Concepts in Calculus, III : Multivariable Calculus

PAGE 1

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

PAGE 3

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

PAGE 4

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

PAGE 5

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 .

PAGE 6

viCONTENTS Chapter14.MultipleIntegrals293 97.DoubleIntegrals29398.PropertiesoftheDoubleIntegral30199.IteratedIntegrals310100.DoubleIntegralsOverGeneralRegions315101.DoubleIntegralsinPolarCoordinates330102.ChangeofVariablesinDoubleIntegrals341103.TripleIntegrals356104.TripleIntegralsinCylindricalandSphericalCoordinates369105.ChangeofVariablesinTripleIntegrals382106.ImproperMultipleIntegrals392107.LineIntegrals403108.SurfaceIntegrals408109.MomentsofInertiaandCenterofMass423Chapter15.VectorCalculus437 110.LineIntegralsofaVectorField437111.FundamentalTheoremforLineIntegrals446112.GreensTheorem458113.FluxofaVectorField470114.StokesTheorem481115.Gauss-Ostrogradsky(Divergence)Theorem490Acknowledgments501

PAGE 7

CHAPTER11 VectorsandtheSpaceGeometry Ourspacemaybeviewedasacollectionofpoints.Everygeometricalgure,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,butthishasyettobeveriedbyexperiments. 71.RectangularCoordinatesinSpace Theelementaryobjectinspaceisapoint.Sothediscussionshould beginwiththequestion:Howcanonedescribeapointinspacebyreal numbers?Thefollowingprocedurecanbeadopted.Selectaparticular pointinspacecalledthe origin andusuallydenoted O .Setupthree mutuallyperpendicularlinesthroughtheorigin.Arealnumberis associatedwitheverypointoneachlineinthefollowingway.The origincorrespondsto0.Distancestopointsononesideoftheline1

PAGE 8

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 speciedasan orderedtripleofreal numbers relativetoagivenrectangularcoordinatesystem.Consider arectangularboxwhosetwooppositevertices(theendpointsofthe largestdiagonal)aretheoriginandapoint P ,whileitssidesthatare adjacentattheoriginlieontheaxesofthecoordinatesystem.For everypoint P ,thereisonlyonesuchrectangularbox.Itisuniquely determinedbyitsthreesidesadjacentattheorigin.Letthenumber x denotethepositionofonesuchsidethatliesontherstaxis;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.Toreect theorderin( x,y,z ),theaxesofthecoordinatesystemwillbedenoted as x y ,and z axes.Thus,tondapointinspacewithrectangular coordinates(1 2 3),onehastoconstructarectangularboxwitha vertexattheoriginsuchthatitssidesadjacentattheoriginoccupythe intervals[0 1],[0 2],and[ 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 denedsimilarly.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 coordinateisxedtoaparticular value z = z0(e.g., z =1).Theyformaplaneparalleltothe xy plane thatlies | z0| unitsoflengthaboveitif z0> 0orbelowitif z0< 0.

PAGE 9

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.Thisisachievedbytranslatingtheoriginrst alongthe x axisbythedistance x0(asshowninthegure), thenalongthe y axisbythedistance y0,andnallyalong the z axisbythedistance z0.Asaresult,apoint P thathad coordinates( x,y,z )intheoldsystemwillhavethecoordinates x= x x0, y= y y0,and z= z 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

PAGE 10

411.VECTORSANDTHESPACEGEOMETRY paralleltothecorrespondingaxesoftheoldcoordinatesystem,but whoseoriginisshiftedtothepoint Owithcoordinates( x0, 0 0).It isstraightforwardtoseethatthepoint P wouldhavethecoordinates ( x x0,y,z )relativetothenewcoordinatesystem(Figure11.1,right panel).Similarly,iftheoriginisshiftedtoapoint Owithcoordinates ( x0,y0,z0),whiletheaxesremainparalleltothecorrespondingaxesof theoldcoordinatesystem,thenthecoordinatesof P aretransformedas (11.1)( x,y,z ) ( x x0,y y0,z 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.Thequestionthereforeishowtondthelengthofthisdiagonal. Considerthreesidesofthisrectangularboxthatareadjacent,say, atthevertex P1.Thesideparalleltothe x axisliesbetweenthe coordinateplanes x = x1and x = x2andisperpendiculartothem.So thelengthofthissideis | x2 x1| .Theabsolutevalueisnecessaryas thedierence x2 x1maybenegative,dependingonthevaluesof x1and x2,whereasthedistancemustbenonnegative.Similararguments

PAGE 11

71.RECTANGULARCOORDINATESINSPACE5 leadtotheconclusionthatthelengthsoftheothertwoadjacentsides are | y2 y1| and | z2 z1| .Ifarectangularboxhasadjacentsidesof length a b ,and c ,thenthelength d ofitslargestdiagonalsatisesthe 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 = | x2 x1| b = | y2 y1| ,and c = | z2 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 = | x1 x2| b = | y1 y2| ,and c = | z1 z2| .Thelengthofthediagonal d isobtainedbythe doubleuseofthePythagoreantheoremineachoftheindicatedrectangles: d2= c2+ f2(topright)and f2= a2+ b2(bottomright).

PAGE 12

611.VECTORSANDTHESPACEGEOMETRY found: (11.2) | P1P2| = ( x2 x1)2+( y2 y1)2+( z2 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( x1 x0,y1 y0,z1 z0)and( x2 x0,y2 y0,z2 z0),respectively.Thenumericalvalueofthedistancedoesnotchangebecause thecoordinatedierences,( x2 x0) ( x1 x0)= x2 x1(similarly forthe y and z coordinates),donotchange. Example 11.1 Apointmoves 3 unitsoflengthparalleltoaline, thenitmoves 6 unitsparalleltothesecondlinethatisperpendicularto therstline,andthenitmoves 6 unitsparalleltothethirdlinethatis perpendiculartotherstandsecondlines.Findthedistancebetween theinitialandnalpositions. Solution: Thedistancebetweenpointsdoesnotdependonthechoice ofthecoordinatesystem.Lettheoriginbepositionedattheinitialpointofthemotionandletthecoordinateaxesbedirectedalong thethreemutuallyperpendicularlinesparalleltowhichthepointhas moved.Inthiscoordinatesystem,thenalpointhasthecoordinates (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( x2 x1,y2 y1,z2 z1).Sincethedistanceshouldnotdependontheorientationof thecoordinateaxes,anyrotationcannowbedescribedalgebraically as alineartransformationofanorderedtriple ( x,y,z ) underwhichthe combination x2+ y2+ z2remainsinvariant .Alineartransformation meansthatthenewcoordinatesarelinearcombinationsoftheoldones. Itshouldbenotedthatreectionsofthecoordinateaxes, x x (similarlyfor y and z ),arelinearandalsopreservethedistance.However, acoordinatesystemobtainedbyanoddnumberofreectionsofthe coordinateaxescannotbeobtainedbyanyrotationoftheoriginalcoordinatesystem.So,intheabovealgebraicdenitionofarotation,the

PAGE 13

71.RECTANGULARCOORDINATESINSPACE7 reectionsshouldbeexcluded.Analgebraicdescriptionofrotationsin aplaneandinspaceisgiveninStudyProblems11.2and11.20.71.5.SpheresinSpace.Inthiscourse,relationsbetweentwoequivalent descriptionsofobjectsinspacethegeometricalandthealgebraic willalwaysbeemphasized.Oneofthecourseobjectivesistolearnhow tointerpretanalgebraicequationbygeometricalmeansandhowtodescribegeometricalobjectsinspacealgebraically.Oneofthesimplest examplesofthiskindisasphere.GeometricalDescriptionofaSphere.Asphereisasetofpointsinspace thatareequidistantfromaxedpoint.Thexedpointiscalledthe center ofthesphere.Thedistancefromthespherecentertoanypoint ofthesphereiscalledthe radius ofthesphere.AlgebraicDescriptionofaSphere.Analgebraicdescriptionofasphere impliesndinganalgebraicconditiononcoordinates( x,y,z )ofpoints inspacethatbelongtothesphere.Soletthecenterofthesphere beapoint P0withcoordinates( x0,y0,z0)(denedrelativetosome 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 x0)2+( y y0)2+( z 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+ z2 2 x +4 y 6 z +5=0 Solution: Inordertondthecoordinatesofthecenterandtheradius ofthesphere,theequationmustbetransformedtothestandardform (11.3)bycompletingthesquares: x2 2 x =( x 1)2 1, y2+4 y = ( y +2)2 4,and z2 6 z =( z 3)2 9.Thentheequationofthe spherebecomes ( x 1)2 1+( y +2)2 4+( z 3)2 9+5=0 ( x 1)2+( y +2)2+( z 3)2=9 .

PAGE 14

811.VECTORSANDTHESPACEGEOMETRY Figure11.3.Left :Asphereisdenedasapointsetin space.Eachpoint P ofthesethasaxeddistance R from axedpoint P0.Thepoint P0isthecenterofthesphere, and R istheradiusofthesphere. Right :Anillustrationto StudyProblem11.2.Transformationofcoordinatesundera rotationofthecoordinatesysteminaplane.Acomparisonwith(11.3)showsthatthecenterisat( x0,y0,z0)= (1 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 x0)2+( y y0)2+( z 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

PAGE 15

71.RECTANGULARCOORDINATESINSPACE9 isthecollectionofallpointsintheportionofaballofradius2that liesintherstoctant.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 twoconditionsdene M asthemidpoint.Considerarectangularbox B1whoselargestdiagonalis P1M .Thelengthofitssideparallelto the x axisis | ( x1+ x2) / 2 x1| = | x2 x1| / 2.Similarly,itssides paralleltothe y and z axeshavethelengths | y2 y1| / 2and | z2 z1| / 2, respectively.Considerarectangularbox B2whoselargestdiagonalis thesegment MP2.Thenitssideparalleltothe x axishasthelength | x2 ( x1+ x2) / 2 | = | x2 x1| / 2.Similarly,thesidesparalleltothe y and z axeshavelengths | y2 y1| / 2and | z2 z1| / 2,respectively.Thus, thesidesof B1and B2paralleltoeachcoordinateaxishavethesame length.Bythedistanceformula(11.2),thediagonalsof B1and B2musthavethesamelength | P2M | = | MP1| .Thelengthsofthesidesof arectangularboxwhoselargestdiagonalis P1P2are | x2 x1| | y2 y1| and | z2 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

PAGE 16

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 = Therefore, x= r cos = r cos( )= r cos cos + r sin sin = x cos + y sin y= r sin = r sin( )= r sin cos r cos sin = y cos x sin Theseequationsdenethetransformation( x,y ) ( x,y)oftheold coordinatestothenewones.Theinversetransformation( x,y) ( x,y )canbefoundbysolvingtheequationsfor( x,y ).Asimplerway istonotethatif( x,y)areviewedasoldcoordinatesand( x,y )as newcoordinates,thenthetransformationthatrelatesthemisthe rotationthroughtheangle (aclockwiserotation).Therefore,the inverserelationscanbeobtainedbyswappingtheoldandnew coordinatesandreplacing by inthedirectrelations.Thisyields x = xcos ysin ,y = ycos + xsin Problem11.3. Giveageometricaldescriptionoftheset S = ( x,y,z ) x2+ y2+ z2 4 z =0 Solution: Theconditiononthecoordinatesofpointsthatbelong tothesetcontainsthesumofsquaresofthecoordinatesjustlikethe equationofasphere.Thedierenceisthat(11.3)containsthesum ofperfectsquares.Sothesquaresmustbecompletedintheabove equationandtheresultingexpressioncomparedwith(11.3).Onehas z2 4 z =( z 2)2 4sothattheconditionbecomes x2+ y2+( z 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+ y2 2 x 4 y 4 Solution: Asinthepreviousproblem,theconditioncanbewritten viathesumofperfectsquares( x 1)2+( y 2)2 9bymeansof therelations x2 2 x =( x 1)2 1and y2 4 y =( y 2)2 4.In the xy plane,theinequalitydescribesadiskofradius3whosecenter

PAGE 17

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 x )=0 Solution: Theconditionissatisedifeither 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) Findthedistancebetweenthefollowingspeciedpoints: (i)(1 2 3)and( 1 0 2) (ii)( 1 3 2)and( 1 2 1) (2) Findthedistancefromthepoint(1 2 3)toeachofthecoordinate planesandtoeachofthecoordinateaxes. (3) Findthelengthofthemediansofthetrianglewithvertices A (1 2 3), B ( 3 2 1),and C ( 1 4 1). (4) Lettheset S consistofpoints( t, 2 t, 3 t ),where 0 (iv) x2+ y2 4 y< 0, z> 0 (v)4 x2+ y2+ z2 9 (vi) x2+ y2 1, x2+ y2+ z2 4 (vii) x2+ y2+ z2 2 z< 0, z> 1 (viii) x2+ y2+ z2 2 z =0, z =1 (ix)( x a )( y b )( z c )=0 (x) | x | 1, | y | 2, | z | 3

PAGE 18

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 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
PAGE 19

72.VECTORSINSPACE13 thenthedistancetraveledis | P1P2| =5m.Therate(orspeed)5m/s doesnotprovideacompletedescriptionofthemotionoftheobjectin spacebecauseitonlyanswersthequestionHowfastdoestheobject move?butitdoesnotsayanythingaboutWheretodoestheobject move?Sincetheinitialandnalpositionsoftheobjectareknown, bothquestionscanbeanswered,ifoneassociatesan orientedsegment P1P2withthemovingobject.Thearrowspeciesthedirection,from P1to P2,andthelength | P1P2| denestherate(speed)atwhichthe objectmoves.So,foreverymovingobject,onecanassignanoriented segmentwhoselengthequalsitsspeedandwhosedirectioncoincides withthedirectionofmotion.Thisorientedsegmentiscalleda velocity .Considertwoobjectsmovingparallelwiththesamespeed.The orientedsegmentscorrespondingtothevelocitiesoftheobjectshave thesamelengthandthesamedirection,buttheyarestilldierent becausetheirinitialpointsdonotcoincide.Ontheotherhand,the velocityshoulddescribeaparticularphysicalpropertyofthemotion itself(howfastandwhereto),andthereforethespatialposition wherethemotionoccursshouldnotmatter.Thisobservationleads totheconceptofa vector asanabstractmathematicalobjectthat representsallorientedsegmentsthatareparallelandhavethesame length Vectorswillbedenotedbyboldfaceletters. Twoorientedsegments AB and CD representthesamevector a iftheyareparalleland | AB | = | CD | ;thatis,theycanbeobtainedfromoneanotherbytransporting themparalleltothemselvesinspace .Arepresentationofanabstract vectorbyaparticularorientedsegmentisdenotedbytheequality a = AB or a = CD .Thefactthattheorientedsegments AB and CD representthesamevectorisdenotedbytheequality AB = CD .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,thefollowingalgebraic denitionofavectorcanbeadopted.

PAGE 20

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 = 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. Thisdenitionagreeswiththegeometricaldenitionofavectorasa classofallorientedsegmentsthatareparallelandhavethesamelength. Indeed,iftwoorientedsegmentsrepresentthesamevector,then,after

PAGE 21

72.VECTORSINSPACE15 paralleltransportsuchthattheirinitialpointscoincidewiththeorigin, theirnalpointscoincidetooandhencehavethesamecoordinates.By virtueofthecorrespondencebetweenvectorsandpointsinspace,this denitionreectsthefactthattwosamepointsshouldhavethesame positionvectors. Example 11.3 Findthecomponentsofavector P1P2ifthecoordinatesof P1and P2are ( x1,y1,z1) and ( x2,y2,z2) ,respectively. Solution: Considerarectangularboxwhoselargestdiagonalcoincideswiththesegment P1P2andwhosesidesareparalleltothecoordinateaxes.Afterparalleltransportofthesegmentsothat P1moves totheorigin,thecoordinatesoftheotherendpointarethecomponentsof P1P2.Alternatively,theoriginofthecoordinatesystemcan bemovedtothepoint P1,keepingthedirectionsofthecoordinateaxes. Therefore, P1P2= x2 x1,y2 y1,z2 z1 accordingtothecoordinatetransformationlaw(11.1),where P0= P1. Thus,inordertondthecomponentsofthevector P1P2,onehasto subtractthecoordinatesoftheinitialpoint P1fromthecorresponding coordinatesofthenalpoint 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 ifandonlyifstatementimpliestwostatements.First,if a = 0 ,then a =0(thedirectstatement).Second,if a =0,then a = 0 (the conversestatement).

PAGE 22

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 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 .

PAGE 23

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.Itspeciesonlythedirectionofavector a Example 11.4 Let A =(1 2 3) and B =(3 1 1) .Find a = AB b = BA ,theunitvectors a and b ,andthevector c = 2 AB andits norm.

PAGE 24

1811.VECTORSANDTHESPACEGEOMETRY Solution: ByExample11.3, a = 3 1 2 1 1 3 = 2 1 2 .The normof a is a = 22+( 1)2+( 2)2= 9=3 Theunitvector inthedirectionof a is a =(1 / 3) a = 2 / 3 1 / 3 2 / 3 .Usingtherule ofmultiplicationofavectorbyanumber, c = 2 a = 2 2 1 2 = 4 2 4 and c = ( 2) a = | 2 | a =2 a =6.Thedirectionof BA istheoppositeto AB ,andbothvectorshavethesamelength. Therefore, b = 2 1 2 b =3,and b = a = 2 / 3 1 / 3 2 / 3 TheParallelogramRule.Supposeapersoniswalkingonthedeckofa shipwithspeed v m / s.In1second,thepersongoesadistance v from point A topoint B ofthedeck.Thevelocityvectorrelativetothedeck is v = AB and v = | AB | = v (thespeed).Theshipmovesrelative tothewatersothatin1seconditcomestoapoint D fromapoint C onthesurfaceofthewater.Theshipsvelocityvectorrelativeto thewateristhen 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 withtheshipsvelocity u because B travelsthedistance u parallelto u .Thissuggestsasimplegeometricalrulefornding ABasshownin Figure11.6.Takethevector AB = v ,placethevector u sothatits initialpointcoincideswith B ,andmaketheorientedsegmentwiththe initialpointof v andthenalpointof u inthisdiagram.Theresulting vectoristhedisplacementvectorofthepersonrelativetothesurface ofthewaterin1secondandhencedenesthevelocityoftheperson 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

PAGE 25

72.VECTORSINSPACE19 Figure11.6.Left :Parallelogramruleforaddingtwo vectors.Iftwovectorsformadjacentsidesofaparallelogramatavertex A ,thenthesumofthevectorsisavector thatcoincideswiththediagonaloftheparallelogramand originatesatthevertex A Right :Addingseveralvectorsbyusingtheparallelogramrule.Giventherstvectorinthesum,allothervectorsaretransportedparallel sothattheinitialpointofthenextvectorinthesumcoincideswiththeterminalpointofthepreviousone.The sumisthevectorthatoriginatesfromtheinitialpointof therstvectorandterminatesattheterminalpointofthe lastvector.Itdoesnotdependontheorderofvectorsin thesum.thattheadditionofvectorsis commutative : a + b = b + a ; thatis,theorderinwhichthevectorsareaddeddoesnotmatter.To addseveralvectors(e.g., a + b + c ),onecanrstnd a + b bythe parallelogramruleandthenadd c tothevector a + b .Alternatively, thevectors b and c canbeaddedrst,andthenthevector a canbe addedto b + c .Accordingtotheparallelogramrule,theresulting vectoristhesame: ( a + b )+ c = a +( b + c ) Thismeansthattheadditionofvectorsis associative .Soseveralvectorscanbeaddedinanyorder.Taketherstvector,thenmovethe secondvectorparalleltoitselfsothatitsinitialpointcoincideswiththe terminalpointoftherstvector.Thethirdvectorismovedparallel sothatitsinitialpointcoincideswiththeterminalpointofthesecond vector,andsoon.Finally,makeavectorwhoseinitialpointcoincides withtheinitialpointoftherstvectorandwhoseterminalpointcoincideswiththeterminalpointofthelastvectorinthesum.Tovisualize thisprocess,imagineamanwalkingalongtherstvector,thengoing paralleltothesecondvector,thenparalleltothethirdvector,andso

PAGE 26

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 Thisdenitionisequivalenttothegeometricaldenitionofadding 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 = c1 a1,c2 a2,c3 a3 ,where therelationbetweenthecomponentsofavectorandthecoordinates ofitsendpointshasbeenused(seeExample11.3).Theequalityoftwo vectorsmeanstheequalityofthecorrespondingcomponents,thatis, b1= c1 a1, b2= c2 a2,and b3= c3 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 Thedierenceoftwovectorscanbedenedas a b = a +( 1) b Intheparallelogramwithadjacentsides a and b ,thesumofvectors a and( 1) b representsthevectorthatoriginatesfromtheendpoint of b andendsattheendpointof a because b +[ a +( 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.

PAGE 27

72.VECTORSINSPACE21 Figure11.7.Left :Subtractionoftwovectors.Thedifference a b isviewedasthesumof a and b ,thevector thathasthedirectionoppositeto b andthesamelengthas b .Theparallelogramruleforadding a and b showsthat thedierence a b = a +( b )isthevectorthatoriginates fromtheterminalpointof b andendsattheterminalof a if a and b areadjacentsidesofaparallelogram;thatis,the sum a + b andthedierence a b arethetwodiagonals oftheparallelogram. Right :IllustrationtoStudyProblem 11.6.Anyvectorinaplanecanalwaysberepresentedasa linearcombinationoftwononparallelvectors.Solution: Lettheinitialandterminalpointsbe A and B ,respectively.Let C bethepointatwhichthevelocitywaschanged.Then AC =3 v and CB =2 u .Therefore, AB = AC + 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 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

PAGE 28

2211.VECTORSANDTHESPACEGEOMETRY parallelogram.Foragiven c ,thereals t and s areuniquelydened bytheproposedgeometricalconstruction.Bytheparallelogramrule, c = t a + s b Problem11.7. Findthecoordinatesofapoint B thatisatadistance of 6 unitsoflengthfromthepoint A (1 1 2) inthedirectionofthe vector v = 2 1 2 Solution: Thepositionvectorofthepoint A is a = OA = 1 1 2 Thepositionvectorofthepoint B is b = a + s v ,where s isapositive numbertobechosensuchthatthelength | AB | = s v equals6.Since v =3,onends s =2.Therefore, b = 1 1 2 +2 2 1 2 = 5 1 2 Problem11.8. Considerastraightlinesegmentwiththeendpoints A (1 2 3) and B ( 2 1 0) .Findthecoordinatesofthepoint C onthe segmentsuchthatitistwiceasfarfrom A asitisfrom B Solution: Let a = 1 2 3 b = 1 0 1 ,and c bepositionvectors of A B ,and C ,respectively.Thequestionisnowtoexpress c via a and b .Onehas c = a + AC .Thevector AC isparallelto AB = 3, 3 3 andhence AC = s AB .Since | AC | =2 | CB | | AC | =2 3| AB | andtherefore s =2 3.Thus, c = a +2 3 AB = a +2 3( b a )= 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 donotdependonthecoordinatesystemrelativetowhichthecomponentsofallthevectorsaredened.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 = 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 ,onends3= t s and3 3= s 3,or s =3 and t =6.Hence, c =6 a +3 b Problem11.10. Supposethethreecoordinateplanesareallmirrored. Alightraystrikesthemirrors.Determinethedirectioninwhichthe reectedraywillgo. Solution: Let u beavectorparalleltotheincidentray.Under areectionfromaplanemirror,thecomponentof u perpendicularto

PAGE 29

72.VECTORSINSPACE23 theplanechangesitssign.Therefore,afterthreeconsecutivereections fromeachcoordinateplane,allthreecomponentsof u changetheir signs,andthereectedraywillgoparalleltotheincidentraybutin theexactoppositedirection.Forexample,supposetherayisreected rstbythe xz plane,thenbythe yz plane,andnallybythe xy plane. Inthiscase, u = u1,u2,u3 u1, u2,u3 u1, u2,u3 u1, u2, u3 = u Remark. Thisprincipleisusedtodesignreectorslikethecatseyesonbicyclesandthosethatmarktheborderlinesofaroad.No matterfromwhichdirectionsuchareectorisilluminated(e.g.,bythe headlightsofacar),itreectsthelightintheoppositedirection(so thatitwillalwaysbeseenbythedriver).72.5.Exercises.(1) Findthecomponentsofeachofthefollowingvectorsandtheir norms: (i)Thevectorthathasendpoints A (1 2 3)and B ( 1 5 1)and isdirectedfrom A to B (ii)Thevectorthathasendpoints A (1 2 3)and B ( 1 5 1)and isdirectedfrom B to A (iii)Thevectorthathastheinitialpoint A (1 2 3)andthenal point C thatisthemidpointofthelinesegment AB ,where B =( 1 5 1) (iv)Thepositionvectorofapoint P obtainedfromthepoint A ( 1 2 1)bytransportingthelatteralongthevector u = 2 2 1 3unitsoflengthandthenalongthevector w = 3 0 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 2 a a 1 2b and2 a +3 b (3) Let a b ,and c bethreevectorsinaplaneanyofwhichisnot paralleltotheothers.Sketcheachofthefollowingvectors: a +( b c ), ( a + b ) c ,2 a 3( b + c ),and(2 a 3 b ) 3 c (4) Let a = 2 1 2 and b = 3 0 4 .Findunitvectors a and b Express6 a 15 b via a and b (5) Let a and b bevectorsinthe xy planesuchthattheirsum c = a + bmakestheangle / 3with a andhaslengthtwicethelengthof a .Find

PAGE 30

2411.VECTORSANDTHESPACEGEOMETRY b if a liesintherstquadrant,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 algebratond 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 ,beunitvectorsasdenedinexercise(7). Let wk= uk +1 ukfor k =1 2 ,...,n 1and wn= u1 un.Findthe limitof w1 + w2 + + wn as n (9) Aplaneiesataspeedof 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 a + r b = k ,where k> a b (12) Letpointlikemassiveobjectsbepositionedat Pi, i =1 2 ,...,n andlet mibethemassat Pi.Thepoint P0iscalledthe centerof mass if m1 P0P1+ m2 P0P2+ + mn P0Pn= 0 Expressthepositionvector r0of P0viathepositionvectors riof Pi. Inparticular,ndthecenterofmassofthreepointmasses, m1= m2= m3= m ,locatedattheverticesofatriangle ABC for A (1 2 3), B ( 1 0 1),and C (1 1 1). (13) Considerthegraph y = f ( x )ofadierentiablefunctionandthe linetangenttoitatapoint x = a .Expresscomponentsofavector paralleltothelinevia f( a )andndavectorperpendiculartotheline. Inparticular,ndsuchvectorsforthegraph y = x2atthepoint x =1. (14) Letthevectors a b ,and c havexedlengths a b ,and c ,respectively,whiletheirdirectionsmaybechanged.Isitalwayspossibleto achieve a + b + c = 0 ?Ifnot,formulatetheconditionunderwhichit ispossible.

PAGE 31

73.THEDOTPRODUCT25 (15) Letthevectors a and b havexedlengths,whiletheirdirections maybechanged.Put c= a b .Isitalwayspossibletoachieve c>c+,or c= c+,or c
PAGE 32

2611.VECTORSANDTHESPACEGEOMETRY 73.1.GeometricalSignicanceoftheDotProduct.Asitstands,thedot productisanalgebraicruleforcalculatinganumberoutofsixgiven numbersthatarecomponentsofthetwovectorsinvolved.Thecomponentsofavectordependonthechoiceofthecoordinatesystem. Naturally,oneshouldaskwhetherthenumericalvalueofthedotproductdependsonthecoordinatesystemrelativetowhichthecomponents ofthevectorsaredetermined.Itturnsoutthatitdoesnot.Therefore, itrepresentsanintrinsicgeometricalquantityassociatedwithtwovectorsinvolvedintheproduct.Toelucidatethegeometricalsignicance ofthedotproduct,notersttherelationbetweenthedotproductand thenorm(length)ofavector: a a = a2 1+ a2 2+ a2 3= a 2or a = a a Thus,if a = b inthedotproduct,thenthelatterdoesnotdepend onthecoordinatesystemwithrespecttowhichthecomponentsof a aredened.Next,considerthetrianglewhoseadjacentsidesarethe vectors a and b asdepictedinFigure11.8(leftpanel).Thentheother sideofthetrianglecanberepresentedbythedierence c = b a .The squaredlengthofthislattersideis (11.4) c c =( b a ) ( b a )= b b + a a 2 a b wherethealgebraicpropertiesofthedotproducthavebeenused. Therefore,thedotproductcanbeexpressedviathegeometricalinvariants,namely,thelengthsofthesidesofthetriangle: (11.5) a b = 1 2 a 2+ b 2 c 2 Thismeansthatthenumericalvalueofthedotproductisindependent ofthechoiceofacoordinatesystem.Inparticular,letustakethe coordinatesysteminwhichthevector a isparalleltothe x axisand thevector b liesinthe xy planeasshowninFigure11.8(rightpanel). Lettheanglebetween a and b be .Bydenition,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)revealsthegeometricalsignicanceofthedotproduct.Itdeterminestheanglebetweentwoorientedsegmentsinspace. Itprovidesasimplealgebraicmethodtoestablishamutualorientation oftwostraightlinesegmentsinspace.

PAGE 33

73.THEDOTPRODUCT27 Figure11.8.Left :Independenceofthedotproductfrom thechoiceofacoordinatesystem.Thedotproductof twovectorsthatareadjacentsidesofatrianglecanbeexpressedviathelengthsofthetrianglesidesasshownin (11.5). Right :Geometricalsignicanceofthedotproduct.Itdeterminestheanglebetweentwovectorsasstated in(11.6).Twononzerovectorsareperpendicularifandonly iftheirdotproductvanishes.Thisfollowsfrom(11.5)and thePythagoreantheorem: a 2+ b 2= c 2forarightangledtriangle.Theorem 11.1 (GeometricalSignicanceoftheDotProduct) 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+ b2 2 ab cos Foraright-angledtriangle, c2= a2+ b2(thePythagoreantheorem). Example 11.7 Consideratrianglewhoseverticesare A (1 1 1) B ( 1 2 3) ,and C (1 4 3) .Findalltheanglesofthetriangle. Solution: Lettheanglesatthevertices A B ,and C be ,and respectively.Then + + =180.Soitissucienttondanytwo angles.Tondtheangle ,denethevectors a = AB = 2 1 2 and b = AC = 0 3 4 .Theinitialpointofthesevectorsis A ,and hencetheanglebetweenthevectorscoincideswith .Since a =3

PAGE 34

2811.VECTORSANDTHESPACEGEOMETRY and b =5,bythegeometricalpropertyofthedotproduct, cos = a b a b = 0+3 8 15 = 1 3 = =cos 1( 1 / 3) 109 5. Tondtheangle ,denethevectors a = BA = 2 1 2 and b = BC = 2 2 6 withtheinitialpointatthevertex B .Thenthe anglebetweenthesevectorscoincideswith .Since a =3, b = 2 11,and a b =4 2+12=14,onendscos =14 / (6 11)and =cos 1(7 / (3 11)) 45 3.Therefore, 180 109 5 45 3= 25 2.Notethattherangeofthefunctioncos 1mustbetakenfrom 0to180inaccordancewiththedenitionoftheanglebetweentwo vectors. 73.2.FurtherGeometricalPropertiesoftheDotProduct.Corollary 11.1 (OrthogonalDecompositionofaVector). Givenanonzerovector a ,anyvector b canbeuniquelydecomposed intothesumoftwoorthogonalvectors,oneofwhichisparallelto a : b = b+ b, b= b s a b= s a ,s = b a a 2. Indeed,given a and b ,put b= b s a andassumethat bis orthogonalto a ,thatis, a b=0.Thisconditionuniquelydetermines thecoecient s : a b 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 .Itisalsoeasytoseefromthegure that b = b sin Example 11.8 Let a = 1 2 1 and b = 5 1 9 .Findthe orthogonaldecomposition b = b+ brelativetothevector a Solution: Onehas a b =5 2+9=12and a 2= a a =1+( 2)2+ 1=6.Therefore, s =12 / 6=2, b= s a =2 1 2 1 = 2 4 2 and b= b b= 5 1 9 2 4 2 = 3 5 7 .Theresultcanalso beveried: a b=3 10+7=0;thatis, a isorthogonalto bas required.

PAGE 35

73.THEDOTPRODUCT29 Theorem 11.2 (Cauchy-SchwarzInequality) Foranytwovectors a and b | a b | a b wheretheequalityisreachedonlyifthevectorsareparallel. Thisinequalityisadirectconsequenceoftherstrelationin(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 b | .Theabsolutevalue isnecessaryasthelengthof a maybelessthanthelengthof b .This observationcanbestatedinthefollowingalgebraicform: (11.7) a 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).Denetheangle thatiscountedfromthe positivedirectionofthe x axistowardthevector a .Inotherwords, theangle istheanglebetween e1and a .Similarly,theangles and

PAGE 36

3011.VECTORSANDTHESPACEGEOMETRY Figure11.9.Left :Thedirectionanglesofavectoraredenedastheanglesbetweenthevectorandthreecoordinate axes.Eachanglerangesbetween0and andiscounted fromthecorrespondingpositivecoordinatesemiaxistoward thevector.Thecosinesofthedirectionanglesofavectorare componentsoftheunitvectorparalleltothatvector. Right : Thedecompositionofavector a intothesumofthreemutuallyperpendicularvectorsthatareparalleltothecoordinate axesofarectangularcoordinatesystem.Thevectoristhe diagonaloftherectangularboxwhoseedgesareformedby thevectorsinthesum. are,bydenition,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 .

PAGE 37

73.THEDOTPRODUCT31 Inpractice(physics,mechanics,etc.),vectorsareoftenspeciedby 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,andthenormaredenedasinvectoralgebrais 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,anytwononparallelvectorsinaplanedenea(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 = ( y/x ) b ( z/x ) c .Itcanbeprovedthat anythreelinearlyindependentvectorsformabasisinspace (StudyProblems11.11 and11.12).

PAGE 38

3211.VECTORSANDTHESPACEGEOMETRY 73.5.ApplicationsoftheDotProduct. StaticProblems.AccordingtoNewtonsmechanics,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 tworopessothatthesmallestanglebetweentherstropeandtheceiling is 1andtheangle 2isdenedsimilarlyforthesecondrope.Findthe magnitudesofthetensionforcesintheropes. Solution: ThesysteminquestionisshowninFigure11.10(left panel).Theequilibriumconditionis T1+ T2+ G = 0 Let e1beaunitvectorthatishorizontalanddirectedfromlefttoright andlet e2beaunitvectordirectedupward.Theyformanorthonormal basisintheplane.Usingthescalarprojections,theforcescanbe expandedinthisbasisas T1= T1cos 1 e1+ T1sin 1 e2, T2= T2cos 2 e1+ T2sin 2 e2, G = mg e2, where T1and T2arethemagnitudesofthetensionforces.Thescalar projectionsofthetotalforceontothehorizontalandverticaldirections denedby e1and e2shouldvanish: T1cos 1+ T2cos 2=0 ,T1sin 1+ T2sin 2 mg =0 Thissystemisthensolvedfor T1and T2.Bymultiplyingtherst equationbysin 1andthesecondbycos 1andthenaddingthem,one gets T2= mg cos 1/ sin( 1+ 2).Substituting T2intotherstequation, thetension T1= mg cos 2/ sin( 1+ 2)isobtained.

PAGE 39

73.THEDOTPRODUCT33 Figure11.10.Left :IllustrationtoExample11.9.At equilibrium,thevectorsumofallforcesactingontheball vanishes.Thecomponentsoftheforcesareeasytondinthe 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 .Alawofphysicsstatesthatachangeinanobjectskinetic energyisequaltothework W donebythisforce: K2 K1= F P1P2= W, where K1and K2arethekineticenergiesattheinitialandnalpoints ofthemotion,respectively. Example 11.10 Letanobjectslideonaninclinedplanewithout frictionunderthegravitationalforce.Themagnitudeofthegravitationalforceisequalto mg ,where m isthemassoftheobjectand g isauniversalconstantforallobjectsnearthesurfaceoftheEarth, g 9 8 m/s2.Findthenalspeed v oftheobjectiftherelativeheight oftheinitialandnalpointsis h andtheobjectwasinitiallyatrest. Solution: Choosethecoordinatesystemsothatthedisplacement vector P1P2andthegravitationalforceareinthe xy plane.Letthe y axisbeverticalsothatthegravitationalforceis F = 0 mg, 0 where m isthemassand g istheaccelerationofthefreefall.Theinitial pointischosentohavethecoordinates(0 ,h, 0)whilethenalpoint

PAGE 40

3411.VECTORSANDTHESPACEGEOMETRY is( L, 0 0),where L isthedistancetheobjecttravelsinthehorizontal directionwhilesliding.Thedisplacementvectoris P1P2= L, h, 0 Since K1=0,onehas mv2 2 = W = F P1P2= 0 mg, 0 L, 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.

PAGE 41

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= 2 c3c2= c3. Thesubstitutionofthelasttwoequationsintotherstoneyields c3 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 2 s3s2=3 s3. Thesubstitutionofthelasttwoequationsintotherstoneyields s3=4 andhence s1= 6and s2= 1sothat a = 6 u1 u2+4 u3. Problem11.13. Describethesetofallpointsinspacewhoseposition vectors r satisfythecondition ( r a ) ( r b )=0 .Hint: Notethatthe positionvectorsatisfyingthecondition r c = R describesasphere ofradius R whosecenterhasthepositionvector c Solution: Theequationofaspherecanalsobewrittenintheform r c 2=( r c ) ( r c )= R2.Theequation( r a ) ( r b )=0can betransformedintothesphereequationbycompletingthesquares. Usingthealgebraicpropertiesofthedotproduct, ( r a ) ( r b )= r r r ( a + b )+ a b =( r c ) ( r c ) c c + a b c =1 2( a + b ) c c a b = R R R =1 2( a b ) Hence,thesetisasphereofradius R = R ,anditscenterispositioned at c .If a and b arethepositionvectorsofpoints A and B ,then,by theparallelogramrule,thecenterofthesphereisthemidpointofthe

PAGE 42

3611.VECTORSANDTHESPACEGEOMETRY straightlinesegment AB andthesegment AB isadiameterofthe sphere. 73.7.Exercises.(1) Findthedotproduct a b if (i) a = 1 2 3 and b = 1 2 0 (ii) a = e1+3 e2 e3and b =3 e1 2 e2+ e3(iii) a =2 c 3 d and b = c +2 d if c istheunitvectorthatmakes theangle / 3withthevector d and d =2 (2) Forwhatvaluesof b arethevectors 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 ( 2 4 3),and C (1 2 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 2 2 b = 2 2 1 ,and c = 1 1 1 (seethedenitionofaparallelepipedin StudyProblem11.11).Findtheanglesbetweenitslargestdiagonal andtheadjacentsides. (9) Let a = 1 2 2 .Forthevector b = 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 .

PAGE 43

73.THEDOTPRODUCT37 (13) Letthevectors a and b havethesamelength.Showthatthe vectors a + b and a 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 1 0 ,and u3= 2 2 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?

PAGE 44

3811.VECTORSANDTHESPACEGEOMETRY 74.TheCrossProduct74.1.DeterminantofaSquareMatrix.Definition 11.10 Thedeterminantofa 2 2 matrixisthenumber computedbythefollowingrule: det a11a12a21a22 = a11a22 a12a21, thatis,theproductofthediagonalelementsminustheproductofthe o-diagonalelements. Definition 11.11 Thedeterminantofa 3 3 matrix A isthe numberobtainedbythefollowingrule: det a11a12a13a21a22a23a31a32a33 = a11det A11 a12det A12+ a13det A13=3k =1( 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( 1)k + mamkdet Amkforany m =1 2 3 det A =3m =1( 1)k + mamkdet Amkforany k =1 2 3 wherethematrix Amkisobtainedfrom A byremovingtherowand columncontaining amk.Thisdenitionofthedeterminantisextended 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

PAGE 45

74.THECROSSPRODUCT39 foranynumbers b c ,and d .Also,itfollowsfromtheexpansionofthe determinantoveranycolumnorrowthat, ifanytworowsoranytwo columnsareswappedinthematrix,itsdeterminantchangessign .For 2 2matrices,thisiseasytoseedirectlyfromDenition11.10.In general,ifthematrix B isobtainedfrom A byswappingtherstand 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( 1)2+ kb2 kdet B2 k=3k =1( 1)2+ ka1 kdet A1 k= 3k =1( 1)1+ ka1 kdet A1 k= det A. Thisargumentcanbeappliedtoanytworowsorcolumnsinasquare matrixofanydimension. Example 11.11 Calculate det A ,where A = 123 013 121 Solution: Expandingthedeterminantovertherstrowyields det A =1(1 1 2 3) 2(0 1 ( 1) 3)+3(0 2 ( 1) 1)= 8 Alternatively,expandingthedeterminantoverthesecondrowyields thesameresult: det A = 0(2 1 3 2)+1(1 1 ( 1) 3) 3(1 2 ( 1) 2)= 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

PAGE 46

4011.VECTORSANDTHESPACEGEOMETRY determinantoftheformalmatrixexpandedovertherstrow: a b =det e1 e2 e3a1a2a3b1b2b3 =det a2a3b2b3 e1 det a1a3b1b3 e2+det a1a2b1b2 e3= a2b3 a3b2,a3b1 a1b3,a1b2 a2b1 (11.8) Notethattherstrowofthematrixconsistsoftheunitvectors 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: Bydenition, 1 2 3 2 0 1 =det e1 e2 e3123 201 =det 23 01 e1 det 13 21 e2+det 12 20 e3=(2 0) e1 (1 6) e2+(0 4) e3=2 e1+5 e2 4 e3= 2 5 4 Thecrossproducthasthefollowingpropertiesthatfollowfromits denition: a b = b a ( a + c ) b = a b + c b ( s a ) b = s ( a b ) Therstpropertyisobtainedbyswappingthecomponentsof b and a in(11.8).Alternatively,recallthatthedeterminantofamatrix changesitssigniftworowsareswappedinthematrix(therows a and b inDenition11.12).Sothecrossproductisskew-symmetric; thatis,itis notcommutative ,andtheorderinwhichthevectorsare multipliedisessential.Changingtheorderleadstotheoppositevector. Inparticular,if b = a ,then a a = a a or2( a a )= 0 or a a = 0 .

PAGE 47

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 computedrstandthen scaledby s (change aito saiin(11.8)andthenfactorout s ).The doublecrossproductsatisesthesocalled bac cab rule (11.9) a ( b c )= b ( a c ) c ( a b ) andtheJacobiidentity (11.10) a ( b c )+ b ( c a )+ c ( a b )= 0 Notethatthesecondandthirdtermsontheleftsideof(11.10)are obtainedfromtherstbycyclicpermutationsofthevectors.The proofsofthe bac cab ruleandtheJacobiidentityaregiveninStudy Problems11.16and11.17.TheJacobiidentityimpliesthat a ( b c ) =( a b ) c Thismeansthatthemultiplicationofvectorsdenedbythecrossproductis notassociative incontrasttomultiplicationofnumbers.This observationisfurtherdiscussedinStudyProblem11.18.74.3.GeometricalSignicanceoftheCrossProduct.Theabovealgebraicdenitionofthecrossproductusesaparticularcoordinatesystemrelativetowhichthecomponentsofthevectorsaredened.Does thecrossproductdependonthechoiceofthecoordinatesystem?To answerthisquestion,oneshouldinvestigatewhetherbothits direction andits magnitude dependonthechoiceofthecoordinatesystem.Let usrstinvestigatethemutualorientationoftheorientedsegments a b ,and a b .Asimplealgebraiccalculationleadstothefollowing result: a ( a b )= a1( a2b3 a3b2)+ a2( a3b1 a1b3)+ a3( a1b2 a2b1)=0 Bytheskewsymmetryofthecrossproduct,itisalsoconcludedthat b ( a b )= 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 .

PAGE 48

4211.VECTORSANDTHESPACEGEOMETRY Letuscalculatethelengthofthecrossproduct.Bythedenition (11.8), a b 2=( a b ) ( a b ) =( a2b3 a3b2)2+( a3b1 a1b3)2+( a1b2 a2b1)2=( a2 1+ a2 2+ a2 3)( b2 1+ b2 2+ b2 3) ( a1b1+ a2b2+ a3b3)2= a 2 b 2 ( a b )2, wherethethirdequalityisobtainedbycomputingthesquaresofthe componentsofthecrossproductandregroupingtermsintheobtained expression.Thelastequalityusesthedenitionsofthenormandthe dotproduct.Next,recallthegeometricalpropertyofthedotproduct (11.6).If istheanglebetweenthevectors a and b ,then a b 2= a 2 b 2 a 2 b 2cos2 = a 2 b 2(1 cos2 )= a 2 b 2sin2. Since0 ,sin 0andthesquarerootofbothsidesofthis equationcanbetakenwiththeresultthat a b = a b sin Thisrelationshowsthatthelengthofthecrossproductdenedby (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 eithertherstorsecondquadrantofthe xy planeand b2= b sin if b liesineitherthethirdorfourthquadrant.Intheformercase, a b = 0 0 ,A ,where A istheareaoftheparallelogram.Inthe lattercase,thedenition(11.8)yields a b = 0 0 A .Itturnsout thatthedirectionofthecrossproductinbothcasescanbedescribed byasimpleruleknownasthe right-handrule : Ifthengersoftheright handcurlinthedirectionofarotationfrom a toward b throughthe smallestanglebetweenthem,thenthethumbpointsinthedirectionof a b .

PAGE 49

74.THECROSSPRODUCT43 Figure11.11.Left :Geometricalinterpretationofthe crossproductoftwovectors.Thecrossproductisavectorthatisperpendiculartobothvectorsintheproduct.Its lengthequalstheareaoftheparallelogramwhoseadjacent sidesarethevectorsintheproduct.Ifthengersoftheright handcurlinthedirectionofarotationfromtherstvectorto thesecondvectorthroughthesmallestanglebetweenthem, thenthethumbpointsinthedirectionofthecrossproduct ofthevectors. Right :IllustrationtoStudyProblem11.15.Inparticular,byDenition11.12, 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 (GeometricalSignicanceoftheCrossProduct) 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 .

PAGE 50

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 productisincalculationsoftheareasofplanarguresinspace. 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 1 3) ,and C = ( 1 3 1) .Findtheareaofthetriangle ABC andavectororthogonaltotheplanethatcontainsthetriangle. Solution: Taketwovectorswiththeinitialpointatanyofthevertices ofthetrianglethatformtheadjacentsidesofthetriangleatthatvertex. Forexample, a = AB = 1 2 2 and b = AC = 2 2 0 .Then a b = 4 4 6 .Since 4 4 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 = CB and b = CA or a = BA and b = BC wouldgive thesameanswer(modulothesignchangeinthecrossproduct). ApplicationsinPhysics:Torque.Torque, or momentofforce ,isthe tendencyofaforcetorotateanobjectaboutanaxisorapivot.Just asaforceisapushorapull,atorquecanbethoughtofasatwist.

PAGE 51

74.THECROSSPRODUCT45 If r isthevectorfromapivotpointtothepointwhereaforce F is applied,thenthetorqueisdenedasthecrossproduct = 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 beaxed(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

PAGE 52

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 fortherstandsecondballs,respectively( g isthefree fallacceleration).Theyaredirecteddownwardandthereforeliein theplanethatcontainsthepositionvectorsoftheballsrelativetothe pivotpoint.Sothetorquesofthegravitationalforcesareorthogonal tothisplane,andtheequilibriumcondition 1+ 2= 0 isequivalent to 1 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.Theanglebetweenthepositionvectoroftherstballand thegravitationalforceactingonitis 1= / 2 1,andsimilarly 2= / 2 2istheanglebetweenthepositionvectorofthesecond ballandthegravitationalforceactingonit.Therefore, 1= L1F1sin 1and 2= L2F2sin 2.Owingtotheidentitysin( / 2 )=cos ,it followsthat 1= 2 m1L1cos 1= m2L2cos 2 tan 1= m1L1 m2L2, wheretherelation 2= / 2 1hasbeenused. 74.4.StudyProblems.Problem11.14. Findthemostgeneralvector r thatsatisestheequations 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

PAGE 53

74.THECROSSPRODUCT47 Problem11.15. Usegeometricalmeanstondthecrossproductsof 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= e3). Sinceacyclicpermutationofthreeindices { ijk }{ kij } (andsoon) consistsoftwopermutationsofanytwoindices,therelationbetween theunitvectorscanbecastintheform ei= ej ek, { ijk } = { 123 } andcyclicpermutations Problem11.16. Provethebac 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 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 )= c b 2= b ( b c ) c ( b b ) Byswappingthevector b and c inthisequation,onealsoobtains c ( b c )= c ( c b )= b c 2= c ( c b )+ b ( c c ) Itfollowsfromtheserelationsthat a ( b c )= t b ( b c )+ p c ( b c ) = b [( t b + p c ) c ] c [( t b + p c ) b ] = b ( a c ) c ( a b )

PAGE 54

4811.VECTORSANDTHESPACEGEOMETRY Problem11.17. ProvetheJacobiidentity(11.10). Solution: Bythe bac cab rule(11.9)appliedtoeachterm, a ( b c )= b ( a c ) c ( a b ) b ( c a )= c ( b a ) a ( b c ) c ( a b )= a ( c b ) 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) .Denethevectorproductoftwovectors 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=vector)canbe denedinaEuclideanspaceofmorethantwodimensions. 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 isxed,while themagnitudeof u doesnotchangewhenrotating.Therefore,the absolutemaximumofthecross-productmagnitudeisreachedwhen sin =1orcos =0(i.e.,whenthevectorsareorthogonal).The

PAGE 55

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 = v1/v2.Thisequationhastwosolutionsinthe range0 < 2 : = tan 1( v1/v2)and = tan 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 = 1 0 1 (ii) a = 1 1 2 and b = 3 2 1 (iii) a = e1+3 e2 e3and b =3 e1 2 e2+ e3(iv) a =2 c d and b =3 c +4 d ,where c d = 1 2 3 (2) Let a = 3 2 1 b = 2 1 1 ,and c = 1 0 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 itsterminalpointisinrstquadrantofthe 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 =( 1 0 2)beverticesofaparallelogram.Iftheothertwoverticesareobtainedbymoving A and B by

PAGE 56

5011.VECTORSANDTHESPACEGEOMETRY 3unitsoflengthalongthevector a = 2 1 2 ,ndtheareaofthe parallelogram. (11) Considerfourpointsinspace.Supposethatthecoordinatesofthe pointsareknown.Describeaprocedurebasedonthepropertiesofthe crossproducttodeterminewhetherthepointsareinoneplane.Inparticular,arethepoints(1 2 3),( 1 0 1),(1 3 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: Denethesidesasvectorsandexpresstheareaofthetriangle viathevectorsateachvertexofthetriangle. (13) Considerapolygonwithfourvertices A B C ,and D .Ifthe coordinatesoftheverticesarespecied,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 adjacentsidesarediagonalsoftherstparallelogram.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 andnd 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

PAGE 57

75.THETRIPLEPRODUCT51 computedbythedeterminantrulegiveninDenition11.12where eiarereplacedby ui. Hint: Usethe bac cab ruletondallpairwise 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 ) Itfollowsfromthealgebraicdenitionofthecrossproductandthe denitionofthedeterminantofa3 3matrixthat a ( b c )= a1det b2b3c2c3 a2det b1b3c1c3 + a3det b1b2c1c2 =det a1a2a3b1b2b3c1c2c3 Thisprovidesaconvenientwaytocalculatethenumericalvalueof thetripleproduct.Iftworowsofamatrixareswapped,thenits determinantchangessign.Therefore, a ( b c )= b ( a c )= c ( b a ) Thismeans,inparticular,thattheabsolutevalueofthetripleproduct isindependentoftheorderofthevectorsinthetripleproduct.Also, thevalueofthetripleproductisinvariantundercyclicpermutations ofvectorsinit: a ( b c )= b ( c a )= c ( a b ).

PAGE 58

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 (GeometricalSignicanceoftheTripleProduct) Thevolume V ofaparallelepipedwhoseadjacentsidesarethevectors a b ,and c istheabsolutevalueoftheirtripleproduct: V = | a ( b c ) | Thus,thetripleproductisaconvenientalgebraictoolforcalculatingvolumes.Notealsothatthevectorscanbetakeninanyorderin

PAGE 59

75.THETRIPLEPRODUCT53 thetripleproducttocomputethevolumebecausethetripleproduct onlychangesitssignwhentwovectorsareswappedinit. Example 11.15 Findthevolumeofaparallelepipedwithadjacent sides a = 1 2 3 b = 2 0 1 ,and c = 2 1 2 Solution: Theexpansionofthedeterminantovertherstrowyields b ( a c )=det 201 123 212 = 2(4 3)+1(1 4)= 5 Takingtheabsolutevalueofthetripleproduct,thevolumeisobtained, V = | 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,

PAGE 60

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 1) ,and D (0 2 3) areinthesameplane. Solution: Considerthevectors a = AB = 1 1 1 b = AC = 2 0 2 ,and c = AD = 1 1 2 .Thepointsinquestionareinthe sameplaneifandonlyifthevectors a b ,and c arecoplanar,or a ( b c )=0.Theevaluationofthetripleproductyields a ( b c )=det 1 11 202 112 =1(0 2)+1(4+2)+1(2 0)=6 =0 Therefore,thepointsarenotinthesameplane. Example 11.17 Let u1= 1 2 3 u2= 2 1 6 ,and u3= 1 1 1 .Canthevector a = 1 1 1 berepresentedasalinearcombinationofthevectors u1, u2,and u3? Solution: Anyvectorinspaceisalinearcombinationof u1, u2,and u3iftheyformabasis.Letusverifyrstwhetherornottheyforma basis.ByCorollary11.5, u1 ( u2 u3)=det 123 21 6 11 1 =1( 1+6) 2( 2+6)+3(2 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 1 =1( 2 3) 1( 1 3)+1(1 2)= 2 thevector a isnotintheplaneinwhichthevectors u1, u2,and u3lie, andtherefore a isnotalinearcombinationofthem.

PAGE 61

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)= 1 iscalled left-handed .Aright-handedsystemcanbevisualizedasfollows.Withthethumb,index,andmiddlengersofthe righthand at rightanglestoeachother(withtheindexngerpointedstraight),the middlengerpointsinthedirectionof e1= e2 e3whenthethumb represents e2andtheindexngerrepresents e3.Aleft-handedsystem isobtainedbythereection e1 e1andthereforeisvisualizedby thengersofthe lefthand inthesameway.Sincethedotproductcannotbechangedbyrotationsandtranslationsinspace,thehandedness ofthecoordinatesystemdoesnotchangeundersimultaneousrotations andtranslationsofthetripleofvectors ei(threemutuallyorthogonal ngersofthelefthandcannotbemadepointinginthesamedirection asthecorrespondingngersoftherighthandbyanyrotationofthe hand).Thereectionofallthreevectors ei eiorjustoneofthem turnsaright-handedsystemintoaleft-handedoneandviceversa.A mirrorreectionofaright-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,thenhowcanonendthedistancebetween 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.

PAGE 62

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 | x1 x2| ,where0 0canbemadesmallerthananypreassigned positivenumberbytaking x1and x2closeenoughto1.Sincethe distance D 0,theonlypossiblevalueis D =0.Intuitively,thesets areseparatedbyasinglepointthatisnotanextendedobject,and hencethedistancebetweenthemshouldvanish.Inotherwords,there aresituationsinwhichtheminimumof | A1A2| isnotattainedforsome A1S1,orsome A2S2,orboth.Nevertheless,thedistancebetween thesetsisstillwelldenedasthelargestnumberthatislessthanor equaltoallnumbers | A1A2| .Suchanumberiscalledthe inmum of thesetofnumbers | A1A2| anddenotedinf | A1A2| .Thus, D =inf | A1A2| ,A1S1,A2S2. Thenotation A1S1standsforapoint A1belongstotheset S1, orsimply A1isanelementof S1.Thedenitionisillustratedin Figure11.13(leftpanel). Figure11.13.Left :Distancebetweentwopointsets S1and S2denedasthelargestnumberthatislessthanor 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.

PAGE 63

75.THETRIPLEPRODUCT57 Corollary 11.6 (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 arenotparallel,andtheircrossproduct isavectorperpendiculartotheplanes(seeFigure11.13,rightpanel). Considertheparallelepipedwithadjacentsides a = 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 = AB and b = 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 = AB AC 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.

PAGE 64

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).Denethevectors 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 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 = | AC ( AB CP ) | AB CP where A and B areanytwodistinctpointson L1,while C and P are anytwodistinctpointson L2.

PAGE 65

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

PAGE 66

6011.VECTORSANDTHESPACEGEOMETRY notedinSection74.1,thecoordinates a iand aiarerelatedbyalinear transformationthatpreservesthelength, a i= vi 1a1+ vi 2a2+ vi 3a3,i =1 2 3 a = a andexcludesthereectionsofallcoordinateaxesorjustoneofthem. 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.Foraxed 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

PAGE 67

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. Foraxed 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. Therstrelationin(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) det V12( e3 e1)+det V13( e1 e2) V11= v22v23v23v33 ,V12= v21v23v31v33 ,V13= v22v22v31v32 wheretheskewsymmetryofthecrossproduct ei ej= ej eiand thedenitionofthedeterminantofa2 2matrixhavebeenused; thematrices V1 i, i =1 2 3,areobtainedbyremovingfrom V therow andcolumnthatcontain v1 i.Usingthesymmetryofthetripleproduct undercyclicpermutationsofthevectors,onehas u1 ( u2 u3)= v11det V11 v12det V12+ v13det V13 e1 ( e2 e3) Equation(11.14)followsfromthisrelationandthedenitionofthe determinantofa3 3matrix.Nowrecallthatthehandednessof acoordinatesystemispreservedunderrotations, u1 ( u2 u3)= e1 ( e2 e3)= 1,andthereforedet V =1.Itisalsoworthnoting thatacombinationofrotationsandreectionsisdescribedbymatrices V whoserowsandcolumnsareorthonormal,butdet V = 1.The handednessofacoordinatesystemischangedifdet V = 1. Thevector u3isdeterminedbyitsthreedirectionalanglesinthe originalcoordinatesystem.Onlytwooftheseanglesareindependent. Arotationabouttheaxiscontainingthevector u3doesnotaect u3andcanbespeciedbyarotationangleinaplaneperpendicularto theaxis.Thisangledeterminesthevectors u1and u2relativetothe

PAGE 68

6211.VECTORSANDTHESPACEGEOMETRY originalbasis.Soageneralrotationmatrix V hasthreeindependent parameters. Inparticular,thematrix V forcounterclockwiserotationsabout the z axisthroughanangle (seeStudyProblem11.2)is V = cos sin 0 sin cos 0 001 Relations(11.12)for V anditsrowsandcolumns v1= cos sin 0 v2= sin cos 0 v3= 0 0 1 w1= cos 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 thatsatisestheequation 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 = 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.

PAGE 69

75.THETRIPLEPRODUCT63 Problem11.23.(SystemsofLinearEquations). Considerasystemoflinearequationsforthevariables x y ,and z : a1x + b1y + c1z = d1a2x + b2y + c2z = d2a3x + b3y + c3z = d3. Denevectors 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 1 2 b = 2 1 2 ,and c = 2 1 3 (ii) a = u1+2 u2, b = u1 u2+2 u3,and c = u2 3 u3if u1 ( u2 u3)=2 (2) Verifywhetherthevectors a = e1+2 e2 e3, b =2 e1 e2+ e3, and c =3 e1+ e2 2 e3arecoplanar. (3) Findthevalueof s ,ifany,forwhichthevectors a = 1 2 3 b = 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 t b ,and a p c ,

PAGE 70

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 = ( 1 1 2),and D =( 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 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 = 1 2 2 (13) Considertwolines.Therstlinepassesthroughthepoints (1 2 3)and(2 1 1),whiletheotherpassesthroughthepoints( 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

PAGE 71

76.PLANESINSPACE65 (1 2 1). Hint: Ifthelineisnotparalleltotheplane,thentheyintersectandthedistanceis0.Socheckrstwhetherthelineisparallel totheplane.Howcanthisbedone? (15) Considerthelinethroughthepoints(1 2 3)and(2 1 2).Ifa secondlinepassesthroughthepoints(1 1 ,s )and(2 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 =( 1 2 5), C =(5 1 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, aplaneinspaceisdenedasa 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 denedbyapoint P0thatbelongstoitandanormal n .Insome coordinatesystem,thepoint P0hascoordinates( x0,y0,z0)andthe vector n isspeciedbyitscomponents n = n1,n2,n3 .Ageneric pointinspace P hascoordinates( x,y,z ).Analgebraicdescriptionof aplaneamountstospecifyingconditionsonthevariables( x,y,z )such

PAGE 72

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 P0P = r r0= x x0,y y0,z z0 .This vectorliesintheplane P ifitisorthogonaltothenormal n ,according tothegeometricaldescriptionofaplane(seeFigure11.15,leftpanel). Thealgebraicconditionequivalenttothegeometricalone, n P0P reads n 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 x0)+ n2( y y0)+ n3( z 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 n a .If n a =0,eachpointoftheplaneistranslatedwithin theplanebecause a isorthogonalto n .Theplaneasapointsetdoes notchangeandneitherdoesthenumber d .Ifthedisplacementvector a isnotorthogonalto n ,then d changesbytheamount 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 2 y z =5and 2 x +4 y +2 z =1areparallelbecausetheirnormals, n1= 1 2 1 and n2= 2 4 2 ,areproportional n2= 2 n1.

PAGE 73

76.PLANESINSPACE67 Figure11.15.Left :Algebraicdescriptionofaplane.If r0isapositionvectorofaparticularpointintheplaneand r isthepositionvectorofagenericpointintheplane,then thevector r r0liesintheplaneandisorthogonaltoits normal,thatis, n ( r 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 ( 1 0 3) Solution: Aplaneisspeciedbyaparticularpoint 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 = AB = 1 2 1 and b = AC = 2 1 2 .Then onecantake n = a b = 3 0 3 .Anequationoftheplaneis 3( x 1)+0( y 1)+( 3)( z 1)=0,or x 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.

PAGE 74

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 n ).Sotherangeof canalways berestrictedtotheinterval[0 ,/ 2].Indeed,if happenstobeinthe interval[ / 2 ](i.e.,cos 0),thentheangle / 2canalsobe viewedastheanglebetweentheplanesbecauseonecanalwaysreverse thedirectionofoneofthenormals n1 n1or n2 n2sothat cos cos Theanglebetweentheplanesisusefulfordeterminingtheirrelative orientation.Twoplanesintersectiftheanglebetweenthemisnot0. Twoplanesareparalleliftheanglebetweenthemvanishes.Theplanes areperpendiculariftheirnormalsareorthogonal.Forexample,the planes x + y + z =1and x +2 y 3 z =4areperpendicularbecause theirnormals n1= 1 1 1 and n2= 1 2 3 areorthogonal: n1 n2= 1+2 3=0(i.e., n1 n2).76.3.TheDistanceBetweenaPointandaPlane.Considertheplane throughapoint P0andnormaltoavector n .Let P1beapointin space.Whatisthedistancebetween P1andtheplane?Lettheangle between n andthevector P0P1be (seeFigure11.16,rightpanel).

