproductofthenormals. Example 11.22 Findthelinethatistheintersectionoftheplanes x + y + z =1 and 2 x Š y + z =2 Solution: Thenormalsoftheplanesare n1= 1 1 1 and n2= 2 Š 1 1 .Sothevector v = n1 n2= 2 1 Š 3 isparalleltotheline. To“ndaparticularpointoftheline,notethatits three coordinates ( x0,y0,z0)shouldsatisfy two equationsoftheplanes.Soonecanchoose oneofthecoordinatesatwilland“ndtheothertwofromtheequations oftheplanes.Itfollowsfromtheparametricequations(11.19)thatif, forexample, v3 =0,thenthereisavalueof t atwhich z vanishes, meaningthatthelinealwayscontainsapointwiththevanishing z coordinate.Since v3 =0forthelineinquestion,put z0=0.Then

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7611.VECTORSANDTHESPACEGEOMETRY x0+ y0=1and2 x0Š y0=2.Byaddingtheseequations, x0=1and hence y0=0.Theparametricequationsofthelineofintersectionare x =1+2 t y = t ,and z = Š 3 t DistanceBetweenaPointandaLine.Let L bealinethrough P0and parallelto v .Whatisthedistancebetweenagivenpoint P1andthe line L ?Consideraparallelogramwithvertex P0andwhoseadjacent sidesarethevectors v and Š Š P0P1asdepictedinFigure11.17(right panel).Thedistanceinquestionistheheightofthisparallelogram, whichisitsareadividedbythelengthofthebase v .If r0and r1are positionvectorsof P0and P1,then Š Š P0P1= r1Š r0andhence D = v Š Š P0P1 v = v ( r1Š r0) v .77.3.RelativePositionsofLinesinSpace.Twolinesinspacecanbe intersecting,parallel,orskew.Thecriterionforrelativepositionsofthe linesinspaceisstatedinCorollary11.9.Givenanalgebraicdescription ofthelinesestablishedhere,itcannowberestatedasfollows. Corollary 11.10 (RelativePositionsofLinesinSpace). Let L1bealinethrough P1andparalleltoavector v1andlet L2bea linethrough P2andparalleltoavector v2.Put r12= Š Š P1P2.Then (1) L1and L2areparallelif v1 v2= 0 or v1= s v2. (2) L1and L2areskewif r12 ( v1 v2) =0 (3) L1and L2intersectif r12 ( v1 v2)=0 and v1 v2 = 0 (4) L1and L2coincideif r12 ( v1 v2)=0 and v1 v2= 0 Indeed,thevector v1canalwaysbeviewedasavectorwithinitial andterminalpointsontheline L1.Thesameobservationistruefor thevector v2andtheline L2. Let L1and L2beintersecting.Howcanone“ndthepointof intersection?Tosolvethisproblem,considerthevectorequationsfor thelines rt= r1+ t v1and rs= r2+ s v2.Whenchangingtheparameter t ,theterminalpointof rtslidesalongtheline L1,whiletheterminal pointof rsslidesalongtheline L2whenchangingtheparameter s as depictedinFigure11.18(leftpanel).Notethattheparametersofboth linesarenotrelatedinanywayaccordingtothegeometricaldescription ofthelines.Iftwolinesareintersecting,thenthereshouldexistapair ofnumbers( t,s )=( t0,s0)atwhichtheterminalpointsofvectors rtand rscoincide, rt= rs.Let vi= ai,bi,ci i =1 2.Writingthis vectorequationincomponents,thefollowingsystemofequationsis

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77.LINESINSPACE77 Figure11.18.Left :Intersectionpointoftwolines L1and L2.Theterminalpointofthevector rttraverses L1as t rangesoverallrealnumbers,whiletheterminalpointofthe vector rstraverses L2as s rangesoverallrealnumbersindependentlyof t .Ifthelinesareintersecting,thenthereshould existapairofnumbers( t,s )=( t0,s0)suchthatthevectors rtand rscoincide,whichmeansthattheircomponentsmust bethesame.Thisde“nesthreeequationsontwovariables t and s Right :Intersectionpointofaline L andaplane P Theterminalpointofthevector rttraverses L as t ranges overallrealnumbers.Ifthelineintersectstheplanede“ned bytheequation r n = d ,thenthereshouldexistaparticular valueof t atwhichthevector rtsatis“estheequationofthe plane: rt n = d .obtained: x1+ ta1= x2+ sa2, y1+ tb1= y2+ sb2, z1+ tc1= z2+ sc2. Thissystemofequationsissolvedinaconventionalmanner,for example,byexpressing t via s fromthe“rstequation,substituting itintothesecondandthirdones,andverifyingthattheresultingtwo equationshavethe same solutionfor s .Notethatthesystemhasthree equationsforonlytwovariables.Itisan overdetermined system,which mayormaynothaveasolution.Iftheconditionsofpart(1)or(2) ofCorollary11.10aresatis“ed,thenthesystemhasnosolution(the linesareparallelorskew).Iftheconditionsofpart(3)ofCorollary 11.10areful“lled,thenthereisauniquesolution.Naturally,ifthelines coincide,therewillbein“nitelymanysolutions.Let( t,s )=( t0,s0)be auniquesolution.Thenthepositionvectorofthepointofintersection is r1+ t0v1or r2+ s0v2.77.4.RelativePositionsofLinesandPlanes.Consideraline L anda plane P .Thequestionofinterestistodeterminewhethertheyare

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7811.VECTORSANDTHESPACEGEOMETRY intersectingorparallel.Ifthelinedoesnotintersecttheplane,then theymustbeparallel.Inthelattercase,thelinemustbeorthogonal tothenormaloftheplane. Corollary 11.11 (CriterionforaLineandaPlanetoBeParallel) Aline L isparalleltoaplane P ifanynonzerovector v paralleltothe lineisorthogonaltoanormal n of P : LP v n v n =0 Ifalineandaplanearenotparallel,theymustintersect.Inthis case,thereshouldexistaparticularvalueoftheparameter t forwhich thepositionvector rt= r0+ t v ofapointof L alsosatis“esanequation oftheplane r n = d (seeFigure11.18,rightpanel).Thevalueofthe parameter t thatcorrespondstothepointofintersectionisdetermined bytheequation rt n = d r0 n + t v n = d t = d Š r0 n v n Thepositionvectorofthepointofintersectionisfoundbysubstituting thisvalueof t intothevectorequationoftheline rt= r0+ t v Example 11.23 Apointobjectistravelingalongtheline x Š 1= y/ 2=( z +1) / 2 withaconstantspeed v =6 meterspersecond.Ifall coordinatesaremeasuredinmetersandtheinitialpositionvectorofthe objectis r0= 1 0 Š 1 ,whendoesitreachtheplane 2 x + y + z =13 ? Whatisthedistancetraveledbytheobject? Solution: Parametricequationsofthelineare x =1+ s y =2 s z = Š 1+2 s .Thevalueoftheparameter s atwhichthelineintersects andtheplaneisdeterminedbythesubstitutionoftheseequationsinto theequationoftheplane: 2(1+ s )+2 s +( Š 1+2 s )=13 6 s =12 s =2 Sothepositionvectorofthepointofintersectionis r = 3 4 3 .The distancebetweenitandtheinitialpointis D = r Š r0 = 2 4 4 = 2 1 2 2 =6metersandthetraveltimeis T = D/v =1sec. Remark. Inthisexample,theparameter s doesnotcoincidewith thephysicaltime.Ifanobjecttravelswithaconstantspeed v along thelinethrough r0andparalleltoa unit vector v ,thenitsvelocity vectoris v = v v anditspositionvectoris r = r0+ v t ,where t is thephysicaltime.Indeed,thevector r Š r0isthedisplacementvector oftheobjectalongitstrajectory,andhenceitslengthdeterminesthe distancetraveledbytheobject: r Š r0 = v t = vt ,whichshows thattheparameter t> 0isthetraveltime.

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77.LINESINSPACE79 Example 11.24 Findanequationoftheplane P thatisperpendiculartotheplane P1, x + y Š z =1 ,andcontainstheline x Š 1= y/ 2= z +1 Solution: Theplane P mustbeparalleltotheline( P containsit) andthenormal n1= 1 1 Š 1 of P1(as PP1).Sothenormal n of P isorthogonaltoboth n1andthevector v = 1 2 1 thatis paralleltotheline.Therefore,onecantake n = n1 v = 3 Š 2 1 Thelineliesin P ,andthereforeanyofitspointscanbetakenasa particularpointof P ,forexample, P0(1 0 Š 1).Anequationof P reads 3( x Š 1) Š 2 y +( z +1)=0or3 x Š 2 y + z =2. Example 11.25 Findtheplanesthatareperpendiculartotheline x = y/ 2= Š z/ 2 andhavethedistance 3 fromthepoint ( Š 1 Š 2 2) on theline. Solution: Thelineisparalleltothevector v = 1 2 Š 2 .Sothe planeshavethesamenormal n = v .Particularpointsintheplanes arethepointsofintersectionofthelinewiththeplanes.Thesepoints areatthedistance3from r0= Š 1 Š 2 2 ,andtheirpositionvectors r shouldsatisfythecondition r Š r0 =3.Ontheotherhand,bythe vectorequationoftheline, r = r0+ t v andhence t v =3or3 | t | =3 or t = 1.Sothepositionvectorsofparticularpointsintheplanes are r = r0 v or r = 0 0 0 and r = Š 2 Š 4 4 .Equationsofthe planesare x +2 y Š 2 z =0and( x +2)+2( y +4) Š 2( z Š 4)=0or x +2 y Š 2 z = Š 18. 77.5.StudyProblems.Problem11.28. Let L1bethelinethrough P1(1 1 1) andparallelto v1= 1 2 Š 1 andlet L2bethelinethrough P2(4 0 Š 2) andparallel to v2= 2 1 0 .Determinewhetherthelinesareparallel,intersecting, orskewand“ndtheline L thatisperpendiculartoboth L1and L2and intersectsthem. Solution: Thevectors v1and v2arenotproportional,andhence thelinesarenotparallel.Onehas r12= Š Š P1P2= 3 Š 1 Š 3 and v1 v2= 1 Š 2 Š 3 .Therefore, r12 ( v1 v2)=14 =0,andthelines areskewbyCorollary11.10.Let rt= r1+ t v1beapositionvectorofa pointof L1andlet rs= r2+ s v2beapositionvectorofapointof L2asshowninFigure11.19(leftpanel).Theline L isorthogonaltoboth vectors v1and v2.Asitintersectsthelines L1and L2,thereshould existapairofvalues( t,s )oftheparametersatwhichthevector rsŠ rtisparallelto L ;thatis,thevector rsŠ rtbecomesorthogonalto v1

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8011.VECTORSANDTHESPACEGEOMETRY Figure11.19.Left :IllustrationtoStudyProblem11.28. Thevectors rsand rttraceouttwogivenskewedlines L1and L2,respectively.Thereareparticularvaluesof t and s at whichthedistance rtŠ rs becomesminimal.Therefore,the line L throughsuchpoints rtand rsisperpendiculartoboth L1and L2. Right :Intersectionofaline L andasphere S AnillustrationtoStudyProblem11.29.Theterminalpoint ofthevector rttraversesthelineas t rangesoverallreal numbers.Ifthelineintersectsthesphere,thenthereshould existaparticularvalueof t atwhichthecomponentsofthe vector rtsatisfytheequationofthesphere.Thisequationis quadraticin t ,andhenceitcanhavetwodistinctrealroots, oronemultiplerealroot,ornorealroots.Thesethreecases correspondtotwo,one,ornopointsofintersection.The existenceofjustonepointofintersectionmeansthattheline istangenttothesphere.and v2.Thecorrespondingalgebraicconditionsare rsŠ rt v1 ( rsŠ rt) v1=4+4 s Š 6 t =0 rsŠ rt v1 ( rsŠ rt) v2=5+5 s Š 4 t =0 Thissystemhasthesolution t =0and s = Š 1.Thus,thepointswith thepositionvectors rt =0= r1and rs = Š 1= r2Š v2= 2 Š 1 Š 2 belong to L .Sothevector v = rs = Š 1Š rt =0= 1 Š 3 Š 1 isparallelto L Takingaparticularpointof L tobe P1(whosepositionvectoris r1), theparametricequationsread x =1+ t y =1 Š 3 t ,and z =1 Š t Problem11.29. Consideralinethroughtheoriginthatisparallelto thevector v = 1 1 1 .Findthepartofthislinethatliesinsidethe sphere x2+ y2+ z2Š x Š 2 y Š 3 z =9 Solution: Theparametricequationsofthelineare x = t y = t z = t Ifthelineintersectsthesphere,thenthereshouldexistparticularand

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77.LINESINSPACE81 valuesof t atwhichthecoordinatesofapointofthelinealsosatisfythe sphereequation(seeFigure11.19,rightpanel).Ingeneral,parametric equationsofalinearelinearin t ,whileasphereequationisquadraticin thecoordinates.Therefore,theequationthatdeterminesthevaluesof t correspondingtothepointsofintersectionisquadratic.Aquadratic equationhastwo,one,ornorealsolutions.Accordingly,thesecases correspondtotwo,one,andnopointsofintersection,respectively.In ourcase,3 t2Š 6 t =9or t2Š 2 t =3andhence t = Š 1and t =3.The pointsofintersectionare( Š 1 Š 1 Š 1)and(3 3 3).Thelinesegment connectingthemcanbedescribedbytheparametricequations x = t y = t ,and z = t ,where Š 1 t 3. 77.6.Exercises.(1) Findparametricequationsofthelinethroughthepoint(1 2 3) andperpendiculartotheplane x + y +2 z =1.Findthepointof intersectionofthelineandtheplane. (2) Findparametricandsymmetricequationsofthelineofintersection oftheplanes x + y + z =1and2 x Š 2 y + z =1. (3) Isthelinethroughthepoints(1 2 3)and(2 Š 1 1)perpendicular tothelinethroughthepoints(0 1 Š 1)and(1 0 2)?Arethelines intersecting?Ifso,“ndthepointofintersection. (4) Determinewhetherthelines x =1+2 t y =3 t ,and z =2 Š t and x +1= y Š 4=( z Š 1) / 3areparallel,skew,orintersecting.Ifthey intersect,“ndthepointofintersection. (5) Findthevectorequationofthestraightlinesegmentfromthepoint (1 2 3)tothepoint( Š 1 1 2). (6) Let r1and r2bepositionvectorsoftwopointsinspace.Findthe vectorequationofthestraightlinesegmentfrom r1to r2. (7) Considertheplane x + y Š z =0andapoint P =(1 1 2)init. Findparametricequationsofthelinesthroughtheoriginthatlieinthe planeandareatadistanceof1unitfrom P Hint: Avectorparallel totheselinescanbetakenintheform v = 1 ,c, 1+ c ,where c isto bedetermined.Explainwhy! (8) Findparametric,symmetric,andvectorequationsofthelinethrough (0 1 2)thatisperpendiculartotheline x =1+ t y = Š 1+ t z =2 Š 2 t andparalleltotheplane x +2 y + z =3. (9) Findparametricequationsofthelinethatisparallelto v = 2 Š 1 2 andgoesthroughthecenterofthesphere x2+ y2+ z2= 2 x +6 z Š 6.Restricttherangeoftheparametertodescribethepart ofthelinethatisinsidethesphere.

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8211.VECTORSANDTHESPACEGEOMETRY (10) Lettheline L1passthroughthepoint A (1 1 0)paralleltothevector v = 1 Š 1 2 andlettheline L2passthroughthepoint B (2 0 2) paralleltothevector w = Š 1 1 2 .Showthatthelinesareintersecting.Findthepoint C ofintersectionandparametricequationsofthe line L3through C thatisperpendicularto L1and L2. (11) Findparametricequationsofthelinethrough(1 2 5)thatisperpendiculartotheline x Š 1=1 Š y = z andintersectsthisline. (12) Findparametricequationsofthelinethatbisectstheangleof thetriangle ABC atthevertex A if A =(1 1 1), B =(2 Š 1 3),and C =(1 4 Š 3). Hint: Seeexercise12inSection73.7. (13) Findthedistancebetweenthelines x = y = z and x +1= y/ 2= z/ 3. (14) Asmallmeteormoveswithspeed v inthedirectionofaunitvector u .Ifthemeteorpassedthepoint r0,“ndtheconditionon u such thatthemeteorhitsanasteroidoftheshapeofasphereofradius R centeredatthepoint r1.Determinethepositionvectoroftheimpact point. (15) Aprojectileis“redinthedirection v = 1 2 3 fromthepoint (1 1 1).Letthetargetbeadiskofradius R centeredat(2 3 6)inthe plane2 x Š 3 y +4 z =19.Ifthetrajectoryoftheprojectileisastraight line,determinewhetherithitsatargetintwocases R =2and R =3. (16) Consideratriangle ABC where A =(1 1 1), B =(3 1 Š 1),and C =(1 3 1).Findtheareaofapolygon DPQB wherethevertices D and Q arethemidpointsof CB and AB ,respectively,andthevertex P istheintersectionofthesegments CQ and AD 78.QuadricSurfaces Definition 11.17 (QuadricSurface) Thesetofpointswhosecoordinatesinarectangularcoordinatesystem satisfytheequation Ax2+ By2+ Cz2+ pxy + qxz + vyz + x + y + z + D =0 where A B C p q v ,and D arerealnumbers,iscalleda quadricsurface Theequationthatde“nesquadricsurfacesisthemostgeneralequation quadratic inallthecoordinates.Thisiswhysurfacesde“nedby itarecalled quadric .Asphereprovidesasimpleexampleofaquadric surface: x2+ y2+ z2Š R2=0,thatis, A = B = C =1, p = q = v =0, = = ,and D = Š R2,where R istheradiusofthesphere.If B = C =1, = Š 1,whiletheotherconstantsvanish,thequadratic equation x = y2+ z2de“nesacircularparaboloidwhosesymmetryaxis

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78.QUADRICSURFACES83 isthe x axis.Ontheotherhand,if A = B =1, = Š 1,whiletheother constantsvanish,theequation z = x2+ y2alsode“nesaparaboloid thatcanbeobtainedfromtheformeronebyarotationaboutthe y axisthroughtheangle / 2underwhich( x,y,z ) ( z,y, Š x )sothat x = y2+ z2 z = y2+ x2.Thus,therearequadricsurfacesofthe sameshapedescribedbydierentequations.Thetaskhereistoclassify alltheshapesofquadricsurfaces.Theshapedoesnotchangeunderits rigidrotationsandtranslations.Ontheotherhand,theequationthat describestheshapewouldchangeundertranslationsandrotationsof thecoordinatesystem.Thefreedominchoosingthecoordinatesystem canbeusedtosimplifytheequationforquadricsurfaceandobtaina classi“cationofdierentshapesdescribedbyit.78.1.QuadricCylinders.Consider“rstasimplerprobleminwhichthe equationofaquadricsurfacedoesnotcontainoneofthecoordinates, say, z (i.e., C = q = v = =0).Thentheset S S = ( x,y,z ) Ax2+ By2+ pxy + x + y + D =0 isthesamecurveineveryhorizontalplane z =const.Forexample, if A = B =1, p =0,and D = Š R2,thecrosssectionofthesurface S byanyhorizontalplaneisacircle x2+ y2= R2.Sothesurface S isacylinderofradius R thatissweptbythecirclewhenthelatter isshiftedupanddownparalleltothe z axis.Similarly,ageneral cylindricalshapeisobtainedbyshiftingacurveinthe xy planeupand downparalleltothe z axis. Theorem 11.8 (Classi“cationofQuadricCylinders) Ageneralequationforquadriccylinders S = ( x,y,z ) Ax2+ By2+ pxy + x + y + D =0 canbebroughttooneofthestandardforms Ax2+ By2+ D=0 or Ax2+ y =0 byrotationandtranslationofthecoordinatesystem, provided A B ,and p donotvanishsimultaneously.Inparticular,these formsde“nethequadriccylinders: y Š ax2=0(paraboliccylinder) =0 x2 a2+ y2 b2=1(ellipticcylinder) A D< 0 B D< 0 ,D =0 x2 a2Š y2 b2=1(hyperboliccylinder) ,AB< 0 ,D =0 TheshapesofquadriccylindersareshowninFigure11.20.Other thanquadriccylinders,thestandardequationsmayde“neplanesora

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8411.VECTORSANDTHESPACEGEOMETRY Figure11.20.Left :Paraboliccylinder.Thecrosssection byanyhorizontalplane z =constisaparabola y = ax2. Middle :Anellipticcylinder.Thecrosssectionbyany horizontalplane z =constisanellipse x2/a2+ y2/b2= 1. Right :Ahyperboliccylinder.Thecrosssectionby anyhorizontalplane z =constisahyperbola x2/a2Š y2/b2=1.lineforsomeparticularvaluesoftheconstants A, B, D,and .For example,for A= Š B=1and D=0,theequation x2= y2de“nes twoplanes x y =0.For A= B=1and D=0,theequation x2+ y2=0de“nestheline x = y =0(the z axis).ProofofTheorem11.8.Let( x,y )becoordinatesinthecoordinatesystemobtainedbyarotationthroughanangle .Theequationof S in thenewcoordinatesystemisobtainedbythetransformation: ( x,y ) ( x cos Š y sin ,y cos + x sin ) accordingtoStudyProblem11.2.Theangle canbechosensothat theequationfor S doesnotcontainthemixedŽterm xy .Indeed, considerthetransformationofquadratictermsintheequationfor S : x2 cos2x2+sin2y2Š 2sin cos xy =1 2(1+cos(2 )) x2+1 2(1 Š cos(2 )) y2Š sin(2 ) xy, y2 sin2x2+cos2y2+2sin cos xy =1 2(1 Š cos(2 )) x2+1 2(1+cos(2 )) y2+sin(2 ) xy, xy sin cos ( x2Š y2)+(cos2 Š sin2 ) xy =1 2sin(2 )( x2Š y2)+cos(2 ) xy. Afterthetransformation,thecoecientat xy becomes: p p=( B Š A )sin(2 )+ p cos(2 ) .

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78.QUADRICSURFACES85 Theangle issetsothat p=0or (11.20)tan(2 )= p A Š B and = 4 if A = B. Similarly,thecoecients A and B (thefactorsat x2and y2)and and (thefactorsat x and y )become A A=1 2[ A + B +( A Š B )cos(2 )+ p sin(2 )] B B=1 2[ A + B Š ( A Š B )cos(2 ) Š p sin(2 )] = cos + sin = cos Š sin where satis“es(11.20).Dependingonthevaluesof A B ,and p ,the followingthreecasescanoccur. First, A= B=0,whichisonlypossibleif A = B = p =0.Note thatthecombination Ax2+ By2+ pxy becomes Ax2+ By2+ pxy in arotatedcoordinatesystem.If A= B= p=0foraparticular (chosentomake p=0),thenthiscombinationshouldbeidentically0 inanyothercoordinatesystemobtainedbyrotation.Inthiscase, S is de“nedbytheequation x + y + D =0,whichisaplaneparallelto the z axis. Second,onlyoneof Aand Bvanishes.Forestablishingtheshape, itisirrelevanthowthehorizontalandverticalcoordinatesintheplane arecalled.So,withoutlossofgenerality,put B=0.Inthiscase, theequationfor S assumestheform Ax2+ x + y + D =0or,by completingthesquares, A x Š x02+ ( y Š y0)=0 ,x0= 2 A,y0= Ax2 0Š D. Afterthe translation ofthecoordinatesystem x x + x0and y y + y0,theequationisreducedto Ax2+ y =0.If =0,itde“nes aparabola y Š ax2=0,where a = Š A/. Third,both Aand Bdonotvanish.Then,afterthecompletion ofsquares,theequation Ax2+ By2+ x + y + D =0hastheform A( x Š x0)2+ B( y Š y0)2+ D=0 where x0= Š 2 A, y0= Š 2 B,and D= Š D +1 2( Ax2 0+ By2 0).Finally, afterthetranslationoftheorigintothepoint( x0,y0),theequation becomes Ax2+ By2+ D=0 If D=0,thenthisequationde“nestwostraightlines y = mx ,where m =( Š A/B)Š 1 / 2,provided Aand Bhaveoppositesigns(otherwise, theequationhasthesolution x = y =0(aline)).If D =0,thenthe equationcanbewrittenas( Š A/D) x2+( Š B/D) y2=1.Onecan

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8611.VECTORSANDTHESPACEGEOMETRY alwaysassumethat A/D< 0.Notethattherotationofthecoordinate systemthroughtheangle / 2swapstheaxes,( x,y ) ( y, Š x ),which canbeusedtoreversethesignof A/D.Nowput Š A/D=1 /a2and B/D= 1 /b2(dependingonwhether B/Dispositiveornegative) sothattheequationbecomes x2 a2 y2 b2=1 Whentheplusistaken,thisequationde“nesanellipse.Whenthe minusistaken,thisequationde“nesahyperbola.78.2.ClassicationofGeneralQuadricSurfaces.Theclassi“cationof generalquadricsurfacescanbecarriedoutinthesameway.The generalquadraticequationcanbewritteninthenewcoordinatesystem thatisobtainedbyatranslation(11.1)andarotation(11.13).The rotationalfreedom(threeparameters)canbeusedtoeliminatethe mixedŽterms: p p=0, q q=0,and v v=0.After thisrotation,thelineartermsareeliminatedbyasuitabletranslation, provided A, B,and Cdonotvanish.Thecorrespondingtechnicalities canbecarriedoutbestbylinearalgebramethods.Sothe“nalresult isgivenwithoutaproof. Theorem 11.9 (Classi“cationofQuadricSurfaces) Byrotationandtranslationofacoordinatesystem,ageneralequation forquadricsurfacescanbebroughtintooneofthestandardforms: Ax2+ By2+ Cz2+ D=0or Ax2+ By2+ z =0 Inparticular,thestandardformsdescribequadriccylindersandthe followingsixsurfaces: x2 a2+ y2 b2+ z2 c2=1(ellipsoid) z2 c2= x2 a2+ y2 b2(ellipticdoublecone) x2 a2+ y2 b2Š z2 c2=1(hyperboloidofonesheet) Š x2 a2Š y2 b2+ z2 c2=1(hyperboloidoftwosheets) z c = x2 a2+ y2 b2(ellipticparaboloid) z c = x2 a2Š y2 b2(hyperbolicparaboloid) .

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78.QUADRICSURFACES87 Figure11.21.Left :Anellipsoid.Acrosssectionbyany coordinateplaneisanellipse. Right :Anellipticdouble cone.Acrosssectionbyahorizontalplane z =constisan ellipse.Acrosssectionbyanyverticalplanethroughthe z axisistwolinesthroughtheorigin.Thesixshapesarethecounterpartsinthreedimensionsoftheconic sectionsintheplanediscussedinCalculusII.Otherthanquadriccylindersortheabovesixshapes,aquadraticequationmayalsode“ne planesandlinesforparticularvaluesofitsparameters.78.3.VisualizationofQuadricSurfaces.Theshapeofaquadricsurface canbeunderstoodbystudyingintersectionsofthesurfacewiththe coordinateplanes x = x0, y = y0,and z = z0.Theseintersectionsare alsocalled traces .AnEllipsoid.If a2= b2= c2= R2,thentheellipsoidbecomesasphere ofradius R .So,intuitively,anellipsoidisaspherestretchedŽalong thecoordinateaxes(seeFigure11.21,leftpanel).Tracesofanellipsoid intheplanes x = x0, | x0|
0 Astheplane x = x0getscloserto x = a or x = Š a k becomessmaller andsodoestheellipsebecauseitsmajoraxes b k and c k decrease. Apparently,thetracesintheplanes x = a consistofasinglepoint ( a, 0 0),andthereisnotraceinanyplane x = x0if | x0| >a .Traces intheplanes y = y0and z = z0arealsoellipsesandexistonlyif

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8811.VECTORSANDTHESPACEGEOMETRY Figure11.22.Left :Ahyperboloidofonesheet.Across sectionbyahorizontalplane z =constisanellipse.Across sectionbyaverticalplane x =constor y =constisahyperbola. Right :Ahyperboloidoftwosheets.Anonemptycross sectionbyahorizontalplaneisanellipse.Acrosssectionby averticalplaneisahyperbola. Figure11.23.Left :Anellipticparaboloid.Anonempty crosssectionbyahorizontalplaneisanellipse.Acrosssectionbyaverticalplaneisaparabola. Right :Ahyperbolic paraboloid(asaddleŽ).Acrosssectionbyahorizontalplane isahyperbola.Acrosssectionbyaverticalplaneisa parabola.

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78.QUADRICSURFACES89 | y0| b and | z0| c .Thus, thecharacteristicgeometricalpropertyof anellipsoidisthatitstracesareellipses .AParaboloid.Suppose c> 0.Thentheparaboloidliesabovethe xy planebecauseithasnotraceinallhorizontalplanesbelowthe xy plane, z = z0< 0.Inthe xy plane,itstracecontainsjusttheorigin. Horizontaltraces(intheplanes z = z0)oftheparaboloidareellipses: x2 a2+ y2 b2= k or x2 ( a k )2+ y2 ( b k )2=1 ,k = z0 c ,c> 0 Theellipsesbecomewideras z0getslargerbecausetheirmajoraxes a k and b k growwithincreasing k .Verticaltraces(tracesinthe planes x = x0and y = y0)areparabolas: z Š kc = c b2y2,k = x2 0 a2and z Š kc = c a2x2,k = y2 0 b2. Similarly,aparaboloidwith c< 0liesbelowthe xy plane.So the characteristicgeometricalpropertyofaparaboloidisthatitshorizontaltracesareellipses,whileitsverticalonesareparabolas (seeFigure11.23,leftpanel).If a = b ,theparaboloidisalsocalleda circular paraboloid becauseitshorizontaltracesarecircles.ADoubleCone.Thehorizontaltracesareellipses: x2 a2+ y2 b2= k2or x2 ( ak )2+ y2 ( bk )2=1 ,k = z0 c Soas | z0| grows,thatis,asthehorizontalplanemovesawayfromthe xy plane( z =0),theellipsesbecomewider.Inthe xy plane,thecone hasatracethatconsistsofasinglepoint(theorigin).Thevertical tracesintheplanes x =0and y =0areapairoflines z = ( c/b ) y and z = ( c/a ) x. Furthermore,thetraceinanyplanethatcontainsthe z axisisalsoa pairofstraightlines.Indeed,takeparametricequationsofalineinthe xy planethroughtheorigin, x = v1t y = v2t .Thenthe z coordinateof anypointofthetraceoftheconeintheplanethatcontainsthe z axis andthislinesatis“estheequation z2/c2=[( v1/a )2+( v2/b )2] t2or z = v3t ,where v3= c ( v1/a )2+( v2/b )2.Sothepointsofintersection, x = v1t y = v2t z = v3t forallreal t ,formtwostraightlines throughtheorigin.Givenanellipseinaplane,consideralinethrough thecenteroftheellipsethatisperpendiculartotheplane.Fixapoint P onthislinethatdoesnotcoincidewiththepointofintersectionof thelineandtheplane.Thenadoubleconeisthesurfacethatcontains

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9011.VECTORSANDTHESPACEGEOMETRY alllinesthrough P andpointsoftheellipse.Thepoint P iscalled the vertex ofthecone.So thecharacteristicgeometricalpropertyofa coneisthathorizontaltracesareellipses;itsverticaltracesinplanes throughtheaxisoftheconearestraightlines (seeFigure11.21,right panel). Verticaltracesintheplanes x = x0 =0and y = y0 =0are hyperbolas y2/b2Š z2/c2= k ,where k = Š x2 0/a2,and x2/a2Š z2/c2= k where k = Š y2 0/b2.Recallinthisregard conicsections studiedin CalculusII. If a = b ,theconeiscalleda circularcone .Inthiscase,vertical tracesintheplanescontainingtheconeaxisareapairoflineswith thesameslopethatisdeterminedbytheangle betweentheaxisof theconeandanyoftheselines: c/b = c/a =cot .Theequationofa circulardoubleconecanbewrittenas z2=cot2( )( x2+ y2) 0 < 0.If z0> 0(horizontalplanesbelowthe xy plane),then k> 0.Inthiscase,thehyperbolasaresymmetricaboutthe x axis,and theirbrancheslieeitherin x> 0orin x< 0(i.e.,theydonotintersect the y axis)because x2/a2= y2/b2+ k> 0( x doesnotvanishforany y ).If z0< 0,then k< 0.Inthiscase,thehyperbolasaresymmetric aboutthe y axis,andtheirbrancheslieeitherin y> 0orin y< 0 (i.e.,theydonotintersectthe x axis)because y2/b2= x2/a2Š k> 0 ( y cannotvanishforany x ).Verticaltracesintheplanes x = x0and y = y0are upward and downward parabolas,respectively: z Š z0= Š c b2y2,z0= cx2 0 a2and z Š z0= c a2x2,z0= Š cy2 0 b2. Taketheparabolictraceinthe zx plane z =( c/a2) x2(i.e.,inthe plane y = y0=0).Thetracesintheperpendicularplanes x = x0are parabolaswhoseverticesare( x0, 0 ,z0),where z0=( c/a2) x2 0,andhence lieontheparabola z =( c/a2) x2inthe zx plane.Thisobservation suggeststhatthehyperbolicparaboloidissweptbytheparabolain the zy plane, z = Š ( c/b2) y2,whenthelatterismovedparallelsothat itsvertexremainsontheparabola z =( c/a2) x2intheperpendicular

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78.QUADRICSURFACES91 plane.TheobtainedsurfacehasthecharacteristicshapeofasaddleŽ (seeFigure11.23,rightpanel).AHyperboloidofOneSheet.Tracesinhorizontalplanes z = z0are ellipses: x2 a2+ y2 b2= k2or x2 ( ka )2+ y2 ( kb )2=1 ,k = 1+ z2 0/c2 1 Theellipseisthesmallestinthe xy plane( z0=0and k =1).The majoraxesoftheellipse, ka and kb ,growasthehorizontalplanegets awayfromthe xy planebecause k increases.Thesurfacelookslike atubewithever-expandingellipticcrosssection.Theverticalcross sectionofthetubeŽbytheplanes x =0and y =0arehyperbolas: y2 b2Š z2 c2=1and x2 a2Š z2 c2=1 So thecharacteristicgeometricalpropertyofahyperboloidofonesheet isthatitshorizontaltracesareellipsesanditsverticaltracesarehyperbolas (seeFigure11.22,leftpanel).AHyperboloidofTwoSheets.Adistinctivefeatureofthissurfaceis thatitconsistsoftwosheets(seeFigure11.22,rightpanel).Indeed, thetraceintheplane z = z0satis“estheequation x2 a2+ y2 b2= z2 0 c2Š 1 whichhasnosolutionif z2 0/c2Š 1 < 0or Š cc or z = z0< Š c areellipseswhose majoraxesincreasewithincreasing | z0| .Theuppersheettouchesthe plane z = c atthepoint(0 0 ,c ),whilethelowersheettouchesthe plane z = Š c atthepoint(0 0 Š c ).Thesepointsarecalled vertices ofahyperboloidoftwosheets.Verticaltracesintheplanes x =0and y =0arehyperbolas: z2 c2Š y2 b2=1and z2 c2Š x2 a2=1 Thus,thecharacteristicgeometricalpropertiesofhyperboloidsofone sheetandtwosheetsaresimilar,apartfromthefactthatthelatter oneconsistsoftwosheets.Also,intheasymptoticregion | z | c ,the hyperboloidsapproachthesurfaceofthedoublecone.Indeed,inthis case, z2/c2 1,andhencetheequations x2/a2+ y2/b2= 1+ z2/c2canbewellapproximatedbythedouble-coneequation( 1canbe

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9211.VECTORSANDTHESPACEGEOMETRY neglectedontherightsideoftheequations).Intheregion z> 0,the hyperboloidofonesheetapproachesthedoubleconefrombelow,while thehyperboloidoftwosheetsapproachesitfromabove.For z< 0, theconverseholds.Inotherwords,thehyperboloidoftwosheetslies insideŽthecone,whilethehyperboloidofonesheetliesoutsideŽit.78.4.ShiftedQuadricSurfaces.Iftheoriginofacoordinatedsystem isshiftedtoapoint( x0,y0,z0)withoutanyrotationofthecoordinate axes,thenthecoordinatesofapointinspacearetranslated( x,y,z ) ( x Š x0,y Š y0,z Š z0).Therefore,anyequationoftheform f ( x,y,z )=0 becomes f ( x Š x0,y Š y0,z Š z0)=0inthenewcoordinatesystem.If theequation f ( x,y,z )=0de“nesasurfaceinspace,thentheequation f ( x Š x0,y Š y0,z Š z0)=0de“nestheverysamesurfacethathasbeen translatedasthewhole(eachpointofthesurfaceisshiftedbythesame vector x0,y0,z0 ).Forexample,theequation ( x Š x0)2 a2+ ( y Š y0)2 b2= ( z Š z0)2 c2describesadoubleellipticconewhoseaxisisparalleltothe z axisand whosevertexisat( x0,y0,z0).Equationsofshiftedquadricsurfacescan bereducedtothestandardformbycompletingthesquares. Example 11.26 Classifythequadricsurface 9 x2+36 y2+4 z2Š 18 x +72 y +16 z +25=0 Solution: Letuscompletethesquaresforeachofthevariables: 9 x2Š 18 x =9( x2Š 2 x )=9[( x Š 1)2Š 1]=9( x Š 1)2Š 9 36 y2+72 y =36( y2+2 y )=36[( y +1)2Š 1]=36( y +1)2Š 36 4 z2+16 z =4( z2+4 z )=4[( z +2)2Š 4]=4( z +2)2Š 16 Theequationbecomes9( x Š 1)2+36( y +1)2+4( z +2)2=36,and,by dividingitby36,thestandardformisobtained ( x Š 1)2 16 +( y +1)2+ ( z +2)2 9 =1 Thisequationdescribesanellipsoidwiththecenterat(1 Š 1 Š 2)and majoraxes a =4, b =1,and c =3. Example 11.27 Classifythesurface x2+2 y2Š 4 y Š 2 z =0 Solution: Bycompletingthesquares 2 y2Š 4 y =2( y2Š 2 y )=2[( y Š 1)2Š 1] ,

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78.QUADRICSURFACES93 theequationcanwrittenintheform x2+2( y Š 1)2Š 2 Š 2 z =0or z +1= x2 2 +( y Š 1)2, whichisanellipticparaboloidwiththevertexat(0 1 Š 1)becauseit isobtainedfromthestandardequation z = x2/ 2+ y2bytheshiftof thecoordinatesystem( x,y,z ) ( x,y Š 1 ,z +1). Example 11.28 Classifythesurface x2Š 4 y2+ z2Š 2 x Š 8 z +1=0 Solution: Bycompletingthesquares,theequationistransformedto ( x Š 1)2Š 1 Š 4 y2+4( z +1)2Š 4+1=0 ( x Š 1)2 4 +( z +1)2Š y2=1 whichisahyperboloidofonesheetwhoseaxisisthelinethrough (1 0 Š 1)thatisparalleltothe y axis. Example 11.29 Useanappropriaterotationinthe xy planeto reducetheequation z =2 xy tothestandardformandclassifythe surface. Solution: Let( x,y)becoordinatesintherotatedcoordinatesystem throughtheangle asdepictedinFigure11.3(rightpanel).InStudy Problem11.2,theoldcoordinates( x,y )areexpressedviathenewones ( x,y): x = xcos Š ysin ,y = ycos + xsin Inthenewcoordinatesystem,theequation z =2 xy =2 x 2cos sin Š 2 y 2cos sin +2 xy(cos2 Š sin2 ) wouldhavethestandardformifthecoecientat xyvanishes.Soput = / 4.Then2sin cos =sin(2 )=1and z = x 2Š y 2,whichis thehyperbolicparaboloid. 78.5.StudyProblems.Problem11.30. Classifythequadricsurface 3 x2+3 z2Š 2 xz =4 Solution: Theequationdoesnotcontainonevariable(the y coordinate).Thesurfaceisacylinderparalleltothe y axis.Todetermine thetypeofcylinder,considerarotationofthecoordinatesysteminthe xz planeandchoosetherotationanglesothatthecoecientatthe xz termvanishesinthetransformedequation.Accordingto(11.20), A = B =3, p = Š 2,andhence = / 4.Then A=( A + B Š p ) / 2=4

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9411.VECTORSANDTHESPACEGEOMETRY and B=( A + B + p ) / 2=2.So,inthenewcoordinates,theequationbecomes4 x2+2 z2=4or x2+ z2/ 2=1,whichisanellipsewith semiaxes a =1and b = 2.Thesurfaceisanellipticcylinder. Problem11.31. Classifythequadricsurface x2Š 2 x + y + z =0 Solution: Bycompletingthesquares,theequationcanbetransformedintotheform( x Š 1)2+( y Š 1)+ z =0.Aftershiftingtheorigin tothepoint(1 1 0),theequationbecomes x2+ y Š z =0.Considerrotationsofthecoordinatesystemaboutthe x axis: y cos y +sin z z cos z Š sin y .Underthisrotation, y Š z (cos +sin ) y + (sin Š cos ) z .Therefore,for = / 4,theequationassumesoneof thestandardforms x2+ 2 y =0,whichcorrespondstoaparabolic cylinder. Problem11.32. Classifythequadricsurface x2+ z2Š 2 x +2 z Š y =0 Solution: Bycompletingthesquares,theequationcanbetransformedintotheform( x Š 1)2+( z +1)2Š ( y +2)=0.Thelatter canbebroughtintooneofthestandardformsbyshiftingtheoriginto thepoint(1 Š 2 Š 1): x2+ z2= y ,whichisacircularparaboloid.Its symmetryaxisisparalleltothe y axis(thelineofintersectionofthe planes x =1and z = Š 1),anditsvertexis(1 Š 2 Š 1). Figure11.24.AnillustrationtoStudyProblem11.33. Thevector urotatesabouttheverticallinesothattheline through(1 2 0)andparallelto vsweepsadoubleconewith thevertexat(1 2 0).Problem11.33. Sketchand/ordescribethesetofpointsinspace formedbyafamilyoflinesthroughthepoint (1 2 0) andparallelto v= cos sin 1 ,where [0 2 ] labelslinesinthefamily.

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78.QUADRICSURFACES95 Solution: Theparametricequationsofeachlineare x =1+ t cos y =2+ t sin ,and z = t .Therefore,( x Š 1)2+( y Š 2)2= z2forall valuesof t and .Thus,thelinesformadoubleconewhoseaxisis paralleltothe z axisandwhosevertexis(1 2 0).Alternatively,one couldnoticethatthevector vrotatesaboutthe z axisas changes. Indeed,put v= u + ez,where u = cos sin 0 istheunitvectorin the xy planeasshowninFigure11.24.Itrotatesas changes,making afullturnas increasesfrom0to2 .Sothesetinquestioncanbe obtainedbyrotatingaparticularline,say,theonecorrespondingto =0,abouttheverticallinethrough(1 2 0).Thelinesweepsthe doublecone. 78.6.Exercises.(1) Usetracestosketchandidentifyeachofthefollowingsurfaces: (i) y2= x2+9 z2(ii) y = x2Š z2(iii)4 x2+2 y2+ z2=4 (iv) x2Š y2+ z2= Š 1 (v) y2+4 z2=16 (vi) x2Š y2+ z2=1 (vii) x2+4 y2Š 9 z2+1=0 (viii) x2+ z =0 (ix) x2+9 y2+ z =0 (x) y2Š 4 z2=16 (2) Reduceeachofthefollowingequationstooneofthestandardform, classifythesurface,andsketchit: (i) x2+ y2+4 z2Š 2 x +4 y =0 (ii) x2Š y2+ z2+2 x Š 2 y +4 z +2=0 (iii) x2+4 y2Š 6 x + z =0 (iv) y2Š 4 z2+2 y Š 16 z =0 (v) x2Š y2+ z2Š 2 x +2 y =0 (3) Userotationsintheappropriatecoordinateplanetoreduceeach ofthefollowingequationstooneofthestandardformandclassifythe surface: (i)6 xy + x2+ y2=1 (ii)3 y2+3 z2Š 2 yz =1 (iii) x Š yz =0 (iv) xy Š z2=0 (v)2 xz + x2Š y2=0

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9611.VECTORSANDTHESPACEGEOMETRY (4) Findanequationforthesurfaceobtainedbyrotatingtheline y =2 x aboutthe y axis.Classifythesurface. (5) Findanequationforthesurfaceobtainedbyrotatingthecurve y =1+ z2aboutthe y axis.Classifythesurface. (6) Findequationsforthefamilyofsurfacesobtainedbyrotatingthe curves x2Š 4 y2= k aboutthe y axiswhere k isreal.Classifythe surfaces. (7) Findanequationforthesurfaceconsistingofallpointsthatare equidistantfromthepoint(1 1 1)andtheplane z =2. (8) Sketchthesolidregionboundedbythesurface z = x2+ y2from belowandby x2+ y2+ z2Š 2 z =0fromabove. (9) Sketchthesolidregionboundedbythesurfaces y =2 Š x2Š z2, y = x2+ z2Š 2,andliesinsidethecylinder x2+ z2=1. (10) Sketchthesolidregionboundedbythesurfaces x2+ y2= R2and x2+ z2= R2. (11) Findanequationforthesurfaceconsistingofallpoints P for whichthedistancefrom P tothe y axisistwicethedistancefrom P tothe zx plane.Identifythesurface. (12) Showthatifthepoint( a,b,c )liesonthehyperbolicparaboloid z = y2Š x2,thenthelinesthrough( a,b,c )andparallelto v = 1 1 2( b Š a ) and u = 1 Š 1 Š 2( b Š a ) bothlieentirelyon thisparaboloid.Deducefromthisresultthatthehyperbolicparaboloidcanbegeneratedbythemotionofastraightline.Showthat hyperboloidsofonesheet,cones,andcylinderscanalsobeobtainedby themotionofastraightline. Remark. Thefactthathyperboloidsofonesheetaregeneratedbythe motionofastraightlineisusedtoproducegeartransmissions.The cogsofthegearsarethegeneratinglinesofthehyperboloids. (13) Findanequationforthecylinderofradius R whoseaxisgoes throughtheoriginandisparalleltoavector v (14) Showthatthecurveofintersectionofthesurfaces x2Š 2 y2+3 z2Š 2 x + y Š z =1and2 x2Š 4 y2+6 z2+ x Š y +2 z =4liesinaplane. (15) Whatarethecurvesthatboundtheprojectionsoftheellipsoid x2+ y2+ z2Š xy =1onthecoordinateplanes?