PAGE 75

76.PLANESINSPACE69 Thenthedistanceinquestionis D = P0P1 cos if / 2(the lengthofthestraightlinesegmentconnecting P1andtheplanealong thenormal n ).For >/ 2,cos mustbereplacedby cos because D 0.So (11.15) D = P0P1| cos | = n P0P1| cos | n = | n P0P1| n Let r0and r1bepositionvectorsof P0and P1,respectively.Then P0P1= r1 r0,and (11.16) D = | n ( r1 r0) | n = | n r1 d | n whichisabitmoreconvenientthan(11.15)iftheplaneisdenedby anequation n r = d .DistanceBetweenParallelPlanes.Equation(11.16)allowsustoobtain asimpleformulaforthedistancebetweentwoparallelplanesdened bytheequations n r = d1and n r = d2(seeFigure11.15,rightpanel): (11.17) D = | d2 d1| n Indeed,thedistancebetweentwoparallelplanesisthedistancebetween therstplaneandapoint r2inthesecondplane.By(11.16),this distanceis D = | n r2 d1| / n = | d2 d1| / n because n r2= d2foranypointinthesecondplane. Example 11.20 Findanequationofaplanethatisparalleltothe plane 2 x y +2 z =2 andatadistanceof3unitsfromit. Solution: Thereareafewwaystosolvethisproblem.Fromthe geometricalpointofview,aplaneisdenedbyaparticularpointinit anditsnormal.Sincetheplanesareparallel,theymusthavethesame normal n = 2 1 2 .Notethatthecoecientsatthevariablesinthe planeequationdenethecomponentsofthenormalvector.Therefore, theproblemisreducedtondingaparticularpoint.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.Tondaparticular pointonthegivenplane,onecansettwocoordinatesto0andnd

PAGE 76

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

PAGE 77

76.PLANESINSPACE71 Problem11.27. Findafamilyofallplanesthatcontainsthestraight linesegment AB if A =(1 2 1) and B =(2 4 1) Solution: Alltheplanesinquestioncontainthepoint A .Soitcan bechosenasaparticularpointineveryplane.Sincethesegment AB liesineveryplaneofthefamily,thequestionamountstodescribingall vectorsorthogonalto a = AB = 1 2 2 thatdeterminethenormalsof theplanesinthefamily.Itiseasytondaparticularvectororthogonal to a .Forexample, b = 0 1 1 isorthogonalto a because a b =0. Next,thevector a b = 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 r0)=0,where0 < 2 and r0= 1 2 1 (thepositionvectorof A ). 76.5.Exercises.(1) Findanequationoftheplanethroughtheoriginandparalleltothe plane2 x 2 y + z =4.Whatisthedistancebetweenthetwoplanes? (2) Dotheplanes2 x + y z =1and4 x +2 y 2 z =10intersect? (3) Determinewhethertheplanes2 x + y z =3and x + y + z =1 areintersecting.Iftheyare,ndtheanglebetweenthem. (4) Consideraparallelepipedwithonevertexattheorigin O atwhich theadjacentsidesarethevectors a = 1 2 3 b = 2 1 1 ,and c = 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 , 1 1)and(2 0 1)andthatareatthedistance2fromthe point(1 2 3).

PAGE 78

7211.VECTORSANDTHESPACEGEOMETRY (7) Findanequationforthesetofpointsthatareequidistantfrom thepoints(1 2 3)and( 1 2 1).Giveageometricaldescriptionofthe set. (8) Findanequationoftheplanethatisperpendiculartotheplane x + y + z =1andcontainsthelinethroughthepoints(1 2 3)and ( 1 1 0). (9) Towhichoftheplanes x + y + z =1and x +2 y 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 1 .Findanequationoftheplane thatcontainsthepoint(1 2 1),thevector a ,andavectororthogonal toboth a and b (15) Considertheplane P throughthreepoints A (1 1 1), B (2 0 1), and C ( 1 3 2).Findalltheplanesthatcontainthesegment AB and havetheangle / 3withtheplane P Hint: SeeStudyProblem11.27. (16) Findanequationoftheplanethatcontainsthelinethrough (1 2 3)and(2 1, 1)andcutsthesphere x2+ y2+ z2 2 x +4 y 6 z =0 intotwohemispheres. (17) Findanequationoftheplanethatistangenttothesphere x2+ y2+ z2 2 x 4 y 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: Showrstthatanormalof

PAGE 79

77.LINESINSPACE73 theplanecanalwaysbewrittenintheform n = r1+ c ( r2 r1).Then ndaconditiontodeterminetheconstant c 77.LinesinSpace77.1.AGeometricalDescriptionofaLineinSpace.Considertheline thatcoincideswithacoordinateaxisofarectangularsystem,say,the x axis.Anypointonithasthecharacteristicpropertythatitsposition vectorisproportionaltothepositionvectorofaparticularpoint(e.g., to e1).Sincethecoordinatesystemcanbearbitrarilychosenbytranslatingtheoriginandrotatingthecoordinateaxes, alineinspaceis denedasasetofpointswhosepositionvectorsrelativetoaparticular pointinthesetareparalleltoagivennonzerovector v .Thus,the geometricaldescriptionofaline L inspaceentailsspecifyingapoint P0thatbelongsto L andavector v thatisparallelto L Remark. Considertwopointsinspace.Theycanbeconnectedby apath.Amongallthecontinuouspathsthatconnectthetwopoints, thereisadistinctone,namely,theonethathasthesmallestlength. Thispathiscalleda straightlinesegment .Alineinspacecanalso bedenedas asetofpointsinspacesuchthattheshortestpathconnectinganypairofpointsofthesetbelongstoit .Thisdenitionof thelineisdeeplyrootedintheverystructureofspaceitself.How canalineberealizedinthespaceinwhichwelive?Onecanusea pieceofrope,asintheancientworld,orthelineofsight(i.e.,the pathtraveledbylightfromonepointtoanother).Einsteinstheory ofgravitystatesthatstraightlinesdenedastrajectoriestraversed bylightarenotexactlythesameasstraightlinesinaEuclidean space.SoaEuclideanspacemayonlybeviewedasamathematical approximation(ormodel)ofourspace.Agoodanalogywouldbeto comparetheshortestpathsinaplaneandonthesurfaceofasphere; theyarenotthesame,asthelatteraresegmentsofcirclesandhence arebentorcurved.Theconceptofcurvatureofapathisdiscussed inthenextchapter.Theshortestpathbetweentwopointsinaspace iscalleda geodesic (byanalogywiththeshortestpathonthesurface oftheEarth).ThegeodesicsofaEuclideanspacearestraightlines anddonothavecurvature,whereasthegeodesicsofourspace(i.e.,the pathstraversedbylight)dohavecurvaturethatisdeterminedbythe distributionofgravitatingmasses(planets,stars,etc.).Adeviationof thegeodesicsfromstraightlinesnearthesurfaceoftheEarthisvery hardtonotice.However,adeviationofthetrajectoryoflightfrom astraightlinehasbeenobservedforthelightcomingfromadistant

PAGE 80

7411.VECTORSANDTHESPACEGEOMETRY startotheEarthandpassingneartheSun.Einsteinstheoryofgeneralrelativityassertsthatabettermodelofourspaceisa Riemann space .AsucientlysmallneighborhoodinaRiemannspacelookslike aportionofaEuclideanspace.77.2.AnAlgebraicDescriptionofaLine.Insomecoordinatesystems, aparticularpointofaline L hascoordinates P0( x0,y0,z0),andavectorparallelto L isdenedbyitscomponents, v = v1,v2,v3 .Let r = x,y,z beapositionvectorofagenericpointof L andlet r0= x0,y0,z0 bethepositionvectorof P0.Thenthevector r r0is thepositionvectorof P relativeto P0.Bythegeometricaldescription oftheline,itmustbeparallelto v .Sinceanytwoparallelvectorsare proportional,apoint( x,y,z )belongsto L ifandonlyif r r0= t v forsomereal t (seeFigure11.17,leftpanel). Figure11.17.Left :Algebraicdescriptionofaline L through r0andparalleltoavector v .If r0and r arepositionvectorsofparticularandgenericpointsoftheline, thenthevector r r0isparalleltothelineandhencemust beproportionaltoavector v ,thatis, r r0= t v forsome realnumber t Right :Distancebetweenapoint P1anda line L throughapoint P0andparalleltoavector v .Itis theheightoftheparallelogramwhoseadjacentsidesarethe vectors 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
PAGE 81

77.LINESINSPACE75 Theparametricequationsofthelinecanbesolvedfor t .Asaresult, oneinfersequationsforthecoordinates x y ,and z : t = x x0 v1= y y0 v2= z z0 v3. Theseequationsarecalled symmetric equationsofaline.Notethat theseequationsmakesenseonlyifallthecomponentsof v donot vanish.If,say, v1=0,thentherstequationin(11.19)doesnot containtheparameter t atall.Sothesymmetricequationsarewritten intheform x = x0, y y0 v2= z z0 v3. Example 11.21 Findthevector,parametric,andsymmetricequationsofthelinethroughthepoints A (1 1 1) and B (1 2 3) Solution: Take v = 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 1= z 1 2 arethevector,parametric,andsymmetricequationsoftheline, respectively. Thelineinthisexamplecanalsobeviewedasthesetofpointsof intersectionoftheplanes x =1and2 y z =1asfollowsfromthe symmetricequations.Clearly,alinecanalwaysbedescribedastheset ofpointsofintersectionoftwononparallelplanes.Sincethelinelies ineachplane,itmustbeorthogonaltothenormalsoftheseplanes. Therefore,avectorparalleltothelinecanalwaysbechosenasthe crossproductofthenormals. Example 11.22 Findthelinethatistheintersectionoftheplanes x + y + z =1 and 2 x y + z =2 Solution: Thenormalsoftheplanesare n1= 1 1 1 and n2= 2 1 1 .Sothevector v = n1 n2= 2 1 3 isparalleltotheline. Tondaparticularpointoftheline,notethatits three coordinates ( x0,y0,z0)shouldsatisfy two equationsoftheplanes.Soonecanchoose oneofthecoordinatesatwillandndtheothertwofromtheequations oftheplanes.Itfollowsfromtheparametricequations(11.19)thatif, forexample, v3 =0,thenthereisavalueof t atwhich z vanishes, meaningthatthelinealwayscontainsapointwiththevanishing z coordinate.Since v3 =0forthelineinquestion,put z0=0.Then

PAGE 82

7611.VECTORSANDTHESPACEGEOMETRY x0+ y0=1and2 x0 y0=2.Byaddingtheseequations, x0=1and hence y0=0.Theparametricequationsofthelineofintersectionare x =1+2 t y = t ,and z = 3 t DistanceBetweenaPointandaLine.Let L bealinethrough P0and parallelto v .Whatisthedistancebetweenagivenpoint P1andthe line L ?Consideraparallelogramwithvertex P0andwhoseadjacent sidesarethevectors v and P0P1asdepictedinFigure11.17(right panel).Thedistanceinquestionistheheightofthisparallelogram, whichisitsareadividedbythelengthofthebase v .If r0and r1are positionvectorsof P0and P1,then P0P1= r1 r0andhence D = v P0P1 v = v ( r1 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= 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.Howcanonendthepointof 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

PAGE 83

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.Thisdenesthreeequationsontwovariables t and s Right :Intersectionpointofaline L andaplane P Theterminalpointofthevector rttraverses L as t ranges overallrealnumbers.Ifthelineintersectstheplanedened bytheequation r n = d ,thenthereshouldexistaparticular valueof t atwhichthevector rtsatisestheequationofthe plane: rt n = d .obtained: x1+ ta1= x2+ sa2, y1+ tb1= y2+ sb2, z1+ tc1= z2+ sc2. Thissystemofequationsissolvedinaconventionalmanner,for example,byexpressing t via s fromtherstequation,substituting itintothesecondandthirdones,andverifyingthattheresultingtwo equationshavethe same solutionfor s .Notethatthesystemhasthree equationsforonlytwovariables.Itisan overdetermined system,which mayormaynothaveasolution.Iftheconditionsofpart(1)or(2) ofCorollary11.10aresatised,thenthesystemhasnosolution(the linesareparallelorskew).Iftheconditionsofpart(3)ofCorollary 11.10arefullled,thenthereisauniquesolution.Naturally,ifthelines coincide,therewillbeinnitelymanysolutions.Let( t,s )=( t0,s0)be auniquesolution.Thenthepositionvectorofthepointofintersection is r1+ t0v1or r2+ s0v2.77.4.RelativePositionsofLinesandPlanes.Consideraline L anda plane P .Thequestionofinterestistodeterminewhethertheyare

PAGE 84

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 alsosatisesanequation oftheplane r n = d (seeFigure11.18,rightpanel).Thevalueofthe parameter t thatcorrespondstothepointofintersectionisdetermined bytheequation rt n = d r0 n + t v n = d t = d r0 n v n Thepositionvectorofthepointofintersectionisfoundbysubstituting thisvalueof t intothevectorequationoftheline rt= r0+ t v Example 11.23 Apointobjectistravelingalongtheline x 1= y/ 2=( z +1) / 2 withaconstantspeed v =6 meterspersecond.Ifall coordinatesaremeasuredinmetersandtheinitialpositionvectorofthe objectis r0= 1 0 1 ,whendoesitreachtheplane 2 x + y + z =13 ? Whatisthedistancetraveledbytheobject? Solution: Parametricequationsofthelineare x =1+ s y =2 s z = 1+2 s .Thevalueoftheparameter s atwhichthelineintersects andtheplaneisdeterminedbythesubstitutionoftheseequationsinto theequationoftheplane: 2(1+ s )+2 s +( 1+2 s )=13 6 s =12 s =2 Sothepositionvectorofthepointofintersectionis r = 3 4 3 .The distancebetweenitandtheinitialpointis D = r 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 r0isthedisplacementvector oftheobjectalongitstrajectory,andhenceitslengthdeterminesthe distancetraveledbytheobject: r r0 = v t = vt ,whichshows thattheparameter t> 0isthetraveltime.

PAGE 85

77.LINESINSPACE79 Example 11.24 Findanequationoftheplane P thatisperpendiculartotheplane P1, x + y z =1 ,andcontainstheline x 1= y/ 2= z +1 Solution: Theplane P mustbeparalleltotheline( P containsit) andthenormal n1= 1 1 1 of P1(as PP1).Sothenormal n of P isorthogonaltoboth n1andthevector v = 1 2 1 thatis paralleltotheline.Therefore,onecantake n = n1 v = 3 2 1 Thelineliesin P ,andthereforeanyofitspointscanbetakenasa particularpointof P ,forexample, P0(1 0 1).Anequationof P reads 3( x 1) 2 y +( z +1)=0or3 x 2 y + z =2. Example 11.25 Findtheplanesthatareperpendiculartotheline x = y/ 2= z/ 2 andhavethedistance 3 fromthepoint ( 1 2 2) on theline. Solution: Thelineisparalleltothevector v = 1 2 2 .Sothe planeshavethesamenormal n = v .Particularpointsintheplanes arethepointsofintersectionofthelinewiththeplanes.Thesepoints areatthedistance3from r0= 1 2 2 ,andtheirpositionvectors r shouldsatisfythecondition r 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 = 2 4 4 .Equationsofthe planesare x +2 y 2 z =0and( x +2)+2( y +4) 2( z 4)=0or x +2 y 2 z = 18. 77.5.StudyProblems.Problem11.28. Let L1bethelinethrough P1(1 1 1) andparallelto v1= 1 2 1 andlet L2bethelinethrough P2(4 0 2) andparallel to v2= 2 1 0 .Determinewhetherthelinesareparallel,intersecting, orskewandndtheline L thatisperpendiculartoboth L1and L2and intersectsthem. Solution: Thevectors v1and v2arenotproportional,andhence thelinesarenotparallel.Onehas r12= P1P2= 3 1 3 and v1 v2= 1 2 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 rs rtisparallelto L ;thatis,thevector rs rtbecomesorthogonalto v1

PAGE 86

8011.VECTORSANDTHESPACEGEOMETRY Figure11.19.Left :IllustrationtoStudyProblem11.28. Thevectors rsand rttraceouttwogivenskewedlines L1and L2,respectively.Thereareparticularvaluesof t and s at whichthedistance rt 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 rs rt v1 ( rs rt) v1=4+4 s 6 t =0 rs rt v1 ( rs rt) v2=5+5 s 4 t =0 Thissystemhasthesolution t =0and s = 1.Thus,thepointswith thepositionvectors rt =0= r1and rs = 1= r2 v2= 2 1 2 belong to L .Sothevector v = rs = 1 rt =0= 1 3 1 isparallelto L Takingaparticularpointof L tobe P1(whosepositionvectoris r1), theparametricequationsread x =1+ t y =1 3 t ,and z =1 t Problem11.29. Consideralinethroughtheoriginthatisparallelto thevector v = 1 1 1 .Findthepartofthislinethatliesinsidethe sphere x2+ y2+ z2 x 2 y 3 z =9 Solution: Theparametricequationsofthelineare x = t y = t z = t Ifthelineintersectsthesphere,thenthereshouldexistparticularand

PAGE 87

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 t2 6 t =9or t2 2 t =3andhence t = 1and t =3.The pointsofintersectionare( 1 1 1)and(3 3 3).Thelinesegment connectingthemcanbedescribedbytheparametricequations x = t y = t ,and z = t ,where 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 2 y + z =1. (3) Isthelinethroughthepoints(1 2 3)and(2 1 1)perpendicular tothelinethroughthepoints(0 1 1)and(1 0 2)?Arethelines intersecting?Ifso,ndthepointofintersection. (4) Determinewhetherthelines x =1+2 t y =3 t ,and z =2 t and x +1= y 4=( z 1) / 3areparallel,skew,orintersecting.Ifthey intersect,ndthepointofintersection. (5) Findthevectorequationofthestraightlinesegmentfromthepoint (1 2 3)tothepoint( 1 1 2). (6) Let r1and r2bepositionvectorsoftwopointsinspace.Findthe vectorequationofthestraightlinesegmentfrom r1to r2. (7) Considertheplane x + y 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 = 1+ t z =2 2 t andparalleltotheplane x +2 y + z =3. (9) Findparametricequationsofthelinethatisparallelto v = 2 1 2 andgoesthroughthecenterofthesphere x2+ y2+ z2= 2 x +6 z 6.Restricttherangeoftheparametertodescribethepart ofthelinethatisinsidethesphere.

PAGE 88

8211.VECTORSANDTHESPACEGEOMETRY (10) Lettheline L1passthroughthepoint A (1 1 0)paralleltothevector v = 1 1 2 andlettheline L2passthroughthepoint B (2 0 2) paralleltothevector w = 1 1 2 .Showthatthelinesareintersecting.Findthepoint C ofintersectionandparametricequationsofthe line L3through C thatisperpendicularto L1and L2. (11) Findparametricequationsofthelinethrough(1 2 5)thatisperpendiculartotheline x 1=1 y = z andintersectsthisline. (12) Findparametricequationsofthelinethatbisectstheangleof thetriangle ABC atthevertex A if A =(1 1 1), B =(2 1 3),and C =(1 4 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) Aprojectileisredinthedirection v = 1 2 3 fromthepoint (1 1 1).Letthetargetbeadiskofradius R centeredat(2 3 6)inthe plane2 x 3 y +4 z =19.Ifthetrajectoryoftheprojectileisastraight line,determinewhetherithitsatargetintwocases R =2and R =3. (16) Consideratriangle ABC where A =(1 1 1), B =(3 1 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 Theequationthatdenesquadricsurfacesisthemostgeneralequation quadratic inallthecoordinates.Thisiswhysurfacesdenedby itarecalled quadric .Asphereprovidesasimpleexampleofaquadric surface: x2+ y2+ z2 R2=0,thatis, A = B = C =1, p = q = v =0, = = ,and D = R2,where R istheradiusofthesphere.If B = C =1, = 1,whiletheotherconstantsvanish,thequadratic equation x = y2+ z2denesacircularparaboloidwhosesymmetryaxis

PAGE 89

78.QUADRICSURFACES83 isthe x axis.Ontheotherhand,if A = B =1, = 1,whiletheother constantsvanish,theequation z = x2+ y2alsodenesaparaboloid thatcanbeobtainedfromtheformeronebyarotationaboutthe y axisthroughtheangle / 2underwhich( x,y,z ) ( z,y, x )sothat x = y2+ z2 z = y2+ x2.Thus,therearequadricsurfacesofthe sameshapedescribedbydierentequations.Thetaskhereistoclassify alltheshapesofquadricsurfaces.Theshapedoesnotchangeunderits rigidrotationsandtranslations.Ontheotherhand,theequationthat describestheshapewouldchangeundertranslationsandrotationsof thecoordinatesystem.Thefreedominchoosingthecoordinatesystem canbeusedtosimplifytheequationforquadricsurfaceandobtaina classicationofdierentshapesdescribedbyit.78.1.QuadricCylinders.Considerrstasimplerprobleminwhichthe 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 = 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 (ClassicationofQuadricCylinders) 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 formsdenethequadriccylinders: y ax2=0(paraboliccylinder) =0 x2 a2+ y2 b2=1(ellipticcylinder) A D< 0 B D< 0 ,D =0 x2 a2 y2 b2=1(hyperboliccylinder) ,AB< 0 ,D =0 TheshapesofquadriccylindersareshowninFigure11.20.Other thanquadriccylinders,thestandardequationsmaydeneplanesora

PAGE 90

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/a2 y2/b2=1.lineforsomeparticularvaluesoftheconstants A, B, D,and .For example,for A= B=1and D=0,theequation x2= y2denes twoplanes x y =0.For A= B=1and D=0,theequation x2+ y2=0denestheline x = y =0(the z axis).ProofofTheorem11.8.Let( x,y )becoordinatesinthecoordinatesystemobtainedbyarotationthroughanangle .Theequationof S in thenewcoordinatesystemisobtainedbythetransformation: ( x,y ) ( x cos y sin ,y cos + x sin ) accordingtoStudyProblem11.2.Theangle canbechosensothat theequationfor S doesnotcontainthemixedterm xy .Indeed, considerthetransformationofquadratictermsintheequationfor S : x2 cos2x2+sin2y2 2sin cos xy =1 2(1+cos(2 )) x2+1 2(1 cos(2 )) y2 sin(2 ) xy, y2 sin2x2+cos2y2+2sin cos xy =1 2(1 cos(2 )) x2+1 2(1+cos(2 )) y2+sin(2 ) xy, xy sin cos ( x2 y2)+(cos2 sin2 ) xy =1 2sin(2 )( x2 y2)+cos(2 ) xy. Afterthetransformation,thecoecientat xy becomes: p p=( B A )sin(2 )+ p cos(2 ) .

PAGE 91

78.QUADRICSURFACES85 Theangle issetsothat p=0or (11.20)tan(2 )= p A 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 B )cos(2 )+ p sin(2 )] B B=1 2[ A + B ( A B )cos(2 ) p sin(2 )] = cos + sin = cos sin where satises(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 denedbytheequation 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 x02+ ( y y0)=0 ,x0= 2 A,y0= Ax2 0 D. Afterthe translation ofthecoordinatesystem x x + x0and y y + y0,theequationisreducedto Ax2+ y =0.If =0,itdenes aparabola y ax2=0,where a = A/. Third,both Aand Bdonotvanish.Then,afterthecompletion ofsquares,theequation Ax2+ By2+ x + y + D =0hastheform A( x x0)2+ B( y y0)2+ D=0 where x0= 2 A, y0= 2 B,and D= D +1 2( Ax2 0+ By2 0).Finally, afterthetranslationoftheorigintothepoint( x0,y0),theequation becomes Ax2+ By2+ D=0 If D=0,thenthisequationdenestwostraightlines y = mx ,where m =( A/B) 1 / 2,provided Aand Bhaveoppositesigns(otherwise, theequationhasthesolution x = y =0(aline)).If D =0,thenthe equationcanbewrittenas( A/D) x2+( B/D) y2=1.Onecan

PAGE 92

8611.VECTORSANDTHESPACEGEOMETRY alwaysassumethat A/D< 0.Notethattherotationofthecoordinate systemthroughtheangle / 2swapstheaxes,( x,y ) ( y, x ),which canbeusedtoreversethesignof A/D.Nowput A/D=1 /a2and B/D= 1 /b2(dependingonwhether B/Dispositiveornegative) sothattheequationbecomes x2 a2 y2 b2=1 Whentheplusistaken,thisequationdenesanellipse.Whenthe minusistaken,thisequationdenesahyperbola.78.2.ClassicationofGeneralQuadricSurfaces.Theclassicationof generalquadricsurfacescanbecarriedoutinthesameway.The generalquadraticequationcanbewritteninthenewcoordinatesystem thatisobtainedbyatranslation(11.1)andarotation(11.13).The rotationalfreedom(threeparameters)canbeusedtoeliminatethe mixedterms: p p=0, q q=0,and v v=0.After thisrotation,thelineartermsareeliminatedbyasuitabletranslation, provided A, B,and Cdonotvanish.Thecorrespondingtechnicalities canbecarriedoutbestbylinearalgebramethods.Sothenalresult isgivenwithoutaproof. Theorem 11.9 (ClassicationofQuadricSurfaces) 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 b2 z2 c2=1(hyperboloidofonesheet) x2 a2 y2 b2+ z2 c2=1(hyperboloidoftwosheets) z c = x2 a2+ y2 b2(ellipticparaboloid) z c = x2 a2 y2 b2(hyperbolicparaboloid) .

PAGE 93

78.QUADRICSURFACES87 Figure11.21.Left :Anellipsoid.Acrosssectionbyany coordinateplaneisanellipse. Right :Anellipticdouble cone.Acrosssectionbyahorizontalplane z =constisan ellipse.Acrosssectionbyanyverticalplanethroughthe z axisistwolinesthroughtheorigin.Thesixshapesarethecounterpartsinthreedimensionsoftheconic sectionsintheplanediscussedinCalculusII.Otherthanquadriccylindersortheabovesixshapes,aquadraticequationmayalsodene 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,anellipsoidisaspherestretchedalong thecoordinateaxes(seeFigure11.21,leftpanel).Tracesofanellipsoid intheplanes x = x0, | x0| 0 Astheplane x = x0getscloserto x = a or x = 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

PAGE 94

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(asaddle).Acrosssectionbyahorizontalplane isahyperbola.Acrosssectionbyaverticalplaneisa parabola.

PAGE 95

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 kc = c b2y2,k = x2 0 a2and z 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 andthislinesatisestheequation 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

PAGE 96

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/b2 z2/c2= k ,where k = x2 0/a2,and x2/a2 z2/c2= k where k = 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/a2 k> 0 ( y cannotvanishforany x ).Verticaltracesintheplanes x = x0and y = y0are upward and downward parabolas,respectively: z z0= c b2y2,z0= cx2 0 a2and z z0= c a2x2,z0= 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 = ( c/b2) y2,whenthelatterismovedparallelsothat itsvertexremainsontheparabola z =( c/a2) x2intheperpendicular

PAGE 97

78.QUADRICSURFACES91 plane.Theobtainedsurfacehasthecharacteristicshapeofasaddle (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 sectionofthetubebytheplanes x =0and y =0arehyperbolas: y2 b2 z2 c2=1and x2 a2 z2 c2=1 So thecharacteristicgeometricalpropertyofahyperboloidofonesheet isthatitshorizontaltracesareellipsesanditsverticaltracesarehyperbolas (seeFigure11.22,leftpanel).AHyperboloidofTwoSheets.Adistinctivefeatureofthissurfaceis thatitconsistsoftwosheets(seeFigure11.22,rightpanel).Indeed, thetraceintheplane z = z0satisestheequation x2 a2+ y2 b2= z2 0 c2 1 whichhasnosolutionif z2 0/c2 1 < 0or cc or z = z0< c areellipseswhose majoraxesincreasewithincreasing | z0| .Theuppersheettouchesthe plane z = c atthepoint(0 0 ,c ),whilethelowersheettouchesthe plane z = c atthepoint(0 0 c ).Thesepointsarecalled vertices ofahyperboloidoftwosheets.Verticaltracesintheplanes x =0and y =0arehyperbolas: z2 c2 y2 b2=1and z2 c2 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

PAGE 98

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

PAGE 99

78.QUADRICSURFACES93 theequationcanwrittenintheform x2+2( y 1)2 2 2 z =0or z +1= x2 2 +( y 1)2, whichisanellipticparaboloidwiththevertexat(0 1 1)becauseit isobtainedfromthestandardequation z = x2/ 2+ y2bytheshiftof thecoordinatesystem( x,y,z ) ( x,y 1 ,z +1). Example 11.28 Classifythesurface x2 4 y2+ z2 2 x 8 z +1=0 Solution: Bycompletingthesquares,theequationistransformedto ( x 1)2 1 4 y2+4( z +1)2 4+1=0 ( x 1)2 4 +( z +1)2 y2=1 whichisahyperboloidofonesheetwhoseaxisisthelinethrough (1 0 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 ysin ,y = ycos + xsin Inthenewcoordinatesystem,theequation z =2 xy =2 x 2cos sin 2 y 2cos sin +2 xy(cos2 sin2 ) wouldhavethestandardformifthecoecientat xyvanishes.Soput = / 4.Then2sin cos =sin(2 )=1and z = x 2 y 2,whichis thehyperbolicparaboloid. 78.5.StudyProblems.Problem11.30. Classifythequadricsurface 3 x2+3 z2 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 = 2,andhence = / 4.Then A=( A + B p ) / 2=4

PAGE 100

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 x2 2 x + y + z =0 Solution: Bycompletingthesquares,theequationcanbetransformedintotheform( x 1)2+( y 1)+ z =0.Aftershiftingtheorigin tothepoint(1 1 0),theequationbecomes x2+ y z =0.Considerrotationsofthecoordinatesystemaboutthe x axis: y cos y +sin z z cos z sin y .Underthisrotation, y z (cos +sin ) y + (sin cos ) z .Therefore,for = / 4,theequationassumesoneof thestandardforms x2+ 2 y =0,whichcorrespondstoaparabolic cylinder. Problem11.32. Classifythequadricsurface x2+ z2 2 x +2 z y =0 Solution: Bycompletingthesquares,theequationcanbetransformedintotheform( x 1)2+( z +1)2 ( y +2)=0.Thelatter canbebroughtintooneofthestandardformsbyshiftingtheoriginto thepoint(1 2 1): x2+ z2= y ,whichisacircularparaboloid.Its symmetryaxisisparalleltothe y axis(thelineofintersectionofthe planes x =1and z = 1),anditsvertexis(1 2 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.

PAGE 101

78.QUADRICSURFACES95 Solution: Theparametricequationsofeachlineare x =1+ t cos y =2+ t sin ,and z = t .Therefore,( x 1)2+( y 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 = x2 z2(iii)4 x2+2 y2+ z2=4 (iv) x2 y2+ z2= 1 (v) y2+4 z2=16 (vi) x2 y2+ z2=1 (vii) x2+4 y2 9 z2+1=0 (viii) x2+ z =0 (ix) x2+9 y2+ z =0 (x) y2 4 z2=16 (2) Reduceeachofthefollowingequationstooneofthestandardform, classifythesurface,andsketchit: (i) x2+ y2+4 z2 2 x +4 y =0 (ii) x2 y2+ z2+2 x 2 y +4 z +2=0 (iii) x2+4 y2 6 x + z =0 (iv) y2 4 z2+2 y 16 z =0 (v) x2 y2+ z2 2 x +2 y =0 (3) Userotationsintheappropriatecoordinateplanetoreduceeach ofthefollowingequationstooneofthestandardformandclassifythe surface: (i)6 xy + x2+ y2=1 (ii)3 y2+3 z2 2 yz =1 (iii) x yz =0 (iv) xy z2=0 (v)2 xz + x2 y2=0

PAGE 102

9611.VECTORSANDTHESPACEGEOMETRY (4) Findanequationforthesurfaceobtainedbyrotatingtheline y =2 x aboutthe y axis.Classifythesurface. (5) Findanequationforthesurfaceobtainedbyrotatingthecurve y =1+ z2aboutthe y axis.Classifythesurface. (6) Findequationsforthefamilyofsurfacesobtainedbyrotatingthe curves x2 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+ z2 2 z =0fromabove. (9) Sketchthesolidregionboundedbythesurfaces y =2 x2 z2, y = x2+ z2 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 = y2 x2,thenthelinesthrough( a,b,c )andparallelto v = 1 1 2( b a ) and u = 1 1 2( b a ) bothlieentirelyon thisparaboloid.Deducefromthisresultthatthehyperbolicparaboloidcanbegeneratedbythemotionofastraightline.Showthat hyperboloidsofonesheet,cones,andcylinderscanalsobeobtainedby themotionofastraightline. Remark. Thefactthathyperboloidsofonesheetaregeneratedbythe motionofastraightlineisusedtoproducegeartransmissions.The cogsofthegearsarethegeneratinglinesofthehyperboloids. (13) Findanequationforthecylinderofradius R whoseaxisgoes throughtheoriginandisparalleltoavector v (14) Showthatthecurveofintersectionofthesurfaces x2 2 y2+3 z2 2 x + y z =1and2 x2 4 y2+6 z2+ x y +2 z =4liesinaplane. (15) Whatarethecurvesthatboundtheprojectionsoftheellipsoid x2+ y2+ z2 xy =1onthecoordinateplanes?

PAGE 103

CHAPTER12 VectorFunctions 79.CurvesinSpaceandVectorFunctions Todescribethemotionofapointlikeobjectinspace,itsposition vectorsmustbespeciedateverymomentoftime.Avectorisdened 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. Themostcommonlyusedrulestodeneavectorfunctionarealgebraicrulesthatspecifycomponentsofavectorfunctioninacoordinatesystemasfunctionsofarealvariable: r ( t )= x ( t ) ,y ( t ) ,z ( t ) .For example, r ( t )= 1 t, ln( t ) ,t2 or x ( t )= 1 t,y ( t )=ln( t ) ,z ( t )= t2. Unlessspeciedotherwise,thedomainofthevectorfunctionistheset D ofallvaluesof t atwhichthealgebraicrulemakessense;thatis, allthreecomponentscanbecomputedforany t from D .Intheabove example,thedomainof x ( t )is
PAGE 104

9812.VECTORFUNCTIONS Figure12.1.Left :Theterminalpointofavector r ( t ) whosecomponentsarecontinuousfunctionsof t tracesout acurveinspace. Right :Graphingaspacecurve.Drawa curveinthe xy planedenedbytheparametricequations x = x ( t ), y = y ( t ).Itistracedoutbythevector R ( t )= x ( t ) ,y ( t ) 0 .Thisplanarcurvedenesacylindricalsurface 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 ofavectorfunctiondenesacurvein 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. Considerthecurvedenedbytheparametricequations 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

PAGE 105

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 ),likeawalloftheshapedenedbythiscurve.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 ) r0 =0; thelimitisdenotedas limt t0r ( t )= r0. Theleftandrightlimits,limt t 0r ( t )andlimt t+ 0r ( t ),aredened similarly.Thisdenitionsaysthatthelengthornormofthevector r ( t ) r0approaches0as t tendsto t0.Thenormofavectorvanishes ifandonlyifthevectoristhezerovector.Therefore,thefollowing theoremholds.

PAGE 106

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 isastraightlinethatdenestheheightofhelix 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. Thistheoremreducestheproblemofndingthelimitofavector functiontotheproblemofndingthelimitsofthreeordinaryfunctions. Example 12.2 Let r ( t )= sin( t ) /t,t ln t, ( et 1 t ) /t2 .Find thelimitof r ( t ) as t 0+orshowthatitdoesnotexist. Solution: Theexistenceofthelimitsofthecomponentsofthegiven vectorfunctioncanbeinvestigatedbylHospitalsrule: 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 t 1=limt 0+(ln t ) ( t 1)=limt 0+t 1 t 2= limt 0+t =0 limt 0+et 1 t t2=limt 0+et 1 2 t =limt 0+et 2 = 1 2 wherelHospitalsrulehasbeenusedtwicetocalculatethelastlimit. Therefore,limt 0+r ( t )= 1 0 1 / 2

PAGE 107

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.BylHospitalsrule, 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 ofacontinuousvectorfunctiondenesacurveinspace .Conversely, givenacurve C asapointsetinspace,onemightaskthequestion: Whatisavectorfunctionthattracesoutagivencurveinspace?The answertothisquestionisnotunique.Forexample,aline L asapoint setinspaceisuniquelydenedbyitsparticularpointandavector v paralleltoit.If r1and r2arepositionvectorsoftwoparticularpoints of L ,thenbothvectorfunctions r1( t )= r1+ t v and r2( t )= r2 2 t v traceoutthesameline L becausethevectors 2 v and v areparallel. Thefollowing,moresophisticatedexampleisalsoofinterest.Supposeonewantstondavectorfunctionthattracesoutasemicircleof

PAGE 108

10212.VECTORFUNCTIONS radius R .Letthesemicirclebepositionedintheupperpartofthe xy plane: x2+ y2= R2and y 0.Thefollowingthreevectorfunctions traceoutthesemicircle: r1( t )= t, R2 t2, 0 R t R, r2( t )= R cos t,R sin t, 0 0 t r3( t )= R cos t,R sin t, 0 0 t Thisiseasytoseebynotingthatthe y componentsarenonnegativein thespeciedintervalsandthenormofthesevectorfunctionsisconstant 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 manyvectorfunctionswhoserangesdenethesamecurveinspace. Anotherobservationisthattherearevectorfunctionsthattraceout thesamecurveinoppositedirectionsat t increasesfromitssmallest value a toitslargestvalue b .Intheaboveexample,thevectorfunction r2( t )tracesoutthesemicirclecounterclockwise,whilethefunctions r1( t )and r3( t )dosoclockwise.Soavectorfunctiondenesthe 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[ 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

PAGE 109

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
PAGE 110

10412.VECTORFUNCTIONS circleistracedoutclockwise,whereasitistracedoutcounterclockwiseintheformerparameterization.Consequently,thevectorfunction r ( t )= R cos t, 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 rangesoverthespeciedinterval,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 latterequationissatisedforallreal t .Therefore,thecurveremains onthesurfaceoftheellipticcylinderparalleltothe y axis.Oneturn aroundtheellipseoccursas t changesfrom0to2becausethefunctions

PAGE 111

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 .Itdenesthecylindrical surfaceonwhichthecurvelies. Bottomright :Thegraph z = z ( t )=sin(4 t ),whichdenestheheightofpointsofthe curverelativetothecircleinthe xy plane.Thedotsonthe t axisindicatethepointsofintersectionofthecurvewiththe xy planeintherstquadrant.cos( t )andsin( t )havetheperiod2 / =2.Thehelixrisesby2along the y axisperturnbecause y ( t )= t .Now,tosolvetheproblem,onehas tondthevaluesof 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)= 2 3 0 .Theportionofthehelixthatliesinsidethe ellipsoidcorrespondstotherange 3 t 3. Problem12.5. Considertwocurves C1and C2tracedoutbythe vectorfunctions r1( t )= t2,t,t2+2 t 8 and r2( s )= 8 4 s, 2 s,s2+ s 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

PAGE 112

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 4 s t =2 s t2+2 t 8= s2+ s 2 Substitutingthesecondequation t =2 s intotherstequation,one ndsthat(2 s )2=8 4 s whosesolutionsare s = 2and s =1.Onehas yettoverifythatthethirdequationholdsforthepairs( t,s )=( 4 2) and( t,s )=(2 1)(otherwise,the z componentsdonotmatch).Asimplecalculationshowsthatindeedbothpairssatisfytheequation.So thepositionvectorsofthepointsofintersectionare r1( 4)= r2( 2)= 16 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, theseconditionscannotbefullledforany t because,amongallthe

PAGE 113

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: Onehastondthecomponents x ( t ), y ( t ),and z ( t )such thattheysatisfytheequationsoftheparaboloidandplanesimultaneouslyforallvaluesof t .Thisensuresthattheendpointofthevector r ( t ) remainsonbothsurfaces,thatis,tracesouttheircurveofintersection (seeFigure12.5).Considerrstthemotioninthe xy plane.Solving theplaneequationfor z z =2 2 x 2 y ,andsubstitutingthesolution intotheparaboloidequation,onends2 2 x 2 y = x2+ y2.After completingthesquares,thisequationbecomes4=( x +1)2+( y +1)2, whichdescribesacircleofradius2centeredat( 1 1).Byconstruction,thiscircleistheverticalprojectionofthecurveofintersection ontothe xy plane(theplane P0inFigure12.5).Itsparametricequationsmaybechosenas x = x ( t )= 1+2cos t y = y ( t )= 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.

PAGE 114

10812.VECTORFUNCTIONS relativetothe xy planeis z ( t )=2 2 x ( t ) 2 y ( t ).Thus, r ( t )= 1+2cos t, 1+2sin t, 6 2cos t 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 Denition12.2.ThenprovethislawusingTheorem12.1andbasic limitlawsforordinaryfunctions. Solution: Theideaissimilartotheproofofthebasiclimitlawsfor ordinaryfunctionsgiveninCalculusI.Onehastondanupperbound for | v u v0 u0| intermsof v v0 and u u0 .ByDenition12.2, thelatterquantitiesconvergeto0as t t0.Theconclusionshould followfromthesqueezeprinciple.Considertheidentities: v u v0 u0=( v v0) u + v0 u v0 u0=( v v0) u + v0 ( u u0) =( v v0) ( u u0)+( v v0) u0+ v0 ( u u0) Itfollowsfromtheinequality0 | a + b || a | + | b | andtheCauchySchwarzinequality(Theorem11.2) | a b | a b that 0 | v u v0 u0| | ( v v0) ( u u0) | + | ( v v0) u0| + | v0 ( u u0) | v v0 u u0 + v v0 u0 + v0 u u0 ByDenition12.2, v v0 0and u u0 0as t t0.So therightsideoftheaboveinequalityconvergesto0.Bythesqueeze principle,itisthenconcludedthat | v u 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 t2, ln t, cos t

PAGE 115

79.CURVESINSPACEANDVECTORFUNCTIONS109 (iv) r ( t )= ln(9 t2) ln | t | (1+ t ) / (2+ t ) (v) r ( t )= t 1 ln t, 1 t (2) Findeachofthefollowinglimitsorshowthatitdoesnotexist: (i)limt 1 t, 2 t t2, 1 / ( t2 2) (ii)limt 1 t, 2 t t2, 1 / ( t2 1) (iii)limt 0 et, sin t,t/ (1 t ) (iv)limt e t, 1 /t2, 4 (v)limt e t, (1 t2) /t2,3 t/ ( t + t ) (vi)limt 2 ,t2, 1 /3 t (vii)limt 0+ ( e2 t 1) /t, ( 1+ t 1) /t,t ln t (viii)limt 0 sin2(2 t ) /t2,t2+2 (cos t 1) /t2 (ix)limt 0 ( e2 t t ) /t,t cot t, 1+ t (x)limt e2 t/ cosh2t,t2012e t,e 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 t, ( t 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 y =0 (viii) x2+ y2 2 x =0and z = x2+ y2(6) Specifythepartsofthecurve r ( t )= sin t, cos t, 4sin2t thatlie abovetheplane z =1.

PAGE 116

11012.VECTORFUNCTIONS (7) Findthevaluesoftheparameters a and b atwhichthecurve r ( t )= 1+ at2,b 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) Provethelastlimitlawinexercise9directlyfromDenition12.2, thatis,withoutusingTheorem12.1. Hint: SeeStudyProblem12.7. (11) Let v ( t )= ( e2 t 1) /t, ( 1+ t 1) /t,t ln | t | u ( t )= sin2(2 t ) /t2,t2+2 (cos t 1) /t2 w ( t )= t2 / 3, 2 / (1 t ) 1+ t t2+ t3 UsethebasiclawsoflimitsestablishedinExercise(9)tond (i)limt 0(2 v ( t ) 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.

PAGE 117

80.DIFFERENTIATIONOFVECTORFUNCTIONS111 80.DifferentiationofVectorFunctions Definition 12.5 (DerivativeofaVectorFunction) Supposeavectorfunction r ( t ) isdenedonaninterval [ a,b ] and t0 [ a,b ] .Ifthelimit limh 0r ( t0+ h ) 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 ) x ( t ) h y ( t + h ) y ( t ) h z ( t + h ) z ( t ) h = x( t ) ,y( t ) ,z( t ) (12.1) Forexample, r ( t )= sin(2 t ) ,t2 t,e 3 t r( t )= 2cos(2 t ) 2 t 1 3 e 3 t Definition 12.6 (ContinuouslyDierentiableVectorFunction) Ifthederivative r( t ) isacontinuousvectorfunctiononaninterval [ a,b ] ,thenthevectorfunction r ( t ) issaidtobe continuouslydierentiable on [ a,b ] Higher-orderderivativesaredenedsimilarly:thesecondderivative isthederivativeof r( t ), r( t )=( r( t )),thethirdderivativeisthe derivativeof r( t ), r( t )=( r( t )),and r( n )( t )=( r( n 1)( t )),provided theyexist.80.1.DifferentiationRules.Thefollowingrulesofdierentiationofvectorfunctionscanbededucedfrom(12.1). Theorem 12.2 (DierentiationRules) Suppose u ( t ) and v ( t ) aredierentiablevectorfunctionsand f ( t ) isa

PAGE 118

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 Findtherstandsecondderivativesofthevector function r ( t )=( a + t2b ) ( c t d ) ,where a b c ,and d areconstant vectors. Solution: Bytheproductrule, r( t )=( a + t2b ) ( c t d )+( a + t2b ) ( c t d )=2 t b ( c t d ) ( a + t2b ) d r( t )=(2 t b ) ( c t d )+2 t b ( c t d ) ( a + t2b ) d =2 b ( c t d ) 2 t b d 2 t b d =2 b c 6 t b d Alternatively,thecrossproductcanbecalculatedrstandthendierentiated: r ( t )= a c t a d + t2b c t3b d r( t )= a d +2 t b c 3 t2b d r( t )=2 b c 6 t b d 80.2.DifferentialofaVectorFunction.If r ( t )isdierentiable,then (12.2) r ( t )= r ( t + t ) r ( t )= r( t ) t + u ( t ) t, where u ( t ) 0 as t 0.Indeed,bythedenitionofthederivative, u ( t )= r / t r( t ) 0 as t 0.Therefore,the componentsofthedierence r r t convergeto0fasterthan t .

PAGE 119

80.DIFFERENTIATIONOFVECTORFUNCTIONS113 Supposethat r( t0)doesnotvanish.Consideralinearvectorfunction L ( t )withtheproperty L ( t0)= r ( t0).Itsgeneralformis L ( t )= r ( t0)+ v ( t t0),where v isaconstantvector.For t closeto t0, L ( t ) isa linearapproximation of r ( t )inthesensethattheapproximation error r ( t ) L ( t ) becomessmallerwithdecreasing | t t0| .Itfollows from(12.2)that r ( t ) L ( t )=( r( t0) v ) t + u ( t ) t, t = t t0. Bythetriangleinequality | a b | a + b a + b ,the approximationerrorisboundedas r( t0) v u ( t ) r ( t ) L ( t ) | t | r( t0) v + u ( t ) If r( t0) v =0or v = r( t0),then u ( t ) r( t0) v fora sucientlysmall t because u ( t ) convergesto0as t 0(the sign meansmuchsmallerthan).Therefore,theapproximation errordecreaseslinearlywithdecreasing t : r ( t ) L ( t ) r( t0) v | t | .When v = r( t0),theapproximationerrordecreasesfaster than t : r ( t ) L ( t ) | t | = u ( t ) 0as t 0 Thus,thelinearvectorfunction L ( t )= r ( t0)+ r( t0)( t t0) isthe bestlinearapproximation of r ( t )near t = t0.Providedthe derivativedoesnotvanish, r( t0) = 0 ,thelinearvectorfunction L ( t ) denesalinepassingthroughthepoint 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 x0)(recallCalculusI).Thegraphisacurvein the xy planewhoseparametricequationsare x = t y = f ( t )orinthe vectorform r ( t )= t,f ( t ) .Theparametricequationsofthetangent linecanthereforebewrittenintheform x = t = x0+( t t0), y = y0+ f( t0)( t t0),where x0= t0.Put r0= x0,y0 .Thenthetangent lineistraversedbythelinearvectorfunction L ( t )= r0+ r( t0)( t 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 ) .

PAGE 120

11412.VECTORFUNCTIONS Inparticular,thederivativeistheratioofthedierentials, r( t )= d r /dt .Recallthatthedierential dt isanindependentvariablethat describesinnitesimalvariationsof t suchthathigherpowersof dt canbeneglected.Inthissense,thedenitionofthedierentialis thelinearizationof(12.2)in dt = t .Atanyparticular t = t0,the dierential d r ( t0)= r( t0) dt = 0 denesthetangentline 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 P0Ph= r ( t0+ h ) 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).Thedirectionofthetangentvectoralsodenestheorientation Figure12.6.Left :Asecantlinethroughtwopointsof thecurve, P0and Ph.As h getssmaller,thedirectionof thevector P0P1= r ( t0+ h ) r ( t0)becomesclosertothe tangenttothecurveat P0. Right :Thederivative r( t ) denesatangentvectortothecurveatthepointwiththe positionvector r ( t ).Italsospeciesthedirectioninwhich r ( t )traversesthecurvewithincreasing t T ( t )istheunit tangentvector.

PAGE 121

80.DIFFERENTIATIONOFVECTORFUNCTIONS115 ofthecurve,thatis,thedirectioninwhichthecurveistracedout by r ( t ). Example 12.6 Findthelinetangenttothecurve r ( t )= 2 t,t2 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 canbedenedby T ( t )= r( t ) r( t ) InSection79.3,spatialcurveswereidentiedwithcontinuousvector 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,withthedenitionabove, asmoothcurvedoesindeedhaveacontinuousunittangentvector. Therefore,ifacurvedoesnothaveacontinuousunittangentvector, itcannotbesmooth.Thisenablesustoconcludethatsomecurvesare notsmooth,basedonpropertiesdeducedfromasingleparametrization. ThisisimportantbecauseonecannotpossiblytestallparameterizationstoseewhetheroneofthemmeetstheconditionsinDenition12.8. Considertheplanarcurve r ( t )= t3,t2, 0 .Thevectorfunctionis dierentiableeverywhere, r( t )= 2 t, 3 t2, 0 ,andthederivativevanishesattheorigin, r(0)= 0 .Theunittangentvector T ( t )isnot denedat 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

PAGE 122

11612.VECTORFUNCTIONS curveisnotsmoothattheorigin.Thetangentlineistheverticalline x =0because y( x )=(2 / 3) x 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(ithasaninectionpoint 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 sin t ) y = a (1 cos t ) issmooth,where a> 0 isaparameter. Ifitisnotsmoothatparticularpoints,investigateitsbehaviornear thosepoints. Solution: FollowingtheremarkafterDenition12.8,theexistence ofacontinuousunittangentvectorhastobeveried.Let r ( t )= x ( t ) ,y ( t ) .Since x( t )= a (1 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 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 )

PAGE 123

80.DIFFERENTIATIONOFVECTORFUNCTIONS117 doesnotexist.Indeed,bylHospitalsruletheleftandrightlimitsare dierent: limt 0+T2( t )=limt 0+sin t 2sin( t/ 2) =limt 0+cos t cos( t/ 2) =1 limt 0T2( t )=limt 0sin t 2sin( t/ 2) = limt 0+cos t cos( t/ 2) = 1 Therefore T ( t ) 0 1 as t 0+,but T ( t ) 0 1 as t 0. 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 t3/ 6andcos t 1 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 ofanypointofthespheresatisesthe equation r r0 = R or( r r0) ( r r0)= R2(because a 2= a a foranyvector a ).Let r ( t )beavectorfunctionthattracesoutacurve onthesphere.Then,forallvaluesof t ,( r ( t ) r0) ( r ( t ) r0)= R2. Dierentiatingbothsidesofthelatterrelation,oneinfers r( t ) ( r ( t ) r0)=0 r( t ) r ( t ) r0. If r ( t )isthepositionvectorof P and O isthecenterofthesphere,then OP = r ( t ) r0,andhencethetangentvector r( t )at P isorthogonal to 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,te t

PAGE 124

11812.VECTORFUNCTIONS (iv) r ( t )= 3 t 2 t2 4 ,t (v) r ( t )= a + b t2 c et(vi) r ( t )= t a ( b c et) (2) Sketchthecurvetraversedbythevectorfunction r ( t )= 2 ,t 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 ( t )= 2 t 1 ,t2+1 for t = 2, 1,0. (3) Determineifthecurvetracedoutbyeachofthefollowingvector functionsissmoothforaspeciedintervaloftheparameter.Ifthe curveisnotsmoothataparticularpoint,graphitnearthatpoint. (i) r ( t )= t,t2,t3 0 t 1 (ii) r ( t )= t2,t3, 2 1 t 1 (iii) r ( t )= t1 / 3,t,t3 1 t 1 (iv) r ( t )= t5,t3,t4 1 t 1 (v) r ( t )= sin3t, 1 ,t2 / 2 t / 2 (4) Findtheparametricequationsofthetangentlinetoeachofthe followingcurvesataspeciedpoint: (i) r ( t )= t2 t,t3/ 3 2 t P0=(6 9 6) (ii) r ( t )= ln t, 2 t,t2 P0=(0 2 1) (5) Findtheunittangentvectortothecurvetraversedbythespecied vectorfunctionatthegivenpoint P0: (i) r ( t )= 2 t +1 2tan 1t,e t ,P0(1 0 1) (ii) r ( t )= cos( t ) cos(3 t ) sin( t ) ,P0(1 / 2 1 1 / 3) (6) Find r( t ) r( t )and r( t ) r( t )if r ( t )= t,t2 1 ,t3+2 (7) Isthereapointonthecurve r ( t )= t2 t,t3/ 3 2 t atwhichthe tangentlineisparalleltothevector v = 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) 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 t2 intersecttheplane x + y + z = 3?Ifnot,ndapointonthecurvethatisclosesttotheplane.What isthedistancebetweenthecurveandtheplane. Hint: Expressthe distancebetweenapointonthecurveandtheplaneasafunctionof t thensolvetheextremevalueproblem.

PAGE 125

80.DIFFERENTIATIONOFVECTORFUNCTIONS119 (11) Findthepointofintersectionoftwocurves r1( t )= 1 1 t, 3+ t2 and r2( s )= 3 s,s 2 ,s2 .Iftheangleatwhichtwocurves intersectisdenedastheanglebetweentheirtangentlinesatthepoint 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) Aspacewarshipcanrealasercannonforwardalongthetangent 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 normalplaneataspeciedpoint: (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 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 .Denethepointsofintersectionofthe lineandthecurveandndtheangleatwhichtheyintersect.

PAGE 126

12012.VECTORFUNCTIONS 81.IntegrationofVectorFunctions Definition 12.9 (DeniteIntegralofaVectorFunction) Let r ( t ) bedenedontheinterval [ a,b ] .Thevectorwhosecomponents arethedeniteintegralsofthecorrespondingcomponentsof r ( t )= x ( t ) ,y ( t ) ,z ( t ) iscalledthe deniteintegral 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 ] Bythisdenition,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, cos t 0, sin t 0 = / 2 2 0 Definition 12.10 (IndeniteIntegralofaVectorFunction) Avectorfunction R ( t ) iscalledan indeniteintegral 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 ,

PAGE 127

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 )andsatisesthe 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 ) 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 ),onends 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 1 1 3 = 1 0 3 and r ( t )= t2+1 ,t, 3 t3 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.

PAGE 128

12212.VECTORFUNCTIONS 81.1.ApplicationstoMechanics.Let r ( t )bethepositionvectorofa particleasafunctionoftime t .Therstderivative r( t )= v ( t )is calledthe velocity oftheparticle.Themagnitudeofthevelocityvector v ( t )= v ( t ) iscalledthe speed .Thespeedofacarisanumbershown onthespeedometer.Thevelocitydenesthedirectioninwhichthe particletravelsandtheinstantaneousrateatwhichitmovesinthat direction.Thesecondderivative r( t )= v( t )= a ( t )iscalledthe acceleration .If m isthemassofaparticleand F istheforceactingon theparticle,accordingtoNewtonssecondlaw,theaccelerationand forcearerelatedas F = m a Ifthetimeismeasuredinseconds,thelengthinmeters,andthemass inkilograms,thentheforceisgiveninnewtons,1N=1kg m / s2. Iftheforceisknownasavectorfunctionoftime,thenNewtons secondlawdeterminesaparticlestrajectory.Theproblemofndingthetrajectoryamountstoreconstructingthevectorfunction 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 pointisspecied.So thetrajectoryofmotionisuniquelydetermined byNewtonsequation,providedthepositionandvelocityvectorsare speciedataparticularmomentoftime ,forexample, r ( t0)= r0and v ( t0)= v0.Thelatterconditionsarecalled initialconditions .Given theinitialconditions,thetrajectoryofmotionisuniquelydenedby therelations: v ( t )= v0+ 1 m t t0F ( u ) du, r ( t )= r0+ t t0v ( u ) du iftheforceisacontinuousvectorfunctionoftime. Remark. Iftheforceisafunctionofaparticlesposition,thenNewtonsequationbecomesasystemof ordinarydierentialequations ,that is,asetofsomerelationsbetweencomponentsofthevectorfunctions, itsderivatives,andtime. Example 12.10 (MotionUnderaConstantForce) Provethatthetrajectoryofmotionunderaconstantforceisaparabola iftheinitialvelocityisnotparalleltotheforce.

PAGE 129

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 )denesthepositionoftheparticleonthe lineasafunctionoftime.Otherwise,thevector r ( t ) r0isalinear combinationoftwononparallelvectors v0and F andhencemustbe orthogonalto n = v0 F bythegeometricalpropertyofthecross product.Therefore,theparticleremainsintheplanethrough r0that isparallelto F and v0ororthogonalto n ,thatis,( r ( t ) 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 F and v0= 0 ,v0 y,v0 z .Theparametricequationsofthetrajectoryof motionassumetheform x =0, y = v0 yt ,and z = v0 zt t2F/ (2 m ).The substitutionof t = y/v0 yintothelatterequationyields z = ay2+ by where a = Fv2 0 y/ (2 m )and b = v0 z/v0 y,whichdenesaparabolain 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

PAGE 130

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.Thepointofimpactdenestherange L ( ).elevationangledoestherangeattainaspeciedvalue(e.g.,tohita target)? Toanswertheseandrelatedquestions,choosethecoordinatesystemsuchthatthe z axisisdirectedupwardfromthegroundandthe parabolictrajectoryliesinthe zy plane.Theprojectileisredfromthe point(0 0 ,h ),where h istheinitialelevationoftheprojectileabove theground(ringfromahill).InthenotationofExample12.10, F = mg ( F isnegativebecausethegravitationalforceisdirected towardtheground,whilethe z axispointsupward), v0 y= v0cos ,and v0 z= v0sin .Thetrajectoryis y = tv0cos ,z = h + tv0sin 1 2 gt2,t 0 Itisinterestingtonotethatthetrajectoryisindependentofthemass oftheprojectile.Lightandheavyprojectileswouldfollowthesame parabolictrajectory,providedtheyareredfromthesameposition, 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 .

PAGE 131

81.INTEGRATIONOFVECTORFUNCTIONS125 Thedistance L traveledbytheprojectileinthehorizontaldirection untilitlandsisthe range : L = y ( tL)= tLv0cos Forexample,iftheprojectileisredfromtheground, h =0,then tL=2 v0sin /g andtherangeis L = v2 0sin(2 ) /g .Therangeattainsits maximalvalue v2 0/g whentheprojectileisredatanangleofelevation = / 4.Theangleofelevationatwhichtheprojectilehitsatargetata givenrange L = L0is =(1 / 2)sin 1( L0g/v2 0).For h =0,theangleat which L = L ( )attainsitsmaximalvaluescanbefoundbysolvingthe equation L( )=0,whichdenescriticalpointsofthefunction 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 ( 1 4 1) andhadthevelocity 0 2 1 Solution: Thevelocityvectoris v ( t )= a ( t ) dt + c = 2 t, 3 t2, 0 + c Theconstantvector c isxedbytheinitialcondition 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)= 1 4 1 ,whichyields c = 1 4 1 .Thus, r ( t )= t2 1 ,t3+ 2 t 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 remainsataxeddistance R fromtheoriginallthetime.

PAGE 132

12612.VECTORFUNCTIONS Problem12.11. Achargedparticlemovinginamagneticeld B is subjecttotheLorentzforce F =( e/c ) v B ,where e istheelectric chargeoftheparticleand c isthespeedoflightinvacuum.Assume thatthemagneticeldisaconstantvectorparalleltothe 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 isthemagnitudeofthemagneticeldand m isthe particlemass. Solution: Newtonssecondlawreads m v= e c v B Put B = 0 0 ,B .Then v = r= R cos( t ) R sin( t ) ,v v B = RB sin( t ) RB cos( t ) 0 v= 2R sin( t ) 2R cos( t ) 0 ThesubstitutionoftheserelationsintoNewtonssecondlawyields m2R = eBR/c andhence =( eB ) / ( mc ).Since v (0)= R, 0 ,v = v, 0 ,v ,itfollowsthat R = v/ Remark. Therateatwhichthehelixrisesalongthemagneticeld isdeterminedbythemagnitude(speed)oftheinitialvelocitycomponent vparalleltothemagneticeld,whereastheradiusofthehelix isdeterminedbythemagnitudeoftheinitialvelocitycomponent vperpendiculartothemagneticeld.Aparticlemakesonefullturn aboutthemagneticeldintime T =2 / =2 mc/ ( eB );thatis,the largerthemagneticeld,thefastertheparticlerotatesaboutit.81.4.Exercises.(1) Findtheindeniteanddeniteintegralsoverspeciedintervalsfor eachofthefollowingfunctions: (i) r ( t )= 1 2 t, 3 t2 0 t 2 (ii) r ( t )= sin t,t3, cos t t (iii) r ( t )= t2,t t 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 b (2) Find r ( t )ifthederivatives r( t )and r ( t0)aregiven:

PAGE 133

81.INTEGRATIONOFVECTORFUNCTIONS127 (i) r( t )= 1 2 t, 3 t2 r (0)= 1 2 3 (ii) r( t )= t 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 1 (ii) r( t )= t1 / 3,t1 / 2, 6 t r (1)= 1 0 1 r(0)= 1 2 0 (iii) r( t )= sin t, cos t, 1 /t r ( )= 1 1 0 r( )= 1 0 2 (iv) r( t )= 0 2 6 t r (0)= 1 2 3 r (1)= 1 0 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 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,t2 16 t .Find r ( t )whenthespeedoftheparticleismaximal. (11) Aprojectileisredataninitialspeedof400m/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,

PAGE 134

12812.VECTORFUNCTIONS where vgisthevelocityoftheexhaustgasesrelativetotherocket. Showthat v ( t )= v (0) 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 2 10 .Itis supposedtodestroyatargetlocatedat(2 0 12).Thetargetcanbe destroyediftheprojectilesspeedisatleast3.1atimpact.Willthe targetbedestroyed? 82.ArcLengthofaCurve Letavectorfunction r ( t ), a t b ,traverseaspacecurve C Considerapartitionoftheinterval[ a,b ], a = t0N .Underarenementofapartition, DN 0 as N .Arenementisobtainedbyaddingapartitionpointin 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 1Pk| bethedistancebetweentwoneighboringpartitionpoints.The arclengthofacurve C isthelimit L =limN Nk =1| Pk 1Pk| wherethepartitionisrenedas N ,provideditexistsandis independentofthechoiceofpartition.If L< ,thecurveiscalled measurable or rectiable Thegeometricalmeaningofthisdenitionisrathersimple.Here thesumof | Pk 1Pk| isthelengthofapolygonalpathwithverticesat P0, P1,..., PNinthisorder.Asthepartitionbecomesnerandner, thispolygonalpathapproachesthecurvemoreandmoreclosely(see Figure12.8,leftpanel).Incertaincases,thearclengthisgivenbythe Riemannintegral.

PAGE 135

82.ARCLENGTHOFACURVE129 Figure12.8.Left :Thearclengthofacurveisdened 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 ) ,whichdenesaone-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 arenementofthepartition, DN=maxk tk 0as N and therefore tk 0forall k as N .Let r k 1= r( tk 1).The dierentiabilityof r ( t )impliesthat rk rk 1= r k 1 tk+ uk tk,

PAGE 136

13012.VECTORFUNCTIONS where uk 0 as tk 0forevery k (cf.(12.2)).Bythetriangle inequality(11.7), r k 1 tk uk tk rk rk 1 r k 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 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 a ) 0as N because uk 0as tk 0forall k ,andhence MN 0as N Itfollowsfromthesqueezeprinciplethatthelimitof N k =1 rk rk 1 as N existsandequals b a r( t ) dt Remark. Assume r ( t )iscontinuouslydierentiableon[ a,b ]butdoes notnecessarilydeneaone-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 .