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CHAPTER12 VectorFunctions 79.CurvesinSpaceandVectorFunctions Todescribethemotionofapointlikeobjectinspace,itsposition vectorsmustbespeci“edateverymomentoftime.Avectorisde“ned bythreecomponentsinacoordinatesystem.Therefore,themotionof theobjectcanbedescribedbyanorderedtripleofreal-valuedfunctionsoftime.Thisobservationleadstotheconceptofvector-valued functionsofarealvariable. Definition 12.1 (VectorFunction) Let D beasetofrealnumbers.Avectorfunction r ( t ) ofarealvariable t isarulethatassignsavectortoeveryvalueof t from D .Theset D iscalledthe domain ofthevectorfunction. Themostcommonlyusedrulestode“neavectorfunctionarealgebraicrulesthatspecifycomponentsofavectorfunctioninacoordinatesystemasfunctionsofarealvariable: r ( t )= x ( t ) ,y ( t ) ,z ( t ) .For example, r ( t )= 1 Š t, ln( t ) ,t2 or x ( t )= 1 Š t,y ( t )=ln( t ) ,z ( t )= t2. Unlessspeci“edotherwise,thedomainofthevectorfunctionistheset D ofallvaluesof t atwhichthealgebraicrulemakessense;thatis, allthreecomponentscanbecomputedforany t from D .Intheabove example,thedomainof x ( t )is Š
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9812.VECTORFUNCTIONS Figure12.1.Left :Theterminalpointofavector r ( t ) whosecomponentsarecontinuousfunctionsof t tracesout acurveinspace. Right :Graphingaspacecurve.Drawa curveinthe xy planede“nedbytheparametricequations x = x ( t ), y = y ( t ).Itistracedoutbythevector R ( t )= x ( t ) ,y ( t ) 0 .Thisplanarcurvede“nesacylindricalsurface inspaceinwhichthespacecurveinquestionlies.Thespace curveisobtainedbyraisingorloweringthepointsofthe planarcurvealongthesurfacebytheamount z ( t ),thatis, r ( t )= R ( t )+ e3z ( t ).Inotherwords,thegraph z = z ( t )is wrappedaroundthecylindricalsurface.r ( t )= r0+ t v .Thus,the range ofavectorfunctionde“nesacurvein space,andagraphofavectorfunctionisacurveinspace.79.1.GraphingSpaceCurves.Tovisualizetheshapeofacurve C tracedoutbyavectorfunction,itisconvenienttothinkabout r ( t ) asatrajectoryofmotion.Thepositionofaparticleinspacemaybe determinedbyitspositioninaplaneanditsheightrelativetothat plane.Forexample,thisplanecanbechosentobethe xy plane.Then r ( t )= x ( t ) ,y ( t ) ,z ( t ) = x ( t ) ,y ( t ) 0 + 0 0 ,z ( t ) = R ( t )+ z ( t ) e3. Considerthecurvede“nedbytheparametricequations x = x ( t ), y = y ( t )inthe xy plane.Onecanmarkafewpointsalongthe curvecorrespondingtoparticularvaluesof t ,say, Pnwithcoordinates ( x ( tn) ,y ( tn)), n =1 2 ,...,N .Thenthecorrespondingpointsofthe curve C areobtainedfromthembymovingthepoints Pnalongthe directionnormaltotheplane(i.e.,alongthe z axisinthiscase)bythe

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79.CURVESINSPACEANDVECTORFUNCTIONS99 amount z ( tn);thatis, Pngoesupif z ( tn) > 0ordownif z ( tn) < 0.In otherwords,asaparticlemovesalongthecurve x = x ( t ), y = y ( t ), itascendsordescendsaccordingtothecorrespondingvalueof z ( t ). Thecurvecanalsobevisualizedbyusingapieceofpaper.Consider ageneralcylinderwiththehorizontaltracebeingthecurve x = x ( t ), y = y ( t ),likeawalloftheshapede“nedbythiscurve.Thenmakea graphofthefunction z ( t )onapieceofpaper(wallpaper)andglueitto thewallsothatthe t axisofthegraphisgluedtothecurve x = x ( t ), y = y ( t )whileeachpoint t onthe t axiscoincideswiththecorrespondingpoint( x ( t ) ,y ( t ))ofthecurve.Aftersuchaprocedure,thegraph of z ( t )alongthewallwouldcoincidewiththecurve C tracedoutby r ( t ).TheprocedureisillustratedinFigure12.1(rightpanel). Example 12.1 Graphthevectorfunction r = cos t, sin t,t ,where t rangesovertherealline. Solution: Itisconvenienttorepresent r ( t )asthesumofavectorin the xyplaneandavectorparalleltothe z axis.Inthe xy plane,the curve x =cos t y =sin t isthecircleofunitradiustracedoutcounterclockwisesothatthepoint(1 0 0)correspondsto t =0.Thecircular motionisperiodicwithperiod2 .Theheight z ( t )= t riseslinearly asthepointmovesalongthecircle.Startingfrom(1 0 0),thecurve makesoneturnonthesurfaceofthecylinderofunitradiusclimbing upby2 .Thinkofapieceofpaperwithastraightlinedepictedon itthatiswrappedaroundthecylinder.Thus,thecurvetracedby r ( t ) liesonthesurfaceofacylinderofunitradiusandperiodicallywinds aboutitclimbingby2 perturn.Suchacurveiscalleda helix .The procedureisshowninFigure12.2. 79.2.LimitsandContinuityofVectorFunctions.Definition 12.2 (LimitofaVectorFunction) Avector r0iscalledthe limit ofavectorfunction r ( t ) as t t0if limt t0 r ( t ) Š r0 =0; thelimitisdenotedas limt t0r ( t )= r0. Theleftandrightlimits,limt tŠ 0r ( t )andlimt t+ 0r ( t ),arede“ned similarly.Thisde“nitionsaysthatthelengthornormofthevector r ( t ) Š r0approaches0as t tendsto t0.Thenormofavectorvanishes ifandonlyifthevectoristhezerovector.Therefore,thefollowing theoremholds.

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10012.VECTORFUNCTIONS Figure12.2.Graphingahelix. Left :Thecurve R ( t )= cos t, sin t, 0 isacircleofunitradius,tracedoutcounterclockwise.Sothehelixliesonthecylinderofunitradius whosesymmetryaxisisthe z axis. Middle :Thegraph z = z ( t )= t isastraightlinethatde“nestheheightofhelix pointsrelativetothecircletracedoutby R ( t ). Right :The graphofthehelix r ( t )= R ( t )+ z ( t ) e3.As R ( t )traverses thecircle,theheight z ( t )= t riseslinearly.Sothehelixcan beviewedasastraightlinewrappedaroundthecylinder.Theorem 12.1 (LimitofaVectorFunction) Let r ( t )= x ( t ) ,y ( t ) ,z ( t ) andlet r0= x0,y0,z0 .Thenthelimitofa vectorfunctionexistsifandonlyifthelimitsofitscomponentsexist: limt t0r ( t )= r0 limt t0x ( t )= x0, limt t0y ( t )= y0, limt t0z ( t )= z0. Thistheoremreducestheproblemof“ndingthelimitofavector functiontotheproblemof“ndingthelimitsofthreeordinaryfunctions. Example 12.2 Let r ( t )= sin( t ) /t,t ln t, ( etŠ 1 Š t ) /t2 .Find thelimitof r ( t ) as t 0+orshowthatitdoesnotexist. Solution: Theexistenceofthelimitsofthecomponentsofthegiven vectorfunctioncanbeinvestigatedbylHospitalsrule: limt 0+sin t t =limt 0+(sin t ) ( t )=limt 0+cos t 1 =1 limt 0+t ln t =limt 0+ln t tŠ 1=limt 0+(ln t ) ( tŠ 1)=limt 0+tŠ 1 Š tŠ 2= Š limt 0+t =0 limt 0+etŠ 1 Š t t2=limt 0+etŠ 1 2 t =limt 0+et 2 = 1 2 wherelHospitalsrulehasbeenusedtwicetocalculatethelastlimit. Therefore,limt 0+r ( t )= 1 0 1 / 2

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79.CURVESINSPACEANDVECTORFUNCTIONS101 Definition 12.3 (ContinuityofaVectorFunction) Avectorfunction r ( t ) t [ a,b ] ,issaidtobecontinuousat t = t0 [ a,b ] if limt t0r ( t )= r ( t0) Avectorfunction r ( t ) iscontinuousintheinterval [ a,b ] ifitiscontinuousateverypointof [ a,b ] ByTheorem12.1, avectorfunctioniscontinuousifandonlyifall itscomponentsarecontinuousfunctions Example 12.3 Let r ( t )= sin(2 t ) /t,t2,et forall t =0 and r (0)= 1 0 1 .Determinewhetherthisvectorfunctioniscontinuous. Solution: Thecomponents y ( t )= t2and z ( t )= etarecontinuousfor allreal t and y (0)=0and z (0)=1.Thecomponent x ( t )=sin(2 t ) /t iscontinuousforall t =0becausetheratiooftwocontinuousfunctions iscontinuous.BylHospitalsrule, limt 0x ( t )=limt 0sin(2 t ) t =limt 02cos(2 t ) 1 =2 limt 0x ( t ) = x (0)=1; thatis, x ( t )isnotcontinuousat t =0.Thus, r ( t )iscontinuous everywhere,but t =0. 79.3.SpaceCurvesandContinuousVectorFunctions.Acurveconnectingtwopointsinspaceasapointsetcanbeobtainedasacontinuous transformation(oradeformationwithoutbreaking)ofastraightline segmentinspace.Conversely,everysuchspacecurvecanbecontinuouslydeformedtoastraightlinesegment.So acurveconnectingtwo pointsinspaceisacontinuousdeformationofastraightlinesegment, andthisdeformationhasacontinuousinverse Astraightlinesegmentcanbeviewedasaninterval a t b (a setofrealnumbersbetween a and b ).Itscontinuousdeformationcan bedescribedbyacontinuousvectorfunction r ( t )on[ a,b ].So therange ofacontinuousvectorfunctionde“nesacurveinspace .Conversely, givenacurve C asapointsetinspace,onemightaskthequestion: Whatisavectorfunctionthattracesoutagivencurveinspace?The answertothisquestionisnotunique.Forexample,aline L asapoint setinspaceisuniquelyde“nedbyitsparticularpointandavector v paralleltoit.If r1and r2arepositionvectorsoftwoparticularpoints of L ,thenbothvectorfunctions r1( t )= r1+ t v and r2( t )= r2Š 2 t v traceoutthesameline L becausethevectors Š 2 v and v areparallel. Thefollowing,moresophisticatedexampleisalsoofinterest.Supposeonewantsto“ndavectorfunctionthattracesoutasemicircleof

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10212.VECTORFUNCTIONS radius R .Letthesemicirclebepositionedintheupperpartofthe xy plane: x2+ y2= R2and y 0.Thefollowingthreevectorfunctions traceoutthesemicircle: r1( t )= t, R2Š t2, 0 Š R t R, r2( t )= R cos t,R sin t, 0 0 t r3( t )= Š R cos t,R sin t, 0 0 t Thisiseasytoseebynotingthatthe y componentsarenonnegativein thespeci“edintervalsandthenormofthesevectorfunctionsisconstant foranyvalueof t : ri( t ) 2= R2or x2 i( t )+ y2 i( t )= R2,where i =1 2 3. Thelattermeansthattheendpointsofthevectors ri( t )alwaysremain onthecircleofradius R .Itcanthereforebeconcludedthatthereare manyvectorfunctionswhoserangesde“nethesamecurveinspace. Anotherobservationisthattherearevectorfunctionsthattraceout thesamecurveinoppositedirectionsat t increasesfromitssmallest value a toitslargestvalue b .Intheaboveexample,thevectorfunction r2( t )tracesoutthesemicirclecounterclockwise,whilethefunctions r1( t )and r3( t )dosoclockwise.Soavectorfunctionde“nesthe orientation ofacurve.However,thisnotionoftheorientationofa curveshouldberegardedwithcautionbecauseavectorfunctionmay traverseitsrange(orapartofit)severaltimes.Forexample,thevectorfunction r ( t )= R cos t,R | sin t | 0 tracesoutthesemicircletwice, backandforth,when t rangesfrom0to2 .Thevectorfunction r ( t )=( t2,t2,t2)iscontinuousontheinterval[ Š 1 1]andtracesout thestraightlinesegment, x = y = z ,betweenthepoints(0 0 0)and (1 1 1)twice. Toemphasizethenoteddierencesbetweenspacecurvesaspoint setsandcontinuousvectorfunctions,thenotionofa parametric curve isintroduced. Definition 12.4 (ParametricCurve). Acontinuousvectorfunctiononanintervaliscalleda parametric curve Ifacontinuousvectorfunction r ( t )= x ( t ) ,y ( t ) ,z ( t ) a t b establishesaone-to-onecorrespondencebetweenaninterval[ a,b ]anda spacecurve C ,thenthevectorfunctionisalsocalleda parameterization ofthecurve C ,theequations x = x ( t ), y = y ( t ),and z = z ( t )arecalled parametricequations of C ,and t iscalleda parameter .Asnoted, aparameterizationofagivenspacecurveisnotunique,andthere aredierentparametricequationsthatdescribetheverysamespace

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79.CURVESINSPACEANDVECTORFUNCTIONS103 curve.Acurveissaidtobe simple if,looselyspeaking,itdoesnot intersectitself.Tomakethisnotionprecise,itisrephrasedintermsof parametriccurves.Aparametriccurve r ( t )iscalled simple onaclosed interval[ a,b ]if r ( t1) = r ( t2)if t1and t2liein[ a,b ]and t1
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10412.VECTORFUNCTIONS circleistracedoutclockwise,whereasitistracedoutcounterclockwiseintheformerparameterization.Consequently,thevectorfunction r ( t )= R cos t, Š R sin t,ht/ (2 ) alsotracesoutahelixwiththerequiredproperties.Thetwohelicesaredierentdespitetheirsharing thesameinitialandterminalpoints.Onehelixwindsaboutthe z axis clockwisewhiletheothercounterclockwise. Problem12.2. Sketchand/ordescribethecurvetracedoutbythe vectorfunction r ( t )= cos t, sin t, sin(4 t ) if t rangesintheinterval [0 2 ] Solution: Thevectorfunction R ( t )= cos t, sin t, 0 traversesthe circleofunitradiusinthe xy plane,counterclockwise,startingfrom thepoint(1 0 0).As t rangesoverthespeci“edinterval,thecircleis traversedonlyonce.Theheight z ( t )=sin(4 t )hasaperiodof2 / 4= / 2.Therefore,thegraphofsin(4 t )makesfourupsandfourdowns if0 t 2 .Thecurve r ( t )= R ( t )+ e3z ( t )lookslikethegraphof sin(4 t )wrappedaroundthecylinderofunitradius.Itmakesoneup andonedownineachquarterofthecylinder.Theprocedureisshown inFigure12.3. Problem12.3. Sketchand/ordescribethecurvetracedoutbythe vectorfunction r ( t )= t cos t,t sin t,t Solution: Thecomponentsof r ( t )satisfytheequation x2( t )+ y2( t )= z2( t )forallvaluesof t .Therefore,thecurveliesonthedoublecone x2+ y2= z2.Since x2( t )+ y2( t )= t2,theparametriccurve x = x ( t ), y = y ( t )inthe xy planeisaspiral(thinkofarotationalmotionabout theoriginsuchthattheradiusincreaseslinearlywiththeangleof rotation).If t increasesfrom t =0,thecurveinquestionistracedbya pointthatriseslinearlywiththedistancefromtheoriginasittravels alongthespiral.If t decreasesfrom t =0,insteadofrising,thepoint woulddescend( z ( t )= t< 0).Sothecurvewindsabouttheaxisofthe doubleconewhileremainingonitssurface.Theprocedureisshownin Figure12.4. Problem12.4. Findtheportionoftheelliptichelix r ( t )= 2cos( t ) t, sin( t ) thatliesinsidetheellipsoid x2+ y2+4 z2=13 Solution: Thehelixhereiscalled elliptic becauseitliesonthesurface ofanellipticcylinder.Indeed,inthe xz plane,theparametriccurve x =2cos( t ), z =sin( t )traversestheellipse x2/ 4+ z2=1becausethe latterequationissatis“edforallreal t .Therefore,thecurveremains onthesurfaceoftheellipticcylinderparalleltothe y axis.Oneturn aroundtheellipseoccursas t changesfrom0to2becausethefunctions

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79.CURVESINSPACEANDVECTORFUNCTIONS105 Figure12.3.IllustrationtoStudyProblem12.2. Left : Thecurveliesonthecylinderofunitradius.Itmaybe viewedasthegraphof z =sin(4 t )ontheinterval0 t 2 wrappedaroundthecylinder. Topright :Thecircle tracedoutby R ( t )= cos t, sin t, 0 .Itde“nesthecylindrical surfaceonwhichthecurvelies. Bottomright :Thegraph z = z ( t )=sin(4 t ),whichde“nestheheightofpointsofthe curverelativetothecircleinthe xy plane.Thedotsonthe t axisindicatethepointsofintersectionofthecurvewiththe xy planeinthe“rstquadrant.cos( t )andsin( t )havetheperiod2 / =2.Thehelixrisesby2along the y axisperturnbecause y ( t )= t .Now,tosolvetheproblem,onehas to“ndthevaluesof t atwhichthehelixintersectstheellipsoid.The intersectionhappenswhenthecomponentsof r ( t )satisfytheequation oftheellipsoid,thatis,when x2( t )+ y2( t )+4 z2( t )=1or4+ t2=13 andhence t = 3.Thepositionvectorsofthepointsofintersection are r ( 3)= Š 2 3 0 .Theportionofthehelixthatliesinsidethe ellipsoidcorrespondstotherange Š 3 t 3. Problem12.5. Considertwocurves C1and C2tracedoutbythe vectorfunctions r1( t )= t2,t,t2+2 t Š 8 and r2( s )= 8 Š 4 s, 2 s,s2+ s Š 2 ,respectively.Dothecurvesintersect?Ifso,“ndthepointsof intersection.Supposetwoparticleshavethetrajectories r1( t ) and r2( t ) where t istime.Dotheparticlescollide? Solution: Thecurvesintersectiftherearevaluesofthepair( t,s ) suchthat r1( t )= r2( s ).Thisvectorequationisequivalenttothe

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10612.VECTORFUNCTIONS Figure12.4.IllustrationtoStudyProblem12.3. Left : Theheightofthegraphrelativetothe xy plane(top).The curve R ( t )= t cos t,t sin t, 0 .For t 0,itlookslikean unwindingspiral(bottom). Right :For t> 0,thecurveis traversedbythepointmovingalongthespiralwhilerising linearlyabovethe xy planewiththedistancetraveledalong thespiral.Itcanbeviewedasastraightlinewrappedaround thecone x2+ y2= z2.systemofthreeequations r1( t )= r2( s ) x1( t )= x2( s ) y1( t )= y2( s ) z1( t )= z2( s ) t2=8 Š 4 s t =2 s t2+2 t Š 8= s2+ s Š 2 Substitutingthesecondequation t =2 s intothe“rstequation,one “ndsthat(2 s )2=8 Š 4 s whosesolutionsare s = Š 2and s =1.Onehas yettoverifythatthethirdequationholdsforthepairs( t,s )=( Š 4 Š 2) and( t,s )=(2 1)(otherwise,the z componentsdonotmatch).Asimplecalculationshowsthatindeedbothpairssatisfytheequation.So thepositionvectorsofthepointsofintersectionare r1( Š 4)= r2( Š 2)= 16 Š 4 0 and r1(2)= r2(1)= 4 2 0 .Althoughthecurvesalong whichtheparticlestravelintersect,thisdoesnotmeanthattheparticleswouldnecessarilycollidebecausetheymaynotarriveatapoint ofintersectionatthesamemomentoftime,justliketwocarstraveling alongintersectingstreetsmayormaynotcollideatthestreetintersection.Thecollisionconditionismorerestrictive, r1( t )= r2( t )(i.e., thetime t mustsatisfythreeconditions).Fortheproblemathand, theseconditionscannotbeful“lledforany t because,amongallthe

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79.CURVESINSPACEANDVECTORFUNCTIONS107 solutionsof r1( t )= r2( s ),thereisnosolutionforwhich t = s .Thus, theparticlesdonotcollide. Problem12.6. Findavectorfunctionthattracesoutthecurveof intersectionoftheparaboloid z = x2+ y2andtheplane 2 x +2 y + z =2 counterclockwiseasviewedfromthetopofthe z axis. Solution: Onehasto“ndthecomponents x ( t ), y ( t ),and z ( t )such thattheysatisfytheequationsoftheparaboloidandplanesimultaneouslyforallvaluesof t .Thisensuresthattheendpointofthevector r ( t ) remainsonbothsurfaces,thatis,tracesouttheircurveofintersection (seeFigure12.5).Consider“rstthemotioninthe xy plane.Solving theplaneequationfor z z =2 Š 2 x Š 2 y ,andsubstitutingthesolution intotheparaboloidequation,one“nds2 Š 2 x Š 2 y = x2+ y2.After completingthesquares,thisequationbecomes4=( x +1)2+( y +1)2, whichdescribesacircleofradius2centeredat( Š 1 Š 1).Byconstruction,thiscircleistheverticalprojectionofthecurveofintersection ontothe xy plane(theplane P0inFigure12.5).Itsparametricequationsmaybechosenas x = x ( t )= Š 1+2cos t y = y ( t )= Š 1+2sin t As t increasesfrom0to2 ,thecircleistracedoutcounterclockwise asrequired(theclockwiseorientationcanbeobtained,e.g.,byreversingthesignofsin t ).Theheightalongthecurveofintersection Figure12.5.IllustrationtoStudyProblem12.6.The curveisanintersectionoftheparaboloidandtheplane P .It istraversedbythepointmovingcounterclockwiseaboutthe circleinthe xy plane(indicatedby P0)andrisingsothatit remainsontheparaboloid.

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10812.VECTORFUNCTIONS relativetothe xy planeis z ( t )=2 Š 2 x ( t ) Š 2 y ( t ).Thus, r ( t )= Š 1+2cos t, Š 1+2sin t, 6 Š 2cos t Š 2sin t ,where t [0 2 ]. Problem12.7. Let v ( t ) v0and u ( t ) u0as t t0.Provethe limitlawforvectorfunctions: limt t0( v ( t ) u ( t ))= v0 u0usingonly De“nition12.2.ThenprovethislawusingTheorem12.1andbasic limitlawsforordinaryfunctions. Solution: Theideaissimilartotheproofofthebasiclimitlawsfor ordinaryfunctionsgiveninCalculusI.Onehasto“ndanupperbound for | v u Š v0 u0| intermsof v Š v0 and u Š u0 .ByDe“nition12.2, thelatterquantitiesconvergeto0as t t0.Theconclusionshould followfromthesqueezeprinciple.Considertheidentities: v u Š v0 u0=( v Š v0) u + v0 u Š v0 u0=( v Š v0) u + v0 ( u Š u0) =( v Š v0) ( u Š u0)+( v Š v0) u0+ v0 ( u Š u0) Itfollowsfromtheinequality0 | a + b || a | + | b | andtheCauchySchwarzinequality(Theorem11.2) | a b | a b that 0 | v u Š v0 u0| | ( v Š v0) ( u Š u0) | + | ( v Š v0) u0| + | v0 ( u Š u0) | v Š v0 u Š u0 + v Š v0 u0 + v0 u Š u0 ByDe“nition12.2, v Š v0 0and u Š u0 0as t t0.So therightsideoftheaboveinequalityconvergesto0.Bythesqueeze principle,itisthenconcludedthat | v u Š v0 u0| 0as t t0,which provestheassertion.AproofbasedonTheorem12.1issimpler.If vi( t ) and ui( t ), i =1 2 3,arecomponentsof v ( t )and u ( t ),respectively, then,byTheorem12.1, vi( t ) v0 iand ui( t ) u0 ias t t0.Hence, limt t0v ( t ) u ( t )=limt t0 v1( t ) u1( t )+ v2( t ) u2( t )+ v3( t ) u3( t ) =limt t0v1( t ) u1( t )+limt t0v2( t ) u2( t )+limt t0v3( t ) u3( t ) = v01u01+ v02u02+ v03u03= v0 u0, wherethebasiclimitlawsforordinaryfunctionshavebeenused. 79.5.Exercises.(1) Findthedomainofeachofthefollowingvectorfunctions: (i) r ( t )= t,t2,et (ii) r ( t )= t,t2,et (iii) r ( t )= 9 Š t2, ln t, cos t

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79.CURVESINSPACEANDVECTORFUNCTIONS109 (iv) r ( t )= ln(9 Š t2) ln | t | (1+ t ) / (2+ t ) (v) r ( t )= t Š 1 ln t, 1 Š t (2) Findeachofthefollowinglimitsorshowthatitdoesnotexist: (i)limt 1 t, 2 Š t Š t2, 1 / ( t2Š 2) (ii)limt 1 t, 2 Š t Š t2, 1 / ( t2Š 1) (iii)limt 0 et, sin t,t/ (1 Š t ) (iv)limt eŠ t, 1 /t2, 4 (v)limt eŠ t, (1 Š t2) /t2,3 t/ ( t + t ) (vi)limt Š 2 ,t2, 1 /3 t (vii)limt 0+ ( e2 tŠ 1) /t, ( 1+ t Š 1) /t,t ln t (viii)limt 0 sin2(2 t ) /t2,t2+2 (cos t Š 1) /t2 (ix)limt 0 ( e2 tŠ t ) /t,t cot t, 1+ t (x)limt e2 t/ cosh2t,t2012eŠ t,eŠ 2 tsinh2t (3) Sketcheachofthefollowingcurvesandidentifythedirectionin whichthecurveistracedoutastheparameter t increases: (i) r ( t )= t, cos(3 t ) sin(3 t ) (ii) r ( t )= 2sin(5 t ) 4 3cos(5 t ) (iii) r ( t )= 2 t sin t, 3 t cos t,t (iv) r ( t )= sin t, cos t, ln t (v) r ( t )= t, 1 Š t, ( t Š 1)2 (vi) r ( t )= t2,t, sin2( t ) (vii) r ( t )= sin t, sin t, 2cos t (4) Twoobjectsaresaidtocollideiftheyareatthesameposition atthesametime .Twotrajectoriesaresaidtointersectiftheyhave commonpoints.Let t bethephysicaltime.Lettwoobjectstravelalong thespacecurves r1( t )= t,t2,t3 and r2( t )= 1+2 t, 1+6 t, 1+14 t Dotheobjectscollide?Dotheirtrajectoriesintersect?Ifso,“ndthe collisionandintersectionpoints. (5) Findtwovectorfunctionsthattraverseagivencurve C inthe oppositedirectionsif C isthecurveofintersectionoftwosurfaces: (i) y = x2and z =1 (ii) x =sin y and z = x (iii) x2+ y2=9and z = xy (iv) x2+ y2= z2and x + y + z =1 (v) z = x2+ y2and y = x2(vi) x2/ 4+ y2/ 9=1and x + y + z =1 (vii) x2/ 2+ y2/ 2+ z2/ 9=1and x Š y =0 (viii) x2+ y2Š 2 x =0and z = x2+ y2(6) Specifythepartsofthecurve r ( t )= sin t, cos t, 4sin2t thatlie abovetheplane z =1.

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11012.VECTORFUNCTIONS (7) Findthevaluesoftheparameters a and b atwhichthecurve r ( t )= 1+ at2,b Š t,t3 passesthroughthepoint(1 2 8). (8) Findthevaluesof a b ,and c ,ifany,atwhicheachofthefollowing vectorfunctionsiscontinuous: r (0)= a,b,c and,for t =0, (i) r ( t )= t, cos2t, 1+ t + t2 (ii) r ( t )= t, cos2t, 1+ t2 (iii) r ( t )= t, cos2t, ln | t | (iv) r ( t )= sin(2 t ) /t, sinh(3 t ) /t,t ln | t | (v) r ( t )= t cot(2 t ) ,t1 / 3ln | t | ,t2+2 (9) Supposethatthelimitslimt av ( t )andlimt au ( t )exist.Prove thebasiclawsoflimitsforthefollowingvectorfunctions: limt a( v ( t )+ u ( t ))=limt av ( t )+limt au ( t ) limt a( s v ( t ))= s limt av ( t ) limt a( v ( t ) u ( t ))=limt av ( t ) limt au ( t ) limt a( v ( t ) u ( t ))=limt av ( t ) limt au ( t ) (10) Provethelastlimitlawinexercise9directlyfromDe“nition12.2, thatis,withoutusingTheorem12.1. Hint: SeeStudyProblem12.7. (11) Let v ( t )= ( e2 tŠ 1) /t, ( 1+ t Š 1) /t,t ln | t | u ( t )= sin2(2 t ) /t2,t2+2 (cos t Š 1) /t2 w ( t )= t2 / 3, 2 / (1 Š t ) 1+ t Š t2+ t3 UsethebasiclawsoflimitsestablishedinExercise(9)to“nd (i)limt 0(2 v ( t ) Š u ( t )+ w ( t )) (ii)limt 0( v ( t ) u ( t )) (iii)limt 0( v ( t ) u ( t )) (iv)limt 0[ w ( t ) ( v ( t ) u ( t ))] (v)limt 0[ w ( t ) ( v ( t ) u ( t ))] (vi)limt 0[ w ( t ) ( v ( t ) u ( t ))+ v ( t ) ( u ( t ) w ( t ))+ u ( t ) ( w ( t ) v ( t ))] (12) Supposethatthevectorfunction v ( t ) u ( t )iscontinuous.Does thisimplythatbothvectorfunctions v ( t )and u ( t )arecontinuous? Supportyourargumentsbyexamples. (13) Supposethatthevectorfunctions v ( t ) u ( t )and v ( t )arecontinuous.Doesthisimplythatthevectorfunction u ( t )iscontinuous? Supportyourargumentsbyexamples.

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80.DIFFERENTIATIONOFVECTORFUNCTIONS111 80.DifferentiationofVectorFunctions Definition 12.5 (DerivativeofaVectorFunction) Supposeavectorfunction r ( t ) isde“nedonaninterval [ a,b ] and t0 [ a,b ] .Ifthelimit limh 0r ( t0+ h ) Š r ( t0) h = r( t0)= d r dt ( t0) exists,thenitiscalledthe derivativeofavectorfunction r ( t ) at t = t0, and r ( t ) issaidtobe dierentiable at t0.For t0= a or t0= b ,thelimit isunderstoodastheright( h> 0 )orleft( h< 0 )limit,respectively.If thederivativeexistsforallpointsin [ a,b ] ,thenthevectorfunction r ( t ) issaidtobedierentiableon [ a,b ] ItfollowsfromTheorem12.1thatavectorfunctionisdierentiable ifandonlyifallitscomponentsaredierentiable: r( t )=limh 0 x ( t + h ) Š x ( t ) h y ( t + h ) Š y ( t ) h z ( t + h ) Š z ( t ) h = x( t ) ,y( t ) ,z( t ) (12.1) Forexample, r ( t )= sin(2 t ) ,t2Š t,eŠ 3 t r( t )= 2cos(2 t ) 2 t Š 1 Š 3 eŠ 3 t Definition 12.6 (ContinuouslyDierentiableVectorFunction) Ifthederivative r( t ) isacontinuousvectorfunctiononaninterval [ a,b ] ,thenthevectorfunction r ( t ) issaidtobe continuouslydierentiable on [ a,b ] Higher-orderderivativesarede“nedsimilarly:thesecondderivative isthederivativeof r( t ), r( t )=( r( t )),thethirdderivativeisthe derivativeof r( t ), r( t )=( r( t )),and r( n )( t )=( r( n Š 1)( t )),provided theyexist.80.1.DifferentiationRules.Thefollowingrulesofdierentiationofvectorfunctionscanbededucedfrom(12.1). Theorem 12.2 (DierentiationRules) Suppose u ( t ) and v ( t ) aredierentiablevectorfunctionsand f ( t ) isa

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11212.VECTORFUNCTIONS real-valueddierentiablefunction.Then d dt v ( t )+ u ( t ) = v( t )+ u( t ) d dt f ( t ) v ( t ) = f( t ) v ( t )+ f ( t ) v( t ) d dt v ( t ) u ( t ) = v( t ) u ( t )+ v ( t ) u( t ) d dt v ( t ) u ( t ) = v( t ) u ( t )+ v ( t ) u( t ) d dt v ( f ( t )) = f( t ) v( f ( t )) Theproofisbasedonastraightforwarduseoftherule(12.1)and basicrulesofdierentiationforordinaryfunctionsandleftasanexercisetothereader. Example 12.5 Findthe“rstandsecondderivativesofthevector function r ( t )=( a + t2b ) ( c Š t d ) ,where a b c ,and d areconstant vectors. Solution: Bytheproductrule, r( t )=( a + t2b ) ( c Š t d )+( a + t2b ) ( c Š t d )=2 t b ( c Š t d ) Š ( a + t2b ) d r( t )=(2 t b ) ( c Š t d )+2 t b ( c Š t d )Š ( a + t2b ) d =2 b ( c Š t d ) Š 2 t b d Š 2 t b d =2 b c Š 6 t b d Alternatively,thecrossproductcanbecalculated“rstandthendierentiated: r ( t )= a c Š t a d + t2b c Š t3b d r( t )= Š a d +2 t b c Š 3 t2b d r( t )=2 b c Š 6 t b d 80.2.DifferentialofaVectorFunction.If r ( t )isdierentiable,then (12.2) r ( t )= r ( t + t ) Š r ( t )= r( t ) t + u ( t ) t, where u ( t ) 0 as t 0.Indeed,bythede“nitionofthederivative, u ( t )= r / t Š r( t ) 0 as t 0.Therefore,the componentsofthedierence r Š r t convergeto0fasterthan t .

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80.DIFFERENTIATIONOFVECTORFUNCTIONS113 Supposethat r( t0)doesnotvanish.Consideralinearvectorfunction L ( t )withtheproperty L ( t0)= r ( t0).Itsgeneralformis L ( t )= r ( t0)+ v ( t Š t0),where v isaconstantvector.For t closeto t0, L ( t ) isa linearapproximation of r ( t )inthesensethattheapproximation error r ( t ) Š L ( t ) becomessmallerwithdecreasing | t Š t0| .Itfollows from(12.2)that r ( t ) Š L ( t )=( r( t0) Š v ) t + u ( t ) t, t = t Š t0. Bythetriangleinequality | a Š b | a + b a + b ,the approximationerrorisboundedas r( t0) Š v Š u ( t ) r ( t ) Š L ( t ) | t | r( t0) Š v + u ( t ) If r( t0) Š v =0or v = r( t0),then u ( t ) r( t0) Š v fora sucientlysmall t because u ( t ) convergesto0as t 0(the sign meansmuchsmallerthanŽ).Therefore,theapproximation errordecreaseslinearlywithdecreasing t : r ( t ) Š L ( t ) r( t0) Š v | t | .When v = r( t0),theapproximationerrordecreasesfaster than t : r ( t ) Š L ( t ) | t | = u ( t ) 0as t 0 Thus,thelinearvectorfunction L ( t )= r ( t0)+ r( t0)( t Š t0) isthe bestlinearapproximation of r ( t )near t = t0.Providedthe derivativedoesnotvanish, r( t0) = 0 ,thelinearvectorfunction L ( t ) de“nesalinepassingthroughthepoint r ( t0).Thislineiscalledthe tangentline tothecurvetracedoutby r ( t )atthepoint r ( t0).The analogycanbemadewiththetangentlinetothegraph y = f ( x )at apoint( x0,y0),where y0= f ( x0).Theequationofthetangentline is y = y0+ f( x0)( x Š x0)(recallCalculusI).Thegraphisacurvein the xy planewhoseparametricequationsare x = t y = f ( t )orinthe vectorform r ( t )= t,f ( t ) .Theparametricequationsofthetangent linecanthereforebewrittenintheform x = t = x0+( t Š t0), y = y0+ f( t0)( t Š t0),where x0= t0.Put r0= x0,y0 .Thenthetangent lineistraversedbythelinearvectorfunction L ( t )= r0+ r( t0)( t Š t0) because r( t0)= 1 ,f( t0) Definition 12.7 (DierentialofaVectorFunction) Let r ( t ) beadierentiablevectorfunction.Thenthevector d r ( t )= r( t ) dt iscalledthe dierential of r ( t ) .

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11412.VECTORFUNCTIONS Inparticular,thederivativeistheratioofthedierentials, r( t )= d r /dt .Recallthatthedierential dt isanindependentvariablethat describesin“nitesimalvariationsof t suchthathigherpowersof dt canbeneglected.Inthissense,thede“nitionofthedierentialis thelinearizationof(12.2)in dt = t .Atanyparticular t = t0,the dierential d r ( t0)= r( t0) dt = 0 de“nesthetangentline L ( t )= r ( t0)+ d r ( t0)= r ( t0)+ r( t0) dt,t = t0+ dt. Thus, thedierential d r ( t ) atapointofthecurve r ( t ) istheincrement ofthepositionvectoralongthelinetangenttothecurveatthatpoint .80.3.GeometricalSignicanceoftheDerivative.Consideravectorfunctionthattracesoutalineparalleltoavector v r ( t )= r0+ t v .Then r( t )= v ;thatis,thederivativeisavectorparallelortangenttothe line.Thisobservationisofageneralnature;thatis, thevector r( t0) is tangenttothecurvetracedoutby r ( t ) atthepointwhosepositionvector is r ( t0).Let P0and Phhavepositionvectors r ( t0)and r ( t0+ h ).Then ŠŠ P0Ph= r ( t0+ h ) Š r ( t0)isasecantvector.As h 0, ŠŠ P0Phapproaches avectorthatliesonthetangentlineasdepictedinFigure12.6.On theotherhand,itfollowsfrom(12.2)that,forsmallenough h = dt ŠŠ P0Ph= d r ( t0)= r( t0) h ,andthereforethetangentlineisparallelto r( t0).Thedirectionofthetangentvectoralsode“nestheorientation Figure12.6.Left :Asecantlinethroughtwopointsof thecurve, P0and Ph.As h getssmaller,thedirectionof thevector Š ŠŠ P0P1= r ( t0+ h ) Š r ( t0)becomesclosertothe tangenttothecurveat P0. Right :Thederivative r( t ) de“nesatangentvectortothecurveatthepointwiththe positionvector r ( t ).Italsospeci“esthedirectioninwhich r ( t )traversesthecurvewithincreasing t T ( t )istheunit tangentvector.

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80.DIFFERENTIATIONOFVECTORFUNCTIONS115 ofthecurve,thatis,thedirectioninwhichthecurveistracedout by r ( t ). Example 12.6 Findthelinetangenttothecurve r ( t )= 2 t,t2Š 1 ,t3+2 t atthepoint P0(2 0 3) Solution: Bythegeometricalpropertyofthederivative,avector paralleltothelineis v = r( t0),where t0isthevalueoftheparameter t atwhich r ( t0)= 2 0 3 isthepositionvectorof P0.Therefore, t0=1.Then v = r(1)= 2 2 t, 3 t +2 |t =1= 2 2 5 .Parametric equationsofthelinethrough P0andparallelto v are x =2+2 t y =2 t z =3+5 t Ifthederivative r( t )existsanddoesnotvanish,then,atanypoint ofthecurvetracedoutby r ( t ),a unittangentvector canbede“nedby T ( t )= r( t ) r( t ) InSection79.3,spatialcurveswereidenti“edwithcontinuousvector functions.Intuitively,asmoothcurveasapointsetinspaceshould haveaunittangentvectorthatiscontinuousalongthecurve.Recall alsothat,foranycurveasapointsetinspace,therearemanyvector functionswhoserangecoincideswiththecurve. Definition 12.8 (SmoothCurve) Apointset C inspaceiscalleda smoothcurve ifthereisasimple, continuouslydierentiableparametriccurve r ( t ) whoserangecoincides with C andwhosederivativedoesnotvanish. Asmoothparametriccurve r ( t )is oriented bythedirectionofthe unittangentvector T ( t ).Notethatif r( t )iscontinuousandnever 0,then T ( t )iscontinuous.Inparticular,withthede“nitionabove, asmoothcurvedoesindeedhaveacontinuousunittangentvector. Therefore,ifacurvedoesnothaveacontinuousunittangentvector, itcannotbesmooth.Thisenablesustoconcludethatsomecurvesare notsmooth,basedonpropertiesdeducedfromasingleparametrization. ThisisimportantbecauseonecannotpossiblytestallparameterizationstoseewhetheroneofthemmeetstheconditionsinDe“nition12.8. Considertheplanarcurve r ( t )= t3,t2, 0 .Thevectorfunctionis dierentiableeverywhere, r( t )= 2 t, 3 t2, 0 ,andthederivativevanishesattheorigin, r(0)= 0 .Theunittangentvector T ( t )isnot de“nedat t =0.Solvingtheequation x = t3for t t = x1 / 3,and substitutingthelatterinto y = t2,itisconcludedthatthecurvetraversedby r ( t )isthegraph y = x2 / 3,whichhasa cusp at x =0.The

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11612.VECTORFUNCTIONS curveisnotsmoothattheorigin.Thetangentlineistheverticalline x =0because y( x )=(2 / 3) xŠ 1 / 3 as x 0.Thegraph liesinthepositivehalf-plane y 0andapproachesthe y axis,formingahornlikeshapeattheorigin.Acuspdoesnotnecessarilyoccur atapointwherethederivative r( t )vanishes.Forexample,consider r ( t )= t3,t5, 0 suchthat r(0)= 0 .Thisvectorfunctiontracesoutthe graph y = x5 / 3,whichhasnocuspat x =0(ithasanin”ectionpoint at x =0).Thereisanothervectorfunction R ( s )= s,s5 / 3, 0 that tracesoutthesamegraph,but R(0)= 1 0 0 = 0 ,andthecurveis smooth.Sothevanishingofthederivativeismerelyassociatedwitha poorchoiceofthevectorfunction.Notethat r ( t )= R ( s )identically if s = t3.Bythechainrule,d dtr ( t )=d dtR ( s )= R( s )( ds/dt ).This showsthat,evenif R( s )nevervanishes,thederivative r( t )canvanish,provided ds/dt vanishesatsomepoint,whichisindeedthecasein theconsideredexampleas ds/dt =3 t2vanishesat t =0. Example 12.7 Determinewhetherthecycloid C parameterizedby x = a ( t Š sin t ) y = a (1 Š cos t ) issmooth,where a> 0 isaparameter. Ifitisnotsmoothatparticularpoints,investigateitsbehaviornear thosepoints. Solution: FollowingtheremarkafterDe“nition12.8,theexistence ofacontinuousunittangentvectorhastobeveri“ed.Let r ( t )= x ( t ) ,y ( t ) .Since x( t )= a (1 Š cos t ) 0forall t ,and x( t )=0 onlywhen t isamultipleof2 x ( t )ismonotonicallyincreasing.In particular, x ( t )isone-to-one,so C issimple.Since y( t )= a cos t thederivatives x( t )and y( t )vanishsimultaneouslyifandonlyif t = 2 n forsomeinteger n .Thus, r( t ) =0unless t =2 n ,so C is smoothexceptpossiblyatthepoints r (2 n )= 2 na, 0 ;thatis,the portionof C betweentwoconsecutivesuchpointsissmooth,butitis notyetknownwhether C issmoothatthosepoints.Since r( t ) = a 2(1 Š cos t )= a 4sin2( t/ 2)=2 a | sin( t/ 2) | ,thecomponentsofthe unittangentvectorfor t =2 n are T1( t )= x( t ) r( t ) = | sin( t/ 2) | ,T2( t )= x( t ) r( t ) = sin t 2 | sin( t/ 2) | Owingtotheperiodicityofthesineandcosinefunctions,itissucienttoinvestigatethepointcorrespondingto t =0.Ifthereexistsa continuousunittangentvector,thenthelimitlimt 0 T ( t )shouldexist andbetheunittangentvectoratthepointcorrespondingto t =0.By Theorem12.1,thelimitsofthecomponents T1( t )and T2( t )shouldexistas t 0.Evidently, T1( t ) 0as t 0,butthelimitlimt 0T2( t )

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80.DIFFERENTIATIONOFVECTORFUNCTIONS117 doesnotexist.Indeed,bylHospitalsruletheleftandrightlimitsare dierent: limt 0+T2( t )=limt 0+sin t 2sin( t/ 2) =limt 0+cos t cos( t/ 2) =1 limt 0ŠT2( t )=limt 0Šsin t Š 2sin( t/ 2) = Š limt 0+cos t cos( t/ 2) = Š 1 Therefore T ( t ) 0 1 as t 0+,but T ( t ) 0 Š 1 as t 0Š. Thus T ( t )cannotbecontinuouslyextendedacrossthepoint(0 0),so C isnotsmooththere(aswellasat(2 n, 0)),and,infact,hasacusp there. Alocalbehaviorofthecycloidnear(0 0)maybeinvestigatedas follows.UsingtheTaylorpolynomialapproximationnear t =0,sin t t Š t3/ 6andcos t 1 Š t2/ 2,thecycloidisapproximatedbythecurve x = at3/ 6, y = at2/ 2.Expressing t =(6 x/a )1 / 3andsubstitutingitinto theotherequation,itisconcludedthat y = cx2 / 3,where c =(9 a/ 2)1 / 3. Thiscurvehasacuspat x =0asnotedabove.80.4.StudyProblem.Problem12.8. Provethat,foranysmoothcurveonasphere,atangentvectoratanypoint P isorthogonaltothevectorfromthesphere centerto P Solution: Let r0bethepositionvectorofthecenterofasphereof radius R .Thepositionvector r ofanypointofthespheresatis“esthe equation r Š r0 = R or( r Š r0) ( r Š r0)= R2(because a 2= a a foranyvector a ).Let r ( t )beavectorfunctionthattracesoutacurve onthesphere.Then,forallvaluesof t ,( r ( t ) Š r0) ( r ( t ) Š r0)= R2. Dierentiatingbothsidesofthelatterrelation,oneinfers r( t ) ( r ( t ) Š r0)=0 r( t ) r ( t ) Š r0. If r ( t )isthepositionvectorof P and O isthecenterofthesphere,then Š OP = r ( t ) Š r0,andhencethetangentvector r( t )at P isorthogonal to Š OP forany t oratanypoint P ofthecurve. 80.5.Exercises.(1) Findthederivativesanddierentialsofeachofthefollowingvector functions: (i) r ( t )= 1 1+ t, 1+ t3 (ii) r ( t )= cos t, sin2( t ) ,t2 (iii) r ( t )= ln( t ) ,e2 t,teŠ t

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11812.VECTORFUNCTIONS (iv) r ( t )= 3 t Š 2 t2Š 4 ,t (v) r ( t )= a + b t2Š c et(vi) r ( t )= t a ( b Š c et) (2) Sketchthecurvetraversedbythevectorfunction r ( t )= 2 ,t Š 1 ,t2+1 .Indicatethedirectioninwhichthecurveistraversedby r ( t ) withincreasing t .Sketchthepositionvectors r (0), r (1), r (2)andthe vectors r(0), r(1), r(2).Repeattheprocedureforthevectorfunction R ( t )= r ( Š t )= 2 Š t Š 1 ,t2+1 for t = Š 2, Š 1,0. (3) Determineifthecurvetracedoutbyeachofthefollowingvector functionsissmoothforaspeci“edintervaloftheparameter.Ifthe curveisnotsmoothataparticularpoint,graphitnearthatpoint. (i) r ( t )= t,t2,t3 0 t 1 (ii) r ( t )= t2,t3, 2 Š 1 t 1 (iii) r ( t )= t1 / 3,t,t3 Š 1 t 1 (iv) r ( t )= t5,t3,t4 Š 1 t 1 (v) r ( t )= sin3t, 1 ,t2 Š / 2 t / 2 (4) Findtheparametricequationsofthetangentlinetoeachofthe followingcurvesataspeci“edpoint: (i) r ( t )= t2Š t,t3/ 3 2 t P0=(6 9 6) (ii) r ( t )= ln t, 2 t,t2 P0=(0 2 1) (5) Findtheunittangentvectortothecurvetraversedbythespeci“ed vectorfunctionatthegivenpoint P0: (i) r ( t )= 2 t +1 2tanŠ 1t,eŠ t ,P0(1 0 1) (ii) r ( t )= cos( t ) cos(3 t ) sin( t ) ,P0(1 / 2 Š 1 1 / 3) (6) Find r( t ) r( t )and r( t ) r( t )if r ( t )= t,t2Š 1 ,t3+2 (7) Isthereapointonthecurve r ( t )= t2Š t,t3/ 3 2 t atwhichthe tangentlineisparalleltothevector v = Š 5 / 2 2 1 ?Ifso,“ndthe point. (8) Let r ( t )= et, 2cos t, sin(2 t ) .Usethebestlinearapproximation L ( t )near t =0toestimate r (0 2).Useacalculatortoassessthe accuracy r (0 2) Š L (0 2) oftheestimate.Repeattheprocedurefor r (0 7)and r (1 2).Comparetheerrorsinallthreecases. (9) Findthepointofintersectionoftheplane y + z =3andthecurve r ( t )= ln t,t2, 2 t .Findtheanglebetweenthenormaloftheplane andthetangentlinetothecurveatthepointofintersection. (10) Doesthecurve r ( t )= 2 t2, 2 t, 2 Š t2 intersecttheplane x + y + z = Š 3?Ifnot,“ndapointonthecurvethatisclosesttotheplane.What isthedistancebetweenthecurveandtheplane. Hint: Expressthe distancebetweenapointonthecurveandtheplaneasafunctionof t thensolvetheextremevalueproblem.