PAGE 137

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 ) = 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 spacecurvedenedasapointsetinspacecanbetraversedbydierent vectorfunctions.Thesevectorfunctionsaredierentparameterizations ofthesamecurve.Forexample,asemicircleofradius R istraversed bythevectorfunctions r ( t )= R cos t,R sin t, 0 ,t [0 ] R ( u )= u, R2 u2, 0 ,u [ R,R ] Theyarerelatedtooneanotherbythecompositionrule: R ( u )= r ( t ( u )) ,t ( u )=cos 1( u/R ) or r ( t )= R ( u ( t )) ,u ( t )= R cos t. Thisexampleillustratestheconceptofa reparameterization ofacurve. Areparameterizationofacurveisachangeoftheparameterthatlabels

PAGE 138

13212.VECTORFUNCTIONS pointsofthecurve.Itmerelyreectsasimplefactthattherearemany 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 ofacurveisatechnicaltooltondparametricequationsofthecurve convenientforparticularapplications.82.2.ANaturalParameterizationofaSmoothCurve.Supposeoneis travelingalongahighwayfromtown A totown B andcomesupon anaccident.Howcanthelocationoftheaccidentbereportedtothe police?IfonehasaGPSnavigator,onecanreportcoordinateson thesurfaceoftheEarth.Thisimpliesthatthepoliceshouldusea specic(GPS)coordinatesystemtolocatetheaccident.Isitpossible toavoidanyreferencetoacoordinatesystem?Asimplerwaytodene 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 )

PAGE 139

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,thersttaskistond R ( s ).Onehas r( u )= 4 sin( u ) 4 cos( u ) 3 r( u ) =5 .

PAGE 140

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 helixatthespecieddistancefrom(4 0 0).Onesuchpointisupward alongthehelix,andtheotherisdownwardalongit.Notethat s ( t ) denedaboveisthearclengthparametercountedfrom(4 0 0)inthe directionof increasing t (upwardalongthehelix, t> 0).Accordingly, s ( t )canbecountedinthedirectionof decreasing t (downwardalong thehelix, t< 0).Inthiscase, s ( t )= 5 t> 0.Hence,theposition vectoroftheotherpointis R ( 5 / 3)= 2 2 3 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. Bydenition,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 .

PAGE 141

82.ARCLENGTHOFACURVE135 82.3.Exercises.(1) Findthearclengthofeachofthefollowingcurves: (i) r ( t )= 3cos t, 2 t, 3sin t 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,e t 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 ) t w ( t )+ t2 e3,where v w and e3aremutuallyorthogonaland v( t )= w ( t ), w( t )= v ( t ).Use thePythagoreantheoremtocalculate r( t ) (2) Findthearclengthofthecurve r ( t )= e tcos t,e tsin t,e t 0 t< Hint: Put r ( t )= e 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 2 z +3=0. (6) Let C bethecurveofintersectionofthesurfaces z2=2 y and 3 x = yz .Findthelengthof C fromtheorigintothepoint(36 18 6). (7) Foreachofthefollowingcurvesdenedbythegivenequationswith aparameter a ,ndsuitableparametricequationsandevaluatethearc lengthbetweenagivenpoint A andandagenericpoint B =( x0,y0,z0): (i) y = a sin 1( x/a ), z =( a/ 4)ln[( a x ) / ( a + x )], A =(0 0 0) (ii)( x y )2= a ( x + y ), x2 y2=9 z2/ 8, A =(0 0 0) Hint: Usethenewvariables u = x + y and v = x y tondthe parametricequations. (iii) x2+ y2= az y/x =tan( z/a ), A =(0 0 0) Hint: Usethepolarcoordinatesinthe xy planetondtheparametric equations. (iv) x2+ y2+ z2= a2, x2+ y2cosh(tan 1( y/x ))= a A =( a, 0 0) Hint: Representthesecondequationasapolargraph. (8) Reparameterizeeachofthefollowingcurveswithrespecttothe arclengthmeasurefromthepointwhere t =0inthedirectionof increasing t : (i) r = t, 1 2 t, 5+3 t (ii) r =2 t t2+1 e1+(2 t2+1 1) e3

PAGE 142

13612.VECTORFUNCTIONS (iii) r ( t )= cosh t, sinh t,t (iv) x = a ( t sin t ), y = a (1 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 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 curve)soithasthesameunittangentvectoratallitspoints.If acurvebends,thenitsunittangentvectorbecomesafunctionofits positiononthecurve.Thepositiononthecurvecanbespeciedina coordinate-andparameterization-independentwaybythearclength s countedfromaparticularpointofthecurve.If T ( s )istheunit tangentvectorasafunctionof s ,thenitsderivative T( s )vanishesfor astraightline(seeFigure12.9),whilethisisnotthecaseforageneral curve.Fromthedenitionofthederivative T( s0)=lims s0 T ( s ) T ( s0) s s0, itfollowsthatthemagnitude T( s0) becomeslargerwhenthecurve bendsmore.Foraxeddistance s s0betweentwoneighboring

PAGE 143

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 ) 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 .

PAGE 144

13812.VECTORFUNCTIONS where L = (2 R )2+ h2isthearclengthofoneturn.Dierentiating thisvectorfunctiontwicewithrespecttothearclengthparameter s r( s )= (2 /L )2R cos(2 s/L ) (2 /L )2R sin(2 s/L ) 0 = (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 (seeDenition12.13), 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 ) .

PAGE 145

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)= t 1, 2 t, 2 t =1= 1 2 2 r(1) =3 r(1)= t 2, 2 0 t =1= 1 2 0 r(1) r(1)= 4 2 4 =2 2 1 2 (1)= r(1) r(1) r(1) 3= 2 2 1 2 33= 6 27 = 2 9 Equation(12.5)canbesimpliedintwoparticularlyinteresting 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:

PAGE 146

14012.VECTORFUNCTIONS r r= 0 0 ,xy xy .Thesubstitutionofthisrelationinto(12.5) leadstothefollowingresult. Corollary 12.1 (CurvatureofaPlanarCurve) Thecurvatureofaplanarsmoothcurve r ( t )= x ( t ) ,y ( t ) 0 is = | xy xy| [( x)2+( y)2]3 / 2. Afurthersimplicationoccurswhentheplanarcurveisagraph 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 )= R sin t y( t )= R cos t x( t )= R cos t y( t )= R sin t ByCorollary12.1, ( x)2+( y)2= R2xy xy= R2 = R2 R3= 1 R Therefore,thecurvatureisconstantalongthecircleandequalsareciprocalofitsradius.Thefactthatthecurvatureisindependentofits positiononthecirclecanbeanticipatedfromtherotationalsymmetry ofthecircle(itbendsuniformly).Naturally,iftwocirclesofdierent radiipassthroughthesamepoint,thenthecircleofsmallerradius bendsmore.Notealsothatthecurvaturehasthedimensionofthe inverselength.Thismotivatesthefollowingdenition. 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

PAGE 147

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.Onesaysthatthecurvebendsinthe 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 ) R ( s ) (ortheapproximationerror)inasmallxedintervaldependsonthe orientationoftheplane.Thinkaboutaportionofthecurveoflength s andaportionofthecircleofthesamelengththathaveacommon endpoint(thepoint P )sothatthepartofthecirclecanberigidly rotatedaboutthetangentlinethrough P .For s smallenough,theerror isroughlydeterminedbythedistancebetweentheotherends.Now

PAGE 148

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 ) T2( s )) /s2 0as s 0.UsingtheTaylorapproximation foreachcomponentofthevectorfunctions r ( s )and R ( s ),onends 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 ) R ( s ) r(0) 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 x0)+ n2( y y0)+ n3( z z0)=0 n = n1,n2,n3 .

PAGE 149

83.CURVATUREOFASPACECURVE143 Proof. Itfollowsfrom(12.6)thatthesecondderivative r( t0)lies intheosculatingplanebecauseitisalinearcombinationof T ( t0)and T( t0).Hence,theosculatingplanecontainstherstandsecondderivatives 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 6 2 .Theosculatingplaneis6( x 1) 6( y 1)+2( z 1)=0or3 x 3 y + z =1. Example 12.18 Findtheosculatingcircleforthegraph y =cos(2 x ) atthepoint (0 1) Solution: Foraplanarcurve,theosculatingplaneistheplanein whichthecurvelies.Since y(0)= 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 )= 2sin(2 x ), y( x )= 4cos(2 x ),and (0)= | 4 | =4. Sothecurvatureradiusis (0)=1 / (0)=1 / 4.Thecenterofthe osculatingcirclelies1/4ofunitlengthdownfrom y =1,thatis,at (0 3 / 4).Theequationoftheosculatingcircleis x2+( y 3 / 4)2=(1 / 4)2. 83.2.StudyProblems.Problem12.12. Showthatanysmoothcurvewhosecurvaturevanishesisastraightline. Solution: Let r ( s )beanaturalparameterizationofasmoothcurve. Thenthederivativeisaunittangentvectortothecurve, T ( s )= r( s ).Bythedenitionofthecurvature, ( s )= T( s ) = r( s ) .If ( s )=0,then r( s )= 0 forall s .Therefore,theunittangentvector r( s )= T isaconstantvector.Theintegrationofthisrelationyields

PAGE 150

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 x 1] (1+ e2 x)3=0 2 e2 x 1=0 x = ln2 2 Fromtheshapeofthegraphoftheexponential,itisclearthatthe foundcriticalpointcorrespondstothe(absolute)maximumof ( x ) (maximalbending)and max= ( 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 ) R0liesintheosculatingplaneandhencemustbea linearcombinationof T0and N0,thatis, R ( t ) R0= a ( t ) N0+ b ( t ) T0. Since R ( t )traversesacircleofradius 0centeredat R0,thelengthof R ( t ) R0mustbe 0forany t .Owingtotheorthogonalityoftheunit vectors T0and N0,thisconditionyields: R ( t ) R02= 2 0 a2( t )+ b2( t )= 2 0bythePythagoreantheorem.Parametricequationsofthecirclecan betakenintheform a ( t )= 0cos t and b ( t )= 0sin t .Thevector functionthattracesouttheosculatingcircleis R ( t )= r0+ 0(1 cos t ) N0+ 0sin t T0, where t [0 2 ].Theabovechoiceof a ( t )and b ( t )hasbeenmadeso that R (0)= r0.

PAGE 151

83.CURVATUREOFASPACECURVE145 Problem12.15. Considerahelix r ( t )= R cos( t ) ,R sin( t ) ,ht where and h arenumericalparameters.Thearclengthofoneturnof thehelixisafunctionoftheparameter L = L ( ) ,andthecurvature atanyxedpointofthehelixisalsoafunctionof = ( ) .Use onlygeometricalarguments(nocalculus)tondthelimitsof 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 theparameterandthecurvatureradiusataspeciedpoint P : (i) r ( t )= t, 1 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, t2,t ,P (4 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,e t ,P (1 0 1) (x) r ( t )= sin t 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.

PAGE 152

14612.VECTORFUNCTIONS (3) Findthemaximalandminimalcurvaturesofthegraph y =cos( ax ) andthepointsatwhichtheyoccur.Sketchthegraphfor a =1and markthepointsofthemaximalandminimalcurvatures,localmaxima andminimaofcos x ,andtheinectionpoints. (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 =0atinectionpointsofthegraph. (iii)Showbyanexamplethattheconverseisnottrue,thatis,thatthe curvaturevanishesat x = x0doesnotimplythatthepoint( x0,f ( x0)) isaninectionpoint. (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 curvesataspeciedpoint: (i) r ( t )= 4 t3 / 2, t2,t ,P (4 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 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

PAGE 153

84.APPLICATIONSTOMECHANICSANDGEOMETRY147 ( x0,y0)onthecurve,where y0= f ( x0),ndtheequationoftheosculatingcircleintheform( x a )2+( y b )2= R2. Hint: Showrstthat thevector 1 ,f( x0) istangenttothegraphandavectororthogonalto itis f( x0) 1 .Thenconsidertwocases f( x0) > 0and f( x0) < 0. (15) Findtheosculatingcircleforthecycloid x = a ( t sin t ), y = a (1 cos t )atthepoint t = / 2. (16) Letasmoothcurve r = r ( t )beplanarandlieinthe xy plane.At apoint( x0,y0)onthecurve,ndtheequationoftheosculatingcircle intheform( x a )2+( y 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,onends(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 aparticlesspeed,whilethenormalaccelerationappearsonlywhen theparticlemakesaturn.Inparticular,acircularmotionwitha constantspeed, v = v0,hasnotangentialacceleration, aT=0,and

PAGE 154

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.Thebinormalisdenedby 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 .ByNewtonslaw,itsmagnitudeis

PAGE 155

84.APPLICATIONSTOMECHANICSANDGEOMETRY149 F = maN= mv2/ ,where m isthemassofamovingobject(e.g.,a car).Whenmakingaturn,thecardoesnotslideotheroadaslong asthefrictionforcebetweenthetiresandtheroadcompensatesfor thecentrifugalforce.Themaximalfrictionforcedependsontheroad andtireconditions(e.g.,awetroadandworntiresreducesubstantially themaximalfrictionforce).Thecentrifugalforceisdeterminedbythe speed(thecurvatureoftheroadisxedbytheroadshape).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] 3 / 2.Themaximalcurvatureandhencethe maximalnormalaccelerationareattainedat x =0.So,ifthespeed issuchthat aN(0)= v2 0/Ram. Thecarcanskidotheroadwhenmovingonitspartcorrespondingto theinterval R ( 1)1 / 2 x R ( 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 )canbedenedasfunctionsofthearclength s counted

PAGE 156

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 denea right-handed coordinatesystem.Theorientation oftheunittangent,normal,andbinormalvectorsrelativetosome coordinatesystemdependsonthepointofthecurve.Thetripleof thesevectorscanonlyrotateasthepointslidesalongthecurve(the vectorsaremutuallyorthogonalandunitatanypoint).Therefore,the rateswithrespecttothearclengthatwhichthesevectorschangemust becharacteristicfortheshapeofthecurve(seeFigure12.11,right panel). Bythedenitionofthecurvature, 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 = ( s ) N ( s ) wherethenumber ( s ) iscalledthe torsionofthecurve Bydenition,thetorsionismeasuredinunitsofareciprocallength, justlikethecurvature,becausetheunitvectors T N ,and B are dimensionless.

PAGE 157

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= N T + B N = B T wherethedenitionsofthetorsionandcurvaturehavebeenused.The obtainedratesoftheunitvectorsareknownasthe Frenet-Serretformulasorequations : T( s )= ( s ) N ( s ) (12.9) N( s )= ( s ) T ( s )+ ( s ) B ( s ) (12.10) B( s )= ( 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,theinitialpointofacurveisspeciedbya 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,insomespecicexamples,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.

PAGE 158

15212.VECTORFUNCTIONS Example 12.20 UsetheFrenet-Serretequationstoprovethata curvewithaconstantcurvature ( s )= 0 =0 andzerotorsion ( s )= 0 isacircle(oritsportion)ofradius R =1 /0. Solution: Avectorfunction r ( s )thatsatisestheFrenet-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= e2 e1= e3,a unitvector N orthogonalto T suchthat T N = B = e3musthave theform N = sin e1+cos e2.Equation(12.9)gives T= 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+ 1 0sin( 0s ) e1 1 0cos( 0s ) e2, where r0isaconstantvector.BythePythagoreantheorem,thedistancebetweenanypointofthecurveandaxedpoint r0isconstant: r ( s ) 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)andthedenitionofthecurvature, (12.12) r= v T + v2 N andby(12.7)andthedenitionofthebinormal, (12.13) r r= v T r= v3 B Dierentiationofbothsidesof(12.12)gives r= v T + v T+( v2+2 vv) N + v2 N.

PAGE 159

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= v T + v B Therefore, (12.14) r=( v 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 forcalculatingtheunitbinormalvectorthanitsdenition.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)= t 1, 1 ,t t =1= 1 1 1 r(1) = 3 r(1)= t 2, 0 1 t =1= 1 0 1 r(1)= 2 t 3, 0 0 t =1= 2 0 0 r(1) r(1)= 1 2 1 r(1) r(1) = 6 ,

PAGE 160

15412.VECTORFUNCTIONS T (1)= 1 3 1 1 1 B (1)= 1 6 1 2 1 N (1)= 1 6 3 1 2 1 1 1 1 = 1 3 2 3 0 3 = 1 2 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

PAGE 161

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 )= 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 = 1 2 0 .Sothespeedis v = v =3. Thetangentialaccelerationis aT= v a /v =1.As v a =2 2 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) 2 1 2 astheunitvectoralong v a Therefore,theunitnormalvectoris N = T B =(1 / 9) v ( v a )= (1 / 3) 2 2 1 .Tondthetorsionat 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= 8 / 36= 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

PAGE 162

15612.VECTORFUNCTIONS ( s )= ( s )=0,onehas w= T+ B=( ) 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 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 sin2 B T =(cos2 +sin2 ) N = N Bydierentiatingthevector u andusingtheFrenet-Serretequations, u=cos T 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 e3isdeneduptoageneralrotationabout e3. Thisfreedomisusedtoset e1= u (0),whichimpliesthatthefunction ( s )satisesthecondition (0)=0.Thentheunitnormalvectorin thisbasisis N = u w =cos e1 e3+sin e2 e3= sin e1+cos e2and u= 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 ) e1 R cos( s ) e2+ hs e3,R = 0 2,h = 0 Thisisahelixofradius R whoseaxisgoesthroughthepoint r0parallel to e3;thehelixclimbsalongitsaxisby2 h/ pereachturn.

PAGE 163

84.APPLICATIONSTOMECHANICSANDGEOMETRY157 Problem12.19.(MotioninaConstantMagneticField,Revisited) Theforceactingonachargedparticlemovinginthemagneticeld B is givenby F =( e/c ) v B ,where e istheelectricchargeoftheparticle, c isthespeedoflight,and v isitsvelocity.Showthatthetrajectoryof theparticleinaconstantmagneticeldisahelixwhoseaxisisparallel tothemagneticeld. Solution: IncontrasttoStudyProblem12.11,heretheshapeofthe trajectoryistobeobtaineddirectlyfromNewtonssecondlawwith arbitraryinitialconditions.Choosethecoordinatesystemsothatthe magneticeldisparalleltothe z axis, B = B e3,where B isthe magnitudeofthemagneticeld.Newtonslawofmotion, 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 0 v2 )1 / 2.Thevelocityvectorcan thereforebewrittenintheform v = vcos ,vsin ,v ,wherethe function = ( t )istobedeterminedbytheequationsofmotion: a = B v e3= B vsin ,vcos 0 a = v= 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) R cos( t + 0) ,vt where R = v/ .Thisequationdescribesahelixofradius R whose axisgoesthrough r0paralleltothe z axis.Soachargedparticlemoves alongahelixthatwindsaboutforcelinesofthemagneticeld.The particlerevolvesintheplaneperpendiculartothemagneticeldwith frequency = eB/ ( mc ).Ineachturn,theparticlemovesalongthe magneticeldadistance h =2 v/ .Inparticular,iftheinitial

PAGE 164

15812.VECTORFUNCTIONS velocityisorthogonaltothemagneticeld(i.e., v=0),thenthe trajectoryisacircleofradius R ThePolarLights .TheSunproducesastreamofchargedparticles (thesolarwind).ThemagneticeldoftheEarthplaystheroleof 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 .ByNewtonssecondlaw, 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.(KeplersLawsofPlanetaryMotion). Newtonslawofgravitystatesthattwomasses m and M atadistance r areattractedbyaforceofmagnitude GmM/r2,where G istheuniversal constant(called Newtonsconstant ).ProveKeplerslawsofplanetary 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

PAGE 165

84.APPLICATIONSTOMECHANICSANDGEOMETRY159 theunitvectorparallelto r .Thenthegravitationalforceis F = GMm r2 r = GMm r3r where M isthemassoftheSunand m isthemassofaplanet.The minussignisnecessarybecauseanattractiveforcemustbeopposite tothepositionvector.ByNewtonssecondlaw,thetrajectoryofa planetsatisestheequation m a = F andhence a = 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 = GM r2 r l = 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,

PAGE 166

16012.VECTORFUNCTIONS Figure12.12.Left :Thesetupofthecoordinatesystem forthederivationofKeplersrstlaw. Right :AnillustrationtothederivationofKeplerssecondlaw.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 notconnedbythegravitationalpulloftheSun)shouldfolloweither parabolicorhyperbolictrajectories. ToproveKeplerssecondlaw,put r = cos sin 0 andhence r= 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 .

PAGE 167

84.APPLICATIONSTOMECHANICSANDGEOMETRY161 Foranymomentsoftime t1and t2,theareaofthesectorbetween r ( t1) and r ( t2)is A12= t2t1dA dt dt = t2t1l 2 dt = l 2 ( t2 t1) Thus,thepositionvector r sweepsoutequalareasinequaltimes(see Figure12.12,rightpanel). Keplersthirdlawfollowsfromthelastequation.Indeed,theentire areaoftheellipse A issweptwhen t2 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,Keplerslawsare universal forallmassive objectstrappedbytheSun(planets,asteroids,andcomets). 84.5.Exercises.(1) Foreachofthefollowingtrajectoriesofaparticle,ndthevelocity, speed,andnormalandtangentialaccelerationsasfunctionsoftimeand theirvaluesataspeciedpoint P : (i) r ( t )= t, 1 t,t2+1 ,P (1 0 2) (ii) r ( t )= t2,t, 1 ,P (4 2 1) (iii) r ( t )= 4 t3 / 2, t2,t ,P (4 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,e t ,P (1 0 1) (vii) r ( t )= sin t t cos t,t2, cos t + t sin t ,P (0 0 1) (2) Findthenormalandtangentialaccelerationsofaparticlewiththe positionvector r ( t )= t2+1 ,t,t2 1 whentheparticleisclosestto theorigin. (3) Findthetangentialandnormalaccelerationsofaparticlewiththe positionvector r ( t )= R sin( t + 0) 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

PAGE 168

16212.VECTORFUNCTIONS aspeedlimitforthisportionoftheroad. (5) Aparticlemovesalongthecurve y = x2+ x3.Iftheacceleration oftheparticleatthepoint(1 2)is a = 3 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, andbinormalvectorsandthetorsionataspeciedpoint P : (i) r ( t )= t, 1 t,t2+1 ,P (1 0 2) (ii) r ( t )= t3,t2, 1 ,P (8 4 1) (iii) r ( t )= 4 t3 / 2, t2,t ,P (4 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 ) t w ( t )+ t2 e3,where v w ,and e3aremutuallyorthogonalunitvectors suchthat v( t )= w ( t ), w( t )= v ( t ).Usethepropertiesofthecross productsofmutuallyorthogonalunitvectors. (12) Let C bethecurveofintersectionofanellipsoid x2/a2+ y2/b2+ z2/c2=1withtheplane2 x 2 y + z =0.Findthetorsionandthe binormal B along C .

PAGE 169

CHAPTER13 DifferentiationofMultivariable Functions 85.FunctionsofSeveralVariables Theconceptofafunctionofseveralvariablescanbequalitatively understoodfromsimpleexamplesineverydaylife.Thetemperaturein aroommayvaryfrompointtopoint.Apointinspacecanbedenedby 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 axedpoint(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

PAGE 170

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 b =0.Finally,thedot product a b oftwoelementsisdenedinthesamewayfortwo-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,andthenormaredenedbytheserules iscalledan n dimensionalEuclideanspace TwopointsofaEuclideanspacearesaidtocoincide, a = b ,ifthe correspondingcomponentsareequal,thatis, ai= bifor i =1 2 ,...,n Itfollowsthat a = b ifandonlyif a b =0.Indeed,bythe denitionofthenorm, c =0ifandonlyif c =(0 0 ,..., 0).Put c = a b .Then a b =0ifandonlyif a = b .Thenumber a b iscalledthe distance betweenpoints a and b ofaEuclideanspace. ThedotproductinaEuclideanspacehasthesamegeometrical propertiesasintwoandthreedimensions.TheCauchy-SchwarzinequalitycanbeextendedtoanyEuclideanspace(cf.Theorem11.2). Theorem 13.1 (Cauchy-SchwarzInequality) | a b | a b

PAGE 171

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 t b 2=( a t b ) ( a t b ) 0. Therefore, a2 2 tc + t2b2 0,where c = a b .Completingthesquares ontheleftsideofthisinequality, bt c b 2 c2 b2+ a2 0 showsthattheleftsideattainsitsabsoluteminimumwhentheexpressionintheparenthesesvanishes,thatis,at t = c/b2.Sincetheinequalityisvalidforany t ,itissatisedfor t = c/b2,thatis, a2 c2/b2 0 or c2 a2b2or | c | ab ,whichistheconclusionofthetheorem.The inequalitybecomesanequalityifandonlyif a 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 onetodene 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 } ThisdenitionisillustratedinFigure13.1.Therulemaybedened bydierentmeans.If D isaniteset,afunction f canbedenedby 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 denedgeometrically.Forexample,theheightofamountainrelative tosealevelisafunctionofitspositionontheglobe.Sotheheightis afunctionoftwovariables,thelongitudeandlatitude.Afunctioncan bedenedbyanalgebraicrulethatprescribesalgebraicoperationsto

PAGE 172

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 )= x2 y + z3.Thevalueofthis functionat(1 2 3)is f (1 2 3)=12 2+33=28.Unlessspecied otherwise,thedomain D ofafunctiondenedbyanalgebraicruleis thesetof n -tuplesforwhichtherulemakessense. Example 13.1 Findthedomainandtherangeofthefunctionof twovariables f ( x,y )=ln(1 x2 y2) Solution: Thelogarithmisdenedforanystrictlypositivenumber. Therefore,thedoublet( x,y )mustbesuchthat1 x2 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 x2 y2< 1.Sotherangeof f isthesetofvaluesofthe logarithmintheinterval(0 1],whichis
PAGE 173

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 :Thegraphofafunctionoftwovariablesisthesurfacedenedbytheequation 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 x2 1 x2 2 x2 n)1 / 2=(1 r 2)1 / 2isthesetofpointsinthe n -dimensionalEuclideanspacewhosedistance fromtheorigin(thezerovector)doesnotexceed1, D = { r | r 1 } ; thatis,itisan n -dimensionalballofradius1.Sothedomainofa multivariablefunctiondenedbyanalgebraicrulecanbedescribedby 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 } .

PAGE 174

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).Onecanthinkofthegraphasmountainsof height f ( x,y )onthe xy plane. Example 13.3 Sketchthegraphofthefunction f ( x,y )= 1 ( x/ 2)2 ( y/ 3)2. Solution: Thedomainistheportionofthe xy plane( x/ 2)2+( y/ 3)2 1;thatis,itisboundedbytheellipsewithsemiaxes2and3.Thegraph isthesurfacedenedbytheequation z = 1 ( x/ 2)2 ( y/ 3)2.By squaringbothsidesofthisequation,onends( x/ 2)2+( y/ 3)2+ z2= 1,whichdenesanellipsoid.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 .Thecurveofintersectionisdenedbytheequation f ( x,y )= k .Continuingtheanalogythat f ( x,y )denestheheightof amountain,ahikertravelingalongthepath f ( x,y )= k doesnothave toclimbordescendastheheightalongthepathremainsconstant. Definition 13.4 (LevelSets) Thelevelsetsofafunction f aresubsetsofthedomainof f onwhich thefunctionhasaxedvalue;thatis,theyaredeterminedbytheequation f ( r )= k ,where k isanumberfromtherangeof f Forfunctionsoftwovariables,theequation f ( x,y )= k generally denesacurve,butnotnecessarilyso.Forexample,if f ( x,y )= x2+ y2, thentheequation x2+ y2= k denesconcentriccirclesofradii 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 denesacurve.Recallthat

PAGE 175

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 denesacurveifthere 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 ( x/ 2)2 ( y/ 3)2= k2or( x/a )2+( y/b )2=1,where a =2 1 k2and b =3 1 k2.The levelsetfor k =1consistsofasinglepoint,theorigin. Acontourmapisausefultoolforstudyingthequalitativebehavior ofafunction.Considerthecontourmapthatconsistsoflevelcurves Ci, i =1 2 ,... f ( x,y )= ki,where ki +1 ki= k isxed.Thevalues ofthefunctionalongtheneighboringcurves Ciand Ci +1dierby k So,intheregionwherethelevelcurvesaredense(closetooneanother),

PAGE 176

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 )=( a2 ( x2+ y2/ 4))2. Solution: Thefunctiondependsonthecombination u2= x2+ y2/ 4, f ( x,y )=( a2 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 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 denesasurfaceinspace,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

PAGE 177

85.FUNCTIONSOFSEVERALVARIABLES171 in D Apointset S isasurfaceinspaceifeverypointof S hasa neighborhoodthatcanbeobtainedbyacontinuousdeformation(ora deformationwithoutbreaking)ofanopensetinaplaneandthisdeformationhasacontinuousinverse .Thisisanalogoustothedenitionof acurveasapointsetinspacegiveninSection79.3. Whenthelevelsetsofafunctionofthreevariablesaresurfaces, theyarecalled levelsurfaces .Theshapeofthelevelsurfacesmaybe studied,forexample,bythemethodofcrosssectionswithcoordinate planes.Acollectionoflevelsurfaces Si, f ( x,y,z )= ki, ki +1 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 k = k ( x2+ y2),whichdenesacircularparaboloidwhose 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/ ( y2 4 x2) (iv) f ( x,y )=ln(9 x2 ( y/ 2)2) (v) f ( x,y )= 1 ( x/ 2)2 ( y/ 3)2(vi) f ( x,y )= 4 x2 y2+2 x ln y (vii) f ( x,y )= 4 x2 y2+ x ln y2(viii) f ( x,y )= 4 x2 y2+ln(1 x2 ( y/ 2)2) (ix) f ( x,y,z )= x/ ( yz ) (x) f ( x,y,z )= x/ ( x y2 z2) (xi) f ( x,y,z )=ln(1 z + x2+ y2) (xii) f ( x,y,z )= x2 y2 z2+ln(1 x2 y2 z2)

PAGE 178

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,itisdenedbytheequation f ( x,y,z )= k .Herethreelevelsurfacesaredepicted. Right :LevelsurfacesofthefunctionstudiedinExample 13.5.Here k2>k1> 0and k4
PAGE 179

86.LIMITSANDCONTINUITY173 (ii) f ( x,y )= | x | + | y || 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 1forpositiveandnegative a ,respectively (vi) f ( x,y,z )=( x + y )2+ z2(vii) f ( x,y )=tan 12 ay x2+ y2 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 a,y 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 )denedinexercise5. Analyzecarefullyvariouscasesforvaluesoftheconstants,forexample, m> 0, m< 0, p> 1,0 0. (8) Find f ( x,y )if f ( x + y,y/x )= x2 y2. (9) Let z = y + f ( x 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
PAGE 180

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 accuratelydened. Asnotedbefore,thedomainofafunction f ofseveralvariablesisa setinan n -dimensionalEuclideanspace.Twopoints x =( x1,x2,...,xn) and y =( y1,y2,...,yn)coincideifandonlyifthedistancebetweenthem x y = ( x1 y1)2+( x2 y2)2+ +( xn yn)2vanishes. Definition 13.6 Apoint r issaidtoapproachaxedpoint r0inaEuclideanspaceifthedistance r r0 tendsto0.Thelimit r 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 )istherstquadrant,includingitsboundariesexceptthepoint(0 0).Thepoints(0 0) and( 1 1)arenotinthedomainofthefunction.However,the limitof f as( x,y ) (0 0)canbedened,whereasthelimitof f as ( x,y ) ( 1 1)doesnotmakeanysense.Thedierencebetween thesetwopointsisthatanyneighborhoodof(0 0)containspointsof thedomain,whilethisisnotsofor( 1 1).Sothelimitcanbe denedonlyforsomespecialclassofpointscalled limitpoints ofa set D Definition 13.7 (LimitPointofaSet) Apoint r0issaidtobea limitpoint ofaset D ifanyopenball N( r0)= { r | 0 < r r0 < } (withthecenter r0removed)containsapointof D Alimitpoint r0of D mayormaynotbein D ,butitcanalways beapproachedfromwithintheset D inthesensethat r r0and r D because,nomatterhowsmall is,onecanalwaysndapoint r D thatdoesnotcoincidewith r0andwhosedistancefrom r0is lessthan .Inotherwords,anintersectionofanyball N( r0)centered atalimitpointof D withtheset D ,denotedas N( r0) D ,isalways

PAGE 181

86.LIMITSANDCONTINUITY175 nonempty.Intheaboveexampleof D beingtherstquadrant,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 r0 < then | f ( r ) f0| < .Inthiscase,onewrites limr r0f ( r )= f0. Thenumber | f ( r ) f0| determinesthedeviationofthevalueof f fromthenumber f0.Theexistenceofthelimitmeansthatnomatter howsmallthenumber is,thereisaneighborhood N( r0) D ,which containsallpointsof D whosedistancefrom r0islessthananumber andinwhichthevaluesofthefunction f deviatefromthelimitvalue f0nomorethan ,thatis, f0 0.Toestablishtheexistenceof > 0,notethat theinequality8 R3< or R<3 / 2guaranteesthat | f ( r ) f0| < .Therefore,forallpoints r = 0 inthedomainofthefunctionfor which R< =3 / 2,thefunctiondiersfrom0nomorethan Forexample,put =10 6.Then,intheinteriorofaballofradius

PAGE 182

17613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS z y x f0+ f0f0 D r0y x C1C2r0 = ( r0)Figure13.5.Left :AnillustrationofDenition13.8in thecaseofafunction f oftwovariables.Givenapositive number ,considertwohorizontalplanes z = f0 and z = f0+ .Thenonecanalwaysndacorrespondingnumber > 0andthedisk Ncenteredat r0suchthattheportion ofthegraph z = f ( r )abovetheintersection N D lies betweentheplanes: f0 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 than10 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 x0)2+( y y0)2.Therefore, | x x0| R and | y y0| R Considertheidentity xy x0y0=( x x0)( y y0)+ x0( y y0)+( x x0) y0.

PAGE 183

86.LIMITSANDCONTINUITY177 Put a =( | x0| + | y0| ) / 2.Thenthedeviationof f fromthelimitvalue f0= x0y0isboundedas | f ( x,y ) f0|| x x0|| y y0| + | x0|| y y0| + | x x0|| y0| R2+( | x0| + | y0| ) R = R2+2 aR =( R + a )2 a2. Nowx > 0andassumethat R issuchthat( R + a )2 a2< or 0 0.Then,bytheexistenceofthelimit f0,thereisaballof radius = ( r0) > 0centeredat r0suchthatthevaluesof f liein theinterval f0
PAGE 184

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 ) g ( r ) h ( r ) g ( r ).Put F ( r )= f ( r ) g ( r )and H ( r )= h ( r ) g ( r ). Then0 F ( r ) H ( r )implies | F ( r ) || H ( r ) | (thepositivityof F isessentialforthisconclusion).Bythehypothesisofthetheoremand thebasicpropertiesofthelimit, H ( r )= h ( r ) g ( r ) f0 f0=0as r r0.Hence,forany > 0,thereisacorrespondingnumber such that0 | F ( r ) || H ( r ) | < whenever0 < r r0 < .ByDenition 13.8,thismeansthatlimr r0F ( r )=0.Bythebasicpropertiesofthe limit,itisthenconcludedthat f ( r )= F ( r )+ g ( r ) 0+ f0= f0as r r0. Aparticularcaseofthesqueezeprincipleisalsouseful. Corollary 13.1 (SimpliedSqueezePrinciple) Ifthereexistsafunction h ofonevariablesuchthat | f ( r ) f0| h ( R ) 0as r r0 = R 0+, then limr r0f ( r )= f0. Thecondition | f ( r ) f0| h ( R )isequivalentto f0 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
PAGE 185

86.LIMITSANDCONTINUITY179 Solution: Let R = x2+ y2(thedistancefromthelimitpoint (0 0)).Then | x | R and | y | R .Therefore, | x3y 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 4( x2+ y2) f ( x,y ) 4( x2+ y2),and,bythesqueezeprinciple, f ( x,y )musttendto0because 4( x2+ y2)= 4 R2 0as R 0.InDenition13.8,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 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 suchpointonecanndaneighborhood( x x0)+( y y0)2<2(a diskofradius > 0centeredat( x0,y0))thatliesintheregion y>x Therefore, | f ( r ) f ( r0) | =1 1=0 < forany > 0inthisdisk, thatis,limr r0f ( r )= f ( r0)=1.Thesamelineofreasoningapplies toestablishthecontinuityof f atanypoint( x0,y0),where y0
PAGE 186

18013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Inonepart( y x ), f ( r )=1,whereasintheotherpart( y 0inwhich | f ( r ) f ( r0) | = | f ( r ) 1 | < because | f ( r ) 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 ) f ( r0) | = | x1|k1| x2|k2| xn|kn Rk1+ k2+ + kn= RN 0

PAGE 187

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 ruledenesalargeclassofcontinuousfunctionsofseveralvariables, whichissucientformanypracticalapplications. Example 13.10 Findthelimit limr 0exzcos( xy + z2) x + yz +3 xz4+( xyz 2)2. Solution: Thefunctionisaratio.Thedenominatorisapolynomial andhencecontinuous.Itslimitvalueis( 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) Usethedenitionofthelimittoverifyeachofthefollowinglimits (i.e.,given > 0,ndthecorresponding ( )): (i)limr 0x3 4 y2x +5 y3 x2+ y2=0 (ii)limr 0x3 4 y2x +5 y3 3 x2+4 y2=0 (iii)limr 0x3 4 y4+5 y3x2 3 x2+4 y2=0

PAGE 188

18213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (iv)limr 0x3 4 y2x +5 y3 3 x2+4 y2+ y4=0 (v)limr 03 x3+4 y4 5 z5 x2+ y2+ z2=0 (2) Usethesqueezeprincipletoprovethefollowinglimitsandnda neighborhoodofthelimitpointinwhichthedeviationofthefunction fromthelimitvaluedoesnotexceedasmallgivennumber : (i)limr 0y sin( x/ y )=0 (ii)limr 0[1 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 )=( x2 y2) / ( x y )if x = y and f ( x,x )=2 x (viii) f ( x,y )=( x2 y2) / ( x y )if x = y and f ( x,x )= x (ix) f ( x,y,z )=1 / [sin( x )sin( z 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

PAGE 189

87.AGENERALSTRATEGYFORSTUDYINGLIMITS183 (iii) f ( x,y )= xnymsin(1 / ( x2+ y2)), n 0, m 0,and n + m> 0 (6) Usethepropertiesofcontinuousfunctionstondthefollowing limits (i)limr 0(1+ x + yz2)1 / 3 2+3 x 4 y +5 z2(ii)limr 0sin( x y ) (iii)limr 0sin( x y ) cos( x2y ) (iv)limr 0[ exyz 2cos( yz )+3sin( xy )] (v)limr 0ln(1+ x2+ y2z2) 87.AGeneralStrategyforStudyingLimits Thedenitionofthelimitgivesonlythecriterionforwhethera number f0isthelimitof f ( r )as r r0.Inpractice,however,a possiblevalueofthelimitistypicallyunknown.Somestudiesare neededtomakeaneducatedguessforapossiblevalueofthelimit. 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 x3 y2= 2 3 becausethefunctioninquestionisarationalfunctionthatiscontinuous if x3 y2 =0.Thelatterisindeedthecaseforthelimitpoint(1 2). Ifthecontinuityargumentdoesnotapply,thenitishelpfultocheck thefollowing.

PAGE 190

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 signicanceofthistheoremisthat,underthehypothesesofthetheorem,atoughproblemofstudyingamultivariablelimitisreducedto theproblemofthelimitofafunctionofasingleargument.Thelatter problemcanbestudiedby,forexample,lHospitalsrule.Itmustbe emphasizedthat thereisnoanalogoflHospitalsruleformultivariable limits Example 13.11 Find lim( x,y ) (0 0)cos( xy ) 1 x2y2. Solution: Thefunctioninquestionis g ( t )=(cos t 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 =0isnotdened.UsinglHospitals ruletwice, limt 0cos t 1 t2=limt 0 sin t 2 t =limt 0 cos t 2 = 1 2 So,bysetting g (0)= 1 / 2,thefunction g ( t )becomescontinuousat t = 0,andthehypothesesofthecompositionrulearefullled.Therefore, thetwodimensionallimitinquestionexistsandequals 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 x 0f ( x )= f0 limx x0f ( x )= f0. Inotherwords,ifthelimitexists,itdoesnotdependonthedirection fromwhichthelimitpointisapproached.Iftheleftandrightlimits existbutdonotcoincide,thenthelimitdoesnotexist.

PAGE 191

87.AGENERALSTRATEGYFORSTUDYINGLIMITS185 Forfunctionsofseveralvariables,thereareinnitelymanypaths 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 inaplaneorspaceasavectorfunctiondenedbytheparametric 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 r0 < } inwhichthevaluesof f deviatefrom f0nomorethan | f ( r ) f0| < Sincethecurve C passesthrough r0,thereshouldbeaportionofitthat liesin N( r0);thatis,thereisanumber suchthat r ( t ) r0 < for all t ( t0,t0+ ),whichismerelythedenitionofthelimit r ( t ) r0as t t+ 0.Hence,forany > 0,thedeviationofvaluesof f along thecurve, F ( t )= f ( r ( t )),doesnotexceed | F ( t ) f0| < whenever 0 < | t t0| <.Bythedenitionoftheone-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.

PAGE 192

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 isxed).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 isdenedonlyforstrictlypositive x and y sothatthehalf-lines x =0, y> 0and y =0, x> 0arelimitpoints ofthedomain).Inthiscase,thehypothesesofCorollary13.2arenot fullled,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

PAGE 193

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: Thedomainofthefunctionconsistsoftherstandthird quadrantsas xy 0excepttheorigin.Linesapproaching(0 0)from withinthedomainare x = t y = at a 0and t 0.Theline x =0, y = t alsoliesinthedomain(thelinewithaninniteslope).The limitalongastraightlineapproachingtheoriginfromwithintherst 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 wherelHospitalsrulehasbeenusedtocalculatethelimit.Thelimit dependsontheslopeoftheline,andhencethetwo-variablelimitdoes notexist. LimitsAlongPowerCurves(Optional).Ifthelimitalongstraightlines existsandisindependentofthechoiceoftheline,thenumericalvalue ofthislimitprovidesadesirededucatedguessfortheactualmultivariablelimit.However,thishasyettobeprovedbymeansofeither thedenitionofthemultivariablelimitor,forexample,thesqueeze principle.Thiscomprisesthelaststepoftheanalysisoflimits(Step4; seebelow).

PAGE 194

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,thenaneducatedguess forapossiblevalueofthelimitishelpful.ThisistheoutcomeofStep 3.Iflimitsalongafamilyofcurves(e.g.,straightlines)happentobe thesamenumber f0,thenthisnumberisthesought-aftereducated guess.Thedenitionofthemultivariablelimitorthesqueezeprinciple canbeusedtoproveordisprovethat f0isthemultivariablelimit.

PAGE 195

87.AGENERALSTRATEGYFORSTUDYINGLIMITS189 Example 13.15 Findthelimitorprovethatitdoesnotexist: lim( x,y ) (0 0)sin( xy2) x2+ y2. Solution: Step1 .Thefunctionisnotdenedattheorigin.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 = t2andlHospitalsrulehavebeen usedtocalculatethelimit). Step4 .Ifthetwo-variablelimitexists,thenitmustbeequalto0. Thiscanbeveriedbymeansofthesimpliedsqueezeprinciple;that is,onehastoverifythatthereexists h ( R )suchthat | f ( x,y ) f0| = | f ( x,y ) | h ( R ) 0as R = x2+ y2 0.Akeytechnicaltrickhere istheinequality | sin u || u | ,whichholdsforanyreal u .Onehas | f ( x,y ) 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 x0= R cos y y0= R sin Theideaistondoutwhetherthedeviationofthefunction f ( x,y ) from f0(theeducatedguessfromStep3)canbeboundedby h ( R ) uniformly forall [0 2 ]: | f ( x,y ) f0| = | f ( x0+ R cos ,y0+ R sin ) f0| h ( R ) 0 as R 0+.Thistechnicaltaskcanbeaccomplishedwiththehelpof thebasicpropertiesoftrigonometricfunctions,forexample, | sin | 1, | cos | 1,andsoon.

PAGE 196

19013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS InExample13.15,Step3gives f0=0ifonlythelimitsalong straightlineshavebeenstudied.Then | f ( x,y ) 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 ) f0| < isviolated inthatsectorofanydisk0 < r 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) f0| h ( R ) 0 ,R 0+. Thistechnical,oftenratherdicult,taskmaybeaccomplishedusing theinequalities | ui| 1andsomespecicpropertiesofthefunction f Asnoted,thevariables uiarethedirectionalcosines.Theycanalsobe trigonometricfunctionsoftheanglesinthesphericalcoordinatesystem inan n -dimensionalEuclideanspace.87.5.InniteLimitsandLimitsatInnity.Supposethatthelimitofa multivariablefunction f doesnotexistas r r0.Therearetwo particularcases,whichareofinterest,when f tendstoeitherpositive ornegativeinnity. Definition 13.12 (InniteLimits). Thelimitof f ( r ) as r r0issaidtobethe positiveinnity if,for anynumber M> 0 ,thereexistsanumber > 0 suchthat f ( r ) >M whenever 0 < r r0 < .Similarly,thelimitissaidtobethe negative innity if,foranynumber M< 0 ,thereexistsanumber > 0 such that f ( r )
PAGE 197

87.AGENERALSTRATEGYFORSTUDYINGLIMITS191 respectively, limr r0f ( r )= andlimr r0f ( r )= 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,ifthelimitisinnite,thefunction f approachesthe innitevaluealonganycurvethatleadstothelimitpoint.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 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 ,onecandenethelimitsatinnity (i.e.,when x + or x ).Boththelimitshaveacommon propertythatthedistance | x | oftheinnitepoints fromthe origin x =0isinnite.Similarly,inaEuclideanspace,thelimitat innityisdenedinthesensethat r .If D isanunbounded region,thenaneighborhoodoftheinnitepointin D consistsofall pointsof D whosedistancefromtheoriginexceedsanumber r > Asmallerneighborhoodisobtainedbyincreasing Definition 13.13 (LimitatInnity). Let f beafunctiononanunboundedregion D .Anumber f0isthe limitofafunction f atinnity, limr f ( r )= f0if,foranynumber > 0 ,thereexists > 0 suchthat | f ( r ) f0| < whenever r > in D Innitelimitsatinnitycanbedenedsimilarly.Thesqueezeprinciplehasanaturalextensiontotheinnitelimitsandlimitsatinnity. Forexample,if g ( r ) f ( r )and g ( r ) as r r0(or r ),then f ( r ) .

PAGE 198

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 denedat r0. Step2 .Nosubstitutionispossibletotransformthelimittoaonevariablelimit. Step3 .Put r ( t )=( at,bt,ct )forsomeconstants a b ,and c thatdo notvanishsimultaneouslysothattheydenethedirectionoftheline throughtheorigin.Then f ( r ( t ))= At2ln( Bt2)=2 At2ln t + At2ln B, where A = a2+2 b2+4 c2and B = a2+ b2+ c2> 0.BylHospitals rule, limt 0+t2ln t =limt 0+ln t t 2=limt 0+t 1 2 t 3= 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 cos( x2y ) x2+2 y2, r0=(0 0) andndadiskcenteredat r0inwhichvaluesof f deviatefromthe limitbymorethan =0 5 10 4. Solution: Step1 .Thecontinuityargumentdoesnotapplybecause f isnot denedat r0. Step2 .Nosubstitutionispossibletotransformthelimittoaonevariablelimit.

PAGE 199

87.AGENERALSTRATEGYFORSTUDYINGLIMITS193 Step3 .Put r ( t )=( t,at ).Then limt 0+f ( r ( t ))=limt 0+1 cos( at3) t2(1+2 a2) = 1 1+2 a2limu 0+1 cos( au3 / 2) u = 1 1+2 a2limu 0+au1 / 2sin( au3 / 2) 1 =0 wherethesubstitution u = t2andlHospitalsrulehavebeenusedto evaluatethelimit.Therefore,ifthelimitexists,itmustbeequalto0. Step4 .Noterstthat1 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 ) 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 r0 < ( )=(2 )1 / 4=0 1. Problem13.3. Findthelimit limr r0f ( r ) orshowthatitdoesnot exist,where f ( r )= f ( x,y )= x2y x2 y2, r0=(0 0) Solution: Step1 .Thecontinuityargumentdoesnotapplybecause f isnot denedat 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 a2)= a (1 a2) 1t 0as t 0+.So,ifthelimit exists,thenitmustbeequalto0. Step4 .Inpolarcoordinates, x = R cos and y = R sin ,where r r0 = R f ( r )= R3cos2 sin R2(cos2 sin2 ) = 1 2 R cos sin(2 ) cos(2 ) = R cos 2 tan(2 ) Therefore,inanydisk0 < r r0 < ,thereisasectorcorresponding tothepolarangle / 4 < 0small enoughbecausetan(2 )isnotboundedinthisinterval.Hence,forany

PAGE 200

19413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS > 0,thereisno > 0suchthat | f ( r ) | < whenever r D liesin thedisk0 < r r0 < .Thus,thelimitdoesnotexist. Step3(Optional) .ThenonexistenceofthelimitestablishedinStep 4impliesthatthereshouldexistcurvesalongwhichthelimitdiers from0.Itisinstructivetodemonstratethisexplicitly.Anysuchcurve shouldapproachtheoriginfromwithinoneofthenarrowsectorscontainingthelines y = x (wheretan(2 )takeslargevalues).Soput, forexample, r ( t )=( t,t 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 atnin r ( t )modelsasmalldeviationofthecurvefromtheline y = x inthe vicinityofwhichthefunction f isexpectedtobeunbounded.Then f ( r ( t ))=( t3 atn +2) / (2 atn +1 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 ))=( t3 at4) / (2 at3 a2t4)=(1 at2) / (2 a 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 (bylHospitalsrule).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 ) 0 | ln(4 R4) R2= 4ln R +ln4 R2 0as R Thus,bythesqueezeprinciplethelimitisindeed0.

PAGE 201

87.AGENERALSTRATEGYFORSTUDYINGLIMITS195 87.7.Exercises.(1) Provethefollowingstatements: (i)Let f ( x,y )=( x y ) / ( x + y ).Then limx 0 limy 0f ( x,y ) =1 limy 0 limx 0f ( x,y ) = 1 butthelimitof f ( x,y )as( x,y ) (0 0)doesnotexist. (ii)Let f ( x,y )= x2y2/ ( x2y2+( x 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 ) xy ( xy )3(iii)limr 0 xy2+1 1 xy2(iv)limr 0sin( xy3) x2(v)limr 0x3+ y5 x2+2 y2(vi)limr 0e r 1 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( x2 y2) x2 y22(xiii)limr 0 xy +1 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 ,

PAGE 202

19613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS y ( t )= t sin foranyxed ,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) e| 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. Itdenestheslopeofthetangentline L ( x )= f ( x0)+ f( x0)( x 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.Asignicant dierenceisthat,givenapointinthedomain,therateofchangeis goingtodependonthedirectioninwhichitismeasured.Forexample, if f ( r )istheheightofahillasafunctionofposition r ,thentheslopes fromwesttoeastandfromsouthtonorthmaybedierent.This

PAGE 203

88.PARTIALDERIVATIVES197 observationleadstotheconceptofpartialderivatives.If x and y are thecoordinatesfromwesttoeastandfromsouthtonorth,respectively, thenthegraphof f isthesurface z = f ( x,y ).Ataxedpoint r0= ( x0,y0),theheightchangesas h ( x )= f ( x,y0)alongthewesteast directionandas g ( y )= f ( x0,y )alongthesouthnorthdirection.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 ).Theslopealongthewesteastdirectionis h( x0),and theslopealongthesouthnorthdirectionis 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 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
PAGE 204

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.Geometricalsignicanceofpartialderivatives. 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) f ( r0) h exists,thenitiscalledthepartialderivativeof f withrespectto xiat r0.

PAGE 205

88.PARTIALDERIVATIVES199 Thereasonthepoint r0needstobeaninteriorpointissimple.By thedenitionoftheone-variablelimit, h canbenegativeorpositive. Sothepoints r0+ h ei, i =1 2 ,...,n ,mustbeinthedomainofthe functionbecauseotherwise f ( r0+ h ei)isnotevendened.Thisis guaranteedif r0isaninteriorpointbecauseallpoints r intheball B( r0)ofsucientlysmallradius = | h | arein D Remark. Itisalsocommontoomitprimeinthenotationsforpartialderivatives.Forexample,thepartialderivativeof f withrespect to x isdenotedas fx.Inwhatfollows,thenotationintroducedinDefinition13.16willbeused. Let r0=( a1,a2,...,an),where aiarexednumbers.Considerthe function F ( xi)ofonevariable xi( i isxed),whichisobtainedfrom f ( r )byxingallthevariables xj= ajexceptthe i thone(i.e., xj= ajforall j = i ).Bythedenitionoftheordinaryderivative,the partialderivative f xi( r0)existsifandonlyifthederivative F( ai)exists because (13.1) f xi( r0)=limh 0F ( ai+ h ) F ( ai) h = dF ( xi) dxi xi= aijustlikeinthecaseoftwovariablesdiscussedatthebeginningofthis section.Thisruleispracticalforcalculatingpartialderivativesasit reducestheproblemtocomputingordinaryderivatives. Example 13.16 Findthepartialderivativesof f ( x,y,z )= x3 y2z atthepoint (1 2 3) Solution: Bytherule(13.1), f x(1 2 3)= d dx f ( x, 2 3) x =1= d dx ( x3 12) x =1=3 f y(1 2 3)= d dy f (1 ,y, 3) y =2= d dy (1 3 y2) y =2= 12 f z(1 2 3)= d dz f (1 2 ,z ) z =3= d dz (1 4 z ) z =3= 4 GeometricalSignicanceofPartialDerivatives.Fromtherule(13.1), itfollowsthat thepartialderivative f xi( r0) denestherateofchange ofthefunction f whenonlythevariable xichangeswhiletheother variablesarekeptxed .If,forinstance,thefunction f inExample13.16 denesthetemperatureindegreesCelsiusasafunctionoftheposition whosecoordinatesaregiveninmeters,then,atthepoint(1 2 3),the temperatureincreasesatrateof4degreesCelsiuspermeterinthe

PAGE 206

20013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS directionofthe x axis,anditdecreasesattherates 12and 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 arenotsettoanyspecicvaluesbutratherviewedasindependentof xi(i.e., dxj/dxi=0forall j = i ).Thisagreementisreectedbythe 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: Assumingrstthat 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 fg xi g2. Let h ( u )beadierentiablefunctionofonevariableandlet g ( r )bea functionofseveralvariableswhoserangeliesinthedomainof f .Then

PAGE 207

88.PARTIALDERIVATIVES201 onecandenethecomposition 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 1,where r =( x1,x2,...,xn) Solution: Put h ( u )= u 1 / 2and g ( r )= x2 1+ x2 2+ + x2 n= r 2. Then f ( r )= h ( g ( r )).Since h( u )=( 1 / 2) u 3 / 2and g/xi=2 xi, thechainrulegives xi r 1= xi r 3. 88.4.Exercises.(1) Findthespeciedpartialderivativesofeachofthefollowingfunctions: (i) f ( x,y )=( x 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 1)sin 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 )=tan 1( x/y ) (ix) f ( x,y )=sin 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

PAGE 208

20213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (3) Determinewhetherthefunction f ( x,y )increasesordecreaseswhen x increases,while y isxed,andwhen y increases,while x isxed,at aspeciedpoint P0: (i) f ( x,y )= xy/ ( x + y ), P0(1 2) (ii) f ( x,y )=( x2 2 y2)1 / 3, P0(1 1) (iii) f ( x,y )= x2sin( xy ), P0( 1 ) 89.Higher-OrderPartialDerivatives Sincepartialderivativesofafunctionarealsofunctionsofseveral variables,theycanbedierentiatedwithrespecttoanyvariable.For example,forafunctionoftwovariables,allpossiblesecondpartial derivativesare f x x f x = 2f x2, y f x = 2f yx f y 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 denedaspartialderivativesofsecond-orderpartialderivatives,and soon. Example 13.19 Forthefunction f ( x,y )= x4 x2y + y2,ndall second-andthird-orderpartialderivatives. Solution: Therstpartialderivativesare f x=4 x3 2 xy and f y= x2+2 y .Thenthesecondpartialderivativesare f xx=(4 x3 2 xy ) x=12 x2 2 y,f yy=( x2+2 y ) y=2 f xy=(4 x3 2 xy ) y= 2 x,f yx=( x2+2 y ) x= 2 x. Thethirdpartialderivativesarefoundsimilarly: fxxx=(12 x2 2 y ) x=24 x,fyyy=(2) y=0 fxxy=(12 x2 2 y ) y= 2 ,fxyx= fyxx=( 2 x ) x= 2 fyyx=(2) x=0 ,fyxy= fxyy=( 2 x ) y=0 Incontrasttotheone-variablecase,therearehigher-orderpartial derivativesofanewtypethatareobtainedbydierentiatingwith

PAGE 209

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 (ClairautsTheorem) Let f beafunctionofseveralvariables ( x1,x2,...,xn) thatisdened onanopenball D inaEuclideanspace.Ifthesecondpartialderivatives f xixjand f xjxi,where j = i ,arecontinuousfunctionson D ,then f xixj= f xjxiatanypointof D AconsequenceofClairautstheoremcanbeproved.Itassertsthat theresultofpartialdierentiationdoesnotdependontheorderin whichthederivativeshavebeentakenifallhigher-orderpartialderivativesinquestionarecontinuous .Itisnotalwaysnecessarytocalculatehigher-orderpartialderivativesinallpossibleorderstoverify thehypothesisofClairautstheorem(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

PAGE 210

20413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS continuousintheentirespace.ThehypothesisofClairautstheoremis satised,andhenceallthepartialderivativesinquestioncoincideand areequalto fxyz=( f x) yz=(2 x cos( x2+ yz )) yz=( 2 xz sin( x2+ yz )) z= 2 x sin( x2+ yz ) 2 xyz cos( x2+ yz ) 89.1.ReconstructionofaFunctionfromItsPartialDerivatives.Oneof thestandardproblemsincalculusisndingafunction 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. Theindeniteintegralisgivenbythesumofaparticularantiderivative 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 Clairautstheorem,oneinfersthat (13.4) f xixj= f xjxi= Fi xj= Fj xi. Thus,theconditions(13.4)onthefunctions Fimustbefullled;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 x .Doessuchafunction f exist? Solution: Therstpartialderivativesof f F1( x,y )=2 x + y and F2( x,y )=2 y x ,arepolynomials,andhencetheirderivativesare continuousintheentireplane.Inorderfor f toexist,theintegrability condition F1/y = F2/x mustholdintheentireplane.Thisis notsobecause F1/y =1,whereas F2/x = 1.Thus,nosuch f exists. Supposenowthattheintegrabilityconditions(13.4)aresatised. Howisasolution f to(13.3)tobefound?Evidently,onehasto

PAGE 211

89.HIGHER-ORDERPARTIALDERIVATIVES205 calculateanantiderivativeofthepartialderivative.Intheone-variable case,anantiderivativeisdeneduptoanadditiveconstant.Thisis notsointhemultivariablecase.Forexample,let f x( x,y )=3 x2y Anantiderivativeof f xwithrespectto x isafunctionwhose partial derivativewithrespectto x is3 x2y .Itiseasytoverifythat x3y satises 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 .Theintegrabilityconditionsarefullled:( 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 isadirectconsequenceofthefullledintegrabilitycondition.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 x yields x + g( y )=2 y x or g( y )=2 y 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) zaresatised(theirvericationislefttothereader). So f exists.Takingtheantiderivativewithrespectto x intherst equation,onends 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 ) ,

PAGE 212

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 fromitsrstpartialderivatives aswellastheintegrabilityconditions(13.4)willbeimportantwhen discussing conservativevectorelds andthe potential ofaconservative vectoreld.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 istheamplitudeofelectricormagneticelds.

PAGE 213

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 equationsareusedtodeterminestaticelectromagneticeldscreated bystaticelectricchargesandcurrents. Example 13.23 Let h ( q ) beatwice-dierentiablefunctionofa variable q .Showthat u ( r ,t )= h ( ct n r ) isasolutionofthewave equationforanyxedunitvector n Solution: Let n =( n1,n2,n3),where n2 1+ n2 2+ n2 3=1as n isthe unitvector.Put q = ct n r = ct n1x n2y n3z .Bythechain rule(13.2), u t= q th( q )andsimilarlyfortheotherpartialderivatives. Therefore, u t= ch( q ), u tt= c2h( q ), u x= 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,meaningthatthewaveequationissatisedforany h Considerthelevelsurfacesofthesolutionofthewaveequation discussedinthisexample.Theycorrespondtoaxedvalueof q = q0. So,foreachmomentoftime t ,thedisturbanceoftheairdensity u ( r ,t ) hasaconstantvalue h ( q0)intheplane n r = ct q0= d ( t ).All planeswithdierentvaluesoftheparameter d areparallelasthey havethesamenormalvector n .Sincehere d ( t )isafunctionoftime, theplaneonwhichtheairdensityhasaxedvaluemovesalongthe 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.

PAGE 214

20813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 89.3.StudyProblems.Problem13.5. Findthevalueofaconstant a forwhichthefunction u ( r ,t )= t 3 / 2e ar2/t,r = r satisesthediusionequationforall t> 0 Solution: Notethat u dependsonthecombination r2= x2+ y2+ z2. Tondthepartialderivativesof u ,itisconvenienttousethechain rule: u x = u r2r2 x =2 x u r2= 2 ax t u, u xx= x u x = 2 a t u 2 ax t u x = 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)= 6 ka t + 4 ka2r2 t2 u. Usingtheproductruletocalculatethepartialderivativewithrespect totime,onendsfortheleftside u t= 3 2 t 5 / 2e ar2/t+ t 3 / 2e ar2/tar2 t2= 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 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)= 1 and f yx(0 0)=1 ,thatis, f xy(0 0) = f yx(0 0) .Doesthisresultcontradict Clairautstheorem?

PAGE 215

89.HIGHER-ORDERPARTIALDERIVATIVES209 Solution: Usingthequotientrulefordierentiation,onends f x( x,y )= x4y +4 x2y3 y5 ( x2+ y2)2,f y( x,y )= x5 4 x3y2 xy4 ( x2+ y2)2if( x,y ) =(0 0).Notethat,owingtothesymmetry f ( x,y )= 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 Therst-orderpartialderivativesarecontinuousfunctions(theproof islefttothereaderasanexercise).Next,onehas f xy(0 0)= d dy f x(0 ,y ) y =0=limh 0f x(0 ,h ) f x(0 0) h =limh 0 h 0 h = 1 f yx(0 0)= d dx f y( x, 0) x =0=limh 0f y( h, 0) f y(0 0) h =limh 0h 0 h =1 TheresultdoesnotcontradictClairautstheorembecause 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)= 1, and f xyisnotcontinuousat(0 0).Thetechnicaldetailsarelefttothe reader. 89.4.Exercises.(1) Findallsecondpartialderivativesofeachofthefollowingfunctions andverifyClairautstheorem: (i) f ( x,y )=tan 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 )=cos 1( x/y ) (vi) f ( x,y )= xy

PAGE 216

21013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (2) Explainwithoutexplicitcalculationofhigher-orderpartialderivativesthatthehypothesisofClairautstheoremissatisedforthefollowingfunctions (i) f ( x,y,z )=sin( x2+ y 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) / ( x2 y2), x2 = y2(v) f ( x,y,z )=( x + yz2 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 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 )=tan 1x + y + z xyz 1 xy xz yz f(3) abc,where abc denotesall permutationsof xyz (vii) f ( x,y,z,t )=ln(( x y )2+( z t )2) 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 x )+ B cos( ct + x ), c 2f tt f xx=0 (ii) f ( x,y,t )= g ( ct ax 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 yy ( 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 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)

PAGE 217

90.LINEARIZATIONOFMULTIVARIABLEFUNCTIONS211 (6) Findarelationbetweentheconstants a b ,and c suchthatthe function u ( x,y,t )=sin( ax + by + ct )satisesthewaveequation u tt u xx 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+ z3 3 xyz (ii) f =( x2+ y2+ z2) 1 / 2(9) Lettheactionof K onafunction f bedenedby 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 )thatsatises: (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 x0)orthe tangentline.Put x = x0+ x .Then,bythedenitionofthederivative f( x0), lim x 0f ( x ) L ( x ) x =lim x 0f ( x0+ x ) f ( x0) x f( x0) = f( x0) f( x0)=0 Thisrelationimpliesthattheerrorofthelinearapproximationgoesto 0fasterthanthedeviation x = x x0of x from x0,thatis, (13.5) f ( x )= L ( x )+ ( x ) x, where ( x ) 0as x 0 .

PAGE 218

21213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Forexample,if f ( x )= x2,thenitslinearizationat x =1is L ( x )= 1+2( x 1).Itfollowsthat f (1+ x ) 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 x0)= f ( x0)+ n x and lim x 0f ( x ) L ( x ) x =lim x 0f ( x0+ x ) f ( x0) x n =0 Bythedenitionofthederivative 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 ) L ( x )) / x =( | x | nx ) /x = | x | /x 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 x0)+ n2( y y0),where n1and n2are arbitrarynumbers.Itdenesalinearapproximationto 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 r0) issaidtobea linearapproximation to f near r0inthesensethat L ( r0)= f ( r0).Thedotproductisdenedinan 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 r0=( x x0,y y0)sothat n ( r r0)= n1( x x0)+ n2( y y0). Definition 13.17 (DierentiableFunctions) Thefunction f ofseveralvariables r =( x1,x2,...,xm) onanopenset D issaidtobe dierentiable atapoint r0 D ifthereexistsagood

PAGE 219

90.LINEARIZATIONOFMULTIVARIABLEFUNCTIONS213 linearapproximation L ( r ) ,i.e.alinearapproximation L ( r ) forwhich (13.6)limr r0f ( r ) L ( r ) r r0 =0 If f isdierentiableatallpointsof D ,then f issaidtobedierentiable on D Bythisdenition, thedierentiabilityofafunctionisindependent ofthecoordinatesystem chosentolabelpointsof D (alinearfunction remainslinearundergeneralrotationsandtranslationsofthecoordinatesystemandthedistance r r0 isalsoinvariantunderthese transformations).Forfunctionsofasinglevariable f ( x ),theexistence ofalinearapproximationat x0withtheproperty(13.6)isequivalentto theexistenceofthederivative f( x0).Indeed,put x x0= x .Then [ f ( x ) L ( x )] / | x | = [ f ( x ) 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 r0thatsatisestheproperty(13.6)isuniqueifit exists. Proof. Let L1( r )= f ( r0)+ n1 ( r r0)and L2( r )= f ( r0)+ n2 ( r r0) betwolinearapproximationsthatsatisfythecondition(13.6)forwhich n2 = n1.Makinguseoftheidentity L2( r ) L1( r )=[ f ( r ) L1( r )] [ f ( r ) L2( r )] itisconcludedthat limr r0L2( r ) L1( r ) r r0 =limr r0f ( r ) L1( r ) r r0 limr r0f ( r ) L2( r ) r r0 =0 Notethatowingtotheexistenceofthelimit(13.6)forbothlinear functions L1and L2,thelimitofthedierenceequalsthedierenceof thelimits(thebasiclawoflimits).Ontheotherhand, L2( r ) L1( r )= ( n2 n1) ( r r0).Put n = n2 n1.Byassumption, n =0.Then 0=limr r0L2( r ) L1( r ) r r0 =limr r0n ( r r0) r 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,

PAGE 220

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 r0 : (13.7) f ( r )= L ( r )+ ( r ) r 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 r0 = | t | 0as t 0alongtheline,and f ( r ( t )) L ( r ( t ))= f ( r0+ t ej) f ( r0) n ejt = f ( r0+ t ej) f ( r0) njt, where njisthe j thcomponentofthevector n .Bythesamereasoning asintheone-variablecasewith x = t (givenafterDenition13.17), thecondition(13.6)implies limt 0f ( r0+ t ej) f ( r0) t nj=0 nj= f xj( r0) accordingtoDenition13.16ofpartialderivativesatapoint.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) .

PAGE 221

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+. Tondthepartialderivativesinquestion,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: Thecontinuityisveriedbythesqueezeprinciple.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 ) 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).

PAGE 222

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( x1 a1)+ n2( x2 a2)+ + nm( xm 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

PAGE 223

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). Nowonecandeneaplanethroughthepoint r0ofthegraphthat containsthetwotangentlines.Thisplaneiscalledthe tangentplane to

PAGE 224

21813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS thegraph.Itsnormalmustbeperpendiculartobothvectors v1and v2and,bythegeometricalpropertiesofthecrossproduct,maybetaken as n = v1 v2=( f x( x0,y0) ,f y( x0,y0) 1).Thestandardequationof theplane n ( r r0)=0canthenbewrittenintheform z = z0+ n1( x x0)+ n2( y 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)denesthetangentplanetothegraphof 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= 1.Anequationofthetangent planeis4( x 2)+6( y 1) ( z 7)=0or4 x +6 y 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 2=0 03, y = y 2= 0 03,and z =0 94 1= 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

PAGE 225

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) . Solution: Bythedenitionofpartialderivatives, f x(0 0)=limh 0f ( h, 0) 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 )=0satisesthe condition(13.6): lim( x,y ) (0 0)f ( x,y ) 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 2 x x2+ y2cos 1 x2+ y2 Therstterminthisexpressionconvergesto0,0 | 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)=( 1)n,sin(1 /t2 n)=0,and f x( tn, 0)= 2 tnsin(1 /t2 n)+(2 /tn)cos(1 /t2 n)=2( 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)?

PAGE 226

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+ y2 z2(v) f ( r )=ln(1 r ),where r =( x1,x2,...,xm) (vi) f ( x,y )=3 x3+ y3(vii) f ( r )= e 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 thefollowingsurfacesatthespeciedpoint: (i) z = x2+3 y y4x ,(1 2 1) (ii) z = x3y ,(1 4 2) (iii) z = y ln( x2 3 y ),(2 1 0) (iv) y =tan 1( xz2),(1 4, 1) (v) x = z cos( y z ),(1 1 1) (vi) z = y +ln( x/z ),(1 1 1) (7) Findthelinearizationofeachofthefollowingfunctionsatthe speciedpoint: (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 isaxedvectororthogonal to r0and r =( x1,x2,...,xm) (8) Usethelinearizationtoapproximatethefollowingnumbers.Then useacalculatortondthenumbers.Comparetheresults.

PAGE 227

91.CHAINRULESANDIMPLICITDIFFERENTIATION221 (i) 20 7 x2 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 ) foreveryxedpair( 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 ) 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 ateachxed 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 = (3 x2+ y2) r sin +2 xyr cos = r3cos where x and y havebeenexpressedinthepolarcoordinatestoobtain thenalexpressions.Thelatterrelationsareanexampleofa chain rule forfunctionsoftwovariables.Supposethattherates f x( x0,y0) and f y( x0,y0)areknownataparticularpoint( x0,y0).Then,byusing

PAGE 228

22213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS thechainrule, anexplicitformofthefunction f inthenewvariablesis notrequiredtonditsrateswithrespecttothenewvariables 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 ) 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 ) denes 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/ujexistanddeneagoodlinearapproximationinthesense (13.7).Inparticular,foraxedvalueof u andforevery i xi= xi( u + ejh ) xi( u )= xi ujh + i( h ) | h | ,i( h ) 0as h 0 Denethevector rh=( x1, x2,..., xn).Ithasthepropertythat rh 0 as h 0.If F ( u )= f ( x1( u ) ,x2( u ) ,...,xn( u )),then,bythe

PAGE 229

91.CHAINRULESANDIMPLICITDIFFERENTIATION223 denitionofthepartialderivativesatapoint u f uj=limh 0F ( u + ejh ) F ( u ) h =limh 0f ( r + rh) 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) 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,therst 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)= 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

PAGE 230

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+( 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+( 2) 3= 3 Example 13.30 Let f ( x,y,z )= z2(1+ x2+2 y2) 1.Findtherate ofchangeof f alongthecurve r ( t )=(sin t, cos t,et) inthedirectionof increasing t Solution: Thefunction f isdierentiableastheratiooftwopolynomials(itspartialderivativesarecontinuous): f x= 2 xz2 (1+ x2+2 y2)2,f y= 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 )= 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 ) = 2 e2 tsin t (2+cos2t )2 (cos t ) 4 e2 tcos t (2+cos2t )2 ( 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 =( u2 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= v y u= v ,and y v= u .Then g u= f xx u+ f yy u= f xu + f yv.

PAGE 231

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 ( vf xx+ uf xy)+ v ( vf yx+ uf yy)+ f y= uv ( f yy f xx)+( u2 v2) f xy+ f y= y ( f yy f xx)+2 xf xy+ f y, where f xy= f yxhasbeenused.Thevalueof g uvatthepointinquestion is2 (2 2)+2 3+1=7. 91.2.ImplicitDifferentiation.Considerthefunctionofthreevariables, F ( x,y,z )= x2+ y4 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 )isdenedasarootof 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 )=0isnotalwayspossibletond. Puttingasidethequestionabouttheveryexistenceofasolutionand itsuniqueness,supposethatthisequationisprovedtohaveaunique solutionwhen( x,y ) D .Inthiscase,thefunction z ( x,y )withthe property(13.9)forall( x,y ) D issaidtobedened implicitly on D Althoughananalyticformofanimplicitlydenedfunctionisunknown,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

PAGE 232

22613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS and y = v .Therefore, u F ( x,y,z ( x,y ))= F x + F z z x =0= z x= F x F z, v F ( x,y,z ( x,y ))= F y + F z z y =0= z y= F y F z. Theseequationsdeterminetheratesofchangeofanimplicitlydened 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 )= 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 )= 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 y )= sin( xyz ) hasauniquesolution z = z ( x,y ) inaneighborhoodof (1 1) suchthat z (1 1)= / 2 andndtheratesofchange z x(1 1) and z y(1 1) .

PAGE 233

91.CHAINRULESANDIMPLICITDIFFERENTIATION227 Solution: Put F ( x,y,z )= sin( xyz ) z (3 x y ).Thentheexistenceanduniquenessofthesolutioncanbeestablishedbyverifying thehypothesesoftheimplicitfunctiontheoreminwhich r =( x,y ), r0=(1 1),and z0= / 2.First,notethatthefunction F isthesum ofapolynomialandthesinefunctionofapolynomial.Soitspartial derivatives F x= yz cos( xyz ) 3 z,F y= xz cos( xyz )+ z, F z= xy cos( xyz ) 3 x + y arecontinuousforall( x,y,z );hence, F isdierentiableeverywhere. Next, F (1 1 ,/ 2)=0asrequired.Finally, F z(1 1 ,/ 2)= 2 =0. Therefore,bytheimplicitfunctiontheorem,thereisanopendiskin the xy planecontainingthepoint(1 1)inwhichtheequationhasa uniquesolution z = z ( x,y ).Bytheimplicitdierentiationformulas, z x(1 1)= F x(1 1 ,/ 2) F z(1 1 ,/ 2) = 3 4 ,z y(1 1)= 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 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)= 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 3 x 2 + y 2 Putting x =0 04and y = 0 04,thisequationyieldstheestimate z (1 04 0 96) 0 45

PAGE 234

22813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Problem13.9. Letthefunction z ( x,y ) bedenedimplicitlyby z5+ zx y =0 inaneighborhoodof (1 2 1) .Findallitsrstandsecondpartialderivatives.Inparticular,givethevaluesofthesepartial derivativesat ( x,y )=(1 2) Solution: Let F ( x,y,z )= z5+ zx 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.Therst andsecondpartialderivativesof F arecontinuouseverywhere: F x= z,F y= 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= F x F z= z 5 z4+ x ,z y= F y F z= 1 5 z4+ x Takingthepartialderivativesoftheserelationswithrespectto x and y andusingthequotientrulefordierentiation,thesecondpartial derivativesareobtained: z xx= z x(5 z4+ x ) z (20 z3z x+1) (5 z4+ x )2= (15 z4 x ) z x+ z (5 z4+ x )2, z xy= z yx=( z y) x= 20 z3z x+1 (5 z4+ x )2,z yy= 20 z3z y (5 z4+ x )2. Theexplicitformof z xand z ymaybesubstitutedintotheserelations toexpressthesecondpartialderivativesvia x y ,and z .Atthepoint (1 2),thevaluesoftherstpartialderivativesare z x(1 2)= 1 / 6and z y(1 2)=1 / 6.Usingthesevalues,thevaluesofthesecondpartial derivativesareevaluated: z xx(1 2)= 1 / 27, z xy(1 2)=7 / 108,and z yy(1 2)= 5 / 54. 91.4.Exercises.(1) Usethechainruletond dz/dt if z = 1+ x2+2 y2and x =2 t3, y =ln t (2) Usethechainruletond z/s and z/t if z = e 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 )

PAGE 235

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 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= e 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 v ,showthat ( z x)2+( z y)2= z uz v. (9) Findalltherstandsecondpartialderivativesofthefollowing 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 )satisestheLaplaceequation f xx+ f yy=0,thenthefunction g ( x,y )= f ( x/ ( x2+ y2) ,y/ ( x2+ y2)), x2+ y2> 0,alsosatisestheLaplaceequation. (14) Provethatifafunction f ( x,t )satisesthediusionequation f t= a2f xx,thenthefunction g ( x,t )= 1 a t e x2/ (4 a2t )f x a2t x a4t ,t> 0 alsosatisesthediusionequation.

PAGE 236

23013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (15) Provethatif f ( x,y,z )satisestheLaplaceequation 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 alsosatisestheLaplaceequation. (16) Showthatthefunction g ( x,y )= xnf ( y/x2),where f isadierentiablefunction,satisestheequation xg x+2 yg y= ng (17) Showthatthefunction g ( x,y,z )= xnf ( y/xa,z/xb),where f isa dierentiablefunction,satisestheequation xg x+ ayg y+ bzg z= ng (18) Letthefunction z = f ( x,y )bedenedimplicitly.Finditsrst andsecondpartialderivativesif (i) x +2 y +3 z = ez(ii) x z =tan 1( yz ) (iii) x/z =ln( z/y )+1 (19) Let z ( x,y )beasolutionoftheequation z3 xz + y =0suchthat z (3 2)=2.Findthelinearizationof z ( x,y )near(3 2)anduseit toestimate z (2 8 2 3). (20) Find f xand f y,where f =( x + z ) / ( y + z )and z isdenedbythe equation zez= xex+ yey. (21) Showthatthefunction z ( x,y )denedbytheequation F ( x az,y bz )=0,where F isadierentiablefunctionoftwovariables and a and b areconstants,satisestheequation 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 = t2 1.If T x(2 6 8)=2, T y(2 6 8)= 1,and T z(2 6 8)=1,howfastisthetemperaturerising (ordecreasing)ontheinsectspathasitiesthroughthepoint(2 6 8)? (23) Considerafunction f = f ( x,y,z )andthechangeofvariables: x =2 uv,y = u2 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

PAGE 237

92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS231 homogeneousdierentiablefunctionsatisestheequation xf x+ yf y= nf .Showalsothat f x( tx,ty )= tn 1f ( x,y ). (26) Supposethattheequation F ( x,y,z )=0denesimplicitly 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 )= 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 sin y ),where f isadierentiablefunction. 92.TheDifferentialandTaylorPolynomials Justlikeintheone-variablecase,givenvariables r =( x1,x2,...,xm), onecanintroduceindependentvariables d r =( dx1,dx2,...,dxm)that areinnitesimalvariationsof 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 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,iftheinnitesimalvariations(ordierentials) d r arereplaced bythedeviations r = r r0ofthevariables r from r0,then the dierential df atthepoint r0denesthelinearizationof f at r0.Accordingto(13.6),thedierence f ( r ) f ( r0) df ( r0)tendsto0faster than r as r 0 ,andhencethedierentialcanbeusedtostudy variationsofadierentiablefunction f undersmallvariationsofits arguments.

PAGE 238

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.Geometricalsignicanceofthedierential. Left :Thedierentialofafunctionofonevariable.Itdenes 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 df ) / x = y/ x f( x0) 0;thatis,thedierence y df tendsto0 fasterthan x Right :Thedierentialofafunctionoftwo variables.Itdenestheincrementof 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 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 0 2) .Whatisthesignicance 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 0 05=0 1 Thenumber f (1 3)+ df (1 3)denesthevalueofthelinearization L ( x,y )of f at(1 3)for( x,y )=(1+ dx, 3+ dy ).Itcanbeusedto approximate f (1+ dx, 3+ dy ) f (1 3) df (1 3)when | dx | and | dy | aresmallenough.Inparticular, f (1+0 1 3 0 2) f (1 3)=0 09476 (acalculatorvalue),whichistobecomparedwith df (1 3)=0 1. 92.1.ErrorAnalysis.Supposeaquantity f dependsonseveralother quantities,say, x y ,and z ,fordeniteness,thatis, f isafunction f ( x,y,z ).Supposemeasurementsshowthat x = x0, y = y0,and z = z0.Since,inpractice,allmeasurementscontainerrors,thevalue f ( x0,y0,z0)doesnothavemuchpracticalsignicanceuntilitserroris determined.

PAGE 239

92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS233 Forexample,thevolumeofarectanglewithdimensions x y ,and z isthefunctionofthreevariables V ( x,y,z )= xyz .Inpractice, repetitivemeasurementsgivethevaluesof x y ,and z fromintervals x [ x0 x,x0+ x ], y [ y0 y,y0+ y ],and z [ z0 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 [ 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 ) V ( r0)= dV ( r0)iftermstendingto0fasterthan d r canbeneglected.Thecomponentsof d r areindependentvariablestakingtheirvaluesinthespeciedintervals.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 [ 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

PAGE 240

23413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS islinearin dxi,and r0isxed,themaximumisattainedbysetting dxiequalto xiforall i forwhichthecoecient f xi( r0)ispositive,and equalto 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 ) |
PAGE 241

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 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 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 isdenedsuccessively dnf = dn 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

PAGE 242

23613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS dierentiationisirrelevant(Clairautstheorem),thisobservationholds ingeneral: dnf =nk =0Bn knf xn kyk( dx )n k( dy )k,Bn k= n k !( n k )! where Bn karethebinomialcoecients:( a + b )n= n k =0Bn kan 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 kyk= an 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 x0and dy = y y0. Forexample,put r =( x,y ), r0=( x0,y0), dx = x x0,and dy = y y0.TherstfourTaylorpolynomialsare 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.

PAGE 243

92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS237 Thelinearortangentplaneapproximation f ( r ) L ( r )= T1( r )is aparticularcaseoftheTaylorpolynomialapproximationoftherst degree. Example 13.35 Let Pn( x,y ) beapolynomialofdegree n .Findits Taylorpolynomialsabout (0 0) .Inparticular,ndTaylorpolynomials for P3( x,y )=1+2 x xy + y2+4 x3 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,Denition13.22denes Tkasthesumofhomogeneouspolynomialsin x and y if( x0,y0)=(0 0). Twopolynomialsareequalonlyifthecoecientsatthecorresponding monomialsmatch.Itfollowsfrom Tn= Pnthat Tn= Tn 1+( Tn Tn 1)= Q0+ Q1+ + Qn 1+ Qn.Since Qnand Tn Tn 1containsonly monomialsofdegree n ,theequalityispossibleonlyif Qn= Tn Tn 1, andhence Tn 1= Q0+ Q1+ + Qn 1.Continuingtheprocess recursivelybackward,itisconcludedthat Tk= Q0+ Q1+ + Qk,k =0 1 ,...,n. Inparticular,forthegivenpolynomial P3,onehas Q0=1, Q1=2 x Q2= xy + y2,and Q3=4 x3 y2x .Therefore,itsTaylorpolynomials abouttheoriginare T0=1, T1= T0 2 x T2= T1 xy + y2,and Tk= P3for k 3. Theorem 13.15 (TaylorTheorem). Let D beanopendiskcenteredat r0andletthepartialderivatives ofafunction f becontinuousuptoorder n 1 on D .Then f ( r )= Tn 1( r )+ n( r ) ,wherethereminder ( r ) satisesthecondition | n( r ) | hn( r ) r r0n 1, where hn( r ) 0as r r0. InSection92.4,Taylorpolynomialsforfunctionsofanynumber ofvariableswillbedened.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 ) T1( r )decreasesto0fasterthan r r0 as r r0.For n> 2,itstatesthattheapproximationof f bytheTaylor

PAGE 244

23813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS polynomial Tn 1isagoodapproximationinthesensethattheerror decreasesfasterthan r r0n 1. ApracticalsignicanceoftheTaylor 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 0 2) Solution: Put( dx,dy )=(0 1 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 / 2 x2y2(1+ x2y ) 1 / 2 1+ x2y (1 3)= 3 8 f xy(1 3)= 2 x (1+ x2y )1 / 2 x3y (1+ x2y ) 1 / 2 2(1+ x2y ) (1 3)= 5 16 f yy(1 3)= x4 4 (1+ x2y ) 3 / 2 (1 3)= 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 1 8 ( dy )2= 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 0 01375 / 2=2 093125,while acalculatorvalueof f (1+ dx, 3+ dy )is2 094755(roundedtothe samesignicantdigit).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 kxky | Mnkfor

PAGE 245

92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS239 all r D ,thentheremaindersatises | n( r ) |nk =0Bn kMnk n | x x0|n k| y y0|kforall ( x,y ) D ,where Bn k= n / ( k !( n k )!) arebinomialcoecients. Next,note | x x0| r r0 and | y y0| r r0 andhence | x x0|n k| y y0|k r r0n.Makinguseofthisinequality,one infersthat (13.12) | n( r ) | Mn n r r0n, wheretheconstant Mn= n k =0Bn kMnk.Inparticular,forthelinear approximation n =2, (13.13) | f ( r ) L ( r ) | M2 2 r 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 x0and dy = y 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 ) 1 / 2and f y=(1+ x +2 y ) 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= 1 4(1+ x +2 y ) 3 / 2, f xy= 1 2(1+ x +2 y ) 3 / 2,and f yy= (1+ x +

PAGE 246

24013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 2 y ) 3 / 2.Theirabsolutevaluesaremaximalifthecombination1+ x +2 y isminimalonthesquare.Setting x = y = 1 / 4,1 / 4 < 1+ x +2 y inthesquare.Therefore, | f xx|
PAGE 247

92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS241 So,forpracticalpurposes,theerror | n| oftheapproximation f Tn 1maybeestimatedby dnf ( r0) /n !,where dxjarereplacedbytheirabsolutevalues | dxj| andthevaluesofpartialderivativesarealsoreplaced bytheirabsolutevalues(justlikeithasbeendonewhenestimating | 2| inthecaseoftwovariables),providedthepartialderivativesof f oforder n orhigherarecontinuousinaneighborhoodof r0. Calculationofhigher-orderderivativestondTaylorpolynomials 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 1 6u3+ 5( u ),where 5containsonlymonomialsofdegree5 andhigher.Sincethepolynomial u vanishesattheorigin,itspowers unmaycontainonlymonomialsofdegree n andhigher.Therefore, T3isobtainedfrom u 1 6u3=( xy + z ) 1 6( xy + z )3= z + xy 1 6 z3+3( xy ) z2+3( xy )2z +( xy )3 byretaininginthelatterallmonomialsuptodegree3,whichyields T3( r )= z + xy 1 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.ItisthereforesucienttondasuitableTaylorpolynomialforthefunction(1+ x + y2+ z3) 1andthenmultiplyitbythepolynomial1+ xy ,retainingonlymonomialsuptodegree3.Put u = x + y2+ z3.Then (1+ u ) 1=1 u + u2 u3+ (asageometricseries).Notethat,for n 4,theterms uncontainonlymonomialsofdegree4andhigherand

PAGE 248

24213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS hencecanbeomitted.Uptodegree3,onehas u2= x2+2 xy2+ and u3= x3+ .Therefore, (1+ xy )(1 u + u2 u3)=(1+ xy )(1 x y2 z3+ x2+2 xy2 x3+ ) Carryingoutthemultiplicationandarrangingthemonomialsinthe orderofincreasingdegrees,oneinfers: T1( x,y,z )=1 x, T2( x,y,z )= T1( x,y,z )+ x2+ xy y2, T3( x,y,z )= T2( x,y,z ) x3 x2y +2 xy2 z3. Problem13.11.(MultivariableTaylorandMaclaurinSeries) Supposethatafunction f hascontinuouspartialderivativesofany orderandthereminderintheTaylorpolynomialapproximation f = Tn 1+ nnear r0convergesto0as n (i.e., n 0 ).Thenthe functioncanberepresentedbytheTaylorseriesaboutapoint r0: f ( r )= f ( r0)+n =11 n dnf ( r0) where d r = r r0.TheTaylorseriesabout r0= 0 iscalledtheMaclaurinseries.FindtheMaclaurinseriesof sin( xy2) Solution: Sincetheargumentofthesineisthepolynomial xy2,the Maclaurinseriesof f canbeobtainedfromtheMaclaurinseriesofsin u bysetting u = xy2init.FromCalculusII, f ( r )=sin u =n =1( 1)n +1 (2 n 1)! u2 n 1=n =1( 1)n +1 (2 n 1)! x2 n 1y4 n 2. 92.6.Exercises.(1) Findthedierential df ofeachofthefollowingfunctions: (i) f ( x,y )= x3+ y3 3 xy ( x y ) (ii) f ( x,y )= y cos( x2y ) (iii) f ( x,y )=sin( x2+ y2) (iv) f ( x,y,z )= x + yz + yexyz(v) f ( x,y,z )=ln( xxyyzz) (vi) f ( x,y,z )= y/ (1+ xyz ) (vii) f ( r )= a2 r 2,where a isaconstantand r =( x1,x2,...,xm)

PAGE 249

92.THEDIFFERENTIALANDTAYLORPOLYNOMIALS243 (2) Fourpositivenumbers,eachlessthan100,areroundedandthen multipliedtogether.Usedierentialstoestimatethemaximumpossible errorinthecomputedproductthatmightresultfromtherounding. (3) Aboundarystripe10cmwideispaintedaroundarectanglewhose dimensionsare50mby100m.Usedierentialstoapproximatethe numberofsquaremetersofpaintinthestripe.Assesstheaccuracyof theapproximation. (4) Arectanglehassidesof x =6mand y =8m.Usedierentialsto estimatethechangeofthelengthofthediagonalandtheareaofthe rectangleif x isincreasedby2cmand y isdecreasedby5cm.Assess theaccuracyoftheestimates. (5) Considerasectorofadiskwithradius R =20cmandtheangle = / 3.Usethedierentialtodeterminehowmuchtheradiusshould bedecreasedinorderfortheareaofthesectortoremainthesame whentheangleisincreasedby1.Assesstheaccuracyoftheestimate. (6) Letthequantities f and g bemeasuredwithrelativeerrors Rfand Rg.Showthattherelativeerroroftheproduct fg isthesum Rf+ Rg. (7) Measurementsoftheradius r andtheheight h ofacylinderare r =2 2 0 1and h =3 1 0 2,inmeters.Findtheabsoluteand relativeerrorsofthevolumeofthecylindercalculatedfromthesedata. (8) Theadjacentsidesofatrianglehavelengths a =100 2and b =200 5,inmeters,andtheanglebetweenthemis =60 1. Findtherelativeandabsoluteerrorsincalculationofthelengthofthe thirdsideofthetriangle. (9) If R isthetotalresistanceof n resistors,connectedinparallel,with resistances Rj, j =1 2 ,...,n ,then R 1= R 1 1+ R 1 2+ + R 1 n.If eachresistance Rjisknownwitharelativeerrorof0 5%,whatisthe relativeerrorof R ? (10) UsetheTaylortheoremtoassessthemaximalerrorofthelinear approximationofthefollowingfunctionsabouttheoriginintheballof radius R (i.e.,for r R ): (i) f ( x,y )= 1+sin( x + y ), R =1 2(ii) f ( x,y )=1+3 x 2+ y, R =1 (iii) f ( x,y,z )=ln(1+ x +2 y 3 z ), R =0 1 (11) Findtheindicateddierentialsofthefollowingfunctions: (i) f ( x,y )= x y + x2y dnf n =1 2 ,... (ii) f ( x,y )=ln( x + y ), dnf n =1 2 ,... (iii) f ( x,y )=sin( x )cosh( y ), d3f (iv) f ( x,y,z )= xyz dnf n =1 2 ,... (v) f ( x,y,z )=1 / (1+ xyz ), d2f (vi) f ( r )= r df and d2f r =( x1,x2,...,xm)

PAGE 250

24413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (12) Let f ( r )= g ( u ),where u isalinearfunctionof r u = c + n r where c isaconstantand n isaconstantvector.Showthat dnf = g( n )( u )( n d r )n. (13) Let Qn( x,y,z )beahomogeneouspolynomialofdegree n (it containsonlymonomialsofdegree n ).Showthat dnQn( x,y,z )= n Qn( dx,dy,dz ). (14) FindtheTaylorpolynomial T2aboutaspeciedpointandagiven function: (i) f ( x,y )= y + x3+2 xy2 x2y2,(1 1) (ii) f ( x,y )=sin( xy ),( / 2 1) (iii) f ( x,y )= xy,(1 1) (15) Let f ( x,y )= xy.UseTaylorpolynomials T1, T2,and T3toapproximate f (1 2 0 7).Comparetheresultsofthethreeapproximations withacalculatorvalueof f (1 2 0 7). (16) UsethemethodofExample13.39andStudyProblem13.10to ndtheindicatedTaylorpolynomialsabouttheoriginforeachofthe followingfunctions: (i) f ( x,y )= 1+ x +2 y Tn( x,y ), n 2 (ii) f ( x,y )=xy 1 x2 y2, Tn( x,y ), n 4 (iii) f ( x,y,z )=sin( x +2 y + z2), Tn( x,y,z ), n 3 (iv) f ( x,y,z )= exycos( zy ), Tn( x,y,z ), n 4 (v) f ( x,y )=ln(1+ x +2 y ) / (1+ x2+ y2), Tn( x,y ), n 2 (17) Findpolynomialsofdegree2tocalculateapproximatevalues ofthefollowingfunctionsintheregioninwhich x2+ y2issmallas comparedwith1: (i) f ( x,y )=cos y/ cos x (ii)tan 11+ x + y 1 x + y (18) Findanonzeropolynomialofthesmallestdegreetoapproximate alocalbehaviorofthefunctioncos( x + y + z ) cos( x )cos( y )cos( z ) neartheorigin. (19) Let g ( r )= 1 2 2 0f ( x0+ r cos ,y0+ r sin ) d, where f hascontinuouspartialderivativesuptoorder4and x0and y0areconstants.Find T4about r =0for g ( r ). (20) Considertheroots z = z ( x,y )oftheequation F ( x,y,z )= z5+ xz y =0near(1 2 1).UseTaylorpolynomials T1( x,y )and T2( x,y )about(1 2)toapproximate z ( x,y ).Inparticular,calculate theapproximations z1= T1(0 7 2 5)and z2= T2(0 7 2 5)of z (0 7 2 5). Useacalculatortond F (0 7 2 5 ,z1)and F (0 7 2 5 ,z2).Theirdeviationfrom0determinesanerroroftheapproximations z1and z2.Which

PAGE 251

93.DIRECTIONALDERIVATIVEANDTHEGRADIENT245 oftheapproximationsismoreaccurate? Hint: UsetheresultofStudy Problem13.9. 93.DirectionalDerivativeandtheGradient93.1.DirectionalDerivative.Let f beafunctionofseveralvariables r =( x1,x2,...,xm).Thepartialderivative f xi( r0)istherateofchange inthedirectionofthe i thcoordinateaxis.Thisdirectionisdenedby theunitvector eiparalleltothecorrespondingcoordinateaxis.Let u beaunitvectorthatdoesnotcoincidewithanyofthevectors ei.What istherateofchangeof f at r0inthedirectionof u ?Forexample,if f ( x,y )istheheightofamountain,wherethe x and y axesareoriented alongthewesteastandsouthnorthdirections,respectively,thenitis reasonabletoaskabouttheslopes,forexample,inthesoutheastor northwestdirections.Naturally,theseslopesgenerallydierfromthe slopes f xand f y. P0 tan = D uf ( x0,y0) z = f ( x,y ) u ( x0,y0) DFigure13.9.Geometricalsignicanceofthedirectional derivativeofafunctionoftwovariables.Considerthegraph z = f ( x,y )overaregion D .Theverticalplane(i.e.,paralleltothe z axis)through( x0,y0)andparalleltotheunit vector u intersectsthegraphalongacurve.Theslopeof thetangentlinetothecurveofintersectionatthepoint P0=( x0,y0,f ( x0,y0))isdeterminedbythedirectionalderivative D uf ( x0,y0).Sothedirectionalderivativedenesthe rateofchangeof f at( x0,y0)inthedirection u .

PAGE 252

24613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Toanswerthequestionabouttheslopeinthedirectionofaunit vector u ,considerastraightlinethrough r0parallelto u .Itsvector equationis r ( h )= r0+ h u ,where h isaparameterthatlabelspoints oftheline.Thevaluesof f alongthelinearegivenbythecomposition F ( h )= f ( r ( h )).Thenumbers F (0)and F ( h )arethevaluesof f ata givenpoint r0andthepoint r ( h ), h> 0,thatisatthedistance h from r0inthedirectionof u .Sotheslopeisgivenbythederivative F(0). Therefore,thefollowingdenitionisnatural. Definition 13.23 (DirectionalDerivative) Let f beafunctiononanopenset D .Thedirectionalderivativeof f at r0 D inthedirectionofaunitvector u isthelimit Duf ( r0)=limh 0f ( r0+ h u ) f ( r0) h ifthelimitexists. Thenumber Duf ( r0)istherateofchangeof f at r0inthedirectionof u .Supposethat f isa dierentiable function.Bydenition, Duf ( r0)= df ( r ( h )) /dh takenat h =0,where r ( h )= r0+ h u .So,by thechainrule, 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. Remark If f haspartialderivativesat r0,butisnotdierentiable at r0,thentherelation(13.14)isfalse .AnexampleisgiveninStudy Problem13.12.Notethat(13.14)followsfromthechainrule,butthe mereexistenceofpartialderivativesis not sucientforthechainrule tohold.Furthermore, evenifafunctionhasdirectionalderivativesat apointineverydirection,itmaynotbedierentiableatthatpoint (no goodlinearapproximationexistsatthatpoint). Equation(13.14)providesaconvenientwaytocomputethedirectionalderivativeif f isdierentiable.Recallalsothatifthedirection isspeciedbyanonunitvector u ,thenthecorrespondingunitvector canbeobtainedbydividingitbyitslength u ,thatis, u = u / u Example 13.40 Theheightofahillis f ( x,y )=(9 3 x2 y2)1 / 2, wherethe x and y axesaredirectedfromwesttoeastandfromsouth tonorth,respectively.Ahikerisatthepoint r0=(1 2) .Supposethe hikerisfacinginthenorthwestdirection.Whatistheslopethehiker sees?

PAGE 253

93.DIRECTIONALDERIVATIVEANDTHEGRADIENT247 Solution: Aunitvectorintheplanecanalwaysbewrittenintheform u =(cos sin ),wheretheangle iscountedcounterclockwisefrom thepositive x axis;thatis, =0correspondstotheeastdirection, = / 2tothenorthdirection, = tothewestdirection,andsoon. So,forthenorthwestdirection, =3 / 2and u =( 1 / 2 1 / 2)= ( u1,u2).Thepartialderivativesare f x= 3 x/ (9 3 x2 y2)1 / 2and f y= y/ (9 3 x2 y2)1 / 2.Theirvaluesat r0=(1 2)are f x(1 2)= 3 / 2and f y(1 2)= 2 / 2.By(13.14),theslopeis Duf ( r0)= f x( r0) u1+ f y( r0) u2=3 / 2 1 / 2=1 Ifthehikergoesnorthwest,hehastoclimbatanangleof45relative tothehorizon. Example 13.41 Findthedirectionalderivativeof f ( x,y,z )= x2+3 xz + z2y atthepoint (1 1 1) inthedirectiontowardthepoint (3 1 0) .Doesthefunctionincreaseordecreaseinthisdirection? Solution: Put r0=(1 1 1)and r1=(3 1 0).Thenthevector u = r1 r0=(2 2 1)pointsfromthepoint r0towardthepoint r1accordingtotherulesofvectoralgebra.Butitisnotaunitvector becauseitslengthis u =3.Sotheunitvectorinthesamedirection is u = u / 3=(2 / 3 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)= 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= 2 / 3 2 / 3+1 / 3= 1 Sincethedirectionalderivativeisnegative,thefunctiondecreasesat r0inthedirectiontoward r1(therateofchangeisnegativeinthat direction). 93.2.TheGradientandItsGeometricalSignicance.Definition 13.24 (TheGradient) Let f beadierentiablefunctionofseveralvariables r =( x1,x2,...,xm) onanopenset D andlet r0 D .Thevectorwhosecomponentsare partialderivativesof f at r0, f ( r0)=( f x1( r0) ,f x2( r0) ,...,f xm( r0)) iscalledthe gradient of 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)withthedenitionofthegradientand

PAGE 254

24813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS recallingthedenitionofthedotproduct,thedirectionalderivative cannowbewritteninthecompactform (13.15) Duf ( r0)= f ( r0) u Thisequationisthemostsuitableforanalyzingthesignicanceofthe gradient. Considerrstthecasesoftwo-andthree-variablefunctions.The gradientisavectorineitheraplaneorspace,respectively.InExample 13.40,thegradientat(1,2)is f (1 2)=( 3 / 2 2 / 2).InExample13.41,thegradientat(1,1, 1)is f (1 1 1)=( 1 1 1).Recall thegeometricalpropertyofthedotproduct a b = a b cos ,where [0 ]istheanglebetweenthenonzerovectors a and b .Thevalue =0correspondstoparallelvectors a and b .When = / 2,the vectorsareorthogonal.Thevectorspointintheoppositedirections if = .Assumethat f ( r0) = 0 .Let betheanglebetweenthe gradient f ( r0)andtheunitvector u .Then (13.16) Duf ( r0)= f ( r0) u = f ( r0) u cos = f ( r0) cos because u =1(theunitvector).Asthecomponentsofthegradient arexednumbers(thevaluesofthepartialderivativesataparticular point r0),thedirectionalderivativeat r0variesonlyifthevector u changes.Thus,theratesofchangeof f inalldirectionsthathavethe sameangle withthegradientarethesame.Inthetwo-variablecase, onlytwosuchdirectionsarepossibleif u isnotparalleltothegradient, whileinthethree-variablecasetheraysfrom r0inallsuchdirections formaconewhoseaxisisalongthegradientasdepictedintheleftand rightpanelsofFigure13.10,respectively.Itisthenconcludedthatthe y x u2 u1 P0 f x y z P0 u fFigure13.10.Left :Thesamerateofchangeofafunction oftwovariablesatapoint P0occursintwodirectionsthat havethesameanglewiththegradient f ( P0). Right :The samerateofchangeofafunctionofthreevariablesatapoint P0occursininnitelymanydirectionsthathavethesame anglewiththegradient f ( P0).

PAGE 255

93.DIRECTIONALDERIVATIVEANDTHEGRADIENT249 P f ( P ) f ( x,y )= k f ( x,y )= k f f f C f ( x,y,z )= k f ( P ) PFigure13.11.Left :Thegradientatapoint P isnormal toalevelcurve f ( x,y )= k through P ofafunction f of twovariables. Middle :Acurve C ofsteepestdescentor ascentforafunction f hasthecharacteristicpropertythat thegradient f istangenttoit.Thelevelcurves(surfaces) of f arenormalto C .Thefunction f increasesmostrapidly along C inthedirectionof f ,and f decreasesmostrapidly along C intheoppositedirection f Right :Thegradient ofafunctionofthreevariablesisnormaltoanycurvethrough P inthelevelsurface f ( x,y,z )= k .So f ( P )isanormal tothetangentplanethrough P tothelevelsurface.rateofchangeof f attainsitsabsolutemaximumorminimumwhen cos does.Therefore, themaximalrateisattainedinthedirection ofthegradient( =0 )andisequaltothemagnitudeofthegradient f ( r0) ,whereastheminimalrateofchange f ( r0) occursin thedirectionof f ( r0) ,thatis,oppositetothegradient ( = ). Thegraphofafunctionoftwovariables z = f ( x,y )maybeviewed astheshapeofahill.Thenthegradientataparticularpointshows thedirectionofthe steepestascent ,whileitsoppositepointsinthe directionofthe steepestdescent .InExample13.40,themaximalslope atthepoint(1,2)is f ( r0) =(1 / 2) ( 3 2) = 13 / 2.Itoccurs inthedirectionof( 3 / 2 2 / 2)or( 3 2)(themultiplicationofa vectorbyapositiveconstantdoesnotchangeitsdirection).If isthe anglebetweenthepositive x axis(orthevector e1)andthegradient, thentan = 2 / 3or 146.Ifthehikergoesinthisdirection,he hastoclimbupatanangleoftan 1( 13 / 2) 69withthehorizon. Also,notethehikersoriginaldirectionwas =135,whichmakes theangle11withthedirectionofthesteepestascent.Sotheslopein thedirection =146+11=157hasthesameslopeasthehikers originalone.Ashasbeenargued,inthetwo-variablecase,therecan onlybetwodirectionswiththesameslope. Next,consideralevelcurve f ( x,y )= k ofadierentiablefunction oftwovariables.Supposethatthereisadierentiablevectorfunction r ( t )=( x ( t ) ,y ( t ))thattracesoutthelevelcurve.Thisvectorfunction

PAGE 256

25013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS shouldsatisfytheconditionthat f ( x ( t ) ,y ( t ))= k forallvaluesofthe parameter t .Bythedenitionoflevelcurves,thefunction f hasa constantvalue 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)isatangentvector tothecurveatthepoint r0.Thus, thegradient f ( r0) isorthogonal toatangentvectoratthepoint r0tothelevelcurveof f throughthat point.Thisisoftenexpressedbysayingthatthegradientof f isalways normaltosmoothlevelcurvesof f Theexistenceofasmoothvectorfunctionthattraversesalevel curveof f canbeestablishedbytheimplicitfunctiontheorem.Suppose that f hascontinuouspartialderivativesand f ( r0) = 0 atapoint r0=( x0,y0)onthelevelsurface f ( x,y )= k .Thelevelsurfacecanalso bedenedasthesetofrootsofthefunction F ( x,y )= f ( x,y ) k ,that is,thesetofsolutionof F ( x,y )=0.Inparticular,thepoint r0isa root, F ( x0,y0)=0.Thefunction F hascontinuouspartialderivatives and F ( r0)= f ( r0) = 0 .Thecomponentsofthegradientdonot vanishsimultaneously,andwithoutlossofgenerality,onecanassume that F y( r0)= f y( r0) =0.Then,bytheimplicitfunctiontheorem, thereisafunction y = g ( x )suchthat F ( x,g ( x ))=0insomeopen intervalcontaining x0;thatis,thegraph y = g ( x )coincideswiththe levelcurve f ( x,y )= k inaneighborhoodof( x0,y0),where y0= g ( x0). Furthermore,thederivative g( x )= F x/F y= f x/f yexistsinthat interval.Hence,thevectorfunction r ( t )=( t,g ( t ))traversesthegraph y = g ( x )andthelevelcurvenear r0= r ( t0),where t0= x0.Itis smoothbecause r( t )=(1 ,g( t ))sothat r( t ) = 0 Recallthatafunction f ( x,y )canbedescribedbyacontourmap, whichisacollectionoflevelcurves.Iflevelcurvesaresmoothenoughto havetangentvectorseverywhere,thenonecandeneacurvethrougha particularpointthatisnormaltoalllevelcurvesinsomeneighborhood ofthatpoint.Thiscurveiscalledthe curveofsteepestdescentorascent throughthatpoint.Thetangentvectorofthiscurveatanypoint isparalleltothegradientatthatpoint.Thevaluesofthefunction increase(ordecrease)mostrapidlyalongthiscurve.Ifahikerfollows thedirectionofthegradientoftheheight,hewouldgoalongthepath ofsteepestascentordescent. Thecaseoffunctionsofthreevariablescanbeanalyzedalongsimilarlines.Letafunction f ( x,y,z )havecontinuouspartialderivatives

PAGE 257

93.DIRECTIONALDERIVATIVEANDTHEGRADIENT251 inaneighborhoodof r0=( x0,y0,z0)suchthat f ( r0) = 0 .Consideralevelsurfaceof f through r0, f ( x,y,z )= k ,whichisasetof rootsofthefunctions F ( x,y,z )= f ( x,y,z ) k .Sincethecomponents ofthegradient F ( r0)= f ( r0)donotvanishsimultaneously,one canassumethat,say, F z( r0)= f z( r0) =0.Bytheimplicitfunction theorem,thereisafunction g ( x,y )suchthatthegraph z = g ( x,y )coincideswiththelevelsurfacenear r0;thatis, F ( x,y,g ( x,y ))=0forall ( x,y )near( x0,y0).Thefunction g hascontinuouspartialderivatives, anditslinearizationat( x0,y0)denesaplanetangenttothegraph z = g ( x,y )at r0,where z0= g ( x0,y0).Anormalofthetangentplane is n =( g x,g y, 1),wherethederivativesaretakenatthepoint( x0,y0). Using g x= F x/F z= f x/f zand g y= F x/F z= f x/f z,wherethe derivativesof f aretakenat r0,itfollowsthat n = 1 f z( r0) f x( r0) ,f y( r0) ,f z( r0) = 1 f z( r0) f ( r0) Thus,thegradient f ( r0)isproportionalto n andhence isalsonormal tothetangentplane .Furthermore,if r ( t )=( x ( t ) ,y ( t ) ,z ( t ))isasmooth curveonthelevelsurface,thatis, f ( r ( t ))= k forallvaluesof t ,then df/dt =0(thevaluesof f donotchangealongthecurve),and,bythe chainrule, df/dt = f xx+ f yy+ f zz= f ( r ( t )) r( t )=0 Sothegradient f ( r0)isorthogonaltoatangentvectorto any such curvethrough r0,andalltangentvectorstocurvesthrough r0liein thetangentplanetothelevelsurfacethrough r0. Allthesendingsaresummarizedinthefollowingtheorem. Theorem 13.16 (GeometricalPropertiesoftheGradient) Let f bedierentiableat r0.Let S bethelevelsetthroughthepoint r0, andassume f ( r0) = 0 .Then (1) Themaximalrateofchangeof f at r0occursinthedirection ofthegradient f ( r0) andisequaltoitsmagnitude f ( r0) (2) Theminimalrateofchangeof f at r0occursinthedirection oppositetothegradient f ( r0) andequals f ( r0) (3) IffhascontinuouspartialderivativesonanopenballDcontaining r0,thentheportionof S inside D isasmoothsurface (orcurve),and f isnormalto S at r0. Example 13.42 Findanequationofthetangentplanetotheellipsoid x2+2 y2+3 z2=11 atthepoint (2 1 1) Solution: Theequationoftheellipsoidcanbeviewedasthelevelsurface f ( x,y,z )=11ofthefunction f ( x,y,z )= x2+2 y2+3 z2through

PAGE 258

25213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS thepoint r0=(2 1 1)because f (2 1 1)=11.Bythegeometrical propertyofthegradient,thevector n = f ( r0)isnormaltotheplane inquestionbecausethecomponentsof f =(2 x, 4 y, 6 z )arecontinuous.Onehas n =(4 4 6).Anequationoftheplanethroughthe point(2,1,1)andnormalto n is4( x 2)+4( y 1)+6( z 1)=0or 2 x +2 y +3 z =9. Theorem13.16holdsforfunctionsofmorethanthreevariablesas well.Equation(13.15)wasobtainedforanynumberofvariables,and therepresentationofthedotproduct(13.16)holdsinanyEuclidean space.Thus,thersttwopropertiesofthegradientarevalidinany multivariablecase.Thethirdpropertyishardertovisualizeasthe levelsurfaceofafunctionof m variablesisan( m 1)-dimensional surfaceembeddedinan m -dimensionalEuclideanspace.Tothisend, itisonlynotedthatif r ( t )isasmoothcurveinthelevelset f ( r )= k ofadierentiablefunction,then f hasaconstantvaluealonganysuch curve,and,bythechainrule,itfollowsthat df ( r ( t )) /dt = f ( r ( t )) r( t )=0forany t .Atanyparticularpoint r0= r ( t0),alltangent vectors r( t0)tosuchcurvesthrough r0areorthogonaltoa single vector f ( r0).Intuitively,thesevectorsshouldforman( m 1)-dimensional Euclideanspace(calleda tangentspace tothelevelsurfaceat r0),just likeallvectorsinaplaneinthree-dimensionalEuclideanspaceare orthogonaltoanormaloftheplane. Remark. Thegradientcanbeviewedastheresultoftheactionof theoperator =( /x1,/x2,...,/xm)onafunction f f is understoodinthesenseofmultiplicationofthevector byascalar f .Withthisnotation,thedierentialoperator d hasacompactform d = d r .Thelinearization L ( r )of f ( r )at r0andthedierentialsof f alsohaveasimpleformforanynumberofvariables: L ( r )= f ( r0)+ f ( r0) ( r r0) ,dnf ( r )=( d r )nf ( r ) .93.3.StudyProblems.Problem13.12.(DierentiabilityandDirectionalDerivative). Let f ( x,y )= y3/ ( x2+ y2) if ( x,y ) =(0 0) and f (0 0)=0 .Showthat Duf (0 0) existsforany u ,butitisnotgivenbytherelation(13.14). Showthatthisfunctionisnotdierentiableat (0 0) .Thus, theexistenceofalldirectionalderivativesatapointdoesnotimplydierentiabilityatthatpoint.Inotherwords,despitethatthefunctionhasarate ofchangeatapointineverydirection,agoodlinearapproximation maynotexistatthatpoint .

PAGE 259

93.DIRECTIONALDERIVATIVEANDTHEGRADIENT253 Solution: Put u =(cos sin )for0 < 2 .Bythedenitionof thedirectionalderivative, Duf (0 0)=limh 0f ( h cos ,h sin ) f (0 0) h =limh 0h3sin3 h3=sin3. Inparticular,for =0, u =(1 0),and Duf (0 0)= f x(0 0)=0; similarlyfor = / 2, u =(0 1),and Duf (0 0)= f y(0 0)=1.If therelation(13.14)wereused,onewouldhavefoundthat Duf (0 0)= sin ,whichcontradictstheaboveresult.Ifagoodlinearapproximation exists,thenitshouldbe L ( x,y )= f x(0 0) x + f y(0 0) y = y .But ( f ( x,y ) L ( x,y )) / ( x2+ y2)1 / 2= yx2/ ( x2+ y2)3 / 2doesnotvanish as( x,y ) (0 0)becauseithasanonzeroconstantvaluealongany straightline( x,y )=( t,at ), a =0.So f isnotdierentiableatthe origin. Problem13.13. Supposethatthreelevelsurfaces f ( x,y,z )=1 g ( x,y,z )=2 ,and h ( x,y,z )=3 areintersectingalongasmoothcurve C .Let P beapointon C inwhoseneighborhood f g ,and h have continuouspartialderivativesandtheirgradientsdonotvanishat P 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 beinaplaneperpendiculartothevectorv .Thetripleproductfor anythreecoplanarvectorsvanishes,andhence f ( g h )=0 at P Problem13.14.(EnergyConservationinMechanics). ConsiderNewtonssecondlaw m a = F .Supposethattheforceisthe gradient F = U ,where U = U ( r ) .Let r = r ( t ) bethetrajectory satisfyingNewtonslaw.Provethatthequantity E = mv2/ 2+ U ( r ) where v = r( t ) isthespeed,isaconstantofmotion,thatis, dE/dt = 0 .Thisconstantiscalledthe totalenergy ofaparticle. 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 F )=0 Sothetotalenergyisconservedforthetrajectoryofthemotion.

PAGE 260

25413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 93.4.Exercises.(1) Let f ( x,y )beadierentiablefunction.Howwouldyouspecifythe directionsataparticularpointinwhichthefunctiondoesnotchange atall?Howmanysuchdirectionsexistiftherstpartialderivatives donotvanishatthatpoint?Answerthesamequestionsforafunction f ( x,y,z ). (2) Foreachofthefollowingfunctions,ndthegradientandthedirectionalderivativeataspeciedpointinthedirectionparalleltoagiven vector v .Indicatewhetherthefunctionincreasesordecreasesinthat direction. (i) f ( x,y )= x2y ,(1 2), v =(4 5) (ii) f ( x,y )= x/ (1+ xy ),(1 1), v =(2 1) (iii) f ( x,y,z )= x2y zy2+ xz2,(1 2 1), v =(1 2 2) (iv) f ( x,y,z )=tan 1(1+ x + y2+ z3),(1 1 1), v =(1 1 1) (v) f ( x,y,z )= x + yz ,(1 1 3), v =(2,6,3) (vi) f ( x,y,z )=( x + y ) /z ,(2 1 1), v =(2 1 2) (3) Findthemaximalandminimalratesofchangeofeachofthefollowingfunctionsataspeciedpointandthedirectionsinwhichthey occur.Findthedirectionsinwhichthefunctiondoesnotchange. (i) f ( x,y )= x/y2,(2 1) (ii) f ( x,y )= xy,(2 1) (iii) f ( x,y,z )= xz/ (1+ yz ),(1 2 3) (iv) f ( x,y,z )= x sin( yz ),(1 2 ,/ 3) (v) f ( x,y,z )= xyz,(2 2 1) (4) Let f ( x,y )= y/ (1+ x2+ y ).Findallunitvectors u alongwhich therateofchangeof f at(2 3)isanumber 1 p 1timesthe maximalrateofchangeof f at(2 3). (5) Forthefunction f ( x,y,z )=1 2x21 2y2x + z3y atthepoint P0(1 2 1) nd: (i)Themaximalrateofchangeof f andthedirectioninwhichit occurs; (ii)Adirectioninwhichtherateofchangeishalfofthemaximal rateofchange.Howmanysuchdirectionsexist? (iii)Therateofchangeinthedirectiontothepoint P1(3 1 1) (6) If f and u aredierentiablefunctions,provethat f ( u )= f( u ) u (7) Find c r 2,where c isaconstantvector. (8) If f u ,and v aredierentiablefunctions,provethat f ( u,v )= f u u + f v v (9) Findthedirectionalderivativeof f ( r )=( x/a )2+( y/b )2+( z/c )2at apoint r inthedirectionof r .Findthepointsatwhichthisderivative isequalto f .

PAGE 261

93.DIRECTIONALDERIVATIVEANDTHEGRADIENT255 (10) Findtheanglebetweenthegradientsof f = x/ ( x2+ y2+ z2)at thepoints(1 2 2)and( 3 1 0). (11) Let f = z/ x2+ y2+ z2.Sketchthelevelsurfacesof f and f .Whatisthesignicanceofthelevelsurfacesof f ?Find themaximalandminimalvaluesof f and f intheregion1 z 2. (12) Letacurve C bedenedastheintersectionoftheplane sin ( x x0) cos ( y y0)=0,where isaparameter,andthe graph z = f ( x,y ),where f isdierentiable.Findtan ,where isthe anglebetweenthetangentlineto C at( x0,y0,f ( x0,y0))andthe xy plane. (13) Considerthefunction f ( x,y,z )=2 z + xy andthreepoints P0(1 2 2), P1( 1 4 1),and P2( 2 2 2).Inwhichdirectiondoes f changefasterat P0,toward P1ortoward P2?Whatisthedirectionin which f increasesmostrapidlyat P0? (14) Forthefunction f ( x,y,z )= xy + zy + zx atthepoint P0(1 1 0), nd: (i)Themaximalrateofchange (ii)Therateofchangeinthedirection v =( 1 2 2) (iii)Theangle between v andthedirectioninwhichthemaximal rateof f occurs (15) Let f ( x,y,z )= x/ ( x2+ y2+ z2)1 / 2.Findtherateofchangeof f inthedirectionofthetangentvectortothecurve r ( t )=( t, 2 t2, 2 t2) atthepoint(1 2 2). (16) Findthepointsatwhichthegradientof f = x3+ y3+ z3 3 xyz is (i)Orthogonaltothe z axis (ii)Paralleltothe z axis (iii)Zero (17) Let f =ln r r0 ,where r0isaxedvector.Findpointsin spacewhere f =1. (18) Foreachofthefollowingsurfaces,ndthetangentplaneandthe normallineataspeciedpoint: (i) x2+ y2+ z2=169,(3 4 12) (ii) x2 2 y2+ z2+ yz =2,(2 1 1) (iii) x =tan 1( y/z ),( / 4 1 1) (iv) z = y +ln( x/z ),(1 1 1) (v)2x/z+2y/z=8,(2 2 1) (vi) x2+4 y2+3 z2=5,(1 1 / 2 1) (19) Findthepointsofthesurface x2+2 y2+3 z2+2 xy +2 zx +4 yz =8 atwhichthetangentplanesareparalleltothecoordinateplanes. (20) Findthetangentplanestothesurface x2+2 y2+3 z2=21that areparalleltotheplane x +4 y +6 z =0.

PAGE 262

25613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (21) Findthepointsontheellipsoid x2/a2+ y2/b2+ z2/c2=1atwhich thenormallinemakesequalangleswiththecoordinateaxes. (22) Considertheparaboloid z = x2+ y2. (i)Givetheparametricequationsofthenormallinethrougha point P0( x0,y0,z0)ontheparaboloid; (ii)Considerallnormallinesthroughpointswithaxedvalueof z0(say, z0=2).Showthatallsuchlinesintersectatonesingle pointthatliesonthe z axisandndthecoordinatesofthis point. (23) Findthepointsonthehyperboloid x2 y2+2 z2=5,wherethe normallineisparalleltothelinethatjoinsthepoints(3 1 0)and (5 3 8). (24) Findanequationoftheplanetangenttothesurface x2+ y2 4 z2= 1atagenericpoint( x0,y0,z0)ofthesurface. (25) Findtherateofchangeofthefunction h ( x,y )= 10 x2y2atthepoint P0(1 1)inthedirectiontowardthepoint P ( 2 5).Let h ( x,y )betheheightinaneighborhoodof P0.Wouldyoubeclimbing uporgettingdownwhenyougofrom P0toward P ? (26) YourMarsroveriscaughtontheslopeofamountainsbyadust storm.Thevisibilityis0.Yourcurrentpositionis P0(1 2).You canescapeinthedirectionofacavelocatedat P1(4 2)orinthe directionofthebaselocatedat P2(17 14).Whichwaywouldyoudrive toavoidsteepclimbingordescendingiftheheightintheareacanbe approximatedbythefunction h ( x,y )= xy + x2? (27) YouareyingasmallaircraftontheplanetWeirdo.Youhave disturbedanestofnastyeverything-eatingbugs.Theonboardradar indicatesthattheconcentrationofthebugsis C ( x,y,z )=100 x2 2 y2 3 z2and C ( x,y,z )=0if x2+2 y2+3 z2> 100.Ifyourcurrent positionis(2 3 1),inwhichdirectionwouldyoureamass-destruction microwavelasertokillasmanypoorbugsaspossiblenearyou?Find theoptimalescapetrajectory. (28) Lettwolevelcurves f ( x,y )=0and g ( x,y )=0offunctions f and g ,whosepartialderivativesarecontinuous,intersectatsomepoint P0. Therateofchangeofthefunction f at P0alongthecurve g ( x,y )=0 ishalfofitsmaximalrateofchangeat P0.Whatistheangleatwhich thecurvesintersect(theanglebetweenthetangentlines)? (29) Supposethatthedirectionalderivatives Duf = a and Dvf = b ofadierentiablefunction f ( x,y )areknownataparticularpoint P0fortwounitnonparallelvectors u and v thatmaketheangles and withthe x axiscountedcounterclockwisefromthelatter,respectively. Findthegradientof f at P0. (30) Threetestsofdrillingintorockalongthedirections u =(1 2 2), v =(0 4 3),and w =(0 0 1)showedthatthegoldconcentration

PAGE 263

94.MAXIMUMANDMINIMUMVALUES257 increasesattherates3g/m,3g/m,and1g/m,respectively.Assume thattheconcentrationisadierentiablefunction.Inwhatdirection wouldyoudrilltomaximizethegoldyieldandatwhatratedoesthe goldconcentrationincreaseinthatdirection?Iftheconcentrationis notdierentiable,wouldyoufollowyourpreviousndingaboutthe drillingdirection?Explain. (31) Alevelsurfaceofadierentiablefunction f ( x,y,z )containsthe curves r1( t )=(2+3 t, 1 t2, 3 4 t + t2)and r2( t )=(1+ t2, 2 t3 1 2 t +1). Canthisinformationbeusedtondthetangentplanetothesurface at(2 1 3)?Ifso,ndanequationoftheplane. (32) Provethattangentplanestothesurface xyz = a3> 0andthe coordinateplanesformtetrahedronsofequalvolumes. (33) Provethetotallengthofintervalsfromtheorigintothepoints ofintersectionoftangentplanestothesurface x + y + z = a a> 0,withthecoordinateaxesisconstant. (34) Twosurfacesarecalled orthogonal atapointofintersectionifthe normallinestothesurfacesatthatpointareorthogonal.Showthat thesurfaces x2+ y2+ z2= r2, x2+ y2= z2tan2 ,and y cos = x sin arepairwiseorthogonalattheirpointsofintersectionforanyvaluesof theconstants r> 0,0 << ,and0 < 2 (35) Findthedirectionalderivativeof f ( x,y,z )inthedirectionof thegradientof g ( x,y,z ).Whatisthegeometricalsignicanceofthis derivative? (36) Findtheangleatwhichthecylinder x2+ y2= a2intersectsthe surface bz = xy atagenericpointofintersection( x0,y0,z0). (37) Arayoflightreectsfromamirroredsurfaceatapoint P justas itwouldreectfromthemirroredplanetangenttothesurfaceat P (if thelighttravelsalongavector u ,thenthereectedlighttravelsalong avectorobtainedfrom u byreversingthedirectionofthecomponent paralleltothenormaltothesurface).Showthatthelightcomingfrom thetopofthe z axisandparalleltoitwillbefocusedbytheparabolic mirror az = x2+ y2, a> 0,toasinglepoint.Finditscoordinates. Thispropertyofparabolicmirrorsisusedtodesigntelescopes. (38) Let f ( x,y )= y if y = x2and f ( x,y )=0if y = x2.Find Dnf (0 0) forallunitvectors n .Showthat f ( x,y )isnotdierentiableat(0 0). Isthefunctioncontinuousat(0 0)? 94.MaximumandMinimumValues94.1.CriticalPointsofMultivariableFunctions.Thepositionsofthelocalmaximaandminimaofaone-variablefunctionplayanimportant rolewhenanalyzingitsoverallbehavior.InCalculusI,itwasshown

PAGE 264

25813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS howthederivativescanbeusedtondlocalmaximaandminima.Here thisanalysisisextendedtomultivariablefunctions. Thefollowingnotationwillbeused.Anopenballofradius centeredatapoint r0isdenoted B= { r | r r0 < } ;thatis,itisaset ofpointswhosedistancefrom r0islessthan > 0.Aneighborhood Nofapoint r0inaset D isasetofcommonpointsof D and B;that is, N= D Bcontainsallpointsin D whosedistancefrom r0isless than Definition 13.25 (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 r0,andthenumber f ( r0) iscalleda localminimumvalue .Iftheinequality f ( r0) f ( r ) or f ( r0) f ( r ) holdsforallpoints r inthedomainof f ,then f hasan absolutemaximumorabsoluteminimumat r0,respectively. Minimalandmaximalvaluesarealsocalled extremumvalues .In theone-variablecase,Fermatstheoremassertsthatifadierentiable functionhasalocalextremumat x0,thenitsderivativevanishesat x0. Thetangentlinetothegraphof f at x0ishorizontal: y = f ( x0)+ df ( x0)= f ( x0)+ f( x0) dx = f ( x0).ThereisanextensionofFermats theorem. Theorem 13.17 (NecessaryConditionforaLocalExtremum) Ifadierentiablefunction f hasalocalextremumataninteriorpoint r0ofitsdomain D ,then df ( r0)=0 or f ( r0)=0 (allpartialderivativesof f vanishat r0). Proof. Considerasmoothcurve r ( t )throughthepoint r0suchthat r ( t0)= r0.Then d r ( t0)= r( t0) dt = 0 (thecurveissmoothand hencehasanonzerotangentvector).Thefunction F ( t )= f ( r ( t )) denesthevaluesof f alongthecurve.Therefore, F ( t )musthave alocalextremumat t = t0.Since f isdierentiable,thedierential dF ( t0)= F( t0) dt existsbythechainrule: dF ( t0)= d r ( t0) f ( r0). ByFermatstheorem dF ( t0)=0andhence d r ( t0) f ( r0)=0.This relationmeansthatthevectors d r ( t0)and f ( r0)areorthogonalfor allsmoothcurvesthrough r0.Theonlyvectorthatisorthogonalto anyvectoristhezerovector,andtheconclusionofthetheoremfollows: f ( r0)= 0 Inparticular,foradierentiablefunction f oftwovariables,this theoremstatesthatthetangentplanetothegraphof f atalocal extremumishorizontal.