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80.DIFFERENTIATIONOFVECTORFUNCTIONS119 (11) Findthepointofintersectionoftwocurves r1( t )= 1 1 Š t, 3+ t2 and r2( s )= 3 Š s,s Š 2 ,s2 .Iftheangleatwhichtwocurves intersectisde“nedastheanglebetweentheirtangentlinesatthepoint ofintersection,“ndtheangleatwhichtheabovetwocurvesintersect. (12) Statetheconditionunderwhichthetangentlinestothecurve r ( t )attwodistinctpoints r ( t1)and r ( t2)areintersecting,orskew,or parallel.Let r ( t )= 2sin( t ) cos( t ) sin( t ) t1=0,and t2=1 / 2. Determinewhetherthetangentlinesatthesepointsareintersecting and,ifso,“ndthepointofintersection. (13) Supposeasmoothcurve r ( t )doesnotintersectaplanethrough apoint P0andorthogonaltoavector n .Whatistheanglebetween n andthetangentlinetothecurveatthepointthatistheclosesttothe plane? (14) Suppose r ( t )istwicedierentiable.Showthat( r ( t ) r( t ))= r ( t ) r( t ). (15) Supposethat r ( t )isdierentiablethreetimes.Showthat [ r ( t ) ( r( t ) r( t ))]= r ( t ) ( r( t ) r( t )). (16) Let r ( t )beadierentiablevectorfunction.Showthat ( r ( t ) )= r ( t ) r( t ) / r ( t ) (17) Aspacewarshipcan“realasercannonforwardalongthetangent linetoitstrajectory.Ifthetrajectoryistraversedbythevectorfunction r ( t )= t,t,t2+4 inthedirectionofincreasing t andthetarget isthesphere x2+ y2+ z2=1,“ndthepartofthetrajectoryinwhich thelasercannoncanhitthetarget. (18) Aplane normal toacurveatapoint P0istheplanethrough P0whosenormalistangenttothecurveat P0.Foreachofthefollowing curves,“ndsuitableparametricequations,thetangentline,andthe normalplaneataspeci“edpoint: (i) y = x z = x2, P0=(1 1 1) (ii) x2+ z2=10, y2+ z2=10, P0=(1 1 3) (iii) x2+ y2+ z2=6, x + y + z =0, P0=(1 Š 2 1) (19) Showthattangentlinestoacircularhelixhaveaconstantangle withtheaxisofthehelix. (20) Consideralinethroughtheorigin.Anysuchlinesweepsacircular conewhenrotatedaboutthe z axisand,forthisreason,iscalleda generating lineofacone.Provethatthecurve r ( t )=( etcos t,etsin t,et) intersectsallgeneratinglinesofthecone x2+ y2= z2atthesame angle. Hint: Showthatparametricequationsofageneratinglineare x = s cos y = s sin z = s .De“nethepointsofintersectionofthe lineandthecurveand“ndtheangleatwhichtheyintersect.

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12012.VECTORFUNCTIONS 81.IntegrationofVectorFunctions Definition 12.9 (De“niteIntegralofaVectorFunction) Let r ( t ) bede“nedontheinterval [ a,b ] .Thevectorwhosecomponents arethede“niteintegralsofthecorrespondingcomponentsof r ( t )= x ( t ) ,y ( t ) ,z ( t ) iscalledthe de“niteintegral of r ( t ) overtheinterval [ a,b ] anddenotedas (12.3) b ar ( t ) dt = b ax ( t ) dt, b ay ( t ) dt, b ax ( t ) dt Iftheintegral(12.3)exists,then r ( t ) issaidtobeintegrableon [ a,b ] Bythisde“nition,avectorfunctionisintegrableifandonlyifall itscomponentsareintegrablefunctions.Recallthatacontinuousrealvaluedfunctionisintegrable.Therefore,thefollowingtheoremholds. Theorem 12.3 Ifavectorfunctioniscontinuousontheinterval [ a,b ] ,thenitisintegrableon [ a,b ] Example 12.8 Findtheintegralof r ( t )= t/, sin t, cos t over theinterval [0 ] Solution: Thecomponentsof r ( t )arecontinuouson[0 ].Therefore, bythefundamentaltheoremofcalculus, 0r ( t ) dt = 0( t/ ) dt, 0sin tdt, 0cos tdt = t2 2 0, Š cos t 0, sin t 0 = / 2 2 0 Definition 12.10 (Inde“niteIntegralofaVectorFunction) Avectorfunction R ( t ) iscalledan inde“niteintegral oran antiderivative of r ( t ) if R( t )= r ( t ) If R ( t )= X ( t ) ,Y ( t ) ,Z ( t ) and r ( t )= x ( t ) ,y ( t ) ,z ( t ) .Then,accordingto(12.1),thefunctions X ( t ), Y ( t ),and Z ( t )areantiderivatives of x ( t ), y ( t ),and z ( t ),respectively, X ( t )= x ( t ) dt + c1,Y ( t )= y ( t ) dt + c2,Z ( t )= z ( t ) dt + c3, where c1, c2,and c3areconstants.Thelatterrelationscanbecombined intoasinglevectorrelation: R ( t )= r ( t ) dt + c ,

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81.INTEGRATIONOFVECTORFUNCTIONS121 where c isanarbitraryconstantvector. Recallthat,forafunction x ( t )continuouson[ a,b ],itsparticular antiderivativederivativeisgivenby X ( t )= t ax ( u ) du,a t b. Therefore,aparticularantiderivativeofacontinuousvectorfunction r ( t )is R ( t )= t ar ( u ) du,a t b. Thevectorfunction R ( t )is dierentiable on( a,b )andsatis“esthe condition R ( a )=0.Ageneralantiderivativeisobtainedbyaddinga constantvector, R ( t ) R ( t )+ c .Thisobservationallowsustoextend thefundamentaltheoremofcalculustovectorfunctions. Theorem 12.4 (FundamentalTheoremofCalculusforVector Functions) If r ( t ) iscontinuouson [ a,b ] ,then b ar ( t ) dt = R ( b ) Š R ( a ) where R ( t ) isanyantiderivativeof r ( t ) ,thatis,avectorfunctionsuch that R( t )= r ( t ) Example 12.9 Find r ( t ) if r( t )= 2 t, 1 6 t2 and r (1)= 2 1 0 Solution: Takingtheantiderivativeof r( t ),one“nds r ( t )= 2 t, 1 6 t2 dt + c = t2,t, 3 t3 + c Theconstantvector c isdeterminedbythecondition r (1)= 2 1 0 whichgives 1 1 3 + c = 2 1 0 .Hence, c = 2 1 0 Š 1 1 3 = 1 0 Š 3 and r ( t )= t2+1 ,t, 3 t3Š 3 Ingeneral,thesolutionoftheequation r( t )= v ( t )satisfyingthe condition r ( t0)= r0canbewrittenintheform r( t )= v ( t )and r ( t0)= r0 r ( t )= r0+ t t0v ( u ) du if v ( t )isacontinuousvectorfunction.Asnotedabove,iftheintegrand isacontinuousfunction,thenthederivativeoftheintegralwithrespect toitsupperlimitisthevalueoftheintegrandatthatlimit.Therefore, r( t )=( d/dt ) t t0v ( u ) du = v ( t ),andhence r ( t )isanantiderivativeof v ( t ).When t = t0,theintegralvanishesand r ( t0)= r0asrequired.

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12212.VECTORFUNCTIONS 81.1.ApplicationstoMechanics.Let r ( t )bethepositionvectorofa particleasafunctionoftime t .The“rstderivative r( t )= v ( t )is calledthe velocity oftheparticle.Themagnitudeofthevelocityvector v ( t )= v ( t ) iscalledthe speed .Thespeedofacarisanumbershown onthespeedometer.Thevelocityde“nesthedirectioninwhichthe particletravelsandtheinstantaneousrateatwhichitmovesinthat direction.Thesecondderivative r( t )= v( t )= a ( t )iscalledthe acceleration .If m isthemassofaparticleand F istheforceactingon theparticle,accordingtoNewtonssecondlaw,theaccelerationand forcearerelatedas F = m a Ifthetimeismeasuredinseconds,thelengthinmeters,andthemass inkilograms,thentheforceisgiveninnewtons,1N=1kg m / s2. Iftheforceisknownasavectorfunctionoftime,thenNewtons secondlawdeterminesaparticlestrajectory.Theproblemof“ndingthetrajectoryamountstoreconstructingthevectorfunction r ( t ) ifitssecondderivative r( t )=(1 /m ) F ( t )isknown;thatis, r ( t )is givenbythesecondantiderivativeof(1 /m ) F ( t ).Indeed,thevelocity v ( t )isanantiderivativeof(1 /m ) F ( t ),andthepositionvector r ( t )is anantiderivativeofthevelocity v ( t ).Asshownintheprevioussection,anantiderivativeisnotunique,unlessitsvalueataparticular pointisspeci“ed.So thetrajectoryofmotionisuniquelydetermined byNewtonsequation,providedthepositionandvelocityvectorsare speci“edataparticularmomentoftime ,forexample, r ( t0)= r0and v ( t0)= v0.Thelatterconditionsarecalled initialconditions .Given theinitialconditions,thetrajectoryofmotionisuniquelyde“nedby therelations: v ( t )= v0+ 1 m t t0F ( u ) du, r ( t )= r0+ t t0v ( u ) du iftheforceisacontinuousvectorfunctionoftime. Remark. Iftheforceisafunctionofaparticlesposition,thenNewtonsequationbecomesasystemof ordinarydierentialequations ,that is,asetofsomerelationsbetweencomponentsofthevectorfunctions, itsderivatives,andtime. Example 12.10 (MotionUnderaConstantForce) Provethatthetrajectoryofmotionunderaconstantforceisaparabola iftheinitialvelocityisnotparalleltotheforce.

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81.INTEGRATIONOFVECTORFUNCTIONS123 Solution: Let F beaconstantforce.Withoutlossofgenerality,the initialconditionscanbesetat t =0, r (0)= r0,and v (0)= v0.Then v ( t )= v0+ 1 m t 0F du = v0+ t m F r ( t )= r0+ t 0v ( u ) du = r0+ t v0+ t2 2 m F Ifthevectors v0and F areparallel,thentheyareproportional, v0= c F .Inthisparticularcase,thetrajectory r ( t )= r0+( ct + t2/ (2 m )) F = r0+ s F liesinthestraightlinethrough r0andparallelto F .The parameter s = ct + t2/ (2 m )de“nesthepositionoftheparticleonthe lineasafunctionoftime.Otherwise,thevector r ( t ) Š r0isalinear combinationoftwononparallelvectors v0and F andhencemustbe orthogonalto n = v0 F bythegeometricalpropertyofthecross product.Therefore,theparticleremainsintheplanethrough r0that isparallelto F and v0ororthogonalto n ,thatis,( r ( t ) Š r0) n =0(see Figure12.7,leftpanel).Theshapeofaspacecurvedoesnotdepend onthechoiceofthecoordinatesystem.Letuschoosethecoordinate systemsuchthattheoriginisattheinitialposition r0andtheplane inwhichthetrajectoryliescoincideswiththe zy planesothat F is paralleltothe z axis.Inthiscoordinatesystem, r0= 0 F = 0 0 Š F and v0= 0 ,v0 y,v0 z .Theparametricequationsofthetrajectoryof motionassumetheform x =0, y = v0 yt ,and z = v0 zt Š t2F/ (2 m ).The substitutionof t = y/v0 yintothelatterequationyields z = ay2+ by where a = Š Fv2 0 y/ (2 m )and b = v0 z/v0 y,whichde“nesaparabolain the zy plane.Thus,thetrajectoryofmotionunderaconstantforceis aparabolathroughthepoint r0thatliesintheplanecontainingthe forceandinitialvelocityvectors F and v0.Theparabolaisconcave inthedirectionoftheforce.InFigure12.7,theforcevectorpoints downwardandthetrajectoryisconcavedownward. 81.2.MotionUnderaConstantGravitationalForce.Themagnitudeof thegravitationalforcethatactsonanobjectofmass m nearthesurface oftheEarthis mg ,where g 9 8m / s2isauniversalconstantcalledthe accelerationofafreefall .AccordingtoExample12.10,anyprojectile “redfromsomepointfollowsaparabolictrajectory.Thisfactallows onetopredicttheexactpositionsoftheprojectileand,inparticular, thepointatwhichitimpactstheground.Inpractice,theinitialspeed v0oftheprojectileandangleofelevation atwhichtheprojectileis “redareknown(seeFigure12.7,rightpanel).Somepracticalquestions are:Atwhatelevationangleisthemaximalrangereached?Atwhat

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12412.VECTORFUNCTIONS Figure12.7.Left :Motionunderaconstantforce F .The trajectoryisaparabolathatliesintheplanethroughthe initialpointofthemotion r0andorthogonaltothevector n = v0 F ,wheretheinitialvelocity v0isassumedtobe nonparalleltotheforce F Right :Motionofaprojectile thrownatanangle andaninitialheight h .Thetrajectory isaparabola.Thepointofimpactde“nestherange L ( ).elevationangledoestherangeattainaspeci“edvalue(e.g.,tohita target)? Toanswertheseandrelatedquestions,choosethecoordinatesystemsuchthatthe z axisisdirectedupwardfromthegroundandthe parabolictrajectoryliesinthe zy plane.Theprojectileis“redfromthe point(0 0 ,h ),where h istheinitialelevationoftheprojectileabove theground(“ringfromahill).InthenotationofExample12.10, F = Š mg ( F isnegativebecausethegravitationalforceisdirected towardtheground,whilethe z axispointsupward), v0 y= v0cos ,and v0 z= v0sin .Thetrajectoryis y = tv0cos ,z = h + tv0sin Š 1 2 gt2,t 0 Itisinterestingtonotethatthetrajectoryisindependentofthemass oftheprojectile.Lightandheavyprojectileswouldfollowthesame parabolictrajectory,providedtheyare“redfromthesameposition, atthesamespeed,andatthesameangleofelevation.Theheightof theprojectilerelativetothegroundisgivenby z ( t ).Thehorizontal displacementis y ( t ).Let tL> 0bethemomentoftimewhenthe projectilelands;thatis,when t = tL,theheightvanishes, z ( tL)=0. Apositivesolutionofthisequationis tL= v0sin + v2 0sin2 +2 gh g .

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81.INTEGRATIONOFVECTORFUNCTIONS125 Thedistance L traveledbytheprojectileinthehorizontaldirection untilitlandsisthe range : L = y ( tL)= tLv0cos Forexample,iftheprojectileis“redfromtheground, h =0,then tL=2 v0sin /g andtherangeis L = v2 0sin(2 ) /g .Therangeattainsits maximalvalue v2 0/g whentheprojectileis“redatanangleofelevation = / 4.Theangleofelevationatwhichtheprojectilehitsatargetata givenrange L = L0is =(1 / 2)sinŠ 1( L0g/v2 0).For h =0,theangleat which L = L ( )attainsitsmaximalvaluescanbefoundbysolvingthe equation L( )=0,whichde“nescriticalpointsofthefunction L ( ). Theangleofelevationatwhichtheprojectilehitsatargetatagiven rangeisfoundbysolvingtheequation L ( )= L0.Thetechnicalities arelefttothereader. Remark. Inreality,thetrajectoryofaprojectiledeviatesfroma parabolabecausethereisanadditionalforceactingonaprojectile movingintheatmosphere,thefrictionforce.Thefrictionforcedependsonthevelocityoftheprojectile.Soamoreaccurateanalysisof theprojectilemotionintheatmosphererequiresmethodsofordinary dierentialequations.81.3.StudyProblems.Problem12.9. Theaccelerationofaparticleis a = 2 6 t, 0 .Find thepositionvectoroftheparticleanditsvelocityin2unitsoftime t if theparticlewasinitiallyatthepoint ( Š 1 Š 4 1) andhadthevelocity 0 2 1 Solution: Thevelocityvectoris v ( t )= a ( t ) dt + c = 2 t, 3 t2, 0 + c Theconstantvector c is“xedbytheinitialcondition v (0)= 0 2 1 whichyields c = 0 2 1 .Thus, v ( t )= 2 t, 3 t2+2 1 and v (2)= 4 14 1 .Thepositionvectoris r ( t )= v ( t ) dt + c = t2,t3+2 t,t + c Heretheconstantvector c isdeterminedbytheinitialcondition r (0)= Š 1 Š 4 1 ,whichyields c = Š 1 Š 4 1 .Thus, r ( t )= t2Š 1 ,t3+ 2 t Š 4 ,t +1 and r (2)= 3 8 3 Problem12.10. Showthatifthevelocityandpositionvectorsofa particleremainorthogonalduringthemotion,thenthetrajectorylies onasphere. Solution: If v ( t )= r( t )and r ( t )areorthogonal,then r( t ) r ( t )=0 forall t .Since( r r )= r r + r r=2 r r =0,oneconcludesthat r ( t ) r ( t )= R2=constor r ( t ) = R forall t ;thatis,theparticle remainsata“xeddistance R fromtheoriginallthetime.

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12612.VECTORFUNCTIONS Problem12.11. Achargedparticlemovinginamagnetic“eld B is subjecttotheLorentzforce F =( e/c ) v B ,where e istheelectric chargeoftheparticleand c isthespeedoflightinvacuum.Assume thatthemagnetic“eldisaconstantvectorparalleltothe z axisand theinitialvelocityis v (0)= v, 0 ,v .Showthatthetrajectoryisa helix: r ( t )= R sin( t ) ,R cos( t ) ,vt = eB mc ,R = v where B = B isthemagnitudeofthemagnetic“eldand m isthe particlemass. Solution: Newtonssecondlawreads m v= e c v B Put B = 0 0 ,B .Then v = r= R cos( t ) Š R sin( t ) ,v v B = Š RB sin( t ) Š RB cos( t ) 0 v= Š 2R sin( t ) Š 2R cos( t ) 0 ThesubstitutionoftheserelationsintoNewtonssecondlawyields m2R = eBR/c andhence =( eB ) / ( mc ).Since v (0)= R, 0 ,v = v, 0 ,v ,itfollowsthat R = v/ Remark. Therateatwhichthehelixrisesalongthemagnetic“eld isdeterminedbythemagnitude(speed)oftheinitialvelocitycomponent vparalleltothemagnetic“eld,whereastheradiusofthehelix isdeterminedbythemagnitudeoftheinitialvelocitycomponent vperpendiculartothemagnetic“eld.Aparticlemakesonefullturn aboutthemagnetic“eldintime T =2 / =2 mc/ ( eB );thatis,the largerthemagnetic“eld,thefastertheparticlerotatesaboutit.81.4.Exercises.(1) Findtheinde“niteandde“niteintegralsoverspeci“edintervalsfor eachofthefollowingfunctions: (i) r ( t )= 1 2 t, 3 t2 0 t 2 (ii) r ( t )= sin t,t3, cos t Š t (iii) r ( t )= t2,t t Š 1 t 0 t 1 (iv) r ( t )= t ln t,t2,e2 t 0 t 1 (v) r ( t )= 2sin t cos t, 3sin t cos2t, 3sin2t cos t 0 t / 2 (vi) r ( t )= a +cos( t ) b 0 t (vii) r ( t )= a ( u( t )+ b ),0 t 1if u (0)= a and u (1)= a Š b (2) Find r ( t )ifthederivatives r( t )and r ( t0)aregiven:

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81.INTEGRATIONOFVECTORFUNCTIONS127 (i) r( t )= 1 2 t, 3 t2 r (0)= 1 2 3 (ii) r( t )= t Š 1 ,t2, t r (1)= 1 0 1 (iii) r( t )= sin(2 t ) 2cos t, sin2t r ( )= 1 2 3 (3) Find r ( t )if (i) r( t )= 0 2 6 t r (0)= 1 2 3 r(0)= 1 0 Š 1 (ii) r( t )= t1 / 3,t1 / 2, 6 t r (1)= 1 0 Š 1 r(0)= 1 2 0 (iii) r( t )= Š sin t, cos t, 1 /t r ( )= 1 Š 1 0 r( )= Š 1 0 2 (iv) r( t )= 0 2 6 t r (0)= 1 2 3 r (1)= 1 0 Š 1 (4) Solvetheequation r( t )= a ,where a isaconstantvectorif r (0)= b and r ( t0)= c forsome t = t0 =0. (5) Findthemostgeneralvectorfunctionwhose n thderivativevanishes, r( n )( t )=0. (6) Showthatacontinuouslydierentiablevectorfunction r ( t )satisfyingtheequation r( t ) r ( t )= 0 traversesastraightline(orapart ofit). (7) Ifaparticlewasinitiallyatpoint(1 2 1)andhadvelocity v = 0 1 Š 1 ,“ndthepositionvectoroftheparticleafterithasbeenmovingwithacceleration a ( t )= 1 0 ,t for2unitsoftime. (8) Aparticleofunitmassmovesunderaconstantforce F .Ifaparticlewasinitiallyatthepoint r0andpassedthroughthepoint r1after 2unitsoftime,“ndtheinitialvelocityoftheparticle.Whatwasthe velocityoftheparticlewhenitpassedthrough r1? (9) Aparticleofmass1kgwasinitiallyatrest.Thenduring2seconds aconstantforceofmagnitude3Nwasappliedtotheparticleinthe directionof 1 2 2 .Howfaristheparticlefromitsinitialpositionin 4seconds? (10) Thepositionvectorofaparticleis r ( t )= t2, 5 t,t2Š 16 t .Find r ( t )whenthespeedoftheparticleismaximal. (11) Aprojectileis“redataninitialspeedof400m/sandatanangle ofelevationof30.Findtherangeoftheprojectile,themaximum heightreached,andthespeedatimpact. (12) Aballofmass m isthrownsouthwardintotheairataninitial speedof v0atanangleof totheground.Aneastwindappliesa steadyforceofmagnitude F totheballinawesterlydirection.Find thetrajectoryoftheball.Wheredoestheballlandandatwhatspeed? Findthedeviationoftheimpactpointfromtheimpactpoint A when nowindispresent.Isthereanywaytocorrectthedirectioninwhich theballisthrownsothattheballstillhits A ? (13) Arocketburnsitsonboardfuelwhilemovingthroughspace.Let v ( t )and m ( t )bethevelocityandmassoftherocketattime t .Itcan beshownthattheforceexertedbytherocketjetenginesis m( t ) vg,

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12812.VECTORFUNCTIONS where vgisthevelocityoftheexhaustgasesrelativetotherocket. Showthat v ( t )= v (0) Š ln( m (0) /m ( t )) vg.Therocketistoaccelerate inastraightlinefromresttotwicethespeedofitsownexhaustgases. Whatfractionofitsinitialmasswouldtherockethavetoburnasfuel? (14) Theaccelerationofaprojectileis a ( t )= 0 2 6 t .Theprojectile isshotfrom(0 0 0)withaninitialvelocity v (0)= 1 Š 2 Š 10 .Itis supposedtodestroyatargetlocatedat(2 0 Š 12).Thetargetcanbe destroyediftheprojectilesspeedisatleast3.1atimpact.Willthe targetbedestroyed? 82.ArcLengthofaCurve Letavectorfunction r ( t ), a t b ,traverseaspacecurve C Considerapartitionoftheinterval[ a,b ], a = t0N .Underare“nementofapartition, DN 0 as N .Are“nementisobtainedbyaddingapartitionpointin eachpartitionintervalwhoselengthis DN(atleastonesuchinterval isalwayspresent). Definition 12.11 (ArcLengthofaCurve) Let r ( t ) a t b ,beavectorfunctiontraversingacurve C .Let acollectionofpoints Pkbeapartitionof C k =0 1 ,...,N ,andlet | Pk Š 1Pk| bethedistancebetweentwoneighboringpartitionpoints.The arclengthofacurve C isthelimit L =limN Nk =1| Pk Š 1Pk| wherethepartitionisre“nedas N ,provideditexistsandis independentofthechoiceofpartition.If L< ,thecurveiscalled measurable or recti“able Thegeometricalmeaningofthisde“nitionisrathersimple.Here thesumof | Pk Š 1Pk| isthelengthofapolygonalpathwithverticesat P0, P1,..., PNinthisorder.Asthepartitionbecomes“nerand“ner, thispolygonalpathapproachesthecurvemoreandmoreclosely(see Figure12.8,leftpanel).Incertaincases,thearclengthisgivenbythe Riemannintegral.

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82.ARCLENGTHOFACURVE129 Figure12.8.Left :Thearclengthofacurveisde“ned asthelimitofthesequenceoflengthsofpolygonalpaths throughpartitionpointsofthecurve. Right :Naturalparameterizationofacurve.Givenapoint A ofthecurve,the arclength s iscountedfromittoanypoint P ofthecurve. Thepositionvectorof P isavector R ( s ).Ifthecurveis tracedoutbyanothervectorfunction r ( t ),thenthereisa relation s = s ( t )suchthat r ( t )= R ( s ( t )).Theorem 12.5 (ArcLengthofaCurve) Let C beacurvetracedoutbyacontinuouslydierentiablevectorfunction r ( t ) ,whichde“nesaone-to-onecorrespondencebetweenpointsof C andtheinterval t [ a,b ] .Then L = b a r( t ) dt. Proof .Owingtotheone-to-onecorrespondencebetween[ a,b ]and C ,givenapartition tkof[ a,b ]suchthat t0= a 0, k =1 2 ,...,N .Under are“nementofthepartition, DN=maxk tk 0as N and therefore tk 0forall k as N .Let r k Š 1= r( tk Š 1).The dierentiabilityof r ( t )impliesthat rkŠ rk Š 1= r k Š 1 tk+ uk tk,

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13012.VECTORFUNCTIONS where uk 0 as tk 0forevery k (cf.(12.2)).Bythetriangle inequality(11.7), r k Š 1 tkŠ uk tk rkŠ rk Š 1 r k Š 1 tk+ uk tk. Thelowerandupperboundsforthelengthofthepolygonalpathare obtainedbytakingthesumover k inthisinequality.Next,itisshown thattheseboundsconvergetotheRiemannintegralof r( t ) over [ a,b ],andtheassertionfollowsfromthesqueezeprinciple. Bythecontinuityofthederivative,thefunction r( t ) iscontinuousandhenceintegrable.Therefore,itsRiemannsumconverges:Nk =1 r k Š 1 tk b a r( t ) dt as N Putmaxk uk = MN(thelargest uk foragivenpartitionsize N ). ThenNk =1 uk tk MN Nk =1 tk= MN( b Š a ) 0as N because uk 0as tk 0forall k ,andhence MN 0as N Itfollowsfromthesqueezeprinciplethatthelimitof N k =1 rkŠ rk Š 1 as N existsandequals b a r( t ) dt Remark. Assume r ( t )iscontinuouslydierentiableon[ a,b ]butdoes notnecessarilyde“neaone-to-onecorrespondencewithitsrange C Thentheintegral b a r( t ) dt isnotthelengthofthecurve C asa pointsetinspacebecause r ( t )maytraverseapartof C severaltimes. Suppose r ( t )isatrajectoryofaparticle.Thenitsvelocityis v ( t )= r( t )anditsspeedis v ( t )= v ( t ) .Thedistancetraveledbythe particleinthetimeinterval[ a,b ]isgivenby D = b av ( t ) dt = b a r( t ) dt. Ifaparticletravelsalongthesamespacecurve(orsomeofitsparts) severaltimes,thenthedistancetraveleddoesnotcoincidewiththearc length L ofthecurve, D L Example 12.11 Findthearclengthofthecurve r ( t )= t2, 2 t, ln t 1 t 2 Solution: Thederivative r( t )= 2 t, 2 1 /t iscontinuouson[1 2]. Itsnormis r( t ) = 4 t2+4+ 1 t2= 2 t + 1 t 2=2 t + 1 t .

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82.ARCLENGTHOFACURVE131 Therefore,byTheorem12.5, L = 2 1 r( t ) dt = 2 1 2 t + 1 t dt = t2 2 1+ln t 2 1=3+ln2 Example 12.12 Findthearclengthofoneturnofahelixofradius R thatrisesby h pereachturn. Solution: Letthehelixaxisbethe z axis(seeStudyProblem12.1). Thehelixistracedoutbythevectorfunction r ( t )= R cos t,R sin t,th/ (2 ) .Oneturncorrespondstotheinterval t [0 2 ].Therefore, r( t ) = Š R sin t,R cos t,h/ (2 ) = R2+( h/ (2 ))2. Sothenormofthederivativeturnsouttobeconstant.Thearclength is L = 2 0 r( t ) dt = R2+( h/ (2 ))22 0dt = (2 R )2+ h2. Thisresultisrathereasytoobtainwithoutcalculus.Thehelixlieson acylinderofradius R .Ifthecylinderiscutparalleltoitsaxisand unfoldedintoastrip,thenoneturnofthehelixbecomesthehypotenuse oftheright-angledtrianglewithcatheti2 R and h .Theresultfollows fromthePythagoreantheorem.Thisconsiderationalsoshowsthat thelengthdoesnotdependonwhetherthehelixwindsaboutitsaxis clockwiseorcounterclockwise. 82.1.ReparameterizationofaCurve.InSection79,itwasshownthata spacecurvede“nedasapointsetinspacecanbetraversedbydierent vectorfunctions.Thesevectorfunctionsaredierentparameterizations ofthesamecurve.Forexample,asemicircleofradius R istraversed bythevectorfunctions r ( t )= R cos t,R sin t, 0 ,t [0 ] R ( u )= u, R2Š u2, 0 ,u [ Š R,R ] Theyarerelatedtooneanotherbythecompositionrule: R ( u )= r ( t ( u )) ,t ( u )=cosŠ 1( u/R ) or r ( t )= R ( u ( t )) ,u ( t )= R cos t. Thisexampleillustratestheconceptofa reparameterization ofacurve. Areparameterizationofacurveisachangeoftheparameterthatlabels

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13212.VECTORFUNCTIONS pointsofthecurve.Itmerelyre”ectsasimplefactthattherearemany dierentvectorfunctionsthattraversethesamespacecurve. Definition 12.12 (ReparameterizationofaCurve) Let r ( t ) traverseacurve C if t [ a,b ] .Let g ( u ) beacontinuousoneto-onefunctiononaninterval [ a,b] whoserangeistheinterval [ a,b ] ; thatis,forevery a t b ,thereisjustone a u bsuchthat t = g ( u ) andviceversa.Thevectorfunction R ( u )= r ( g ( u )) iscalled a reparameterization of C Thegeometricalpropertiesofthecurve(e.g.,itsshapeorlength) donotdependonaparameterizationofthecurvebecausethevector functions r ( t )and R ( u )havethe same range.Areparameterization ofacurveisatechnicaltoolto“ndparametricequationsofthecurve convenientforparticularapplications.82.2.ANaturalParameterizationofaSmoothCurve.Supposeoneis travelingalongahighwayfromtown A totown B andcomesupon anaccident.Howcanthelocationoftheaccidentbereportedtothe police?IfonehasaGPSnavigator,onecanreportcoordinateson thesurfaceoftheEarth.Thisimpliesthatthepoliceshouldusea speci“c(GPS)coordinatesystemtolocatetheaccident.Isitpossible toavoidanyreferencetoacoordinatesystem?Asimplerwaytode“ne thepositionoftheaccidentistoreportthedistancetraveledfrom A alongthehighwaytothepointwheretheaccidenthappened(byusing, e.g.,milemarkers).Nocoordinatesystemisneededtouniquelylabel allpointsofthehighwaybyspecifyingthedistancefromaparticular point A tothepointofinterestalongthehighway.Thisobservation canbeextendedtoallsmoothcurves(seeFigure12.8,rightpanel). Definition 12.13 (NaturalorArcLengthParameterization) Let C beasmoothcurveoflength L betweenpoints A and B .Let r ( t ) t [ a,b ] ,beaone-to-onevectorfunctionthattracesout C sothat r ( a ) and r ( b ) arepositionvectorsof A and B ,respectively.Thenthearc length s = s ( t ) oftheportionofthecurvebetween r ( a ) and r ( t ) isa functionoftheparameter t : s = s ( t )= t a r( u ) du,s [0 ,L ] Thevectorfunction R ( s )= r ( t ( s )) iscalleda naturalorarclength parameterization of C ,where t ( s ) istheinversefunctionof s ( t ) Forasmoothcurve,thefunction r ( t )iscontinuouslydierentiable, andhence r( t ) iscontinuouson[ a,b ].Therefore,thederivative s( t )

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82.ARCLENGTHOFACURVE133 existsandisobtainedbydierentiatingtheintegralwithrespecttoits upperlimit: s( t )= r( t ) > 0.Itispositivebecause r( t ) = 0 fora smoothcurve.Theexistenceoftheinversefunction s ( t )isguaranteed bytheinversefunctiontheoremprovedinCalculusI: Theorem 12.6 (InverseFunctionTheorem) Let s ( t ) a t b ,haveacontinuousderivativesuchthat s( t ) > 0 for a 0guaranteestheexistence ofaone-to-onecorrespondencebetweenthevariables s and t andthe existenceofthedierentiableinversefunction t = t ( s ).Let r ( t )= x ( t ) ,y ( t ) ,z ( t ) beparametricequationsofasmoothcurve C .Then theparametricequationsof C inthenaturalparameterizationhavethe form R ( s )= x ( t ( s )) ,y ( t ( s )) ,z ( t ( s )) Example 12.13 ReparameterizethehelixfromExample12.12, r ( t )= R cos t,R sin t,th/ (2 ) ,withrespecttothearclengthmeasured fromthepoint ( R, 0 0) inthedirectionofincreasing t Solution: Thepoint( R, 0 ,0)correspondsto t =0.Then s ( t )= t 0 r( u ) du = L 2 t 0du = Lt 2 t ( s )= 2 s L where L = (2 R )2+ h2isthearclengthofoneturnofthehelix(see Example12.12).Therefore, R ( s )= r ( t ( s ))= R cos(2 s/L ) ,R sin(2 s/L ) ,hs/L Inparticular, R (0)= R, 0 0 and R ( L )= R, 0 ,h aretheposition vectorsoftheendpointsofoneturnofthehelixasrequired. Example 12.14 Findthecoordinatesofapoint P thatis 5 / 3 unitsoflengthawayfromthepoint (4 0 0) alongthehelix r ( t )= 4cos( t ) 4sin( t ) 3 t Solution: If R ( s )isthenaturalparameterizationofthehelixwhere s iscountedfromthepoint(4 0 0),thenthepositionvectorofthe pointinquestionisgivenby R (5 / 3).Thus,the“rsttaskisto“nd R ( s ).Onehas r( u )= Š 4 sin( u ) 4 cos( u ) 3 r( u ) =5 .

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13412.VECTORFUNCTIONS Theinitialpointofthehelixcorrespondsto t =0.Sothearclength countedfrom(4 0 0)asafunctionof t is s ( t )= t 0 r( u ) du = t 05 du =5 t t ( s )= s 5 Thenaturalparameterizationreads R ( s )= r ( t ( s ))= 4cos( s/ 5) 4sin( s/ 5) 3 s/ 5 Thepositionvectorof P is R (5 / 3)= 2 2 3 .However,thisisnot acompleteanswertotheproblembecausetherearetwopointsofthe helixatthespeci“eddistancefrom(4 0 0).Onesuchpointisupward alongthehelix,andtheotherisdownwardalongit.Notethat s ( t ) de“nedaboveisthearclengthparametercountedfrom(4 0 0)inthe directionof increasing t (upwardalongthehelix, t> 0).Accordingly, s ( t )canbecountedinthedirectionof decreasing t (downwardalong thehelix, t< 0).Inthiscase, s ( t )= Š 5 t> 0.Hence,theposition vectoroftheotherpointis R ( Š 5 / 3)= 2 Š 2 3 Š ItfollowsfromTheorem12.6that thederivativeofavectorfunction thattraversesasmoothcurve C withrespecttothenaturalparameter, thearclength,isa unit tangentvectortothecurve .Indeed,bythe chainruleappliedtothecomponentsofthevectorfunction: d r ( t ) ds = dx ( t ) ds dy ( t ) ds dx ( t ) ds = x( t ) t( s ) ,y( t ) t( s ) ,z( t ) t( s ) = t( s ) x( t ) ,y( t ) ,z( t ) = 1 s( t ) r( t )= 1 r( t ) r( t )= T ( t ) Thus,foranaturalparameterization r ( s )ofasmoothcurve C ,the derivative r( s )isaunittangentvectorto C r( s ) =1. Byde“nition,thearclengthisindependentofaparameterizationof aspacecurve.Forsmoothcurves,thiscanalsobeestablishedthrough thechangeofvariablesintheintegralthatdeterminesthearclength. Indeed,let r ( t ), t [ a,b ],beaone-to-onecontinuouslydierentiable vectorfunctionthattracesoutacurve C oflength L .Considerthe changeoftheintegrationvariable t = t ( s ), s [0 ,L ].Then,bythe inversefunctiontheorem, s( t )= r( t ) and ds = s( t ) dt = r( t ) dt Thus, L = b a r( t ) dt = L 0ds foranyparameterizationofthecurve C .

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82.ARCLENGTHOFACURVE135 82.3.Exercises.(1) Findthearclengthofeachofthefollowingcurves: (i) r ( t )= 3cos t, 2 t, 3sin t Š 2 t 2 (ii) r ( t )= 2 t,t3/ 3 ,t2 0 t 1 (iii) r ( t )= 3 t2, 4 t3 / 2, 3 t 0 t 2 (iv) r ( t )= et, 2 t,eŠ t Š 1 t 1 (v) r ( t )= cosh t, sinh t,t 0 t 1 (vi) r ( t )= cos t + t sin t, sin t + t cos t,t2 ,0 t 2 Hint: Findthedecomposition r ( t )= v ( t ) Š t w ( t )+ t2 e3,where v w and e3aremutuallyorthogonaland v( t )= w ( t ), w( t )= Š v ( t ).Use thePythagoreantheoremtocalculate r( t ) (2) Findthearclengthofthecurve r ( t )= eŠ tcos t,eŠ tsin t,eŠ t 0 t< Hint: Put r ( t )= eŠ tu ( t ),dierentiate,showthat u ( t ) isorthogonalto u( t ),andusethePythagoreantheoremtocalculate r( t ) (3) Findthearclengthoftheportionofthehelix r ( t )= cos t, sin t,t thatliesinsidethesphere x2+ y2+ z2=2. (4) Findthearclengthoftheportionofthecurve r ( t )= 2 t, 3 t2, 3 t3 thatliesbetweentheplanes z =3and z =24. (5) Findthearclengthoftheportionofthecurve r ( t )= ln t,t2, 2 t thatliesbetweenthepointsofintersectionofthecurvewiththeplane y Š 2 z +3=0. (6) Let C bethecurveofintersectionofthesurfaces z2=2 y and 3 x = yz .Findthelengthof C fromtheorigintothepoint(36 18 6). (7) Foreachofthefollowingcurvesde“nedbythegivenequationswith aparameter a ,“ndsuitableparametricequationsandevaluatethearc lengthbetweenagivenpoint A andandagenericpoint B =( x0,y0,z0): (i) y = a sinŠ 1( x/a ), z =( a/ 4)ln[( a Š x ) / ( a + x )], A =(0 0 0) (ii)( x Š y )2= a ( x + y ), x2Š y2=9 z2/ 8, A =(0 0 0) Hint: Usethenewvariables u = x + y and v = x Š y to“ndthe parametricequations. (iii) x2+ y2= az y/x =tan( z/a ), A =(0 0 0) Hint: Usethepolarcoordinatesinthe xy planeto“ndtheparametric equations. (iv) x2+ y2+ z2= a2, x2+ y2cosh(tanŠ 1( y/x ))= a A =( a, 0 0) Hint: Representthesecondequationasapolargraph. (8) Reparameterizeeachofthefollowingcurveswithrespecttothe arclengthmeasurefromthepointwhere t =0inthedirectionof increasing t : (i) r = t, 1 Š 2 t, 5+3 t (ii) r =2 t t2+1 e1+(2 t2+1Š 1) e3

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13612.VECTORFUNCTIONS (iii) r ( t )= cosh t, sinh t,t (iv) x = a ( t Š sin t ), y = a (1 Š cos t ), a> 0 (9) Aparticletravelsalongahelixofradius R thatrises h unitsof lengthperturn.Letthe z axisbethesymmetryaxisofthehelix.If aparticletravelsthedistance4 R fromthepoint( R, 0 0),“ndthe positionvectoroftheparticle. (10) Aparticletravelsalongacurvetraversedbythevectorfunction r ( u )= u, cosh u, sinh u fromthepoint(0 1 0)withaconstantspeed 2m/ssothatits x coordinateincreases.Findthepositionofthe particlein1second. (11) Let C beasmoothclosedcurvewhosearclengthis L .Let r ( t ) beavectorfunctionthattraverses C onlyoncefor a t b .Prove thatthereisanumber a t b suchthat r( t) = L/ ( b Š a ). Hint: Recalltheintegralmeanvaluetheorem. (12) Aparticletravelsinspaceadistance D intime T .Showthat thereisamomentoftime0 t T atwhichthespeedoftheparticle coincideswiththeaveragespeed D/T 83.CurvatureofaSpaceCurve Considertwocurvespassingthroughapoint P .Bothcurvesbend at P .Whichonebendsmorethantheotherandhowmuchmore? Theanswertothisquestionrequiresanumericalcharacterizationof bending,thatis,anumbercomputedat P foreachcurvewiththe propertythatitbecomeslargerasthecurvebendsmore.Naturally,this numbershouldnotdependonaparameterizationofacurve.Suppose thatacurveissmoothsothataunittangentvectorcanbeattached toeverypointofthecurve.Astraightlinedoesnotbend(doesnot curveŽ)soithasthesameunittangentvectoratallitspoints.If acurvebends,thenitsunittangentvectorbecomesafunctionofits positiononthecurve.Thepositiononthecurvecanbespeci“edina coordinate-andparameterization-independentwaybythearclength s countedfromaparticularpointofthecurve.If T ( s )istheunit tangentvectorasafunctionof s ,thenitsderivative T( s )vanishesfor astraightline(seeFigure12.9),whilethisisnotthecaseforageneral curve.Fromthede“nitionofthederivative T( s0)=lims s0 T ( s ) Š T ( s0) s Š s0, itfollowsthatthemagnitude T( s0) becomeslargerwhenthecurve bendsmore.ŽFora“xeddistance s Š s0betweentwoneighboring

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83.CURVATUREOFASPACECURVE137 Figure12.9.Left :Astraightlinedoesnotbend.The unittangentvectorhaszerorateofchangerelativetothearc lengthparameter s Right :Curvatureofasmoothcurve. Themoreasmoothcurvebends,thelargertherateofchange oftheunittangentvectorrelativetothearclengthparameterbecomes.Sothemagnitudeofthederivative(curvature) T( s ) = ( s )canbetakenasageometricalmeasureof bending.pointsofthecurve,themagnitude T ( s ) Š T ( s0) becomeslarger whenthecurvebendsmoreatthepointcorrespondingto s0.Sothe number T( s0) canbeusedasanumericalmeasureofthebending or curvature ofacurve. Definition 12.14 (CurvatureofaSmoothCurve) Let C beasmoothcurveandletitsunittangentvector T ( s ) bea dierentiablefunctionofthearclengthcountedfromaparticularpoint of C .Thenumber ( s )= d ds T ( s ) iscalledthe curvature of C atthepointcorrespondingtothevalue s ofthearclength. Let r ( s )bethenaturalparameterizationofasmoothcurve(the parameter s isthearclengthmeasuredfromaparticularpointonthe curve).Then r( s )= T ( s ),asshownintheprevioussection,and therefore ( s )= r( s ) Example 12.15 Findthecurvatureofahelixofradius R thatrises adistance h perturn. Solution: InExample12.13,thenaturalparameterizationofthe helixisobtained r ( s )= R cos(2 s/L ) ,R sin(2 s/L ) ,hs/L .