PAGE 265

94.MAXIMUMANDMINIMUMVALUES259 Theconverseofthistheoremisnottrue.Let f ( x,y )= xy .It isdierentiableeverywhere,anditspartialderivativesare f x= y and f y= x .Theyvanishattheorigin, f (0 0)= 0 .However,thefunction hasneitheralocalmaximumnoralocalminimum.Indeed,considera straightlinethroughtheorigin, x = at y = bt .Thenthevaluesof f alongthelineare F ( t )= f ( x ( t ) ,y ( t ))= abt2.So F ( t )hasaminimum at t =0if ab> 0oramaximumif ab< 0.Eachcaseispossible. Forexample,if a = b =1,then ab =1 > 0;if a = b =1,then ab = 1 < 0.Thus, f cannothavealocalextremumat(0 0).The graph z = xy isahyperbolicparaboloidrotatedthroughanangle / 4 aboutthe z axis(seeExample11.29).Itlookslikeasaddle.Ifthe graphof f ( x,y )hasahorizontaltangentplaneat( x0,y0)andlooks likeahyperbolicparaboloidinasmallneighborhoodof( x0,y0),then thepoint( x0,y0)iscalleda saddlepoint (seeFigure13.12,rightpanel). Remark. Theaboveanalysismightmaketheimpressionthatif f ( r0) = 0 andthevaluesof f alonganystraightlinethrough r0havealocalextremum(i.e., F ( t )= f ( r0+ v t )haseitheralocalmaximumor alocalminimumat t =0forallvectors v ),then f hasalocalextremumat r0.Thisconjectureis false !AnexampleisgiveninStudy Problem13.16. Alocalextremummayoccuratapointatwhichthefunctionisnot dierentiable.Forexample, f ( x,y )= | x | + | y | iscontinuouseverywhere z = f ( x,y ) y0x0Nz = f ( x,y ) y0x0Nz = f ( x,y ) y0x0NFigure13.12.Left :Thegraph z = f ( x,y )nearalocal minimumof f .Thevaluesof f arenolessthan f ( x0,y0)for all( x,y )inasucientlysmallneighborhood Nof( x0,y0). Middle :Thegraph z = f ( x,y )nearalocalmaximumof f .Thevaluesof f donotexceed f ( x0,y0)forall( x,y )ina sucientlysmallneighborhood Nof( x0,y0). Right :The graph z = f ( x,y )nearasaddlepointof f .Inasuciently smallneighborhood Nof( x0,y0),thevaluesof f havea localmaximumalongsomelinesthrough( x0,y0)andalocal minimumalongtheotherlinesthrough( x0,y0).

PAGE 266

26013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS andhasanabsoluteminimumat(0 0).However,thepartialderivatives f x(0 0)and f y(0 0)donotexist(e.g., f x( x,y )=( | x | ),whichis1if x> 0andis 1if x< 0,so f x(0 ,y )doesnotexistandneitherdoes f x(0 0)). Definition 13.26 (CriticalPoints) Aninteriorpoint r0ofthedomainofafunction f issaidtobea critical point of f ifeither f ( r0)= 0 orthegradientdoesnotexistat r0. Thus, if f hasalocalmaximumorminimumat r0,then r0isa criticalpointof f .However,notallcriticalpointscorrespondtoeither alocalmaximumoralocalminimum .94.2.Concavity.RecallfromCalculusIthatifthegraphofafunction f ( x )liesaboveallitstangentlinesinaninterval I ,then f isconcave upwardon I .Ifthegraphliesbelowallitstangentlinesin I ,then f isconcavedownwardon I .Furthermore,if f( x0)=0(thetangent lineishorizontalat x0)and f isconcaveupwardinsmallopeninterval I containing x0,then f ( x ) >f ( x0)forall x = x0in I ,andhence f hasalocalmaximum.Similarly, f hasalocalminimumat x0, where f( x0)=0,ifitisconcavedownwardinaneighborhoodof x0.Ifthefunction f istwicedierentiableon I ,thenitisconcave upwardif f( x ) > 0on I anditisconcavedownwardif f( x ) < 0on I .Theconcavitytestcanberestatedintheformofthesecond-order dierential d2f ( x )= f( x )( dx )2,whichisafunctionoftwoindependent variables x and dx .If d2f ( x ) > 0for dx =0, f isconcaveupward;if d2f ( x ) < 0for dx =0, f isconcavedownward.Supposethat f( x0)= 0, f( x0) =0,and fiscontinuousat x0.Thecontinuityof fensures that d2f ( x )hasthesamesignas d2f ( x0)forall x near x0andall dx =0.Hence,thegraphof f hasaxedconcavityinaneighborhood of x0.Thus,if d2f ( x0) < 0( dx =0),then f hasalocalmaximum at x0;if d2f ( x0) > 0( dx =0),then f hasalocalminimumat x0.It turnsoutthatthissucientconditionforafunctiontohavealocal extremumhasanaturalextensiontofunctionsofseveralvariables. Theorem 13.18 (SucientConditionforaLocalExtremum). Supposethatafunction f hascontinuoussecondpartialderivativesin anopenballcontainingapoint r0and f ( r0)= 0 .Then f hasalocalmaximumat r0if d2f ( r0) < 0 f hasalocalminimumat r0if d2f ( r0) > 0 forall d r suchthat d r =0 Theproofofthistheoremisomitted.However,ananalogycanbe madewiththeone-variablecase.BytheTaylortheorem(Theorem

PAGE 267

94.MAXIMUMANDMINIMUMVALUES261 13.15),valuesofafunction f inasucientlysmallneighborhoodof apoint r0arewellapproximatedas f ( r )= f ( r0)+ df ( r0)+1 2d2f ( r0), where d r = r r0.Thersttwotermsdenealinearization L ( r )= f ( r0)+ df ( r0)(ortangentplane)of f at r0.Therefore, f ( r ) L ( r )=1 2d2f ( r0) wherethecontributionsoftermssmallerthan d r 2havebeenneglectedaccordingtoTheorem13.15.Thisequationshowsthatif d2f ( r0) < 0forall d r =0(asafunctionofindependentvariables d r ),thenthevaluesof f arestrictlylessthanthevaluesofitslinearizationinaneighborhood0 < r r0 < if > 0issmallenough. Thecontinuityofthesecondpartialderivativesinaneighborhoodof r0ensuresthat,forall r near r0andall d r d2f ( r )isacontinuousfunctionoftwoindependentvariables r and d r .Therefore,if d2f ( r0) < 0, d r =0,then d2f ( r ) < 0forall r near r0.Thevaluesof f along any smoothcurvethrough r0havealocalmaximumat r0if,inaddition, df ( r0)=0.Inthetwo-variablecase,onecansaythatthegraphof f is concavedownwardnear r0;itlookslikeaparaboloidconcavedownward (seeFigure13.12,middlepanel).Thefunctionhasalocalmaximum. Similarly,if d2f ( r0) > 0forall d r =0,thenthegraphof f lies above itstangentplanesforall0 < r r0 < .Thefunctionhasalocal minimumat r0.Thegraphof f near r0lookslikeaparaboloidconcave upward(seeFigure13.12,leftpanel).94.3.Second-DerivativeTest.Thedierential d2f ( r0)isahomogeneous quadraticpolynomialinthevariables d r .Itssignisdeterminedbyits coecients,whicharethesecond-orderpartialderivativesof f at r0. Thecaseoffunctionsoftwovariablesisdiscussedrst. Supposethatafunction f ( x,y )hascontinuoussecondderivatives inanopenballcenteredat r0.Thesecondderivatives a = f xx( r0), b = f yy( r0),and c = f xy( r0)= f yx( r0)(Clairautstheorem)canbe arrangedintoa2 2symmetricmatrixwhosediagonalelementsare a and b andwhoseo-diagonalelements c .Thequadraticpolynomialof avariable P2( )=det a c cb =( a )( b ) c2, iscalledthe characteristicpolynomial ofthematrixofsecondpartial derivativesof f at r0. Theorem 13.19 (Second-DerivativeTest) Let r0beacriticalpointofafunction f .Supposethatthesecondorderpartialderivativesof f arecontinuousinanopendiskcontaining r0.Let P2( ) bethecharacteristicpolynomialofthematrixofsecond derivativesat r0.Let i, i =1 2 ,betherootsof P2( ) .Then

PAGE 268

26213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (1) Iftherootsarestrictlypositive, i> 0 ,then f hasalocal minimumat r0. (2) Iftherootsarestrictlynegative, i< 0 ,then f hasalocal maximumat r0. (3) Iftherootsdonotvanishbuthavedierentsigns,then f has neitheralocalmaximumnoralocalminimumat r0. (4) Ifatleastoneoftherootsvanishes,then f mayhavealocal maximum,alocalminimum,ornoneoftheabove(thesecondderivativetestisinconclusive). Incase(3),thecriticalpointissaidtobea saddlepoint of f and thegraphof f crossesitstangentplaneat r0. Proof. Considerarotation ( dx,dy )=( dxcos dysin ,dycos + dxsin ) FollowingtheproofofTheorem11.8(classicationofquadriccylinders),thesecond-orderdierentialiswritteninthenewvariables ( dx,dy)as d2f ( r0)= a ( dx )2+2 cdxdy + b ( dy )2= a( dx)2+2 cdxdy+ b( dy)2, a=1 2 a + b +( a b )cos(2 )+2 c sin(2 ) b=1 2 a + b ( a b )cos(2 ) 2 c sin(2 ) 2 c=2 c cos(2 ) ( a b )sin(2 ) Therotationangleischosensothat c=0.Put A2=( a b )2+4 c2.If cos(2 )=( a b ) /A andsin(2 )=2 c/A ,then c=0.Withthischoice, a=1 2( a + b + A ) ,b=1 2( a + b A ) Nextnotethat a+ b= a + b and ab=1 4(( a + b )2 A2)= ab c2. Ontheotherhand,therootsofthequadraticequation P2( )=0also satisfythesameconditions 1+ 2= a + b and 12= ab c2.Thus, a= 1, b= 2,and d2f ( r0)= 1( dx)2+ 2( dy)2. If 1and 2arestrictlypositive,then d2f ( r0) > 0forall( dx,dy ) = (0 0),andbyTheorem13.18thefunctionhasalocalminimumat r0.If 1and 2arestrictlynegative,then d2f ( r0) > 0forall( dx,dy ) =(0 0) andbyTheorem13.18thefunctionhasalocalmaximumat r0.If 1and 2donotvanishbuthaveoppositesigns, 12< 0,thenina neighborhoodof r0,thegraphof f lookslike z = f ( r0)+ 1( x x 0)2+ 2( y y 0)2,

PAGE 269

94.MAXIMUMANDMINIMUMVALUES263 wherethecoordinates( x,y)areobtainedfrom( x,y )byrotationthrough anangle .When 1and 2havedierentsigns,thissurfaceisahyperbolicparaboloid(asaddle),and f hasneitheralocalminimumnor alocalmaximum.Case(4)iseasilyprovedbyexamples(seeStudy Problem13.17). Corollary 13.4 Letthefunction f satisfythehypothesesofTheorem13.19.Put D = ab c2,where a = f xx( r0) b = f yy( r0) ,and c = f xy( r0) .Then (1) If D> 0 and a> 0 ( b> 0 ),then f ( r0) isalocalminimum. (2) If D> 0 and a< 0 ( b< 0 ),then f ( r0) isalocalmaximum. (3) If D< 0 ,then f ( r0) isnotalocalextremum. Thiscorollaryisasimpleconsequenceofthesecond-derivativetest. Notethat 12= D .So D< 0if 12< 0or r0isasaddlepoint. Similarly,theconditions(1)and(2)areequivalenttothecaseswhen 1and 2arestrictlypositiveornegative,respectively. Example 13.43 Findallcriticalpointsofthefunction f ( x,y )=1 3x3+ xy2 x2 y2anddeterminewhether f hasalocalmaximum, minimum,orsaddleatthem. Solution: CriticalPoints .Thefunctionisapolynomial,andtherefore ithascontinuouspartialderivativeseverywhereofanyorder.Soits criticalpointsaresolutionsofthesystemofequations f x= x2+ y2 2 x =0 f y=2 xy 2 y =0 Itisimportantnottolosesolutionswhentransformingthesystemof equations f ( r )= 0 forthecriticalpoints .Itfollowsfromthesecondequationthat y =0or x =2.Therefore,theoriginalsystemof equationsis equivalent totwosystemsofequations: f x= x2+ y2 2 x =0 x =1 or f x= x2+ y2 2 x =0 y =0 Solutionsoftherstsystemare(1 1)and(1 1).Solutionsofthe secondsystemare(0 0)and(2 0).Thus,thefunctionhasfourcritical points. Second-DerivativeTest .Thesecondderivativesare f xx=2 x 2 ,f yy=2 x 2 ,f xy=2 y. Forthepoints(1 1), a = b =0and c = 2.Thecharacteristic polynomialis P2( )= 2 4.Itsroots = 2donotvanishand haveoppositesigns.Therefore,thefunctionhasasaddleatthepoints

PAGE 270

26413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (1 1).Forthepoint(0 0), a = b = 2and c =0.Thecharacteristic polynomialis P2( )=( 2 )2.Ithasonerootofmultiplicity2,that is, 1= 2= 2 < 0,and f hasalocalmaximumat(0 0).Finally,for thepoint(2 0), a = b =2and c =0.Thecharacteristicpolynomial P2( )=(2 )2hasonerootofmultiplicity2, 1= 2=2 > 0;that is,thefunctionhasalocalminimumat(2 0). Example 13.44 Investigatethefunction f ( x,y )= ex2 y(5 2 x + y ) forextremevalues. Solution: Thefunctionisdenedonthewholeplane,and,asthe productofanexponentialandapolynomial,ithascontinuouspartial derivativesofanyorder.Soitsextremevalues,ifany,canbeinvestigatedbythesecond-derivativetest. CriticalPoints .Usingtheproductruleforderivatives, f x= ex2 y 2 x (5 2 x + y ) 2 =0 x (5 2 x + y )=1 f y= ex2 y ( 1)(5 2 x + y )+1 =0 5 2 x + y =1 Thesubstitutionofthesecondequationintotherstoneyields x =1. Thenitfollowsfromthesecondequationthat y = 2.Sothefunction hasjustonecriticalpoint(1 2). Second-DerivativeTest .Usingtheproductruleforderivatives, f xx=( f x) x= ex2 y 2 x 2 x (5 2 x + y ) 2 +2(5 2 x + y ) 4 f yy=( f y) y= ex2 y ( 1) ( 1)(5 2 x + y )+1 1 f xy=( f y) x= ex2 y 2 x ( 1)(5 2 x + y )+1 +2 Therefore, a = f xx(1 2)= 2 e3, b = f yy(1 2)= e3,and c = f xy(1 2)=2 e3.Therefore, D = ab c2= 2 e6< 0.ByCorollary 13.4,theonlycriticalpointisasaddlepoint.Thefunctionhasno extremevalues. 94.4.StudyProblems.Problem13.15. Findallcriticalpointsofthefunction f ( x,y )= sin( x )sin( y ) anddeterminewhethertheyarealocalmaximum,alocal minimum,asaddlepoint. Solution: Thefunctionhascontinuouspartialderivativesofanyorderonthewholeplane.Sothesecond-derivativetestappliestostudy criticalpoints.

PAGE 271

94.MAXIMUMANDMINIMUMVALUES265 CriticalPoints .If n and m areintegers,then f x=cos( x )sin( y )=0 x = 2 + n or y = m, f y=sin( x )cos( y )=0 If x = 2+ n ,thenitfollowsfromthesecondequationthat y = 2+ m If y = m ,thenitfollowsfromthesecondequationthat x = n .Thus, foranypairofintegers n and m ,thepoints rnm=( 2+ n, 2+ m ) and r nm=( n,m )arecriticalpointsofthefunction. Second-DerivativeTest hastobeappliedtoeachcriticalpoint.The secondpartialderivativesare f xx= sin( x )sin( y ) ,f yy= sin( x )sin( y ) ,f xy=cos( x )cos( y ) Forthecriticalpoints rnm,onehas a = f xx( rnm)= ( 1)n + m, b = f yy( rnm)= ( 1)n + m= a ,and c = f xy( rnm)=0.Thecharacteristic equationis( a )2=0andhence 1= 2= ( 1)n + m.If n + m is even,thentherootsarenegativeand f ( rnm)=1isalocalmaximum. If n + m isodd,thentherootsarepositiveand f ( rnm)= 1isa localminimum.Forthecriticalpoints r nm,onehas a = f xx( r nm)=0, b = f yy( r nm)=0,and c = f xy( r nm)=( 1)n + m.Thecharacteristic equation 2 c2= 2 1=0hastworoots = 1ofoppositesigns. Thus, r nmaresaddlepointsof f .Infact,thelocalextremaofthis functionarealsoitsabsoluteextrema. Problem13.16. Dene f (0 0)=0 and f ( x,y )= x2+ y2 2 x2y 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 ) 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 b )2= a2 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.

PAGE 272

26613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Thecontinuityofthelasttermin f at(0 0)hastobeveried.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 = 2cos2 sin ,b = 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)= t4, whichattainsan absolutemaximum at t =0.Thus,alongtheparabola, f hasamaximumvalueattheoriginandhencecannothavealocal minimumthere.Theproblemillustratestheremarkgivenearlierin thissection. Problem13.17. Supposethat 1=0 or 2=0 (orboth)inthe second-derivativetestforafunction f .Giveexamplesof f when f has alocalmaximum,oralocalminimum,oritsgraphlookslikeasaddle, ornoneoftheabove. Solution: Considerthefunction f ( x,y )= x2+ sy4,where s isa number.Ithasacriticalpoint(0 0)because f x(0 0)= f y(0 0)=0 and a = f xx(0 0)=2, b = f yy(0 0)=0,and c = f xy(0 0)=0. Therefore, P2( )= (2 ) hastheroots 1=2and 2=0.If s> 0,then f ( x,y ) 0forall( x,y )and f hasaminimumat(0 0). Let s = 1.Thenthecurves x = y2dividetheplaneintofour sectorswiththevertexatthecriticalpoint(theorigin).Inthesectors containingthe x axis, f ( x,y ) 0,whereasinthesectorscontaining the y axis, f ( x,y ) 0.Thus,thegraphof f hastheshapeofasaddle. Thefunction f ( x,y )= ( x2+ sy4)hasamaximumat(0 0)if s> 0.

PAGE 273

94.MAXIMUMANDMINIMUMVALUES267 If s< 0,thegraphof f hastheshapeofasaddle.So,ifoneofthe rootsvanishes,then f mayhavealocalmaximumoralocalminimum, orasaddle.Thesameconclusionisreachedwhen 1= 2=0by studyingthefunctions f ( x,y )= ( x4+ sy4)alongthesimilarlinesof arguments. Furthermore,considerthefunction f ( x,y )= xy2.Italsohasa criticalpointattheorigin,andallitssecondderivativesvanishat(0 0), thatis, P2( )= 2and 1= 2=0.Thefunctionvanishesalongthe coordinateaxes.Sotheplaneisdividedintofoursectors(quadrants) ineachofwhichthefunctionhasaxedsign.Thefunctionispositive intherstandfourthquadrants( x> 0)andnegativeinthesecond andthirdquadrants( x< 0).Itthenfollowsthat f hasnomaximum orminimum,anditsgraphdoesnothavetheshapeofasaddle.Next, put f ( x,y )= x2 y3.Ithasacriticalpoint(0 0)and a =2, b =0, and c =0,thatis, 1=2and 2=0.Thezerosof f formthecurve y = x2 / 3,whichdividestheplaneintotwopartssothat f isnegative abovethiscurveand f ispositivebelowit.Therefore,thegraphof f doesnothavetheshapeofasaddle,and f doesnothaveaminimumor maximumat(0 0).Thebehaviorofthefunctions xy2and x2 y3near theircriticalpointresemblesthebehaviorofafunctionofonevariable nearitscriticalpointthatisalsoan inectionpoint 94.5.Exercises.(1) Foreachofthefollowingfunctions,ndallcriticalpointsand determineiftheyarearelativemaximum,arelativeminimum,ora saddlepoint: (i) f ( x,y )= x2+( y 2)2(ii) f ( x,y )= x2 ( y 2)2(iii) f ( x,y )=( x y +1)2(iv) f ( x,y )= x2 xy + y2 2 x + y (v) f ( x,y )=1 3x3+ y2 x2 3 x y +1 (vi) f ( x,y )= x2y3(6 x y ) (vii) f ( x,y )= x3+ y3 3 xy (viii) f ( x,y )= x4+ y4 x2 2 xy 2 y2(ix) f ( x,y )= xy +50 /x +20 /y x> 0, y> 0 (x) f ( x,y )= x2+ y2+1 x2y2(xi) f ( x,y )=cos( x )cos( y ) (xii) f ( x,y )=cos x + y2(xiii) f ( x,y )= y3+6 xy +8 x3(xiv) f ( x,y )= x3 2 xy + y2(xv) f ( x,y )= xy (1 x y )

PAGE 274

26813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (xvi) f ( x,y )= x cos y (xvii) f ( x,y )= xy 1 x2/a2 y2/b2(xviii) f ( x,y )=( ax + by + c ) / 1+ x2+ y2, a2+ b2+ c2 =0 (xix) f ( x,y )=(5 x +7 y 25) e x2 xy y2(xx) f ( x,y )=sin x +sin y +cos( x + y ) (xxi) f ( x,y )=1 3x3+ xy2 x2 y2(xxii) f ( x,y )=1 3y3+ xy +8 3x3(xxiii) f ( x,y )= x2+ xy + y2 4ln x 10ln y (xxiv) f ( x,y )= xy ln( x2+ y2) (xxv) f ( x,y )= x + y +sin( x )sin( y ) (xxvi) f ( x,y )=sin( x )+cos( y )+cos( x y ) (xxvii) f ( x,y )= x 2 y +ln( x2+ y2)+3tan 1( y/x ) (2) Letthefunction z = z ( x,y )bedenedimplicitlybythegiven equation.Usetheimplicitdierentiationtondextremevaluesof z ( x,y ): (i) x2+ y2+ z2 2 x +2 y 4 z 10=0 (ii) x2+ y2+ z2 xz yz +2 x +2 y +2 z 2=0 (iii)( x2+ y2+ z2)2= a2( x2+ y2 z2) 95.MaximumandMinimumValues(Continued)95.1.Second-DerivativeTestforMultivariableFunctions.Theorem13.18 holdsforanynumberofvariables,andthereisamultivariableanalog ofthesecond-derivativetest(Theorem13.19).Asinthetwo-variable case,thenumbers f xixj( r0)= Dijcanbearrangedintoan m m matrix.ByClairautstheorem,thismatrixissymmetric Dij= Dji.The polynomialofdegree m Pm( )=det " " D11 D12D13 D1 mD21D22 D23 D2 mD31D32D33 D3 m. . . . . . . . Dm 1Dm 2Dm 3 Dmm # # # # iscalledthe characteristic polynomialofthematrixofsecondderivatives.Thefollowingfactsareestablishedbymethodsoflinearalgebra: (1)Thecharacteristicpolynomialofarealsymmetric m m matrixhas m realroots 1, 2,..., m(arootofmultiplicity k counted k times).

PAGE 275

95.MAXIMUMANDMINIMUMVALUES(CONTINUED)269 (2)Thereexistsarotation d r =( dx1,dx2,...,dxm) d r=( dx 1,dx2,...,dx m) whichisalinearhomogeneoustransformationthatpreserves thelength d r = d r ,suchthat d2f ( r0)=mi =1 mj =1f xixj( r0) dxidxj=mi =1 mj =1Dijdxidxj= 1( dx 1)2+ 2( dx 2)2+ + n( dx m)2. (3)Therootsofthecharacteristicpolynomialsatisfytheconditions: 1+ 2+ + m= D11+ D22+ + Dmm, 12 m=det D. Fact(2)impliesthatifallrootsofthecharacteristicpolynomialare strictlypositive,then d2f ( r0) > 0forall d r = d r =0,andhence f ( r0)isalocalminimumbyTheorem13.18.Similarly,ifalltheroots arestrictlynegative,then f ( r0)isalocalmaximum.Corollary13.4 followsfromfact(2)for m =2.For m> 2,thesepropertiesofthe rootsareinsucienttoestablishamultivariableanalogofCorollary 13.4.Fact(3)alsoimpliesthatifdet D =0,thenoneormoreroots are0.Hence, d2f ( r0)=0forsome d r = 0 ,andthehypothesesof Theorem13.18arenotfullled. Theorem 13.20 (Second-DerivativeTestfor m Variables) Let r0beacriticalpointof f andsupposethat f hascontinuoussecondorderpartialderivativesinsomeopenballcenteredat r0.Let i, i = 1 2 ,...,m ,berootsofthecharacteristicpolynomial Pm( ) ofthematrix ofsecondderivatives Dij= f xixj( r0) (1) Ifalltherootsarestrictlypositive, i> 0 ,then f hasalocal minimum. (2) Ifalltherootsarestrictlynegative, i< 0 ,then f hasalocal maximum. (3) Ifalltherootsdonotvanishbuthavedierentsigns,then f hasnolocalminimumormaximumat r0. (4) Ifsomeoftherootsvanish,then f mayhavealocalmaximum,oralocalminimum,ornoneoftheabove(thetestis inconclusive). Incase(3),thedierence f ( r ) f ( r0)changesitssigninaneighborhoodof r0.Itisan m -dimensionalanalogofa saddlepoint .Case (4)holdsifdet D =0,whichiseasytoverify.Ingeneral,rootsof

PAGE 276

27013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Pm( )arefoundnumerically.Ifsomeoftherootsareguessed,thena syntheticdivisioncanbeusedtoreducetheorderoftheequation.If Pm( 1)=0,thenthereisapolynomial Qm 1ofdegree m 1suchthat Pm( )=( 1) Qm 1( )sothattheotherrootssatisfy Qm 1( )=0. Thesignsoftherootscanalsobeestablishedbyagraphicalmethod (anexampleisgivenStudyProblem13.18). Example 13.45 Investigatethefunction f ( x,y,z )=1 3x3+1 2y2+ z2+ xy +2 z forextremevalues. Solution: Thefunctionisapolynomialsoithascontinuouspartial derivativesofanyordereverywhere.Soitscriticalpointssatisfythe equations: f x= x2+ y =0 f y= y + x =0 f z=2 z +2= x2= x y = x z = 1 Therstequationhastwosolutions x =0and x =1.Sothefunctionhastwocriticalpoints r1=(0 0 1)and r2=(1 1 1).The second-orderpartialderivativesare f xx=2 x,f xy= f yy=1 ,f xz= f yz=0 ,f zz=2 Forthecriticalpoint r1,thecharacteristicpolynomial det 10 11 0 002 =(2 )( 2 1) hastheroots2and(1 5) / 2.Theydonotvanishbuthavedierent signs.So r1isasaddlepointof f (noextremevalue).Forthecritical point r2,thecharacteristicpolynomial det 2 10 11 0 002 =(2 )( 2 3 +1) haspositiveroots2 > 0and(3 5) / 2 > 0.So f (1 1 1)= 7 / 6 isalocalminimum. 95.2.WhentheSecond-DerivativeTestIsInconclusive.Ifatleastone oftherootsofthecharacteristicpolynomialsvanishes,thesecondderivativetestisinconclusive.Howcanthelocalbehaviorofafunctionbeanalyzednearitscriticalpoint?Ifthefunctioninquestionhas continuouspartialderivativeofsucientlyhighordersinaneighborhoodofacriticalpoint r0,thentheTaylortheoremprovidesauseful

PAGE 277

95.MAXIMUMANDMINIMUMVALUES(CONTINUED)271 techniqueforansweringthisquestion.Thelocalbehaviorofafunctionnear r0isdeterminedbyhigher-orderdierentials dnf ( r ),where d r = r r0.Itisgenerallyeasiertostudytheconcavityofapolynomialthanthatofageneralfunction.Theconceptisillustratedbythe followingexample. Example 13.46 Investigatealocalbehaviorofthefunction f ( x,y )=sin( xy ) / ( xy ) if x =0 and y =0 and f (0 ,y )= f ( x, 0)=1 Solution: Since u = xy issmallneartheorigin,bytheTaylortheoremsin u = u u3/ 6+ ( u ) u3,where ( u ) 0as u 0.Therefore, f ( x,y )=1 u2 6 1 6 ( u ) =1 x2y2 6 1 6 ( xy ) Inadisk x2+ y2<2ofasucientlysmallradius > 0,1 6 ( xy ) > 0because ( xy ) 0as( x,y ) (0 0).Sothefunctionattainsa localmaximumat(0 0)because x2y2 0inthedisk.Theinequality | sin u || u | suggeststhat f attainsthemaximumvalue1alsoalong thecoordinateaxes(criticalpointsarenotisolated).Theestablished localbehaviorofthefunctionimpliesthatitssecondpartialderivatives vanishat(0 0),andhencebothrootsofthecharacteristicpolynomial P2( )= 2vanish(thesecond-derivativetestisinconclusive). 95.3.AbsoluteMaximalandMinimalValues.Forafunction f ofone variable,the extremevaluetheorem saysthatif f iscontinuousona closedinterval[ a,b ],then f hasanabsoluteminimumvalueandan absolutemaximumvalue(seeCalculusI).Forexample,thefunction f ( x )= x2on[ 1 2]attainsanabsoluteminimumvalueat x =0and anabsolutemaximumvalueat x =2.Thefunctionisdierentiablefor all x ,andthereforeitscriticalpointsaredeterminedby f( x )=2 x =0. Sotheabsoluteminimumvalueoccursatthecriticalpoint x =0inside theinterval,whiletheabsolutemaximumvalueoccursontheboundary oftheintervalthatisnotacriticalpointof f .Thus,tondtheabsolute maximumandminimumvaluesofafunction f inaclosedintervalin thedomainof f ,thevaluesof f mustbeevaluatedandcomparednot onlyatthecriticalpointsbutalsoattheboundariesoftheinterval. Thesituationformultivariablefunctionsissimilar.Forexample, thefunction f ( x,y )= x2+ y2whoseargumentsarerestrictedtothe square D =[0 1] [0 1]attainsitsabsolutemaximumandminimum valuesontheboundaryof D asshownintheleftpanelofFigure13.13. Definition 13.27 (ClosedSet) Aset D inaEuclideanspaceissaidtobeclosedifitcontainsallits limitpoints.

PAGE 278

27213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS x y z 1 1 x y z z = xy min min max maxFigure13.13.Left :Thegraph z = x2+ y2(acircular paraboloid)overthesquare D =[0 1] [0 1].Thefunction f ( x,y )= x2+ y2attainsitsabsolutemaximumandminimum valuesontheboundaryof D : f (0 0) f ( x,y ) f (1 1)for allpointsin D Right :Thegraph z = xy .Thevaluesof f ( x,y )= xy alongthecircle x2+ y2=4areshownbythe curveonthegraph.Thefunctionhastwolocalmaximaand minimaonthedisk x2+ y2 4,whileithasnomaximum andminimumvaluesontheentireplane.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 13.28 (BoundedSet) Aset D inaEuclideanspaceissaidtobeboundedifitiscontainedin someball. Inotherwords,foranytwopointsinaboundedset,thedistance betweenthemcannotexceedsomevalue(thediameteroftheballthat containstheset). Theorem 13.21 (ExtremeValueTheorem) If f iscontinuousonaclosed,boundedset D inaEuclideanspace,then f attainsanabsolutemaximumvalue f ( r1) andanabsoluteminimum value f ( r2) atsomepoints r1 D and r2 D .

PAGE 279

95.MAXIMUMANDMINIMUMVALUES(CONTINUED)273 Theclosednessof D isessential.Forexample,ifthefunction f ( x,y )= x2+ y2isrestrictedtothe open square D =(0 1) (0 1),then ithasnoextremevalueson D .Forall( x,y )in D f (0 0)
PAGE 280

27413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS functionontheboundaryare 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 = / 4and t = / 4+ .Thesevaluesof t correspondtothepoints( 2 2)and( 2 2).Similarly, F ( t )attains itsminimalvalue2on[0 2 ]whensin(2 t )= 1or t =3 / 4and t =3 / 4+ .Thesevaluesof t correspondtothepoints( 2 2) and( 2 2). Step3 .Thelargestnumberof0,2,and6is6.Sotheabsolutemaximumvalueof f is6;itoccursatthepoints( 2 2)and( 2 2). Thesmallestnumberof0,2,and6is0.Sotheabsoluteminimum valueof f is0;itoccursatthepoint(0 0). Example 13.48 Findtheabsolutemaximumandminimumvalues of f ( x,y,z )= x2+ y2 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 hascontinuouspartialderivativesofanyorderonthewhole space. Step1 .Since f isdierentiableeverywhere,itscriticalpointssatisfy theequations f x=2 x =0, f y=2 y =0,and f z= 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.Theproblemnowistond themaximalandminimalvaluesofatwo-variablefunction F1on thedisk.Inprinciple,atthispoint,Steps1,2,and3havetobe appliedto F1.Thesetechnicalitiescanbeavoidedinthisparticularcasebynotingthat F1( x,y )= x2+ y2 8= r2 8,where r2= x2+ y2 4.Therefore,themaximalvalueof F1isreached when r2=4,anditsminimalvalueisreachedwhen r2=0.Sothe maximalandminimalvaluesof f on S1are 4and 8.Thevalues of f on S2are F2( x,y )= f ( x,y,x2+ y2)=3 r2 r4= g ( r ),where r2= x2+ y2 4or r [0 2].Thecriticalpointsof g ( r )satisfythe equation g( r )=6 r 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)= 4,andtheminimalvalueis 4 asthesmallestofthesenumbers.

PAGE 281

95.MAXIMUMANDMINIMUMVALUES(CONTINUED)275 Step3 .Theabsolutemaximumvalueof f on D ismax { 1 8 4 9 / 4 } = 9 / 4,andtheabsoluteminimumvalueof f on D ismin { 1 8 4 9 / 4 } = 8.Bothextremevaluesof f occurontheboundaryof D : f (0 0 4)= 8,andtheabsolutemaximalvalueisattainedalongthecircleof intersectionoftheplane z =3 / 2withtheparaboloid z = x2+ y2. 95.4.StudyProblems.Problem13.18. Investigatetheextremevaluesofthefunction f ( x,y,z )= x + y2/ (4 x )+ z2/y +2 /z if x> 0 y> 0 ,and z> 0 Solution: Thecriticalpointsaredeterminedbytheequations f x=1 y2 4 x2=0 ,f y= y 2 x z2 y2=0 ,f z= 2 z y 2 z2=0 Therstequationisequivalentto y =2 x (since x> 0and y> 0). Thesubstitutionofthisrelationintothesecondequationgives z = y because y> 0and z> 0.Thesubstitutionofthisrelationintothe thirdequationyields z =1as z> 0.Thereisonlyonecriticalpoint r0=(1 2, 1 1)inthepositiveoctant.Thesecondpartialderivativesat r0are f xx( r0)=y2 2 x3 r0=4 ,f xy( r0)= y 2 x2 r0= 2 f xz( r0)=0 ,f yy( r0)=1 2 x+2 z2 y3 r0=3 f yz( r0)= 2 z y2 r0= 2 ,f zz( r0)=2 y+4 z3 r0=6 Thecharacteristicequationofthesecondderivativematrixis det 4 20 23 2 0 26 =(4 )[(3 )(6 ) 4] 4(6 ) = 3+13 2 46 +32=0 Firstofall, =0isnotaroot.Toanalyzethesignsoftheroots,the followingmethodisemployed.Thecharacteristicequationiswritten intheform ( 2 13 +46)=32 Thisequationdeterminesthepointsofintersectionofthegraph y = g ( )= ( 2 13 +46)withthehorizontalline y =32.Thepolynomial g ( )hasonesimpleroot g (0)=0becausethequadraticequation 2 13 +46=0hasnorealroots.Therefore, g ( ) > 0if > 0and g ( ) < 0if < 0.Thisimpliesthattheintersectionofthehorizontal line y =32 > 0withthegraph y = g ( )occursonlyfor > 0.Thus,

PAGE 282

27613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS allthreerootsofthecharacteristicpolynomial P3( )arepositive,and hence f (1 / 2 1 1)=4isaminimum. Problem13.19.(TheLeastSquaresMethod). Supposethatascientisthasareasontobelievethattwoquantities x and y arerelatedlinearly, y = mx + b ,where m and b areunknown constants.Thescientistperformsanexperimentandcollectsdataas pointsontheplane ( xi,yi) i =1 2 ,...,N .Sincethedatacontain errors,thepointsdonotlieonastraightline.Let di= yi ( mxi+ b ) be theverticaldeviationofthepoint ( xi,yi) fromtheline y = mb + x .The methodofleastsquaresdeterminestheconstants m and b bydemanding thatthesumofsquares d2 iattainsitsminimalvalue,thusproviding thebestttothedatapoints.Find m and b Solution: Considerthefunction f ( m,b )= N i =1d2 i.Itscriticalpoints satisfytheequations f b= 2 N i =1di=0and f m= 2 N i =1xidi=0 because( di) b= 1and( di) m= xi.Thesubstitutionoftheexplicit formof diintotheseequation,yieldsthefollowingsystem: mNi =1xi+ bN =Ni =1yi,mNi =1x2 i+ bNi =1xi=Ni =1xiyiwhosesolutiondeterminestheslope m andtheconstant b oftheleast squareslinearttothedatapoints.Notethatthesecond-derivative testhereisnotreallynecessarytoconcludethat f hasaminimumat thecriticalpoint.Explainwhy! 95.5.Exercises.(1) Foreachofthefollowingfunctions,ndallcriticalpointsand determineiftheyarearelativemaximum,arelativeminimum,ora saddlepoint: (i) f ( x,y,z )= x2+ y2+ z2+2 x +4 y 8 z (ii) f ( x,y,z )= x2+ y3+ z2+12 xy 2 z (iii) f ( x,y,z )= x2+ y3+ z2+12 xy +2 z (iv) f ( x,y,z )=sin x + z sin y (v) f ( x,y,z )= x2+5 3y3+ z2 2 xy 4 zy (vi) f ( x,y,z )= x + y2/ (4 x )+ z2/y +2 /z (vii) f ( x,y,z )= a2/x + x2/y + y2/z + z2/b x> 0, y> 0, z> 0, b> 0 (viii) f ( x,y,z )=sin x +sin y +sin z +sin( x + y + z ),where( x,y,z ) [0 ] [0 ] [0 ] (ix) f ( x1,...,xm)= m k =1sin xk(x) f ( r )=( R2 r 2)2,where r =( x1,...,xm)and R isaconstant

PAGE 283

95.MAXIMUMANDMINIMUMVALUES(CONTINUED)277 (xi) f ( x1,...,xm)= x1+ x2/x1+ x3/x2+ + xm/xm 1+2 /xm, xi> 0, i =1 2 ,...,m (2) Giventwopositivenumbers a and b ,nd m numbers xi, i = 1 2 ,...,m ,between a and b sothattheratio x1x2 xm ( a + x1)( x1+ x2) ( xm+ b ) ismaximal. (3) UsemultivariableTaylorpolynomialstoshowthattheoriginisa criticalpointofeachofthefollowingfunctions.Determineifthereis alocalmaximum,alocalminimum,ornoneoftheabove. (i) f ( x,y )= x2+ xy2+ y4(ii) f ( x,y )=ln(1+ x2y2) (iii) f ( x,y )= x2ln(1+ x2y2) (iv) f ( x,y )= xy (cos( x2y ) 1) (v) f ( x,y )=( x2+2 y2)tan 1( x + y ) (vi) f ( x,y )=cos( exy 1) (vii) f ( x,y )=ln( y2sin2x +1) (viii) f ( x,y )= ex + y2 1 sin( x y2) (ix) f ( x,y,z )=sin( xy + z2) / ( xy + z2) (x) f ( x,y,z )=2 2cos( x + y + z ) x2 y2 z2(4) Let f ( x,y,z )= xy2z3( a x 2 y 3 z ), a> 0.Findallitscritical pointsanddetermineiftheyarearelativemaximum,arelativeminimum,orasaddlepoint. (5) Giveexamplesofafunction f ( x,y )oftwovariablesonaregion D thatattainsitsextremevaluesandhasthefollowingproperties: (i) f iscontinuouson D ,and D isnotclosed. (ii) f isnotcontinuouson D ,and D isboundedandclosed. (iii) f isnotcontinuouson D ,and D isnotboundedandnotclosed. Dosuchexamplescontradicttheextremevaluetheorem?Explain. (6) Foreachofthefollowingfunctions,ndtheextremevaluesonthe speciedset D : (i) f ( x,y )=1+2 x 3 y D istheclosedtrianglewithvertices (0 0),(1 2),and(2 4) (ii) f ( x,y )= x2+ y2+ xy2 1, D = { ( x,y ) || x | 1 | y | 2 } (iii) f ( x,y )= yx2, D = { ( x,y ) | x 0 ,y 0 ,x2+ y2 4 } (iv) f ( x,y,z )= xy2+ z D = { ( x,y,z ) || x | 1 | y | 1 | z | 1 } (v) f ( x,y,z )= xy2+ z D = { ( x,y,z ) | 1 x2+ y2 4 4 x z 4+ x }

PAGE 284

27813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (7) Findthepointontheplane x + y z =1thatisclosesttothe point(1 2 3). Hint: Letthepointinquestionhavethecoordinates ( x,y,z ).Thenthesquareddistancebetweenitand(1 2 3)is f ( x,y )= ( x 1)2+( y 2)2+( z 3)2,where z = x + y 1because( x,y,z )is intheplane. (8) Findthepointonthecone z2= x2+ y2thatisclosesttothepoint (1 2 3). (9) Findanequationoftheplanethatpassesthroughthepoint(3 2 1) andcutsothesmallestvolumeintherstoctant. (10) Findtheextremevaluesof f ( x,y )= ax2+2 bxy + cy2onthecircle x2+ y2=1. (11) Findtheextremevaluesof f ( x,y,z )= x2/a2+ y2/b2+ z2/c2on thesphere x2+ y2+ z2=1. (12) Findtwopositivenumberswhoseproductisxed,whilethesum oftheirreciprocalsisminimal. (13) Find m positivenumberswhoseproductisxed,whilethesum oftheirreciprocalsisminimal. (14) Asolidobjectconsistsofasolidcylinderandasolidcircularcone suchthatthebaseoftheconecoincideswiththebaseofthecylinder. Ifthetotalsurfaceareaoftheobjectisxed,whatarethedimensions oftheconeandcylinderatwhichtheobjecthasmaximalvolume? (15) Findalinearapproximation y = mx + b totheparabola y = x2suchthatthedeviation=max | x2 mx b | isminimalintheinterval 1 x 3. (16) Let N pointsofmasses mj, j =1 2 ,...,N ,bepositionedina planeat Pj=( xj,yj).RecallfromCalculusIIthatthemomentof inertiaofthissystemaboutapoint P =( x,y )is I ( x,y )=Nj =1mj| PPj|2. Find P aboutwhichthemomentofinertiaisminimal. 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 ].

PAGE 285

96.LAGRANGEMULTIPLIERS279 Example 13.49 Theheightasafunctionofpositionis f ( x,y )= xy .Findthelocalmaximaandminimaoftheheightalongthecircular path x2+ y2=4 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 function sin(2 t )attainsitsabsoluteminimumvalueat t = / 4and t = / 4+ anditsabsolutemaximumvalueat t =3 / 4and t =3 / 4+ .So,alongthepath,thefunction f attainstheabsolutemaximum value2at( 2 2)and( 2 2)andtheabsoluteminimumvalue 2at( 2 2)and( 2 2).Thesolutionisillustratedinthe rightpanelofFigure13.13. However,inmanysimilarproblems,anexplicitformof r ( t )isnot knownornoteasytond.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 .Nothingprecludesusfromaskingaboutthemaximumandminimumtemperaturesalongacurvedenedasanintersectionoftwo 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 13.29 (LocalMaximaandMinimaSubjectto Constraints) Afunction f ( r ) hasalocalmaximum(orminimum)at r0ontheset denedbytheconstraints ga( r )=0 if f ( r ) f ( r0) (or f ( r ) f ( r0) ) forall r insomeneighborhoodof r0thatsatisfytheconstraints,that is, ga( r )=0 Notethatafunction f maynothavelocalmaximaorminimainits domain.However,whenitsargumentsbecomesubjecttoconstraints, itmaywellhavelocalmaximaandminimaonthesetdenedbythe

PAGE 286

28013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS constraints.Intheexampleconsidered, f ( x,y )= xy hasnolocal maximaorminima,but,whenitisrestrictedonthecirclebyimposing theconstraint g ( x,y )= x2+ y2 4=0,ithappenstohavetwolocal minimaandmaxima.96.1.CriticalPointsofaFunctionSubjecttoaConstraint.Theextreme valueproblemwithconstraintsamountstondingthecriticalpoints ofafunctionwhoseargumentsaresubjecttoconstraints.Theexample discussedaboveshowsthattheequation f = 0 nolongerdetermines thecriticalpointsfordierentiablefunctionsifitsargumentsareconstrained.Considerrstthecaseofasingleconstraintfortwovariables r =( x,y ).Let r0beapointatwhich f ( r )hasalocalextremumon theset S denedbytheconstraint g ( r )=0.Letusassumethatthe function g hascontinuouspartialderivativesinaneighborhoodof r0and g ( r0) = 0 .Thentheequation g ( r )=0denesa smooth curve throughthepoint r0(recalltheargumentgivenbeforeTheorem13.16). Let r ( t )beparametricequationsofthiscurveinaneighborhoodof r0, thatis,forsome t = t0, r ( t0)= r0.Thefunction F ( t )= f ( r ( t ))denes valuesof f alongthecurveandhasalocalextremumat t0.Let f be dierentiableat r0and f ( r0) = 0 .Sincethecurveissmooth,the vectorfunction r ( t )isdierentiable,anditisconcludedthat F hasno rateofchangeat t = t0, F( t0)=0.Thechainruleyields F( t0)= f x( r0) x( t0)+ f y( r0) y( t0) = f ( r0) r( t0)=0= f ( r0) r( t0) provided r( t0) = 0 ( thecurveissmooth ).Thegradient f ( r0)is orthogonaltoatangentvectortothecurveat thepointwhere f hasa localextremumonthecurve .ByTheorem13.16,thegradient g ( r )at any pointisnormaltothelevelcurve g ( r )=0,thatis, g ( r ( t )) r( t ) forany t provided g ( r0) = 0 .Therefore,thegradients f ( r0)and g ( r0)mustbeparallelat r0(seeFigure13.14).Thecharacteristic geometricalpropertyofthepoint r0isthat thelevelcurveof f andthe curve g ( x,y )=0 intersectat r0andaretangentialtooneanother .For thisveryreason, f hasnorateofchangealongthecurve g ( x,y )=0 at r0.Thisgeometricalstatementcanbetranslatedintoanalgebraic one:thereshouldexistanumber suchthat f ( r0)= g ( r0).This provesthefollowingtheorem. Theorem 13.22 (CriticalPointsSubjecttoaConstraint) Supposethat f hasalocalextremevalueatapoint r0onthesetdened byaconstraint g ( r )=0 .Supposethat g hascontinuouspartialderivativesinaneighborhoodof r0and g ( r0) = 0 .If f isdierentiableat

PAGE 287

96.LAGRANGEMULTIPLIERS281 g f P P0 f g y x g ( x,y )=0 f g P g f P0g ( x,y,z )=0Figure13.14.Left :Relativeorientationsofthegradients f and g alongthecurve g ( x,y )=0.Atthepoint P0,the function f hasalocalextremevaluealongthecurve g =0. Atthispoint,thegradientsareparallel,and thelevelcurve of f through P0andthecurve g =0 haveacommontangent line Right :Relativeorientationsofthegradients f and g alonganycurveintheconstraintsurface g ( x,y,z )=0. Atthepoint P0,thefunction f hasalocalextremevalueon thesurface g =0.Atthispoint,thegradientsareparallel, and thelevelsurfaceof f through P0andthesurface g =0 haveacommontangentplane .r0,thenthereexistsanumber 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.On theotherhand,bythepropertiesofthegradient,thevector g ( r0)is orthogonalto r( t0)foreverysuchcurve.Therefore,atthepoint r0, thegradientsof f and g mustbeparallel.Asimilarlineofreasoning provesthetheoremforanynumberofvariables. Thistheoremprovidesapowerfulmethodtondcriticalpointsof f subjecttoaconstraint g =0.Itiscalledthe methodofLagrange multipliers .Tondthecriticalpointsof f ,thefollowingsystemof equationsmustbesolved: (13.17) 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

PAGE 288

28213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS thesystem.Inthethree-variablecase,thesystemcontainsfourequationsforfourvariables( x,y,z, ).Foreachsolution( x0,y0,z0,0),the correspondingcriticalpointof f is( x0,y0,z0). Example 13.50 UsethemethodofLagrangemultiplierstosolve theprobleminExample13.33. Solution: Put g ( x,y )= x2+ y2 4.Thefunctions f ( x,y )= xy and g havecontinuouspartialderivativeseverywhereastheyarepolynomials.Then f x= g xf y= g yg =0 = y =2 x x =2 y x2+ y2=4 Thesubstitutionoftherstequationintothesecondonegives x = 4 2x .Thismeansthateither x =0or = 1 / 2.If x =0,then y =0 bytherstequation,whichcontradictstheconstraint.For =1 / 2, x = y andtheconstraintgives2 x2=4or x = 2.Thecritical pointscorrespondingto =1 / 2are( 2 2)and( 2 2).If = 1 / 2, x = y andtheconstraintgives2 x2=4or x = 2.Thecritical pointscorrespondingto = 1 / 2are( 2 2)and( 2 2).So f ( 2 2)=2isthemaximalvalueand f ( 2 2)= 2is theminimalone. Example 13.51 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 .Thequestionistondtheminimalvalue of f subjecttoconstraint g ( x,y,z )= xyz V =0.TheLagrange 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 wherethelastsystemhasbeenobtainedbymultiplyingtherstequationby x ,thesecondoneby y ,andthethirdoneby z withthesubsequentuseoftheconstraint.Combiningthersttwoequations,one infersthat2 z ( y x )=0.Since z =0( V =0),onehas y = x .Combiningtherstandthirdequations,oneinfersthat y ( x 2 z )=0and hence x =2 z .Thesubstitutionof y = x =2 zintotheconstraint

PAGE 289

96.LAGRANGEMULTIPLIERS283 yields4 z3= V .Hence,theoptimaldimensionsare x = y =(2 V )1 / 3and z =(2 V )1 / 3/ 2.Theamountofcardboardminimizingthecostis 3(2 V )2 / 3(thevalueof f atthecriticalpoint).Fromthegeometryof theproblem,itisclearthat f attainsitsminimumvalueattheonly criticalpoint. ThemethodofLagrangemultiplierscanbeusedtodetermineextremevaluesofafunctiononaset D .Recallthattheextremevalues mayoccurontheboundaryof D .InExample13.47,explicitparametricequationsoftheboundaryof D havebeenused(Step2).Instead, analgebraicequationoftheboundary, g ( x,y )= x2+ y2 4=0,canbe usedincombinationwiththemethodofLagrangemultipliers.Indeed, if f ( x,y )= x2+ y2+ xy ,thenitscriticalpointsalongtheboundary circlesatisfythesystemofequations: f x= g xf y= g yg =0 = 2 x + y =2 x 2 y + x =2 y x2+ y2=4 Bysubtractingthesecondequationfromtherstone,itfollowsthat x y =2 ( x y ).Hence,either x = y or =1 / 2.Intheformercase, theconstraintyields2 x2=4or x = 2.Thecorrespondingcritical pointsare( 2 2).If =1 / 2,thenfromthersttwoequations inthesystem,oneinfersthat x = y .Theconstraintbecomes2 x2=4 or x = 2,andthecriticalpointsare( 2 2). Remark. Thecondition g ( r0) = 0 iscrucialforthemethodof Lagrangemultiplierstowork.If f ( r0) = 0 (i.e. r0isnotacriticalpointof f withouttheconstraint),then(13.17)havenosolution when g ( r0)= 0 ,andthemethodofLagrangemultiplersfails.Recallthatthederivationof(13.17)requiresthatacurvedenedbythe equation g ( r )=0issmoothnear r0,whichmaynolongerbethe caseif g ( r0)= 0 (seetheimplicitfunctiontheorem).So,if f attainsitslocalextremevalueatapointwhichisacuspofthecurve g ( x,y )=0,thenitcannotbedeterminedby(13.17).Forexample, theequation g ( x,y )= x3 y2=0denesacurvethathasacusp at(0 0)and g (0 0)= 0 ,i.e.,thecurvehasnonormalvectoratthe origin.Since x = y2 / 3 0onthecurve,thefunction f ( x,y )= x attainsitsabsoluteminimumvalue0alongthiscurveattheorigin. However, f (0 0)=(1 0)andthemethodofLagrangemultipliers failstodetectthispointbecausethereisno atwhichEqs.(13.17) aresatised.Ontheotherhand,thefunction f ( x,y )= x2alsoattains itsabsoluteminimumvalueattheorigin.Equations(13.17)havethe

PAGE 290

28413.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS solution( x,y )=(0 0)and =0.Thedierencebetweenthetwocases isthatinthelattercase f (0 0)= 0 ,i.e.,(0 0)isalsoacriticalpoint of f withouttheconstraint.ThemethodofLagrangemultipliersalso becomesinapplicableif g isnotdierentiableat r0(seetheexercises). Thebehaviorofafunction f atthepointswhere g doesnotexistor vanisheshastobeinvestigatedbydierentmeans.96.2.TheCaseofTwoorMoreConstraints.Letafunctionofthree variables f havealocalextremevalueatpoint r0onthesetdenedby twoconstraints g1( r )=0and g2( r )=0.Provided g1and g2havecontinuouspartialderivativesinaneighborhoodof r0andtheirgradients donotvanishat r0,eachconstraintdenesasurfaceinthedomainof f (levelsurfacesof g1and g2).Sothesetdenedbytheconstraintsisthe curveofintersectionofthelevelsurfaces g1=0and g2=0.Let v bea tangentvectortothecurveat r0.Sincethecurveliesinthelevelsurface g1=0,bytheearlierarguments, f ( r0) v and g1( r0) v .On theotherhand,thecurvealsoliesinthelevelsurface g2=0andhence g2( r0) v .Itfollowsthatthegradients f g1,and g2become coplanar atthepoint r0astheylieintheplanenormalto v .Suppose that thevectors g1( r0) and g2( r0) arenotparallelor,equivalently, g1( r0) isnotproportionalto g2( r0).Thenanyvectorintheplane normalto v isalinearcombinationofthem(seeStudyProblem11.6). Therefore,thereexistnumbers 1and 2suchthat f ( r )= 1 g1( r )+ 2 g2( r ) ,g1( r )= g2( r )=0 when r = r0.Thisisasystemofveequationsforvevariables ( x,y,z,1,2).Foranysolution( x0,y0,z0,10,20),thepoint( x0,y0,z0) isacriticalpointof f onthesetdenedbytheconstraints.Ingeneral, thefollowingresultcanbeprovedbyasimilarlineofreasoning. Theorem 13.23 (CriticalPointsSubjecttoConstraints) Supposethatfunctions ga, a =1 2 ,...,M ,of m variables, m>M havecontinuouspartialderivativesinaneighborhoodofapoint r0and afunction f hasalocalextremevalueat r0inthesetdenedbythe constraints ga( r )=0 .Supposethat ga( r0) arenonzerovectorsanyof whichcannotbeexpressedasalinearcombinationoftheothersand f isdierentiableat r0.Thenthereexistnumbers asuchthat f ( r0)= 1 g1( r0)+ 2 g2( r0)+ + M gM( r0) Example 13.52 Findextremevaluesofthefunctions f ( x,y,z )= xyz onthecurvethatisanintersectionofthesphere x2+ y2+ z2=1 andtheplane x + y + z =0 .

PAGE 291

96.LAGRANGEMULTIPLIERS285 Solution: Put g1( x,y,z )= x2+ y2+ z2 1and g2( x,y,z )= x + y + z Onehas g1=(2 x, 2 y, 2 z ),whichcanonlyvanishat(0 0 0),and hence g1 = 0 onthesphere.Also, g2=(1 1 1) = 0 .Therefore, criticalpointsof f onthesurfaceofconstraintsaredeterminedbythe equations: f x= 1g x+ 2g xf y= 1g y+ 2g yf z= 1g z+ 2g zg1=0 g2=0 yz =2 1x + 2xz =2 1y + 2xy =2 1z + 21= x2+ y2+ z20= x + y + z Subtractthesecondequationfromtherstonetoobtain( y x ) z = 2 1( x y ).Itfollows,then,thateither x = y or z = 2 1.Suppose rstthat y = x .Then z = 2 x bythefthequation.Thesubstitution of x = y and z = 2 x intothefourthequationyields6 x2=1or x = 1 / 6.Therefore,thepoints r1=(1 / 6 1 / 6 2 / 6)and r2=( 1 / 6 1 / 6 2 / 6)arecriticalpointsprovidedthereexist thecorrespondingvalues 1and 2suchthatallequationsaresatised. Forexample,take r1.Thenthesecondandthirdequationsbecome 2 6=2 61+ 2 1 6= 4 61+ 2 3 6=6 61 1 6= 4 61+ 2. So 1= 1 / (2 6)and 2= 1 / 6.Theexistenceof 1and 2forthe point r2isveriedsimilarly.Next,supposethat z = 2 1.Subtract thethirdequationfromthesecondonetoobtain( z y ) x =2 1( y z ). Itfollowsthateither y = z or x = 2 1.Let y = z .Thefthequation yields x = 2 y ,andthefourthequationisreducedto6 y2=1.Therefore,therearetwomorecriticalpoints: r3=( 2 / 6 1 / 6 1 / 6) and r4=(2 / 6 1 / 6 1 / 6).Thereaderistoverifytheexistenceof 1and 2inthesecases(notethat 1= z/ 2).Finally,let x = 2 1and z = 2 1.Theseconditionsimplythat x = z and,by thefthequation, y = 2 x .Thefourthequationyields6 x2=1sothat thereisanotherpairofcriticalpoints: r5=(1 / 6 2 / 6 1 / 6)and r6=( 1 / 6 2 / 6 1 / 6)(thereaderistoverifytheexistenceof 1and 2).Theintersectionofthesphereandtheplaneisacircle.Soit issucienttocomparevaluesof f atthecriticalpointstodetermine theextremevaluesof f .Itfollowsthat f attainsthemaximumvalue 2 / 6at r2, r4,and r6andtheminimumvalue 2 / 6at r1, r3,and r5.

PAGE 292

28613.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS Let f ( r )beafunctionsubjecttoaconstraint g ( r ).Denethe function F ( r )= f ( r ) g ( r ) where isviewedasanadditionalindependentvariable.Thencriticalpointsof F aredeterminedby(13.17).Indeed,thecondition F/ =0yieldstheconstraint g ( r )=0,whilethedierentiation withrespecttothevariables r gives F = f g =0,whichcoincideswiththerstequationin(13.17).Similarly,ifthereareseveral constraints,criticalpointsofthefunctionwithadditionalvariables a, a =1 2 ,...,M (13.18) F ( r ,1,2,...,n)= f ( r ) Ma =1aga( r ) coincidewiththecriticalpointsof f subjecttotheconstraints ga=0as statedinTheorem13.23.Thefunctions F and f havethesamevalues onthesetdenedbytheconstraints ga=0becausetheydierbya linearcombinationofconstraintfunctionswiththecoecientsbeing the Lagrangemultipliers .Theaboveobservationprovidesasimpleway toformulatetheequationsforcriticalpointssubjecttoconstraints.96.3.FindingLocalMaximaandMinima.Inthesimplestcaseofafunction f oftwovariablessubjecttoaconstraint,thenatureofcritical points(localmaximumorminimum)canbedeterminedbygeometrical means.Supposethatthelevelcurve g ( x,y )=0isclosed.Then,bythe extremevaluetheorem, f attainsitsmaximumandminimumvalueson itatsomeofthecriticalpoints.Suppose f attainsitsabsolutemaximumatacriticalpoint r1.Then f shouldhaveeitheralocalminimum oraninectionattheneighboringcriticalpoint r2alongthecurve. Let r3bethecriticalpointnextto r2alongthecurve.Then f hasa localminimumat r2if f ( r2) f ( r3). Thisproceduremaybecontinueduntilallcriticalpointsareexhausted. Comparethispatternofcriticalpointswiththebehaviorofaheight alongaclosedhikingpath. Remark. Iftheconstraintscanbesolved,thenanexplicitformof f onthesetdenedbytheconstraintscanbefound,andthestandardsecond-derivativetestapplies!Forinstance,inExample13.51, 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 criticalpoint x = y =(2 V )1 / 3(and z = V/ ( xy )=(2 V )1 / 3/ 2).So

PAGE 293

96.LAGRANGEMULTIPLIERS287 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 hascontinuouspartialderivativesinaneighborhood of r0and g ( r0) = 0 .Then g xand g ycannotsimultaneouslyvanish atthecriticalpoint.Withoutlossofgenerality,assumethat g y =0at r0=( x0,y0).Bytheimplicitfunctiontheorem,thereisaneighborhood of r0inwhichtheequation g ( x,y )=0hasauniquesolution y = h ( x ). Thevaluesof f onthelevelcurve g =0nearthecriticalpointare F ( x )= f ( x,h ( x )).Bythechainrule,oneinfersthat F= f x+ f yhand (13.19) F=( d/dx )( f x+ f yh)= f xx+2 f xyh+ f yy( h)2+ f yh. So,inordertond F( x0),onehastocalculate h( x0)and h( x0).This taskisaccomplishedbytheimplicitdierentiation.Bythedenition of h ( x ), G ( x )= g ( x,h ( x ))=0forall x inanopenintervalcontaining x0.Therefore, G( x )=0,whichdenes hbecause G= g x+ g yh=0 and h= g x/g y.Similarly, G( x )=0yields (13.20) G= g xx+2 g xyh+ g yy( h)2+ g yh=0 whichcanbesolvedfor h,where h= g x/g y.Thesubstitutionof h( x0), h( x0),andallthevaluesofallthepartialderivativesof f atthe criticalpoint( x0,y0)into(13.19)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 ysatisedatthecriticalpoint. 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 13.53 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 )= exy x y 1.Then g (0 0)=0;thatis, thepoint(0 0)satisestheconstraint.Therstpartialderivativesof f and g are f x=2 xy +1, f y= x2+1, g x= yexy 1,and g y= xexy 1. Therefore,bothequations f x(0 0)= g x(0 0)and f y(0 0)= g y(0 0)

PAGE 294

28813.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS aresatisedat = 1.Thus,thepoint(0 0)isacriticalpointof f subjecttotheconstraint g =0. Second-DerivativeTest .Since g y(0 0)= 1 =0,thereisafunction y = h ( x )near x =0suchthat G ( x )= g ( x,h ( x ))=0.Bytheimplicit dierentiation, h(0)= g x(0 0) /g y(0 0)= 1 Thesecondpartialderivativesof g are g xx= y2exy,g yy= x2exy,g xy= exy+ xyexy. Thederivative h(0)isfoundfrom(13.20),where g xx(0 0)= g yy(0 0)= 0, g xy(0 0)=1, h(0)= 1,and g y(0 0)= 1: h(0)= [ g xx(0 0)+2 g xy(0 0) h(0)+ g yy(0 0)( h(0))2] /g y(0 0)= 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)= 1, f y(0 0)=1,and h(0)= 2into(13.19)gives F(0)= 2 < 0. Therefore, f attainsalocalmaximumat(0 0)alongthecurve g = 0.Notealsothat F(0)= f x(0 0)+ f y(0 0) h(0)=1 1=0as required. Theimplicitdierentiationandtheimplicitfunctiontheoremcan beusedtoestablishthesecond-derivativetestforthemultivariable casewithconstraints(seeanotherexampleinStudyProblem13.21).96.4.StudyProblems.Problem13.20. Anaxiallysymmetricsolidconsistsofacircular cylinderandaright-angledcircularconeattachedtooneofthecylindersbases.Whatarethedimensionsofthesolidatwhichithasa maximalvolumeifthesurfaceareaofthesolidhasaxedvalue S ? Solution: Let r and h betheradiusandheightofthecylinder.Since theconeisright-angled,itsheightis r .Thesurfaceareaisthesum ofthreeterms:theareaofthebase(disk) r2,theareaofthesideof thecylinder2 rh ,andthesurfacearea Scofthecone.Aconewithan angle atthevertexisobtainedbyrotationofastraightline y = mx where m =tan( / 2),aboutthe x axis.Inthepresentcase, = / 2 and m =1.If a istheheightofthecone,thenthesurfaceareaofthe coneis(seeCalculusII) Sc= a 02 ydx = r 02 xdx = r2

PAGE 295

96.LAGRANGEMULTIPLIERS289 because a = r inthepresentcase.Similarly,thevolumeoftheconeis Vc= a 0y2dx = r 0x2dx = r3 3 Therefore,theproblemisreducedtondingthemaximalvalueofthe function(volume) V ( r,h )= r2h + r3/ 3subjecttotheconstraint 2 rh +2 r2= S .Put g ( r,h )=2 rh +2 r2 S .Thencriticalpoints of V satisfytheequations: V r= g rV h= g h0= g ( r,h ) 2 rh + r2h = (2 h +4 r ) r2=2 rS 2 = rh + r2. Since r =0(thethirdequationisnotsatisedif r =0),thesecond equationimpliesthat = r/ 2.Thesubstitutionofthelatterintothe rstequationyields rh = r2or h = r .Thenitfollowsfromthethird equationthatthesought-afterdimensionsare h = r = S/ (4 ). Problem13.21. Letfunctions f and g ofthreevariables r =( x,y,z ) havecontinuouspartialderivativesuptoorder 2 .Usetheimplicit dierentiationtoestablishthesecond-derivativetestforcriticalpoints of 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 therstpartialderivativesof h : g x+ g zh x=0= h x= g x/g z; g y+ g zh y=0= h y= g y/g z. Thesecondpartialderivatives h xx, h xy,and h yyarefoundfromthe equations 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

PAGE 296

29013.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS 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, wheretherstandsecondpartialderivativesof h havebeenfoundearlier.If( x0,y0,z0)isthecriticalpointfoundbytheLagrangemultiplier method,then a = F xx( x0,y0), b = F yy( x0,y0),and c = F xy( x0,y0)in thesecond-derivativetestforthetwo-variablefunction F 96.5.Exercises.(1) UseLagrangemultiplierstondthemaximumandminimumvalues ofeachofthefollowingfunctionssubjecttothespeciedconstraints: (i) f ( x,y )= xy x + y =1 (ii) f ( x,y )= x2+ y2, x/a + y/b =1 (iii) f ( x,y )= xy2,2 x2+ y2=6 (iv) f ( x,y )= y2, x2+ y2=4 (v) f ( x,y )= x + y x2/ 16+ y2/ 9=1 (vi) f ( x,y )=2 x2 2 y2, x4+ y4=32 (vii) f ( x,y )= Ax2+2 Bxy + Cy2, x2+ y2=1 (viii) f ( x,y,z )= xyz ,3 x2+2 y2+ z2=6 (ix) f ( x,y,z )= x 2 y +2 z x2+ y2+ z2=1 (x) f ( x,y,z )= x2+ y2+ z2, x2/a2+ y2/b2+ z2/c2=1 (xi) f ( x,y,z )= x +3 y 3 z x + y z =0, y2+2 z2=1 (xii) f ( x,y,z )= xy + yz xy =1, y2+2 z2=1 (xiii) f ( x,y,z )= xy + yz x2+ y2=2, y + z =2( x> 0, y> 0, z> 0) (xiv) f ( x,y,z )=sin( x )sin( y )sin( z ), x + y + z = / 2( x> 0, y> 0, z> 0) (xv) f ( x,y,z )= x2/a2+ y2/b2+ z2/c2, x2+ y2+ z2=1, n1x + n2y + n3z =0,where n =( n1,n2,n3)isaunitvector (xvi) f ( r )= u r r = R ,where r =( x1,...,xm), u isaconstant unitvector,and R isaconstant (xvii) f ( r )= r r n r =1,where n hasstrictlypositivecomponents and r =( x1,x2,...,xm) (xviii) f ( r )= xn 1+ xn 2+ + xn m, x1+ x2+ + xm= a ,where n> 0 and a> 0 (2) Provetheinequality xn+ yn 2 x + y 2 n

PAGE 297

96.LAGRANGEMULTIPLIERS291 if n 1, x 0,and y 0. Hint: Minimizethefunction f =( xn+ yn) / 2 underthecondition x + y = s (3) Findtheminimalvalueofthefunction f ( x,y )= y onthecurve x2+ y4 y3=0.ExplainwhythemethodofLagrangemultipliersfails. Hint: Sketchthecurveneartheorigin. (4) UsethemethodofLagrangemultiplierstomaximizethefunction f ( x,y )=3 x +2 y onthecurve x + y =5.Comparetheobtained valuewith f (0 25).ExplainwhythemethodofLagrangemultipliers fails. (5) Findthreepositivenumberswhosesumisaxednumber c> 0 andwhoseproductismaximal. (6) UsethemethodofLagrangemultiplierstosolvethefollowingexercisesfromSection95.5: (i)Exercise10 (ii)Exercise11 (iii)Exercise12 (iv)Exercise13 (v)Exercise14 (7) Thecrosssectionofacylindricaltabisahalf-disk.Ifthetabhas totalarea S ,whatarethedimensionsatwhichthetabhasmaximal volume? (8) Findarectanglewithaxedperimeter2 p thatformsasolidofthe maximalvolumeunderrotationaboutoneofitssides. (9) Findatrianglewithaxedperimeter2 p thatformsasolidofthe maximalvolumeunderrotationaboutoneofitssides. (10) Findarectangularboxwiththemaximalvolumethatiscontainedinahalf-ballofradius R (11) Findarectangularboxwiththemaximalvolumethatiscontainedinanellipsoid x2/a2+ y2/b2+ z2/c2=1. (12) Consideracircularconeobtainedbyrotationofastraightline segmentoflength l abouttheaxisthroughanendpointofthesegment. Iftheanglebetweenthesegmentandtheaxisis ,ndarectangular boxwithintheconethathasamaximalvolume. (13) Thesolidconsistsofarectangularboxandtwoidenticalpyramids whosebasesareoppositefacesofthebox.Theedgesofthepyramid adjacentatthevertexoppositetoitsbasehaveequallengths.Ifthe solidhasaxedvolume V ,atwhatanglebetweentheedgesofthe pyramidanditsbaseisthesurfaceareaofthesolidminimal? (14) UseLagrangemultiplierstondthedistancebetweentheparabola y = x2andtheline x y =2.

PAGE 298

29213.DIFFERENTIATIONOFMULTIVARIABLEFUNCTIONS (15) Findthemaximumvalueofthefunction f ( r )=m x1x2 xmgiventhat x1+ x2+ + xm= c ,where c isapositiveconstant. Deducefromtheresultthatif xi> 0, i =1 2 ,...,m ,thenm x1x2 xm 1 m ( x1+ x2+ + xm); thatis,the geometricalmean of m numbersisnolargerthanthe arithmeticmean .Whenistheequalityreached? (16) GiveanalternativeproofoftheCauchy-Schwarzinequalityina Euclideanspace(Theorem13.1)usingthemethodofLagrangemultiplierstomaximizethefunctionof2 m variables f ( x y )= x y subject tothecontraints x x =1and y y =1,where x =( x1,...,xm)and y =( y1,...,ym). Hint: Aftermaximizingthefunction,put x = a / a and y = b / b foranytwononzerovectors a and b .

PAGE 299

CHAPTER14 MultipleIntegrals 97.DoubleIntegrals97.1.TheVolumeProblem.Supposeoneneedstodeterminethevolumeofahillwhoseheight f ( r )asafunctionofposition r =( x,y )is known.Forexample,thehillmustbeleveledtoconstructahighway. Itsvolumeisrequiredtoestimatethenumberoftruckloadsneededto movethesoilaway.Thefollowingprocedurecanbeusedtoestimate thevolume.Thebase D ofthehillisrstpartitionedintosmallpieces Dpofarea Ap,where p =1 2 ,...,N enumeratesthepieces;thatis, theunionofallthepieces Dpistheregion D .Thepartitionelements shouldbesmallenoughsothattheheight f ( r )hasnosignicantvariationwhen r isin Dp.Thevolumeoftheportionofthehillaboveeach partitionelement Dpisapproximately Vp f ( rp) Ap,where rpis apointin Dp(seetheleftpanelofFigure14.1).Theapproximation becomesbetterforsmaller Dp.Thevolumeofthehillcanthereforebe estimatedas 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 alongthestripmaybesignicantand theaccuracyoftheapproximation Vp f ( rp) Apispoor.Onecan293

PAGE 300

29414.MULTIPLEINTEGRALS x y z zpz = f ( x,y ) D rp Apx y d ykyk 1c a xj 1xjb RjkRDDFigure14.1.Left :Thevolumeofasolidregionbounded fromabovebythegraph z = f ( x,y )andfrombelowby aportion D ofthe xy planeisapproximatedbythesum ofvolumes Vp= zp Apofcolumnswiththebasearea Apandtheheight zp= f ( rp),where rpisasamplepoint withinthebaseand p enumeratesthecolumns. Right :A rectangularpartitionofaregion D isobtainedbyembedding D intoarectangle RD.Thentherectangle RDispartitioned intosmallerrectangles Rkj.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
PAGE 301

97.DOUBLEINTEGRALS295 Sothelimit(14.1)seemstodeneanintegraloveratwo-dimensional region D (i.e.,withrespecttobothvariables x and y usedtolabel pointsin D ).Thisobservationleadstotheconceptofa doubleintegral .However,thequalitativeconstructionusedtoanalyzethevolume problemstilllacksthelevelofrigorusedtodenethesingle-variable integration.Forexample,howdoesonechoosetheshapeofthepartitionelements Dp,orhowdoesonecalculatetheirareas?Thesekinds ofquestionswerenotevenpresentinthesingle-variablecaseandhave tobeaddressed.97.2.TheDoubleIntegral.Let D beaclosed,boundedregion.The boundariesof D areassumedtobepiecewise-smoothcurves.Let f ( r ) bea bounded functionon D ,thatis, m f ( r ) M forsomenumbers M and m andall r D .Thenumbers m and M arecalled lowerand upperbounds of f on D .Evidently,upperandlowerboundsarenot uniquebecauseanynumbersmallerthan m isalsoalowerbound,and, similarly,anynumbergreaterthan M isanupperbound.However, thesmallestupperboundandthelargestlowerboundareunique. Definition 14.1 (SupremumandInmum) Let f beboundedon D .Thesmallestupperboundof f on D iscalled the supremum of f on D anddenotedby supDf .Thelargestlower boundof f on D iscalledthe inmum of f on D anddenotedby infDf Asaboundedregion, D canalwaysbeembeddedinarectangle RD= { ( x,y ) | x [ a,b ] ,y [ c,d ] } (i.e., D isasubsetof RD).The function f isthen extended totherectangle RDbysettingitsvaluesto0 forallpointsoutside D ,thatis, f ( r )=0if r RDand r / D .Consider apartition xj, j =0 1 ,...,N1,oftheinterval[ a,b ],where xj= a + j x x =( b a ) /N1,andapartition yk, k =0 1 ,...,N2,oftheinterval [ c,d ],where yk= c + k y and y =( d c ) /N2.Thesepartitions induceapartitionoftherectangle RDbyrectangles Rjk= { ( x,y ) | x [ xj 1,xj] ,y [ yk 1,yk] } ,where j =1 2 ,...,N1and k =1 2 ,...,N2. Theareaofeachpartitionrectangle Rkjis A = x y .Thispartition iscalleda rectangularpartition of RD.Itisdepictedintherightpanel ofFigure14.1.Foreverypartitionrectangle Rjk,therearenumbers Mjk=sup f ( r )and mjk=inf f ( r ),thesupremumandinmumof f on Rjk. Definition 14.2 (UpperandLowerSums) Let f beaboundedfunctiononaclosed,boundedregion D .Let RDbea rectanglethatcontains D andletthefunction f bedenedtohavezero

PAGE 302

29614.MULTIPLEINTEGRALS valueforallpointsof RDthatdonotbelongto D .Givenarectangular partition Rjkof RD,let Mjk=sup f and mjk=inf f bethesupremum andinmumof f on Rjk.Thesums U ( f,N1,N2)=N1j =1 N2k =1Mjk A,L ( f,N1,N2)=N1j =1 N2k =1mjk A arecalledthe upperandlowersums Theupperandlowersumsareexamplesofdoublesequences. Definition 14.3 (DoubleSequence) Adoublesequenceisarulethatassignsanumber anmtoanordered pairofintegers ( n,m ) n,m =1 2 ,... Inotherwords,adoublesequenceisafunction f oftwovariables ( x,y )whosedomainconsistsofpointswithinteger-valuedcoordinates, anm= f ( n,m ).Similarlytoordinarynumericalsequences,onecan denealimitofadoublesequence. Definition 14.4 (LimitofaDoubleSequence) If,foranypositivenumber ,thereexistsaninteger N suchthat | anm a | < forall n,m>N ,thenthesequenceissaidtoconvergeto a andthenumber a iscalledthe limit ofthesequenceand denoted limn,m anm= a Thelimitofadoublesequenceisanalogoustothelimitofafunctionoftwovariables.Alimitofadoublesequencecanbefoundby studyingthecorrespondinglimitofafunctionoftwovariableswhose rangecontainsthedoublesequence.Suppose anm= f (1 /n, 1 /m )and f ( x,y ) 0as( x,y ) (0 0).Thelattermeansthat,forany > 0, thereisanumber > 0suchthat | f ( x,y ) | < forall r < where r =( x,y ).Inparticular,for r =(1 /n, 1 /m ),thecondition r 2=1 /n2+1 /m2<2issatisedforall n,m>N> 2 / .Hence, forallsuch n,m | anm| < ,whichmeansthat anm 0as n,m Continuingtheanalogywiththevolumeproblem, theupperand lowersumsrepresentthesmallestupperestimateandthegreatestlower estimateofthevolume .Theyshouldbecomecloserandclosertothe volumeasthepartitionbecomesnerandner.Thisleadstothe followingdenitionofthedoubleintegral. Definition 14.5 (DoubleIntegral) Ifthelimitsoftheupperandlowersumsexistas N1 2 (or

PAGE 303

97.DOUBLEINTEGRALS297 ( 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 Itshouldbeemphasizedthatthedoubleintegralisdenedasthe two-variablelimit( x, y ) (0 0)orasthelimitofdoublesequences. Theexistenceofthelimitanditsvaluemustbeestablishedaccording toDenition14.4. LetusdiscussDenition14.5fromthepointofviewofthevolume problem.First,notethataspecicpartitionof D byrectangleshas beenused.Inthisway,thearea Apofthepartitionelementhas beengivenaprecisemeaningastheareaofarectangle.Later,itwill beshownthatifthedoubleintegralexistsinthesenseoftheabove denition,thenitexistsiftherectangularpartitionisreplacedbyany partitionof D byelements Dpofanarbitraryshapesubjecttocertain conditionsthatallowforapreciseevaluationoftheirarea.Second,the volume(14.1)isindeedgivenbythedoubleintegralof f ,anditsvalue is independentofthechoiceofsamplepoints rp.Thisisanextremely usefulpropertythatallowsonetoapproximatethedoubleintegralwith anydesiredaccuracybyevaluatingasuitable Riemannsum Definition 14.6 (RiemannSum) Let f beaboundedfunctionon D thatiscontainedinarectangle RD. Let f bedenedbyzerovaluesoutsideof D in RD.Let r jkbeapoint inapartitionrectangle Rjk,where Rjkformarectangularpartitionof RD.Thesum R ( f,N1,N2)=N1j =1 N2k =1f ( r jk) A iscalleda Riemannsum 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

PAGE 304

29814.MULTIPLEINTEGRALS upperandlowersumsexistandcoincide.Theconclusionofthetheorem followsfromthesqueezeprincipleforlimits. ApproximationofDoubleIntegrals.If f isintegrable,itsdoubleintegral canbeapproximatedbyasuitableRiemannsum.Acommonlyused choiceofsamplepointsistotake r jktobetheintersectionofthe diagonalsofpartitionrectangles Rjk,thatis, r jk=( xj, yk),where xjand ykarethemidpointsoftheintervals[ xj 1,xj]and[ yk 1,yk], respectively.Thisruleiscalledthe midpointrule .Theaccuracyofthe midpointruleapproximationcanbeassessedbyndingtheupperand lowersums;theirdierencegivestheupperboundontheabsoluteerror oftheapproximation.Alternatively,iftheintegralistobeevaluated uptosomesignicantdecimals,thepartitionintheRiemannsum hastobereneduntilitsvaluedoesnotchangeinthesignicant digits.Theintegrabilityof f guaranteestheconvergenceofRiemann sumsandtheindependenceofthelimitfromthechoiceofsample points.97.3.ContinuityandIntegrability.Noteveryboundedfunctionisintegrable.Therearefunctionswhosebehaviorissoirregularthatone cannotgiveanymeaningtothevolumeundertheirgraphbyconverging upperandlowersums.AnExampleofaNonintegrableFunction.Let f bedenedonthesquare x [0 1]and y [0 1]sothat f ( x,y )=1ifboth x and y arerational, f ( x,y )=2ifboth x and y areirrational,and f ( x,y )=0otherwise. Thisfunctionisnotintegrable.Recallthatanyinterval[ a,b ]contains bothrationalandirrationalnumbers.Therefore,anypartitionrectangle Rjkcontainspointswhosecoordinatesarebothrational,orboth irrational,orpairsofrationalandirrationalnumbers.Hence, Mjk=2 and mjk=0.Thelowersumvanishesforanypartitionandtherefore itslimitis0,whereastheuppersumis2 jk A =2 A =2forany partition,where A istheareaofthesquare.Thelimitsoftheupper andlowersumsdonotcoincide,2 =0,andthedoubleintegralof f doesnotexist.TheRiemannsumforthisfunctioncanconvergetoany numberbetween2and0,dependingonthechoiceofsamplepoints. Forexample,ifthesamplepointshaverationalcoordinates,thenthe Riemannsumequals1.Ifthesamplepointshaveirrationalcoordinates,thentheRiemannsumequals2.Ifthesamplepointsaresuch thatonecoordinateisrationalwhiletheotherisirrational,thenthe Riemannsumvanishes.

PAGE 305

97.DOUBLEINTEGRALS299 z x M m A1A2y x z = f ( x,y ) D1D2Figure14.2.Left :Thegraphofapiecewise-constant function.Thefunctionhasajumpdiscontinuityalonga straightline.Thevolumeunderthegraphis V = MA1+ mA2.Despitethejumpdiscontinuity,thefunctionisintegrableandthevalueofthedoubleintegralcoincideswiththe volume V Right :Additivityofthedoubleintegral.Ifaregion D issplitbyacurveintotworegions D1and D2,then thedoubleintegralof f over D isthesumofintegralsover D1and D2.Theadditivityofthedoubleintegralisanalogoustotheadditivityofthevolume:Thevolumeunderthe graph z = f ( x,y )andabove D isthesumofvolumesabove D1and D2.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 )bedenedon 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

PAGE 306

30014.MULTIPLEINTEGRALS piecewiseconstantandhasajumpdiscontinuityalongtheline x =1in D .ItsgraphisshownintheleftpanelofFigure14.2.Thevolumebelow thegraph z = f ( x,y )andabove D iseasytond;itisthesumofthe volumesoftworectangularboxeswiththesamebasearea A1= A2=1 anddierentheights M and m V = MA1+ mA2= M + m .Thedouble integralof f existsandalsoequals M + m .Indeed,foranyrectangular partition,thenumbers Mjkand mjkdieronlyforpartitionrectangles intersectedbythediscontinuityline x =1,thatis, Mjk mjk= M m forallsuchrectangles.Therefore,thedierencebetweentheupperand lowersumsis l x ( M m ),where l =1isthelengthofthediscontinuity curve.Inthelimit x 0,thedierencevanishes.Asnotedearlier, theupperandlowersumsaretheupperandlowerestimatesofthe volumeandshouldthereforeconvergetoitastheirlimitscoincide. Usingasimilarlineofarguments,onecanprovethefollowing. Corollary 14.1 Let D beaclosed,boundedregionwhoseboundariesarepiecewise-smoothcurves.Ifafunction f isboundedon D anddiscontinuousonlyonanitenumberofsmoothcurves,thenitis integrableon D .97.4.Exercises.(1) Foreachofthefollowingfunctionsandthespeciedrectangular domain D ,ndthedoubleintegralusingitsdenition: (i) f ( x,y )= k =const, D = { ( x,y ) | a x b,c y d } (ii) f ( x,y )= k1=constif y> 0and f ( x,y )= k2=constif y 0, D = { ( x,y ) | 0 x 1 1 y 1 } (iii) f ( x,y )= xy D = { ( x,y ) | 0 x 1 0 y 1 } Hint: 1+2+ + N = N ( N +1) / 2. (2) Let D betherectangle1 x 2,1 y 3.Considera rectangularpartitionof D bylines x =1+ j/N and y =1+2 k/N j,k =1 2 ,...,N .Forthefunction f ( x,y )= x2+ y2,nd (i)Theloweranduppersums, U and L (ii)Thelimitofthedierence U L as N (iii)Thelimitofthesumsas N (3) Foreachofthefollowingfunctions,useaRiemannsumwithspecied N1and N2andsamplepointsatlowerrightcornerstoestimate thedoubleintegraloveragivenregion D : (i) f ( x,y )= x + y2,( N1,N2)=(2 2), D = { ( x,y ) | 0 x 2 0 y 4 } (ii) f ( x,y )=sin( x + y ),( N1,N2)=(3 3), D = { ( x,y ) | 0 x 0 y }

PAGE 307

98.PROPERTIESOFTHEDOUBLEINTEGRAL301 (4) Approximatetheintegralof f ( x,y )=(24+ x2+ y2) 1 / 2overthe disk x2+ y2 25byaRiemannsum.Useapartitionbysquareswhose verticeshaveinteger-valuedcoordinatesandsamplepointsatvertices ofthesquaresthatarefarthestfromtheorigin. (5) Evaluateeachofthefollowingdoubleintegralsbyrstidentifying itasthevolumeofasolid: (i) DkdA if D isthedisk x2+ y2 1and k isaconstant (ii) D 1 x2 y2dA if D isthedisk x2+ y2 1 (iii) D(1 x y ) dA if D isthetrianglewithvertices(0 0),(0 1), and(1 0) (iv) D( k x ) dA if D istherectangle0 x k and0 y a (v) D(2 x2+ y2) dA if D isthepartofthedisk x2+ y2 1in therstquadrant( Hint: Thevolumeofacircularsolidcone withthebasebeingthediskofradius R andtheheight h is R2h/ 3.) (6) Let I betheintegralofsin( x + y )overthedisk x2+ y2 1.Suppose thattheintegrationregionispartitionedbyrectanglesofarea A .If R isaRiemannsum,nd A suchthat | I R | < 0 001foranychoice ofsamplepoints. 98.PropertiesoftheDoubleIntegral Thepropertiesofthedoubleintegralaresimilartothoseofan ordinaryintegralandcanbeestablisheddirectlyfromthedenition.Linearity.Let f and g befunctionsintegrableon D andlet c bea number.Then D( f + g ) dA = DfdA + DgdA, DcfdA = c DfdA.Area.Thedoubleintegral (14.2) A ( D )= DdA iscalledthe area of D (ifitexists).If D isboundedbypiecewise-smooth curves,thenitexistsbecausetheunitfunction f =1iscontinuouson D .Bythegeometricalinterpretationofthedoubleintegral,thenumber A ( D )isthevolumeofthesolidcylinderwiththecrosssection D and the unit height( f =1).Intuitively,theregion D canalwaysbecovered bytheunionofadjacentrectanglesofarea A = x y .Inthelimit ( x, y ) (0 0),thetotalareaoftheserectanglesconvergestothe areaof D .

PAGE 308

30214.MULTIPLEINTEGRALS Additivity.Supposethat D istheunionof D1and D2suchthatthearea oftheirintersectionis0;thatis, D1and D2mayonlyhavecommon pointsattheirboundariesornocommonpointsatall.If f isintegrable on D ,then DfdA = D1fdA + D2fdA. Thispropertyisdiculttoprovedirectlyfromthedenition.However,itappearsrathernaturalwhenmakingtheanalogyofthedouble integralandthevolume.Iftheregion D iscutintotwopieces D1and D2,thenthesolidabove D isalsocutintotwosolids,oneabove D1andtheotherabove D2.Naturally,thevolumeisadditive(seethe rightpanelofFigure14.2). Supposethat f isnonnegativeon D1andnonpositiveon D2.The doubleintegralover D1isthevolumeofthesolidabove D1andbelow thegraphof f .Since f 0on D2,thedoubleintegralover D2is the negative volumeofthesolid below D2and above thegraphof f When f becomesnegative,itsgraphgoesbelowtheplane z =0(the xy plane).Sothedoubleintegralisthedierenceofthevolumesaboveand belowthe xy plane.Therefore,itmayvanishortakenegativevalues, dependingonwhichvolumeislarger.Thispropertyisanalogousto thefamiliarrelationbetweentheordinaryintegralandtheareaunder thegraph.ItisillustratedinFigure14.3(leftpanel).Positivity.If f ( r ) 0forall r D ,then DfdA 0 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 .Fromthepositivitypropertyforthe doubleintegralsof f ( x,y ) m 0and M f ( x,y ) 0over D and (14.2),itfollowsthat mA ( D ) DfdA MA ( D ) Thisinequalityiseasytovisualize.If f ispositive,thenthedouble integralisthevolumeofthesolidbelowthegraphof f .Thesolid

PAGE 309

98.PROPERTIESOFTHEDOUBLEINTEGRAL303 x y z z = f ( x,y ) D2D1x y z z = M z = f ( x,y ) z = m DFigure14.3.Left :Afunction f isnonnegativeonthe region D1andnonpositiveon D2.Thedoubleintegralof f ( x,y )overtheunionofregions D1and D2isthedierence oftheindicatedvolumes.Thevolumebelowthe xy plane andabovethegraphof f contributestothedoubleintegral withthenegativesign. Right :Anillustrationtotheupper andlowerboundsofthedoubleintegralofafunction f over aregion D .If A ( D )istheareaof D and m f ( x,y ) M in D ,thenthevolumeunderthegraphof f isnolessthan thevolume mA ( D )andnolargerthan MA ( D ).liesinthecylinderwithcrosssection D .Thegraphof f liesbetween theplanes z = m and z = M .Therefore,thevolumeofthecylinder ofheight m cannotexceedthevolumeofthesolid,whereasthelatter cannotexceedthevolumeofthecylinderofheight M asshowninthe rightpanelofFigure14.3. 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 h ) dA = DfdA 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(Extreme ValueTheorem12.21).Therefore,forany m h0 M ,thereis r0 D suchthat f ( r0)= h0. Ageometricalinterpretationoftheintegralmeanvaluetheoremis rathersimple.Imaginethatthesolidbelowthegraphof f ismadeof

PAGE 310

30414.MULTIPLEINTEGRALS z = f ( x,y ) h0D x y kk 1x y z 1 1Figure14.4.Left :Aclaysolidwithanonattop(the graphofacontinuousfunction f )maybedeformedtothe solidofthesamevolumeandwiththesamehorizontalcross section D ,butwithaattop h0.Thefunction f takes thevalue h0atsomepointof D .Thisillustratestheintegralmeanvaluetheorem. Middle :Apartitionofadisk byconcentriccirclesofradii r = rpandrays = kasdescribedinExample14.1.Apartitionelementistheregion rp 1 r rpand k 1 k. Right :Thevolumebelowthe graph z = x2+ y2andabovethedisk D x2+ y2 1.The correspondingdoubleintegralisevaluatedinExample14.1 bytakingthelimitofRiemannsumsforthepartitionof D showninthemiddlepanel.clay(seetheleftpanelofFigure14.4).Theshapeofapieceofclay maybedeformedwhilethevolumeispreservedunderdeformation.The nonattopofthesolidcanbedeformedsothatitbecomesat,turning thesolidintoacylinderofheight h0,which,byvolumepreservation, shouldbebetweenthesmallestandthelargestheightsoftheoriginal solid.Theintegralmeanvaluetheoremmerelystatestheexistenceof suchan average heightatwhichthevolumeofthecylindercoincides withthevolumeofthesolidwithanonattop.Thecontinuityof thefunctionissucient(butnotnecessary)toestablishthatthereis apointatwhichtheaverageheightcoincideswiththevalueofthe function. Definition 14.7 (AverageValueofaFunction) Let f beintegrableon D andlet A ( D ) betheareaof D .The average value of f on D istheintegral: 1 A ( D ) DfdA.

PAGE 311

98.PROPERTIESOFTHEDOUBLEINTEGRAL305 If f iscontinuouson D ,thentheintegralmeanvaluetheorem assertsthat f attainsitsaveragevalueatsomepointin D .Thecontinuityhypothesisiscrucialhere.Forexample,thefunctiondepicted intheleftpanelofFigure14.2isdiscontinuous.Itsaveragevalueis ( MA1+ mA2) / ( A1+ A2),whichgenerallydoesnotcoincidewitheither M or m .IntegrabilityoftheAbsoluteValue.Supposethat f isintegrableona bounded,closedregion D .Thenitsabsolutevalue | f | isalsointegrable and 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 V1 V2,while thedoubleintegralof | f | is V1+ V2.Naturally, | V1 V2| V1+ V2forpositive V1 2.Theconverseisnottrue.Theintegrabilityofthe absolutevalue | f | doesnotgenerallyimplytheintegrabilityof f .The readerisadvisedtoconsiderthefunction f ( x,y )=1if x and y are rationaland f ( x,y )= 1otherwisewhere( x,y )spantherectangle [0 1] [0 1].Notethat | f ( x,y ) | =1isintegrable.IndependenceofPartition.Ithasbeenarguedthatthevolumeofasolid underthegraphof f andabovearegion D canbecomputedby(14.1) inwhichtheRiemannsumisdenedforan arbitrary (nonrectangular) partitionof D .Canthedoubleintegralof f over D becomputedinthe sameway?Theanalysisislimitedtothecasewhen f iscontinuous. Definition 14.8 (UniformContinuity) Let f beafunctiononaregion D inaEuclideanspace.If,forany number > 0 ,thereexistsanumber > 0 suchthat | f ( r ) f ( r) | < whenever r r < forany r and rin D ,then f iscalled uniformly continuous on D Denition14.8impliescontinuityof f at r: f ( r ) f ( r)as r raccordingtoDenition13.9.Sotheuniformcontinuityimpliescontinuity.Theconverseisnottrue.Theuniformcontinuityimposesa strongerconditiononthebehaviorofthefunction.Thedierenceis thatthenumber canbechosen independently ofapoint r0,whereas

PAGE 312

30614.MULTIPLEINTEGRALS foracontinuousfunction, generallydependsonboth and r0.In otherwords,theuniformcontinuitymeansthat,foranypreassigned positivenumber ,thereisapositivenumber suchthatthevaluesof thefunctiondonotdiermorethan withinaballofradius nomatter wherein D theballiscentered .Therelationbetweencontinuousand uniformlycontinuousfunctionsisestablishedinthefollowingtheorem. Theorem 14.4 If f iscontinuousonabounded,closedregion D inaEuclideanspace,then f isuniformlycontinuouson D Theproofisomitted.Thehypothesisoftheclosednessof D is essential.Take f ( x,y )=1 /x ,whichiscontinuousintherectangle D =(0 1] [0 1].Note D isnotclosed.Theninadiskwhosecenteris sucientlyclosetotheline x =0,thevaluesof f canhavevariationsas largeasdesiredwithinthisdiskbecause1 /x divergesas x approaches 0.Forexample,takeaninterval[ x1,x2]oflength = x2 x1.Then 1 /x1 1 /x2canbemadearbitrarilylargebymoving x1closerto0for anychoiceof > 0.Sothevariationsof f cannotbeboundedbya xednumber uniformlyinanydiskofsomenonzeroradiusin D ,and f isnotuniformlycontinuousin D .Similarly,take f ( x,y )= x2,which iscontinuousintheunboundedrectangle D =[0 ) [0 1].Thenin adiskwhosecenterissucientlyfarfromtheline x =0,thevalues of f canhavevariationsaslargeasdesiredwithinthisdisk.Foran interval[ x1,x2]oflength > 0,thevariation x2 2 x2 1= ( x2+ x1)can bemadeaslargeasdesiredbytaking x2largeenoughnomatterhow small is.Hence, f isnotuniformlycontinuousin D Let f becontinuousinaclosed,boundedregion D .Let D be partitionedbypiecewise-smoothcurvesintopartitionelements Dp, p = 1 2 ,...,N ,sothattheunionof Dpis D and A ( D )= N p =1 Ap,where Apistheareaof Dpdenedby(14.2).If Rpisthesmallestradiusof adiskthatcontains Dp,put RN=max Rp;thatis, Rpcharacterizes thesizeofthepartitionelement Dp,and RNisthesizeofthelargest partitionelement.Recallthatthelargestpartitionelementdoesnot necessarilyhavethelargestarea.Thepartitionissaidtoberenedif RN
PAGE 313

98.PROPERTIESOFTHEDOUBLEINTEGRAL307 Proof. As f iscontinuouson D ,therearepoints rp Dpsuchthat DfdA =Np =1DpfdA =Np =1f ( rp) Ap. Therstequalityfollowsfromtheadditivityofthedoubleintegral,and thesecondoneholdsbytheintegralmeanvaluetheorem.Considerthe Riemannsum R ( f,N )=Np =1f ( r p) Ap, where r p Dparesamplepoints.If r p = rp,thentheRiemannsum doesnotcoincidewiththedoubleintegral.However,itslimitas N equalsthedoubleintegral.Indeed,put cp= | f ( r p) f ( rp) | and cN= max cp, p =1 2 ,...,N .ByTheorem14.4, f isuniformlycontinuouson D .Forany > 0,thereis > 0suchthatvariationsof f inanydisk ofradius in D donotexceed .Since RN 0as N RN< forall N largerthansome N0.Hence, cN< becauseanypartition element Dpiscontainedinadiskofradius Rp RN< ,whichimplies that cN 0as N .Therefore,thedeviationoftheRiemannsum fromthedoubleintegralconvergesto0: fdA R ( f,N ) = Np =1( f ( rp) f ( r p)) Ap Np =1| f ( rp) f ( r p) | Ap=Np =1cp Ap cN Np =1 Ap= cNA ( D ) 0 as N Apracticalsignicanceofthistheoremisthatthedoubleintegral canbeapproximatedbyRiemannsumsfor anyconvenientpartition of theintegrationregion.Notethattheregion D isnolongerrequiredto beembeddedinarectangleand f doesnothavetobeextendedoutside of D .Thispropertyisusefulforevaluatingdoubleintegralsbymeans of changeofvariables discussedlaterinthischapter.Itisalsouseful tosimplifycalculationsofRiemannsums. Example 14.1 Findthedoubleintegralof f ( x,y )= x2+ y2over thedisk Dx2+ y2 1 usingthepartitionof D byconcentriccircles andraysfromtheorigin.

PAGE 314

30814.MULTIPLEINTEGRALS Solution: Considercircles x2+ y2= r2 p,where rp= p r r =1 /N and p =0 1 2 ,...,N .If isthepolarangleintheplane,thenpoints withaxedvalueof formarayfromtheorigin.Letthedisk D be partitionedbycirclesofradii rpandrays = k= k =2 /n k =1 2 ,...,n .Eachpartitionelementliesinthesectorofangle and isboundedbytwocircleswhoseradiidierby r (seethemiddlepanel ofFigure14.4).Theareaofasectorofradius rpis r2 p / 2.Therefore, theareaofapartitionelementbetweencirclesofradii rpand rp +1is Ap= r2 p +1 / 2 r2 p / 2=( r2 p +1 r2 p) / 2 =( rp +1+ rp) r / 2. IntheRiemannsum,usethemidpointrule;thatis,thesamplepoints areintersectionsofthecirclesofradius rp=( rp +1+ rp) / 2andthe rayswithangles k=( k +1+ k) / 2.Thevaluesof f atthesample pointsare f ( r p)= r2 p,theareaelementsare Ap= rp r ,andthe correspondingRiemannsumreads 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 .98.1.Exercises.(1) Evaluateeachofthefollowingdoubleintegralsbyusingthepropertiesofthedoubleintegralanditsinterpretationasthevolumeofa solid:

PAGE 315

98.PROPERTIESOFTHEDOUBLEINTEGRAL309 (i) DkdA ,where k isaconstantand D isthesquare 2 x 2, 2 y 2withacircularholeofradius1(i.e., x2+ y2 1 in D ) (ii) DfdA ,where D isadisk x2+ y2 4and f isapiecewiseconstantfunction: f ( x,y )=2if1 x2+ y2 4and f ( x,y )= 3if0 x2+ y2< 1 (iii) D(4+3 x2+ y2) dA ,where D isthedisk x2+ y2 1 (iv) D( 4 x2 y2 2) dA if D isthepartofthedisk x2+ y2 4intherstquadrant (v) D(4 x y ) dA ,where D istherectanglewithvertices(0 0), (1 0),and(0 1)( Hint: Usetheidentity4 x y =3+(1 x y ) andthelinearityofthedoubleintegral.) (2) Usethepositivityofthedoubleintegraltoshowthat (i) Dsin( xy ) / ( xy ) dA A ( D ),where D isaboundedregionin which x> 0and y> 0 (ii) D( ax2+ by2) dA ( a + b ) / 2,where D isthedisk x2+ y2 1 ( Hint: Put r2= x2+ y2.Thenuse x2 r2and y2 r2and applytheresultofExample14.1.) (3) Findthelowerandupperboundsforeachofthefollowingintegrals: (i) Dxy3dA ,where D isthesquare1 x 2,1 y 2 (ii) D 1+ xe ydA ,where D isthesquare0 x 1,0 y 1 (iii) Dsin( x + y ) dA ,where D isthetrianglewithvertices(0 0), (0 ),and( / 4 0)( Hint: Graph D .Thengraphthesetof pointatwhichsin( x + y )attainsitsmaximalvalue.) (iv) D(100+cos2x +cos2y ) 1dA ,where D isdenedby | x | + | y | 10 (4) Let f becontinuousonaboundedregion D withanonzeroarea. Ifthedoubleintegralof f over D vanishes,provethatthereisapoint in D atwhich f vanishes. (5) UsethemethodofExample14.1tond Dex2+ y2dA ,where D is thepartofthedisk x2+ y2 1. (6) UseaRiemannsumtoapproximatethedoubleintegralof f ( x,y )= x + y overthetriangleboundedbythelines x =0, y =0,and x + y =1.Partitiontheintegrationregionintofourequaltriangles bythelines x =const, y =const,and x + y =const.Choosesample pointstobecentroidsofthetriangle. (7) Determinethesignofeachofthefollowingintegrals: (i) Dln( x2+ y2) dA ,where D isdenedby | x | + | y | 1 (ii) D3 1 x2 y2dA ,where D isdenedby x2+ y2 4

PAGE 316

31014.MULTIPLEINTEGRALS (iii) Dsin 1( x + y ) dA ,where D isdenedby0 x 1and 0 y 1 x 99.IteratedIntegrals Hereapracticalmethodforevaluatingdoubleintegralswillbedeveloped.Tosimplifythetechnicalities,thederivationofthemethodis givenforcontinuousfunctions.Incombinationwiththepropertiesof thedoubleintegral,itissucientformanyapplications. RecallSection87.3whereithasbeenshownthatifamultivariablelimitexists,thenthe repeatedlimits existandcoincide.Asimilar statementistruefordoublesequences. Theorem 14.6 Supposethatadoublesequence anmconvergesto a as n,m .Then limn limm anm =limm limn anm = a. Aproofofthissimpletheoremislefttothereaderasanexercise. Thelimitlimm anm= bnistakenforaxedvalueof n .Similarly, thelimitlimn anm= cmistakenforaxed m .Thetheoremstates thatthelimitsoftwogenerally dierent sequences bnand cmcoincide andareequaltothelimitofthedoublesequence.Thispropertyof doublesequenceswillbeappliedtoRiemannsumstoreduceadouble integraltoordinary iterated integrals.99.1.RectangularDomains.Letafunction f becontinuouson D .Supposerstthat D isarectangle x [ a,b ]and y [ c,d ].Inwhatfollows, abriefnotationforrectangleswillbeused: D =[ a,b ] [ c,d ].Let Rjkbearectangularpartitionof D asdenedearlier.Foranychoice ofsamplepoints( x j,y k),where x j [ xj 1,xj]and y k [ yk 1,yk],the Riemannsum R ( f,N1,N2)convergestothedoubleintegralof f over D byTheorem14.1.Sincethelimitofthedoublesequence R ( f,N1,N2) exists,itshouldnotdependontheorderinwhichthelimits N1 (or x 0)and N2 (or y 0)arecomputed(Theorem14.6). Supposethelimit y 0istobeevaluatedrst: DfdA =limN1 2R ( f,N1,N2) =limN1 N1j =1% limN2 N2k =1f ( x j,y k) y & x. TheexpressioninparenthesesisnothingbuttheRiemannsumforthe single-variablefunction gj( y )= f ( x j,y )ontheinterval y [ c,d ].So,

PAGE 317

99.ITERATEDINTEGRALS311 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. Deneafunction A ( x )by (14.5) A ( x )= d cf ( x,y ) dy. Thevalueof A at x isgivenbytheintegralof f withrespectto y ; theintegrationwithrespectto y iscarriedoutasif x wereaxed 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 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 (seetherightpanelofFigure14.5). 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, wheretheintegralexistsbythecontinuityof A .Theintegralonthe rightsideofthisequalityiscalledthe iteratedintegral .Inwhatfollows, theparenthesesintheiteratedintegralwillbeomitted.Theorder inwhichtheintegralsareevaluatedisspeciedbytheorderofthe dierentialsinit;forexample, dydx meansthattheintegrationwith respectto y istobecarriedoutrst.Inasimilarfashion,bycomputing thelimit x 0rst,thedoubleintegralcanbeexpressedasan iteratedintegralinwhichtheintegrationiscarriedoutwithrespect to x andthenwithrespectto y .Sothefollowingresulthasbeen established.

PAGE 318

31214.MULTIPLEINTEGRALS z = f ( x,y ) A ( y ) y R y z = f ( x,y ) R x A ( x ) xFigure14.5.AnillustrationtoFubinistheorem.The volumeofasolidbelowthegraph z = f ( x,y )andabovea rectangle R isthesumofthevolumesoftheslices. Left : Theslicingisdoneparalleltothe x axissothatthevolume ofeachsliceis yA ( y ),where A ( y )istheareaofthecross sectionbyaplanewithaxedvalueof y Right :Theslicing isdoneparalleltothe y axissothatthevolumeofeachslice is xA ( x ),where A ( x )istheareaofthecrosssectionbya planewithaxedvalueof x asgivenin(14 5).Theorem 14.7 (FubinisTheorem) If f iscontinuousontherectangle D =[ a,b ] [ c,d ] ,then Df ( x,y ) dA = d cb af ( x,y ) dxdy = b ad cf ( x,y ) dydx. Thinkofaloafofbreadwitharectangularbaseandwithatop havingtheshapeofthegraph z = f ( x,y ).Itcanbeslicedalongeither ofthetwodirectionsparalleltoadjacentsidesofitsbase.Fubinis theoremsaysthatthevolumeoftheloafisthesumofthevolumesof theslicesandisindependentofhowtheslicingisdone. Example 14.2 Findthevolumeofthesolidboundedfromabove bytheportionoftheparaboloid z =4 x2 2 y2andfrombelowbythe portionoftheparaboloid z = 4+ x2+2 y2,where ( x,y ) [0 1] [0 1] Solution: Iftheheightofthesolidatany( x,y ) D is h ( x,y )= ztop( x,y ) zbot( x,y ),wherethegraphs z = ztop( x,y )and z = zbot( x,y )

PAGE 319

99.ITERATEDINTEGRALS313 arethetopandbottomboundariesofthesolid,thenthevolumeis V = Dh ( x,y ) dA = D[ ztop( x,y ) zbot( x,y )] dA = D(8 2 x2 4 y2) dA = 1 01 0(8 2 x2 4 y2) dydx = 1 0[(8 2 x2) y 4 y3/ 3] 1 0dx = 1 0(8 2 x2 4 / 3) dx =6 Corollary 14.2 (FactorizationofIteratedIntegrals) Let D bearectangle [ a,b ] [ c,d ] .Suppose f ( x,y )= g ( x ) h ( y ) ,where thefunctions g and h areintegrableon [ a,b ] and [ c,d ] ,respectively. Then Df ( x,y ) dA = b ag ( x ) dx d ch ( y ) dy. Sothedoubleintegralbecomestheproductoftwoordinaryintegrals inthiscase.ThissimpleconsequenceofFubinistheoremisquite useful. Example 14.3 Evaluatethedoubleintegralof f ( x,y )=sin( x + y ) overtherectangle [0 ] [ / 2 ,/ 2] Solution: Onehassin( x + y )=sin x cos y +cos x sin y .Theintegral ofsin y over[ / 2 ,/ 2]vanishesbysymmetry.So,bythefactorization propertyoftheiteratedintegral,onlythersttermcontributestothe doubleintegral: Dsin( x + y ) dA = 0sin xdx / 2 / 2cos ydy =4 Thefollowingexampleillustratestheuseoftheadditivityofadoubleintegral. Example 14.4 Evaluatethedoubleintegralof f ( x,y )=15 x4y2overtheregion D ,whichistherectangle [ 2 2] [ 2 2] withtherectangularhole [ 1 1] [ 1 1] Solution: Let D1=[ 2 2] [ 2 2]andlet D2=[ 1 1] [ 1 1]. Therectangle D1istheunionof D and D2suchthattheirintersection

PAGE 320

31414.MULTIPLEINTEGRALS hasnoarea.Hence, D2fdA = DfdA + D1fdA DfdA = D2fdA D1fdA. Byevaluatingthedoubleintegralsover D1 2, D115 x4y2dA =15 2 2x4dx 2 2y2dy =210, D215 x4y2dA =15 1 1x4dx 1 1y2dy =4 thedoubleintegralover D isobtained,1024 4=1020. 99.2.StudyProblem.Problem14.1. Supposeafunction f hascontinuoussecondderivativesontherectangle D =[0 1] [0 1] .Find Df xydA if f (0 0)=1 f (0 1)=2 f (1 0)=3 ,and f (1 1)=5 Solution: ByFubinistheorem, Df xydA = 1 01 0 x f y( x,y ) dxdy = 1 0f y( x,y ) 1 0dy = 1 0[ f y(1 ,y ) f y(0 ,y )] dy = 1 0d dy [ f (1 ,y ) f (0 ,y )] dy =[ f (1 ,y ) f (0 ,y )] 1 0=[ f (1 1) f (0 1)] [ f (1 0) f (0 0)]=1 ByClairautstheorem, f xy= f yx,andthevalueoftheintegralis independentoftheorderofintegration. 99.3.Exercises.(1) Evaluatethefollowingdoubleintegralsoverspeciedrectangular regions: (i) D( x + y ) dA D =[0 1] [0 2] (ii) Dxy2dA D =[0 1] [ 1 1] (iii) D x +2 ydA D [1 2] [0 1] (iv) D(1+3 x2y ) dA D =[0 1] [0 2] (v) DxeyxdA D =[0 1] [0 1] (vi) Dcos( x +2 y ) dA D =[0 ] [0 ,/ 4]

PAGE 321

100.DOUBLEINTEGRALSOVERGENERALREGIONS315 (vii) D 1+2 x 1+ y2dA D =[0 1] [0 1] (viii) D y x2+ y2dA D =[0 1] [1 2] (ix) D( x y )ndA D =[0 1] [0 1],where n isapositiveinteger (x) Dex y + exdA D =[0 1] [0 2] (xi) Dsin2( x )sin2( y ) dA D =[0 ] [0 ] (xii) Dln( x + y ) dA D =[1 2] [1 2] (xiii) D 1 2 x + ydA D [0 1] [1 2] (xiv) Dx 2x ydA D =[0 1] [0 1] (xv) D 1 ( xy )3 1 xydA D =[0 ,1 2] [0 ,1 2] (2) Findthevolumeofeachofthefollowingsolids E : (i) E liesundertheparaboloid z =1+3 x2+6 y2andabovethe rectangle[ 1 1] [0 2]. (ii) E liesintherstoctantandisboundedbythecylinder z = 4 y2andtheplane x =3. (iii) E liesintherstoctantandisboundedbytheplanes x + y z =0, y =2,and x =1. (3) Evaluate DxydA ,where D isthepartofthesquare[ 1 1] [ 1 1]thatdoesnotlieintherstquadrant. (4) Let f becontinuouson[ a,b ] [ c,d ]andlet g ( u,v )= Duvf ( x,y ) dA where Duv=[ a,u ] [ b,v ]for a
PAGE 322

31614.MULTIPLEINTEGRALS u v D D DFigure14.6.Left :Aregion D issimpleinthedirection u Middle :Aregion D isnotsimpleinthedirection v Right :Aregion D issimpleorconvex.Anystraightline intersectsitalongatmostonesegment,orastraightline segmentconnectinganytwopointsof D liesin D .tothevector u intersects D alongatmostonestraightlinesegment.A region D iscalled convex ifitissimpleinanydirection. ThisdenitionisillustratedinFigure14.6.Suppose D issimple inthedirectionofthe y axis.Itwillbereferredtoas y simple or verticallysimple .Since D isbounded,thereisaninterval[ a,b ]such thatverticallines x = x0intersect D if x0 [ a,b ].Inotherwords, theregion D lieswithintheverticalstrip a x b .Takeavertical line x = x0 [ a,b ]andconsiderallpointsof D thatalsobelongto theline,thatis,pairs( x0,y ) D ,wheretherstcoordinateisxed. Sincethelineintersects D alongasegment,thevariable y rangesover aninterval.Theendpointsofthisintervaldependonthelineorthe valueof x0;thatis,forevery x0 [ a,b ], ybot y ytop,wherethe numbers ybotand ytopdependon x0.Soallverticallysimpleregions admitthefollowingalgebraicdescription.AlgebraicDescriptionofVerticallySimpleRegions.If D isvertically simple,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 Example 14.5 Giveanalgebraicdescriptionofthehalf-disk x2+ y2 1 y 0 ,asaverticallysimpleregion.

PAGE 323

100.DOUBLEINTEGRALSOVERGENERALREGIONS317 x y y = ytop( x ) y D y = ybot( x ) ax b x y d y c D x x = xbot( y ) x = xtop( y )Figure14.7.Left :Analgebraicdescriptionofaverticallysimpleregionasgivenin(14.6):forevery x [ a,b ],the y coordinaterangesovertheinterval ybot( x ) y ytop( x ). Right :Analgebraicdescriptionofahorizontallysimpleregion D asgivenin(14.7):forevery y [ c,d ],the x coordinate rangesovertheinterval xbot( y ) x xtop( y ).Solution: The x coordinateofanypointinthediskliesintheinterval [ a,b ]=[ 1 1](seeFigure14.8,leftpanel).Takeaverticallinecorrespondingtoaxedvalueof x inthisinterval.Thislineintersectsthe half-diskalongthesegmentwhoseoneendpointliesonthe x axis;that is, y =0= ybot( x ).Theotherendpointliesonthecircle.Solvingthe equationofthecirclefor y ,onends y = 1 x2.Since y 0inthe half-disk,thepositivesolutionhastobetaken, y = 1 x2= ytop( x ). Sotheregionisboundedbytwographs y =0and y = 1 x2.For every 1 x 1,0 y 1 x2. 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 secondcoordinateisxed.Sincethelineintersects D alongasegment, thevariable x rangesoveraninterval.Theendpointsofthisinterval dependonthelineorthevalueof y0;thatis,forevery y0 [ c,d ], xbot x xtop,wherethenumbers xbotand xtopdependon y0.Soall horizontallysimpleregionsadmitthefollowingalgebraicdescription.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 ] } .

PAGE 324

31814.MULTIPLEINTEGRALS x y y = 1 x2x y 1 1 y =0 x y x = 1 y21 x y x = 1 y2Figure14.8.Thehalf-disk D x2+ y2 1, y 0,isa simpleregion. Left :Analgebraicdescriptionof D asaverticallysimpleregionasgivenin(14.6).Themaximalrange of x in D is[ 1 1].Foreverysuch x ,the y coordinatein D hastherange0 y 1 x2. Right :Analgebraic descriptionof D asahorizontallysimpleregionasgivenin (14.7).Themaximalrangeof y in D is[0 1].Foreverysuch y ,the x coordinatein D hastherange 1 y2 x 1 y2.Thenumbers c and d are,respectively,thesmallestandthelargest valuesofthe y coordinateofpointsof D .Thetermsbelowand abovearenowdenedrelativetothelineofsightinthedirectionof the x axis. Example 14.6 Giveanalgebraicdescriptionofthehalf-disk x2+ y2 1 y 0 ,asahorizontallysimpleregion. Solution: The y coordinateofanypointinthediskliesintheinterval [ c,d ]=[0 1].Takeahorizontallinecorrespondingtoaxedvalueof y fromthisinterval.Thelineintersectsthehalf-diskalongasegment whoseendpointslieonthecircle.Solvingtheequationofthecircle for x ,the x coordinatesoftheendpointsareobtained: x = 1 y2. So,forevery0 y 1, 1 y2 x 1 y2.Whenviewedin thehorizontaldirection,thetopboundaryoftheregionisthegraph x = 1 y2= xtop( y )andthebottomboundaryisthegraph x = 1 y2= xbot( y )(seeFigure14.8,rightpanel). 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 anddenedbyzerovaluesoutside D ;thatis, f ( x,y )=0if c y
PAGE 325

100.DOUBLEINTEGRALSOVERGENERALREGIONS319 ConsideraRiemannsumforarectangularpartitionof RDwithsamplepoints( x j,y k)justlikeinthecaseofrectangulardomainsdiscussed earlier.Since f isintegrable,thedoubleintegralexists,andthedouble limitoftheRiemannsumshouldnotdependontheorderinwhichthe limits x 0and y 0aretaken(Theorem14.6).Foravertically simple D ,thelimit y 0istakenrst.Similarlyto(14.4),oneinfers that limN2 N2k =1f ( x j,y k) y = d cf ( x j,y ) dy = ytop( x j) ybot( x j)f ( x j,y ) dy becausethefunction f vanishesoutsidetheinterval ybot( x ) y ytop( x )forany x [ a,b ]. Supposethat f ( x,y ) 0andconsiderthesolidboundedfromabove bythegraph z = f ( x,y )andfrombelowbytheregion D .Theareaof thecrosssectionofthesolidbythecoordinateplanecorrespondingto axedvalueof x isisgivenby(14.5): A ( x )= d cf ( x,y ) dy = ytop( x ) ybot( x )f ( x,y ) dy. Sojustlikeinthecaseofrectangulardomains,theabovelimitequals A ( x j).Thattheareaisgivenbyanintegralovera single intervalis onlypossibleforaverticallysimplebase D ofthesolid.If D were notverticallysimple,thensuchaslicewouldnothavebeenasingle slicebutratherafewdisjointslices,dependingonhowmanydisjoint intervalsareintheintersectionofaverticallinewith D .Inthiscase, theintegrationwithrespectto y wouldhaveyieldedasumofintegrals overallsuchintervals.Thereasontheintegrationwithrespectto y istobecarriedoutrstonlyforverticallysimpleregionsisexactly toavoidthenecessitytointegrateoveraunionofdisjointintervals. Finally,thevalueofthedoubleintegralisgivenbytheintegralof A ( x ) overtheinterval[ a,b ].Recallthatthevolumeofasliceofwidth dx andcrosssectionarea A ( x )is dV = A ( x ) dx sothatthetotalvolume ofthesolidisgivenbytheintegral V = b aA ( x ) dx (asthesumof volumesofallslicesinthesolid).IteratedIntegralforVerticallySimpleRegions.Let D beavertically simpleregion;thatis,itadmitsthealgebraicdescription(14.6).The doubleintegralof f over D isthengivenbytheiteratedintegral (14.8) Df ( x,y ) dA = b aytop( x ) ybot( x )f ( x,y ) dydx.

PAGE 326

32014.MULTIPLEINTEGRALS x y 1 1 1 x y D y =1 y = x2x y D x y 1 x = yx = y y =1Figure14.9.IllustrationtoExample14.8. Left :Theintegrationregionasaverticallysimpleregion: 1 x 1 and,foreverysuch x x2 y 1. Right :Theintegration regionasahorizontallysimpleregion:0 y 1and,for everysuch y y x y .IteratedIntegralforHorizontallySimpleRegions.Naturally,forhorizontallysimpleregions,theintegrationwithrespectto x shouldbecarried outrst.Therefore,thelimit x 0shouldbetakenrstintheRiemannsum.Thetechnicalitiesaresimilartothecaseofverticallysimple regions.Let D beahorizontallysimpleregion;thatis,itadmitsthe algebraicdescription(14.7).Thedoubleintegralof f over D isthen givenbytheiteratedintegral (14.9) Df ( x,y ) dA = d cxtop( y ) xbot( y )f ( x,y ) dxdy.IteratedIntegralsforNonsimpleRegions.Iftheintegrationregion D is notsimple,howcanoneevaluatethedoubleintegral?Anynonsimple regioncanbecutbysuitablesmoothcurvesintosimpleregions Dp, p =1 2 ,...,n .Thedoubleintegraloversimpleregionscanthenbe evaluated.Thedoubleintegralover D isthenthesumofthedouble integralsover Dpbytheadditivityproperty.Sometimes,itisalso convenienttocuttheintegrationregionintotwoormorepieceseven iftheregionissimple(seeExample14.8). Example 14.7 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).Tondanalgebraic descriptionof D asaverticallysimpleregion,onehastorstspecify

PAGE 327

100.DOUBLEINTEGRALSOVERGENERALREGIONS321 themaximalrangeofthe x coordinatein D .Itisdeterminedbythe intersectionoftheline y =1andtheparabola y = x2,thatis,1= x2, andhence x [ a,b ]=[ 1 1]forallpointsof D (seetheleftpanel ofFigure14.9).Forany x [ 1 1],the y coordinateofpointsof D attainsthesmallestvalueontheparabola(i.e., ybot( x )= x2),andthe largestvalueontheline(i.e., ytop( x )=1).Onehas D6 yx2dA =6 1 1x21 x2ydydx =3 1 1x2(1 x4) dx =8 / 7 Itisalsoinstructivetoobtainthisresultusingthereverseorderofintegration.Tondanalgebraicdescriptionof D asahorizontallysimple region,onehastorstspecifythemaximalrangeofthe y coordinate in D .Thesmallestvalueof y is0andthelargestvalueis1;thatis, y [ c,d ]=[0 1]forallpointsof D .Foranyxed y [0 1],the x coordinateofpointsof D attainsthesmallestandlargestvaluesonthe parabola y = x2or x = y ,thatis, xbot( y )= y and xtop( y )= y (seetherightpanelofFigure14.9).Onehas D6 yx2dA =6 1 0y y yx2dxdy =2 1 0y (2 y3 / 2) dy =4 1 0y5 / 2dy =8 / 7 100.3.ReversingtheOrderofIntegration.Byreversingtheorderofintegration,asimplicationoftechnicalitiesinvolvedinevaluatingdouble integralscanbeachieved,butnotalways,though. Example 14.8 Evaluatethedoubleintegralof f ( x,y )=2 x over theregion D boundedbytheline x =2 y +2 andtheparabola x = y2 1 Solution: Theregion D isbothverticallyandhorizontallysimple. However,theiteratedintegralbasedonthealgebraicdescriptionof D asaverticallysimpleregionismoreinvolved.Indeed,thelargestvalue ofthe x coordinatein D occursatoneofthepointsofintersectionof thelineandtheparabola,2 y +2= y2 1or( y 1)2=4,andhence y = 1 3.Thelargestvalueof x in D is x =32 1=8.Thesmallest valueof x occursatthepointofintersectionoftheparabolawiththe x axis, x = 1.So[ a,b ]=[ 1 8].Foranyxed x [ 1 0],the rangeofthe y coordinateisdeterminedbytheparabola x = y2 1.

PAGE 328

32214.MULTIPLEINTEGRALS x y y = x +1 y x y = x/ 2 1 x y 1 8 y = x +1 y 3 x y x = y2 1 x =2 y +2 1Figure14.10.IllustrationtoExample14.8. Left :The integrationregion D asaverticallysimpleregion.Analgebraicdescriptionrequirestosplitingthemaximalrangeof x intotwointervals.Forevery 1 x 0,the y coordinate rangesovertheinterval x +1 y x +1,whereasfor every0 x 8, x/ 2 1 y x +1.Accordingly,when convertingthedoubleintegraltotheiteratedintegral,the region D hastobesplitintotwopartsinwhich x 0and x 0. Right :Theintegrationregion D asahorizontally simpleregion.Forevery 1 y 3,the x coordinateranges overtheinterval y2 1 x 2 y +2.Sothedoubleintegral canbeconvertedtoasingleiteratedintegral.Solutionsofthisequationare y = x +1,andtherangeofthe y coordinateis x +1 y x +1.Foranyxed x [0 8],the largestvalueof y stilloccursontheparabola, y = x +1,whilethe smallestvalueoccursontheline, x =2 y +2or y =( x 2) / 2,sothat x +1 y ( x 2) / 2.Theboundariesof D are y = ytop( x )= x +1 ,y = ybot( x )= x +1if 1 x 0 x/ 2 1if0 x 8 Thatthebottomboundaryconsistsoftwographsdictatesthenecessity tosplittheregion D intotworegions D1and D2suchthat x [ 1 0] forallpointsin D1and x [0 8]forallpointsin D2.Thecorresponding

PAGE 329

100.DOUBLEINTEGRALSOVERGENERALREGIONS323 iteratedintegralreads D2 xdA = D12 xdA + D22 xdA =2 0 1x x +1 x +1dydx +2 8 0x x/ 2+1 x +1dydx. Ontheotherhand,iftheiteratedintegralcorrespondingtothealgebraicdescriptionof D asahorizontallysimpleregionisused,thetechnicalitiesaregreatlysimplied.Thesmallestandlargestvaluesof y in D occuratthepointsofintersectionofthelineandtheparabolafound above, y = 1 3,thatis,[ c,d ]=[ 1 3].Foranyxed y [ 1 3], the x coordinaterangesfromitsvalueontheparabolatoitsvalueon theline, xbot( y )= y2 1 x 2 y +2= xbot( y ).Thecorresponding iteratedintegralreads D2 xdA =2 3 12 y +2 y2 1xdxdy = 3 1( y4+6 y2+8 y +3) dy =256 / 5 whichissimplertoevaluatethanthepreviousone. Sometimestheiteratedintegrationcannotevenbecarriedoutin oneorder,butitcanstillbedoneintheotherorder. Example 14.9 Evaluatethedoubleintegralof f ( x,y )=sin( y2) overtheregion D ,whichisthetriangleboundedbythelines x =0 y = x ,and y = Solution: Supposethattheiteratedintegralforverticallysimpleregionsisused.Therangeofthe x coordinateis x [0 ]=[ a,b ], and,foreveryxed 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 ].Foreveryxed y [0 ],therangeofthe x coordinateis xbot( y )=0 x y = xtop( y ) in D .Therefore,theiteratedintegralreads Dsin( y2) dA = 0sin( y2) y 0dxdy = 0sin( y2) ydy = 1 2 cos( y2) 0=1

PAGE 330

32414.MULTIPLEINTEGRALS 100.4.TheUseofSymmetry.Thesymmetrypropertyhasbeenestablishedinsingle-variableintegration: f ( x )= f ( x ) a af ( x ) dx =0 whichisquiteuseful.Forexample,anindeniteintegralofsin( x2011) cannotbeexpressedinelementaryfunctions.Nevertheless,tondits deniteintegraloverany symmetric interval[ a,a ],anexplicitformof theindeniteintegralisnotnecessary.Indeed,thefunctionsin( x2011) isantisymmetric,andhenceitsintegraloveranysymmetricinterval vanishes.Asimilarpropertycanbeestablishedfordoubleintegrals. Consideratransformationthatmapseachpoint( x,y )oftheplane toanotherpoint( xs,ys).Aregion D issaidtobe symmetric underatransformation( x,y ) ( xs,ys)iftheimage Dsof D coincides with D (i.e., Ds= D ).Forexample,let D beboundedbyanellipse x2/a2+ y2/b2=1.Then D issymmetricunderreectionsaboutthe x axis,the y axis,ortheircombination,thatis,( x,y ) ( xs,ys)= ( x,y ),( x,y ) ( xs,ys)=( x, y ),or( x,y ) ( xs,ys)=( x, y ). Atransformationoftheplane( x,y ) ( xs,ys)issaidtobe areapreserving iftheimage Dsofanyregion D underthistransformation hasthesamearea,thatis, A ( D )= A ( Ds).Forexample,translations,rotations,reectionsaboutlines,andtheircombinationsare area-preservingtransformations. Theorem 14.8 (SymmetryProperty) Letaregion D besymmetricunderanarea-preservingtransformation ( x,y ) ( xs,ys) suchthat f ( xs,ys)= f ( x,y ) .Thentheintegralof f over D vanishes: Df ( x,y ) dA =0 Aproofispostponeduntilthechangeofvariablesindoubleintegralsisdiscussed.Herethesimplestcaseofareectionaboutaline isconsidered.If D issymmetricunderthisreection,thentheline cuts D intotwoequal-arearegions D1and D2sothat Ds 1= D2and Ds 2= D1.Thedoubleintegralisindependentofthechoiceofpartition (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)byareapreservation. Chooseelements D1 pand Ds 1 ptopartitiontheregion D asshowninthe leftpanelofFigure14.11.Nowrecallthatthedoubleintegralisalso independentofthechoiceofsamplepoints.Suppose( xp,yp)aresample pointsin D1 p.Choosesamplepointsin Ds 1 ptobetheimages( xps,yps)

PAGE 331

100.DOUBLEINTEGRALSOVERGENERALREGIONS325 D1D2D1 pD2 p= Ds 1 pD1D2V1V2z = f ( x,y )Figure14.11.Left :Theregion D issymmetricrelativeto thereectionabouttheline.Underthisreection, D1 D2and D2 D1.Anypartitionof D1byelements D1 pinduces thepartitionof D2bytakingtheimagesof D1 punderthe reection. Right :Thegraphofafunction f thatisskewsymmetricunderthereection.If f ispositivein D2,thenit isnegativein D1.Thevolume V2ofthesolidbelowthegraph andabove D2isexactlythesameasthevolume V1= V2of thesolidabovethegraphandbelow D1.Butthelattersolid liesbelowthe xy plane,andhencethedoubleintegralover D is V2 V1=0.of( xp,yp)underthereection.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 thereectionandtherefore f ( xps,yps)= f ( xp,yp)forall p .Froma geometricalpointofview,theportionofthesolidboundedbythegraph z = f ( x,y )thatliesabovethe xy planehasexactlythesameshapeas thatbelowthe xy plane,andthereforetheirvolumescontributewith oppositesignstothedoubleintegralandcanceleachother(seethe rightpanelofFigure14.11). Example 14.10 Evaluatethedoubleintegralof sin[( x y )3] over theportion D ofthedisk x2+ y2 1 thatliesintherstquadrant ( x,y 0 ). Solution: Theregion D issymmetricunderthereectionabout theline y = x (seetheleftpanelofFigure14.12),thatis,( x,y ) ( xs,ys)=( y,x ),whereasthefunctionisantisymmetric, f ( xs,ys)= f ( y,x )=sin[( y x )3]=sin[ ( x y )3]= sin[( x y )3]= f ( x,y ). Bythesymmetryproperty,thedoubleintegralvanishes.