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13812.VECTORFUNCTIONS where L = (2 R )2+ h2isthearclengthofoneturn.Dierentiating thisvectorfunctiontwicewithrespecttothearclengthparameter s r( s )= Š (2 /L )2R cos(2 s/L ) Š (2 /L )2R sin(2 s/L ) 0 = Š (2 /L )2R cos(2 s/L ) sin(2 s/L ) 0 ( s )= r( s ) =(2 /L )2R = R R2+( h/ 2 )2, wheretherelation cos u, sin u, 0 =1hasbeenused.Sothehelix hasaconstantcurvature. Inpractice,“ndingthenaturalparameterizationofasmoothcurve mightbeatedioustechnicaltask.Therefore,aquestionofinterestisto developamethodtocalculatethecurvatureinanyparameterization. Let r ( t )beavectorfunctionin[ a,b ]thattracesoutasmoothcurve C Theunittangentvectorasafunctionoftheparameter t hastheform T ( t )= r( t ) / r( t ) .So,tocalculatethecurvatureasafunctionof t ,therelationbetweenthederivatives d/ds and d/dt hastobefound. If s = s ( t )isthearclengthasafunctionof t (seeDe“nition12.13), then,bytheinversefunctiontheorem(Theorem12.6),thereexistsan inversedierentiablefunction t = t ( s )thatexpressestheparameter t as afunctionofthearclength s and dt ( s ) /ds =1 / ( ds ( t ) /dt )=1 / r( t ) Bythechainrule, d ds T = dt ds d dt T = 1 r( t ) d dt T andtherefore (12.4) ( t )= T( t ) r( t ) Notethattheexistenceofthecurvaturerequiresthat r ( t )betwice dierentiablebecause T ( t )isproportionalto r( t ).Dierentiationof theunitvector T cansometimesbearathertechnicaltask,too.The followingtheoremprovidesamoreconvenientwaytocalculatethe curvature. Theorem 12.7 (CurvatureofaCurve) Letasmoothcurvebetracedoutbyatwice-dierentiablevectorfunction r ( t ) .Thenthecurvatureis (12.5) ( t )= r( t ) r( t ) r( t ) 3. Proof. Put v ( t )= r( t ) .Withthisnotation, r( t )= v ( t ) T ( t ) .

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83.CURVATUREOFASPACECURVE139 Dierentiatingbothsidesofthisrelation,oneinfers (12.6) r( t )= v( t ) T ( t )+ v ( t ) T( t )= v( t ) v ( t ) r( t )+ v ( t ) T( t ) Sincethecrossproductoftwoparallelvectorsvanishes,itfollowsfrom (12.6)that (12.7) r( t ) r( t )= v ( t ) r( t ) T( t ) Nowrecallthat a b = a b sin ,where istheanglebetween vectors a and b .Therefore, (12.8) r( t ) r( t ) = v ( t ) r( t ) T( t ) = r( t ) 2 T( t ) sin where istheanglebetween T( t )andthetangentvector r( t ).Since T ( t )isaunitvector,onehas T ( t ) 2= T ( t ) T ( t )=1.Bytakingthe derivativeofbothsidesofthelatterrelation,itisconcludedthatthe vectors T( t )and r( t )areorthogonal: T( t ) T ( t )=0 T( t ) T ( t ) T( t ) r( t ) = 2 because r( t )isparallelto T ( t ).Hence,sin =1.Thesubstitution ofthelatterrelationand T( t ) = ( t ) r( t ) (see(12.4))into(12.8) yields r r = r3 fromwhich(12.5)follows. Example 12.16 Findthecurvatureofthecurve r ( t )= ln t,t2, 2 t atthepoint P0(0 1 2) Solution: Thepoint P0correspondsto t =1because r (1)= 0 1 2 coincideswiththepositionvectorof P0.Hence,onehastocalculate (1): r(1)= tŠ 1, 2 t, 2 t =1= 1 2 2 r(1) =3 r(1)= Š tŠ 2, 2 0 t =1= Š 1 2 0 r(1) r(1)= Š 4 Š 2 4 =2 Š 2 Š 1 2 (1)= r(1) r(1) r(1) 3= 2 2 1 Š 2 33= 6 27 = 2 9 Equation(12.5)canbesimpli“edintwoparticularlyinteresting cases.Ifacurveisplanar(i.e.,itliesinaplane),then,bychoosing thecoordinatesystemsothatthe xy planecoincideswiththeplane inwhichthecurvelies,onehas r ( t )= x ( t ) ,y ( t ) 0 .Since rand rareinthe xy plane,theircrossproductisparalleltothe z axis:

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14012.VECTORFUNCTIONS r r= 0 0 ,xyŠ xy .Thesubstitutionofthisrelationinto(12.5) leadstothefollowingresult. Corollary 12.1 (CurvatureofaPlanarCurve) Thecurvatureofaplanarsmoothcurve r ( t )= x ( t ) ,y ( t ) 0 is = | xyŠ xy| [( x)2+( y)2]3 / 2. Afurthersimpli“cationoccurswhentheplanarcurveisagraph y = f ( x ).Thegraphistracedoutbythevectorfunction r ( t )= t,f ( t ) 0 Then,inCorollary12.1, x( t )=1, x( t )=0,and y( t )= f( t )= f( x ),whichleadstothefollowingresult. Corollary 12.2 (CurvatureofaGraph) Thecurvatureofthegraph y = f ( x ) is ( x )= | f( x ) | [1+( f( x ))2]3 / 2.83.1.GeometricalSignicanceoftheCurvature.Letuscalculatethe curvatureofacircleofradius R .Onehas x ( t )= R cos t y ( t )= R sin t x( t )= Š R sin t y( t )= R cos t x( t )= Š R cos t y( t )= Š R sin t ByCorollary12.1, ( x)2+( y)2= R2xyŠ xy= R2 = R2 R3= 1 R Therefore,thecurvatureisconstantalongthecircleandequalsareciprocalofitsradius.Thefactthatthecurvatureisindependentofits positiononthecirclecanbeanticipatedfromtherotationalsymmetry ofthecircle(itbendsuniformly).Naturally,iftwocirclesofdierent radiipassthroughthesamepoint,thenthecircleofsmallerradius bendsmore.Notealsothatthecurvaturehasthedimensionofthe inverselength.Thismotivatesthefollowingde“nition. Definition 12.15 (CurvatureRadius) Thereciprocalofthecurvatureofacurveiscalledthe curvatureradius ( t )=1 / ( t ) Thecurvatureradiusisafunctionofapointonthecurve.Leta planarcurvehaveacurvature atapoint P .Consideracircleofradius =1 / throughthesamepoint P .Thecurveandthecirclehavethe samecurvatureat P ;thatis,inasucientlysmallneighborhoodof P ,thecircleapproximatesthecurvebetterthanthetangentlineat

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83.CURVATUREOFASPACECURVE141 Figure12.10.Left :Curvatureradius.Asmoothcurve nearapoint P canbeapproximatedbyaportionofacircle ofradius =1 / .Thecurvebendsinthesamewayasacircleofradiusthatisthereciprocalofthecurvature.Alarge curvatureatapointcorrespondstoasmallcurvatureradius. Middle :Osculatingplaneandosculatingcircle.Theosculatingplaneatapoint P containsthetangentvector T and itsderivative Tat P andhenceisorthogonalto n = T T. Theosculatingcircleliesintheosculatingplane,ithasradius =1 / ,anditscenterisatadistance from P in thedirectionof T.OnesaysthatthecurvebendsŽinthe osculatingplane. Right :Foracurvetracedoutbyavector function r ( t ),thederivatives rand ratanypoint P0lie intheosculatingplanethrough P0.Sothenormaltothe osculatingplanecanalsobecomputedas n = r ( t0) r( t0), where r ( t0)isthepositionvectorof P0.P becausethecircleandthecurveareequallybentat P anddonot justhavethesameunittangentvectorat P .So,ifonesaysthatthe curvatureofacurveatapoint P is inversemeters,thenthecurve lookslikeacircleofradius1 / metersnear P Forageneralspacecurve,noteverycircleofradius =1 / that passesthrough P wouldapproximatewellthecurvenear P .Let r ( s )be anaturalparameterizationofacurvesuchthatthepositionvectorof P is r (0)(i.e.,thearclengthparameterismeasuredfrom P ).Consider aplanecontainingthetangentlinetothecurveat P .Let R ( s )bea naturalparameterizationofthecircleofradius thatliesintheplane andpassesthrough P sothat r (0)= R (0).Theplaneandthecirclein itcanberotatedaboutthetangentline.Thedeviation r ( s ) Š R ( s ) (ortheapproximationerror)inasmall“xedintervaldependsonthe orientationoftheplane.Thinkaboutaportionofthecurveoflength s andaportionofthecircleofthesamelengththathaveacommon endpoint(thepoint P )sothatthepartofthecirclecanberigidly rotatedaboutthetangentlinethrough P .For s smallenough,theerror isroughlydeterminedbythedistancebetweentheotherends.Now

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14212.VECTORFUNCTIONS recallfromCalculusIthatatwice-dierentiablefunctioncanbewell approximatedbyitssecondTaylorpolynomialinasucientlysmall neighborhoodofanypoint: x ( s ) x (0)+ x(0) s + x(0) s2/ 2= T2( s )and theapproximationerrordecreasesto0fasterthan s2withdecreasing s thatis,( x ( s ) Š T2( s )) /s2 0as s 0.UsingtheTaylorapproximation foreachcomponentofthevectorfunctions r ( s )and R ( s ),one“nds r ( s ) r (0)+ r(0) s + r(0) s2/ 2 R ( s ) R (0)+ R(0) s + R(0) s2/ 2 Sincethecurveandthecirclehavehaveacommonpointat s =0, r (0)= R (0).Furthermore,theyhavethesameunittangentvectorat thecommonpointand,hence r(0)= R(0).Foranaturalparameterization,theunittangentvectorstothecurveandthecirclearegiven bythederivatives r( s )and R( s ).Consequently,theapproximation error r ( s ) Š R ( s ) r(0) Š R(0) s2/ 2willbeminimalifthecurve andthecirclehavethesamederivativeoftheunittangentvectorat thecommonpoint, r(0)= R(0).Moreaccurately,theapproximation errorwilldecreaseto0fasterthan s2withdecreasing s inthiscase. Sincetheunittangentvectoranditsderivativeareorthogonal(seethe proofofTheorem12.7),theylieinoneparticularplanethroughthe point r (0).Thus,thebestapproximationofthecurvebyacircleof radius1 / (0)at P isachievedwhenthecircleliesintheplanethrough P thatisparalleltothetangentvector T (0)= r(0)anditsderivative T(0)= r(0)at P .Thenormalofthisplaneis n = T (0) T(0).By thegeometricalinterpretationofthederivative, T(0)shouldpointin thedirectioninwhichthecurvebends(seeFigure12.9).Therefore, thecenterofthecirclemustbeinthedirectionof T(0)from P ,not intheoppositeone. Definition 12.16 (OsculatingPlaneandCircle) Theplanethroughapoint P ofacurvethatisparalleltotheunit tangentvector T anditsderivative T = 0 at P iscalledthe osculating plane at P .Thecircleofradius =1 / ,where isthecurvatureat P ,through P thatliesintheosculatingplaneandwhosecenterisin thedirectionof Tfrom P iscalledthe osculatingcircle at P Theorem 12.8 (EquationoftheOsculatingPlane) Letacurve C betracedoutbyatwice-dierentiablevectorfunction r ( t ) Let P0beapointof C suchthatitspositionvectoris r ( t0)= x0,y0,z0 atwhichthevector n = r( t0) r( t0) doesnotvanish.Anequationof theosculatingplanethrough P0is n1( x Š x0)+ n2( y Š y0)+ n3( z Š z0)=0 n = n1,n2,n3 .

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83.CURVATUREOFASPACECURVE143 Proof. Itfollowsfrom(12.6)thatthesecondderivative r( t0)lies intheosculatingplanebecauseitisalinearcombinationof T ( t0)and T( t0).Hence,theosculatingplanecontainsthe“rstandsecondderivatives r( t0)and r( t0).Therefore,theircrossproduct n = r( t0) r( t0) isperpendiculartotheosculatingplane,andtheconclusionofthetheoremfollows. Example 12.17 Forthecurve r ( t )= t,t2,t3 ,“ndtheosculating planethroughthepoint (1 1 1) Solution: Thepointinquestioncorrespondsto t =1.Then r(1)= 1 2 t, 3 t2 t =1= 1 2 3 r(1)= 0 2 6 t t =1= 0 2 6 Therefore,thenormaloftheosculatingplaneis n = r(1) r(1)= 1 2 3 0 2 6 = 6 Š 6 2 .Theosculatingplaneis6( x Š 1) Š 6( y Š 1)+2( z Š 1)=0or3 x Š 3 y + z =1. Example 12.18 Findtheosculatingcircleforthegraph y =cos(2 x ) atthepoint (0 1) Solution: Foraplanarcurve,theosculatingplaneistheplanein whichthecurvelies.Since y(0)= Š 2sin0=0,thetangentlineto thegraphishorizontal, y =1,andthe y axisisthelinenormaltothe tangentlineat(0 1).Therefore,thecenteroftheosculatingcirclelies onthe y axisdownfrom y =1becausethegraphofcos(2 x )isconcave downward.Thecurvatureofthegraphat x =0isfoundbyCorollary 12.2: y( x )= Š 2sin(2 x ), y( x )= Š 4cos(2 x ),and (0)= |Š 4 | =4. Sothecurvatureradiusis (0)=1 / (0)=1 / 4.Thecenterofthe osculatingcirclelies1/4ofunitlengthdownfrom y =1,thatis,at (0 3 / 4).Theequationoftheosculatingcircleis x2+( y Š 3 / 4)2=(1 / 4)2. 83.2.StudyProblems.Problem12.12. Showthatanysmoothcurvewhosecurvaturevanishesisastraightline. Solution: Let r ( s )beanaturalparameterizationofasmoothcurve. Thenthederivativeisaunittangentvectortothecurve, T ( s )= r( s ).Bythede“nitionofthecurvature, ( s )= T( s ) = r( s ) .If ( s )=0,then r( s )= 0 forall s .Therefore,theunittangentvector r( s )= T isaconstantvector.Theintegrationofthisrelationyields

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14412.VECTORFUNCTIONS r ( s )= r0+ T s ,whichisthevectorequationofalinethroughthepoint r0andparallelto T Problem12.13. Findthemaximalcurvatureofthegraphoftheexponential, y = ex,andthepoint(s)atwhichitoccurs. Solution: ThecurvatureofthegraphiscalculatedbyCorollary12.2: ( x )= ex/ (1+ e2 x)3 / 2.Criticalpointsaredeterminedby ( x )=0or ( x )= ex(1+ e2 x)1 / 2[2 e2 xŠ 1] (1+ e2 x)3=0 2 e2 xŠ 1=0 x = Š ln2 2 Fromtheshapeofthegraphoftheexponential,itisclearthatthe foundcriticalpointcorrespondstothe(absolute)maximumof ( x ) (maximalbending)and max= ( Š ln(2) / 2)=2 / 33 / 2. Problem12.14.(EquationoftheOsculatingCircle) Findavectorfunctionthattracesouttheosculatingcircleofacurve r ( t ) atapoint r ( t0) Solution: Put r0= r ( t0)and T0= T ( t0)(theunittangentvector tothecurveatthepointwiththepositionvector r0).Put N0= T( t0) / T( t0) ;itisaunitvectorinthedirectionof T( t0).Let 0= 1 / ( t0)bethecurvatureradiusatthepoint r0.Thecenterofthe osculatingcirclemustlie 0unitsoflengthfromthepoint r0inthe directionof N0.Thus,itspositionvectoris R0= r0+ 0 N0.Let R ( t ) bethepositionvectorofagenericpointoftheosculatingcircle.Then thevector R ( t ) Š R0liesintheosculatingplaneandhencemustbea linearcombinationof T0and N0,thatis, R ( t ) Š R0= a ( t ) N0+ b ( t ) T0. Since R ( t )traversesacircleofradius 0centeredat R0,thelengthof R ( t ) Š R0mustbe 0forany t .Owingtotheorthogonalityoftheunit vectors T0and N0,thisconditionyields: R ( t ) Š R02= 2 0 a2( t )+ b2( t )= 2 0bythePythagoreantheorem.Parametricequationsofthecirclecan betakenintheform a ( t )= Š 0cos t and b ( t )= 0sin t .Thevector functionthattracesouttheosculatingcircleis R ( t )= r0+ 0(1 Š cos t ) N0+ 0sin t T0, where t [0 2 ].Theabovechoiceof a ( t )and b ( t )hasbeenmadeso that R (0)= r0.

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83.CURVATUREOFASPACECURVE145 Problem12.15. Considerahelix r ( t )= R cos( t ) ,R sin( t ) ,ht where and h arenumericalparameters.Thearclengthofoneturnof thehelixisafunctionoftheparameter L = L ( ) ,andthecurvature atany“xedpointofthehelixisalsoafunctionof = ( ) .Use onlygeometricalarguments(nocalculus)to“ndthelimitsof L ( ) and ( ) as Solution: Thevectorfunction r ( t )tracesoutoneturnofthehelix when t rangesovertheperiodofcos( t )orsin( t )(i.e.,overtheinterval oflength2 / ).Thus,thehelixrisesby2 h/ = H ( )alongthe z axispereachturn.When ,theheight H ( )tendsto0so thateachturnofthehelixbecomescloserandclosertoacircleofradius R .Therefore, L ( ) 2 R (thecircumference)and ( ) 1 /R (the curvatureofthecircle)as Acalculusapproachrequiresalotmoreworktoestablishthisresult: L ( )= 2 / 0 r( t ) dt = 2 ( R )2+ h2=2 R2+( h/ )2 2 R, ( )= r( t ) r( t ) r( t ) 3= R2[( R )2+ h2]1 / 2 [( R )2+ h2]3 / 2= R R2+( h/ )2 1 R as 83.3.Exercises.(1) Findthecurvatureofeachofthefollowingcurvesasafunctionof theparameterandthecurvatureradiusataspeci“edpoint P : (i) r ( t )= t, 1 Š t,t2+1 ,P (1 0 2) (ii) r ( t )= t2,t, 1 ,P (4 2 1) (iii) y =sin( x/ 2) ,P ( 1) (iv) r ( t )= 4 t3 / 2, Š t2,t ,P (4 Š 1 1) (v) x =1+ t2,y =2+ t3,P (2 1) (vi) x = etcos t,y =0 ,z = etsin t,P (1 0 0) (vii) r ( t )= ln t, t,t2 ,P (0 1 1) (viii) r ( t )= cosh t, sinh t, 2+ t ,P (1 0 2) (ix) r ( t )= et, 2 t,eŠ t ,P (1 0 1) (x) r ( t )= sin t Š t cos t,t2, cos t + t sin t ,P (0 0 1) (2) Findthecurvatureof r ( t )= t,t2/ 2 ,t3/ 3 atthepointofitsintersectionwiththesurface z =2 xy +1 / 3.

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14612.VECTORFUNCTIONS (3) Findthemaximalandminimalcurvaturesofthegraph y =cos( ax ) andthepointsatwhichtheyoccur.Sketchthegraphfor a =1and markthepointsofthemaximalandminimalcurvatures,localmaxima andminimaofcos x ,andthein”ectionpoints. (4) Useageometricalinterpretationofthecurvaturetoguessthepoint onthegraphs y = ax2and y = ax4wherethemaximalcurvatureoccurs.Thenverifyyourguessbycalculations. (5) Let f ( x )beatwicecontinuouslydierentiablefunctionandlet ( x )bethecurvatureofthegraph y = f ( x ). (i)Does attainalocalmaximumvalueateverylocalminimumand maximumof f ?Ifnot,stateanadditionalconditionon f underwhich theanswertothisquestionisarmative. (ii)Provethat =0atin”ectionpointsofthegraph. (iii)Showbyanexamplethattheconverseisnottrue,thatis,thatthe curvaturevanishesat x = x0doesnotimplythatthepoint( x0,f ( x0)) isanin”ectionpoint. (6) Let f betwicedierentiableat x0.Let T2( x )beitsTaylorpolynomialofthesecondorderabout x = x0.Comparethecurvaturesofthe graphs y = f ( x )and y = T2( x )at x = x0. (7) Findtheequationoftheosculatingcirclestotheparabola y = x2atthepoints(0 0)and(1 1). (8) Findthemaximalandminimalcurvatureoftheellipse x2/a2+ y2/b2=1, a>b ,andthepointswheretheyoccur.Givetheequations oftheosculatingcirclesatthesepoints. (9) Let r ( t )= t3,t2, 0 .Thiscurveisnotsmoothandhasacuspat t =0.Findthecurvaturefor t =0andinvestigateitslimitas t 0. (10) Findanequationoftheosculatingplaneforeachofthefollowing curvesataspeci“edpoint: (i) r ( t )= 4 t3 / 2, Š t2,t ,P (4 Š 1 1) (ii) r ( t )= ln t, t,t2 ,P (0 1 1) (11) Findanequationfortheosculatingandnormalplanesforthe curve r ( t )= ln( t ) 2 t,t2 atthepoint P0ofitsintersectionwiththe plane y Š z =1.Aplaneisnormaltoacurveatapointifthetangent tothecurveatthatpointisnormaltotheplane. (12) Isthereapointonthecurve r ( t )= t,t2,t3 wheretheosculatingplaneisparalleltotheplane3 x +3 y + z =1? (13) Provethatthetrajectoryofaparticlehasaconstantcurvature iftheparticlemovessothatthemagnitudesofitsvelocityandaccelerationvectorsareconstant. (14) Consideragraph y = f ( x )suchthat f( x0) =0.Atapoint

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84.APPLICATIONSTOMECHANICSANDGEOMETRY147 ( x0,y0)onthecurve,where y0= f ( x0),“ndtheequationoftheosculatingcircleintheform( x Š a )2+( y Š b )2= R2. Hint: Show“rstthat thevector 1 ,f( x0) istangenttothegraphandavectororthogonalto itis Š f( x0) 1 .Thenconsidertwocases f( x0) > 0and f( x0) < 0. (15) Findtheosculatingcircleforthecycloid x = a ( t Š sin t ), y = a (1 Š cos t )atthepoint t = / 2. (16) Letasmoothcurve r = r ( t )beplanarandlieinthe xy plane.At apoint( x0,y0)onthecurve,“ndtheequationoftheosculatingcircle intheform( x Š a )2+( y Š b )2= R2. Hint: UsetheresultofStudy Problem12.14toexpresstheconstants a b ,and R via x0, y0,andthe curvatureat( x0,y0). 84.ApplicationstoMechanicsandGeometry84.1.TangentialandNormalAccelerations.Let r ( t )bethetrajectory ofaparticle( t istime).Then v ( t )= r( t )and a ( t )= v( t )arethe velocityandaccelerationoftheparticle.Themagnitudeofthevelocity vectoristhespeed, v ( t )= v ( t ) .If T ( t )istheunittangentvectorto thetrajectory,then T( t )isorthogonaltoit.Theunitvector N ( t )= T( t ) / T( t ) iscalledaunit normal tothetrajectory.Inparticular, theosculatingplaneatanypointofthetrajectorycontains T ( t )and N ( t ).Thedierentiationoftherelation v ( t )= v ( t ) T ( t )(see(12.6)) showsthatthataccelerationalwaysliesintheosculatingplane: a = v T + v T= v T + v T N Furthermore,substitutingtherelations = T /v and =1 / into thelatterequation,one“nds(seeFigure12.11,leftpanel)that a = aT T + aN N aT= v= T a = v a v aN= v2= v2 = v a v Definition 12.17 (TangentialandNormalAccelerations) Scalarprojections aTand aNoftheaccelerationvectorontotheunit tangentandnormalvectorsatanypointofthetrajectoryofmotionare called tangentialandnormalaccelerations ,respectively. Thetangentialacceleration aTdeterminestherateofchangeof aparticlesspeed,whilethenormalaccelerationappearsonlywhen theparticlemakesaturn.ŽInparticular,acircularmotionwitha constantspeed, v = v0,hasnotangentialacceleration, aT=0,and

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14812.VECTORFUNCTIONS Figure12.11.Left :Decompositionoftheacceleration a ofaparticleintonormalandtangentialcomponents.The tangentialcomponent aTisthescalarprojectionof a onto theunittangentvector T .Thenormalcomponentisthe scalarprojectionof a ontotheunitnormalvector N .The vectors r and v arethepositionandvelocityvectorsofthe particle. Right :Thetangent,normal,andbinormalvectors associatedwithasmoothcurve.Thesevectorsaremutually orthogonalandhaveunitlength.Thebinormalisde“nedby B = T N .Theshapeofthecurveisuniquelydetermined bytheorientationofthetripleofvectors T N ,and B as functionsofthearclengthparameteruptogeneralrigid rotationsandtranslationsofthecurveasthewhole.thenormalaccelerationisconstant, aN= v2 0/R ,where R isthecircle radius.Indeed,takingthederivativeoftherelation v v = v2 0,itis concludedthat v v =0or a v =0or aT=0.Sincethecurvatureof acircleisthereciprocalofitsradius, aN= v2= v2 0/R Togainanintuitiveunderstandingofthetangentialandnormal accelerations,consideracarmovingalongaroad.Thespeedofthecar canbechangedbypressingthegasorbrakepedals.Whenoneofthese pedalsissuddenlypressed,onecanfeelaforcealongthedirectionof motionofthecar(thetangentialdirection).Thecarspeedometeralso showsthatthespeedchanges,indicatingthatthisforceisduetotheaccelerationalongtheroad(i.e.,thetangentialacceleration aT= v =0). Whenthecarmovesalongastraightroadwithaconstantspeed,its accelerationis0.Whentheroadtakesaturn,thesteeringwheelmust beturnedinordertokeepthecarontheroad,whilethecarmaintainsaconstantspeed.Inthiscase,onecanfeelaforcenormalto theroad.Itislargerforsharperturns(largercurvatureorsmaller curvatureradius)andalsogrowswhenthesameturnispassedwitha greaterspeed.Thisforceisduetothenormalacceleration, aN= v2/ andiscalleda centrifugalforce .ByNewtonslaw,itsmagnitudeis

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84.APPLICATIONSTOMECHANICSANDGEOMETRY149 F = maN= mv2/ ,where m isthemassofamovingobject(e.g.,a car).Whenmakingaturn,thecardoesnotslideotheroadaslong asthefrictionforcebetweenthetiresandtheroadcompensatesfor thecentrifugalforce.Themaximalfrictionforcedependsontheroad andtireconditions(e.g.,awetroadandworntiresreducesubstantially themaximalfrictionforce).Thecentrifugalforceisdeterminedbythe speed(thecurvatureoftheroadis“xedbytheroadshape).So,fora highenoughspeed,thecentrifugalforcecannolongerbecompensated forbythefrictionforceandthecarwouldskidotheroad.Forthis reason,suggestedspeedlimitsignsareoftenplacedathighwayexits. Ifonedrivesacaronahighwayexitwithaspeedtwiceashighas thesuggestedspeed, theriskofskiddingotheroadisquadrupled,not doubled ,becausethenormalacceleration aN= v2/ quadrupleswhen thespeed v isdoubled. Example 12.19 Aroadhasaparabolicshape, y = x2/ (2 R ) ,where ( x,y ) arecoordinatesofpointsoftheroadand R isaconstant(all measuredinunitsoflength,e.g.,meters).Asafetyassessmentrequires thatthenormalaccelerationontheroadshouldnotexceedathres hold value am(e.g.,meterspersecondsquared)toavoidskiddingothe road.Ifacarmoveswithaconstantspeed v0alongtheroad,“ndthe portionoftheroadwherethecarmightskidotheroad. Solution: Thenormalaccelerationofthecarasafunctionof position (nottime!)is aN( x )= ( x ) v2 0.Thecurvatureofthegraph y = x2/ (2 R ) is ( x )=(1 /R )[1+( x/R )2]Š 3 / 2.Themaximalcurvatureandhencethe maximalnormalaccelerationareattainedat x =0.So,ifthespeed issuchthat aN(0)= v2 0/Ram. Thecarcanskidotheroadwhenmovingonitspartcorrespondingto theinterval Š R ( Š 1)1 / 2 x R ( Š 1)1 / 2.Conversely,asuggested speedlimitsigncanbeplaced v0< Ramforapartoftheroadthat contains x =0. 84.2.Frenet-SerretFormulas.Theshapeofaspacecurveasapointset isindependentofaparameterizationofthecurve.Anaturalquestion arises:Whatparametersofthecurvedetermineitsshape?Suppose thecurveissmoothenoughsothattheunittangentvector T ( s )andits derivative T( s )canbede“nedasfunctionsofthearclength s counted

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15012.VECTORFUNCTIONS fromanendpointofthecurve.Let N ( s )betheunitnormalvectorof thecurve. Definition 12.18 (BinormalVector) Let T and N betheunittangentandnormalvectorsatapointofa curve.Theunitvector B = T N iscalledthe binormal(unit)vector So,witheverypointofasmoothcurve,onecanassociateatripleof mutuallyorthogonalunitvectorssothatoneofthemistangenttothe curvewhiletheothertwospantheplanenormaltothetangentvector (normaltothecurve).Byasuitablerotation,thetripleofvectors T N ,and B canbeorientedparalleltotheaxesofanygivencoordinate system,thatis,parallelto e1, e2,and e3,respectively.Indeed, T and N canalwaysbemadeparallelto e1and e2.Then,owingtotherelation e1 e2= e3,thebinormalmustbeparallelto e3.Inotherwords, T N ,and B de“nea right-handed coordinatesystem.Theorientation oftheunittangent,normal,andbinormalvectorsrelativetosome coordinatesystemdependsonthepointofthecurve.Thetripleof thesevectorscanonlyrotateasthepointslidesalongthecurve(the vectorsaremutuallyorthogonalandunitatanypoint).Therefore,the rateswithrespecttothearclengthatwhichthesevectorschangemust becharacteristicfortheshapeofthecurve(seeFigure12.11,right panel). Bythede“nitionofthecurvature, T( s )= ( s ) N ( s ).Next,considertherate: B=( T N )= T N + T N= T Nbecause T( s )isparallelto N ( s ).Itfollowsfromthisequationthat Bisperpendicularto T ,and,since B isaunitvector,itsderivative mustalsobeperpendicularto B .Thus, Bmustbeparallelto N Thisconclusionestablishestheexistenceofanotherscalarquantity thatcharacterizesthecurveshape. Definition 12.19 (TorsionofaCurve) Let N ( s ) and B ( s ) beunitnormalandbinormalvectorsofthecurve asfunctionsofthearclength s .Then d B ( s ) ds = Š ( s ) N ( s ) wherethenumber ( s ) iscalledthe torsionofthecurve Byde“nition,thetorsionismeasuredinunitsofareciprocallength, justlikethecurvature,becausetheunitvectors T N ,and B are dimensionless.

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84.APPLICATIONSTOMECHANICSANDGEOMETRY151 Atanypointofacurve,thebinormal B isperpendiculartothe osculatingplane.So,ifthecurveisplanar,then B doesnotchange alongthecurve, B( s )= 0 ,becausetheosculatingplaneatanypoint coincideswiththeplaneinwhichthecurvelies. Aplanarcurvehas notorsion .Thus,thetorsionisalocalnumericalcharacteristicthat determineshowfastthecurvedeviatesfromtheosculatingplanewhile bendinginitwithsomecurvatureradius. Itfollowsfromtherelation N = B T (compare e2= e3 e1)that N=( B T )= B T + B T= Š N T + B N = B Š T wherethede“nitionsofthetorsionandcurvaturehavebeenused.The obtainedratesoftheunitvectorsareknownasthe Frenet-Serretformulasorequations : T( s )= ( s ) N ( s ) (12.9) N( s )= Š ( s ) T ( s )+ ( s ) B ( s ) (12.10) B( s )= Š ( s ) N ( s ) (12.11) TheFrenet-Serretequationsformasystemof dierentialequations for thecomponentsof T ( s ), N ( s ),and B ( s ).Ifthecurvatureandtorsion arecontinuousfunctionsonaninterval0 s L ,thenthesystemcan beprovedtohaveauniquesolutiononthisintervalforeverygivenset ofthevectors T (0), N (0),and B (0)ataninitialpointofthecurve. Givenacoordinatesystem,theinitialpointofacurveisspeci“edbya translationoftheorigin,andtheorientationof T (0), N (0),and B (0) isdeterminedbyarotationofunitcoordinatevectors.Therefore,the followingassertionabouttheshapeofacurveholds. Theorem 12.9 (ShapeofaSmoothCurveinSpace) Giventhecurvatureandtorsionascontinuousfunctionsalongacurve, thecurveisuniquelydeterminedbythemuptorigidrotationsand translationsofthecurveasawhole. AproofofTheorem12.9requiresaproofoftheuniquenessofa solutiontotheFrenet-Serretequations,whichgoesbeyondthescope ofthiscourse.However,insomespeci“cexamples,theFrenet-Serret equationscanbeexplicitlyintegrated.Forexample,considercurves withthevanishingcurvatureandtorsion, ( s )= ( s )=0.Then T ( s )= T (0), N ( s )= N (0),and B( s )= B (0).If r ( s )isanatural parameterizationofacurve,then r( s )= T ( s )= T (0).Theintegration ofthisequationyields r ( s )= r0+ T (0) s ,where r0isaconstantvector, whichisastraightline.

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15212.VECTORFUNCTIONS Example 12.20 UsetheFrenet-Serretequationstoprovethata curvewithaconstantcurvature ( s )= 0 =0 andzerotorsion ( s )= 0 isacircle(oritsportion)ofradius R =1 /0. Solution: Avectorfunction r ( s )thatsatis“estheFrenet-Serretequationsissoughtinthebasisoftheinitialtangent,normal,andbinormal vectors: e1= T (0), e2= N (0),and e3= B (0).Sincethetorsionis0, thebinormaldoesnotchangealongthecurve, B ( s )= e3.Thecurve isplanarandliesinaplaneorthogonalto e3.Anyunitvector T orthogonalto e3canbewrittenas T =cos e1+sin e2where = ( s ) suchthat (0)=0.Owingtotherelations e1 e2= Š e2 e1= e3,a unitvector N orthogonalto T suchthat T N = B = e3musthave theform N = Š sin e1+cos e2.Equation(12.9)gives T= Š sin e1+ cos e2= N = 0 N ( s )= 0, andtherefore ( s )= 0s because (0)=0.Foranaturalparameterizationofthecurve, r( s )= T ( s ).Hence, r( s )=cos( 0s ) e1+sin( 0s ) e2, r ( s )= r0+ Š 1 0sin( 0s ) e1Š Š 1 0cos( 0s ) e2, where r0isaconstantvector.BythePythagoreantheorem,thedistancebetweenanypointofthecurveanda“xedpoint r0isconstant: r ( s ) Š r02=1 /2 0= R2.Sincethecurveisplanar,itisacircle(or itsportion)ofradius R =1 /0. Theorem 12.10 (TorsionofaCurve) Let r ( t ) beathreetimesdierentiablevectorfunctionthattraversesa smoothcurvewhosecurvaturedoesnotvanish.Thenthetorsionofthe curveis ( t )= ( r( t ) r( t )) r( t ) r( t ) r( t ) 2. Proof. Put r( t ) = v ( t )(if s = s ( t )isthearclengthasafunction of t ,then s= v ).By(12.6)andthede“nitionofthecurvature, (12.12) r= v T + v2 N andby(12.7)andthede“nitionofthebinormal, (12.13) r r= v T r= v3 B Dierentiationofbothsidesof(12.12)gives r= v T + v T+( v2+2 vv) N + v2 N.

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84.APPLICATIONSTOMECHANICSANDGEOMETRY153 Thederivatives T( t )and N( t )arefoundbymakinguseofthedierentiationrule d/ds =(1 /s( t ))( d/dt )=(1 /v )( d/dt )intheFrenet-Serret equations(12.9)and(12.10): T= v N N= Š v T + v B Therefore, (12.14) r=( vŠ 2v3) T +(3 vv+ v2) N + v3 B Sincethetangent,normal,andbinormalvectorsareunitandorthogonaltoeachother,( r r) r= v3( r r) B = 2v6 .Therefore, = ( r r) r 2v6, andtheconclusionofthetheoremfollowsfromTheorem12.7, = r r /v3. Remark. Relation(12.13)showsthat B istheunitvectorinthe directionof r r.Thisobservationsoersamoreconvenientway forcalculatingtheunitbinormalvectorthanitsde“nition.Theunit tangent,normal,andbinormalvectorsataparticularpoint r ( t0)ofthe curve r ( t )are T ( t0)= r( t0) r( t0) B ( t0)= r( t0) r( t0) r( t0) r( t0) N ( t0)= B ( t0) T ( t0) Example 12.21 Findtheunittangent,normal,andbinormalvectorsandthetorsionofthecurve r ( t )= ln t,t,t2/ 2 atthepoint (0 1 1 / 2) Solution: Thepointinquestioncorrespondsto t =1.Therefore, r(1)= tŠ 1, 1 ,t t =1= 1 1 1 r(1) = 3 r(1)= Š tŠ 2, 0 1 t =1= Š 1 0 1 r(1)= 2 tŠ 3, 0 0 t =1= 2 0 0 r(1) r(1)= 1 Š 2 1 r(1) r(1) = 6 ,

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15412.VECTORFUNCTIONS T (1)= 1 3 1 1 1 B (1)= 1 6 1 Š 2 1 N (1)= 1 6 3 1 Š 2 1 1 1 1 = 1 3 2 Š 3 0 3 = 1 2 Š 1 0 1 (1)= ( r(1) r(1)) r(1) r(1) r(1) 2= 2 6 = 1 3 84.3.ApproximationsofaSmoothSpaceCurve.Asmoothcurve C has aunittangentvectoratapoint P .Soasmallpartofthecurve(a partofasmallarclength s )containing P canbeapproximatedbya pieceofthetangentlineofthesamelength s .Ifthecurve C hasa nonzerocurvatureat P ,thenabetterapproximationcanbeobtained byapartoftheosculatingcircleofarclength s (seeStudyProblem 12.14).Ifthecurve C hasanonzerotorsionat P ,anevenmoreaccurateapproximationisprovidedbyacurvethrough P thathasthe sameunittangentvectorat P ,andconstantcurvatureandtorsion equaltothecurvatureandtorsionofthecurve C at P .ByTheorem 12.9,suchacurveisunique.AsshowninStudyProblem12.18,itis ahelixwhoseradiusandlengthofeachturnareuniquelydetermined bythecurvatureandtorsion.Thesethreesuccessivelymoreaccurate approximationsdonotrefertoanyparticularcoordinatesystemorany particularparameterizationof C astheapproximationcurvesarefully determinedasthepointsetsinspacebythegeometricalinvariantsof thecurve C at P :theunittangentvector,curvature,andtorsion.An analogycanbemadewiththeTaylorpolynomialapproximationofa functionataparticularpoint.Thetangentlineistheanalogofthe “rst-orderTaylorpolynomial(alinearapproximation),theosculating circleistheanalogofthesecond-orderTaylorpolynomial(aquadratic approximation),andthehelixistheanalogofthethird-orderTaylor polynomial(acubicapproximation).Given T N ,and B ofthecurve C at P ,theFrenet-Serretequationscanusedtoobtainuniquehigherorderapproximationsof C near P byapproximatingthecurvature ( s ) andthetorsion ( s )of C near P .Thehelixapproximationusesthe constantapproximationsofthecurvatureandtorsionbytheirvalues at P .If,forexample,thecurvatureandtorsionof C isknownat

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84.APPLICATIONSTOMECHANICSANDGEOMETRY155 twopointsnear P ,then ( s )and ( s )canbeapproximatedbylinear functionsnear P thatattainthetwoknownvalues.Thecorresponding(unique)solutionoftheFrenet-Serretequationswouldgenerally provideamoreaccurateapproximationthanthehelixapproximation.84.4.StudyProblems.Problem12.16. Findthepositionvector r ( t ) ofaparticleasafunctionoftime t iftheparticlemovesclockwisealongacircularpathofradius R inthe xy planethrough r (0)= R, 0 0 withaconstantspeed v0. Solution: Foracircleofradius R inthe xy planethroughthepoint ( R, 0 0), r ( t )= R cos ,R sin 0 ,where = ( t )suchthat (0)= 0.Thenthevelocityis v ( t )= r( t )= Š R sin ,R cos 0 .Hence, thecondition v ( t ) = v0yields R | ( t ) | = v0or ( t )= ( v0/R ) t and r ( t )= R cos( t ) R sin( t ) 0 where = v0/R istheangularvelocity.Thesecondcomponentmustbe takenwiththeminussignbecausetheparticlerevolvesclockwise(the secondcomponentshouldbecomenegativeimmediatelyafter t =0). Problem12.17. Lettheparticlepositionvectorasafunctionoftime t be r ( t )= ln( t ) ,t2, 2 t t> 0 .Findthespeed,tangentialandnormal accelerations,theunittangent,normal,andbinormalvectors,andthe torsionofthetrajectoryatthepoint P0(0 1 2) Solution: ByExample12.16,thevelocityandaccelerationvectors at P0are v = 1 2 2 and a = Š 1 2 0 .Sothespeedis v = v =3. Thetangentialaccelerationis aT= v a /v =1.As v a =2 Š 2 Š 1 2 thenormalaccelerationis aN= v a /v =6 / 3=2.Theunittangent vectoris T = v /v =(1 / 3) 1 2 2 ,andtheunitbinormalvectoris B = v a / v a =(1 / 3) Š 2 Š 1 2 astheunitvectoralong v a Therefore,theunitnormalvectoris N = T B =(1 / 9) v ( v a )= (1 / 3) Š 2 2 Š 1 .To“ndthetorsionat P0,thethirdderivativeat t =0 hastobecalculated, r(1)= 2 /t2, 0 0 |t =1= 2 0 0 = b .Therefore, (1)=( v a ) b / v a 2= Š 8 / 36= Š 2 / 9. Problem12.18.(CurveswithConstantCurvatureandTorsion) Provethatallcurveswithaconstantcurvature ( s )= 0 =0 anda constanttorsion ( s )= 0 =0 arehelicesbyintegratingtheFrenetSerretequations. Solution: Itfollowsfrom(12.9)and(12.11)thatthevector w = T + B doesnotchangealongthecurve, w( s )=0.Indeed,because

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15612.VECTORFUNCTIONS ( s )= ( s )=0,onehas w= T+ B=( Š ) N = 0 .By thePythagoreantheorem, w =( 2 0+ 2 0)1 / 2.Considertwonewunit vectorsorthogonalto N : w = 1 w w =sin T +cos B u =cos T Š sin B wherecos = 0/ ,sin = 0/ ,and =( 2 0+ 2 0)1 / 2.Byconstruction,theunitvectors u w ,and N aremutuallyorthogonalunit vectors,whichiseasytoverifybycalculatingthecorrespondingdot products, u u = w w =1and u w =0.Also, u w =cos2 T B Š sin2 B T =(cos2 +sin2 ) N = N Bydierentiatingthevector u andusingtheFrenet-Serretequations, u=cos TŠ sin B=( 0cos + 0sin ) N = N Since w ( s )= w (0)isaconstantunitvector,itisconvenienttoseeka solutioninanorthonormalbasissuchthat e3= w (0)and e1 e2= e3. Inthisbasis, u =cos e1+sin e2,where = ( s ),asaunitvector intheplaneorthogonalto e3.Theorientationofthebasisvectorsin theplaneorthogonalto e3isde“neduptoageneralrotationabout e3. Thisfreedomisusedtoset e1= u (0),whichimpliesthatthefunction ( s )satis“esthecondition (0)=0.Thentheunitnormalvectorin thisbasisis N = u w =cos e1 e3+sin e2 e3= Š sin e1+cos e2and u= Š sin e1+ cos e2= N Hence, ( s )= or ( s )= s owingtothecondition (0)=0. Expressingthevector T via u and w T =cos u +sin w oneinfers(compareExample12.20) r( s )= T ( s )= 0 cos( s ) e1+ 0 sin( s ) e2+ 0 e3, where r ( s )isanaturalparameterizationofthecurve.Theintegration ofthisequationgives r ( s )= r0+ R sin( s ) e1Š R cos( s ) e2+ hs e3,R = 0 2,h = 0 Thisisahelixofradius R whoseaxisgoesthroughthepoint r0parallel to e3;thehelixclimbsalongitsaxisby2 h/ pereachturn.

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84.APPLICATIONSTOMECHANICSANDGEOMETRY157 Problem12.19.(MotioninaConstantMagneticField,Revisited) Theforceactingonachargedparticlemovinginthemagnetic“eld B is givenby F =( e/c ) v B ,where e istheelectricchargeoftheparticle, c isthespeedoflight,and v isitsvelocity.Showthatthetrajectoryof theparticleinaconstantmagnetic“eldisahelixwhoseaxisisparallel tothemagnetic“eld. Solution: IncontrasttoStudyProblem12.11,heretheshapeofthe trajectoryistobeobtaineddirectlyfromNewtonssecondlawwith arbitraryinitialconditions.Choosethecoordinatesystemsothatthe magnetic“eldisparalleltothe z axis, B = B e3,where B isthe magnitudeofthemagnetic“eld.Newtonslawofmotion, m a = F where m isthemassoftheparticle,determinestheacceleration, a = v B = B v e3,where = e/ ( mc ).First,notethat v 3= a3= e3 a =0.Hence, v3= v=const.Second,bythegeometricalproperty ofthecrossproducttheaccelerationandvelocityremainorthogonal duringthemotion,andthereforethetangentialaccelerationvanishes, aT= v a =0.Hence,thespeedoftheparticleisaconstantofmotion, v = v0(because v= aT=0).Put v = v+ v e3,where visthe projectionof v ontothe xy plane.Since v = v0,themagnitudeof visalsoconstant, v = v=( v2 0Š v2 )1 / 2.Thevelocityvectorcan thereforebewrittenintheform v = vcos ,vsin ,v ,wherethe function = ( t )istobedeterminedbytheequationsofmotion: a = B v e3= B Š vsin ,vcos 0 a = v= Š vsin ,vcos 0 Itfollowsfromthecomparisonoftheseexpressionsthat ( t )= B or ( t )= Bt + 0= t + 0,where = eB/ ( mc )istheso-called cyclotronfrequencyandtheintegrationconstant 0isdeterminedby theinitialvelocity: v (0)= vcos 0,vsin 0,v ,thatis,tan 0= v2(0) /v1(0).Integrationoftheequation r( t )= v ( t )= vcos( t + 0) ,vsin( t + 0) ,v yieldsthetrajectoryofmotion: r ( t )= r0+ R sin( t + 0) Š R cos( t + 0) ,vt where R = v/ .Thisequationdescribesahelixofradius R whose axisgoesthrough r0paralleltothe z axis.Soachargedparticlemoves alongahelixthatwindsaboutforcelinesofthemagnetic“eld.The particlerevolvesintheplaneperpendiculartothemagnetic“eldwith frequency = eB/ ( mc ).Ineachturn,theparticlemovesalongthe magnetic“eldadistance h =2 v/ .Inparticular,iftheinitial

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15812.VECTORFUNCTIONS velocityisorthogonaltothemagnetic“eld(i.e., v=0),thenthe trajectoryisacircleofradius R ThePolarLights .TheSunproducesastreamofchargedparticles (thesolarwind).Themagnetic“eldoftheEarthplaystheroleof ashieldfromthesolarwindasittrapstheparticles,forcingthem totravelalongitsforcelinesthatarearcsconnectingthemagnetic polesoftheEarth(whichapproximatelycoincidewiththesouthand northpoles).Asaresult,thesolarwindparticlescanpenetratethe loweratmosphereonlynearthemagneticpolesoftheEarth,causinga spectacularphenomenon,thepolarlights,bycollidingwithmolecules oftheoxygenandnitrogenintheatmosphere. Problem12.20. Supposethattheforceactingonaparticleofmass m isproportionaltothepositionvectoroftheparticle(suchforces arecalled central ).Provethattheangularmomentumoftheparticle, L = m r v ,isaconstantofmotion(i.e., d L /dt =0 ). Solution: Sinceacentralforce F isparalleltothepositionvector r ,theircrossproductvanishes, r F = 0 .ByNewtonssecondlaw, m a = F andhence m r a = 0 .Therefore, d L dt = m ( r v )= m ( r v + r v)= m r a = 0 where r= v v= a ,and v v = 0 havebeenused. Problem12.21.(KeplersLawsofPlanetaryMotion). Newtonslawofgravitystatesthattwomasses m and M atadistance r areattractedbyaforceofmagnitude GmM/r2,where G istheuniversal constant(called Newtonsconstant ).ProveKeplerslawsofplanetary motion: 1.AplanetrevolvesaroundtheSuninanellipticalorbitwiththeSun atonefocus. 2.ThelinejoiningtheSuntoaplanetsweepsoutequalareasinequal times. 3.Thesquareoftheperiodofrevolutionofaplanetisproportionalto thecubeofthelengthofthemajoraxisofitsorbit. Solution: LettheSunbeattheoriginofacoordinatesystemand let r bethepositionvectorofaplanet.ThemassoftheSunismuch largerthanthemassofaplanet;therefore,adisplacementoftheSun duetothegravitationalpullfromaplanetcanbeneglected(e.g.,the Sunisabout332,946timesheavierthantheEarth).Let r = r /r be

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84.APPLICATIONSTOMECHANICSANDGEOMETRY159 theunitvectorparallelto r .Thenthegravitationalforceis F = Š GMm r2 r = Š GMm r3r where M isthemassoftheSunand m isthemassofaplanet.The minussignisnecessarybecauseanattractiveforcemustbeopposite tothepositionvector.ByNewtonssecondlaw,thetrajectoryofa planetsatis“estheequation m a = F andhence a = Š GM r3r Thegravitationalforceisacentralforce,and,byStudyProblem12.20, thevector r v = l isaconstantofmotion.Onehas v = r=( r r )= r r + r r.Usingthisidentity,theconstantofmotioncanalsobewritten as l = r v = r r v = r ( r r r + r r r)= r2( r r) Usingtheruleforthedoublecrossproduct(seeStudyProblem11.17), oneinfersthat a l = Š GM r2 r l = Š GM r ( r r)= GM r, where r r =1hasbeenused.Ontheotherhand, ( v l )= v l + v l= a l because l= 0 .Itfollowsfromthesetwoequationsthat (12.15)( v l )= GM r= v l = GM r + c where c isaconstantvector.Themotionischaracterizedbytwo constantvectors l and c .Itoccursintheplanethroughtheorigin thatisorthogonaltotheconstantvector l because l = r v mustbe orthogonalto r .Italsofollowsfrom(12.15)and l r =0thatthe constantvectors l and c areorthogonalbecause l c =0.Itistherefore convenienttochoosethecoordinatesystemsothat l isparalleltothe z axisand c tothe x axisasshowninFigure12.12(leftpanel). Thevector r liesinthe xy plane.Let bethepolarangleof r (i.e., r c = rc cos ,where c = c isthelengthof c ).Then r ( v l )= r ( GM r + c )= GMr + rc cos .Ontheotherhand,usingacyclicpermutationinthetripleproduct, r ( v l )= l ( r v )= l l = l2,

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16012.VECTORFUNCTIONS Figure12.12.Left :Thesetupofthecoordinatesystem forthederivationofKeplers“rstlaw. Right :AnillustrationtothederivationofKeplerssecondlaw.where l = l isthelengthof l .Thecomparisonofthelasttwo equationsyieldstheequationforthetrajectory: l2= r ( GM + b cos )= r = ed 1+ e cos where d = l2/c and e = c/ ( GM ).Thisisthepolarequationofa conicsectionwithfocusattheoriginandeccentricity e (seeCalculus II).Thus, allpossibletrajectoriesofanymassivebodyinasolarsystemareconicsections! Thisisaquiteremarkableresult.Parabolas andhyperbolasdonotcorrespondtoaperiodicmotion.Soaplanet mustfollowanelliptictrajectorywiththeSunatonefocus.Allobjectscomingtothesolarsystemfromouterspace(i.e.,thosethatare notcon“nedbythegravitationalpulloftheSun)shouldfolloweither parabolicorhyperbolictrajectories. ToproveKeplerssecondlaw,put r = cos sin 0 andhence r= Š sin ,cos 0 .Therefore, l = r2( r r)= 0 0 ,r2 = l = r2. Theareaofasectorwithangle d sweptby r is dA =1 2r2d (see CalculusII;theareaboundedbyapolargraph r = r ( )).Hence, dA dt = 1 2 r2d dt = l 2 .

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84.APPLICATIONSTOMECHANICSANDGEOMETRY161 Foranymomentsoftime t1and t2,theareaofthesectorbetween r ( t1) and r ( t2)is A12= t2t1dA dt dt = t2t1l 2 dt = l 2 ( t2Š t1) Thus,thepositionvector r sweepsoutequalareasinequaltimes(see Figure12.12,rightpanel). Keplersthirdlawfollowsfromthelastequation.Indeed,theentire areaoftheellipse A issweptwhen t2Š t1= T istheperiodofthe motion.Ifthemajorandminoraxesoftheellipseare2 a and2 b respectively, a>b ,then A = ab = lT/ 2and T =2 ab/l .Now recallthat ed = b2/a foranellipticconicsection(seeCalculusII)or b2= eda = l2a/ ( GM ).Hence, T2= 4 2a2b2 l2= 4 2 GM a3. Notethattheproportionalityconstant4 2/ ( GM )isindependentofthe massofaplanet;therefore,Keplerslawsare universal forallmassive objectstrappedbytheSun(planets,asteroids,andcomets). 84.5.Exercises.(1) Foreachofthefollowingtrajectoriesofaparticle,“ndthevelocity, speed,andnormalandtangentialaccelerationsasfunctionsoftimeand theirvaluesataspeci“edpoint P : (i) r ( t )= t, 1 Š t,t2+1 ,P (1 0 2) (ii) r ( t )= t2,t, 1 ,P (4 2 1) (iii) r ( t )= 4 t3 / 2, Š t2,t ,P (4 Š 1 1) (iv) r ( t )= ln t, t,t2 ,P (0 1 1) (v) r ( t )= cosh t, sinh t, 2+ t ,P (1 0 2) (vi) r ( t )= et, 2 t,eŠ t ,P (1 0 1) (vii) r ( t )= sin t Š t cos t,t2, cos t + t sin t ,P (0 0 1) (2) Findthenormalandtangentialaccelerationsofaparticlewiththe positionvector r ( t )= t2+1 ,t,t2Š 1 whentheparticleisclosestto theorigin. (3) Findthetangentialandnormalaccelerationsofaparticlewiththe positionvector r ( t )= R sin( t + 0) Š R cos( t + 0) ,v0t ,where R 0,and v0areconstants(seeStudyProblem12.19). (4) Theshapeofawindingroadcanbeapproximatedbythegraph y = L cos( x/L ),wherethecoordinatesareinmetersand L =40m. Theconditionoftheroadissuchthatifthenormalaccelerationofa caronitexceeds10m / s2,thecarmayskidotheroad.Recommend

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16212.VECTORFUNCTIONS aspeedlimitforthisportionoftheroad. (5) Aparticlemovesalongthecurve y = x2+ x3.Iftheacceleration oftheparticleatthepoint(1 2)is a = 3 Š 1 ,“nditsnormaland tangentialaccelerations. (6) Supposethataparticlemovessothatitstangentialacceleration aTisconstant,whilethenormalacceleration aNremains0.Whatis thetrajectoryoftheparticle? (7) Supposethataparticlemovesinaplanesothatitstangential acceleration aTremains0,whilethenormalacceleration aNisconstant. Whatisthetrajectoryoftheparticle? Hint: Investigatethecurvature ofthetrajectory. (8) Aracecarmoveswithaconstantspeed v0alonganelliptictrack x2/a2+ y2/b2=1, a>b .Findthemaximalandminimalvaluesofthe magnitudeofitsaccelerationandthepointswheretheyoccur. (9) Doesthereexistacurvewithzerocurvatureandnonzerotorsion? Explaintheanswer. (10) Foreachofthefollowingcurves,“ndtheunittangent,normal, andbinormalvectorsandthetorsionataspeci“edpoint P : (i) r ( t )= t, 1 Š t,t2+1 ,P (1 0 2) (ii) r ( t )= t3,t2, 1 ,P (8 4 1) (iii) r ( t )= 4 t3 / 2, Š t2,t ,P (4 Š 1 1) (iv) r ( t )= ln t, t,t2 ,P (0 1 1) (v) r ( t )= cosh t, sinh t, 2+ t ,P (1 0 2) (11) Let r ( t )= cos t + t sin t, sin t + t cos t,t2 .Findthespeed,the tangentialandnormalaccelerations,thecurvatureandtorsion,andthe unittangent,normal,andbinormalvectorsasfunctionsoftime t Hint: Tosimplifycalculations,“ndthedecomposition r ( t )= v ( t ) Š t w ( t )+ t2 e3,where v w ,and e3aremutuallyorthogonalunitvectors suchthat v( t )= w ( t ), w( t )= Š v ( t ).Usethepropertiesofthecross productsofmutuallyorthogonalunitvectors. (12) Let C bethecurveofintersectionofanellipsoid x2/a2+ y2/b2+ z2/c2=1withtheplane2 x Š 2 y + z =0.Findthetorsionandthe binormal B along C .

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CHAPTER13 DifferentiationofMultivariable Functions 85.FunctionsofSeveralVariables Theconceptofafunctionofseveralvariablescanbequalitatively understoodfromsimpleexamplesineverydaylife.Thetemperaturein aroommayvaryfrompointtopoint.Apointinspacecanbede“nedby anorderedtripleofnumbersthatarecoordinatesofthepointinsome coordinatesystem,say,( x,y,z ).Measurementsofthetemperatureat everypointfromaset D inspaceassignarealnumber T (thetemperature)toeverypointof D .Thedependenceof T oncoordinatesofthe pointisindicatedbywriting T = T ( x,y,z ).Similarly,theconcentrationofachemicalcandependonapointinspace.Inaddition,ifthe chemicalreactswithotherchemicals,itsconcentrationatapointmay alsochangewithtime.Inthiscase,theconcentration C = C ( x,y,z,t ) dependsonfourvariables,threespatialcoordinatesandthetime t .In general,ifthevalueofaquantity f dependsonvaluesofseveralother quantities,say, x1, x2,..., xn,thisdependenceisindicatedbywriting f = f ( x1,x2,...,xn).Inotherwords, f = f ( x1,x2,...,xn) indicatesa rulethatassignsanumber f toeachordered n -tupleofrealnumbers ( x1,x2,...,xn).Eachnumberinthe n -tuplemaybeofadierentnature andmeasuredindierentunits.Intheaboveexample,theconcentrationdependsonorderedquadruples( x,y,z,t ),where x y ,and z are thecoordinatesofapointinspace(measuredinunitsoflength)and t istime(measuredinunitsoftime).Allordered n -tuplesform an n -dimensionalEuclideanspace ,muchlikeallordereddoublets( x,y ) formaplane,andallorderedtriples( x,y,z )formaspace.85.1.EuclideanSpaces.Witheveryorderedpairofnumbers( x,y ), onecanassociateapointinaplaneanditspositionvectorrelativeto a“xedpoint(0 0)(theorigin), r = x,y .Witheveryorderedtriple ofnumbers( x,y,z ),onecanassociateapointinspaceanditsposition vector(againrelativetotheorigin(0 0 0)), r = x,y,z .Sotheplane canbeviewedasthesetofalltwo-componentvectors;similarly,space isthesetofallthree-componentvectors.Fromthispointofview, theplaneandspacehavecharacteristiccommonfeatures.First,their elementsarevectors.Second,theyareclosedrelativetoadditionof163

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16413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS vectorsandmultiplicationofvectorsbyarealnumber;thatis,if a and b areelementsofspaceoraplaneand c isarealnumber,then a + b and c a arealsoelementsofspace(orderedtriplesofnumbers)oraplane (orderedpairsofnumbers).Third,thenormorlengthofavector r vanishesifandonlyifthevectorhaszerocomponents.Consequently, twoelementsofspaceoraplanecoincideifandonlyifthenormoftheir dierencevanishes,thatis, a = b a Š b =0.Finally,thedot product a b oftwoelementsisde“nedinthesamewayfortwo-orthreecomponentvectors(planeorspace)sothat a 2= a a .Sincepoints andvectorsaredescribedbythesamemathematicalobject,anordered triple(orpair)ofnumbers,itisnotnecessarytomakeadistinction betweenthem.So,inwhatfollows,thesamenotationisusedfora pointandavector,forexample, r =( x,y,z ).Theseobservationscan beextendedtoordered n -tuplesforany n andleadtothenotionofa Euclideanspace Definition 13.1 (EuclideanSpace) Foreachpositiveinteger n ,considerthesetofallordered n -tuples ofrealnumbers.Foranytwoelements a =( a1,a2,...,an) and b = ( b1,b2,...,bn) andanumber c ,put a + b =( a1+ b1,a2+ b2,...,an+ bn) c a =( ca1,ca2,...,can) a b = a1b1+ a2b2+ + anbn, a = a a = a2 1+ a2 2+ + a2 n. Thesetofallordered n -tuplesinwhichtheaddition,themultiplication byanumber,thedotproduct,andthenormarede“nedbytheserules iscalledan n dimensionalEuclideanspace TwopointsofaEuclideanspacearesaidtocoincide, a = b ,ifthe correspondingcomponentsareequal,thatis, ai= bifor i =1 2 ,...,n Itfollowsthat a = b ifandonlyif a Š b =0.Indeed,bythe de“nitionofthenorm, c =0ifandonlyif c =(0 0 ,..., 0).Put c = a Š b .Then a Š b =0ifandonlyif a = b .Thenumber a Š b iscalledthe distance betweenpoints a and b ofaEuclideanspace. ThedotproductinaEuclideanspacehasthesamegeometrical propertiesasintwoandthreedimensions.TheCauchy-SchwarzinequalitycanbeextendedtoanyEuclideanspace(cf.Theorem11.2). Theorem 13.1 (Cauchy-SchwarzInequality) | a b | a b

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85.FUNCTIONSOFSEVERALVARIABLES165 foranyvectors a and b inaEuclideanspace,andtheequalityisreached ifandonlyif a = t b forsomenumber t Proof. Put a = a and b = b ,thatis, a2= a a andsimilarly for b .If b =0,then b = 0 ,andtheconclusionofthetheoremholds. For b =0andanyrealvariable t a Š t b 2=( a Š t b ) ( a Š t b ) 0. Therefore, a2Š 2 tc + t2b2 0,where c = a b .Completingthesquares ontheleftsideofthisinequality, bt Š c b 2Š c2 b2+ a2 0 showsthattheleftsideattainsitsabsoluteminimumwhentheexpressionintheparenthesesvanishes,thatis,at t = c/b2.Sincetheinequalityisvalidforany t ,itissatis“edfor t = c/b2,thatis, a2Š c2/b2 0 or c2 a2b2or | c | ab ,whichistheconclusionofthetheorem.The inequalitybecomesanequalityifandonlyif a Š t b 2=0andhence ifandonlyif a = t b ItfollowsfromtheCauchy-Schwarzinequalitythat a b = s a b where s isanumbersuchthat | s | 1.Soonecanalwaysput s =cos where [0 ].If =0,then a = t b forsomepositive t> 0(i.e.,the vectorsareparallel),and a = t b t< 0,when = (i.e.,thevectors areantiparallel).Thedotproductvanisheswhen = / 2.Thisallows onetode“ne astheanglebetweentwovectorsinanyEuclidean space:cos = a b / ( a b )muchlikeintwoandthreedimensions. Consequently,thetriangleinequality(11.7)holdsinaEuclideanspace ofanydimension.85.2.Real-ValuedFunctionsofSeveralVariables.Definition 13.2 (Real-ValuedFunctionofSeveralVariables) Let D beasetofordered n -tuplesofrealnumbers ( x1,x2,...,xn) .A function f of n variablesisarulethatassignstoeach n -tupleinthe set D auniquerealnumberdenotedby f ( x1,x2,...,xn) .Theset D is thedomainof f ,anditsrangeisthesetofvaluesthat f takesonit, thatis, { f ( x1,x2,...,xn) | ( x1,x2,...,xn) D } Thisde“nitionisillustratedinFigure13.1.Therulemaybede“ned bydierentmeans.If D isa“niteset,afunction f canbede“nedby atable( Pi,f ( Pi)),where Pi D i =1 2 ,...,N ,areelements(ordered n -tuples)of D ,and f ( Pi)isthevalueof f at Pi.Afunction f canbe de“nedgeometrically.Forexample,theheightofamountainrelative tosealevelisafunctionofitspositionontheglobe.Sotheheightis afunctionoftwovariables,thelongitudeandlatitude.Afunctioncan bede“nedbyanalgebraicrulethatprescribesalgebraicoperationsto

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16613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS y x P D f f ( P ) R f D P f f ( P ) R fFigure13.1.Left :Afunction f oftwovariablesisarule thatassignsanumber f ( P )toeverypoint P ofaplanar region D .Theset R ofallnumbers f ( P )istherangeof f Theregion D isthedomainof f Right :Afunction f of threevariablesisarulethatassignsanumber f ( P )toevery point P ofasolidregion D .becarriedoutwithrealnumbersinany n -tupletoobtainthevalueof thefunction.Forexample, f ( x,y,z )= x2Š y + z3.Thevalueofthis functionat(1 2 3)is f (1 2 3)=12Š 2+33=28.Unlessspeci“ed otherwise,thedomain D ofafunctionde“nedbyanalgebraicruleis thesetof n -tuplesforwhichtherulemakessense. Example 13.1 Findthedomainandtherangeofthefunctionof twovariables f ( x,y )=ln(1 Š x2Š y2) Solution: Thelogarithmisde“nedforanystrictlypositivenumber. Therefore,thedoublet( x,y )mustbesuchthat1 Š x2Š y2> 0or x2+ y2< 1.Hence, D = { ( x,y ) | x2+ y2< 1 } .Sinceanydoublet( x,y ) canbeuniquelyassociatedwithapointonaplane,theset D canbe givenageometricaldescriptionasadiskofradius1whoseboundary, thecircle x2+ y2=1,isnotincludedin D .Foranypointinthe interiorofthedisk,theargumentofthelogarithmliesintheinterval 0 1 Š x2Š y2< 1.Sotherangeof f isthesetofvaluesofthe logarithmintheinterval(0 1],whichis Š
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85.FUNCTIONSOFSEVERALVARIABLES167 z y x ( x,y,z ) z = f ( x,y ) ( x,y, 0) D x y z 1 2 3 DFigure13.2.Left :Thegraphofafunctionoftwovariablesisthesurfacede“nedbytheequation z = f ( x,y ).It isobtainedfromthedomain D of f bymovingeachpoint ( x,y, 0)in D alongthe z axistothepoint( x,y,f ( x,y )). Right :ThegraphofthefunctionstudiedinExample13.3.Ingeneral,thedomainofafunctionof n variablesisviewedasa subsetofan n -dimensionalEuclideanspace.Itisalsoconvenientto adoptthevectornotationoftheargument: f ( x1,x2,...,xn)= f ( r ) r =( x1,x2,...,xn) Forexample,thedomainofthefunction f ( r )=(1 Š x2 1Š x2 2ŠŠ x2 n)1 / 2=(1 Š r 2)1 / 2isthesetofpointsinthe n -dimensionalEuclideanspacewhosedistance fromtheorigin(thezerovector)doesnotexceed1, D = { r | r 1 } ; thatis,itisan n -dimensionalballofradius1.Sothedomainofa multivariablefunctionde“nedbyanalgebraicrulecanbedescribedby conditionsonthecomponents(coordinates)oftheordered n -tuple r underwhichtherulemakessense.85.3.TheGraphofaFunctionofTwoVariables.Thegraphofafunction ofonevariable f ( x )isthesetofpointsofaplane { ( x,y ) | y = f ( x ) } Thedomain D isasetofpointsonthe x axis.Thegraphisobtained bymovingapointofthedomainparalleltothe y axisbyanamount determinedbythevalueofthefunction y = f ( x ).Thegraphprovidesa usefulpictureofthebehaviorofthefunction.Theideacanbeextended tofunctionsoftwovariables. Definition 13.3 (GraphofaFunctionofTwoVariables) Thegraphofafunction f ( x,y ) withdomain D isthepointsetinspace { ( x,y,z ) | z = f ( x,y ) ( x,y ) D } .

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16813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Thedomain D isasetofpointsinthe xy plane.Thegraphisthen obtainedbymovingeachpointof D paralleltothe z axisbyanamount equaltothecorrespondingvalueofthefunction z = f ( x,y ).If D is aportionoftheplane,thenthegraphof f isgenerallyasurface(see Figure13.2,leftpanel).OnecanthinkofthegraphasmountainsŽof height f ( x,y )onthe xy plane. Example 13.3 Sketchthegraphofthefunction f ( x,y )= 1 Š ( x/ 2)2Š ( y/ 3)2. Solution: Thedomainistheportionofthe xy plane( x/ 2)2+( y/ 3)2 1;thatis,itisboundedbytheellipsewithsemiaxes2and3.Thegraph isthesurfacede“nedbytheequation z = 1 Š ( x/ 2)2Š ( y/ 3)2.By squaringbothsidesofthisequation,one“nds( x/ 2)2+( y/ 3)2+ z2= 1,whichde“nesanellipsoid.Thegraphisitsupperportionwith z 0asdepictedintherightpanelofFigure13.2. Theconceptofthegraphisobviouslyhardtoextendtofunctions ofmorethantwovariables.Thegraphofafunctionofthreevariables wouldbeathree-dimensionalsurfaceinfour-dimensionalspace.Sothe qualitativebehaviorofafunctionofthreevariablesshouldbestudied bydierentgraphicalmeans.85.4.LevelSets.Whenvisualizingtheshapeofquadricsurfaces,the methodofcrosssectionsbycoordinateplaneshasbeenhelpful.It canalsobeappliedtovisualizetheshapeofthegraph z = f ( x,y ). Inparticular,considerthecrosssectionsofthegraphwithhorizontal planes z = k .Thecurveofintersectionisde“nedbytheequation f ( x,y )= k .Continuingtheanalogythat f ( x,y )de“nestheheightof amountain,ahikertravelingalongthepath f ( x,y )= k doesnothave toclimbordescendastheheightalongthepathremainsconstant. Definition 13.4 (LevelSets) Thelevelsetsofafunction f aresubsetsofthedomainof f onwhich thefunctionhasa“xedvalue;thatis,theyaredeterminedbytheequation f ( r )= k ,where k isanumberfromtherangeof f Forfunctionsoftwovariables,theequation f ( x,y )= k generally de“nesacurve,butnotnecessarilyso.Forexample,if f ( x,y )= x2+ y2, thentheequation x2+ y2= k de“nesconcentriccirclesofradii k for any k> 0.However,for k =0,thelevelsetconsistsofasinglepoint ( x,y )=(0 0).If f isaconstantfunctionon D ,thenithasjustone levelsetthatcoincideswiththeentiredomain D .Alevelsetiscalled a levelcurve iftheequation f ( x,y )= k de“nesacurve.Recallthat

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85.FUNCTIONSOFSEVERALVARIABLES169 z = k3z = k2z = k1z = f ( x,y ) x y k2k2k3k3k1Figure13.3.Left :Crosssectionsofthegraph z = f ( x,y ) byhorizontalplanes z = ki, i =1 2 3,arelevelcurves f ( x,y )= kiofthefunction f Right :Thecontourmap ofthefunction f consistsoflevelcurves f ( x,y )= ki.The number kiindicatesthevalueof f alongeachlevelcurve.acurveinaplanecanbedescribedbyparametricequations x = x ( t ), y = y ( t ),where x ( t )and y ( t )arecontinuousfunctionsonaninterval a t b .Therefore,theequation f ( x,y )= k de“nesacurveifthere existcontinuousfunctions x ( t )and y ( t )suchthat f ( x ( t ) ,y ( t ))= k for allvaluesof t fromaninterval.Ingeneral,alevelsetofafunctionmay containcurves,isolatedpoints,andevenportionsofthedomainwith nonzeroarea. Definition 13.5 (ContourMap) Acollectionoflevelcurvesiscalleda contourmap ofthefunction f Theconceptsoflevelcurvesandacontourmapofafunctionoftwo variablesareillustratedinFigure13.3.Thecontourmapofthefunction inExample13.3consistsofellipses.Indeed,therangeistheinterval [0 1].Forany0 k< 1,alevelcurveisanellipse,1 Š ( x/ 2)2Š ( y/ 3)2= k2or( x/a )2+( y/b )2=1,where a =2 1 Š k2and b =3 1 Š k2.The levelsetfor k =1consistsofasinglepoint,theorigin. Acontourmapisausefultoolforstudyingthequalitativebehavior ofafunction.Considerthecontourmapthatconsistsoflevelcurves Ci, i =1 2 ,... f ( x,y )= ki,where ki +1Š ki= k is“xed.Thevalues ofthefunctionalongtheneighboringcurves Ciand Ci +1dierby k So,intheregionwherethelevelcurvesaredense(closetooneanother),

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17013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS thefunction f ( x,y )changesrapidly.Indeed,let P beapointof Ciand let s bethedistancefrom P to Ci +1alongthenormalto Ci.Then theslopeofthegraphof f ortherateofchangeof f at P inthat directionis k/ s .Thus,thecloserthecurves Ciaretooneanother, thefasterthefunctionchanges.Contourmapsareusedintopography toindicatethesteepnessofmountainsonmaps. Example 13.4 Describethelevelsets(acontourmap)ofthefunction f ( x,y )=( a2Š ( x2+ y2/ 4))2. Solution: Thefunctiondependsonthecombination u2= x2+ y2/ 4, f ( x,y )=( a2Š u2)2,andthereforethelevelsets f ( x,y )= k 0 areellipses.Thelevelset k =0istheellipse x2/a2+ y2/ (2 a )2=1. Thelevelsets0 0.For k = a4,thelevelsetconsistsoftheellipse u2=2 a2andthepoint ( x,y )=(0 0).Thelevelsetsfor k>a4areellipses u2= k Š a2.So thecontourmapcontainstheellipse x2/a2+ y2/ (2 a )2=1alongwhich thefunctionattainsitsabsoluteminimum f ( x,y )=0.Asthevalue of f increases,thisellipsesplitsintosmallerandlargerellipses.At f ( x,y )= a4(alocalmaximumof f attainedattheorigin),thesmaller ellipsescollapsetoapointanddisappear,whilethelargerellipseskeep expandinginsize.Thegraphof f lookslikeaMexicanhat. 85.5.LevelSurfaces.Incontrasttothegraph,themethodoflevel curvesusesonlythedomainofafunctionoftwovariablestostudyits behavior.Therefore,theconceptoflevelsetscanbeusefultostudy thequalitativebehavioroffunctionsofthreevariables.Ingeneral,the equation f ( x,y,z )= k de“nesasurfaceinspace,butnotnecessarilyso asinthecaseoffunctionsoftwovariables.Thelevelsetsofthefunction f ( x,y,z )= x2+ y2+ z2areconcentricspheres x2+ y2+ z2= k for k> 0,butthelevelsetfor k =0containsjustonepoint,theorigin. Intuitively,asurfaceinspacecanbeobtainedbyacontinuousdeformation(withoutbreaking)ofapartofaplane,justlikeacurveis obtainedbyacontinuousdeformationofalinesegment.Let S bea nonemptypointsetinspace.A neighborhood ofapoint P of S isa collectionofallpointsof S whosedistancefrom P islessthananumber > 0.Inparticular,aneighborhoodofapointinaplaneisa diskcenteredatthatpoint,andtheboundarycircledoesnotbelong totheneighborhood.Ifeverypointofasubset D ofaplanehasa neighborhoodthatiscontainedin D ,thentheset D iscalled open .In otherwords,foreverypoint P ofanopenregion D inaplane,thereis adiskofasucientlysmallradiusthatiscenteredat P andcontained

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85.FUNCTIONSOFSEVERALVARIABLES171 in D Apointset S isasurfaceinspaceifeverypointof S hasa neighborhoodthatcanbeobtainedbyacontinuousdeformation(ora deformationwithoutbreaking)ofanopensetinaplaneandthisdeformationhasacontinuousinverse .Thisisanalogoustothede“nitionof acurveasapointsetinspacegiveninSection79.3. Whenthelevelsetsofafunctionofthreevariablesaresurfaces, theyarecalled levelsurfaces .Theshapeofthelevelsurfacesmaybe studied,forexample,bythemethodofcrosssectionswithcoordinate planes.Acollectionoflevelsurfaces Si, f ( x,y,z )= ki, ki +1Š ki= k i =1 2 ,... ,canbedepictedinthedomainof f .If P0isapointon Siand P isthepointon Si +1thatistheclosestto P0,thentheratio k/ | P0P | determinesthemaximalrateofchangeof f at P .Sothe closerthelevelsurfaces Siaretooneanother,thefasterthefunction changes(seetherightpanelofFigure13.4). Example 13.5 Sketchand/ordescribethelevelsurfacesofthe function f ( x,y,z )= z/ (1+ x2+ y2) Solution: Thedomainistheentirespace,andtherangecontains allrealnumbers.Theequation f ( x,y,z )= k canbewritteninthe form z Š k = k ( x2+ y2),whichde“nesacircularparaboloidwhose symmetryaxisisthe z axisandwhosevertexisat(0 0 ,k ).Forlarger k ,theparaboloidrisesfaster.For k =0,thelevelsurfaceisthe xy plane.For k> 0,thelevelsurfacesareparaboloidsabovethe xy plane; thatis,theyareconcaveupward(seetherightpanelofFigure13.4). For k< 0,theparaboloidsarebelowthe xy plane(i.e.,theyareconcave downward). 85.6.Exercises.(1) Findandsketchthedomainofeachofthefollowingfunctions: (i) f ( x,y )= x/y (ii) f ( x,y )= x/ ( x2+ y2) (iii) f ( x,y )= x/ ( y2Š 4 x2) (iv) f ( x,y )=ln(9 Š x2Š ( y/ 2)2) (v) f ( x,y )= 1 Š ( x/ 2)2Š ( y/ 3)2(vi) f ( x,y )= 4 Š x2Š y2+2 x ln y (vii) f ( x,y )= 4 Š x2Š y2+ x ln y2(viii) f ( x,y )= 4 Š x2Š y2+ln(1 Š x2Š ( y/ 2)2) (ix) f ( x,y,z )= x/ ( yz ) (x) f ( x,y,z )= x/ ( x Š y2Š z2) (xi) f ( x,y,z )=ln(1 Š z + x2+ y2) (xii) f ( x,y,z )= x2Š y2Š z2+ln(1 Š x2Š y2Š z2)

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17213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS f ( x,y,z )= k3f ( x,y,z )= k2f ( x,y,z )= k1k2k1k3k4Figure13.4.Left :Alevelsurfaceofafunction f ofthree variablesisasurfaceinthedomainof f onwhichthefunction attainsaconstantvalue k ;thatis,itisde“nedbytheequation f ( x,y,z )= k .Herethreelevelsurfacesaredepicted. Right :LevelsurfacesofthefunctionstudiedinExample 13.5.Here k2>k1> 0and k4
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86.LIMITSANDCONTINUITY173 (ii) f ( x,y )= | x | + | y |Š| x + y | (iii) f ( x,y )=min( x,y ) (iv) f ( x,y )=max( | x | | y | ) (v) f ( x,y )=sign(sin( x )sin( y ));heresign( a )isthesignfunction, ithasthevalues1and Š 1forpositiveandnegative a ,respectively (vi) f ( x,y,z )=( x + y )2+ z2(vii) f ( x,y )=tanŠ 12 ay x2+ y2Š a2 a> 0 (5) Explainhowthegraph z = g ( x,y )canbeobtainedfromthegraph of f ( x,y )if (i) g ( x,y )= k + f ( x,y ),where k isaconstant (ii) g ( x,y )= mf ( x,y ),where m isanonzeroconstant (iii) g ( x,y )= f ( x Š a,y Š b ),where a and b areconstants (iv) g ( x,y )= f ( px,qy ),where p and q arenonzeroconstants (6) Givenafunction f ,sketchthegraphsof g ( x,y )de“nedinexercise5. Analyzecarefullyvariouscasesforvaluesoftheconstants,forexample, m> 0, m< 0, p> 1,0 0. (8) Find f ( x,y )if f ( x + y,y/x )= x2Š y2. (9) Let z = y + f ( x Š 1).Findthefunctions z and f if z = x when y =1. (10) Graphthefunction F ( t )= f (cos t, sin t ),where f ( x,y )=1if y x and f ( x,y )=0if y
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17413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS example,thedomainofthefunction f ( x,y )=sin( x2+ y2) / ( x2+ y2)is theentireplaneexceptthepoint( x,y )=(0 0).Incontrasttotheonedimensionalcase,thepoint( x,y )mayapproach(0 0)alongvarious paths.Sotheverynotionthat( x,y )approaches(0 0)needstobe accuratelyde“ned. Asnotedbefore,thedomainofafunction f ofseveralvariablesisa setinan n -dimensionalEuclideanspace.Twopoints x =( x1,x2,...,xn) and y =( y1,y2,...,yn)coincideifandonlyifthedistancebetweenthem x Š y = ( x1Š y1)2+( x2Š y2)2+ +( xnŠ yn)2vanishes. Definition 13.6 Apoint r issaidtoapproacha“xedpoint r0inaEuclideanspaceifthedistance r Š r0 tendsto0.Thelimit r Š r0 0 isalsodenotedby r r0. Intheaboveexample,thelimit( x,y ) (0 0)meansthat x2+ y2 0or x2+ y2 0.Therefore, sin( x2+ y2) x2+ y2= sin u u 1as x2+ y2= u 0 Notethatherethelimitpoint(0 0)canbeapproachedfromanydirectionintheplane.Thisisnotalwaysso.Forexample,thedomainofthefunction f ( x,y )=sin( xy ) / ( x + y )isthe“rstquadrant,includingitsboundariesexceptthepoint(0 0).Thepoints(0 0) and( Š 1 Š 1)arenotinthedomainofthefunction.However,the limitof f as( x,y ) (0 0)canbede“ned,whereasthelimitof f as ( x,y ) ( Š 1 Š 1)doesnotmakeanysense.Thedierencebetween thesetwopointsisthatanyneighborhoodof(0 0)containspointsof thedomain,whilethisisnotsofor( Š 1 Š 1).Sothelimitcanbe de“nedonlyforsomespecialclassofpointscalled limitpoints ofa set D Definition 13.7 (LimitPointofaSet) Apoint r0issaidtobea limitpoint ofaset D ifanyopenball N( r0)= { r | 0 < r Š r0 < } (withthecenter r0removed)containsapointof D Alimitpoint r0of D mayormaynotbein D ,butitcanalways beapproachedfromwithintheset D inthesensethat r r0and r D because,nomatterhowsmall is,onecanalways“ndapoint r D thatdoesnotcoincidewith r0andwhosedistancefrom r0is lessthan .Inotherwords,anintersectionofanyball N( r0)centered atalimitpointof D withtheset D ,denotedas N( r0) D ,isalways

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86.LIMITSANDCONTINUITY175 nonempty.Intheaboveexampleof D beingthe“rstquadrant,the limit( x,y ) (0 0)isunderstoodas x2+ y2 0while( x,y ) =(0 0) and x 0, y 0.Theintersection N D isthepartofthedisk 0 0 ,thereexists acorrespondingnumber > 0 suchthatif r D and 0 < r Š r0 < then | f ( r ) Š f0| < .Inthiscase,onewrites limr r0f ( r )= f0. Thenumber | f ( r ) Š f0| determinesthedeviationofthevalueof f fromthenumber f0.Theexistenceofthelimitmeansthatnomatter howsmallthenumber is,thereisaneighborhood N( r0) D ,which containsallpointsof D whosedistancefrom r0islessthananumber andinwhichthevaluesofthefunction f deviatefromthelimitvalue f0nomorethan ,thatis, f0Š 0.Toestablishtheexistenceof > 0,notethat theinequality8 R3< or R<3 / 2guaranteesthat | f ( r ) Š f0| < .Therefore,forallpoints r = 0 inthedomainofthefunctionfor which R< =3 / 2,thefunctiondiersfrom0nomorethan Forexample,put =10Š 6.Then,intheinteriorofaballofradius

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17613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS z y x f0+ f0f0Š D r0y x C1C2r0 = ( r0)Figure13.5.Left :AnillustrationofDe“nition13.8in thecaseofafunction f oftwovariables.Givenapositive number ,considertwohorizontalplanes z = f0Š and z = f0+ .Thenonecanalways“ndacorrespondingnumber > 0andthedisk Ncenteredat r0suchthattheportion ofthegraph z = f ( r )abovetheintersection N D lies betweentheplanes: f0Š 0 of Ndependsonthechoiceof and,generally,onthelimit point r0. Right :Theindependenceofthelimitofapath alongwhichthelimitpoint r0isapproached.Foreverypath leadingto r0,thereisapartofitthatliesin N.Thevalues of f alongthispartofthepathdeviatefrom f0nomorethan anypreassignednumber > 0. =0 005,thevaluesofthefunctioncandeviatefrom f0=0nomore than10Š 6. Theradius ofaneighborhoodinwhichafunction f deviatesno morethan fromthevalueofthelimitdependson and,ingeneral, onthelimitpoint r0. Example 13.7 Let f ( x,y )= xy .Showthat lim( x,y ) ( x0,y0)f ( x,y )= x0y0foranypoint ( x0,y0) Solution: Thedistancebetween r =( x,y )and r0=( x0,y0)is R = ( x Š x0)2+( y Š y0)2.Therefore, | x Š x0| R and | y Š y0| R Considertheidentity xy Š x0y0=( x Š x0)( y Š y0)+ x0( y Š y0)+( x Š x0) y0.

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86.LIMITSANDCONTINUITY177 Put a =( | x0| + | y0| ) / 2.Thenthedeviationof f fromthelimitvalue f0= x0y0isboundedas | f ( x,y ) Š f0|| x Š x0|| y Š y0| + | x0|| y Š y0| + | x Š x0|| y0| R2+( | x0| + | y0| ) R = R2+2 aR =( R + a )2Š a2. Now“x > 0andassumethat R issuchthat( R + a )2Š a2< or 0 0.Then,bytheexistenceofthelimit f0,thereisaballof radius = ( r0) > 0centeredat r0suchthatthevaluesof f liein theinterval f0Š
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17813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS SqueezePrinciple.ThesolutiontoExample13.6employsarathergeneralstrategytoverifywhetheraparticularnumber f0isthelimitof f ( r )as r r0. Theorem 13.3 (SqueezePrinciple) Letthefunctionsofseveralvariables g f ,and h haveacommondomain D andlet g ( r ) f ( r ) h ( r ) forany r D .Ifthelimitsof g ( r ) and h ( r ) as r r0existandequalanumber f0,thenthelimitof f ( r ) as r r0existsandequals f0,thatis, g ( r ) f ( r ) h ( r )andlimr r0g ( r )=limr r0h ( r )= f0 limr r0f ( r )= f0. Proof. Fromthehypothesisofthetheorem,itfollowsthat0 f ( r ) Š g ( r ) h ( r ) Š g ( r ).Put F ( r )= f ( r ) Š g ( r )and H ( r )= h ( r ) Š g ( r ). Then0 F ( r ) H ( r )implies | F ( r ) || H ( r ) | (thepositivityof F isessentialforthisconclusion).Bythehypothesisofthetheoremand thebasicpropertiesofthelimit, H ( r )= h ( r ) Š g ( r ) f0Š f0=0as r r0.Hence,forany > 0,thereisacorrespondingnumber such that0 | F ( r ) || H ( r ) | < whenever0 < r Š r0 < .ByDe“nition 13.8,thismeansthatlimr r0F ( r )=0.Bythebasicpropertiesofthe limit,itisthenconcludedthat f ( r )= F ( r )+ g ( r ) 0+ f0= f0as r r0. Aparticularcaseofthesqueezeprincipleisalsouseful. Corollary 13.1 (Simpli“edSqueezePrinciple) Ifthereexistsafunction h ofonevariablesuchthat | f ( r ) Š f0| h ( R ) 0as r Š r0 = R 0+, then limr r0f ( r )= f0. Thecondition | f ( r ) Š f0| h ( R )isequivalentto f0Š h ( R ) f ( r ) f0+ h ( R ),whichisaparticularcaseofthehypothesisinthesqueeze principle.InExample13.6, h ( R )=8 R3.Ingeneral,thecondition h ( R ) 0as R 0+impliesthat,forany > 0,thereisaninterval 0
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86.LIMITSANDCONTINUITY179 Solution: Let R = x2+ y2(thedistancefromthelimitpoint (0 0)).Then | x | R and | y | R .Therefore, | x3y Š 3 x2y2| x2+ y2+ x4 | x |3| y | +3 x2y2 x2+ y2+ x4 4 R4 R2+ x4= 4 R2 1+( x4/R2) 4 R2. Itfollowsfromthisinequalitythat Š 4( x2+ y2) f ( x,y ) 4( x2+ y2),and,bythesqueezeprinciple, f ( x,y )musttendto0because 4( x2+ y2)= 4 R2 0as R 0.InDe“nition13.8,given > 0, thecorrespondingnumber is = / 2. 86.3.ContinuityofFunctionsofSeveralVariables.Supposethat limr r0f ( r )= f0.Ifthelimitpoint r0liesinthedomainofthefunction f ,thenthefunctionhasavalue f ( r0),whichmayormaynotcoincide withthelimitvalue f0.Infact,thelimitvalue f0doesnotgenerally giveanyinformationaboutthepossiblevalueofthefunctionatthe limitpoint.Forexample,if f ( r )=1everywhereexceptonepoint r0atwhich f ( r0)= c ,then,ineveryneighborhood0 < r Š r0 < f ( r )=1andhencethelimitof f as r r0existsandequals f0=1. When c =1,thelimitvaluedoesnotcoincidewiththevalueofthe functionatthelimitpoint.Thevaluesof f suerajump discontinuity when r reaches r0,andonesaysthat f is discontinuous at r0.A discontinuityalsooccurswhenthelimitof f as r r0doesnotexist while f hasavalueatthelimitpoint. Definition 13.9 (Continuity) Afunction f ofseveralvariableswithdomain D issaidtobe continuous atapoint r0 D if limr r0f ( r )= f ( r0) Thefunction f issaidtobe continuouson D ifitiscontinuousatevery pointof D Example 13.9 Let f ( x,y )=1 if y x andlet f ( x,y )=0 if yx0,then f ( x0,y0)=1.Ontheotherhand,forevery suchpointonecan“ndaneighborhood( x Š x0)+( y Š y0)2<2(a diskofradius > 0centeredat( x0,y0))thatliesintheregion y>x Therefore, | f ( r ) Š f ( r0) | =1 Š 1=0 < forany > 0inthisdisk, thatis,limr r0f ( r )= f ( r0)=1.Thesamelineofreasoningapplies toestablishthecontinuityof f atanypoint( x0,y0),where y0
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18013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Inonepart( y x ), f ( r )=1,whereasintheotherpart( y 0inwhich | f ( r ) Š f ( r0) | = | f ( r ) Š 1 | < because | f ( r ) Š 1 | =1for y 0vanishesattheorigin r0= 0 f ( r0)=0.Put R = x2 1+ x2 2+ + x2 n.Then | xi| R forany elementofthe n -tuple.Hence, | f ( r ) Š f ( r0) | = | x1|k1| x2|k2| xn|kn Rk1+ k2+ + kn= RN 0

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86.LIMITSANDCONTINUITY181 as R 0.Bythesqueezeprinciple, f ( r ) 0= f ( r0).Therational function f ( r ) /g ( r )iscontinuousastheratiooftwocontinuousfunctions if g ( r ) =0. Theorem 13.6 (ContinuityofaComposition) Let g ( u ) becontinuousontheinterval u [ a,b ] andlet h beafunction ofseveralvariablesthatiscontinuouson D andhastherange [ a,b ] Thecomposition f ( r )= g ( h ( r )) iscontinuouson D Theprooffollowsthesamelineofreasoningasinthecaseofthe compositionoftwofunctionsofonevariableinCalculusIandisleft tothereaderasanexercise. Inparticular,somebasicfunctionsstudiedinCalculusI,sin u ,cos u eu,ln u ,andsoon,arecontinuousfunctionsontheirdomains.If f ( r ) isacontinuousfunctionofseveralvariables,theelementaryfunctions whoseargumentisreplacedby f ( r )arecontinuousfunctions.Incombinationwiththepropertiesofcontinuousfunctions,thecomposition rulede“nesalargeclassofcontinuousfunctionsofseveralvariables, whichissucientformanypracticalapplications. Example 13.10 Findthelimit limr 0exzcos( xy + z2) x + yz +3 xz4+( xyz Š 2)2. Solution: Thefunctionisaratio.Thedenominatorisapolynomial andhencecontinuous.Itslimitvalueis( Š 2)2=4 =0.Thefunction exzisacompositionoftheexponential euandthepolynomial u = xz .Soitiscontinuous.Itsvalueis1atthelimitpoint.Similarly, cos( xy + z2)iscontinuousasacompositionofcos u andthepolynomial u = xy + z2.Itsvalueis1atthelimitpoint.Theratioofcontinuous functionsiscontinuousandthelimitis1 / 4. 86.4.Exercises.(1) Usethede“nitionofthelimittoverifyeachofthefollowinglimits (i.e.,given > 0,“ndthecorresponding ( )): (i)limr 0x3Š 4 y2x +5 y3 x2+ y2=0 (ii)limr 0x3Š 4 y2x +5 y3 3 x2+4 y2=0 (iii)limr 0x3Š 4 y4+5 y3x2 3 x2+4 y2=0

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18213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (iv)limr 0x3Š 4 y2x +5 y3 3 x2+4 y2+ y4=0 (v)limr 03 x3+4 y4Š 5 z5 x2+ y2+ z2=0 (2) Usethesqueezeprincipletoprovethefollowinglimitsand“nda neighborhoodofthelimitpointinwhichthedeviationofthefunction fromthelimitvaluedoesnotexceedasmallgivennumber : (i)limr 0y sin( x/ y )=0 (ii)limr 0[1 Š cos( y/x )] x =0 (iii)limr 0cos( xy )sin(4 x y ) xy =0 Hint: | sin u || u | (3) Supposethatlimr r0f ( r )=2and r0isinthedomainof f .If nothingelseisknownaboutthefunction,whatcanbesaidaboutthe value f ( r0)?If,inaddition, f isknowntobecontinuousat r0,what canbesaidaboutthevalue f ( r0)? (4) Findthepointsofdiscontinuityofeachofthefollowingfunctions: (i) f ( x,y )= yx/ ( x2+ y2)if( x,y ) =(0 0)and f (0 0)=1 (ii) f ( x,y,z )= yxz/ ( x2+ y2+ z2)if( x,y,z ) =(0 0 0)and f (0 0 0)=0 (iii) f ( x,y )=sin( xy ) (iv) f ( x,y )=cos( xyz ) / ( x2y2+1) (v) f ( x,y )=( x2+ y2)ln( x2+ y2)if( x,y ) =(0 0)and f (0 0)=0 (vi) f ( x,y )=1ifeither x or y isrationaland f ( x,y )=0elsewhere (vii) f ( x,y )=( x2Š y2) / ( x Š y )if x = y and f ( x,x )=2 x (viii) f ( x,y )=( x2Š y2) / ( x Š y )if x = y and f ( x,x )= x (ix) f ( x,y,z )=1 / [sin( x )sin( z Š y )] (x) f ( x,y )=sin 1 xy (5) Eachofthefollowingfunctionshasthevalueattheorigin f (0 0)= c .Determinewhetherthereisaparticularvalueof c atwhichthe functioniscontinuousattheoriginif,for( x,y ) =(0 0), (i) f ( x,y )=sin(1 / ( x2+ y2)) (ii) f ( x,y )=( x2+ y2)sin(1 / ( x2+ y2)), > 0

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87.AGENERALSTRATEGYFORSTUDYINGLIMITS183 (iii) f ( x,y )= xnymsin(1 / ( x2+ y2)), n 0, m 0,and n + m> 0 (6) Usethepropertiesofcontinuousfunctionsto“ndthefollowing limits (i)limr 0(1+ x + yz2)1 / 3 2+3 x Š 4 y +5 z2(ii)limr 0sin( x y ) (iii)limr 0sin( x y ) cos( x2y ) (iv)limr 0[ exyzŠ 2cos( yz )+3sin( xy )] (v)limr 0ln(1+ x2+ y2z2) 87.AGeneralStrategyforStudyingLimits Thede“nitionofthelimitgivesonlythecriterionforwhethera number f0isthelimitof f ( r )as r r0.Inpractice,however,a possiblevalueofthelimitistypicallyunknown.Somestudiesare neededtomakeaneducatedŽguessforapossiblevalueofthelimit. Hereaproceduretostudylimitsisoutlinedthatmightbehelpful.In whatfollows,thelimitpointisoftensettotheorigin r0=(0 0 ,..., 0). Thisisnotalimitationbecauseonecanalwaystranslatetheoriginof thecoordinatesystemtoanyparticularpointbyshiftingthevaluesof theargument,forexample, lim( x,y ) ( x0,y0)f ( x,y )=lim( x,y ) (0 0)f ( x + x0,y + y0) .87.1.Step1:ContinuityArgument.Thesimplestscenarioinstudying thelimithappenswhenthefunction f inquestioniscontinuousatthe limitpoint: limr r0f ( r )= f ( r0) Forexample, lim( x,y ) (1 2)xy x3Š y2= Š 2 3 becausethefunctioninquestionisarationalfunctionthatiscontinuous if x3Š y2 =0.Thelatterisindeedthecaseforthelimitpoint(1 2). Ifthecontinuityargumentdoesnotapply,thenitishelpfultocheck thefollowing.