PAGE 332

32614.MULTIPLEINTEGRALS x y x y y = x x y 12 1 3 2 3 ( x,y ) ( x, y ) D1D2Figure14.12.Left :IllustrationtoExample14.8.The regionissymmetricunderthereectionabouttheline y = x Right :Theintegrationregion D inExample14.9.It canbeviewedasthedierenceoftheellipticregion D1and thesquare D2.Theellipticregionissymmetricunderthe reectionaboutthe x axis,whereasthefunction f ( x,y )= x2y3isskew-symmetric, f ( x, y )= f ( x,y ).Sotheintegral over D1mustvanish,andthedoubleintegralover D isthe negativeoftheintegralover D2.Example 14.11 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 D2asshownintherightpanelofFigurre14.12.Therefore,theadditivity ofthedoubleintegralcanbeused(compareExample14.4)totransformthedoubleintegraloveranonsimpleregion D intotwodouble integralsoversimpleregions: Dx2y3dA = D1x2y3dA D2x2y3dA = D2x2y3dA = 1 0x2dx 1 0y3dy = 1 / 12; theintegralover D1vanishesbecausetheellipticregion D1issymmetric underthereection( x,y ) ( xs,ys)=( x, y ),whereastheintegrand isantisymmetric, f ( x, y )= x2( y )3= x2y3= f ( x,y ).

PAGE 333

100.DOUBLEINTEGRALSOVERGENERALREGIONS327 100.5.StudyProblems.Problem14.2. ProvetheDirichletformula a 0x 0f ( x,y ) dydx = a 0a yf ( x,y ) dxdy,a> 0 Solution: Theleftsideoftheequationisaniteratedintegralforthe doubleintegral DfdA .Letusndtheshapeof D .Accordingto thelimitsofintegration, D admitsthefollowingalgebraicdescription (asaverticallysimpleregion).Forevery0 x a ,the y coordinate changesintheinterval0 y x .Sotheregion D isthetriangle boundedbythelines y =0, y = x ,and x = a .Toreversetheorderof integration,letusndanalgebraicdescriptionof D asahorizontally simpleregion.Themaximalrangeof y in D istheinterval[0 ,a ].For everyxed0 y a ,the x coordinatespanstheinterval y x a in D .SothetwosidesoftheDirichletformularepresentthesamedouble integralasiteratedintegralsindierentordersandhenceareequal. Problem14.3. Reversetheorderofintegration 2 1 2 x x22 xf ( x,y ) dydx. Solution: Thegiveniteratedintegralrepresentsadoubleintegral DfdA ,wheretheintegrationregionadmitsthefollowingdescription (asaverticallysimpleregion).Foreveryxed1 x 2,the y coordinatesspantheinterval2 x y 2 x x2.So D isbounded bythegraphs y =2 x (aline)and y = 2 x x2or y2=2 x x2or,aftercompletingthesquares,( x 1)2+ y2=1(acircleofradius1 centeredat(1 0)).Thecircleandthelineintersectatthepoints(1 1) and(2 0).Thus,theregion D isthepartofthedisk( x 1)2+ y2 1 thatliesabovetheline y =2 x .Thereaderisadvisedtosketchit. Toreversetheorderofintegration,letusndanalgebraicdescription of D asahorizontallysimpleregion.Themaximalrangeof y isthe interval[0 1],whichisdeterminedbythepointsofintersectionofthe circleandtheline.Viewingtheregion D alongthe x axis,onecansee that,foreveryxed0 y 1,thesmallestvalueof x in D isattained ontheline y =2 x or x =2 y = xbot( y ),whileitsgreatestvalue in D isattainedonthecircle( x 1)2+ y2=1or x 1= 1 y2or x =1+ 1 y2= xtop( y )becausethesolutionwiththeplussign correspondstothepartofthecirclethatliesabovetheline.Hence,

PAGE 334

32814.MULTIPLEINTEGRALS theintegralinthereversedorderreads 1 01+ 1 y22 yf ( x,y ) dxdy. 100.6.Exercises.(1) Foreachofthetwoordersofintegration,specifythelimitsin theiteratedintegralsfor Df ( x,y ) dA ,splittingtheintegrationregion whennecessary,if (i) D isthetrianglewithvertices(0 0),(2 1),and( 2 1) (ii) D isathetrapezoidwithvertices(0 0),(1 0),(1 2),and(0 1) (iii) D isthedisk x2+ y2 1 (iv) D isthedisk x2+ y2 y (v) D isthering1 x2+ y2 4 (2) Evaluatethefollowingdoubleintegralsoverthespeciedregion: (i) DxydA ,where D isboundedbythecurves y = x2and y = x (ii) D(2+ y ) dA ,where D istheregionboundedbythegraphs of x =3and x =4 y2(iii) D( x + y ) dA ,where D isboundedbythecurves x = y4and x = y (iv) D(2+ y ) dA ,where D istheregionboundedbythethree linesof x =3, y + x =0,and y x =0;ndthevalueofthe integralbygeometricmeans (v) Dx2ydA ,where D istheregionboundedbythegraphsof y =2+ x2and y =4 x2(vi) D 1 y2dA ,where D isthetrianglewithvertices(0 0), (0 1),and(1 1) (vii) DxydA ,where D isboundedbythelines y =1, x = 3 y and x =2 y (viii) Dy x2 y2dA ,where D isthetrianglewithvertices(0 0), (0 1),and(1 1) (ix) D(2 a x ) 1 / 2dA ,where D isboundedbythecoordinateaxes andbytheshortestarcofthecircleofradius a andcentered at( a,a ) (x) D| xy | dA ,where D isthediskofradius a centeredatthe origin (xi) D( x2+ y2) dA ,where D istheparallelogramwiththesides y = x y = x + a y = a ,and y =3 a ( a> 0)

PAGE 335

100.DOUBLEINTEGRALSOVERGENERALREGIONS329 (xii) Dy2dA ,where D isboundedbythe x axisandbyonearcof thecycloid x = a ( t sin t ), y = a (1 cos t ),0 t 2 (3) Sketchthesolidregionwhosevolumeisgivenbythefollowing integrals: (i) 1 01 x 0( x2+ y2) dydx (ii) D( x + y ) dA ,where D isdenedbytheinequalities0 x + y 1, x 0,and y 0 (iii) D x2+ y2dA ,where D isdenedbytheinequality x2+ y2 x (iv) D( x2+ y2) dA ,where D isdenedbytheinequality | x | + | y | 1 (v) D 1 ( x/ 2)2 ( y/ 3)2dA ,where D isdenedbytheinequality( x/ 2)2+( y/ 3)2 1 (4) Usethedoubleintegraltondthevolumeofthespeciedsolid region E : (i) E isboundedbytheplane x + y + z =1andthecoordinate planes. (ii) E liesundertheparaboloid z =2 x2+ y2andabovetheregion inthe xy planeboundedbythecurves x = y2and x =1. (iii) E isboundedbythecylinder x2+ y2=1andtheplanes y = z x =0,and z =0intherstoctant. (iv) E isboundedbythecylinders x2+ y2= a2and y2+ z2= a2. (v) E isenclosedbytheparaboliccylinders y =1 x2, y = x2 1 andtheplanes x + y + z =2,2 x +2 y z =10. (5) Sketchtheregionofintegrationandreversetheorderofintegration ineachofthefollowingiteratedintegrals.Evaluatetheintegralifthe integrandisspecied: (i) 4 12 yf ( x,y ) dxdy (ii) 0 ycos( x2) dxdy Hint :Afterreversingtheintegrationorder,makethesubstitution u = x2todotheintegral. (iii) 1 0 x x3f ( x,y ) dydx (iv) 1 0y y2f ( x,y ) dxdy (v) 1 0ex1f ( x,y ) dydx (vi) 4 12 yf ( x,y ) dxdy (vii) 3 0y/ 3 0f ( x,y ) dxdy + 6 36 y 0f ( x,y ) dxdy (viii) 4 02 x(1+ y3) 1dydx (ix) 2 62 x ( x2/ 4) 1f ( x,y ) dydx

PAGE 336

33014.MULTIPLEINTEGRALS (x) 1 11 x2 1 x2f ( x,y ) dydx (xi) 2 a 0 2 ax 2 ax x2f ( x,y ) dydx ( a> 0) (xii) 2 0sin x 0f ( x,y ) dydx (6) Usethesymmetryandthepropertiesofthedoubleintegraltond: (i) Dex2sin( y3) dA ,where D isthetrianglewithvertices(0 1), (0 1),and(1 0) (ii) D( y9+ px9) dA ,where p = 1and D = { ( x,y ) | 1 | x | + | y | 2 } (iii) DxdA ,where D isboundedbytheellipse x2/a2+ y2/b2=1 andhasthetriangularholewithvertices(0 ,b ),(0 b ),and ( a, 0) (iv) D(cos( x2)+sin( y2)) dA ,where D isthedisk x2+ y2 a2(7) Findtheareaofthefollowingregions: (i) D isboundedbythecurves xy = a2and x + y =5 a/ 2, a> 0. (ii) D isboundedbythecurves y2=2 px + p2and y2= 2 qx + q2, where p and q arepositivenumbers. (iii) D isboundedby( x y )2+ x2= a2. 101.DoubleIntegralsinPolarCoordinates Thepolarcoordinatesaredenedbythefollowingrelations: x = r cos ,y = r sin or r = x2+ y2, =tan 1( y/x ) where r isthedistancefromtheorigintothepoint( x,y )and isthe anglebetweenthepositive x axisandtherayfromtheoriginthrough thepoint( x,y )countedcounterclockwise.Thevalueoftan 1must betakenaccordingtothegeometricaldenitionof .If( x,y )lies intherstquadrant,thenthevalueoftan 1mustbeintheinterval [0 ,/ 2)andtan 1( )= / 2andsimilarlyfortheotherquadrants. Theseequationsdenea one-to-one correspondencebetweenallpoints ( x,y ) =(0 0)oftheplaneandpointsofthestrip( r, ) (0 ) [0 2 ).Thepairs( r, )=(0 )correspondtotheorigin( x,y )=(0 0). Alternatively,onecanalsosettherangeof tobetheinterval[ ). Theorderedpair( r, )canbeviewedasapointofanauxiliaryplane or polarplane .Inwhatfollows,the r axisinthisplaneissettobe vertical,andthe axisissettobehorizontal. Therelations x = r cos y = r sin denea transformation ofany region Dinthepolarplanetoaregion D inthe xy plane;thatis, toeveryorderedpair( r, )correspondingtoapointof D,anordered pair( x,y )correspondingtoapointof D isassigned.Accordingly,the inversetransformation r = x2+ y2, =tan 1( y/x )mapsaregion

PAGE 337

101.DOUBLEINTEGRALSINPOLARCOORDINATES331 r rj +1rjD jkDkk +1 x y r = rj = x +1 = kD Djkr = rj +1Figure14.13.Left :Apartitionof Dbythecoordinate lines r = rjand = k,where rj +1 rj= r and k +1 k= .Apartitionelementisarectangle D jk.Itsareais A jk= r Right :Apartitionof D bytheimagesof thecoordinatecurves r = rj(concentriccircles)and = k(raysextendedfromtheorigin).Apatritionelement Djkis theimageoftherectangle D jk.Itsareais Ajk=1 2( r2 j +1 r2 j) =1 2( rj +1+ rj) A jk.D inthe xy planetoaregion Dinthepolarplane.Theboundariesof Daremappedontotheboundariesof D by x = r cos and y = r sin Forexample,let D betheportionofthedisk x2+ y2 1intherst quadrant.Thentheshapeof Dcanbefoundfromtheimagesof boundariesof 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. Thebounadryof Dalwayscontains r =0iftheoriginbelongsto D If D isinvariantunderrotationsabouttheorigin(adiskoraring), then takesitsfullrange[0 2 ]in D. Let Dbearegioninthepolarplaneandlet D beitsimagein the xy plane.Let R Dbearectanglecontaining Dsothattheimage of R Dcontains D .Asbefore,afunction f on D isextendedoutside D bysettingitsvaluesto0.Considerarectangularpartitionof R Dsuchthateachpartitionrectangle D jkisboundedbythecoordinate lines r = rj, r = rj +1= rj+ r = k,and = k +1= k+ as showninFigure14.13(leftpanel).Eachpartitionrectanglehasthe area A= r .Theimageofthecoordinateline r = rkinthe xy

PAGE 338

33214.MULTIPLEINTEGRALS planeisthecircleofradius rkcenteredattheorigin.Theimageofthe coordinateline = konthe xy planeistherayfromtheoriginthat makestheangle kwiththepositive x axiscountedcounterclockwise. Theraysandcirclesarecalled coordinatecurves ofthepolarcoordinatesystem,thatis,thecurvesalongwhicheitherthecoordinate r orthecoordinate remainsconstant(concentriccirclesandrays,respectively).Arectangularpartitionof Dinducesapartitionof D by coordinatecurvesofthepolarcoordinates.Eachpartitionelement Djkistheimageoftherectangle D jkandisboundedbytwocirclesand tworays. 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 +1 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, oneobtainsthedoubleintegralofthefunction f ( r cos ,r sin ) J ( r ) overtheregion D(theimageof D ),where J ( r )= r iscalledthe Jacobian ofthepolarcoordinates.TheJacobiandenestheareaelement transformation dA = JdA= rdA. Definition 14.10 (DoubleIntegralinPolarCoordinates) Let D betheimageof Dinthepolarplanespannedbyorderedpairs ( r, ) ofpolarcoordinates.Thedoubleintegralof f over D inpolar coordinatesis Df ( x,y ) dA = Df ( r cos ,r sin ) J ( r ) dA,J ( r )= r.

PAGE 339

101.DOUBLEINTEGRALSINPOLARCOORDINATES333 Inparticular,theareaofaregion D isgivenbythedoubleintegral A ( D )= DdA = DrdAinthepolarcoordinates.Asimilaritybetweenthedoubleintegralin rectangularandpolarcoordinatesisthattheybothusepartitionsby correspondingcoordinatecurves.Notethathorizontalandvertical linesarecoordinatecurvesoftherectangularcoordinates.Sothevery termadoubleintegralinpolarcoordinatesreferstoaspecicpartitioning D intheRiemannsum,namely,by coordinatecurvesofpolar coordinates (bycirclesandrays). Thedoubleintegralover Dcanbe evaluatedbythestandardmeans,thatis,byconvertingittoasuitable iteratedintegralwithrespectto r and .Supposethat Disavertically simpleregionasshowninFigure14.14(rightpanel): D= { ( r, ) | rbot( ) r rtop( ) ,1 2} Then D isboundedbythe polargraphs r = rbot( ), r = rtop( )andby thelines y =tan 1x and y =tan 2x (seetheleftpanelofFigure14.14). Recallthatcurvesdenedbytheequation r = g ( )arecalled polar graphs .Theycanbevisualizedbymeansofasimplegeometricalprocedure.Takearaycorrespondingtoaxedvalueofthepolarangle .Onthisray,markthepointatadistance r = g ( )fromtheorigin. Allsuchpointsobtainedforallvaluesof formthepolargraph.The doubleintegralover D canbewrittenastheiteratedintegralover D: Df ( x,y ) dA = 21rtop( ) rbot( )f ( r cos ,r sin ) rdrd. Example 14.12 Usepolarcoordinatestoevaluatethedoubleintegralof f ( x,y )= xy2 x2+ y2over D ,whichistheportionofthedisk x2+ y2 1 thatliesintherstquadrant. Solution: Whenconvertingadoubleintegraltopolarcoordinates, threeessentialstepsshouldbefollowed. First,onehastondtheregion Dinthepolarplanewhose imageis D underthetransformation x = r cos y = r sin .Usingthe boundarytransformation,asexplainedatthebeginningofthissection, Distherectangle r [0 1]and [0 ,/ 2]. Second,theintegrandhastobewritteninpolarcoordinates.In thisexample, f ( r cos ,r sin )= r4cos sin2 Third,thedoubleintegraloftheintegrandwritteninpolarcoordinatesand multipliedbytheJacobian r hastobeevaluatedover D.

PAGE 340

33414.MULTIPLEINTEGRALS x y = 2D r = rtop( ) r = rbot( ) = 1r r = rtop( ) r = rbot( ) D12Figure14.14.Left :Inpolarcoordinates,theboundary ofaregion D ,whichistheimageofaverticallysimpleregion Dinthepolarplane,canbeviewedaspolargraphsandlines throughtheorigin. Right :Averticallysimpleregion Din thepolarplane.Inthisexample, Disarectangleandhence,byFubinistheorem,the orderofintegrationintheiteratedintegralisirrelevant: DfdA = / 2 0sin2 cos d 1 0r5dr = 1 3 sin3 / 2 0 1 6 r6 1 0= 1 18 Thisexampleshowsthatthetechnicalitiesinvolvedinevaluating thedoubleintegralhavebeensubstantiallysimpliedbyusingpolar coordinates.Thesimplicationistwofold.First,thedomainofintegrationhasbeensimplied;thenewdomainisarectangle,which ismuchsimplertohandleintheiteratedintegralthanaportionof adisk.Second,theevaluationofordinaryintegralswithrespectto r and appearstobesimplerthantheintegrationof f withrespectto either x or y neededintheiteratedintegral.However,thesesimplicationscannotalwaysbeachievedbyconvertingthedoubleintegral topolarcoordinates.Theregion D andtheintegrand f shouldhave someparticularpropertiesthatguaranteetheobservedsimplications andtherebyjustifytheuseofpolarcoordinates.Herearesomeguidingprinciplestodecidewhethertheconversionofadoubleintegralto polarcoordinatescouldbehelpful: Thedomain D isboundedbycircles,linesthroughtheorigin, andpolargraphs. Thefunction f ( x,y )dependsoneitherthecombination x2+ y2= r2or y/x =tan .

PAGE 341

101.DOUBLEINTEGRALSINPOLARCOORDINATES335 x y 12 2 D x2+ y2=4 x2+ y2=2 x r Dr =2 r =2cos / 2Figure14.15.IllustrationtoExample14.13.Indeed,if D isboundedonlybycirclescenteredattheoriginandlines throughtheorigin,then Disarectanglebecausetheboundariesof D are coordinatecurves ofpolarcoordinates.Iftheboundariesof D containcirclesnotcenteredattheoriginor,generally,polargraphs, thatis,curvesdenedbytherelations r = g ( ),thenanalgebraicdescriptionoftheboundariesof Dissimplerthanthatoftheboundaries of D .If f ( x,y )= h ( u ),where u = x2+ y2= r2or u = y/x =tan thenintheiteratedintegraloneoftheintegrations,eitherwithrespect to or r ,becomestrivial. Example 14.13 Evaluatethedoubleintegralof f ( x,y )= xy over theregion D thatliesintherstquadrantandisboundedbythecircles x2+ y2=4 and x2+ y2=2 x Solution: First,theregion Dwhoseimageis D mustbefound.Using theprinciplethattheboundariesof D arerelatedtotheboundaries of Dbythechangeofvariables,theequationsoftheboundariesof Dareobtainedbyconvertingtheequationsfortheboundariesof D intopolarcoordinates.Theboundaryoftheregion D consistsofthree curves: x2+ y2=4 r2=4 r =2 x2+ y2=2 x r2=2 r cos r =2cos x =0 ,y 0 = / 2 So,inthepolarplane,theregion Disboundedbythehorizontalline r =2,thegraph r =2cos ,andtheverticalline = / 2.Itis convenienttouseanalgebraicdescriptionof Dasaverticallysimple region;thatis,( r, ) Dif rbot( )=2cos r 2= rtop( )and [0 ,/ 2]=[ 1,2](because rtop(0)= rbot(0)).Second,thefunctioniswritteninpolarcoordinates, f ( r cos ,r sin )= r2sin cos MultiplyingitbytheJacobian J = r ,theintegrandisobtained.One

PAGE 342

33614.MULTIPLEINTEGRALS has DxydA = Dr3sin cos dA= 21sin cos rtop( ) rbot( )r3drd = / 2 0sin cos 2 2cos r3drd =4 / 2 0(1 cos )4cos sin d =4 1 0(1 u )4udu =4 1 0v4(1 v ) dv = 4 15 wheretwochangesofvariableshavebeenusedtosimplifythecalculations, u =cos and v =1 u Example 14.14 Findtheareaoftheregion D thatisbounded bytwospirals r = and r =2 ,where [0 2 ] ,andthepositive x axis. Beforesolvingtheproblem,letusmakeafewcommentsaboutthe shapeof D .Theboundaries r = and r =2 arepolargraphs.Givena valueof r = (or r =2 )isthedistancefromthepointonthegraph totheorigin.Asthisdistanceincreasesmonotonicallywithincreasing ,thepolargraphsarespiralswindingabouttheorigin.Theregion D liesbetweentwospirals;itisnotsimpleinanydirection(seethe leftpanelofFigure14.16).Byconvertingthepolargraph r = into therectangularcoordinates,onehas x2+ y2=tan 1( y/x )or y = x tan( x2+ y2).Thereisnowaytondananalyticsolutionofthis equationtoexpress y asafunctionof x orviceversa.Therefore,had onetriedtoevaluatethedoubleintegralintherectangularcoordinates bycuttingtheregion D intosimplepieces,onewouldhavefacedan unsolvable problemofndingtheequationsfortheboundariesof D in theform y = ytop( x )and y = ybot( x )! Solution: Theregion D isboundedbythreecurves:twospirals (polargraphs)andtheline y =0 ,x 0.Theyaretheimagesof thelines r = r =2 ,and =2 inthepolarplaneasshown intherightpanelofFigure14.16.Theselinesformtheboundaries of D.Analgebraicdescriptionof Dasaverticallysimpleregionis convenienttouse,( r, ) Dif rbot( )= r 2 = rtop( )and [0 2 ]=[ 1,2].Hence, A ( D )= DdA = DrdA= 2 02 rdrd = 3 2 2 02d =4 3.

PAGE 343

101.DOUBLEINTEGRALSINPOLARCOORDINATES337 x y 2 2 4 D r = r =2 r Dr = r =2 2 Figure14.16.AnillustrationtoExample14.14. Left : Theintegrationregion D liesbetweentwospirals.Itisnot simpleinanydirection. Right :Theregion Dinthepolar planewhoseimageis D .Theregion Dissimpleandis boundedbystraightlines.Example 14.15 Findthevolumeofthepartofthesolidbounded bythecone z = x2+ y2andtheparaboloid z =2 x2 y2thatlies intherstoctant. Solution: ThesolidisshownintheleftpanelofFigure14.17.The intersectionofthecone(bottomboundary)andparaboloid(topboundary)isacircleofunitradius.Indeed,put r = x2+ y2.Thenthe pointsofintersectionsatisfythecondition x2+ y2=2 x2 y2or r =2 r2or r =1.Sotheprojection D ofthesolidontothe xy plane alongthe z axisisthepartofthedisk r 1intherstquadrant.For any( x,y ) D ,theheightis h = ztop( x,y ) zbot( x,y )=2 r2 r (i.e.,independentofthepolarangle ).Theregion D istheimageof therectangle D=[0 1] [0 ,/ 2]inthepolarplane.Thevolumeis Dh ( x,y ) dA = D(2 r2 r ) rdA= / 2 0d 1 0(2 r r3 r2) dr = 5 24 101.1.StudyProblem.Problem14.4. Findtheareaofthefour-leavedroseboundedbythe polargraph r =cos(2 ) Solution: Thepolargraphcomesthroughtheorigin r =0fourtimes when = / 4, = / 4+ / 2, = / 4+ ,and = / 4+3 / 2.These

PAGE 344

33814.MULTIPLEINTEGRALS x y z h ( x,y ) 1 1 D 1 1 D x2+ y2=1 r 1 D/ 4Figure14.17.AnillustrationtoExample14.15. Left : Thesolidwhosevolumeissought.Itsverticalprojectiononto the xy planeis D ,whichisthepartofthedisk r 1inthe rstquadrant.Atapoint( x,y )in D ,theheight h ( x,y )ofthe solidisthedierencebetweenthevaluesofthe z coordinate onthetopandbottomboundaries(theparaboloidandthe cone,respectively). Right :Theregion Dinthepolarplane whoseimageis D .anglesmaybechangedbyaddinganintegermultipleof ,owingto theperiodicityofcos(2 ).Therefore,eachleafoftherosecorresponds totherangeof betweentwoneighboringzerosofcos(2 ).Sinceall leaveshavethesamearea,itissucienttondtheareaofoneleaf, say,for / 4 / 4.Withthischoice,theleafistheimageofthe verticallysimpleregion D= { ( r, ) | 0 r cos(2 ) / 4 / 4 } inthepolarplane.Therefore,itsareaisgivenbythedoubleintegral A ( D )= DdA = DrdA= / 4 / 4cos(2 ) 0rdrd = 1 2 / 4 / 4cos2(2 ) d = 1 4 / 4 / 4(1+cos(4 )) d = 1 4 + 1 4 sin(4 ) / 4 / 4= 8 .

PAGE 345

101.DOUBLEINTEGRALSINPOLARCOORDINATES339 Thus,thetotalareais4 A ( D )= / 2. 101.2.Exercises.(1) Sketchtheregionwhoseareaisgivenbytheiteratedintegralin polarcoordinatesandevaluatetheintegral: (i) 02 1rdrd (ii) / 2 / 22 a cos 0rdrd (2) Convertthedoubleintegral Df ( x,y ) dA toaniteratedintegral inpolarcoordinatesif (i) D isthedisk x2+ y2 R2(ii) D isthedisk x2+ y2 ax a> 0 (iii) D isthering a2 x2+ y2 b2(iv) D istheparabolicsegment a x a x2/a y a (3) Evaluatethedoubleintegralbychangingtopolarcoordinates: (i) DxydA ,where D isthepartofthering a2 x2+ y2 b2intherstquadrant (ii) Dsin( x2+ y2) dA ,where D isthedisk x2+ y2 a2(iii) Darctan( y/x ) dA ,where D isthepartofthering0 0 (v) Dsin( x2+ y2) dA ,where D isthering 2 x2+ y2 4 2(4) If r and arepolarcoordinates,reversetheorderofintegration: (i) / 2 / 2cos 0f ( r, ) drd (ii) / 2 0a sin(2 ) 0f ( r, ) drd a> 0 (iii) 0 a 0f ( r, ) drd ,0
PAGE 346

34014.MULTIPLEINTEGRALS (i) 2 0x 3 xf ( x2+ y2) dydx (ii) 1 0x20f ( x,y ) dydx (7) Convertthedoubleintegraltoaniteratedintegralinpolarcoordinates: (i) Df ( x2+ y2) dA ,where D isthedisk x2+ y2 1 (ii) Df ( x2+ y2) dA ,where D = { ( x,y ) || y || x | | x | 1 } (iii) Df ( y/x ) dA ,where D isthedisk x2+ y2 x (iv) Df ( x2+ y2) dA where D isboundedbythecurve( x2+ y2)2= a2( x2 y2) (8) Findtheareaofthespeciedregion D : (i) D isenclosedbythepolargraph r =1+cos (ii) D istheboundedplaneregionbetweentwospirals r = / 4 and r = / 2,where [0 2 ],andthepositive x axis. (iii) D isthepartoftheregionenclosedbythecardioid r =1+sin thatliesoutsidethedisk x2+ y2 9 / 4 (iv) D isboundedbythecurve( x2+ y2)2=2 a2( x2 y2)and x2+ y2 a2(v) D isboundedbythecurve( x3+ y3)2= x2+ y2andliesinthe rstquadrant (vi) D isboundedbythecurve( x2+ y2)2= a ( x3 3 xy2), a> 0 (vii) D isboundedbythecurve( x2+ y2)2=8 a2xy and( x a )2+ ( y a )2 a2, a> 0. (9) Findthevolumeofthespeciedsolid E : (i) E isboundedbythecones z =3 x2+ y2and z =4 x2+ y2. (ii) E isboundedbythecone z = x2+ y2,theplane z =0,and thecylinders x2+ y2=1, x2+ y2=4. (iii) E isboundedbytheparaboloid z =1 x2 y2andtheplane z = 3. (iv) E isboundedbythehyperboloid x2+ y2 z2= 1andthe plane z =2. (v) E liesundertheparaboloid z = x2+ y2,abovethe xy plane, andinsidethecylinder x2+ y2=2 x (10) Find lima 01 a2Df ( x,y ) dA,D : x2+ y2 a2if f isacontinuousfunction.

PAGE 347

102.CHANGEOFVARIABLESINDOUBLEINTEGRALS341 102.ChangeofVariablesinDoubleIntegrals Withanexampleofpolarcoordinates,itisquiteclearthatasmart choiceofintegrationvariablescansignicantlysimplifythetechnicalitiesinvolvedwhenevaluatingdoubleintegrals.Thesimplicationis twofold:simplifyingtheshapeoftheintegrationregion(arectangular shapeismostdesirable)andndingantiderivativeswhencalculating theiteratedintegral.Itisthereforeofinteresttodevelopatechnique forageneralchangeofvariablesinthedoubleintegralsothatone wouldbeableto design newvariablesspecictothedoubleintegralin questioninwhichthesought-aftersimplicationisachieved.102.1.ChangeofVariables.Letthefunctions x ( u,v )and y ( u,v )be denedonanopenregion D.Then,foreverypair( u,v ) D,one canndapair( 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 )denea transformation ofaregion Dinthe uv planeontoaregion D inthe xy plane.Ifnotwopoints in Dhavethesameimagepointin D ,thenthetransformationis called one-to-one .Foraone-to-onetransformation,onecandenethe 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 )=tan 1( y/x ). Definition 14.11 (ChangeofVariablesinaPlane) Aone-to-onetransformationofanopenregion Ddenedby x = x ( u,v ) and y = y ( u,v ) iscalleda changeofvariables ifthefunctions x ( u,v ) and y ( u,v ) havecontinuousrst-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

PAGE 348

34214.MULTIPLEINTEGRALS 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 hasaxedvaluealongitscoordinatecurve v ( x,y )= vp. Thecoordinatecurvesareimagesofthestraightlines u = upand v = vpin Dundertheinversetransformation.Ifthecoordinatecurves arenotstraightlines(asinarectangularcoordinatesystem),then suchcoordinatesarenaturallycurvilinear.Forexample,thecoordinate curvesofpolarcoordinatesareconcentriccircles(axedvalueof r ) andraysfromtheorigin(axedvalueof ),andeverypointinaplane canbeviewedasanintersectionofonesuchrayandonesuchcircle.102.2.ChangeofVariablesinaDoubleIntegral.Consideradoubleintegralofafunction f ( x,y )overaregion D .Let x = x ( u,v )and y = y ( u,v )deneatransformationofaregion Dto D ,where Dis boundedbypiecewise-smoothcurvesinthe uv plane.Supposethat thetransformationisachangeofvariablesonanopenregionthat includes D.Thenthereisaninversetransformation,thatis,atransformationof D to D,whichisdenedbythefunctions 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 +1 ui= u and vj +1 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 isunderstoodasthelimitofadoublesequence(Denition14.4)or asthetwo-variablelimit( u, v ) (0 0).Asbefore,thevaluesof f ( x ( u,v ) ,y ( u,v ))outside Daresetto0whencalculatingthevalueof f inapartitionrectanglethatintersectstheboundaryof D. Asinthecaseofpolarcoordinates,theaimistoconvertthislimit intoadoubleintegralof f ( x ( u,v ) ,y ( u,v ))overtheregion D.This

PAGE 349

102.CHANGEOFVARIABLESINDOUBLEINTEGRALS343 x y A B C D Dijv = vjv = vj +1u = uiu = ui +1v u v jvj +1vj u DCABD ij v uiu iui +1Figure14.18.Left :Apartitionofaregion D bythe coordinatecurvesofthenewvariables u ( x,y )= uiand v ( x,y )= vj,whicharetheimagesofthestraightlines u = uiand v = vjinthe uv plane.Apartitionelement Dijis boundedbythecoordinatecurvesforwhich ui +1 ui= u and vj +1 vj= v Right :Theregion D,whoseimage istheintegrationregion D underthecoordinatetransformation,ispartitionedbythecoordinatelines u = uiand v = vj. Apartitionelementistherectangle D ijwhoseareais u v Thechangeofvariablesestablishesaone-to-onecorrespondencebetweenpointsof D and D.Inparticular, A B ,and C in D correspondto A, B,and Cin D,respectively.canbeaccomplishedbyndingarelationbetween Aijand A ij, thatis,theruleoftheareaelementtransformationunderachange ofvariables.Considerarectangle D ijinthe uv planeboundedby thelines u = ui, u = ui+ u v = vj,and v = vj+ v .Let Abethevertex( ui,vj), Bbe( ui+ u,vj),and Cbe( ui,vj+ v ). Theimage Dijof D ijinthe xy planeisaregionboundedbythe coordinatecurvesofthevariables u and v asshowninFigure14.18. Theimages A and B ofthepoints Aand Blieonthecoordinate curve v = vj,while A and C (theimageof C)areonthecoordinate curve u = ui.Sincethetransformation( u,v ) ( x,y )isachangeof variables(seeDenition14.11),thefunctions x ( u,v )and y ( u,v )have continuouspartialderivativesandhencearedierentiable.Inasmall neighborhood,adierentiablefunctioncanbewellapproximatedby itslinearization(recallDenition13.17).So,whencalculatingthearea A of Dij,itissucienttoconsidervariationsof x and y within Dijlinear invariationsof u and v within D ij.Consequently,thenumbers u and v canbeviewedasthedierentialsof u and v .Inthelimit

PAGE 350

34414.MULTIPLEINTEGRALS ( u, v ) (0 0),theirhigherpowerscanbeneglected,andthearea transformationlawshouldhavetheform A = J u v = J A, wherethecoecient J istobefound.Recallthat J = r forpolar coordinates. Supposethatthegradients x ( u,v )and y ( u,v )donotvanish. Thenthelevelcurvesof x ( u,v )and y ( u,v ),whicharealsothecoordinatecurves,aresmooth(recallthediscussionofTheorem13.16).A sucientlysmallpartofasmoothcurvebetweentwopointscanbe wellapproximatedbyasecantlinethroughthesepoints(recallSection 80.3).Thisargumentsuggeststhattheareaof Dijcanbeapproximatedbytheareaofaparallelogramwithadjacentsides AB = b and AC = c .Thecoordinatesof A are( x ( ui,vj) ,y ( ui,vj)),whilethecoordinatesof B are( x ( ui+ u,vj) ,y ( ui+ u,vj))becausetheyareimages of Aand B,respectively,undertheinversetransformation x = x ( u,v ) and y = y ( u,v ).Therefore, b = x ( ui+ u,vj) x ( ui,vj) ,y ( ui+ u,vj) y ( ui,vj) 0 = x u( ui,vj) u,y u( ui,vj) u, 0 = u x u( ui,vj) ,y u( ui,vj) 0 where,owingtothesmallnessof u and v ,thevariationsof x and y havebeenlinearized: x ( ui+ u,vj) x ( ui,vj)= x u( ui,vj) u and y ( ui+ u,vj) y ( ui,vj)= y u( ui,vj) u ;thatis,higherpowersof u havebeenneglected.Thethirdcomponentof b issetto0asthevector isplanar.Ananalogouscalculationforthecomponentsof c yields c = v x v( ui,vj) ,y v( ui,vj) 0 Theareaoftheparallelogramreads (14.10) Aij= b c = det x ux vy uy v u v = J ( ui,vj) u v, wherethepartialderivativesaretakenatthepoint( ui,vi).Thevectors b and c areinthe xy plane.Therefore,theircrossproducthasonlyone nonzerocomponent(the z component)givenbythedeterminant.The absolutevalueofthedeterminantisneededbecausethe z component ofthecrossproductmaybenegative, (0 0 ,z ) = z2= | z | .

PAGE 351

102.CHANGEOFVARIABLESINDOUBLEINTEGRALS345 Definition 14.12 (JacobianofaTransformation) TheJacobianofatransformationdenedbydierentiablefunctions 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 v x vy u. TheJacobiancoincideswiththedeterminantin(14.10).Inthis denition,aconvenientnotationhasbeenintroduced.Thematrix whosedeterminantisevaluatedhasthe rst rowcomposedofthepartialderivativesofthe rst variableinthenumeratorwithrespecttoall variablesinthedenominator,andsimilarlyforthesecondrow.This ruleiseasytoremember. Furthermore,thecoecient J in(14.10)isthe absolutevalue ofthe Jacobian.TheJacobianofachangeofvariablesinthedoubleintegral shouldnotvanishon Dbecause A =0.Ifthepartialderivativesof x and y withrespectto u and v arecontinuouson D, J iscontinuouson D,too.Therefore,foranysamplepoint( u i,v j)in D ij,thedierence ( Aij J ( u i,v j) A) / A= J ( ui,vj) J ( u i,v j)vanishesinthelimit ( u, v ) (0 0).So,ifin(14.10)thevalueoftheJacobianistaken atanysamplepoint,thenthecorrespondingchangeinthevalueof Aijdependsonhigherpowersof u and v .Since Aijhasbeen calculatedinthelinearapproximation,suchvariationsof Aijmaybe neglectedandonecanalwaysput Aij= J ( u i,v j) u v intheRiemannsumforanychoiceofsamplepoints.Thelimitofthe Riemannsum DfdA =limN1,N2 N1i =1 N2j =1f ( x ( u i,v j) ,y ( u i,v j)) Aijdenesthedoubleintegralofthefunction f ( x ( u,v ) ,y ( u,v )) J ( u,v )over theregion D.Theforegoingargumentssuggestthatthefollowing theoremistrue(afullproofisgiveninadvancedcalculuscourses). Theorem 14.9 (ChangeofVariablesinaDoubleIntegral) Supposeatransformation x = x ( u,v ) y = y ( u,v ) hascontinuousrstorderpartialderivativesandmapsaregion Dboundedbypiecewisesmoothcurvesontoaregion D .Supposethatthistransformationis one-to-oneandhasanonvanishingJacobian,exceptperhapsonthe

PAGE 352

34614.MULTIPLEINTEGRALS 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 ) Inthecaseofpolarcoordinates,theboundaryof Dmaycontain theline r =0onwhichtheJacobian J = r vanishes.Thisentireline collapsesintoasinglepoint,theorigin( x,y )=(0 0)inthe xy plane, uponthetransformation x = r cos and y = r sin ;thatis,thistransformationisnotone-to-oneonthisline.Afullproofofthetheorem requiresananalysisofsuchsubtletiesinageneralchangeofvariables aswellasarigorousjusticationofthelinearapproximationinthe areatransformationlaw,whichwereexcludedintheaboveanalysis. Thechangeofvariablesinadoubleintegralentailsthefollowing steps: 1.Findingtheregion Dwhoseimageunderthetransformation x = x ( u,v ), y = y ( u,v )istheintegrationregion D .Auseful ruletorememberhereis boundariesof D boundariesof D underthetransformation.Inparticular,ifequationsofboundariesof D aregiven,thenequationsofthecorrespondingboundariesof Dcanbeobtainedbyexpressingtheformerinthenew variablesbythesubstitution x = x ( u,v )and y = y ( u,v ). 2.Transformationofthefunctiontonewvariables f ( x,y )= f ( x ( u,v ) ,y ( u,v )) 3.CalculationoftheJacobianthatdenestheareaelementtransformation: dA = ( x,y ) ( u,v ) dudv = JdA,J = ( x,y ) ( u,v ) 4.Evaluationofthedoubleintegralof fJ over Dbyconverting ittoasuitableiteratedintegral.Thechoiceofnewvariables shouldbemotivatedbysimplifyingtheshape D(arectangular shapeisthemostdesirable). Example 14.16 Usethechangeofvariables x = u (1 v ) y = uv toevaluatetheintegral D( x + y )5y5dA ,where D isthetrianglebounded bythelines y =0 x =0 ,and x + y =1 .

PAGE 353

102.CHANGEOFVARIABLESINDOUBLEINTEGRALS347 Solution: 1.Noterstthattheline u =0ismappedtoasinglepoint, theorigin,inthe xy plane.Sotheline u =0mustbea boundaryof D.Theequation x =0inthenewvariables becomes u (1 v )=0,whichmeansthateither u =0or v =1.Therefore,theline v =1isaboundaryof Dasitis mappedtotheboundaryline x =0.Theequation y =0in thenewvariablesreads uv =0.Therefore,theline v =0is alsoaboundaryof D.Theequation x + y =1inthenew variableshastheform u =1.Thus,theregion Disbounded byfourlines u =0, u =1, v =0,and v =1,whichisthe square[0 1] [0 1]. 2.Since x + y = u ,theintegrandinthenewvariablesis u5( uv )5= u10v5. 3.TheJacobianofthetransformationis ( x,y ) ( u,v ) =det x ux vy uy v =det 1 v u vu = u (1 v )+ uv = u. Thereforetheareaelementtransformationis dA = | u | dA. Theabsolutevaluemaybeomittedbecause u 0in D.Note thattheJacobianvanishesonlyontheboundaryof Dand, hence,thehypothesesofTheorem14.9arefullled. 4.ThedoubleintegralinthenewvariablesisevaluatedbyFubinistheorem: D( x + y )5y5dA = Du11v5dA= 1 0u11du 1 0v5dv = 1 12 1 6 = 1 72 Thisexampleandtheexampleofpolarcoordinateshowthatthe transformationisnotone-to-oneonthesetswheretheJacobianvanishes(theline u =0ismappedtoasinglepoint)andtheinverse transformationfailstoexist.Itturnsoutthatthisobservationisofa generalnature. Theorem 14.10 (InverseFunctionTheorem) Letthetransformation ( u,v ) ( x,y ) bedenedonanopenset Ucontainingapoint ( u0,v0) .Supposethatthefunctions x ( u,v ) and y ( u,v ) havecontinuouspartialderivativesin UandtheJacobianof thetransformationdoesnotvanishatthepoint ( u0,v0) .Thenthere existsaninversetransformation u = u ( x,y ) v = v ( x,y ) inanopen set U containingtheimagepoint ( x0,y0)=( x ( u0,v0) ,y ( u0,v0)) and

PAGE 354

34814.MULTIPLEINTEGRALS thefunctions u ( x,y ) and v ( x,y ) havecontinuouspartialderivativesin U Bythistheorem,theJacobianofthe inversetransformation can becalculatedas ( u,v ) / ( x,y )sothattheareatransformationlawis dudv = | ( u,v ) / ( x,y ) | dxdy andthefollowingstatementholds. Corollary 14.3 If u = u ( x,y ) and v = v ( x,y ) istheinverseof thetransformation x = x ( u,v ) and y = y ( u,v ) ,then (14.11) ( x,y ) ( u,v ) = 1 ( u,v ) ( x,y ) = 1 det u xu yv xv y Theanalogywithachangeofvariablesintheone-dimensionalcase canbemade.If x = f ( u ),where f hascontinuousderivative f( u )that doesnotvanish,then,bytheinversefunctiontheoremforfunctionsof onevariable(Theorem12.6),thereisaninversefunction u = g ( x ) whosederivativeiscontinuousand g( x )=1 /f( u ),where u = g ( x ). Thenthetransformationofthedierential dx canbewrittenintwo equivalentforms,justlikethetransformationoftheareaelement dA = dxdy : dx = f( u ) du = du g( x ) dxdy = ( x,y ) ( u,v ) dudv = dudv ( u,v ) ( x,y ) Equation(14.11)denestheJacobianasafunctionof( x,y ).Sometimesitistechnicallysimplertoexpresstheproduct f ( x,y ) J ( x,y )in thenewvariablesratherthandoingsofor f and J separately.Thisis illustratedbythefollowingexample. Example 14.17 Useasuitablechangeofvariablestoevaluatethe doubleintegralof f ( x,y )= xy3overtheregionDthatliesintherst quadrantandisboundedbythelines y = x and y =3 x andbythe hyperbolas yx =1 and yx =2 Solution: Theequationsofthelinescanbewrittenintheform y/x =1and y/x =3because y,x> 0in D (seeFigure14.19).Note thattheequationsofboundariesof D dependonjusttwoparticular combinations y/x and yx thattakeconstantvaluesontheboundaries of D .So,ifthenewvariablesdenedbytherelations u = u ( x,y )= y/x and v = v ( x,y )= xy ,thentheimageregion Dinthe uv planeisa rectangle u [1 3]and v [1 2].Indeed,theboundaries y/x =1and y/x =3aremappedontotheverticallines u =1and u =3,while thehyperbolas yx =1and yx =2aremappedontothehorizontal

PAGE 355

102.CHANGEOFVARIABLESINDOUBLEINTEGRALS349 1 2 3 4 x y D y =3 x y = x xy =2 xy =1 v u 1 2 3 4 1 2 3 1 DFigure14.19.AnillustrationtoExample14.17.The transformationoftheintegrationregion D .Equationsof theboundariesof D y =3 x y = x xy =2,and xy =1,are writteninthenewvariables u = y/x and v = xy toobtain theequationsoftheboundariesof D, u =3, u =1, v =2, and v =1,respectively.Thecorrespondencebetweenthe boundariesof D and Disindicatedbyencirclednumbers enumeratingtheboundarycurves.lines v =1and v =2.Letusputasideforamomenttheproblemof expressing x and y asfunctionsofnewvariables,whichisneededto express f and J asfunctionsof u and v ,andndrsttheJacobianas afunctionof x and y bymeansof(14.11): J = det u xu yv xv y 1= det y/x21 /x yx 1= 2 y x 1= x 2 y Theabsolutevaluebarsmaybeomittedas x and y arestrictlypositive in D .Theintegrandbecomes fJ = x2y2/ 2= v2/ 2.Sondingthe functions x = x ( u,v )and y = y ( u,v )happenstobeunnecessaryin thisexample!Hence, Dxy3dA = 1 2 Dv2dA= 1 2 3 1du 2 1v2dv = 7 3 Thereaderisadvisedtoevaluatethedoubleintegralintheoriginal rectangularcoordinatestocomparetheamountofworkneededwith thissolution. Thefollowingexampleillustrateshowachangeofvariablescanbe usedtosimplifytheintegrandofadoubleintegral.

PAGE 356

35014.MULTIPLEINTEGRALS x y 1 2 12 D x + y =2 x + y =1 v u 12 u =1 v = u Du =2 v = uFigure14.20.Left :Theintegrationregion D inExample 14.18isboundedbythelines x + y =1, x + y =2, x =0, and y =0. Right :Theimage Dof D underthechange ofvariables u = x + y and v = y x .Theboundariesof Dareobtainedbysubstitutingthenewvariablesintothe equationsforboundariesof D sothat x + y =1 u =1, x + y =2 u =2, x =0 v = u ,and y =0 v = u .Example 14.18 Evaluatethedoubleintegralofthefunction f ( x,y ) =cos[( y x ) / ( y + x )] overthetrapezoidalregionwithvertices (1 0) (2 0) (0 1) ,and (0 2) Solution: Aniteratedintegralintherectangularcoordinateswould containtheintegralofthecosinefunctionofarationalargument(with respecttoeither 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 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 v ) / 2,theline x =0ismappedontotheline v = u whiletheline y =0ismappedontotheline v = u .Thus,theregion D= { ( u,v ) | u v u,u [1 2] } .TheJacobianofthechangeof variablesis J =1 / 2.Hence, Dcos y x y + x dA = 1 2 Dcos v u dA= 1 2 2 1u ucos v u dvdu = 1 2 2 1u sin v u u udu =sin(1) 2 1udu =3sin(1) / 2

PAGE 357

102.CHANGEOFVARIABLESINDOUBLEINTEGRALS351 x y a b D v u 1 1 Dr D2 1Figure14.21.Thetransformationoftheintegrationregion D inExample14.19.Theregion D x2/a2+ y2/b2 1, isrsttransformedintothedisk D, u2+ v2 1,by x = au y = bv ,andthen Distransformedintotherectangle Dby u = r cos v = r sin .Example 14.19 (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. Ofcourse,thearea A ( D)ofthedisk u2+ v2 1canalsobeevaluated byconvertingtheintegralover Dtopolarcoordinates u = r cos v = r sin .Thedisk Distheimageoftherectangle D=[0 1] [0 2 ], andtheJacobianis r .Thetransformationsoftheintegrationregion areshowninFigure14.21. When a = b ,theellipsebecomesacircleofradius R = a = b ,and theareaoftheellipsebecomestheareaofthedisk, A = R2.102.3.SymmetriesandaChangeofVariables.InSection100.4,the symmetrypropertiesofdoubleintegralsareshowntobequitehelpfulfortheirevaluation.Usingtheconceptofachangeofvariables indoubleintegrals,onecangiveanalgebraiccriterionofthesymmetrytransformationofaregion D thatpreservesitsarea. Atransformation x = x ( u,v ) y = y ( u,v ) thatmaps Donto D issaidto beareapreservingiftheabsolutevalueofitsJacobianis1,thatis, dA = dA.Indeed,changingvariablesinthedoubleintegralforthe area, A ( D )= DdA = DdA= A ( D).Forexample,rotations, translations,andreectionsarearea-preservingtransformationsforobviousgeometricalreasons.Thefollowingtheoremholds.

PAGE 358

35214.MULTIPLEINTEGRALS Theorem 14.11 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 ))= f ( u,v ) .Thenthedoubleintegralof f over D vanishes. Proof. Since D= D and dA = dA,thechangeofvariablesyields I = Df ( x,y ) dA = Df ( x ( u,v ) ,y ( u,v )) dA= Df ( u,v ) dA= I, thatis, I = I ,or I =0. 102.4.StudyProblem.Problem14.5.(GeneralizedPolarCoordinates) Generalizedpolarcoordinatesaredenedbythetransformation x = ar cosn,y = br sinn, where a b ,and n areparameters.FindtheJacobianofthetransformation.Usethegeneralizedpolarcoordinateswithasuitablechoice ofparameterstondtheareaoftheregionintherstoctantthatis boundedbythecurve4 x/a +4 y/b =1 Solution: TheJacobianofthegeneralizedpolarcoordinatesis ( x,y ) ( r, ) =det x rx y ry =det a cosn nar sin cosn 1 b sinnnbr cos sinn 1 = nabr cosn +1 sinn 1 +cosn 1 sinn +1 = nabr cosn 1 sinn 1 (cos2 +sin2 ) = nabr cosn 1 sinn 1. Choosingtheparameter n =8,theequationofthecurve4 x/a +4 y/b =1becomes4 r =1or r =1.Sincetheregioninquestionlies intherstquadrant,itisalsoboundedbythelines y =0and x =0, whicharetheimagesofthelines = / 2and =0inthe( r, )plane. Therefore,therectangle D=[0 1] [0 ,/ 2]ismappedontotheregion D inquestion.TheJacobianofthetransformationis positive in Dso theabsolutevalueoftheJacobianintheareaelementtransformation

PAGE 359

102.CHANGEOFVARIABLESINDOUBLEINTEGRALS353 maybeomitted.Theareaof D is A ( D )= DdA = DJdA=8 ab / 2 0cos7 sin7d 1 0rdr = ab 32 / 2 0 sin(2 ) 7d = ab 64 / 2 0 sin(2 ) 6d cos(2 ) = ab 64 1 1 1 u23du = ab 70 whererstthedouble-angleformulacos sin =1 2sin(2 )hasbeen usedandthentheintegrationhasbeencarriedoutwiththehelpofthe substitution u =cos(2 ). 102.5.Exercises.(1) FindtheJacobianofthefollowingtransformations: (i) x =3 u 2 v y = u +3 v (ii) x = ercos y = ersin (iii) x = uv y = u2 v2(iv) x = u cosh v y = u sinh v (2) Consider hyperboliccoordinates intherstquadrant x> 0, y> 0 denedbythetransformation x = veu, y = ve u.CalculatetheJacobian.Determinetherangeof( u,v )inwhichthetransformation isone-to-one.Findtheinversetransformationandsketchcoordinate curvesofhyperboliccoordinates. (3) Findtheconditionsontheparametersofalineartransformation x = a1u + b1v + c1, y = a2u + b2v + c2sothatthetransformationisarea preserving.Inparticular,provethattherotationsdiscussedinStudy Problem11.2areareapreserving. (4) Findtheimage D ofthespeciedregion Dunderthegiventransformation: (i) D=[0 1] [0 1]andthetransformationis x = u y = v (1 u2). (ii) Disthetrianglewithvertices(0 0),(1 0),and(1 1),andthe transformationis x = v2, y = u (iii) Distheregiondenedbytheinequality | u | + | v | 1,and thetransformationis x = u + v y = u v (5) Findalineartransformationthatmapsthetriangle Dwithvertices (0 0),(0 1),and(1 0)ontothetriangle D withvertices(0 0),( a,b ), and( b,a ),where a and b arepositive,nonequalnumbers.Usethis transformationtoevaluatetheintegralof f ( x,y )= bx ay overthe

PAGE 360

35414.MULTIPLEINTEGRALS triangle D (6) Evaluatethedoubleintegralusingthespeciedchangeofvariables: (i) D(8 x +4 y ) dA ,where D istheparallelogramwithvertices (3 1),( 3 1),( 1 3),and(5 1);thechangeofvariablesis x =( v 3 u ) / 4, y =( u + v ) / 4 (ii) D( x2 xy + y2) dA ,where D istheregionboundedbythe ellipse x2 xy + y2=1;thechangeofvariablesis x = u v/ 3, y = u + v/ 3 (iii) D( x2 y2) 1 / 2dA ,where D isintherstquadrantand boundedbyhyperbolas x2 y2=1, x2 y2=4andbythe lines x =2 y x =4 y ;thechangeofvariablesis x = u cosh v y = u sinh v (iv) De( x/y )( x + y )3/y2dA ,where D isboundedbyisthelines y = x,y =2 x,x + y =1,and x + y =2;thechangeofvariables u = x/y,v = x + y Hint: Followtheprocedurebasedon(14.11) asillustratedinExample14.17. (7) Findtheimage Dofthesquare a 0,andthelines x =0, y =0if x = u cos4v and y = u sin4v (10) Evaluatethedoubleintegralbymakingasuitablechangeofvariables: (i) Dyx2dA ,where D isintherstquadrantandboundedby thecurves xy =1, xy =2, yx2=1,and yx2=2 (ii) Dex ydA ,where D isgivenbytheinequality | x | + | y | 1 (iii) D(1+3 x2) dA ,where D isboundedbythelines x + y =1, x + y =2andbythecurves y x3=0, y x3=1 (iv) D( y +2 x2) dA ,wherethedomain D isboundedbytwo parabolas, y = x2,y = x2+2andbytwohyperbolas xy = 1( x< 0), xy =1( x> 0)

PAGE 361

102.CHANGEOFVARIABLESINDOUBLEINTEGRALS355 (v) D( x + y ) /x2dA ,where D isboundedbyfourlines y = x y =2 x y + x =1,and y + x =2 (vi) D y x/ ( x + y ),where D isthesquarewithvertices(0 2 a ), ( a,a ),(2 a, 2 a ),and( a, 3 a )with a> 0 (vii) Dcos( x2/a2+ y2/b2) dA ,where D isboundedbytheellipse x2/a2+ y2/b2=1 (viii) D( x + y ) dA ,where D isboundedby x2+ y2= x + y (ix) D( | x | + | y | ) dA ,where D isdenedby | x | + | y | 1 (x) D(1 x2 a2y2 b2) 1 / 2dA ,where D isboundedbytheellipse x2/a2+ y2/b2=1 (11) Let f becontinuouson[0 1].Showthat Df ( x + y ) dA = 1 0uf ( u ) du if D isthetrianglewithvertices(0 0),(0 1),and(1 0). (12) Useasuitablechangeofvariablestoreducethedoubleintegral toasingleintegral: (i) Df ( x + y ) dA ,where D isdenedby | x | + | y | 1 (ii) Df ( ax + by + c ) dA ,where D isthedisk x2+ y2 1and a2+ b2 =0 (iii) Df ( xy ) dA ,where D liesintherstquadrantandisbounded bythecurves xy =1, xy =2, y = x ,and y =4 x (13) Let n and m bepositiveintegers.Provethatif DxnymdA =0, where D isboundedbyanellipse x2/a2+ y2/b2=1,thenatleastone ofthenumbers n and m isodd. (14) Supposethatthelevelcurvesofafunction f ( x,y )aresimple closedcurvesandtheregion D isboundedbytwolevelcurves f ( x,y )= a and f ( x,y )= b .Provethat Df ( x,y ) dA = b auF( u ) du, where F ( u )istheareaoftheregionbetweenthecurves f ( x,y )= a and f ( x,y )= u Hint: Splittheregion D byinnitesimallycloselevel curvesofthefunction f (15) Usethegeneralizedpolarcoordinateswithasuitablechoiceof parameterstondtheareaofaregion D if (i) D isboundedbythecurves x2/a2+ y3/b3= x2+ y2andlies intherstquadrant. (ii) D isboundedbythecurves x3/a3+ y3/b3= x2/c2 y2/k2and liesintherstquadrant. (iii) D isboundedbythecurve( x/a + y/b )5= x2y2/c4. (16) Usethedoubleintegralandasuitablechangeofvariablestond theareaof D if

PAGE 362

35614.MULTIPLEINTEGRALS (i) D isboundedbythecurves x + y = a x + y = b y = mx ,and y = nx andliesintherstquadrant (ii) D isboundedbythecurves y2=2 ax y2=2 bx x2=2 cy ,and x2=2 ky ,where0 0and b> 0 (iv) D isboundedbythecurves( x/a )2 / 3+( y/b )2 / 3=1,( x/a )2 / 3+ ( y/b )2 / 3=4, x/a = y/b ,and8 x/a = y/b andliesintherst quadrant (v) D isboundedbytheellipses x2/ cosh2u + y2/ sinh2u =1, where u = u1and u = u2>u1,andbythehyperbolas x2/ cos2v y2/ sin2v =1,where v = v1and v = v2>v1. Hint: Considerthetransformation x =cosh u cos v y =sinh u sin v 103.TripleIntegrals Supposeasolidregion E islledwithaninhomogeneousmaterial. Thelattermeansthat,ifasmallvolume V ofthematerialistaken attwodistinctpointsof E ,thenthemassesofthesetwopiecesare dierent,despitetheequalityoftheirvolumes.Theinhomogeneityof thematerialcanbecharacterizedbythe massdensity asafunctionof position.Let m ( r )bethemassofasmallpieceofmaterialofvolume V cutoutaroundapoint r .Thenthemassdensityisdenedby ( 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 hasamassof5g. Supposethatthemassdensityofthematerialinaregion E is known.Thequestionis:Whatisthetotalmassofthematerialin E ?Apracticalanswertothisquestionistopartitiontheregion E sothateachpartitionelement Ep, p =1 2 ,...,N ,hasamass mp. Thetotalmassis M = p mp.Ifapartitionelement Ephasa volume Vp,then mp ( rp) Vpforsome rp Ep(seetheleft panelofFigure14.22).If Rpistheradiusofthesmallestballthat contains Ep,put RN=max Rp.Then,byincreasingthenumber N of

PAGE 363

103.TRIPLEINTEGRALS357 E rp Vp( x,y,z ) ( x,y, z )Figure14.22.Left :Apartitionelementofasolidregion,where rpisthepositionvectorofasamplepointinit. If ( r )isthemassdensity,thenthemassofthepartition elementis m ( rp) ( rp) Vp,where Vpisthevolumeof thepartitionelement.Thetotalmassisthesumof m ( rp) overthepartitionofthesolid E asgivenin(14.12). Right : AnillustrationtoExample14.17.Aballissymmetricunderthereectionaboutthe xy plane:( x,y,z ) ( x,y, z ). Ifthefunction f isskew-symmetricunderthisreection, f ( x,y, z )= f ( x,y,z ),thenthetripleintegralof f over theballvanishes.partitionelementssothat Rp RN 0as N ,theapproximation mp ( rp) Vpbecomesmoreandmoreaccuratebythedenition ofthemassdensitybecause Vp 0forall p .Sothetotalmassis (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.DenitionofaTripleIntegral. SmoothSurface.InSection85.5,asurfacewasdenedasacontinuous deformationofanopensetinaplanethathascontinuousinverse.A smallpieceofasurfacecanbeviewedasthegraphofacontinuous functionoftwovariables.Similarlytothenotionofasmoothcurve, asmoothsurfacecanbedened.Ifthegraphhasatangentplaneat everypointandthenormaltothetangentplanechangescontinuously alongthegraph,thenthesurfaceiscalled smooth .Consideralevelset

PAGE 364

35814.MULTIPLEINTEGRALS ofafunction g ( r )ofthreevariables r =( x,y,z ).Supposethat g has continuouspartialderivativesandthegradient g doesnotvanish.As explainedinSection93.2(seethediscussionofTheorem13.16),alevel set g ( r )= k isasurfacewhosenormalvectoristhegradient g .If g hascontinuouspartialderivatives,thenthecomponentsofthenormal arecontinuous.Thus, asurfaceissaidtobe smooth inaneighborhood ofapoint r0ifitcoincideswithalevelset g ( r )= g ( r0) ofafunction g thathascontinuouspartialderivativesandwhosegradientdoesnot vanishinat r0.Asurfaceissmoothifitissmoothinaneighborhood ofitseverypoint .Asurfaceis piecewisesmooth ifitconsistsofseveral smoothpiecesadjacentalongsmoothcurves.RectangularPartition.Aregion E inspaceisassumedtobeclosedand bounded;thatis,itiscontainedinaballofsome(nite)radius.The boundariesof E areassumedtobepiecewise-smoothsurfaces.The region E isthenembeddedinarectangularbox RE=[ a,b ] [ c,d ] [ s,q ],thatis, x [ a,b ], y [ c,d ],and z [ s,q ].If f ( r )isabounded functionon E ,thenitisextendedto REbysettingitsvaluesto0 outside E .Therectangle REispartitionedbythecoordinateplanes x = xi= a + i x i =0 1 ,...,N1,where x =( b a ) /N1; y = yj= c + j y i =0 1 ,...,N2,where y =( d c ) /N2;and z = zi= s + k z k =0 1 ,...,N3,where z =( q s ) /N3.Thevolumeofeachpartition elementisarectangle Rijkofvolume V = x y z .Thetotal numberofrectanglesis N = N1N2N3.UpperandLowerSums.ByanalogywithDenition14.2,thelower anduppersumsaredened.Put Mijk=sup f ( r )and mijk=inf f ( r ), wherethesupremumandinmumaretakenoverthepartitionrectangle 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).Sotheupperandlowersumsare triple sequences(arulethatassignsanumber anmktoanorderedtripleof integers( n,m,k )isatriplesequence).Thelimitofatriplesequence isdenedsimilarlytothelimitofadoublesequence( anmisreplaced by anmkinDenition14.4). Definition 14.13 (TripleIntegral) Ifthelimitsoftheupperandlowersumsexistas N1 2 3 (or ( x, y, z ) (0 0 0) )andcoincide,then f issaidtobeRiemann

PAGE 365

103.TRIPLEINTEGRALS359 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)orasthelimitofatriplesequence.103.2.PropertiesofTripleIntegrals.Thepropertiesoftripleintegrals arethesameasthoseofthedoubleintegraldiscussedinSection98; thatis,thelinearity,additivity,positivity,integrabilityoftheabsolute value | f | ,andupperandlowerboundsholdfortripleintegrals.ContinuityandIntegrability.Therelationbetweencontinuityandintegrabilityisprettymuchthesameasinthecaseofdoubleintegrals. Theorem 14.12 (IntegrabilityofContinuousFunctions) Let E beaclosed,boundedspatialregionwhoseboundariesarepiecewisesmoothsurfaces.Ifafunction f iscontinuouson E ,thenitisintegrable on E .Furthermore,if f hasboundeddiscontinuitiesonlyonanite numberofsmoothsurfacesin E ,thenitisalsointegrableon E Inparticular,aconstantfunctionisintegrable,andthevolumeof aregion E isgivenbythetripleintegral V ( E )= EdV. If m f ( r ) M forall r in E ,then mV ( E ) EfdV MV ( E ) .TheIntegralMeanValueTheorem.Theintegralmeanvaluetheorem (Theorem14.3)isextendedtotripleintegrals.If f iscontinuousin E thenthereisexistsapoint r0in E suchthat Ef ( r ) dV = V ( E ) f ( r0) Itsprooffollowsthesamelineofreasoningasinthecaseofdouble integrals.