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18413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 87.2.Step2:CompositionRule.Theorem 13.7 (CompositionRuleforLimits) Let g ( t ) beafunctioncontinuousat t0.Supposethatthefunction f is thecomposition f ( r )= g ( h ( r )) sothat r0isalimitpointofthedomain of f and h ( r ) t0as r r0.Then limr r0f ( r )=limt t0g ( t )= g ( t0) Theproofisomittedasitissimilartotheproofofthecompositionruleforlimitsofsingle-variablefunctionsgiveninCalculusI.The signi“canceofthistheoremisthat,underthehypothesesofthetheorem,atoughproblemofstudyingamultivariablelimitisreducedto theproblemofthelimitofafunctionofasingleargument.Thelatter problemcanbestudiedby,forexample,lHospitalsrule.Itmustbe emphasizedthat thereisnoanalogoflHospitalsruleformultivariable limits Example 13.11 Find lim( x,y ) (0 0)cos( xy ) Š 1 x2y2. Solution: Thefunctioninquestionis g ( t )=(cos t Š 1) /t2for t = 0,wheretheargument t isreplacedbythefunction h ( x,y )= xy Thefunction h isapolynomialandhencecontinuous.Inparticular, h ( x,y ) h (0 0)=0as( x,y ) (0 0).Thefunction g ( t )iscontinuous forall t =0anditsvalueat t =0isnotde“ned.UsinglHospitals ruletwice, limt 0cos t Š 1 t2=limt 0Š sin t 2 t =limt 0Š cos t 2 = Š 1 2 So,bysetting g (0)= Š 1 / 2,thefunction g ( t )becomescontinuousat t = 0,andthehypothesesofthecompositionruleareful“lled.Therefore, thetwodimensionallimitinquestionexistsandequals Š 1 / 2. 87.3.Step3:LimitsAlongCurves.Recallthefollowingresultabout thelimitofafunctionofonevariable.Thelimitof f ( x )as x x0existsandequals f0ifandonlyifthecorrespondingrightandleftlimits of f ( x )existandequal f0: limx x+ 0f ( x )=limx xŠ 0f ( x )= f0 limx x0f ( x )= f0. Inotherwords,ifthelimitexists,itdoesnotdependonthedirection fromwhichthelimitpointisapproached.Iftheleftandrightlimits existbutdonotcoincide,thenthelimitdoesnotexist.

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87.AGENERALSTRATEGYFORSTUDYINGLIMITS185 Forfunctionsofseveralvariables,therearein“nitelymanypaths alongwhichthelimitpointcanbeapproached.Theyincludestraight linesandpathsofanyothershape,incontrasttotheone-variable case.Nevertheless,asimilarresultholdsformultivariablelimits(see thesecondremarkattheendofSection86.1);thatis, ifthelimitexists, thenitshouldnotdependonthepathalongwhichthelimitpointmay beapproached Definition 13.10 (ParametricCurveinaEuclideanSpace) AparametriccurveinaEuclideanspaceisasetofpoints r ( t )= ( x1( t ) ,x2( t ) ,...,xn( t )) ,where xi( t ) i =1 2 ,...,n ,arecontinuousfunctionsofavariable t [ a,b ] Thisisanaturalgeneralizationoftheconceptofaparametriccurve inaplaneorspaceasavectorfunctionde“nedbytheparametric equations xi= xi( t ), i =1 2 ,...,n Definition 13.11 (LimitAlongaCurve) Let r0bealimitpointofthedomain D ofafunction f .Let r ( t )= ( x1( t ) ,x2( t ) ,...,xn( t )) t t0,beaparametriccurve C in D suchthat r ( t ) r0as t t+ 0.Let F ( t )= f ( r ( t )) t>t0,bethevaluesof f on thecurve C .Thelimit limt t+ 0F ( t )=limt t+ 0f ( x1( t ) ,x2( t ) ,...,xn( t )) iscalledthe limitof f alongthecurve C ifitexists. Supposethatthelimitof f ( r )as r r0existsandequals f0.Let C beacurvesuchthat r ( t ) r0as t t+ 0.Fix > 0.Bytheexistenceof thelimit,thereisaneighborhood N( r0)= { r | r D, 0 < r Š r0 < } inwhichthevaluesof f deviatefrom f0nomorethan | f ( r ) Š f0| < Sincethecurve C passesthrough r0,thereshouldbeaportionofitthat liesin N( r0);thatis,thereisanumber suchthat r ( t ) Š r0 < for all t ( t0,t0+ ),whichismerelythede“nitionofthelimit r ( t ) r0as t t+ 0.Hence,forany > 0,thedeviationofvaluesof f along thecurve, F ( t )= f ( r ( t )),doesnotexceed | F ( t ) Š f0| < whenever 0 < | t Š t0| <.Bythede“nitionoftheone-variablelimit,thisimplies that F ( t ) f0as t t0foranycurve C through r0.Thisprovesthe following. Theorem 13.8 (IndependenceoftheLimitfromaCurveThrough theLimitPoint) Ifthelimitof f ( r ) existsas r r0,thenthelimitof f alongany curveleadingto r0fromwithinthedomainof f exists,anditsvalueis independentofthecurve.

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18613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Animmediateconsequenceofthistheoremisausefulcriterionfor thenonexistenceofamultivariablelimit. Corollary 13.2 (CriterionforNonexistenceoftheLimit) Let f beafunctionofseveralvariableson D .Ifthereisacurve r ( t ) in D suchthat r ( t ) r0as t t+ 0andthelimit limt t+ 0f ( r ( t )) doesnot exist,thenthemultivariablelimit limr r0f ( r ) doesnotexisteither.If therearetwocurvesin D leadingto r0suchthatthelimitsof f along themexistbutdonotcoincide,thenthemultivariablelimit limr r0f ( r ) doesnotexist.RepeatedLimits.Let( x,y ) =(0 0).Considerapath C1thatconsists oftwostraightlinesegments( x,y ) ( x, 0) (0 0)andapath C2thatconsistsoftwostraightlinesegments( x,y ) (0 ,y ) (0 0). Bothpathsconnect( x,y )withtheorigin.Thelimitsalong C1and C2, limy 0 limx 0f ( x,y ) andlimx 0 limy 0f ( x,y ) arecalledthe repeated limits.If C1and C2are within thedomainof f thenTheorem13.8andCorollary13.2establishtherelationsbetween therepeatedlimitsandthetwo-variablelimitlim( x,y ) (0 0)f ( x,y ).In particular,supposethat f ( x,y ) f (0 ,y )as x 0and f ( x,y ) f ( x, 0)as y 0(thefunctioniscontinuouswithrespectto x if y is “xedanditisalsocontinuouswithrespectto y if x is“xed).Thenthe repeatedlimitsbecome limy 0f (0 ,y )andlimx 0f ( x, 0) Ifatleastoneofthemdoesnotexistortheyexistbutarenotequal, then,byCorollary13.2,thetwo-variablelimitdoesnotexist.Ifthey existandareequal,thenthetwo-variablelimit mayormaynot exist. Afurtherinvestigationisneeded. Ingeneral,thesegment( x, 0) (0 0)or(0 ,y ) (0 0)orboth maynotbeinthedomainof f ,whiletherepeatedlimitsstillmake sense(e.g.,thefunction f isde“nedonlyforstrictlypositive x and y sothatthehalf-lines x =0, y> 0and y =0, x> 0arelimitpoints ofthedomain).Inthiscase,thehypothesesofCorollary13.2arenot ful“lled,and,inparticular, thenonexistenceoftherepeatedlimitsdoes notimplythenonexistenceofthetwo-variablelimit .Anexampleis providedinexercise1,part(iii).LimitsAlongStraightLines.Letthelimitpointbetheorigin r0= (0 0 ,..., 0).Thesimplestcurveleadingto r0isastraightline xi= vit where t 0+forsomenumbers vi, i =1 2 ,...,n ,thatdonotvanish

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87.AGENERALSTRATEGYFORSTUDYINGLIMITS187 simultaneously.Thelimitofafunctionofseveralvariables f alonga straightline,limt 0+f ( v1t,v2t,...,vnt ),shouldexistandbethesame for anychoice ofnumbers vi.Forcomparison,recallthevectorequation ofastraightlineinspacethroughtheorigin: r = t v ,where v isavector paralleltotheline. Example 13.12 Investigatethetwo-variablelimit lim( x,y ) (0 0)xy3 x4+2 y4. Solution: Considerthelimitsalongstraightlines x = t y = at (or y = ax ,where a istheslope)as t 0+: limt 0+f ( t,at )=limt 0+a3t4 t4(1+2 a4) = a3 1+2 a4. Sothelimitalongastraightlinedependsontheslopeoftheline. Therefore,thetwo-variablelimitdoesnotexist. Example 13.13 Investigatethelimit lim( x,y ) (0 0)sin( xy ) x + y Solution: Thedomainofthefunctionconsistsofthe“rstandthird quadrantsas xy 0excepttheorigin.Linesapproaching(0 0)from withinthedomainare x = t y = at a 0and t 0.Theline x =0, y = t alsoliesinthedomain(thelinewithanin“niteslope).The limitalongastraightlineapproachingtheoriginfromwithinthe“rst quadrantis limt 0+f ( t,at )=limt 0+sin( t a ) t (1+ a ) =limt 0+ a cos( t a ) 1+ a = a 1+ a wherelHospitalsrulehasbeenusedtocalculatethelimit.Thelimit dependsontheslopeoftheline,andhencethetwo-variablelimitdoes notexist. LimitsAlongPowerCurves(Optional).Ifthelimitalongstraightlines existsandisindependentofthechoiceoftheline,thenumericalvalue ofthislimitprovidesadesirededucatedŽguessfortheactualmultivariablelimit.However,thishasyettobeprovedbymeansofeither thede“nitionofthemultivariablelimitor,forexample,thesqueeze principle.Thiscomprisesthelaststepoftheanalysisoflimits(Step4; seebelow).

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18813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Thefollowingshouldbestressed. Ifthelimitsalongallstraightlines happentobethesamenumber,thisdoesnotmeanthatthemultivariablelimitexistsandequalsthatnumberbecausetheremightexistother curvesthroughthelimitpointalongwhichthelimitattainsadierent valueordoesnotevenexist Example 13.14 Investigatethelimit lim( x,y ) (0 0)y3 x Solution: Thedomainofthefunctionisthewholeplanewiththe y axisremoved( x =0).Thelimitalongastraightline limt 0+f ( t,at )=limt 0+a3t3 t = a3limt 0+t2=0 vanishesforanyslope;thatis,itisindependentofthechoiceofthe line.However,thetwo-variablelimitdoesnotexist!Considerthe powercurve x = t y = at1 / 3approachingtheoriginas t 0+.The limitalongthiscurvecanattainanyvaluebyvaryingtheparameter a : limt 0+f ( t,at1 / 3)=limt 0+a3t t = a3. Thus,themultivariablelimitdoesnotexist. Ingeneral,limitsalongpowercurvesareconvenientforstudying limitsofrationalfunctionsbecausethevaluesofarationalfunction ofseveralvariablesonapowercurvearegivenbyarationalfunction ofthecurveparameter t .Onecanthenadjust,ifpossible,thepower parameterofthecurvesothattheleadingtermsofthetopandbottom powerfunctionsmatchinthelimit t 0+.Forinstance,intheexample considered,put x = t and y = atn.Then f ( t,atn)=( a3t3 n) /t .The powersofthetopandbottomfunctionsinthisratiomatchif3 n =1; hence,for n =1 / 3,thelimitalongthepowercurvedependsonthe parameter a andcanbeanynumber.87.4.Step4:UsingtheSqueezePrinciple.IfSteps1and2donot applytothemultivariablelimitinquestion,thenaneducatedŽguess forapossiblevalueofthelimitishelpful.ThisistheoutcomeofStep 3.Iflimitsalongafamilyofcurves(e.g.,straightlines)happentobe thesamenumber f0,thenthisnumberisthesought-aftereducatedŽ guess.Thede“nitionofthemultivariablelimitorthesqueezeprinciple canbeusedtoproveordisprovethat f0isthemultivariablelimit.

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87.AGENERALSTRATEGYFORSTUDYINGLIMITS189 Example 13.15 Findthelimitorprovethatitdoesnotexist: lim( x,y ) (0 0)sin( xy2) x2+ y2. Solution: Step1 .Thefunctionisnotde“nedattheorigin.Thecontinuity argumentdoesnotapply. Step2 .Nosubstitutionexiststotransformthetwo-variablelimittoa one-variablelimit. Step3 .Put( x,y )=( t,at ),where t 0+.Thelimitalongstraight lines limt 0+f ( t,at )=limt 0+sin( a2t3) t2=limu 0+sin( a2u3 / 2) u =limu 0+(3 / 2) a2u1 / 2cos( a2u3 / 2) 1 =0 vanishes(herethesubstitution u = t2andlHospitalsrulehavebeen usedtocalculatethelimit). Step4 .Ifthetwo-variablelimitexists,thenitmustbeequalto0. Thiscanbeveri“edbymeansofthesimpli“edsqueezeprinciple;that is,onehastoverifythatthereexists h ( R )suchthat | f ( x,y ) Š f0| = | f ( x,y ) | h ( R ) 0as R = x2+ y2 0.Akeytechnicaltrickhere istheinequality | sin u || u | ,whichholdsforanyreal u .Onehas | f ( x,y ) Š 0 | = | sin( xy2) | x2+ y2 | xy2| x2+ y2 R3 R2= R 0 wheretheinequalities | x | R and | y | R havebeenused.Thus,the two-variablelimitexistsandequals0. Fortwo-variablelimits,itissometimesconvenienttousepolarcoordinatescenteredatthelimitpoint x Š x0= R cos y Š y0= R sin Theideaisto“ndoutwhetherthedeviationofthefunction f ( x,y ) from f0(theeducatedŽguessfromStep3)canbeboundedby h ( R ) uniformly forall [0 2 ]: | f ( x,y ) Š f0| = | f ( x0+ R cos ,y0+ R sin ) Š f0| h ( R ) 0 as R 0+.Thistechnicaltaskcanbeaccomplishedwiththehelpof thebasicpropertiesoftrigonometricfunctions,forexample, | sin | 1, | cos | 1,andsoon.

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19013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS InExample13.15,Step3gives f0=0ifonlythelimitsalong straightlineshavebeenstudied.Then | f ( x,y ) Š f0| = | y |3 | x | = | R sin3 | | R cos | = R2sin2( ) | tan | Despitethatthedeviationof f from0isproportionalto R2 0as R 0+,itcannotbemadeassmallasdesireduniformlyforall by decreasing R becausetan isnotaboundedfunction.Thereisasector intheplanecorrespondingtoanglesnear = / 2,wheretan canbe largerthananynumberwhereassin2 is strictly positiveinitsothat thedeviationof f from0canbeaslargeasdesirednomatterhow small R is.So,forany > 0,theinequality | f ( r ) Š f0| < isviolated inthatsectorofanydisk0 < r Š r0 < ,andhencethelimitdoes notexist. Remark. Formultivariablelimitswith n> 2,asimilarapproach exists.If,forsimplicity, r0=(0 0 ,..., 0).Thenput xi= Rui,where thevariables uisatisfythecondition u2 1+ u2 2+ + u2 n=1.For n =2, u1=cos and u2=sin .For n 3,thevariables uicanbeviewedas thedirectionalcosines,thatis,thecosinesoftheanglesbetween r and unitvectors eiparalleltothecoordinateaxes, ui= r ei/ r .Then onehastoinvestigatewhetherthereis h ( R )suchthat | f ( Ru1,Ru2,...,Run) Š f0| h ( R ) 0 ,R 0+. Thistechnical,oftenratherdicult,taskmaybeaccomplishedusing theinequalities | ui| 1andsomespeci“cpropertiesofthefunction f Asnoted,thevariables uiarethedirectionalcosines.Theycanalsobe trigonometricfunctionsoftheanglesinthesphericalcoordinatesystem inan n -dimensionalEuclideanspace.87.5.InniteLimitsandLimitsatInnity.Supposethatthelimitofa multivariablefunction f doesnotexistas r r0.Therearetwo particularcases,whichareofinterest,when f tendstoeitherpositive ornegativein“nity. Definition 13.12 (In“niteLimits). Thelimitof f ( r ) as r r0issaidtobethe positivein“nity if,for anynumber M> 0 ,thereexistsanumber > 0 suchthat f ( r ) >M whenever 0 < r Š r0 < .Similarly,thelimitissaidtobethe negative in“nity if,foranynumber M< 0 ,thereexistsanumber > 0 such that f ( r )
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87.AGENERALSTRATEGYFORSTUDYINGLIMITS191 respectively, limr r0f ( r )= andlimr r0f ( r )= Š Forexample, limr 01 x2+ y2= Indeed,put R = x2+ y2.Then,forany M> 0,theinequality f ( r ) >M canbewrittenintheform R< 1 / M .Therefore,the valuesof f inthedisk0 < r < =1 / M arelargerthanany preassignedpositivenumber M Naturally,ifthelimitisin“nite,thefunction f approachesthe in“nitevaluealonganycurvethatleadstothelimitpoint.Forexample, thelimitof f ( x,y )= y/ ( x2+ y2)as( x,y ) (0 0)doesnotexist because,alongstraightlines( x,y )=( t,at )approachingtheorigin when t 0+,thefunction f ( t,at )= c/t ,where c = a/ (1+ a2),tends to+ if a> 0,to Š if a< 0,andto0if a =0.If,however,the domainof f is restricted tothehalf-plane y> 0,thenthelimitexists andequals .Indeed,forall x and y> 0, f ( x,y ) y/y2=1 /y as y 0+,andtheconclusionfollowsfromthesqueezeprinciple. Forfunctionsofonevariable x ,onecande“nethelimitsatin“nity (i.e.,when x + or x Š ).Boththelimitshaveacommon propertythatthedistance | x | ofthein“nitepointsŽ fromthe origin x =0isin“nite.Similarly,inaEuclideanspace,thelimitat in“nityisde“nedinthesensethat r .If D isanunbounded region,thenaneighborhoodofthein“nitepointin D consistsofall pointsof D whosedistancefromtheoriginexceedsanumber r > Asmallerneighborhoodisobtainedbyincreasing Definition 13.13 (LimitatIn“nity). Let f beafunctiononanunboundedregion D .Anumber f0isthe limitofafunction f atin“nity, limr f ( r )= f0if,foranynumber > 0 ,thereexists > 0 suchthat | f ( r ) Š f0| < whenever r > in D In“nitelimitsatin“nitycanbede“nedsimilarly.Thesqueezeprinciplehasanaturalextensiontothein“nitelimitsandlimitsatin“nity. Forexample,if g ( r ) f ( r )and g ( r ) as r r0(or r ),then f ( r ) .

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19213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 87.6.StudyProblems.Problem13.1. Findthelimit limr r0f ( r ) orshowthatitdoesnot exist,where f ( r )= f ( x,y,z )=( x2+2 y2+4 z2)ln( x2+ y2+ z2) r0=(0 0 0) Solution: Step1 .Thecontinuityargumentdoesnotapplybecause f isnot de“nedat r0. Step2 .Nosubstitutionispossibletotransformthelimittoaonevariablelimit. Step3 .Put r ( t )=( at,bt,ct )forsomeconstants a b ,and c thatdo notvanishsimultaneouslysothattheyde“nethedirectionoftheline throughtheorigin.Then f ( r ( t ))= At2ln( Bt2)=2 At2ln t + At2ln B, where A = a2+2 b2+4 c2and B = a2+ b2+ c2> 0.BylHospitals rule, limt 0+t2ln t =limt 0+ln t tŠ 2=limt 0+tŠ 1 Š 2 tŠ 3= Š 1 2 limt 0+t =0 andtherefore f ( r ( t )) 0as t 0+.So,ifthelimitexists,thenit mustbeequalto0. Step4 .Put R2= x2+ y2+ z2.Sincethelimit R 0+isofinterest, onecanalwaysassumethat R< 1sothatln R2=2ln R< 0.By makinguseoftheinequalities | x | R | y | R ,and | z | R ,one has R2 x2+2 y2+4 z2 7 R2.Bymultiplyingthelatterinequality byln R2< 0, R2ln R2 f ( r ) 7 R2ln( R2).Since t ln t 0as t = R2 0+,thelimitexistsandequals0bythesqueezeprinciple. Problem13.2. Provethatthelimit limr r0f ( r ) exists,where f ( r )= f ( x,y )= 1 Š cos( x2y ) x2+2 y2, r0=(0 0) and“ndadiskcenteredat r0inwhichvaluesof f deviatefromthe limitbymorethan =0 5 10Š 4. Solution: Step1 .Thecontinuityargumentdoesnotapplybecause f isnot de“nedat r0. Step2 .Nosubstitutionispossibletotransformthelimittoaonevariablelimit.

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87.AGENERALSTRATEGYFORSTUDYINGLIMITS193 Step3 .Put r ( t )=( t,at ).Then limt 0+f ( r ( t ))=limt 0+1 Š cos( at3) t2(1+2 a2) = 1 1+2 a2limu 0+1 Š cos( au3 / 2) u = 1 1+2 a2limu 0+au1 / 2sin( au3 / 2) 1 =0 wherethesubstitution u = t2andlHospitalsrulehavebeenusedto evaluatethelimit.Therefore,ifthelimitexists,itmustbeequalto0. Step4 .Note“rstthat1 Š cos u =2sin2( u/ 2) u2/ 2,wherethe inequality | sin x || x | hasbeenused.Put R2= x2+ y2.Then,by makinguseoftheaboveinequalitywith u = x2y togetherwith | x | R and | y | R ,thefollowingchainofinequalitiesisobtained: | f ( r ) Š 0 | ( x2y )2/ 2 x2+2 y2= ( x2y )2/ 2 R2+ y2 ( x2y )2/ 2 R2 1 2 R6 R2= R4 2 0 as R 0+.Bythesqueezeprinciple,thelimitexistsandequals0. Itfollowsfrom | f ( r ) | R4/ 2that | f ( r ) | < whenever R4/ 2 < or R = r Š r0 < ( )=(2 )1 / 4=0 1. Problem13.3. Findthelimit limr r0f ( r ) orshowthatitdoesnot exist,where f ( r )= f ( x,y )= x2y x2Š y2, r0=(0 0) Solution: Step1 .Thecontinuityargumentdoesnotapplybecause f isnot de“nedat r0. Step2 .Nosubstitutionispossibletotransformthelimittoaonevariablelimit. Step3 .Thedomain D ofthefunctionisthewholeplanewiththe lines y = x excluded.Soput r ( t )=( t,at ),where a = 1.Then f ( r ( t ))= at3/t2(1 Š a2)= a (1 Š a2)Š 1t 0as t 0+.So,ifthelimit exists,thenitmustbeequalto0. Step4 .Inpolarcoordinates, x = R cos and y = R sin ,where r Š r0 = R f ( r )= R3cos2 sin R2(cos2 Š sin2 ) = 1 2 R cos sin(2 ) cos(2 ) = R cos 2 tan(2 ) Therefore,inanydisk0 < r Š r0 < ,thereisasectorcorresponding tothepolarangle / 4 < 0small enoughbecausetan(2 )isnotboundedinthisinterval.Hence,forany

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19413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS > 0,thereisno > 0suchthat | f ( r ) | < whenever r D liesin thedisk0 < r Š r0 < .Thus,thelimitdoesnotexist. Step3(Optional) .ThenonexistenceofthelimitestablishedinStep 4impliesthatthereshouldexistcurvesalongwhichthelimitdiers from0.Itisinstructivetodemonstratethisexplicitly.Anysuchcurve shouldapproachtheoriginfromwithinoneofthenarrowsectorscontainingthelines y = x (wheretan(2 )takeslargevalues).Soput, forexample, r ( t )=( t,t Š atn),where n> 1and a =0isanumber. Observethattheline L ( t )=( t,t )(or y = x )istangenttothecurve r ( t )attheoriginbecause r(0)=(1 1)for n> 1.Theterm Š atnin r ( t )modelsasmalldeviationofthecurvefromtheline y = x inthe vicinityofwhichthefunction f isexpectedtobeunbounded.Then f ( r ( t ))=( t3Š atn +2) / (2 atn +1Š a2t2 n).Thisfunctiontendstoanumber as t 0+if n ischosentomatchtheleading(smallest)powersofthe topandbottomoftheratiointhislimit(i.e.,3= n +1or n =2).Thus, for n =2, f ( r ( t ))=( t3Š at4) / (2 at3Š a2t4)=(1 Š at2) / (2 a Š a2t ) 1 / (2 a )as t 0+and f ( r ( t ))divergesfor n> 2inthislimit. Problem13.4. Find limr ln( x2+ y4) x2+2 y2orshowthatthelimitdoesnotexist. Solution: Step1 .Doesnotapply. Step2 .Nosubstitutionexiststoreducethelimittoaone-variable limit. Step3 .Put( x,y )=( t,at )andlet t .Then r as t .Onehas f ( t,at )=ln( t2+ a4t4) / ( t2+2 a2t2).Forlargevalues of t ,ln( t2+ a4t4) ln( a4t4)=ln( t4)+ln( a4) 4ln t if a =0and f ( t, 0)=2ln t/t2.Therefore, f ( t,at )behavesasln t/t2 0as t (bylHospitalsrule).Sothelimitalongallstraightlinesis0. Step4 .Put R = x2+ y2sothat | x | R and | y | R .Then, owingtothemonotonicityofthelogarithmfunction,ln( x2+ y4) ln( R2+ R4) ln(2 R4)for R 1.Thedenominatoroftheratio f can beestimatedfrombelow: x2+2 y2= x2+ y2+ y2= R2+ y2 R2. Hence,for R> 1, | f ( x,y ) Š 0 | ln(4 R4) R2= 4ln R +ln4 R2 0as R Thus,bythesqueezeprinciplethelimitisindeed0.

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87.AGENERALSTRATEGYFORSTUDYINGLIMITS195 87.7.Exercises.(1) Provethefollowingstatements: (i)Let f ( x,y )=( x Š y ) / ( x + y ).Then limx 0 limy 0f ( x,y ) =1 limy 0 limx 0f ( x,y ) = Š 1 butthelimitof f ( x,y )as( x,y ) (0 0)doesnotexist. (ii)Let f ( x,y )= x2y2/ ( x2y2+( x Š y )2).Then limx 0 limy 0f ( x,y ) =limy 0 limx 0f ( x,y ) =0 butthelimitof f ( x,y )as( x,y ) (0 0)doesnotexist. (iii)Let f ( x,y )=( x + y )sin(1 /x )sin(1 /y ).Thenthelimits limx 0 limy 0f ( x,y ) andlimy 0 limx 0f ( x,y ) donotexist,butthelimitof f ( x,y )existsandequals0as ( x,y ) (0 0).DoestheresultcontradictTheorem13.8? Explain. (2) Findeachofthefollowinglimitsorshowthatitdoesnotexist: (i)limr 0cos( xy + z ) x4+ y2z2+4 (ii)limr 0sin( xy ) Š xy ( xy )3(iii)limr 0 xy2+1 Š 1 xy2(iv)limr 0sin( xy3) x2(v)limr 0x3+ y5 x2+2 y2(vi)limr 0e r Š 1 Š r r 2(vii)limr 0x2+sin2y x2+2 y2(viii)limr 0xy2+ x sin( xy ) x2+2 y2(ix)lim( x,y ) (1 0)ln( x + ey) x2+ y2(x)limr 0( x2+ y2)x2y2(xi)limr 01 xy tan xy 1+ xy (xii)limr 0ln sin( x2Š y2) x2Š y22(xiii)limr 0 xy +1 Š 1 y x (xiv)limr 0xbya xa+ yb, 0 0atwhichthe functioniscontinuousattheorigin. (4) Let f ( x,y )= x2y/ ( x4+ y2)if x2+ y2 =0and f (0 0)=0.Show that f iscontinuousalonganystraightlinethroughtheorigin;that is, F ( t )= f ( x ( t ) ,y ( t ))iscontinuousforall t ,where x ( t )= t cos ,

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19613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS y ( t )= t sin forany“xed ,but f isnotcontinuousat(0 0). Hint: Investigatethelimitsof f alongpowercurvesleadingtotheorigin. (5) Let f ( x,y )becontinuousinarectangle a if islarge enough?Doesthelimitlimr f ( x,y )exist? (7) Findthelimitorshowthatitdoesnotexist: (i)limr 0sin( x2+ y2+ z2) x4+ y4+ z4(ii)limr 0x2+2 y2+3 z2 x4+ y2z4(iii)limr ln( x2y2z2) x2+ y2+ z2(iv)limr e3 x2+2 y2+ z2 ( x2+2 y2+3 z2)2012(v)limr 0z x2+ y2+ z2(vi)limr 0z x2+ y2+ z2if z< 0 (vii)limr x2+ y2 x2+ y4(viii)limr sin x 2 x + y (ix)limr ( x2+ y2) eŠ| x + y |(x)limr xy x2+ y2x2(8) Findtherepeatedlimits limx 1 limy 0logx( x + y ) andlimy 0 limx 1logx( x + y ) Whatcanbesaidaboutthecorrespondingtwo-variablelimit? 88.PartialDerivatives Thederivative f( x0)ofafunction f ( x )at x = x0containsimportantinformationaboutthelocalbehaviorofthefunctionnear x = x0. Itde“nestheslopeofthetangentline L ( x )= f ( x0)+ f( x0)( x Š x0), and,for x closeenoughto x0,valuesof f canbewellapproximatedby thelinearization L ( x ),thatis, f ( x ) L ( x ).Inparticular,if f( x0) > 0, f increasesnear x0,and,if f( x0) < 0, f decreasesnear x0.Furthermore,thesecondderivative f( x0)suppliesmoreinformationabout f near x0,namely,itsconcavity. Itisthereforeimportanttodevelopasimilarconceptforfunctions ofseveralvariablesinordertostudytheirlocalbehavior.Asigni“cant dierenceisthat,givenapointinthedomain,therateofchangeis goingtodependonthedirectioninwhichitismeasured.Forexample, if f ( r )istheheightofahillasafunctionofposition r ,thentheslopes fromwesttoeastandfromsouthtonorthmaybedierent.This

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88.PARTIALDERIVATIVES197 observationleadstotheconceptofpartialderivatives.If x and y are thecoordinatesfromwesttoeastandfromsouthtonorth,respectively, thenthegraphof f isthesurface z = f ( x,y ).Ata“xedpoint r0= ( x0,y0),theheightchangesas h ( x )= f ( x,y0)alongthewest…east directionandas g ( y )= f ( x0,y )alongthesouth…northdirection.Their graphsareintersectionsofthesurface z = f ( x,y )withthecoordinate planes x = x0and y = y0,thatis, z = f ( x0,y )= g ( y )and z = f ( x,y0)= h ( x ).Theslopealongthewest…eastdirectionis h( x0),and theslopealongthesouth…northdirectionis g( y0).Theseslopesare called partialderivatives of f anddenotedas f x ( x0,y0)= d dx f ( x,y0) x = x0, f y ( x0,y0)= d dy f ( x0,y ) y = y0. Thepartialderivativesarealsodenotedas f x ( x0,y0)= f x( x0,y0) f y ( x0,y0)= f y( x0,y0) Thesubscriptof findicatesthevariablewithrespecttowhichthe derivativeiscalculated.TheaboveanalysisofthegeometricalsignificanceofpartialderivativesisillustratedinFigure13.6.Theconcept ofpartialderivativescaneasilybeextendedtofunctionsofmorethan twovariables.88.1.PartialDerivativesofaFunctionofSeveralVariables.Let D bea subsetofan n -dimensionalEuclideanspace. Definition 13.14 (InteriorPointofaSet) Apoint r0issaidtobean interiorpoint of D ifthereisanopenball B( r0)= { r | r Š r0 < } ofradius thatliesin D (i.e., B( r ) D ). Inotherwords, r0isaninteriorpointof D ifthereisapositive number > 0suchthatallpointswhosedistancefrom r0islessthan alsoliein D .Forexample,if D isasetpointsinaplanewhose coordinatesareintegers,then D hasnointeriorpointsatallbecause thepointsofadiskofradius0
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19813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS P Q z = f ( x0,y ) z = f ( x,y0) z = f ( x,y ) y = y0x = x0z x P x0xz = f ( x,y0) tan x= f x( x0,y0) z y y0P tan y= f y( x0,y0) yz = f ( x0,y )Figure13.6.Geometricalsigni“canceofpartialderivatives. Left :Thegraph z = f ( x,y )anditscrosssectionsbythecoordinateplanes x = x0and y = y0.The point Q =( x0,y0, 0)isinthedomainof f andthepoint P =( x0,y0,f ( x0,y0))liesonthegraph. Middle :Thecross section z = f ( x,y0)ofthegraphintheplane y = y0and thetangentlinetoitatthepoint P .Theslopetan xofthe tangentlineisdeterminedbythepartialderivative f x( x0,y0) atthepoint Q Right :Thecrosssection z = f ( x0,y )ofthe graphintheplane x = x0andthetangentlinetoitatthe point P .Theslopetan yofthetangentlineisdeterminedby thepartialderivative f y( x0,y0)atthepoint Q .Here y< 0 asitiscountedclockwise.Anopensetisanextensionofthenotionofanopeninterval( a,b ) tothemultivariablecase.Inparticular,thewholeEuclideanspaceis open. Recallthatanyvectorinspacemaybewrittenasalinearcombinationofthreeunitvectors, r =( x,y,z )= x e1+ y e2+ z e3,where e1=(1 0 0), e2=(0 1 0),and e3=(0 0 1).Similarly,usingthe rulesforadding n -tuplesandmultiplyingthembyrealnumbers,one canwrite r =( x1,x2,...,xn)= x1 e1+ x2 e2+ + xn en, where eiisthe n -tuplewhosecomponentsarezerosexceptthe i thone, whichisequalto1.Obviously, ei =1, i =1 2 ,...,n Definition 13.16 (PartialDerivativesataPoint) Let f beafunctionofseveralvariables ( x1,x2,...,xn) .Let D bethe domainof f andlet r0beaninteriorpointof D .Ifthelimit f xi( r0)=limh 0f ( r0+ h ei) Š f ( r0) h exists,thenitiscalledthepartialderivativeof f withrespectto xiat r0.

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88.PARTIALDERIVATIVES199 Thereasonthepoint r0needstobeaninteriorpointissimple.By thede“nitionoftheone-variablelimit, h canbenegativeorpositive. Sothepoints r0+ h ei, i =1 2 ,...,n ,mustbeinthedomainofthe functionbecauseotherwise f ( r0+ h ei)isnotevende“ned.Thisis guaranteedif r0isaninteriorpointbecauseallpoints r intheball B( r0)ofsucientlysmallradius = | h | arein D Remark. ItisalsocommontoomitprimeŽinthenotationsforpartialderivatives.Forexample,thepartialderivativeof f withrespect to x isdenotedas fx.Inwhatfollows,thenotationintroducedinDefinition13.16willbeused. Let r0=( a1,a2,...,an),where aiare“xednumbers.Considerthe function F ( xi)ofonevariable xi( i is“xed),whichisobtainedfrom f ( r )by“xingallthevariables xj= ajexceptthe i thone(i.e., xj= ajforall j = i ).Bythede“nitionoftheordinaryderivative,the partialderivative f xi( r0)existsifandonlyifthederivative F( ai)exists because (13.1) f xi( r0)=limh 0F ( ai+ h ) Š F ( ai) h = dF ( xi) dxi xi= aijustlikeinthecaseoftwovariablesdiscussedatthebeginningofthis section.Thisruleispracticalforcalculatingpartialderivativesasit reducestheproblemtocomputingordinaryderivatives. Example 13.16 Findthepartialderivativesof f ( x,y,z )= x3Š y2z atthepoint (1 2 3) Solution: Bytherule(13.1), f x(1 2 3)= d dx f ( x, 2 3) x =1= d dx ( x3Š 12) x =1=3 f y(1 2 3)= d dy f (1 ,y, 3) y =2= d dy (1 Š 3 y2) y =2= Š 12 f z(1 2 3)= d dz f (1 2 ,z ) z =3= d dz (1 Š 4 z ) z =3= Š 4 GeometricalSignicanceofPartialDerivatives.Fromtherule(13.1), itfollowsthat thepartialderivative f xi( r0) de“nestherateofchange ofthefunction f whenonlythevariable xichangeswhiletheother variablesarekept“xed .If,forinstance,thefunction f inExample13.16 de“nesthetemperatureindegreesCelsiusasafunctionoftheposition whosecoordinatesaregiveninmeters,then,atthepoint(1 2 3),the temperatureincreasesatrateof4degreesCelsiuspermeterinthe

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20013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS directionofthe x axis,anditdecreasesattherates Š 12and Š 4degrees Celsiuspermeterinthedirectionofthe y and z axes,respectively.88.2.PartialDerivativesasFunctions.Supposethatthepartialderivativesof f existatallpointsofaset D .Theneachpartialderivative canbeviewedasafunctionofseveralvariableson D .Thesefunctions aredenotedas f xi( r ),where r D .Theycanbefoundbythesame rule(13.1)if,whendierentiatingwithrespectto xi,allothervariables arenotsettoanyspeci“cvaluesbutratherviewedasindependentof xi(i.e., dxj/dxi=0forall j = i ).Thisagreementisre”ectedbythe notation f xi( x1,x2,...,xn)= xif ( x1,x2,...,xn); thatis,thesymbol /ximeansdierentiationwithrespectto xiwhile regardingallothervariablesasnumericalparametersindependentof xi. Example 13.17 Find f x( x,y ) and f y( x,y ) if f ( x,y )= x sin( xy ) Solution: Assuming“rstthat y isanumericalparameterindependentof x ,oneobtains f x( x,y )= x f ( x,y )= x x sin( xy )+ x x sin( xy ) =sin( xy )+ xy cos( xy ) bytheproductruleforthederivative.Ifnowthevariable x isviewed asanumericalparameterindependentof y ,oneobtains f y( x,y )= y f ( x,y )= x y sin( xy )= x2cos( xy ) 88.3.BasicRulesofDifferentiation.Sinceapartialderivativeisjust anordinaryderivativewithoneadditionalagreementthatallother variablesareviewedasnumericalparameters,thebasicrulesofdifferentiationapplytopartialderivatives.Let f and g befunctionsof severalvariablesandlet c beanumber.Then xi( cf )= c f xi, xi( f + g )= f xi+ g xi, xi( fg )= f xig + f g xi, xi f g =f xig Š fg xi g2. Let h ( u )beadierentiablefunctionofonevariableandlet g ( r )bea functionofseveralvariableswhoserangeliesinthedomainof f .Then

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88.PARTIALDERIVATIVES201 onecande“nethecomposition f ( r )= h ( g ( r )).Assumingthatthe partialderivativesof g exist,thechainruleholds (13.2) f xi= h( g ) g xi. Example 13.18 Findthepartialderivativesofthefunction f ( r )= r Š 1,where r =( x1,x2,...,xn) Solution: Put h ( u )= uŠ 1 / 2and g ( r )= x2 1+ x2 2+ + x2 n= r 2. Then f ( r )= h ( g ( r )).Since h( u )=( Š 1 / 2) uŠ 3 / 2and g/xi=2 xi, thechainrulegives xi r Š 1= Š xi r 3. 88.4.Exercises.(1) Findthespeci“edpartialderivativesofeachofthefollowingfunctions: (i) f ( x,y )=( x Š y ) / ( x + y ), f x(1 2), f y(1 2) (ii) f ( x,y,z )=( xy + z ) / ( z + y ), f x(1 2 3), f y(1 2 3), f z(1 2 3) (iii) f ( r )=( x1+2 x2+ + nxn) / (1+ r 2), f xi( 0 ), i =1 2 ,...,n (iv) f ( x,y,z )= x sin( yz ), f x(1 2 ,/ 2), f y(1 2 ,/ 2), f z(1 2 ,/ 2) (v) f ( x,y )= x +( y Š 1)sinŠ 1( x/y ), f x(1 1), f y(1 1) (vi) f ( x,y )=( x3+ y3)1 / 3, f x(0 0), f y(0 0) (vii) f ( x,y )= | xy | f x(0 0), f y(0 0) (2) Findthepartialderivativesofeachofthefollowingfunctions: (i) f ( x,y )=( x + y2)n(ii) f ( x,y )= xy(iii) f ( x,y )= xe( x +2 y )2(iv) f ( x,y )=sin( xy )cos( x2+ y2) (v) f ( x,y,z )=ln( x + y2+ z3) (vi) f ( x,y,z )= xy2cos( z2x ) (vii) f ( r )=( a1x1+ a2x2+ + anxn)m=( a r )m(viii) f ( x,y )=tanŠ 1( x/y ) (ix) f ( x,y )=sinŠ 1( x/ x2+ y2) (x) f ( x,y,z )= xyz(xi) f ( x,y,z )= xy/z(xii) f ( x,y )=tan( x2/y ) (xiii) f ( x,y,z )=sin( x sin( y sin z )) (xiv) f ( x,y )=( x + y2) / ( x2+ y ) (xv) f ( x,y,z )= a ( b r ) where a and b areconstantvectors (xvi) f ( x,y,z )= a r where a isaconstantvector

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20213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (3) Determinewhetherthefunction f ( x,y )increasesordecreaseswhen x increases,while y is“xed,andwhen y increases,while x is“xed,at aspeci“edpoint P0: (i) f ( x,y )= xy/ ( x + y ), P0(1 2) (ii) f ( x,y )=( x2Š 2 y2)1 / 3, P0(1 1) (iii) f ( x,y )= x2sin( xy ), P0( Š 1 ) 89.Higher-OrderPartialDerivatives Sincepartialderivativesofafunctionarealsofunctionsofseveral variables,theycanbedierentiatedwithrespecttoanyvariable.For example,forafunctionoftwovariables,allpossiblesecondpartial derivativesare f x Š x f x = 2f x2, y f x = 2f yx f y Š x f y = 2f xy y f y = 2f y2. Throughoutthetext,briefnotationsforhigher-orderpartialderivatives willalsobeused.Forexample, 2f x2=( f x) x= f xx, 2f xy =( f y) x= f yxandsimilarlyfor f yyand f xy.Partialderivativesofthethirdorderare de“nedaspartialderivativesofsecond-orderpartialderivatives,and soon. Example 13.19 Forthefunction f ( x,y )= x4Š x2y + y2,“ndall second-andthird-orderpartialderivatives. Solution: The“rstpartialderivativesare f x=4 x3Š 2 xy and f y= Š x2+2 y .Thenthesecondpartialderivativesare f xx=(4 x3Š 2 xy ) x=12 x2Š 2 y,f yy=( Š x2+2 y ) y=2 f xy=(4 x3Š 2 xy ) y= Š 2 x,f yx=( Š x2+2 y ) x= Š 2 x. Thethirdpartialderivativesarefoundsimilarly: fxxx=(12 x2Š 2 y ) x=24 x,fyyy=(2) y=0 fxxy=(12 x2Š 2 y ) y= Š 2 ,fxyx= fyxx=( Š 2 x ) x= Š 2 fyyx=(2) x=0 ,fyxy= fxyy=( Š 2 x ) y=0 Incontrasttotheone-variablecase,therearehigher-orderpartial derivativesofanewtypethatareobtainedbydierentiatingwith

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89.HIGHER-ORDERPARTIALDERIVATIVES203 respecttodierentvariablesindierentorders,like f xyand f yx.Inthe aboveexample,ithasbeenfoundthat f xy= f yx, fxxy= fxyx= fyxx, fxyy= fyyx= fyxy; thatis,theresultis independentoftheorderinwhichthepartialderivativeshavebeentaken .Isthisapeculiarityofthefunctionconsidered orageneralfeatureofhigher-orderpartialderivatives?Thefollowing theoremanswersthisquestion. Theorem 13.9 (ClairautsTheorem) Let f beafunctionofseveralvariables ( x1,x2,...,xn) thatisde“ned onanopenball D inaEuclideanspace.Ifthesecondpartialderivatives f xixjand f xjxi,where j = i ,arecontinuousfunctionson D ,then f xixj= f xjxiatanypointof D AconsequenceofClairautstheoremcanbeproved.Itassertsthat theresultofpartialdierentiationdoesnotdependontheorderin whichthederivativeshavebeentakenifallhigher-orderpartialderivativesinquestionarecontinuous .Itisnotalwaysnecessarytocalculatehigher-orderpartialderivativesinallpossibleorderstoverify thehypothesisofClairautstheorem(i.e.,thecontinuityofthepartial derivatives).Partialderivativesofpolynomialsarepolynomialsand hencecontinuous.Bythequotientruleforpartialderivatives,rational functionshavecontinuouspartialderivatives(wherethedenominator doesnotvanish).Derivativesofbasicelementaryfunctionslikethesine andcosineandexponentialfunctionsarecontinuous.Socompositions ofthesefunctionswithmultivariablepolynomialsorrationalfunctions havecontinuouspartialderivativesofanyorder.Inotherwords,the continuityofhigher-orderpartialderivativescanoftenbeestablished bydierent,simplermeans. Example 13.20 Findthethirdderivatives fxyz, fyzx, fzxy,andso on,forallpermutationsof x y ,and z ,if f ( x,y,z )=sin( x2+ yz ) Solution: Thesineandcosinefunctionsarecontinuouslydierentiableasmanytimesasdesired.Theargumentofthesinefunctionisa multivariablepolynomial.Bythecompositionrule,(sin g ) x= g xcos g andsimilarlyfortheotherpartialderivatives.Therefore,partialderivativesofanyordermustbeproductsofpolynomialsandthesineand cosinefunctionswhoseargumentisapolynomial.Therefore,theyare

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20413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS continuousintheentirespace.ThehypothesisofClairautstheoremis satis“ed,andhenceallthepartialderivativesinquestioncoincideand areequalto fxyz=( f x) yz=(2 x cos( x2+ yz )) yz=( Š 2 xz sin( x2+ yz )) z= Š 2 x sin( x2+ yz ) Š 2 xyz cos( x2+ yz ) 89.1.ReconstructionofaFunctionfromItsPartialDerivatives.Oneof thestandardproblemsincalculusis“ndingafunction f ( x )onan interval I ifitsderivative f( x )= F ( x )isknown.Asucientcondition fortheexistenceofasolutionisthecontinuityof F ( x )on I .Inthis case, f( x )= F ( x )= f ( x )= F ( x ) dx. Theinde“niteintegralisgivenbythesumofaparticularantiderivative of F andanarbitraryconstant(see ConceptsinCalculusI ).Asimilar problemcanbeposedforafunctionofseveralvariables.Giventhe “rstpartialderivatives (13.3) f xi( r )= Fi( r ) ,i =1 2 ,...,n, “nd f ( r )ifitexists.Theexistenceofsuch f isamoresubtlequestion inthecaseofseveralvariables.Supposethatthepartialderivatives Fi/xjexistandarecontinuousfunctionsinanopenball.Then takingthepartialderivative /xjofbothsidesof(13.3)andapplying Clairautstheorem,oneinfersthat (13.4) f xixj= f xjxi= Fi xj= Fj xi. Thus,theconditions(13.4)onthefunctions Fimustbeful“lled;otherwise, f satisfying(13.3)doesnotexist.Theconditions(13.4)are called integrabilityconditions forthesystemofequations(13.3). Example 13.21 Supposethat f x( x,y )=2 x + y and f y( x,y )= 2 y Š x .Doessuchafunction f exist? Solution: The“rstpartialderivativesof f F1( x,y )=2 x + y and F2( x,y )=2 y Š x ,arepolynomials,andhencetheirderivativesare continuousintheentireplane.Inorderfor f toexist,theintegrability condition F1/y = F2/x mustholdintheentireplane.Thisis notsobecause F1/y =1,whereas F2/x = Š 1.Thus,nosuch f exists. Supposenowthattheintegrabilityconditions(13.4)aresatis“ed. Howisasolution f to(13.3)tobefound?Evidently,onehasto

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89.HIGHER-ORDERPARTIALDERIVATIVES205 calculateanantiderivativeofthepartialderivative.Intheone-variable case,anantiderivativeisde“neduptoanadditiveconstant.Thisis notsointhemultivariablecase.Forexample,let f x( x,y )=3 x2y Anantiderivativeof f xwithrespectto x isafunctionwhose partial derivativewithrespectto x is3 x2y .Itiseasytoverifythat x3y satis“es thisrequirement.Itisobtainedbytakinganantiderivativeof3 x2y with respectto x whileviewing y asanumericalparameterindependentof x .Justlikeintheone-variablecase,onecanalwaysaddaconstantto anantiderivative, x3y + c andobtainanothersolution.Thekeypointto observeisthattheintegrationconstantmaybeafunctionof y !Indeed, ( x3y + g ( y )) x=3 x2y .Thus,thegeneralsolutionof f x( x,y )=3 x2y is f ( x,y )= x3y + g ( y )forsome g ( y ). If,inaddition,theotherpartialderivative f yisgiven,thenan explicitformof g ( y )canbefound.Put,forexample, f y( x,y )= x3+ 2 y .Theintegrabilityconditionsareful“lled:( f x) y=(3 x2y ) y=3 x2and( f y) x=( x3+2 y ) x=3 x2.Soafunctionwiththesaidpartial derivativesdoesexist.Thesubstitutionof f ( x,y )= x3y + g ( y )into theequation f y= x3+2 y yields x3+ g( y )= x3+2 y or g( y )=2 y andhence g ( y )= y2+ c .Notethecancellationofthe x3term.This isadirectconsequenceoftheful“lledintegrabilitycondition.Had onetriedtoapplythisprocedurewithoutcheckingtheintegrability conditions,onecouldhavefoundthat,ingeneral,nosuch g ( y )exists. InExample13.21,theequation f x=2 x + y hasageneralsolution f ( x,y )= x2+ yx + g ( y ).Itssubstitutionintothesecondequation f y=2 y Š x yields x + g( y )=2 y Š x or g( y )=2 y Š 2 x .Thederivative of g ( y )cannotdependon x andhencenosuch g ( y )exists. Example 13.22 Find f ( x,y,z ) if f x= yz +2 x = F1, f y= xz + 3 y2= F2,and f z= xy +4 z3= F3orshowthatitdoesnotexist. Solution: Theintegrabilityconditions( F1) y=( F2) x,( F1) z=( F3) x, and( F2) z=( F3) zaresatis“ed(theirveri“cationislefttothereader). So f exists.Takingtheantiderivativewithrespectto x inthe“rst equation,one“nds f x= yz +2 x = f ( x,y,z )= xyz + x2+ g ( y,z ) forsome g ( y,z ).Thesubstitutionof f intothesecondequationsyields f y= xz +3 y2= xz + g y( y,z )= xz +3 y2= g y( y,z )=3 y2= g ( y,z )= y3+ h ( z ) = f ( x,y,z )= xyz + x2+ y3+ h ( z ) ,

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20613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS forsome h ( z ).Thesubstitutionof f intothethirdequationyields f z= xy +4 z3= xy + h( z )= xy +4 z3= h( z )=4 z3= h ( z )= z4+ c = f ( x,y,z )= xyz + x2+ y3+ z4+ c, where c isaconstant. Theprocedureofreconstructing f fromits“rstpartialderivatives aswellastheintegrabilityconditions(13.4)willbeimportantwhen discussing conservativevector“elds andthe potential ofaconservative vector“eld.89.2.PartialDifferentialEquations.Therelationbetweenafunctionof severalvariablesanditspartialderivatives(ofanyorder)iscalleda partialdierentialequation .Partialdierentialequationsareakey tooltostudyvariousphenomenainnature.Manyfundamentallawsof naturecanbestatedintheformofpartialdierentialequations.DiffusionEquation.Let n ( r ,t ),where r =( x,y,z )isthepositionvector inspaceand t istime,beaconcentrationofasubstance,say,inairorin wateroreveninasolid.Evenifthereisnomacroscopicmotioninthe medium,theconcentrationchangeswithtimeduetothermalmotion ofthemolecules.Thisprocessisknownas diusion .Insomesimple situations,therateatwhichtheconcentrationchangeswithtimeata pointis n t= k ( n xx+ n yy+ n zz) wheretheparameter k isadiusionconstant.Sotheconcentration asafunctionofthespatialpositionandtimemustsatisfytheabove partialdierentialequation.WaveEquation.Soundinairispropagatingdisturbancesoftheair density.If u ( r ,t )isthedeviationoftheairdensityfromitsconstant (nondisturbed)value u0atthespatialpoint r =( x,y,z )andattime t ,thenitcanbeshownthatsmalldisturbances u/u0 1satisfythe waveequation : u tt= c2( u xx+ u yy+ u zz) where c isthespeedofsoundintheair.Lightisanelectromagnetic wave.Itspropagationisalsodescribedbythewaveequation,where c isthespeedoflightinvacuum(orinamedium,iflightgoesthrough amedium)and u istheamplitudeofelectricormagnetic“elds.

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89.HIGHER-ORDERPARTIALDERIVATIVES207 LaplaceandPoissonEquations.Theequation u xx+ u yy+ u zz= f, where f isagivennonzerofunctionofposition r =( x,y,z )inspace, iscalledthe Poissonequation .Inthespecialcasewhen f =0,this equationisknownasthe Laplaceequation .ThePoissonandLaplace equationsareusedtodeterminestaticelectromagnetic“eldscreated bystaticelectricchargesandcurrents. Example 13.23 Let h ( q ) beatwice-dierentiablefunctionofa variable q .Showthat u ( r ,t )= h ( ct Š n r ) isasolutionofthewave equationforany“xedunitvector n Solution: Let n =( n1,n2,n3),where n2 1+ n2 2+ n2 3=1as n isthe unitvector.Put q = ct Š n r = ct Š n1x Š n2y Š n3z .Bythechain rule(13.2), u t= q th( q )andsimilarlyfortheotherpartialderivatives. Therefore, u t= ch( q ), u tt= c2h( q ), u x= Š n1h( q ), u xx= n2 1h( q ), and,inthesamefashion, u yy= n2 2h( q ),and u zz= n2 3h( q ).Then u xx+ u yy+ u zz=( n2 1+ n2 2+ n2 3) h( q )= h( q ),whichcoincideswith u tt/c2,meaningthatthewaveequationissatis“edforany h Considerthelevelsurfacesofthesolutionofthewaveequation discussedinthisexample.Theycorrespondtoa“xedvalueof q = q0. So,foreachmomentoftime t ,thedisturbanceoftheairdensity u ( r ,t ) hasaconstantvalue h ( q0)intheplane n r = ct Š q0= d ( t ).All planeswithdierentvaluesoftheparameter d areparallelasthey havethesamenormalvector n .Sincehere d ( t )isafunctionoftime, theplaneonwhichtheairdensityhasa“xedvaluemovesalongthe vector n attherate d( t )= c .Thus,adisturbanceoftheairdensity propagateswithspeed c .Thisisthereasonthattheconstant c in thewaveequationiscalledthe speedofsound .Evidently,thesame lineofreasoningappliestoelectromagneticwaves;thatis,theymove throughspaceatthespeedoflight.Thespeedofsoundintheairis about342meterspersecond,orabout768mph.Thespeedoflight is3 108meterspersecond,or186milespersecond.Ifalightning strikeoccursamileawayduringath understorm,itcanbeseenalmost instantaneously,whilethethunderwillbeheardabout5secondslater. Conversely,ifoneseesalightningstrikeandstartscountingseconds untilthethunderisheard,thenonecouldestimatethedistancetothe lightning.Thesoundtravels1mileinabout4.7seconds.

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20813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 89.3.StudyProblems.Problem13.5. Findthevalueofaconstant a forwhichthefunction u ( r ,t )= tŠ 3 / 2eŠ ar2/t,r = r satis“esthediusionequationforall t> 0 Solution: Notethat u dependsonthecombination r2= x2+ y2+ z2. To“ndthepartialderivativesof u ,itisconvenienttousethechain rule: u x = u r2r2 x =2 x u r2= Š 2 ax t u, u xx= x u x = Š 2 a t u Š 2 ax t u x = Š 2 a t + 4 a2x2 t2 u. Toobtain u yyand u zz,notethat r2issymmetricwithrespecttopermutationsof x y ,and z .Therefore, u yyand u zzareobtainedfrom u xxbyreplacing,inthelatter, x by y and x by z ,respectively.Hence,the rightsideofthediusionequationreads k ( u xx+ u yy+ u zz)= Š 6 ka t + 4 ka2r2 t2 u. Usingtheproductruletocalculatethepartialderivativewithrespect totime,one“ndsfortheleftside u t= Š 3 2 tŠ 5 / 2eŠ ar2/t+ tŠ 3 / 2eŠ ar2/tar2 t2= Š 3 2 t + ar2 t2 u. Sincebothsidesmustbeequalfor all valuesof t> 0and r2,the comparisonofthelasttwoexpressionsyields two conditions:6 ka = 3 / 2(astheequalityofthecoecientsat1 /t )and a =4 ka2(asthe equalityofthecoecientsat r2/t2).Theonlycommonsolutionof theseconditionsis a =1 / (4 k ). Problem13.6. Considerthefunction f ( x,y )= x3y Š xy3 x2+ y2if( x,y ) =(0 0)and f (0 0)=0 Find f x( x,y ) and f y( x,y ) for ( x,y ) =(0 0) .Usetherule(13.1)to “nd f x(0 0) and f y(0 0) and,thereby,toestablishthat f xand f yexist everywhere.Usetherule(13.1)againtoshowthat f xy(0 0)= Š 1 and f yx(0 0)=1 ,thatis, f xy(0 0) = f yx(0 0) .Doesthisresultcontradict Clairautstheorem?

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89.HIGHER-ORDERPARTIALDERIVATIVES209 Solution: Usingthequotientrulefordierentiation,one“nds f x( x,y )= x4y +4 x2y3Š y5 ( x2+ y2)2,f y( x,y )= x5Š 4 x3y2Š xy4 ( x2+ y2)2if( x,y ) =(0 0).Notethat,owingtothesymmetry f ( x,y )= Š f ( y,x ), thepartialderivative f yisobtainedfrom f xbychangingthesignof thelatterandswapping x and y .Thepartialderivativesat(0 0)are foundbytherule(13.1): f x(0 0)= d dx f ( x, 0) x =0=0 ,f y(0 0)= d dy f (0 ,y ) y =0=0 The“rst-orderpartialderivativesarecontinuousfunctions(theproof islefttothereaderasanexercise).Next,onehas f xy(0 0)= d dy f x(0 ,y ) y =0=limh 0f x(0 ,h ) Š f x(0 0) h =limh 0Š h Š 0 h = Š 1 f yx(0 0)= d dx f y( x, 0) x =0=limh 0f y( h, 0) Š f y(0 0) h =limh 0h Š 0 h =1 TheresultdoesnotcontradictClairautstheorembecause f xy( x,y ) and f yx( x,y )arenotcontinuousat(0 0).Byusingthequotientrule todierentiate f x( x,y )withrespectto y ,anexplicitformof f xy( x,y ) for( x,y ) =(0 0)canbeobtained.Bytakingthelimitof f xy( x,y )as ( x,y ) (0 0)alongthestraightline( x,y )=( t,at ), t 0,oneinfers thatthelimitdependsontheslope a ,andhencethetwo-dimensional limitdoesnotexist,thatis,lim( x,y ) (0 0)f xy( x,y ) = f xy(0 0)= Š 1, and f xyisnotcontinuousat(0 0).Thetechnicaldetailsarelefttothe reader. 89.4.Exercises.(1) Findallsecondpartialderivativesofeachofthefollowingfunctions andverifyClairautstheorem: (i) f ( x,y )=tanŠ 1xy (ii) f ( x,y,z )= x sin( zy2) (iii) f ( x,y,z )= x3+ zy + z2(iv) f ( x,y,z )=( x + y ) / ( x +2 z ) (v) f ( x,y )=cosŠ 1( x/y ) (vi) f ( x,y )= xy

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21013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (2) Explainwithoutexplicitcalculationofhigher-orderpartialderivativesthatthehypothesisofClairautstheoremissatis“edforthefollowingfunctions (i) f ( x,y,z )=sin( x2+ y Š z )cos( xy ) (ii) f ( x,y )=sin( x + y2) / ( x2+ y2), x2+ y2 =0 (iii) f ( x,y,z )= ex2yz( y2+ zx4) (iv) f ( x,y )=ln(1+ x2+ y4) / ( x2Š y2), x2 = y2(v) f ( x,y,z )=( x + yz2Š xz5) / (1+ x2y2z4) (3) Findtheindicatedpartialderivativesofeachofthefollowingfunctions: (i) f ( x,y )= xn+ xy + ym, f xxy, f xyx, f yyx, f xyy(ii) f ( x,y,z )= x cos( yx )+ z3, f xyz, f xxz, f yyz(iii) f ( x,y,z )=sin( xy ) ez, f5/z5, f(4) xyzz, f(4) zyxz, f(4) zxzy(iv) f ( x,y,z,t )=sin( x +2 y +3 z Š 4 t ), f(4) abcd,where abcd denotes allpermutationsof xyzt (v) f ( x,y )= exy( y2+ x ), f(4) abcd,where abcd denotesallpermutationsof xxxy (vi) f ( x,y,z )=tanŠ 1x + y + z Š xyz 1 Š xy Š xz Š yz f(3) abc,where abc denotesall permutationsof xyz (vii) f ( x,y,z,t )=ln(( x Š y )2+( z Š t )2)Š 1 / 2, f(4) abcd,where abcd areallpermutationsof xyzt (viii) f ( x,y )= exsin( y ),n + mf nxmy(0 0) (4) Givenpartialderivatives,“ndthefunctionorshowthatitdoes notexist: (i) f x=3 x2y f y= x3+3 y2(ii) f x= yz +3 x2, f y= xz +4 y f z= xy +1 (iii) f xk= kxk, k =1 2 ,...,n (iv) f x= xy + z f y= x2/ 2, f z= x + y (v) f x=sin( xy )+ xy cos( xy ), f y= x2cos( xy )+1 (5) Verifythatagivenfunctionisasolutionoftheindicateddierential equation: (i) f ( t,x )= A sin( ct Š x )+ B cos( ct + x ), cŠ 2f ttŠ f xx=0 (ii) f ( x,y,t )= g ( ct Š ax Š by )+ h ( ct + ax + by ), f tt= c2( f xx+ f yy) if a2+ b2=1and g and h aretwicedierentiablefunctions (iii) f ( x,y )=ln( x2+ y2), f xx+ f yy=0 (iv) f ( x,y )=ln( ex+ ey), f x+ f y=1and f xxf yyŠ ( f xy)2=0 (v) f ( r )=exp( a r ),where a a =1, f x1x1+ f x2x2+ + f xnxn= f (vi) f ( r )= r 2 Š n, f x1x1+ f x2x2+ + f xnxn=0for r =0 (vii) f ( x,y,z )=sin( k r ) / r f xx+ f yy+ f zz+ k2f =0(Helmholtz equation)

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90.LINEARIZATIONOFMULTIVARIABLEFUNCTIONS211 (6) Findarelationbetweentheconstants a b ,and c suchthatthe function u ( x,y,t )=sin( ax + by + ct )satis“esthewaveequation u ttŠ u xxŠ u yy= 0.Giveageometricaldescriptionofsucharelation,forexample,by settingvaluesof c onaverticalaxisandthevaluesof a and b ontwo horizontalaxes. (7) Let f ( x,y,z )= u ( t ),where t = xyz .Showthat f(3) xyz= F ( t )and “nd F ( t ). (8) Find( f x)2+( f y)2+( f z)2and f xx+ f yy+ f zzif (i) f = x3+ y3+ z3Š 3 xyz (ii) f =( x2+ y2+ z2)Š 1 / 2(9) Lettheactionof K onafunction f bede“nedby Kf = xf x+ yf y. Find Kf K2f = K ( Kf ),and K3f = K ( K2f )if (i) f = x/ ( x2+ y2) (ii) f =ln x2+ y2(10) Let f ( x,y )= xy/ ( x2+ y2)if( x,y ) =(0 0)and f (0 0).Does f xy(0 0)exist? (11) If f = f ( x,y )and g = g ( x,y,z ),solvethefollowingequations: (i) f xx=0 (ii) f xy=0 (iii) nf/yn=0 (iv) g xyz=0 (12) Find f ( x,y )thatsatis“es: (i) f y= x2+2 y f ( x,x2)=1 (ii) f yy=4, f ( x, 0)=2, f y( x, 0)= x (iii) f xy= x + y f ( x, 0)= x f (0 ,y )= y290.LinearizationofMultivariableFunctions Adierentiableone-variablefunction f ( x )canbeapproximated near x = x0byitslinearization L ( x )= f ( x0)+ f( x0)( x Š x0)orthe tangentline.Put x = x0+ x .Then,bythede“nitionofthederivative f( x0), lim x 0f ( x ) Š L ( x ) x =lim x 0f ( x0+ x ) Š f ( x0) x Š f( x0) = f( x0) Š f( x0)=0 Thisrelationimpliesthattheerrorofthelinearapproximationgoesto 0fasterthanthedeviation x = x Š x0of x from x0,thatis, (13.5) f ( x )= L ( x )+ ( x ) x, where ( x ) 0as x 0 .

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21213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Forexample,if f ( x )= x2,thenitslinearizationat x =1is L ( x )= 1+2( x Š 1).Itfollowsthat f (1+ x ) Š L (1+ x )=( x )2or ( x )= x Conversely,consideralinethroughthepoint( x0,f ( x0))andassume thatthecondition(13.5)holds.If n istheslopeoftheline,then L ( x )= f ( x0)+ n ( x Š x0)= f ( x0)+ n x and lim x 0f ( x ) Š L ( x ) x =lim x 0f ( x0+ x ) Š f ( x0) x Š n =0 Bythede“nitionofthederivative f( x0),theexistenceofthislimit impliestheexistenceof f( x0)andtheequality n = f( x0).Thus, amongalllinearapproximationsof f near x0, only thelinewiththe slope n = f( x0)isagoodapproximationinthesensethattheerror oftheapproximationdecreasesfasterthan x withdecreasing x and theveryexistenceofagoodlinearapproximationat x = x0is equivalenttodierentiabilityof f at x0.Forexample,thefunction f ( x )= | x | isnotdierentiableat x =0.Agoodlinearapproximation doesnotexistat x0=0.Indeed,here x = x and L ( x )= nx .Hence, ( f ( x ) Š L ( x )) / x =( | x |Š nx ) /x = | x | /x Š n ,andnonumber n exists atwhichthisdierencevanishesinthelimit x 0.90.1.DifferentiabilityofMultivariableFunctions.Considerafunctionof twovariables f ( x,y )andapoint( x0,y0)initsdomain.Themost generallinearfunction L ( x,y )withtheproperty L ( x0,y0)= f ( x0,y0) reads L ( x,y )= f ( x0,y0)+ n1( x Š x0)+ n2( y Š y0),where n1and n2are arbitrarynumbers.Itde“nesalinearapproximationto f ( x,y )near ( x0,y0)inthesensethat L ( x0,y0)= f ( x0,y0).Moregenerally,givena multivariablefunction f ( r ),alinearfunction L ( r )= f ( r0)+ n ( r Š r0) issaidtobea linearapproximation to f near r0inthesensethat L ( r0)= f ( r0).Thedotproductisde“nedinan m -dimensionalEuclideanspaceif f isafunctionof m variables.Thevector n isan arbitraryvectorsothat L ( r )isthemostgenerallinearfunctionsatisfyingthecondition L ( r0)= f ( r0).Notethatinthecaseoftwovariables x1= x and x2= y n =( n1,n2)and r Š r0=( x Š x0,y Š y0)sothat n ( r Š r0)= n1( x Š x0)+ n2( y Š y0). Definition 13.17 (DierentiableFunctions) Thefunction f ofseveralvariables r =( x1,x2,...,xm) onanopenset D issaidtobe dierentiable atapoint r0 D ifthereexistsagood

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90.LINEARIZATIONOFMULTIVARIABLEFUNCTIONS213 linearapproximation L ( r ) ,i.e.alinearapproximation L ( r ) forwhich (13.6)limr r0f ( r ) Š L ( r ) r Š r0 =0 If f isdierentiableatallpointsof D ,then f issaidtobedierentiable on D Bythisde“nition, thedierentiabilityofafunctionisindependent ofthecoordinatesystem chosentolabelpointsof D (alinearfunction remainslinearundergeneralrotationsandtranslationsofthecoordinatesystemandthedistance r Š r0 isalsoinvariantunderthese transformations).Forfunctionsofasinglevariable f ( x ),theexistence ofalinearapproximationat x0withtheproperty(13.6)isequivalentto theexistenceofthederivative f( x0).Indeed,put x Š x0= x .Then [ f ( x ) Š L ( x )] / | x | = [ f ( x ) Š L ( x )] / x forall x =0.Therefore,the condition(13.6)isequivalentto(13.5),which,inturn,isequivalentto theexistenceof f( x0)asarguedabove. Theorem 13.10 Alinearapproximation L toamultivariablefunction f nearapoint r0thatsatis“estheproperty(13.6)isuniqueifit exists. Proof. Let L1( r )= f ( r0)+ n1 ( r Š r0)and L2( r )= f ( r0)+ n2 ( r Š r0) betwolinearapproximationsthatsatisfythecondition(13.6)forwhich n2 = n1.Makinguseoftheidentity L2( r ) Š L1( r )=[ f ( r ) Š L1( r )] Š [ f ( r ) Š L2( r )] itisconcludedthat limr r0L2( r ) Š L1( r ) r Š r0 =limr r0f ( r ) Š L1( r ) r Š r0 Š limr r0f ( r ) Š L2( r ) r Š r0 =0 Notethatowingtotheexistenceofthelimit(13.6)forbothlinear functions L1and L2,thelimitofthedierenceequalsthedierenceof thelimits(thebasiclawoflimits).Ontheotherhand, L2( r ) Š L1( r )= ( n2Š n1) ( r Š r0).Put n = n2Š n1.Byassumption, n =0.Then 0=limr r0L2( r ) Š L1( r ) r Š r0 =limr r0n ( r Š r0) r Š r0 =limr 0n r r Ifamultivariablelimitexists,thenitsvaluedoesnotdependona pathalongwhichthelimitpointisapproached.Inparticular,takethe straightlineparallelto n r = n t t 0+,intheaboverelation.Then alongthisline, r / r = n / n andhence 0=limt 0+n n n = n n n = n n = 0 n1= n2,

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21413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS whichisacontradiction.Thus,agoodlinearapproximationisunique ifitexists, L1( r )= L2( r ). 90.2.DifferentiabilityandPartialDerivatives.Intheone-variablecase, afunction f ( x )isdierentiableat x0ifandonlyifithasthederivative f( x0).Also,theexistenceofthederivativeat x0impliescontinuityat x0(recallCalculusI).Inthemultivariablecase,therelationsbetween dierentiability,continuity,andtheexistenceofpartialderivativesare moresubtle. Theorem 13.11 (PropertiesofDierentiableFunctions) If f isdierentiableatapoint r0,thenitiscontinuousat r0andits partialderivativesexistat r0. Proof. Theproperty(13.6)requiresthattheerrorofthelinearapproximationdecreasefasterthanthedistance r Š r0 : (13.7) f ( r )= L ( r )+ ( r ) r Š r0 where ( r ) 0as r r0. Alinearfunctioniscontinuous(itisapolynomialofdegree1).Therefore, L ( r ) L ( r0)= f ( r0)as r r0.Bytakingthelimit r r0in (13.7),itisconcludedthat f ( r ) f ( r0).Hence, f iscontinuousat r0.Ifthemultivariablelimit(13.6)exists,thenitdoesnotdependon apathapproachingthelimitpoint.Inparticular,takeastraightline paralleltothe j thcoordinateaxis.If ejistheunitvectorparallelto thisaxis,thenvectorequationofthelineis r = r ( t )= r0+ t ej.Then r Š r0 = | t | 0as t 0alongtheline,and f ( r ( t )) Š L ( r ( t ))= f ( r0+ t ej) Š f ( r0) Š n ejt = f ( r0+ t ej) Š f ( r0) Š njt, where njisthe j thcomponentofthevector n .Bythesamereasoning asintheone-variablecasewith x = t (givenafterDe“nition13.17), thecondition(13.6)implies limt 0f ( r0+ t ej) Š f ( r0) t Š nj=0 nj= f xj( r0) accordingtoDe“nition13.16ofpartialderivativesatapoint.The existenceofpartialderivativesisguaranteedbytheexistenceofthe limit(13.6). Thefollowingimportantremarksareinorder.Incontrasttothe one-variablecase, theexistenceofpartialderivativesatapointdoesnot generallyimplycontinuityatthatpoint Example 13.24 Considerthefunction f ( x,y )= xy x2+ y2if( x,y ) =(0 0) 0if( x,y )=(0 0) .

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90.LINEARIZATIONOFMULTIVARIABLEFUNCTIONS215 Showthatthisfunctionisnotcontinuousat (0 0) ,butthatthepartial derivatives f x(0 0) and f y(0 0) exist. Solution: Inordertocheckthecontinuity,onehastocalculatethe limitlim( x,y ) (0 0)f ( x,y ).Ifitexistsandequals f (0 0)=0,thenthe functioniscontinuousat(0 0).Thislimitdoesnotexist.Alonglines ( x,y )=( t,at ),thefunctionhasconstantvalue f ( t,at )= at2/ ( t2+ a2t2)= a/ (1+ a2)andhencedoesnotapproach f (0 0)=0as t 0+. To“ndthepartialderivativesinquestion,notethat f ( x, 0)=0forall x ,whichimpliesthatitsratealongthe x axisvanishes, f x( x, 0)=0. Similarly,thefunctionvanishesonthe y axis, f (0 ,y )=0andhence f y(0 ,y )=0.Inparticular,thepartialderivativesexistattheorigin, f x(0 0)= fy(0 0)=0. Thisexampleshowsthat boththecontinuityofafunctionandthe existenceofitspartialderivativesatapointarenecessaryconditions fordierentiabilityofthefunctionatthatpoint .Incontrasttotheonevariablecase,theyarenotsucient;thatis,theconverseofTheorem 13.11isfalse.Agoodlinearapproximationinthesenseof(13.6)(or (13.7))maynotexistevenifafunctioniscontinuousandhaspartial derivativesatapoint. Example 13.25 Let f ( x,y )= !xy x2+ y2if( x,y ) =(0 0) 0if( x,y )=(0 0) Showthat f iscontinuousat (0 0) andhasthepartialderivatives f x(0 0) and f y(0 0) ,butitisnotdierentiableat (0 0) Solution: Thecontinuityisveri“edbythesqueezeprinciple.Put r = x2+ y2.Then | xy | = | x || y | r2.Therefore, | f ( x,y ) | r2/r = r 0as r 0,whichmeansthatlim( x,y ) (0 0)f ( x,y )=0= f (0 0), andhence f iscontinuousat(0 0).Thepartialderivativesarefound inthesamefashionasinExample13.24.Since f ( x, 0)=0forall x f x( x, 0)=0.Similarly, f y(0 ,y )=0followsfrom f (0 ,y )=0for all y .Inparticular, f x(0 0)= f y(0 0)=0.Thecontinuityof f andtheexistenceofitspartialderivativesat(0 0)suggestthatifa linearapproximationwiththeproperty(13.6)exists,then L ( x,y )=0 (Theorem13.11).However, L ( x,y )=0doesnotsatisfy(13.6).Indeed, inthiscase,[ f ( x,y ) Š L ( x,y )] / r = f ( x,y ) / r isthefunctionfrom Example13.24,whichisnotcontinuousat(0 0),andthelimit(13.6) doesnotevenexist.Sothefunctionisnotdierentiableat(0 0). Thefollowingtheoremestablishesa sucient conditionfordierentiability(itsproofisomitted).

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21613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Theorem 13.12 (DierentiabilityandPartialDerivatives) Let f beafunctiononanopenset D ofaEuclideanspace.Then f isdierentiableon D ifitspartialderivativesexistandarecontinuous functionson D Thus,if,inadditiontotheirexistence,thepartialderivativeshappentobecontinuousfunctionsina neighborhood ofapoint,thenthe functionisdierentiableatthatpointandthereexistsagoodlinear approximationinthesenseof(13.6). Example 13.26 Findtheregioninwhichthefunction exzcos( yz ) isdierentiable. Solution: Thefunction exzisthecompositionoftheexponential euandthepolynomial u = xz .Soitspartialderivativesarecontinuous everywhere.Similarly,thepartialderivativesofcos( yz )arealsocontinuouseverywhere.Bytheproductruleforpartialderivatives,the partialderivativesof exzcos( yz )arecontinuouseverywhere.ByTheorem13.12,thefunctionisdierentiableeverywhereandagoodlinear approximationexistseverywhere. Remark .Theorem13.12providesonlyasucientconditionfor dierentiability.Therearedierentiablefunctionsatapointwhose partialderivativesexistbutarenotcontinuousatthatpoint.AnexampleisdiscussedinStudyProblem13.7.90.3.TangentPlaneApproximation.Theconceptofdierentiabilityis importantforapproximations.Onlydierentiablefunctionshavea goodlinearapproximation.Owingtotheuniquenessofagoodlinear approximation,itisconvenienttogiveitaname. Definition 13.18 (LinearizationofaMultivariableFunction) Let f beafunctionof m variables r =( x1,x2,...,xm) on D thatis dierentiableataninteriorpoint r0=( a1,a2,...,am) of D .Put ni= f xi( r0) i =1 2 ,...,m .Thelinearfunction L ( r )= f ( r0)+ n1( x1Š a1)+ n2( x2Š a2)+ + nm( xmŠ am) iscalledthe linearization of f at r0. If xidenotesthedeviationof xifrom ai,then (13.8) L ( r )= f ( r0)+ n1 x1+ n2 x2+ + nm xm,ni= f xi( r0) .GeometricalSignicanceofLinearization.Considerthegraph z = f ( x,y ) ofacontinuoustwo-variablefunction.Supposethat f hascontinuous partialderivativesatapoint( x0,y0).Considerthecurveofintersection

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90.LINEARIZATIONOFMULTIVARIABLEFUNCTIONS217 z = L ( x,y ) z = f ( x,y ) z = f ( x0,y ) P0v1v2z = f ( x,y0)Figure13.7.Thetangentplanetothegraph z = f ( x,y ) atthepoint P0=( x0,y0,f ( x0,y0)).Thecurves z = f ( x0,y ) and z = f ( x,y0)arethecrosssectionsofthegraphbythe coordinateplanes x = x0and y = y0,respectively.The vectors v1and v2aretangenttothecurvesofthecrosssectionsatthepoint P0.Theplanethrough P0andparallelto thesevectorsisthetangentplanetothegraph.Itsnormal is n = v1 v2.ofthegraphwiththecoordinateplane x = x0.Itsequationis z = f ( x0,y ).Thenthevectorfunction r ( t )=( x0,t,f ( x0,t ))tracesoutthe curveofintersection.Thecurvegoesthroughthepoint r0=( x0,y0,z0), where z0= f ( x0,y0),because r ( y0)= r0.Itstangentvectoratthepoint r0is v1= r( y0)=(0 1 ,f y( x0,y0))(seeFigure13.7).Thelineparallel to v1throughthepoint r0liesintheplane x = x0andistangentto theintersectioncurve z = f ( x0,y ).Similarly,thegraph z = f ( x,y )intersectsthecoordinateplane y = y0alongthecurve z = f ( x,y0)whose parametricequationsare r ( t )=( t,y0,f ( t,y0)).Thetangentvectorto thiscurveatthepoint r0is v2= r( x0)=(1 0 ,f x( x0,y0)).Theline parallelto v2through r0liesintheplane y = y0andistangenttothe curve z = f ( x,y0). Nowonecande“neaplanethroughthepoint r0ofthegraphthat containsthetwotangentlines.Thisplaneiscalledthe tangentplane to

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21813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS thegraph.Itsnormalmustbeperpendiculartobothvectors v1and v2and,bythegeometricalpropertiesofthecrossproduct,maybetaken as n = v1 v2=( f x( x0,y0) ,f y( x0,y0) Š 1).Thestandardequationof theplane n ( r Š r0)=0canthenbewrittenintheform z = z0+ n1( x Š x0)+ n2( y Š y0) ,n1= f x( x0,y0) ,n2= f y( x0,y0) Thus,thegraph z = L ( x,y )ofthelinearization L ofadierentiable function f at( x0,y0)de“nesthetangentplanetothegraphof f at ( x0,y0).Forthisreason,thelinearizationof f atapointisalsocalled the tangentplaneapproximation ;itisagoodlinearapproximationin thesenseof(13.6).Thetangentplaneapproximation(orthelinearizationingeneral)isamultivariableanalogofthetangentlineapproximationforsingle-variablefunctions. Example 13.27 Findthetangentplanetotheparaboloid z = x2+ 3 y2atthepoint (2 1 7) Solution: Theparaboloidisthegraphofthepolynomialfunction f ( x,y )= x2+3 y2thatiscontinuousandhascontinuouspartialderivativesatanypoint,andhence f isdierentiable.Thecomponents ofanormalofthetangentplaneare n1= f x(2 1)=2 x |(2 1)=4, n2= f y(2 1)=6 y |(2 1)=6,and n3= Š 1.Anequationofthetangent planeis4( x Š 2)+6( y Š 1) Š ( z Š 7)=0or4 x +6 y Š z =7. Example 13.28 Usethelinearizationtoestimatethe number [(2 03)2+(1 97)2+(0 94)2]1 / 2. Solution: Let f ( x,y,z )=[ x2+ y2+ z2]1 / 2.Itiscontinuousandhas continuouspartialderivativeseverywhereexcepttheoriginbecauseitis acompositionofthepolynomial g = x2+ y2+ z2andthepowerfunction: f =( g )1 / 2.Thenumberinquestionisthevalueofthisfunctionat ( x,y,z )=(2 03 1 97 0 94).Thispointiscloseto r0=(2 2 1)atwhich f ( r0)=3.Since f isdierentiableat r0,itslinearizationcanbeusedto approximatevaluesof f near r0.Thedeviationsare x = x Š 2=0 03, y = y Š 2= Š 0 03,and z =0 94 Š 1= Š 0 06.Thepartial derivativesare f x= x/ ( x2+ y2+ z2)1 / 2, f y= y/ ( x2+ y2+ z2)1 / 2,and f z= z/ ( x2+ y2+ z2)1 / 2.Therefore, n1=2 / 3, n2=2 / 3,and n3=1 / 3. Thelinearapproximation(see(13.8))gives f ( x,y,z ) L ( x,y,z )=3+(2 / 3) x +(2 / 3) y +(1 / 3) z =2 98

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90.LINEARIZATIONOFMULTIVARIABLEFUNCTIONS219 90.4.StudyProblems.Problem13.7. Let f ( x,y )= ( x2+ y2)sin 1 x2+ y2 if( x,y ) =(0 0) 0if( x,y )=(0 0) Showthat f isdierentiableat (0 0) (andhencethat f x(0 0) and f y(0 0) exist),butthat f xand f yarenotcontinuousat (0 0) .Ž Solution: Bythede“nitionofpartialderivatives, f x(0 0)=limh 0f ( h, 0) Š f (0 0) h =limh 0h sin(1 /h2)=0 whichfollowsfromthesqueezeprinciple:0 | h sin(1 /h2) || h | 0. Similarly, f y(0 0)=0.Since f (0 0)=0and f x(0 0)= f y(0 0)=0, onehastoverifywhetherthelinearfunction L ( x,y )=0satis“esthe condition(13.6): lim( x,y ) (0 0)f ( x,y ) Š L ( x,y ) x2+ y2=limr 0+r sin(1 /r2)=0 bythesqueezeprinciple,0 | r sin(1 /r2) | r 0as r 0+.Thus, L ( x,y )=0isagoodlinearapproximation,andthefunction f is dierentiableattheorigin. For( x,y ) =(0 0), f x( x,y )=2 x sin 1 x2+ y2 Š 2 x x2+ y2cos 1 x2+ y2 The“rstterminthisexpressionconvergesto0,0 | 2 x sin(1 /r2) | 2 | x | 0as( x,y ) (0 0),whereasthesecondtermcantakearbitrarilylargevaluesinanyneighborhoodoftheorigin.Toseethis,consider asequenceofpointsthatconvergestotheorigin:( x,y )=( tn, 0), n =0 1 ,... ,where1 /t2 n= / 2+ n or tn=1 / ( / 2+ n )1 / 2 0 as n .Thencos(1 /t2 n)=( Š 1)n,sin(1 /t2 n)=0,and f x( tn, 0)= 2 tnsin(1 /t2 n)+(2 /tn)cos(1 /t2 n)=2( Š 1)n( / 2+ n )1 / 2.So f x( tn, 0)can takearbitrarilylargepositiveandnegativevaluesas n ,andthe limitlim( x,y ) (0 0)f x( x,y )doesnotexist,whichmeansthatthepartialderivative f x( x,y )isnotcontinuousattheorigin.Owingtothe symmetry f ( x,y )= f ( y,x ),thesameconclusionholdsfor f y( x,y ). 90.5.Exercises.(1) Let f ( x,y )= xy2if( x,y ) =(0 0)and f (0 0)=1andlet g ( x,y )= | x | + | y | .Arethefunctions f and g dierentiableat(0 0)?

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22013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (2) Let f ( x,y )= xy2/ ( x2+ y2)if x2+ y2 =0and f (0 0)=0.Show that f iscontinuousandhasboundedpartialderivatives f xand f y, butitisnotdierentiableat(0 0).Investigatethecontinuityofthe partialderivativesnear(0 0). (3) Showthatthefunction f ( x,y )= | xy | iscontinuousat(0 0)and hasthepartialderivatives f x(0 0)and f y(0 0),butitisnotdierentiableat(0 0).Investigatethecontinuityofthepartialderivatives f xand f yneartheorigin. (4) Let f ( x,y )= x3/ ( x2+ y2)if( x,y ) =(0 0)and f (0 0)=0.Show that f iscontinuous,haspartialderivativesat(0 0),butisnotdierentiableat(0 0). (5) Findthedomaininwhichthefollowingfunctionsaredierentiable: (i) f ( x,y )= y x (ii) f ( x,y )=xy 2 x + y(iii) f ( x,y,z )=sin( xy + z ) ezy(iv) f ( x,y,z )= x2+ y2Š z2(v) f ( r )=ln(1 Š r ),where r =( x1,x2,...,xm) (vi) f ( x,y )=3 x3+ y3(vii) f ( r )= eŠ 1 / r 2if r = 0 and f ( 0 )=0,where r =( x1,x2,...,xm) Hint: Show f xj( 0 )=0.Investigatethecontinuityofthepartialderivatives. (6) Thelinethroughapoint P0ofasurfaceperpendiculartothetangentplaneat P0iscalledthe normalline .Findanequationofthe tangentplaneandsymmetricequationsofthenormallinetoeachof thefollowingsurfacesatthespeci“edpoint: (i) z = x2+3 y Š y4x ,(1 2 Š 1) (ii) z = x3y ,(1 4 2) (iii) z = y ln( x2Š 3 y ),(2 1 0) (iv) y =tanŠ 1( xz2),(1 4, Š 1) (v) x = z cos( y Š z ),(1 1 1) (vi) z = y +ln( x/z ),(1 1 1) (7) Findthelinearizationofeachofthefollowingfunctionsatthe speci“edpoint: (i) f ( x,y )=2 y +3 4 x +1,(0 0) (ii) f ( x,y,z )= z1 / 3 x +cos2( y ),(0 0 1) (iii) f ( r )=sin( n r ), r = r0,where n isa“xedvectororthogonal to r0and r =( x1,x2,...,xm) (8) Usethelinearizationtoapproximatethefollowingnumbers.Then useacalculatorto“ndthenumbers.Comparetheresults.

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91.CHAINRULESANDIMPLICITDIFFERENTIATION221 (i) 20 Š 7 x2Š x2,where( x,y )=(1 08 1 95) (ii) xy2z3,where( x,y,z )=(1 002 2 003 3 004) (iii)(1 03)2 3 0 984 (1 05)2(iv)(0 97)1 05(9) Considertheequation f ( x,y,z )=0thathasaroot z = z ( x,y ) forevery“xedpair( x,y ).Supposethat f ( x0,y0,z0)=0and f is dierentiableat( x0,y0,z0)sothat f z( x0,y0,z0) =0.If L ( x,y,z )is thelinearizationof f at( x0,y0,z0),theequation L ( x,y,z )=0is acalledalinearizationoftheequation f ( x,y,z )=0.Itssolution determinesanapproximationtotheroot z = z ( x,y )near( x0,y0). Findthisapproximation,andusetheresulttosolvetheequation yz ln(1+ xz ) Š x ln(1+ zy )=0for z = z ( x,y )nearthepoint(1 1 1). Inparticular,estimatetheroot z at( x,y )=(0 8 1 1). (10) Supposethatafunction f ( x,y )iscontinuouswithrespectto x ateach“xed y andhasaboundedpartialderivative f y( x,y ),thatis, | f y( x,y ) | M forsome M> 0andall( x,y ).Provethat f iscontinuous. 91.ChainRulesandImplicitDifferentiation91.1.ChainRules.Considerthefunction f ( x,y )= x3+ xy2whose domainistheentireplane.Pointsoftheplanecanbelabeledina dierentway.Forexample,thepolarcoordinates x = r cos y = r sin maybeviewedasarulethatassignsanorderedpair( x,y )to anorderedpair( r, ).Usingthisrule,thefunctioncanbeexpressed inthenewvariablesas f ( r cos ,r sin )= r3sin = F ( r, ).Onecan computetheratesofchangeof f withrespecttothenewvariables: f r = F r =3 r2sin f = F = r3cos Alternatively,theseratescanbecomputedas f r = f x x r + f y y r =(3 x2+ y2)cos +2 xy sin =3 r2sin f = f x x + f y y = Š (3 x2+ y2) r sin +2 xyr cos = r3cos where x and y havebeenexpressedinthepolarcoordinatestoobtain the“nalexpressions.Thelatterrelationsareanexampleofa chain rule forfunctionsoftwovariables.Supposethattherates f x( x0,y0) and f y( x0,y0)areknownataparticularpoint( x0,y0).Then,byusing

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22213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS thechainrule, anexplicitformofthefunction f inthenewvariablesis notrequiredto“nditsrateswithrespecttothenewvariables because therates x r, x y r,and y at( r0,0)correspondingto( x0,y0)can easilybecomputed. Ontheotherhand,considerthefunction f ( x,y )= y3/ ( x2+ y2)if ( x,y ) =(0 0)and f (0 0)=0.Thisfunctioniscontinuousattheorigin (if R2= x2+ y2,then | f ( x,y ) Š f (0 0) || y3| /R2 R3/R2= R 0 as R 0).Ithaspartialderivativesattheorigin.Indeed, f ( x, 0)=0, andhence f x( x, 0)=0forany x and,inparticular, f x(0 0)=0. Similarly, f (0 ,y )= y ,andhence f y(0 ,y )=1sothat f y(0 0)=1.Let x = t cos and y = t sin ,where isanumericalparameter.Then F ( t )= f ( t cos ,t sin )= t3sin3/t2= t sin3 .Therefore, F( t )= sin3 .Thisimpliesthat F(0)=sin3 .However,thechainrulefails: F(0)= df/dt |t =0= f x(0 0) x(0)+ f y(0 0) y(0)=sin .Itisnot diculttoverifythatthechainrule df/dt = f xx( t )+ f yy( t )istrue forall t =0.Thereaderisadvisedtoverifythatthefunctionisnot dierentiableat(0 0)(seeStudyProblem13.12),whichisthereason forthechainruleisnotvalidatthatpoint.Itappearsthat,incontrast totheone-variablecase,themereexistenceofpartialderivativesisnot sucienttovalidatethechainruleinthemulti-variablecase,anda strongerconditionof f isrequired. Theorem 13.13 (ChainRule) Let f beafunctionof n variables r =( x1,x2,...,xn) .Supposethateach variable xiis,inturn,afunctionof m variables u =( u1,u2,...,um) Thecompositionof xi= xi( u ) with f ( r ) de“nes f asafunctionof u Ifthefunctions xiaredierentiableatapoint u and f isdierentiable atthepoint r =( x1( u ) ,x2( u ) ,...,xn( u )) ,thentherateofchangeof f withrespectto uj, j =1 2 ,...,m ,reads f uj= f x1x1 uj+ f x2x2 uj+ + f xnxn uj=ni =1f xixi uj. Proof. Sincethefunctions xi( u )aredierentiable,thepartialderivatives xi/ujexistandde“neagoodlinearapproximationinthesense (13.7).Inparticular,fora“xedvalueof u andforevery i xi= xi( u + ejh ) Š xi( u )= xi ujh + i( h ) | h | ,i( h ) 0as h 0 De“nethevector rh=( x1, x2,..., xn).Ithasthepropertythat rh 0 as h 0.If F ( u )= f ( x1( u ) ,x2( u ) ,...,xn( u )),then,bythe

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91.CHAINRULESANDIMPLICITDIFFERENTIATION223 de“nitionofthepartialderivativesatapoint u f uj=limh 0F ( u + ejh ) Š F ( u ) h =limh 0f ( r + rh) Š f ( r ) h ifthelimitexists.Bythehypothesis,thefunction f isdierentiableand hencehaspartialderivatives f/xiatthepoint r =( x1( u ) ,x2( u ) ,..., xn( u ))thatdetermineagoodlinearapproximation(13.8)inthesense of(13.7): f ( r + rh) Š f ( r )= f x1 x1+ f x2 x2+ + f xn xn+ ( rh) rh where ( rh) 0as rh 0 oras h 0.Thesubstitutionofthis relationintothelimitshowsthatthelimitexists,andtheconclusion ofthetheoremfollows.Indeed,the“rst n termscontainthelimits limh 0 xi h = xi uj+limh 0i( h ) | h | h = xi ujbecause | h | /h = 1forall h =0and i( h ) 0as h 0.Theratio rh / | h | =[( x1/h )2+( x2/h )2+ +( xn/h )2]1 / 2 M< as h 0,where M isdeterminedbythepartialderivatives xi/uj. Therefore,thelimitofthelasttermvanishes: limh 0 ( rh) rh h =limh 0 ( rh) | h | h rh | h | =limh 0 ( rh) | h | h limh 0 rh | h | =0 M =0 because ( rh) | h | /h = ( rh)if h =0and ( rh) 0as h 0. Itisclearfromtheproofthatthepartialderivatives f/xiinthe chainrulearetakenatthepoint( x1( u ) ,x2( u ) ,...,xn( u )).For n = m =1,thisisthefamiliarchainruleforfunctionsofonevariable df/du = f( x ) x( u ).If n =1and m> 1,itisthechainrule(13.2) establishedearlier.Theexampleofpolarcoordinatescorrespondsto thecase n = m =2,where r =( x,y )and u =( r, ). Example 13.29 Letafunction f ( x,y,z ) bedierentiableat r0= (1 2 3) andhavethefollowingratesofchange: f x( r0)=1 f y( r0)=2 and f z( r0)= Š 2 .Supposethat x = x ( t,s )= t2s y = y ( t,s )= s + t and z = z ( t,s )=3 s .Findtheratesofchange f withrespectto t and s atthepoint r0. Solution: Inthechainrule,put r =( x,y,z )and u =( t,s ).The point r0=(1 2 3)correspondstothepoint u0=(1 1)inthenew variables.Notethat z =3gives3 s =3andhence s =1.Then,from

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22413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS y =2,itfollowsthat s + t =2or1+ t =2or t =1.Also, x (1 1)=1 asrequired.Thepartialderivativesoftheoldvariableswithrespectto thenewonesare x t=2 ts y t=1, z t=0, x s= t2, y s=1,and z s=3. Theyarecontinuousfunctionsandhence x ( t,s ), y ( t,s ),and z ( t,s )are dierentiablebyTheorem13.12.Bythechainrule, f t( r0)= f x( r0) x t( u0)+ f y( r0) y t( u0)+ f z( r0) z t( u0) =1 2+2 1+( Š 2) 0=4 f s( r0)= f x( r0) x s( u0)+ f y( r0) y s( u0)+ f z( r0) z s( u0) =1 1+2 1+( Š 2) 3= Š 3 Example 13.30 Let f ( x,y,z )= z2(1+ x2+2 y2)Š 1.Findtherate ofchangeof f alongthecurve r ( t )=(sin t, cos t,et) inthedirectionof increasing t Solution: Thefunction f isdierentiableastheratiooftwopolynomials(itspartialderivativesarecontinuous): f x= Š 2 xz2 (1+ x2+2 y2)2,f y= Š 4 yz2 (1+ x2+2 y2)2,f z= 2 z 1+ x2+4 y2. Thecomponentsof r ( t )arealsodierentiable: x( t )=cos t y( t )= Š sin t z( t )= et.Bythechainrulefor n =3and m =1, df dt = f x( r ( t )) x( t )+ f y( r ( t )) y( t )+ f z( r ( t )) z( t ) = Š 2 e2 tsin t (2+cos2t )2 (cos t ) Š 4 e2 tcos t (2+cos2t )2 ( Š sin t )+ 2 et 2+cos2t et= e2 t(5+sin(2 t )+cos(2 t )) (2+cos2t )2, where2sin t cos t =sin(2 t )and2cos2t =1+cos(2 t )havebeenused. Thechainrulecanbeusedtocalculatehigher-orderpartialderivatives. Example 13.31 If g ( u,v )= f ( x,y ) ,where x =( u2Š v2) / 2 and y = uv ,“nd g uv.Assumethat f hascontinuoussecondpartialderivatives. If f y(1 2)=1 f xx(1 2)= f yy(1 2)=2 ,and f xy(1 2)=3 ,“ndthe valueof g uvat ( x,y )=(1 2) Solution: Onehas x u= u x v= Š v y u= v ,and y v= u .Then g u= f xx u+ f yy u= f xu + f yv.

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91.CHAINRULESANDIMPLICITDIFFERENTIATION225 Thederivative g uv=( g u) viscalculatedbyapplyingthechainruleto thefunction g u: g uv= u ( f x) v+ v ( f y) v+ f y= u ( f xxx v+ f xyy v)+ v ( f yxx v+ f yyy v)+ f y= u ( Š vf xx+ uf xy)+ v ( Š vf yx+ uf yy)+ f y= uv ( f yyŠ f xx)+( u2Š v2) f xy+ f y= y ( f yyŠ f xx)+2 xf xy+ f y, where f xy= f yxhasbeenused.Thevalueof g uvatthepointinquestion is2 (2 Š 2)+2 3+1=7. 91.2.ImplicitDifferentiation.Considerthefunctionofthreevariables, F ( x,y,z )= x2+ y4Š z .Theequation F ( x,y,z )=0canbesolvedfor oneofthevariables,say, z ,toobtain z asafunctionoftwovariables: F ( x,y,z )=0= z = z ( x,y )= x2+ y4; thatis,thefunction z ( x,y )isde“nedasarootof F ( x,y,z )andhas thecharacteristicpropertythat (13.9) F ( x,y,z ( x,y ))=0forall( x,y ) Intheexampleconsidered,theequation F ( x,y,z )=0canbesolved analytically,andan explicit formofitsrootasafunctionof( x,y )can befound. Ingeneral,givenafunction F ( x,y,z ),anexplicitformofasolutiontotheequation F ( x,y,z )=0isnotalwayspossibleto“nd. Puttingasidethequestionabouttheveryexistenceofasolutionand itsuniqueness,supposethatthisequationisprovedtohaveaunique solutionwhen( x,y ) D .Inthiscase,thefunction z ( x,y )withthe property(13.9)forall( x,y ) D issaidtobede“ned implicitly on D Althoughananalyticformofanimplicitlyde“nedfunctionisunknown,itsratesofchangecanbefoundandprovideimportantinformationaboutitslocalbehavior.Supposethat F isdierentiable. Furthermore,theroot z ( x,y )isalsoassumedtobedierentiableonan opendisk D intheplane.Sincerelation(13.9)holdsforall( x,y ) D thepartialderivativesofitsleftsidemustalsovanishin D .They canbecomputedbythechainrule, n =3, m =2, r =( x,y,z ), and u =( u,v ),wheretherelationsbetweenoldandnewvariablesare x = u, y = v ,and z = z ( u,v ).Onehas x u=1, x v=0, y u=0, y v=1,and z u( u,v )= z x( x,y )and z v( u,v )= z y( x,y )because x = u

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22613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS and y = v .Therefore, u F ( x,y,z ( x,y ))= F x + F z z x =0= z x= Š F x F z, v F ( x,y,z ( x,y ))= F y + F z z y =0= z y= Š F y F z. Theseequationsdeterminetheratesofchangeofanimplicitlyde“ned functionoftwovariables.Notethatinorderfortheseequationsto makesense,thecondition F z =0mustbeimposed.Severalquestions abouttheveryexistenceanduniquenessof z ( x,y )foragiven F ( x,y,z ) andthedierentiabilityof z ( x,y )havebeenleftunansweredinthe aboveanalysis.Thefollowingtheoremaddressesthemall. Theorem 13.14 (ImplicitFunctionTheorem) Let F beafunctionof n +1 variables, F ( r ,z ) ,where r =( x1,x2,...,xn) and z isrealsuchthat F and F zarecontinuousinanopenball B Supposethatthereexistsapoint ( r0,z0) B suchthat F ( r0,z0)=0 and F z( r0,z0) =0 .Thereexistsanopenneighborhood D of r0,an openinterval I ,andauniquefunction z : D I suchthatfor ( r,y ) D I F ( r ,y )=0 ifandonlyif y = z ( r ) .Moreover,thefunction z is continuous.If,inaddition, F isdierentiablein B ,thenthefunction z = z ( r ) isdierentiablein D and z xi( r )= Š F xi( r ,z ( r )) F z( r ,z ( r )) forall r in D Theproofofthistheoremgoesbeyondthescopeofthiscourse.It includesproofsoftheexistenceanduniquenessof z ( r )anditsdierentiability.Oncethesefactsareestablished,aderivationoftheimplicit dierentiationformulafollowsthesamewayasinthe n =2case: F xi+ F z z xi=0= z xi( r )= Š F xi( r ,z ( r )) F z( r ,z ( r )) Remark .Ifthefunction F hassucientlymanycontinuoushigherorderpartialderivatives,thenhigherorderpartialderivativesof z ( r ) canbeobtainedbydierentiationoftheserelations.Anexampleis giveninStudyProblem13.9. Example 13.32 Showthattheequation z (3 x Š y )= sin( xyz ) hasauniquesolution z = z ( x,y ) inaneighborhoodof (1 1) suchthat z (1 1)= / 2 and“ndtheratesofchange z x(1 1) and z y(1 1) .

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91.CHAINRULESANDIMPLICITDIFFERENTIATION227 Solution: Put F ( x,y,z )= sin( xyz ) Š z (3 x Š y ).Thentheexistenceanduniquenessofthesolutioncanbeestablishedbyverifying thehypothesesoftheimplicitfunctiontheoreminwhich r =( x,y ), r0=(1 1),and z0= / 2.First,notethatthefunction F isthesum ofapolynomialandthesinefunctionofapolynomial.Soitspartial derivatives F x= yz cos( xyz ) Š 3 z,F y= xz cos( xyz )+ z, F z= xy cos( xyz ) Š 3 x + y arecontinuousforall( x,y,z );hence, F isdierentiableeverywhere. Next, F (1 1 ,/ 2)=0asrequired.Finally, F z(1 1 ,/ 2)= Š 2 =0. Therefore,bytheimplicitfunctiontheorem,thereisanopendiskin the xy planecontainingthepoint(1 1)inwhichtheequationhasa uniquesolution z = z ( x,y ).Bytheimplicitdierentiationformulas, z x(1 1)= Š F x(1 1 ,/ 2) F z(1 1 ,/ 2) = Š 3 4 ,z y(1 1)= Š F y(1 1 ,/ 2) F z(1 1 ,/ 2) = 4 Inparticular,thisresultimpliesthat,nearthepoint(1 1),theroot z ( x,y )decreasesinthedirectionofthe x axisandincreasesinthe directionofthe y axis.Itshouldbenotedthatthenumericalvaluesofthederivativescanbeusedtoaccuratelyapproximatetheroot z ( x,y )ofanonlinearequationinaneighborhoodof(1 1)bylinearizing thefunction z ( x,y )near(1 1).Thecontinuityofpartialderivatives ensuresthat z ( x,y )isdierentiableat(1 1)andhasagoodlinear approximationinthesenseof(13.6)(seeStudyProblem13.8). 91.3.StudyProblems.Problem13.8. Showthattheequation z (3 x Š y )= sin( xyz ) has auniquesolution z = z ( x,y ) inaneighborhoodof (1 1) suchthat z (1 1)= / 2 .Estimate z (1 04 0 96) Solution: InExample13.21,theexistenceanduniquenessof z ( x,y ) hasbeenestablishedbytheimplicitfunctiontheorem.Thepartial derivativeshavealsobeenevaluated, z x(1 1)= Š 3 / 4and z y(1 1)= / 4.Thelinearizationof z ( x,y )near(1 1)is z (1+ x, 1+ y ) z (1 1)+ z x(1 1) x + z y(1 1) y = 2 1 Š 3 x 2 + y 2 Putting x =0 04and y = Š 0 04,thisequationyieldstheestimate z (1 04 0 96) 0 45

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22813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Problem13.9. Letthefunction z ( x,y ) bede“nedimplicitlyby z5+ zx Š y =0 inaneighborhoodof (1 2 1) .Findallits“rstandsecondpartialderivatives.Inparticular,givethevaluesofthesepartial derivativesat ( x,y )=(1 2) Solution: Let F ( x,y,z )= z5+ zx Š y .Then F z=5 z4+ x .The function z ( x,y )existsinaneighborhoodof(1 2)bytheimplicitfunctiontheorembecause F (1 2 1)=0and F z(1 2 1)=6 =0.The“rst andsecondpartialderivativesof F arecontinuouseverywhere: F x= z,F y= Š 1 ,F z=5 z4+ x, F xx=0 ,F xy=0 ,F xz=1 F yy=0 ,F yz=0 ,F zz=20 z3. Byimplicitdierentiation, z x= Š F x F z= Š z 5 z4+ x ,z y= Š F y F z= 1 5 z4+ x Takingthepartialderivativesoftheserelationswithrespectto x and y andusingthequotientrulefordierentiation,thesecondpartial derivativesareobtained: z xx= Š z x(5 z4+ x ) Š z (20 z3z x+1) (5 z4+ x )2= (15 z4Š x ) z x+ z (5 z4+ x )2, z xy= z yx=( z y) x= Š 20 z3z x+1 (5 z4+ x )2,z yy= Š 20 z3z y (5 z4+ x )2. Theexplicitformof z xand z ymaybesubstitutedintotheserelations toexpressthesecondpartialderivativesvia x y ,and z .Atthepoint (1 2),thevaluesofthe“rstpartialderivativesare z x(1 2)= Š 1 / 6and z y(1 2)=1 / 6.Usingthesevalues,thevaluesofthesecondpartial derivativesareevaluated: z xx(1 2)= Š 1 / 27, z xy(1 2)=7 / 108,and z yy(1 2)= Š 5 / 54. 91.4.Exercises.(1) Usethechainruleto“nd dz/dt if z = 1+ x2+2 y2and x =2 t3, y =ln t (2) Usethechainruleto“nd z/s and z/t if z = eŠ xsin( xy )and x = ts y = s2+ t2. (3) Usethechainruletowritethepartialderivativesof F withrespect tothenewvariables: (i) F = f ( x,y ), x = x ( u,v,w ), y = y ( u,v,w ) (ii) F = f ( x,y,z,t ), x = x ( u,v ), y = y ( u,v ), z = z ( w,s ), t = t ( w,s )

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91.CHAINRULESANDIMPLICITDIFFERENTIATION229 (4) Findtheratesofchange z/u z/v z/w when( u,v,w )= (2 1 1)if z = x2+ yx + y3and x = uv2+ w3, y = u + v ln w (5) Findtheratesofchange f/u f/v f/w when( x,y,z )= (1 / 3 2 0)if x =2 /u Š v + w y = vuw z = ew. (6) If z ( u,v )= f ( x,y ),where x = eucos v and y = eusin v ,showthat z xx+ z yy= eŠ 2 s( z uu+ z vv). (7) If z ( u,v )= f ( x,y ),where x = u2+ v2and y =2 uv ,“ndallthe second-orderpartialderivativesof z ( u,v ). (8) If z ( u,v )= f ( x,y ),where x = u + v and y = u Š v ,showthat ( z x)2+( z y)2= z uz v. (9) Findallthe“rstandsecondpartialderivativesofthefollowing functions: (i) g ( x,y,z )= f ( x2+ y2+ z2) (ii) g ( x,y )= f ( x,x/y ) (iii) g ( x,y,z )= f ( x,xy,xyz ) (iv) g ( x,y )= f ( x/y,y/x ) (v) g ( x,y,z )= f ( x + y + z,x2+ y2+ z2) (vi) g ( x,y )= f ( x + y,xy ) (10) Find g xx+ g yy+ g zzif g ( x,y,z )= f ( x + y + z,x2+ y2+ z2). (11) Let x = r cos and y = r sin .Showthat f xx+ f yy= 1 r r r f r + 1 r22f 2 (12) Let x = sin cos y = sin sin z = cos .Thevariables ( ,, )arecalled sphericalcoordinates anddiscussedinSection104.3. Showthat f xx+ f yy+ f zz= 1 2 2f + 1 2sin sin f + 1 2sin2 2f 2. (13) Provethatifafunction f ( x,y )satis“estheLaplaceequation f xx+ f yy=0,thenthefunction g ( x,y )= f ( x/ ( x2+ y2) ,y/ ( x2+ y2)), x2+ y2> 0,alsosatis“estheLaplaceequation. (14) Provethatifafunction f ( x,t )satis“esthediusionequation f t= a2f xx,thenthefunction g ( x,t )= 1 a t eŠ x2/ (4 a2t )f x a2t Š x a4t ,t> 0 alsosatis“esthediusionequation.

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23013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (15) Provethatif f ( x,y,z )satis“estheLaplaceequation f xx+ f yy+ f zz=0,thenthefunction g ( x,y,z )= 1 r f a2x r2, a2y r2, a2z r2 ,r = x2+ y2+ z2 =0 alsosatis“estheLaplaceequation. (16) Showthatthefunction g ( x,y )= xnf ( y/x2),where f isadierentiablefunction,satis“estheequation xg x+2 yg y= ng (17) Showthatthefunction g ( x,y,z )= xnf ( y/xa,z/xb),where f isa dierentiablefunction,satis“estheequation xg x+ ayg y+ bzg z= ng (18) Letthefunction z = f ( x,y )bede“nedimplicitly.Findits“rst andsecondpartialderivativesif (i) x +2 y +3 z = ez(ii) x Š z =tanŠ 1( yz ) (iii) x/z =ln( z/y )+1 (19) Let z ( x,y )beasolutionoftheequation z3Š xz + y =0suchthat z (3 Š 2)=2.Findthelinearizationof z ( x,y )near(3 Š 2)anduseit toestimate z (2 8 Š 2 3). (20) Find f xand f y,where f =( x + z ) / ( y + z )and z isde“nedbythe equation zez= xex+ yey. (21) Showthatthefunction z ( x,y )de“nedbytheequation F ( x Š az,y Š bz )=0,where F isadierentiablefunctionoftwovariables and a and b areconstants,satis“estheequation az x+ bz y=1. (22) Letthetemperatureoftheairatapoint( x,y,z )be T ( x,y,z ) degreesCelsius.Supposethat T isadierentiablefunction.Aninsect ”iesthroughtheairsothatitspositionasafunctionoftime t ,in seconds,isgivenby x = 1+ t y =2 t z = t2Š 1.If T x(2 6 8)=2, T y(2 6 8)= Š 1,and T z(2 6 8)=1,howfastisthetemperaturerising (ordecreasing)ontheinsectspathasit”iesthroughthepoint(2 6 8)? (23) Considerafunction f = f ( x,y,z )andthechangeofvariables: x =2 uv,y = u2Š v2+ w,z = u3vw .Findthepartialderivatives f u, f v,and f watthepoint u = v = w =1,if f x= a f y= b ,and f z= c at ( x,y,z )=(2 1 1). (24) Letarectangularboxhavethedimensions x y ,and z thatchange withtime.Supposethatatacertaininstantthedimensionsare x =1 m, y = z =2m,and x and y areincreasingattherate2m/sand z is decreasingattherate3m/s.Atthatinstant,“ndtheratesatwhich thevolume,thesurfacearea,andthelargestdiagonalarechanging. (25) Afunctionissaidtobehomogeneousofdegree n if,forany number t ,ithastheproperty f ( tx,ty )= tnf ( x,y ).Giveanexample ofapolynomialfunctionthatishomogeneousofdegree n .Showthata

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92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS231 homogeneousdierentiablefunctionsatis“estheequation xf x+ yf y= nf .Showalsothat f x( tx,ty )= tn Š 1f ( x,y ). (26) Supposethattheequation F ( x,y,z )=0de“nesimplicitly z = f ( x,y )or y = g ( x,z )or x = h ( y,z ).Assumingthatthederivatives F x, F y,and F zdonotvanish,provethat( z/x )( x/y )( y/z )= Š 1. (27) Let x2= vw y2= uw z2= uv ,and f ( x,y,z )= F ( u,v,w ).Show that xf x+ yf y+ zf z= uF u+ vF v+ wF w. (28) Simplify z xsec x + z ysec y if z =sin y + f (sin x Š sin y ),where f isadierentiablefunction. 92.TheDifferentialandTaylorPolynomials Justlikeintheone-variablecase,givenvariables r =( x1,x2,...,xm), onecanintroduceindependentvariables d r =( dx1,dx2,...,dxm)that arein“nitesimalvariationsof r andalsocalled dierentials of r Definition 13.19 (Dierential) Let f ( r ) beadierentiablefunction.Thefunction df ( r )= f x1( r ) dx1+ f x2( r ) dx2+ + f xm( r ) dxmiscalledthe dierential of f Thedierentialisafunctionof2 m independent variables r and d r Considerthegraph y = f ( x )ofafunction f ofasinglevariable x (see Figure13.8,leftpanel).Thedierential df ( x0)= f( x0) dx atapoint x0determinestheincrementof y alongthetangentline y = L ( x )= f ( x0)+ f( x0)( x Š x0)as x changesfrom x0to x0+ x ,where x = dx Similarly,thedierential df ( x0,y0)ofafunctionoftwovariablesata point P0=( x0,y0)determinestheincrementof z = L ( x,y )alongthe tangentplanetothegraph z = f ( x,y )atthepoint( x0,y0,f ( x0,y0)) when( x,y )changesfrom( x0,y0)to( x0+ x,y0+ y ),where dx = x and dy = y ,asdepictedintherightpanelofFigure13.8.In general,thedierential df ( r0)andthelinearizationof f atapoint r0arerelatedas L ( r )= f ( r0)+ df ( r0) ,dxi= xi,i =1 2 ,...,m ; thatis,ifthein“nitesimalvariations(ordierentials) d r arereplaced bythedeviations r = r Š r0ofthevariables r from r0,then the dierential df atthepoint r0de“nesthelinearizationof f at r0.Accordingto(13.6),thedierence f ( r ) Š f ( r0) Š df ( r0)tendsto0faster than r as r 0 ,andhencethedierentialcanbeusedtostudy variationsofadierentiablefunction f undersmallvariationsofits arguments.

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23213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS y = f ( x ) y = L ( x ) f ( x0+ x ) y f ( x0) df dx x0x0+ x z = f ( x,y ) z df z = L ( x,y ) z = z0dx dyFigure13.8.Geometricalsigni“canceofthedierential. Left :Thedierentialofafunctionofonevariable.Itde“nes theincrementof y alongthetangentline y = L ( x )tothe graph y = f ( x )at( x0,y0), y0= f ( x0),when x changesfrom x0to x0+ x ,where dx = x .As x 0,( y Š df ) / x = y/ x Š f( x0) 0;thatis,thedierence y Š df tendsto0 fasterthan x Right :Thedierentialofafunctionoftwo variables.Itde“nestheincrementof z alongthetangent plane z = L ( x,y )tothegraph z = f ( x,y )at( x0,y0,z0), z0= f ( x0,y0),when( x,y )changesfrom( x0,y0)to( x0+ x,y0+ y ),where dx = x and dy = y .Thedierence z Š df tendsto0fasterthan r = ( x )2+( y )2as r 0.Example 13.33 Find df ( x,y ) if f ( x,y )= 1+ x2y .Inparticular,evaluate df (1 3) for ( dx,dy )=(0 1 Š 0 2) .Whatisthesigni“cance ofthisnumber? Solution: Thefunctionhascontinuouspartialderivativesinaneighborhoodof(1 3)andhenceisdierentiableat(1 3).Onehas df ( x,y )= f x( x,y ) dx + f y( x,y ) dy = xydx 1+ x2y + x2dy 2 1+ x2y Then df (1 3)= 3 2 dx + 1 4 dy =0 15 Š 0 05=0 1 Thenumber f (1 3)+ df (1 3)de“nesthevalueofthelinearization L ( x,y )of f at(1 3)for( x,y )=(1+ dx, 3+ dy ).Itcanbeusedto approximate f (1+ dx, 3+ dy ) Š f (1 3) df (1 3)when | dx | and | dy | aresmallenough.Inparticular, f (1+0 1 3 Š 0 2) Š f (1 3)=0 09476 (acalculatorvalue),whichistobecomparedwith df (1 3)=0 1. 92.1.ErrorAnalysis.Supposeaquantity f dependsonseveralother quantities,say, x y ,and z ,forde“niteness,thatis, f isafunction f ( x,y,z ).Supposemeasurementsshowthat x = x0, y = y0,and z = z0.Since,inpractice,allmeasurementscontainerrors,thevalue f ( x0,y0,z0)doesnothavemuchpracticalsigni“canceuntilitserroris determined.

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92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS233 Forexample,thevolumeofarectanglewithdimensions x y ,and z isthefunctionofthreevariables V ( x,y,z )= xyz .Inpractice, repetitivemeasurementsgivethevaluesof x y ,and z fromintervals x [ x0Š x,x0+ x ], y [ y0Š y,y0+ y ],and z [ z0Š z,z0+ z ], where r0=( x0,y0,z0)arethemeanvaluesofthedimensions,while r =( x,y,z )areupperboundsoftheabsoluteerrorsorthemaximaluncertaintiesofthemeasurements.Toindicatethemaximaluncertaintyinthemeasuredquantities,onewrites x = x0 x andsimilarly for y and z .Dierentmethodsofthelengthmeasurementwouldhave dierentabsoluteerrorbounds.Inotherwords,thedimensions x y ,and z andthebounds x y ,and z areallindependentvariables. Sincetheerrorboundsshouldbesmall(atleastonewishesso),thevaluesofthedimensionsobtainedineachmeasurementare x = x0+ dx y = y0+ dy ,and z = z0+ dz ,wherethedierentialscantaketheir valuesintheintervals dx [ Š x,x ]= Ixandsimilarlyfor dy and dz .Thequestionarises:Giventhemeanvalues r0=( x0,y0,z0)and theabsoluteerrorbounds r ,whatistheabsoluteerrorboundofthe volumevaluecalculatedat r0?Foreachparticularmeasurement,the erroris V ( r0+ d r ) Š V ( r0)= dV ( r0)iftermstendingto0fasterthan d r canbeneglected.Thecomponentsof d r areindependentvariablestakingtheirvaluesinthespeci“edintervals.Allsuchtriples d r correspondtopointsoftheerrorrectangle R= Ix Iy Iz.Then theabsoluteerrorboundis V = | max dV ( r0) | ,wherethemaximum istakenoverall d r R.Forexample,if r0=(1 2 3)isincentimeters and r =(1 1 1)isinmillimeters,thentheabsoluteerrorboundof thevolumeis V = | max dV ( r0) | =max( y0z0dx + x0z0dy + x0y0dz )= 0 6+0 3+0 2=1 1cm3,and V =6 1 1cm3.Herethemaximumis reachedat dx = dy = dz =0 1cm.Thisconceptcanbegeneralized. Definition 13.20 (AbsoluteandRelativeErrorBounds) Let f beaquantitythatdependsonotherquantities r =( x1,x2,...,xm) sothat f = f ( r ) isadierentiablefunction.Supposethatthevalues xi= aiareknownwiththeabsoluteerrorbounds xi.Put r0= ( a1,a2,...,am) and f = | max df ( r0) | ,wherethemaximumistaken overall dxi [ Š xi,xi] .Thenumbers f and,if f ( r0) =0 f/ | f ( r0) | arecalled,respectively,the absoluteandrelativeerrorbounds ofthe valueof f at r = r0. Intheaboveexample,therelativeerrorboundofthevolumemeasurementsis1 1 / 6 0 18;thatis,theaccuracyofthemeasurements isabout18%.Ingeneral,since df ( r0)=mi =1f xi( r0) dxi

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23413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS islinearin dxi,and r0is“xed,themaximumisattainedbysetting dxiequalto xiforall i forwhichthecoecient f xi( r0)ispositive,and equalto Š xiforall i forwhichthecoecient f xi( r0)isnegative.So theabsoluteerrorboundcanbewrittenintheform f =mi =1| f xi( r0) | xi.92.2.AccuracyofaLinearApproximation.Ifafunction f ( x )isdierentiablesucientlymanytimes,thenitslinearapproximationcanbe systematicallyimprovedbyusingTaylorpolynomials(seeCalculusI andCalculusII).TheTaylortheoremassertsthatif f ( x )hascontinuousderivativesuptoorder n onaninterval I containing x0and f( n +1)existsandisboundedon I | f( n +1)( x ) |
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92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS235 asaconstant).ThentheTaylorpolynomial Tn( x )about x0is Tn( x )= f ( x0)+ 1 1! df ( x0)+ 1 2! d2f ( x0)+ + 1 n dnf ( x0) where dx = x Š x0.Itrepresentsanexpansionof f ( x0+ dx )in powersofthedierential dx .TheTaylorpolynomialapproximation f ( x0+ dx ) Tn( x )isanapproximationinwhichthecontributionsof higherpowers( dx )k, k>n ,areneglected,provided f isdierentiable sucientlymanytimes.TheTaylortheoremensuresthatthisapproximationisbetterthanalinearapproximationinthesensethatthe approximationerrordecreasesfasterthan( dx )n=( x Š x0)nas x x0. Italsoprovidesmoreinformationaboutalocalbehaviorofthefunction nearaparticularpoint x0(e.g.,theconcavityof f near x0).Naturally, thisconceptshouldbequiteusefulinthemultivariablecase.92.3.TaylorPolynomialsofTwoVariables.Let f ( x,y )beafunctionof twovariables.Thedierentials dx and dy areanothertwoindependent variables.Byanalogywiththeone-variablecase,thedierential df is viewedastheresultoftheactionoftheoperator d on f : df ( x,y )= dx x + dy y f ( x,y )= dxf x( x,y )+ dyf y( x,y ) Definition 13.21 Supposethat f hascontinuouspartialderivativesuptoorder n .Thequantity dnf ( x,y )= dx x + dy y nf ( x,y ) iscalledthe n -thorderdierentialof f ,wheretheactionofpowers dnon f isde“nedsuccessively dnf = dn Š 1( df ) andthevariables dx dy x ,and y areviewedasindependentwhendierentiating. Thedierential dnf isafunctionoffourvariables dx dy x ,and y Forexample, d2f = dx x + dy y 2f = dx x + dy y ( dxf x+ dyf y) = ( dx )2 x + dxdy y f x+ dxdy x +( dy )2 y f y= f xx( dx )2+2 f xydxdy + f yy( dy )2. Bythecontinuityofpartialderivatives,theorderofdierentiationis irrelevant, f xy= f yx.Thenumericalcoecientsateachoftheterms arebinomialcoecients:( a + b )2= a2+2 ab + b2.Sincetheorderof

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23613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS dierentiationisirrelevant(Clairautstheorem),thisobservationholds ingeneral: dnf =nk =0Bn knf xn Š kyk( dx )n Š k( dy )k,Bn k= n k !( n Š k )! where Bn karethebinomialcoecients:( a + b )n= n k =0Bn kan Š kbk. Example 13.34 Find dnf if f ( x,y )= eax + by,where a and b are constants. Solution: Since f x= aeax + byand f y= beax + by, df = aeax + bydx + beax + bydy = eax + by( adx + bdy ) Since f xx= a2eax + by, f xy= abeax + bx,and f yy= b2eax + bx, d2f = a2eax + by( dx )2+2 abeax + bydxdy + b2eax + by( dy )2= eax + by( a2( dx )2+2 abdxdy + b2( dy )2)= eax + by( adx + bdy )2. Furthermore,bynotingthateachdierentiationwithrespectto x bringsdownafactor a ,whilethepartialderivativewithrespectto y bringsdownafactor b ,itisconcludedthat nf/xn Š kyk= an Š kbkeax + by. Usingthebinomialexpansion,oneinfers dnf = eax + by( adx + bdy )nforall n =1 2 ,... Definition 13.22 (TaylorPolynomialsofTwoVariables) Let f havecontinuouspartialderivativesuptoorder n .TheTaylor polynomialoforder n aboutapoint ( x0,y0) is Tn( x,y )= f ( x0,y0)+ 1 1! df ( x0,y0)+ 1 2! d2f ( x0,y0)+ + 1 n dnf ( x0,y0) where dx = x Š x0and dy = y Š y0. Forexample,put r =( x,y ), r0=( x0,y0), dx = x Š x0,and dy = y Š y0.The“rstfourTaylorpolynomialsare T0( r )= f ( r0) T1( r )= f ( r0)+ f x( r0) dx + f y( r0) dy = L ( r ) T2( r )= T1( r )+ f xx( r0) 2 ( dx )2+ f xy( r0) dxdy + f yy( r0) 2 ( dy )2, T3( r )= T2( r )+ f xxx( r0) 6 ( dx )3+ f xxy( r0) 2 ( dx )2dy + f xyy( r0) 2 dx ( dy )2+ f yyy( r0) 6 ( dy )3.

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92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS237 Thelinearortangentplaneapproximation f ( r ) L ( r )= T1( r )is aparticularcaseoftheTaylorpolynomialapproximationofthe“rst degree. Example 13.35 Let Pn( x,y ) beapolynomialofdegree n .Findits Taylorpolynomialsabout (0 0) .Inparticular,“ndTaylorpolynomials for P3( x,y )=1+2 x Š xy + y2+4 x3Š y2x Solution: Allpartialderivativesof Pnoforderhigherthan n vanish. Therefore, dkPn=0for k>n ,andhenceforanypolynomialofdegree n Tn= Pnandalso Tk= Pnif k>n .Anypolynomialcanbe uniquelydecomposedintothesum Pn= Q0+ Q1+ + Qn,where Qkisahomogeneouspolynomialofdegree k ;itcontainsonlymonomialsof degree k .Thedierential dkf isahomogeneouspolynomialofdegree k inthevariables dx and dy .Therefore,De“nition13.22de“nes Tkasthesumofhomogeneouspolynomialsin x and y if( x0,y0)=(0 0). Twopolynomialsareequalonlyifthecoecientsatthecorresponding monomialsmatch.Itfollowsfrom Tn= Pnthat Tn= Tn Š 1+( TnŠ Tn Š 1)= Q0+ Q1+ + Qn Š 1+ Qn.Since Qnand TnŠ Tn Š 1containsonly monomialsofdegree n ,theequalityispossibleonlyif Qn= TnŠ Tn Š 1, andhence Tn Š 1= Q0+ Q1+ + Qn Š 1.Continuingtheprocess recursivelybackward,itisconcludedthat Tk= Q0+ Q1+ + Qk,k =0 1 ,...,n. Inparticular,forthegivenpolynomial P3,onehas Q0=1, Q1=2 x Q2= Š xy + y2,and Q3=4 x3Š y2x .Therefore,itsTaylorpolynomials abouttheoriginare T0=1, T1= T0Š 2 x T2= T1Š xy + y2,and Tk= P3for k 3. Theorem 13.15 (TaylorTheorem). Let D beanopendiskcenteredat r0andletthepartialderivatives ofafunction f becontinuousuptoorder n Š 1 on D .Then f ( r )= Tn Š 1( r )+ n( r ) ,wherethereminder ( r ) satis“esthecondition | n( r ) | hn( r ) r Š r0n Š 1, where hn( r ) 0as r r0. InSection92.4,Taylorpolynomialsforfunctionsofanynumber ofvariableswillbede“ned.Theorem13.15istrue,justaswritten, nomatterhowmanyvariablesthereare;thatis r =( x1,x2,...,xm) foranynumberofvariables m .For n =2,thistheoremisnothing butTheorem13.12.Thecontinuityofpartialderivativesensuresthe existenceofagoodlinearapproximation L ( r )= T1( r )inthesense thatthedierence f ( r ) Š T1( r )decreasesto0fasterthan r Š r0 as r r0.For n> 2,itstatesthattheapproximationof f bytheTaylor

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23813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS polynomial Tn Š 1isagoodapproximationinthesensethattheerror decreasesfasterthan r Š r0n Š 1. Apracticalsigni“canceoftheTaylor theoremisthathigher-orderdierentialsofafunctioncanbeusedto obtainsuccessivelybetterapproximationsofvaluesofafunctionneara pointifthefunctionhascontinuouspartialderivativesofhigherorders inaneighborhoodofthatpoint. Example 13.36 Let f ( x,y )= 1+ x2y .Find df (1 3) and d2f (1 3) andusethemtoapproximate f (1+0 1 3 Š 0 2) Solution: Put( dx,dy )=(0 1 Š 0 2).ItwasfoundinExample13.33 that df (1 3)=0 1.Thesecondpartialderivativesareobtainedbythe quotientrule(see f xand f yinExample13.33): f xx(1 3)= y (1+ x2y )1 / 2Š x2y2(1+ x2y )Š 1 / 2 1+ x2y (1 3)= 3 8 f xy(1 3)= 2 x (1+ x2y )1 / 2Š x3y (1+ x2y )Š 1 / 2 2(1+ x2y ) (1 3)= 5 16 f yy(1 3)= Š x4 4 (1+ x2y )Š 3 / 2 (1 3)= Š 1 8 Therefore, d2f (1 3)= f xx(1 3)( dx )2+2 f xy(1 3) dxdy + f yy(1 3)( dy )2= 3 8 ( dx )2+ 5 8 dxdy Š 1 8 ( dy )2= Š 0 01375 Thelinearapproximationis f (1+ dx, 3+ dy ) f (1 3)+ df (1 3)= 2+0 1=2 1.Thequadraticapproximationis f (1+ dx, 3+ dy ) f (1 3)+ df (1 3)+1 2d2f (1 3)=2 1 Š 0 01375 / 2=2 093125,while acalculatorvalueof f (1+ dx, 3+ dy )is2 094755(roundedtothe samesigni“cantdigit).Evidently,thequadraticapproximation(the approximationbythesecond-degreeTaylorpolynomial)isbetterthan thelinearapproximation. Yet,theTaylortheoremdoesnotallowustoestimatetheaccuracy oftheapproximationbecausethefunction hnremainsunknown.What istheorderofapproximationneededtoobtainanerrorsmallerthan someprescribedvalue? Corollary 13.3 (AccuracyofTaylorPolynomialApproximations) If,inadditiontothehypothesesofTheorem13.15,thefunction f has partialderivativesoforder n thatareboundedon D ,thatis,thereexist numbers Mnk, k =1 2 ,...,n ,suchthat | nf ( r ) /n Š kxky | Mnkfor

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92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS239 all r D ,thentheremaindersatis“es | n( r ) |nk =0Bn kMnk n | x Š x0|n Š k| y Š y0|kforall ( x,y ) D ,where Bn k= n / ( k !( n Š k )!) arebinomialcoecients. Next,note | x Š x0| r Š r0 and | y Š y0| r Š r0 andhence | x Š x0|n Š k| y Š y0|k r Š r0n.Makinguseofthisinequality,one infersthat (13.12) | n( r ) | Mn n r Š r0n, wheretheconstant Mn= n k =0Bn kMnk.Inparticular,forthelinear approximation n =2, (13.13) | f ( r ) Š L ( r ) | M2 2 r Š r02, where M2= M20+2 M11+ M02.Theresults(13.12)and(13.13)are tobecomparedwiththesimilarresults(13.10)and(13.11)inthe one-variablecase.Ifthesecondpartialderivativesarecontinuousand boundednear r0,thenvariationsoftheirvaluesmaybeneglectedina sucientlysmallneighborhoodof r0,andthenumbers M20, M11,and M02maybeapproximatedbytheabsolutevaluesofthecorresponding partialderivativesat r0sothat | 2| 1 2 | f xx( r0) | ( dx )2+2 | f xy( r0) || dxdy | + | f xy( r0) | ( dy )2 forsucientlysmallvariations dx = x Š x0and dy = y Š y0.Suchan estimateisoftensucientforpracticalpurposestoassesstheaccuracy ofthelinearapproximation.Thisestimateworksevenbetterif f has continuouspartialderivativesofthethirdorderbecausethesecond partialderivativeswouldhaveagoodlinearapproximationandvariationsoftheirvaluesnear r0areoforder d r .Consequently,theycan onlyproducevariationsof 2oforder d r 3,whichcanbeneglectedas comparedto d r 2forsucientlysmall d r Example 13.37 Findthelinearapproximationnear (0 0) of f ( x,y )= 1+ x +2 y andassessitsaccuracyinthesquare | x | < 1 / 4 | y | < 1 / 4 Solution: Onehas f x=1 2(1+ x +2 y )Š 1 / 2and f y=(1+ x +2 y )Š 1 / 2sothat f x(0 0)=1 / 2and f y(0 0)=1.Thelinearapproximationis T1( x,y )= L ( x,y )=1+ x/ 2+ y .Thesecondpartialderivativesare f xx= Š1 4(1+ x +2 y )Š 3 / 2, f xy= Š1 2(1+ x +2 y )Š 3 / 2,and f yy= Š (1+ x +

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24013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 2 y )Š 3 / 2.Theirabsolutevaluesaremaximalifthecombination1+ x +2 y isminimalonthesquare.Setting x = y = Š 1 / 4,1 / 4 < 1+ x +2 y inthesquare.Therefore, | f xx|
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92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS241 So,forpracticalpurposes,theerror | n| oftheapproximation f Tn Š 1maybeestimatedby dnf ( r0) /n !,where dxjarereplacedbytheirabsolutevalues | dxj| andthevaluesofpartialderivativesarealsoreplaced bytheirabsolutevalues(justlikeithasbeendonewhenestimating | 2| inthecaseoftwovariables),providedthepartialderivativesof f oforder n orhigherarecontinuousinaneighborhoodof r0. Calculationofhigher-orderderivativesto“ndTaylorpolynomials mightbeatechnicallytediousproblem.Insomespecialcases,however,itcanbeavoided.Theconceptisillustratedbythefollowing example. Example 13.39 Find T3forthefunction f ( x,y,z )=sin( xy + z ) abouttheorigin. Solution: TheTaylorpolynomial T3inquestionisapolynomialof degree3in x y ,and z ,whichisuniquelydeterminedbythecoecients ofmonomialsofdegreelessthanorequalto3.Put u = xy + z .The variable u issmallneartheorigin.SotheTaylorpolynomialapproximationfor f neartheoriginisdeterminedbytheTaylorpolynomials forsin u about u =0.ThelatterisobtainedfromtheMaclaurinseries sin u = u Š1 6u3+ 5( u ),where 5containsonlymonomialsofdegree5 andhigher.Sincethepolynomial u vanishesattheorigin,itspowers unmaycontainonlymonomialsofdegree n andhigher.Therefore, T3isobtainedfrom u Š1 6u3=( xy + z ) Š1 6( xy + z )3= z + xy Š1 6 z3+3( xy ) z2+3( xy )2z +( xy )3 byretaininginthelatterallmonomialsuptodegree3,whichyields T3( r )= z + xy Š1 6z3.Evidently,theprocedureisfarsimplerthan calculating19partialderivatives(uptothethirdorder)! 92.5.StudyProblems.Problem13.10. Find T1, T2,and T3for f ( x,y,z )=(1+ xy ) / (1+ x + y2+ z3) abouttheorigin. Solution: Thefunction f isarationalfunction.Itisthereforesucientto“ndasuitableTaylorpolynomialforthefunction(1+ x + y2+ z3)Š 1andthenmultiplyitbythepolynomial1+ xy ,retainingonlymonomialsuptodegree3.Put u = x + y2+ z3.Then (1+ u )Š 1=1 Š u + u2Š u3+ (asageometricseries).Notethat,for n 4,theterms uncontainonlymonomialsofdegree4andhigherand

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24213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS hencecanbeomitted.Uptodegree3,onehas u2=