PAGE 366

36014.MULTIPLEINTEGRALS RiemannSums.Ifafunction f isintegrable,thenitstripleintegral isthelimitofaRiemannsum,anditsvalueisindependentofthe partitionof E andachoiceofsamplepointsinthepartitionelements: (14.13) Ef ( r ) dV =limN ( RN 0) Np =1f ( rp) Vp. Thisequationcanbeusedforapproximationsoftripleintegrals,when evaluatingthelatternumericallyjustlikeinthecaseofdoubleintegrals.Symmetry.Ifatransformationinspacepreservesthevolumeofany region,thenitiscalled volumepreserving .Obviously,rotations,reections,andtranslationsinspacearevolume-preservingtransformations. Supposethat,underavolume-preservingtransformation,aregion E is mappedontoitself;thatis, E is symmetric relativetothistransformation.If rs E istheimageof r E underthistransformationandthe integrandisskew-symmetric, f ( rs)= f ( r ),thenthetripleintegralof f over E vanishes. Example 14.20 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 ballissymmetricrelativetothereectiontransformation( x,y,z ) ( x,y, z ),whereasthefunction g isskew-symmetric, g ( x,y, z )= 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 thengivenbytripleintegralofthedensityover E .Theelectriccharge canbepositiveandnegative.So,ifthetotalpositivechargein E is exactlythesameasthenegativecharge,thetripleintegralvanishes.

PAGE 367

103.TRIPLEINTEGRALS361 x y z E P Dxy( x,y, 0) z = ztop( x,y ) z = zbot( x,y ) x y z Dxz( x, 0 ,z ) E PFigure14.23.Left :Analgebraicdescriptionofasolid regionsimpleinthedirectionofthe z axis.Thesolid E is verticallyprojectedintothe xy plane:everypoint( x,y,z )of E goesintothepoint( x,y, 0).Theprojectionpointsform theregion Dxy.Since E issimpleinthe z direction,for every( x,y, 0)in Dxy,the z coordinateofthepoint P ( x,y,z ) in E rangesovertheinterval zbot( x,y ) z ztop( x,y ). Inotherwords, E liesbetweenthegraphs z = zbot( x,y ) and z = ztop( x,y ). Right :Anillustrationtothealgebraic description(14.14)ofasolid E assimpleinthe y direction. E isprojectedalongthe y axistothe xz plane,forminga region Dxz.Forevery( x, 0 ,z )in Dxz,the y coordinateof thepoint P ( x,y,z )in E rangesovertheinterval ybot( x,z ) y ytop( x,z ).Inotherwords, E liesbetweenthegraphs y = ybot( x,z )and y = ytop( x,z ).103.3.IteratedTripleIntegrals.Similartoadoubleintegral,atriple integralcanbeconvertedtoatripleiteratedintegral,whichcanthen beevaluatedbymeansofordinarysingle-variableintegration. Definition 14.14 (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} .

PAGE 368

36214.MULTIPLEINTEGRALS 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;theyaredenedanalogouslyto Dxy. Accordingto(14.13),thelimitoftheRiemannsumisindependent ofpartitioning E andchoosingsamplepoints(ageneralizationofTheorem14.5tothethree-dimensionalcaseistrivialasitsproofisbased onTheorem14.4,whichholdsinanynumberofdimensions).Let Dp, p =1 2 ,...,N ,beapartitionoftheregion Dxy.Consideraportion Epof E thatisprojectedonthepartitionelement Dp; Episacolumn with Dpitscrosssectionbyahorizontalplane.Since E isbounded, therearenumbers s and q suchthat s zbot( x,y ) ztop( x,y ) q for all( x,y ) Dxy;thatis, E alwaysliesbetweentwohorizontalplanes z = s and z = q .Considerslicingthesolid E byequispacedhorizontal planes z = s + k z k =0 1 ,...,N3, z =( q s ) /N3.Theneach column Epispartitionedbytheseplanesintosmallregions Epk.The unionofall Epkformsapartitionof E ,whichwillbeusedintheRiemannsum(14.13).Thevolumeof Epkis Vpk= z Ap,where Apistheareaof Dp.Assuming,asusual,that f isdenedbyzerovalues outside E ,samplepointsmaybeselectedsothat,if( xp,yp, 0) Dp, then( xp,yp,z k) Epk,thatis, zk 1 z k zkfor k =1 2 ,...,N3. Thethree-variablelimit(14.13)existsandhencecanbetakeninany particularorder(recallTheorem14.6).Takerstthelimit N3 or z 0.Thedoublelimitofthesumoverthepartitionof Dxyis understoodasbefore;thatis,as N 0,theradii Rpofsmallestdisks

PAGE 369

103.TRIPLEINTEGRALS363 containing 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 )]denesafunction F ( x,y ) whosevalues F ( xp,yp)atsamplepointsinthepartitionelements Dpappearintheparentheses.Acomparisonoftheresultingexpression with(14.3)leadstotheconclusionthat,aftertakingthesecondlimit, oneobtainsthedoubleintegralof F ( x,y )over Dxy. Theorem 14.13 (IteratedTripleIntegral) Let f beintegrableonasolidregion E boundedbyapiecewisesmooth surface.Supposethat E issimpleinthe z directionsothatitisbounded bythegraphs 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.103.4.EvaluationofTripleIntegrals.Inpracticalterms,anevaluation ofatripleintegraloveraregion E iscarriedoutbythefollowingsteps: Step1 .Determinethedirectionalongwhich E issimple.Ifnosuch directionexists,split E intoaunionofsimpleregionsandusethe additivityproperty.Fordenitiveness,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)),therstintegrationis carriedoutwithrespectto y (or x ),andthedoubleintegralisevaluated over Dxz(or Dyz).If E issimpleinanydirection,thenanyofthe

PAGE 370

36414.MULTIPLEINTEGRALS iteratedintegralscanbeused.Inparticular,justlikeinthecaseof doubleintegrals,thechoiceofaniteratedintegralforasimpleregion E shouldbemotivatedbythesimplicityofanalgebraicdescriptionof thetopandbottomboundariesorbythesimplicityoftheintegrations involved.Technicaldicultiesmaystronglydependontheorderin whichtheiteratedintegralisevaluated. Fubinistheoremcanbeextendedtotripleintegrals. Theorem 14.14 (FubinisTheorem) 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 ],andthetopandbottomboundariesare theplanes z = q and z = s .Alternatively,onecantake Dyz=[ c,d ] [ s,q ], xbot( y,z )= a ,and xtop( y,z )= b toobtainaniteratedintegral inadierentorder(wherethe x integrationiscarriedoutrst).In particular,if f ( x,y,z )= g ( x ) h ( y ) w ( z ),then Ef ( x,y,z ) dV = b ag ( x ) dx d ch ( y ) dy s qw ( z ) dz, whichisanextensionofthefactorizationpropertystatedinCorollary 14.2totripleintegrals. Example 14.21 Evaluatethetripleintegralof f ( x,y,z )= xy2z3overtherectangle E =[0 2] [1 2] [0 3] Solution: ByFubinistheorem, Exy2z3dV = 2 0xdx 2 1y2dy 3 0z3dz =2 (7 / 3) 9=42 Example 14.22 Evaluatethetripleintegralof f ( x,y,z )=( x2+ y2) z overtheportionofthesolidboundedbythecone z = x2+ y2andparaboloid z =2 x2 y2intherstoctant. Solution: Followingthestep-by-stepprocedureoutlinedabove,the integrationregionis z simple.Thetopboundaryisthegraphof ztop( x,y )=2 x2 y2,andthegraphof zbot( x,y )= x2+ y2is thebottomboundary.Todeterminetheregion Dxy,notethatithas tobeboundedbytheprojectionofthecurveoftheintersectionofthe coneandparaboloidontothe xy plane.Theintersectioncurveisdenedby zbot= ztopor r =2 r2,where r = x2+ y2,andhence r =1,

PAGE 371

103.TRIPLEINTEGRALS365 y z x 2 2 4 Dyzx = y2+ z2x y z 1 1 1 y = x y =1 ztop= 1 x2zbot=0 x y Dxyy y 1 1Figure14.24.Left :TheintegrationregioninExample 14.23.The x axisisvertical.Theregionisboundedbythe plane x =4(top)andtheparaboloid x = y2+ z2(bottom). Itsprojectionintothe yz planeisthediskofradius2asthe planeandparaboloidintersectalongthecircle4= y2+ z2. Right :AnillustrationtoStudyProblem14.6.whichisthecircleofunitradius.Since E isintherstoctant, Dxyis thequarterofthediskofunitradiusintherstquadrant.Onehas E( x2+ y2) zdV = Dxy( x2+ y2) 2 x2 y2 x2+ y2zdzdA = 1 2 Dxy( x2+ y2)[(2 x2 y2)2 ( x2+ y2)] dA = 1 2 / 2 0d 1 0r2[(2 r2)2 r2] rdr = 8 1 0u [(2 u )2 u ] du = 7 96 wherethedoubleintegralhasbetransformedintopolarcoordinates because Dxybecomestherectangle D xy=[0 1] [0 ,/ 2]inthepolar plane.Theintegrationwithrespectto r iscarriedoutbythesubstitution u = r2. Example 14.23 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

PAGE 372

36614.MULTIPLEINTEGRALS therstintegrationwithrespectto 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(seetheleftpanel ofFigure14.24).Onehas E y2+ z2dV = Dyz y2+ z24 y2+ z2dxdA = Dyz y2+ z2[4 ( y2+ z2)] dA = 2 0d 2 0r [4 r2] rdr = 128 15 wherethedoubleintegralover Dyzhasbeenconvertedtopolarcoordinatesinthe yz plane. 103.5.StudyProblems.Problem14.6. Evaluatethetripleintegralof f ( x,y,z )= z overthe region E boundedbythecylinder x2+ z2=1 andtheplanes z =0 y =1 ,and y = x intherstoctant. Solution: Theregionis z simpleandboundedbythe xy planefrom thebottom(i.e., zbot( x,y )=0)andbythecylinderfromthetop(i.e., ztop( x,y )= 1 x2)(bytakingthepositivesolutionof x2+ z2=1). TheintegrationregionisshownintherightpanelofFigure14.24. Theregion Dxyisboundedbythelinesofintersectionoftheplanes x =0, y = x ,andoftheplanes x =0 ,y = x ,and y =1.Thus, Dxyisthetriangleboundedbythelines x =0, y =1,and y = x Onehas EzdV = Dxy 1 x20zdzdA = 1 2 Dxy(1 x2) dA = 1 2 1 0(1 x2) 1 xdydx = 5 24 wherethedoubleintegralhasbeenevaluatedbyusingthedescription of Dxyasaverticallysimpleregion, ybot= x y 1= ytopforall x [0 1]=[ a,b ].

PAGE 373

103.TRIPLEINTEGRALS367 Problem14.7. 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 E2xy2z3dV =0 1 0xdx 1 0y2dy 1 0z3dz = 1 24 Thetripleintegralover E1vanishesbythesymmetryargument(the ballissymmetricunderthereection( x,y,z ) ( x,y,z )whereas f ( x,y,z )= f ( x,y,z )).ThesecondintegralisevaluatedbyFubinis theorem. 103.6.Exercises.(1) Evaluatethetripleintegraloverthespeciedsolidregionbyconvertingittoanappropriateiteratedintegral: (i) E( xy 3 z2) dV ,where E =[0 1] [1 2] [0 2] (ii) E6 xzdV ,where E isdenedbytheinequalities0 x z 0 y x + z ,and0 z 1 (iii) Ezey2,where E isdenedbytheinequalities0 x y 0 y z ,and0 z 1 (iv) E6 xydV ,where E liesundertheplane x + y z = 1 andabovetheregioninthe xy planeboundedbythecurves x = y x =0,and y =1 (v) ExydV ,where E isboundedbytheparaboliccylinders y = x2, x = y2,andbytheplanes x + y z =0, x + y + z =0 (vi) EdV ,where E isboundedbythecoordinateplanesandthe planethroughthepoints( a, 0 0),(0 ,b, 0),and(0 0 ,c )with a,b,c beingpositivenumbers (vii) EzxdV where E liesintherstoctantbetweentwoplanes x = y and x =0andboundedbythecylinder y2+ z2=1 (viii) E( x2z + y2z ) dV ,where E isenclosedbytheparaboloid z =1 x2 y2andtheplane z =0 (ix) EzdV ,where E isenclosedbytheellipticparaboloid z = 1 x2/a2 y2/b2andtheplane z =0. Hint: Useasuitable changeofvariablesinthedoubleintegral.

PAGE 374

36814.MULTIPLEINTEGRALS (x) Exy2z3dV ,where E isboundedbythesurfaces z = xy y = x x =1,and z =0 (xi) E(1+ x + y + z ) 3dV ,where E isboundedbytheplane x + y + z =1andbythecoordinateplanes (2) Usethetripleintegraltondthevolumeofthespeciedsolid E (i) E isboundedbytheparaboliccylinder x = y2andtheplanes z =0and z + x =1. (ii) E liesintherstoctantandisboundedbytheparabolicsheet z =4 y2andbytwoplanes y = x and y =2 x (iii) E isboundedbythesurfaces z2= xy x + y = a ,and x + y = b where0 0. Hint: Usethegeneralizedpolarcoordinatestoevaluatetheintegral.(seeStudy Problem14.5) (vii) E isboundedbythesurfaces z = x + y z = xy x + y =1, x =0,and y =0. (viii) E isboundedbythesurfaces x2+ z2= a2, x + y = a ,and x y = a (ix) E isboundedbythesurfaces az = x2+ y2,and z = a x y andbythecoordinatesurfaces,where a> 0. (x) E isboundedbythesurfaces z =6 x2 y2and z = x2+ y2. (3) Usesymmetryandotherpropertiesofthetripleintegraltoevaluate: (i) E24 xy2z3dV ,where E isboundedbytheellipticcylinder ( x/a )2+( y/b )2=1andbytheparaboloids z = [ c ( x/a )2 ( y/b )2]andhastherectangularcavity x [0 1], y [ 1 1], and z [0 1].Assumethat a b ,and c arelargerthan2. (ii) E(sin2( xz ) cos2( xy )) dV ,where E liesbetweenthespheres: 1 x2+ y2+ z2 4. (4) Expresstheintegral EfdV asaniteratedintegralinsixdierent ways,where E isthesolidboundedbythespeciedsurfaces: (i) x2+ y2=4, z = 1, z =2 (ii) z + y =1, x =0, x = y2(5) Reversetheorderofintegrationinallpossibleways:

PAGE 375

104.TRIPLEINTEGRALSINCYLINDRICALCOORDINATES369 (i) 1 01 x 0x + y 0f ( x,y,z ) dzdydx (ii) 1 1 1 x2 1 x21 x2+ y2f ( x,y,z ) dzdydx (iii) 1 01 0x2+ y20f ( x,y,z ) dzdydx (6) Reducethefollowingiteratedintegraltoasingleintegral: (i) a 0x 0y 0f ( z ) dzdydx (ii) 1 01 0x + y 0f ( z ) dzdydx Hint: Reversetheorderofintegration inasuitableway. (7) Usetheinterpretationofthetripleintegral f over E asthetotal amountofsomequantityin E distributedwiththedensity f tond E forwhich E(1 x2/a2 y2/b2 z2/c2) dV ismaximal. (8) Provethefollowingrepresentationofthetripleintegralbyiterated integrals: Ef ( x,y,z ) dV = b aDzf ( x,y,z ) dAdz, where Dzisthecrosssectionof E bytheplane z =const. (9) Provethatif f ( x,y,z )iscontinuousin E andforanysubregion W of E WfdV =0,then f ( x,y,z )=0in E 104.TripleIntegralsinCylindricalandSphericalCoordinates Achangeofvariableshasbeenprovedtobequiteusefulinsimplifyingthetechnicalitiesinvolvedinevaluatingdoubleintegrals.An essentialadvantageisasimplicationoftheintegrationregion.The conceptofchangingvariablescanbeextendedtotripleintegrals.104.1.CylindricalCoordinates.Oneofthesimplestexamplesofcurvilinearcoordinatesinspaceiscylindricalcoordinates.Theyaredenedby (14.16) x = r cos ,y = r cos ,z = z. Inanyplaneparalleltothe xy plane,thepointsarelabeledbypolarcoordinates,whilethe z coordinateisnottransformed.Equation(14.16) denesatransformationofanorderedtripleofnumbers( r,,z )toanotherorderedtriple( x,y,z ).Asetoftriples( r,,z )canbeviewedas asetofpoints EinaEuclideanspaceinwhichthecoordinateaxes arespannedby r ,and z .Then,underthetransformation(14.16), theregion Eismappedtoan image region E .Fromthestudyofpolarcoordinates,thetransformation(14.16)isone-to-oneif r (0 ),

PAGE 376

37014.MULTIPLEINTEGRALS x y z r = r0r0x y z = 00x y z z = z0z0Figure14.25.Coordinatesurfacesofcylindricalcoordinates:cylinders r = r0,half-planes = 0boundedbythe z axis,andhorizontalplanes z = z0.Anypointinspacecan beviewedasthepointofintersectionofthreecoordinate surfaces. [0 2 ),and z ( ).Theinversetransformationisgivenby r = x2+ y2, =tan 1( y/x ) ,z = z, wherethevalueoftan 1istakenaccordingtothequadrantinwhich thepair( x,y )belongs(seethediscussionofpolarcoordinates).Itmaps anyregion E intheEuclideanspacespannedby( x,y,z )totheimage region E.Tondtheshapeof E,aswellasitsalgebraicdescription, thesamestrategyasinthetwo-variablecaseshouldbeused: boundariesof E boundariesof Eunderthetransformation(14.16)anditsinverse.Isisparticularlyimportanttoinvestigatetheshapeof coordinatesurfaces ofcylindrical coordinates,thatis,surfacesonwhicheachofthecylindricalcoordinateshasaconstantvalue.If E isboundedbycoordinatesurfaces only,thenitisanimageofarectangularbox E,whichisthesimplest, mostdesirableshapewhenevaluatingamultipleintegral. Thecoordinatesurfacesof r arecylinders, r = x2+ y2= r0or x2+ y2= r2 0.Inthe xy plane,theequation = 0denesarayfromthe 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.

PAGE 377

104.TRIPLEINTEGRALSINCYLINDRICALCOORDINATES371 Sothecoordinatesurfacesofcylindricalcoordinatesare r = r0 x2+ y2= r2 0(cylinder) = 0 y cos 0= x sin 0(half-plane) z = z0 z = z0(plane) ThecoordinatesurfacesofcylindricalcoordinatesareshowninFigure14.25.Apointinspacecorrespondingtoanorderedtriple( r0,0,z0) isanintersectionpointofacylinder,ahalf-planeboundedbythecylinderaxis,andaplaneperpendiculartothecylinderaxis. Example 14.24 Findtheregion Ewhoseimageunderthetransformation(14.16)isthesolidregion E thatisboundedbytheparaboloid z = x2+ y2andtheplanes z =4 y = x ,and y =0 intherstoctant. 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] 104.2.TripleIntegralsinCylindricalCoordinates.Tochangevariables inatripleintegraltocylindricalcoordinates,onehastoconsidera partitionoftheintegrationregion E by coordinatesurfaces ,thatis, bycylinders,half-planes,andhorizontalplanes,whichcorrespondsto arectangularpartitionof E(theimageof E underthetransformationfromrectangulartocylindricalcoordinates).Thenthelimitofthe correspondingRiemannsum(14.13)hastobeevaluated.Inthecase ofcylindricalcoordinates,thistaskcanbeaccomplishedbysimpler means. Suppose E is z simplesothat,byTheorem14.13,thetripleintegral canbewrittenasaniteratedintegralconsistingofadoubleintegral over Dxyandanordinaryintegralwithrespectto z .Thetransformation(14.16)merelydenespolarcoordinatesintheregion Dxy.So, if Dxyistheimageof D xyinthepolarplanespannedbypairs( r, ), then,byconvertingthedoubleintegraltopolarcoordinates,oneinfers

PAGE 378

37214.MULTIPLEINTEGRALS 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 = dzdAisthevolumeofaninnitesimal rectangleinthespacespannedbythetriples( r,,z ).Itsimageinthe spacespannedby( x,y,z )liesbetweentwocylinderswhoseradiidier by dr ,betweentwohalf-planeswiththeangle d betweenthem,and betweentwohorizontalplanesseparatedbythedistance dz asshownin theleftpanelofFigure14.26.Soitsvolumeistheproductofthearea dA ofthebaseandtheheight dz dV = dzdA = rdzdA,accordingto theareatransformationlawforpolarcoordinates, dA = rdA.Sothe volumetransformationlawforcylindricalcoordinatesreads 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 14.25 Evaluatethetripleintegralof f ( x,y,z )= x2z overtheregion E boundedbythecylinder x2+ y2=1 ,theparaboloid z = x2+ y2,andtheplane z =0 Solution: Thesolid E isaxiallysymmetricbecauseitisbounded frombelowbytheplane z =0,bythecircularparaboloidfromabove, andthesideboundaryisthecylinder.Hence, Dxyisadiskofunit radius,and D xyisarectangle,( r, ) [0 1] [0 2 ].Thetopand

PAGE 379

104.TRIPLEINTEGRALSINCYLINDRICALCOORDINATES373 z r r z ztop= x2+ y2zbot=0 1 1 DxyFigure14.26.Left :Apartitionelementofthepartition of E bycylinders,half-planes,andhorizontalplanes(coordinatesurfacesofcylindricalcoordinates).Thepartitionisthe imageofarectangularpartitionof E.Keepingonlyterms linearinthedierentials dr = r d = dz = z ,the volumeofthepartitionelementis dV = dAdz = rdrddz = rdV,where dA = rdrd istheareaelementinthepolar coordinates.SotheJacobianofcylindricalcoordinatesis J = r Right :AnillustrationtoExample14.25.bottomboundariesare 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. 104.3.SphericalCoordinates.Sphericalcoordinatesareintroducedby thefollowinggeometricalprocedure.Let( x,y,z )beapointinspace. Considerarayfromtheoriginthroughthispoint.Anysuchraylies inthehalf-planecorrespondingtoaxedvalueofthepolarangle Therefore,therayisuniquelydeterminedbythepolarangle andthe angle betweentherayandthepositive z axis.If isthedistance fromtheorigintothepoint( x,y,z ),thentheorderedtripleofnumbers ( ,, )denesuniquelyanypointinspace.Thetriples( ,, )are called sphericalcoordinates inspace. Tondthetransformationlawfromsphericaltorectangularcoordinates,considertheplanethatcontainsthe z axisandtheray

PAGE 380

37414.MULTIPLEINTEGRALS x y z P r O r Pz P O r Py Pr O xFigure14.27.Sphericalcoordinatesandtheirrelationto therectangularcoordinates.Apoint P inspaceisdenedby itsdistancetotheorigin ,theangle betweenthepositive z axisandtheray OP ,andthepolarangle .fromtheoriginthrough P =( x,y,z )andtherectanglewithvertices (0 0 0),(0 0 ,z ), P=( x,y, 0),and( x,y,z )inthisplane(seeFigure14.27).Thediagonalofthisrectanglehaslength (thedistance between(0 0 0)and( x,y,z )).Therefore,itsverticalsidehaslength z = cos becausetheanglebetweenthissideandthediagonalis .Itshorizontalsidehaslength sin .Ontheotherhand,itisalso thedistancebetween(0 0 0)and( x,y, 0),thatis, 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 coincideswiththepositive z axis.If = ,therayisthenegative z axis.Anyraywith = / 2liesinthe xy plane.CoordinateSurfacesofSphericalCoordinates.Allpointsthathavethe samevalueof = 0formasphereofradius 0centeredattheorigin

PAGE 381

104.TRIPLEINTEGRALSINSPHERICALCOORDINATES375 becausetheyareatthesamedistance 0fromtheorigin.Naturally, thecoordinatesurfacesof arethehalf-planesdescribedearlierwhen discussingcylindricalcoordinates.Considerarayfromtheoriginthat hastheangle = 0withthepositive z axis.Byrotatingthisrayabout the z axis,allrayswiththexedvalueof areobtained.Therefore, thecoordinatesurface = 0isacircularconewhoseaxisisthe z axis. Forsmallvaluesof ,theconeisanarrowconeaboutthepositive z axis.Theconebecomeswideras increasessothatitcoincideswith the xy planewhen = / 2.For >/ 2,theconeliesbelowthe xy plane,anditeventuallycollapsesintothenegative z axisassoonas reachesthevalue .Thealgebraicequationsofthecoordinatesurfaces followfrom(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 Eisarectangular box.Thus,achangeofvariablesinatripleintegraltosphericalcoordinatesisadvantageouswhen E isboundedbyspheres,cones,and half-planes. Example 14.26 Let E betheportionofthesolidboundedbythe sphere x2+ y2+ z2=4 andthecone z2=3( x2+ y2) thatliesintherst octant.Findtheregion Ethatismappedonto E bythetransformation ( ,, ) ( x,y,z ) Solution: Theregion E hasfourboundaries:thesphere,thecone z = 3 x2+ y2,the xz plane( x 0),andthe yz plane( y 0). Theseboundariesaretheimagesof =2,cot = 3or = / 3, =0,and = / 2,respectively.So Eistherectangularbox[0 2] [0 ,/ 3] [0 ,/ 2].Theregion E isintersectedbyallsphereswithradii 0 2,allconeswithangles0 / 3,andallhalf-planeswith angles0 / 2. 104.4.TripleIntegralsinSphericalCoordinates.Atripleintegralinsphericalcoordinatesisobtainedbypartitioningtheintegrationregion E by spheres,cones,andhalf-planes,constructingtheRiemannsum(14.13), andtakingitslimitunderarenementofthepartition.Let Ebe

PAGE 382

37614.MULTIPLEINTEGRALS 0 = 000x y z = 00 = 1>/ 2 x y z = 00Figure14.28.Coordinatesurfacesofsphericalcoordinates:spheres = 0,circularcones = 0,andhalf-planes = 0boundedbythe z axis.Inparticular, =0and = describethepositiveandnegative z axes,respectively,and theconewiththeangle = / 2becomesthe xy plane. d dA d z r rd d d dA d d dFigure14.29.Left :Thebaseofapartitionelementin sphericalcoordinatesisaportionofasphereofradius cut outbytwoconeswiththeangles and + d = + and bytwohalf-planeswiththeangles and + d = + Itsareais dA =( d ) ( rd )= sin dd Right :A partitionelementhastheheight d = asitliesbetweentwosphereswhoseradiidierby d .Soitsvolume is dV = dAd = 2sin ddd = JdV,andtheJacobian ofsphericalcoordinatesis J = 2sin .

PAGE 383

104.TRIPLEINTEGRALSINSPHERICALCOORDINATES377 mappedontoaregion E underthetransformation(14.18).Considera rectangularpartitionof Ebyequispacedplanes = i, = j,and = ksuchthat i +1 i= j +1 j= ,and k +1 k= where ,and aresmallnumbersthatcanberegardedas dierentials(orinnitesimalvariations)ofthesphericalcoordinates. Eachpartitionelementhasvolume V= .Therectangularpartitionof Einducesapartitionof E byspheres,cones,and half-planes.Eachpartitionelementisboundedbytwosphereswhose radiidierby ,bytwoconeswhoseanglesdierby ,andbytwo half-planestheanglebetweenwhichis asshowninFigure14.29. Thevolumeofanysuchpartitionelementcanbewrittenas 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 Eandtherebydenesthetripleintegral of f over E insphericalcoordinates. Tond J ,considertheimageoftherectangularbox [ 0,0+ ], [ 0,0+ ], [ 0,0+ ]underthetransformation (14.18).Sinceitliesbetweentwospheresofradii 0and 0+ ,its volumecanbewrittenas V = A ,where A istheareaofthe portionofthesphereofradius 0thatliesbetweentwoconesandtwo half-planes.Anyhalf-plane = 0intersectsthesphere = 0along ahalf-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 .

PAGE 384

37814.MULTIPLEINTEGRALS InaEuclideanspacespannedbyorderedtriples( ,, ),theJacobian vanishesontheplanes =0, =0,and = .Thetransformation (14.18)isnotone-to-oneonthem.Allpoints(0 ,, )aremappedtoa singlepoint(0 0 0)by(14.18),andallpoints( 0 )and( ,, )are mappedontothe z axis,thatis,theline(0 0 ,z ),
PAGE 385

104.TRIPLEINTEGRALSINSPHERICALCOORDINATES379 =2cos = / 4 x y z Figure14.30.AnillustrationtoExample14.27.Anyray inspaceisdenedastheintersectionofaconewithanangle andahalf-planewithanangle .Tond Ewhoseimage isthedepictedsolid E ,notethatanysuchrayintersects E alonga single straightlinesegmentif0 / 4,where thecone = / 4isapartoftheboundaryof E .Duetothe axialsymmetryof E ,thereisnorestrictionontherangeof ,thatis,0 2 in E.Therangeof isdeterminedby thelengthofthesegmentofintersectionoftherayatxed and with E :0 2cos ,where =2cos isthe equationofthetopboundaryof E insphericalcoordinates.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. 104.5.Exercises.(1) Sketchthesolid E ontowhichthespeciedregion Eismapped bythetransformation( r,,z ) ( x,y,z ): (i)0 r 3, / 4 / 4,0 z 1 (ii)0 r 1,0 2 r 1 z 1 r

PAGE 386

38014.MULTIPLEINTEGRALS (iii)0 r 2,0 / 2,0 z 4 r2(2) Giventhesolid E ,ndtheregion Ewhoseimageis E underthe transformationtocylindricalcoordinates: (i) E isboundedbythecylinder x2+ y2=1,theparaboloid z = x2+ y2,andtheplane z =0. (ii) E isboundedbythecone( z 1)2= x2+ y2andthecylinder x2+ y2=1. (iii) E isboundedbytheparaboloid z = x2+ y2,thecylinder x2+ y2=2 x ,andtheplane z =0. (iv) E isthepartoftheball x2+ y2+ z2 a2intherstoctant. (3) Evaluatethetripleintegralbyconvertingittocylindricalcoordinates: (i) E| z | dV ,where E isboundedbythesphere x2+ y2+ z2=4 andthecylinder x2+ y2=1 (ii) E( x2y + y3) dV ,where E liesbeneaththeparaboloid z = 1 x2 y2intherstoctant (iii) EydV ,where E isenclosedbytheplanes z =0, x + y z = 5andbythecylinders x2+ y2=1, x2+ y2=4 (iv) EdV ,where E isenclosedbythecylinder x2+ y2=2 x ,by theplane z =0,andbythecone z = x2+ y2(v) EyzdV ,where E liesbeneaththeparaboloid z = a2 x2 y2intherstoctant (vi) E( x2+ y2) dV ,where E isboundedbythesurfaces x2+ y2= 2 z and z =2 (vii) ExyzdV ,where E liesinthepositiveoctantandisbounded bythesurfaces x2+ y2= az x2+ y2= bz xy = c2, xy = k2, y = x y = x ,and0
PAGE 387

104.TRIPLEINTEGRALSINSPHERICALCOORDINATES381 (iii) E boundedbythesphere x2+ y2+ z2= a2andbythehalfplanes y = 3 x y = x/ 3where x 0. (6) Evaluatethetripleintegralbyconvertingittosphericalcoordinates: (i) E( x2+ y2+ z2)3dV ,where E istheballofradius a centered attheorigin (ii) Ey2dV ,where E isboundedbythe yz planeandthehemispheres x = 1 y2 z2and x = 4 y2 z2(iii) ExyzdV ,where E isenclosedbythecone z = 3 x2+ y2andthespheres x2+ y2+ z2= a2, a =1 2 (iv) EzdV ,where E isthepartoftheball x2+ y2+ z2 1that liesbelowthecone z = 3 x2+3 y2(v) EzdV ,where E liesintherstoctantbetweentheplanes y =0and x = 3 y andisalsoboundedbythesurfaces z = x2+ y2and x2+ y2+ z2=4 (vi) E x2+ y2+ z2dV ,where E isboundedbythesphere x2+ y2+ z2= z (7) Sketchtheregionofintegration,writethetripleintegralinspherical coordinates,andthenevaluateit: (i) 1 0 1 x201 x2+ y2zdzdydx (ii) 1 0 1 x20 2 x2 y2 x2+ y2z2dzdydx (8) Sketchthesolidwhosevolumeisgivenbytheiteratedintegralin thesphericalcoordinates: / 2 0/ 4 02 / cos 02sin ddd Writetheintegralinthecylindricalcoordinatesandthencomputeit. (9) Sketchthedomainofintegration,writethetripleintegralincylindricalcoordinates,andthenevaluateit: 1 1 1 x2 1 x21 x2 y20zdzdydx (10) Convertthetripleintegral Ef ( x2+ y2+ z2) dV toiteratedintegralsincylindricalandsphericalcoordinatesif E isboundedbythe surfaces: (i) z = x2+ y2, y = x x =1, y =0, z =0 (ii) z2= x2+ y2, x2+ y2+ z2=2 z x = y/ 3, x = y 3,where x 0and y 0 (11) Usesphericalcoordinatestondthevolumeofasolidbounded bythesurfaces: (i) x2+ y2+ z2= a2, x2+ y2+ z2= b2, z = x2+ y2

PAGE 388

38214.MULTIPLEINTEGRALS (ii)( x2+ y2+ z2)3= a6z2/ ( x2+ y2) (iii)( x2+ y2+ z2)2= a2( x2+ y2 z2) (iv)( x2+ y2+ z2)3=3 xyz (12) Findthevolumeofasolidboundedbythesurfaces x2+ z2= a2, x2+ z2= b2, x2+ y2= z2,where x> 0. 105.ChangeofVariablesinTripleIntegrals Consideratransformationofanopenregion Einspaceintoaregion E denedby x = x ( u,v,w ), y = y ( u,v,z ),and z = z ( u,v,w ); thatis,foreverypoint( u,v,w ) E,thesefunctionsdeneanimagepoint( x,y,z ) E .Ifnotwopointsin Ehavethesameimage point,thetransformationis one-to-one ,andthereisa one-to-onecorrespondence betweenpointsof E and E.Theinversetransformation existsandisdenedbythefunctions u = u ( x,y,z ), v = v ( x,y,z ), and w = w ( x,y,z ).Supposethatthesefunctionshavecontinuous partialderivativessothatthegradientofthesefunctionsdoesnotvanish.Then,asshowninSection103.1,theequations u ( x,y,z )= u0, v ( x,y,z )= v0,and w ( x,y,z )= w0denesmoothsurfaces,called coordinatesurfaces ofthenewvariables.Apoint( x0,y0,z0)= r0isthe intersectionpointofthreecoordinateplanes x = x0, y = y0,and z = z0. Alternatively,itcanbeviewedasthepointofintersectionofthreecoordinatesurfaces, u ( x,y,z )= u0, v ( x,y,z )= v0,and w ( x,y,z )= w0, wherethepoint( u0,v0,w0)in Eismappedto r0bythecoordinate transformation. Definition 14.15 (JacobianofaTransformation) Supposethataone-to-onetransformationofanopenset Eonto E has continuousrst-orderpartialderivatives.Thequantity ( x,y,z ) ( u,v,w ) =det x uy uz ux vy vz vx wy wz w iscalledtheJacobianofthetransformation. Ifthedeterminantisexpandedovertherstcolumn,thenitcan alsobewrittenasthetripleproduct: ( x,y,z ) ( u,v,w ) = x ( y z ) Thetechnicaldetailsarelefttothereaderasanexercise.

PAGE 389

105.CHANGEOFVARIABLESINTRIPLEINTEGRALS383 Definition 14.16 (ChangeofVariables) Letatransformationofanopenset Eonto E havecontinuouspartial derivatives.Itiscalleda changeofvariables (orachangeofcoordinates)ifitsJacobiandoesnotvanishin E. Theinversefunctiontheorem(Theorem14.10)holdsforatransformation( u,v,w ) ( x,y,z ).IftheJacobianofthetransformationdoes notvanishon E,thentheinversetransformation( x,y,z ) ( u,v,w ) existsandhascontinuouspartialderivatives. Asinthecaseofdoubleintegrals,achangeofvariablesinspace canbeusedtosimplifytheevaluationoftripleintegrals.Forexample,ifthereisachangeofvariableswhosecoordinatesurfacesforma boundaryoftheintegrationregion E ,thenthenewintegrationregion Eisarectangularbox,andthelimitsinthecorrespondingiterated integralaregreatlysimpliedinaccordancewithFubinistheorem.105.1.TheVolumeTransformationLaw.Itisconvenienttointroduce thefollowingnotations:( u,v,w )= rand( x,y,z )= r sothatthe changeofvariablesiswrittenas (14.20) r = x ( r) ,y ( r) ,z ( r) or r= u ( r ) ,v ( r ) ,w ( r ) Let E 0bearectangularboxin E, u [ u0,u0+ u ], v [ v0,v0+ v ], and w [ w0,w0+ w ].Underthetransformation( u,v,w ) ( x,y,z ), itsimage E0isboundedbysmoothsurfacesifthetransformationisa changeofvariables.Ifthevaluesof u v ,and w areinnitesimally small,thatis,theycanbeviewedasdierentialsofthenewvariables, thentheboundarysurfacesof E0canbewellapproximatedbytangent planestothem,andthevolumeof E0isthenapproximatedbythe volumeofthepolyhedronboundedbytheseplanes.Thisimplies,in particular,thatwhencalculatingthevolume,onlytermslinearin u v ,and w aretoberetained,whiletheirhigherpowersareneglected. Therefore,thevolumesof E0and E 0mustbeproportional: V = J V, V= u v w. Theobjectiveistocalculate J .Bytheexamplesofcylindricaland sphericalcoordinates, J isafunctionofthepoint( u0,v0,w0)atwhich therectangularbox E 0istaken.Thederivationof J isfullyanalogous tothetwo-variablecase. Aninnitesimalrectangularbox E 0anditsimageunderthecoordinatetransformationareshowninFigure14.31.Let O, A, B,and

PAGE 390

38414.MULTIPLEINTEGRALS VC u v w r cBOr bAw r ar 0u v V C B rcrbA O r0rax y zFigure14.31.Left :Arectangularboxintheregion E 0withinnitesimalsides du = u dv = v dw = w so thatitsvolume V= dudvdw Right :Theimageofthe rectangularboxunderachangeofvariables.Theposition vectors rp,where P =0 ,a,b,c ,areimagesoftheposition vectors r p.Thevolume V oftheimageisapproximated bythevolumeoftheparallelepipedwithadjacentsides OA OB ,and OC .Itiscomputedbylinearizationof V in du dv ,and dw sothat V = Jdudvdw = J V,where J> 0 istheJacobianofthechangeofvariables.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 3areunitvectorsalongtherst,second,andthirdcoordinateaxes.Inotherwords,thesegments OA, OB,and OCarethe adjacentsidesoftherectangularbox E 0.Let O A B ,and C bethe imagesof O, A, B,and Cintheregion E .Owingtothesmoothness oftheboundariesof E0,thevolume V of E0canbeapproximatedby thevolumeoftheparallelepipedwithadjacentsides a = OA b = OB and c = OC .Then a = x ( r a) x ( r 0) ,y ( r a) y ( r 0) ,z ( r a) z ( r 0) =( x u,y u,z u) u, b = x ( r b) x ( r 0) ,y ( r b) y ( r 0) ,z ( r b) z ( r 0) =( x v,y v,z v) v, c = x ( r c) x ( r 0) ,y ( r c) y ( r 0) ,z ( r c) z ( r 0) =( x w,y w,z w) w,

PAGE 391

105.CHANGEOFVARIABLESINTRIPLEINTEGRALS385 whereallthedierenceshavebeenlinearized,forinstance, x ( r a) x ( r 0)= x ( r 0+ e1 u ) x ( r 0)= x u( r 0) u .Becauseofdierentiability ofthefunctions x ( r), y ( r),and z ( r),theerrorofthisapproximation decreasesto0fasterthan u v w asthelatterapproachzero values.Thisjustiestheapproachbasedonretainingonlytermslinearin u v w whencalculatingthevolume.Thevolumeofthe parallelepipedisgivenbytheabsolutevalueofthetripleproduct: (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-orderpartialderivativesarecontinuousforachangeofvariables andsoaretheJacobiananditsabsolutevalue.IftheJacobianofthe transformationdoesnotvanish,thenbytheinversefunctiontheorem (Theorem14.10)thereexistsaninversetransformation,and,similarly tothetwo-dimensionalcase(comparewith(14.11)),itcanbeproved that 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 1(14.22) = u ( v w ) 1. Thisexpressiondenes J asafunctionoftheoldvariables( x,y,z ).105.2.TripleIntegralinCurvilinearCoordinates.Considerapartitionof Ebyequispacedplanes u = ui, v = vj,and w = wk: ui +1 ui= u vj +1 vj= v ,and wk +1 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 ijkistherectangularbox u [ ui,ui +1], vj [ vj,vj +1], and w [ wk,wk +1],thenitsimage,beingthecorrespondingpartition elementof E ,isdenotedby Eijk.ARiemannsumcanbeconstructed forthispartitionof E (assuming,asbefore,that f isdenedbyzeros 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).Bythecontinuityof J ,itsvaluein(14.21)canbetaken

PAGE 392

38614.MULTIPLEINTEGRALS atanysamplepointin E ijk.Accordingtothedenitionofthetriple integral,thelimitoftheRiemannsumisthetripleintegralof fJ over theregion E.Theabovequalitativeconsiderationsuggeststhatthe followingtheoremholds. Theorem 14.15 (ChangeofVariablesinaTripleIntegral) Letatransformation E E denedbyfunctions ( u,v,w ) ( x,y,z ) withcontinuouspartialderivatveshaveanonvanishingJacobian,except perhapsontheboundaryof E.Supposethat f iscontinuouson 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 14.28 (VolumeofanEllipsoid) Findthevolumeofasolidregion E boundedbyanellipsoid x2/a2+ y2/b2+ z2/c2=1 Solution: Theintegrationdomaincanbesimpliedbyascaling transformation x = au y = bv ,and z = cw underwhichtheellipsoid 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. Example 14.29 Let a b ,and c benon-coplanarvectors.Findthe volumeofasolid E boundedbythesurface ( a r )2+( b r )2+( c r )2= R2, where r =( x,y,z ) .

PAGE 393

105.CHANGEOFVARIABLESINTRIPLEINTEGRALS387 x y z a b c E E1 1 1 u v wFigure14.32.AnillustrationtoExample14.28.Theellipsoidalregion x2/a2+ y2/b2+ z2/c2 1ismappedonto theball u2+ v2+ w2 1bythecoordinatetransformation u = x/a v = y/b w = z/c withtheJacobian J = abc .Solution: Denenewvariablesbythetransformation u = a r v = b r w = c r .TheJacobianofthistransformationisobtainedby (14.22): ( x,y,z ) ( u,v,w ) = ( u,v,w ) ( x,y,z ) 1= u ( v w ) 1= a ( b c ) 1. Thevectors a b ,and c arenon-coplanar,andhencetheirtripleproduct isnot0.Sothetransformation,isagenuinechangeofvariables.Under thistransformation,theboundaryof E becomesasphere u2+ v2+ w2= R2.So V ( E )= EdV = EJdV= 1 | a ( b c ) | EdV= V ( E) | a ( b c ) | = 4 R3 3 | a ( b c ) | where V ( E)=4 R3/ 3isthevolumeofaballofradius R 105.3.StudyProblems.Problem14.8.(VolumeofaTetrahedron) Atetrahedronisasolidwithfourverticesandfourtriangularfaces. Letthevectors a b ,and c bethreeadjacentsidesofthetetrahedron. Finditsvolume.

PAGE 394

38814.MULTIPLEINTEGRALS c y z E x a b u v w q q q EabcFigure14.33.AnillustrationtoStudyProblem14.8.A generaltetrahedronistransformedtoatetrahedronwhose faceslieinthecoordinateplanesbyachangeofvariables.Solution: Considerrstatetrahedronwhoseadjacentsidesarealong 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 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)

PAGE 395

105.CHANGEOFVARIABLESINTRIPLEINTEGRALS389 and c=(0 0 ,q ).Thus,thevolumeoftheimageregion Eis V ( E)= | q |3/ 6(theabsolutevalueisneededbecausethetripleproductcanbe negative).Tondthevolume V ( E ),theJacobianofthetransformation hastobefound.Itisconvenienttousetherepresentation(14.22): J = det m1m2m3l1l2l3n1n2n3 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) 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. 105.4.Exercises.(1) FindtheJacobianofthefollowingtransformations: (i) x = u/v y = v/w z = w/u (ii) x = v + w2, y = w + u2, z = u + v2(iii) x = uv cos w y = uv sin w z =( u2 v2) / 2(thesecoordinates arecalled parabolic coordinates) (iv) x + y + z = u y + z = uv z = uvw (2) Findtheregion Ewhoseimage E underthetransformationdened inexercise1,part(iv),isboundedbythecoordinateplanesandbythe plane x + y + z =1.Inparticular,investigatetheimageofthosepoints in EatwhichtheJacobianofthetransformationvanishes. (3) Let E bethesolidregionintherstoctantdenedbytheinequality x + y + z a ,where a> 0.Finditsvolumeusingthetriple integralinthenewvariables u = x v = y w = z (4) Useasuitablechangeofvariablesinthetripleintegraltondthe volumeofasolidboundedbythesurfaces:

PAGE 396

39014.MULTIPLEINTEGRALS (i)( x/a )2 / 3+( y/b )2 / 3+( z/c )2 / 3=1 (ii)( x/a )1 / 3+( y/b )1 / 3+( z/c )1 / 3=1,where x 0, y 0, z 0 (iii) x =0, y =0, z =0,and( x/a )n+( y/b )m+( z/c )k=1,where thenumbers n m ,and k arepositive (iv)( x + y + z )2= ax + by ,where( x,y,z )lieintherstoctant and a and b arepositive (v)( x + y )2+ z2= R2,where( x,y,z )lieintherstoctant (5) Evaluatethetripleintegral EzdV ,where E liesabovethecone z = c x2/a2+ y2/b2andisboundedfromabovebytheellipsoid x2/a2+ y2/b2+ z2/c2=1. (6) Evaluatethetripleintegral E(4 x2 9 y2) dV ,where E isenclosed bytheparaboloid z = x2/ 9+ y2/ 4andtheplane z =10. (7) Consideralineartransformationofthecoordinates x = a r, y = b r, z = c r,where r=( u,v,w )andthevectors a b ,and c haveconstantcomponents.Showthatthistransformationis volume preserving if | a ( b c ) | =1.Thetransformationissaidtobevolume preservingiftheimage E ofany Ehasthesamevolumeas E,that is, V ( E)= V ( E ). (8) If a b ,and c areconstantvectors, r =( x,y,z ),and E isgivenby theinequalities0 a r ,0 b r ,and0 c r ,show that E( a r )( b r )( c r ) dV =1 8( )2/ | a ( b c ) | (9) Considerparaboliccoordinates x = uv cos w y = uv sin w ,and z =( u2 v2).Showthat2 z =( x2+ y2) /v2 v2,2 z = ( x2+ y2) /u2+ u2, andtan w = y/x .Usetheserelationstosketchthecoordinatesurfaces u ( x,y,z )= u0, v ( x,y,z )= v0,and w ( x,y,z )= w0.Evaluatethetriple integralof f ( x,y,z )= xyz overtheregion E thatliesintherstoctant beneaththeparaboloid2 z 1= ( x2+ y2)andabovetheparaboloid 2 z +1= x2+ v2byconvertingtoparaboliccoordinates. (10) Useasuitablechangeofvariablestondthevolumeofasolid thatisboundedbythesurface x2 a2+ y2 b2n+ z2 n c2 n= z h x2 a2+ y2 b2n 2,n> 1 (11) (GeneralizedSphericalCoordinates) Generalizedsphericalcoordinates( ,, )aredenedbytheequations x = a sinn cosm,y = b sinn sinm,z = c cosn, where0 < ,0 < 2 ,0 ,and a b c n ,and m are parameters.FindtheJacobianofthegeneralizedsphericalcoordinates.

PAGE 397

105.CHANGEOFVARIABLESINTRIPLEINTEGRALS391 (12) Usegeneralizedsphericalcoordinateswithasuitablechoiceof parameterstondthevolumeofasolidboundedbythesurfaces: (i)[( x/a )2+( y/b )2+( z/c )2]2=( x/a )2+( y/b )2(ii)[( x/a )2+( y/b )2+( z/c )2]2=( x/a )2+( y/b )2 ( z/c )2(iii)( x/a )2+( y/b )2+( z/c )4=1 (iv)[( x/a )2+( y/b )2]2+( z/c )4=1 (13) (DirichletsIntegral) Let n m p ,and s bepositiveintegers.Use thetransformationdenedby x + y + z = u y + z = uv z = uvw to showthat Exnymzp(1 x y z )sdV = n m p s ( n + m + p + s +3)! where E isthetetrahedronboundedbythecoordinateplanesandthe plane x + y + z =1. (14) (OrthogonalCurvilinearCoordinates) Curvilinearcoordinates( u,v,w ) arecalled orthogonal ifthenormalstotheircoordinatesurfacesare mutuallyorthogonalatanypointoftheirintersection.Inotherwords, thegradients u ( x,y,x ), v ( x,y,z ),and w ( x,y,z )aremutuallyorthogonal.Onecandeneunitvectorsorthogonaltothecoordinate surfaces: (14.23) eu= u u ev= v v ew= w w NotethattheJacobianofachangeofvariablesdoesnotvanishand therelation(14.22)guaranteesthattheseunitvectorsarenotcoplanar andformabasisinspace(anyvectorcanbeuniquelyexpandedintoa linearcombinationofthem). (i)Showthat r =1 = 1 r z =1 (14.24) =1 = 1 = 1 sin (14.25) forthecylindrical( r,,z )andspherical( ,, )coordinates. (ii)Showthatthesphericalandcylindricalcoordinatesareorthogonalcoordinatesand,inparticular, (14.26) er=(cos sin 0) e=( sin cos 0) ez=(0 0 1) forthecylindricalcoordinates,and er=(sin cos sin sin cos ) e=(cos cos cos sin sin ) (14.27) e=( sin cos 0) forthesphericalcoordinates.

PAGE 398

39214.MULTIPLEINTEGRALS 106.ImproperMultipleIntegrals Inthecaseofone-variableintegration,improperintegralsoccur whentheintegrandisnotdenedataboundarypointoftheintegration intervalortheintegrationintervalisnotbounded.Forexample, (14.28) 1 0dx x=lima 01 adx x=lima 01 a1 1 = 1 1 ,< 1 or 01 1+ x2dx =lima a 01 1+ x2dx =lima tan 1a = 2 Impropermultipleintegralsarequitecommoninmanypracticalapplications.106.1.MultipleIntegralsofUnboundedFunctions.Supposeafunction f ( r )isnotdenedatapoint r0thatisalimitpointofthedomainof f (anyneighborhoodof r0containspointsofthedomainof f ).Here r =( x,y,z ) E or r =( x,y ) D .Fordeniteness,thethreedimensionalcaseisconsidered,whilethetwo-dimensionalcasecanbe treatedanalogously.If,inanysmallball Bofradius centeredat r0,thevaluesof | f ( r ) | arenotbounded,thenthefunction f issaid tobe singular at r0.Ifaclosedboundedregion E containssingular pointsofafunction f ,thentheupperandlowersumscannotbedened because,forpartitionrectanglescontainingasingularpoint,sup f or inf f orbothdonotexist,andneitherisdenedamultipleintegralof f Let Bbeanopenballofradius centeredatapoint r0.Suppose thatthefunction f issingularat r0.Denetheregion Ebyremoving allpointsof E thatalsolieinaball B.Supposethat f isintegrable on Eforany > 0(e.g.,itiscontinuous).Then,byanalogywiththe one-variablecase,amultipleintegralof f over E is dened asthelimit (14.29) EfdV =lim 0EfdV or DfdA =lim 0DfdA, provided,ofcourse,thelimitexists.If f issingularinapointset S thenonecanconstructaset Sthatistheunionofballsofradius centeredateachpointof S .Then Eisobtainedbyremoving Sfrom E .Theregularizationprocedureintwodimensionsisillustratedin Figure14.34. Althoughthisdenitionseemsarathernaturalgeneralizationofthe one-variablecase,therearesubtletiesthatarespecictomultivariable integrals.Thisisillustratedbythefollowingexample.Supposethat (14.30) f ( x,y )= y2 x2 ( x2+ y2)2

PAGE 399

106.IMPROPERMULTIPLEINTEGRALS393 BBD D D BD = BS D SFigure14.34.Aregularizationofanimproperintegral. Left : Bisaballcenteredatasingularpointoftheintegrand. BD istheintersectionof Bwith D .Theintegration iscarriedoutovertheregion D with Bremoved.Then thelimit 0istaken. Middle :Thesameregularizationprocedurewhenthesingularpointisaninteriorpointof D Right :Aregularizationprocedurewhensingularpoints formacurve S .Byremovingtheset Sfrom D ,theregion Disobtained.Thedistancebetweenanypointof Dand theset S isnolessthan .istobeintegratedoverthesector0 0ofadisk x2+ y2 1, where isthepolarangle.Ifthedenition(14.29)isapplied,then Distheportionofthering 2 x2+ y2 1correspondingto0 0. Then,byevaluatingtheintegralinpolarcoordinates,onendsthat Dy2 x2 ( x2+ y2)2dA = 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 )= f ( x,y ),whiletheintegrationregionisinvariantunderthis transformation.Theintegrandispositiveinthepartofthedomain where x2x2,andthereisamutualcancellationofcontributionsfromtheseregions.Iftheimproperintegralof theabsolutevalue | f ( x,y ) | isconsidered,thennosuchcancellationcan occur,andtheimproperintegralalwaysdiverges. Furthermore,if D intheaboveexampleisthesector x2+ y2 1, x 0, y 0or,inpolarcoordinates,0 r 1,0 / 2,the improperintegralcouldalsoberegularizedbyreducingtheintegration regionto r 1,0 0withthesubsequent two-variable limit( ,0) (0 ,/ 2).Evidently,thislimitdoesnotexist.Recall thateventhoughthetwo-variablelimitdoesnotexist,thelimitalong aparticularcurvemaystillexist.Forexample,ifthelimit 0 / 2is takenrst,thenthelimitis0,whereasthelimitisinniteifthelimit

PAGE 400

39414.MULTIPLEINTEGRALS 0istakenrst.Thisobservationsuggeststhat thevalueofthe improperintegralmaydependonthewayaregularizationisintroduced .IntegrabilityofUnboundedFunctions.Let E bearegioninspace(possiblyunbounded).An exhaustion of E isasequenceofboundedsimple regions Ek, k =1 2 ,... ,suchthat E1 E2 E andtheunion ofall Ekcoincideswith E .Ifthefunction f denedon E issingular atalimitpoint r0of E ,thenonecanconstructanexhaustionof E suchthatnoneof Ekcontain r0.Forexample,onecantake Ektobe theregionsobtainedfrom E byremovingballscenteredat r0ofradii =1 /k .Itcanbeprovedthat thesequenceofvolumes V ( Ek) convergestothevolumeof E .Owingtotheobservationthatthevalueof animproperintegralmaydependontheregularization,thefollowing denitionisadopted. Definition 14.17 (IntegrabilityofanUnboundedFunction) Let Ekbeanexhaustionof E .Supposethatafunction f on E is integrableoneach Ek.Thenthefunction f isintegrableon E ifthe limit limk EkfdV existsandisindependentofthechoiceof Ek. Thevalueofthelimitiscalledan improperintegral of f over E Animproperdoubleintegralisdenedinthesameway.Theconditionthatthelimitshouldnotdependonthechoiceofanexhaustion meansthattheimproperintegralshouldnotdependonitsregularization.Accordingtothisdenition,thefunction(14.30)isnotintegrable onanyregioncontainingtheoriginbecausethelimitdependsonthe waytheregularizationisimposed.AlthoughDenition14.17eliminatesapotentialambiguityoftherelation(14.29)notedabove,itis ratherdiculttouse.Asimplicationusefulinpracticeisachieved withthehelpoftheconceptof absoluteintegrability Theorem 14.16 Let Ekand E kbetwoexhaustionsof E .Let f beafunctionon E suchthat | f | isintegrableoneach Ekandeach E k. Then limk Ek| f | dV =limk E k| f | dV, wherethelimitmaybe + Inotherwords, thevalueoftheimproperintegral E| f | dV< ifitexists,isindependentoftheregularization .Thesamestatement holdsfordoubleintegrals. Definition 14.18 (AbsoluteIntegrability) Iftheimproperintegraloftheabsolutevalue | f | over E exists,then f iscalled absolutelyintegrable on E .

PAGE 401

106.IMPROPERMULTIPLEINTEGRALS395 Theorem 14.17 (SucientConditionforIntegrability) Let f beacontinuousfunctionon E .If f isabsolutelyintegrableon E ,thenitisintegrableon E Thistheoremimpliesthatif thelimit(14.29)existsfortheabsolutevalue | f | ,thentheimproperintegralofacontinuousfunction f existsandcanbecalculatedbytherule(14.29) .Thelattercomprises apracticalwaytotreatimproperintegrals. Example 14.30 Evaluatethetripleintegralof f ( x,y,z )= ( x2+ y2+ z2) 1overaballofradius R centeredattheoriginifit exists. Solution: Thefunctionissingularonlyattheoriginandcontinuous elsewhere.Lettherestrictedregion Eliebetweentwospheres: 2 x2+ y2+ z2 R2.Since | f | = f> 0in E ,theconvergenceofthe integralover Eas 0alsoimpliestheabsoluteintegrabilityof f andhencetheexistenceoftheimproperintegral(Theorem14.17).By makinguseofthesphericalcoordinates,oneobtains EdV x2+ y2+ z2= 2 0 0R 2sin 2ddd =4 ( R ) 4 R as 0.Sotheimproperintegralexistsandequals4 R Thefollowingtheoremisusefultoassesstheintegrability. Theorem 14.18 (AbsoluteIntegrabilityTest) If | f ( r ) | g ( r ) forall r in E and g ( r ) isintegrableon E ,then f is absolutelyintegrableon E Example 14.31 Investigatetheintegrabilityof f ( x,y )= x/ ( x2+ y2)/ 2, > 0 ,onaboundedregion D .Findtheintegral,if itexists,over D thatisthepartofthediskofunitradiusintherst quadrant. Solution: Thefunctionissingularattheorigin.Since f iscontinuous everywhereexcepttheorigin,itissucienttoinvestigatetheintegrabilityonadiskcenteredattheorigin.Put r = x2+ y2(thepolar radialcoordinate).Then | x | r andhence | f | r/r= r1 = g .In thepolarcoordinates,theimproperintegral(14.29)of g overadiskof unitradiusis 2 0d 1 g ( r ) rdr =2 1 r2 dr =2 ln =3 1 3 3 =3 .

PAGE 402

39614.MULTIPLEINTEGRALS Thelimit 0isniteif < 3.Bytheintegrabilitytest(Theorem14.18), thefunction f isabsolutelyintegrableif < 3.For < 3and D being thepartoftheunitdiskintherstquadrant,oneinfersthat lim 0DfdA =lim 0 / 20 1r cos rrdrd =lim 0 1r2 dr =1 Thetwoexamplesstudiedexhibitacommonfeatureofhowthefunction shouldchangewiththedistancefromthepointofsingularityinorder tobeintegrable. Theorem 14.19 Letafunction f becontinuousonaboundedregion D ofaEuclideanspaceandlet f besingularatalimitpoint r0of D .Supposethat | f ( r ) | M r r0 forall r in D suchthat 0 < r r0 0 and M> 0 .Then f isabsolutely integrableon D if
PAGE 403

106.IMPROPERMULTIPLEINTEGRALS397 x y a b 1 1 x y a b C1C2Figure14.35.Left :Therectangle D =[0 1] [0 1]with onepartitionrectangle[0 ,a ] [0 ,b ]removed,where x = a and y = b .Inthelimit( x, y ) (0 0),theregion becomestherectangle[0 1] [0 1]. Right :Thelimitcan betakenalongalongtwoparticularpaths C1and C2.In theformercase( C1), x istakento0rst.TheRiemann sumbecomesaniteratedintegralinwhichtheintegration withrespectto x iscarriedoutrst.When y istakento0 rst,thentheRiemannsumbecomesanintegratedintegral inwhichtheintegrationwithrespectto y iscarriedrst.tion(14.30)overtherectangle D =[0 1] [0 1].Itisnotintegrable asargued.Ontheotherhand,considerarectangularpartitionof D whereeachpartitionrectanglehasthearea x y .TheRiemannsum isregularizedbyremovingtherectangle[0 x ] [0 y ]asshownin Figure14.35.ThelimitoftheRiemannsumisthe two-variable limit ( x, y ) (0 0).Bytakingrst y 0andthen x 0,one obtainsaniteratedintegralinwhichtheintegrationwithrespectto y iscarriedoutrst: lima 01 alimb 01 bx2 y2 ( x2+ y2)2dydx =lima 01 alimb 01 b y y x2+ y2dydx =lima 01 alimb 0 1 1+ x2 b x2+ b2 dx =lima 01 adx 1+ x2= 1 0dx 1+ x2= 4 Here( a,b )=( x, y ).Alternatively,thelimit x 0canbetaken rstandthen y 0,whichresultsintheiteratedintegralinthe reverseorder:

PAGE 404

39814.MULTIPLEINTEGRALS limb 01 blima 01 ax2 y2 ( x2+ y2)2dxdy = limb 01 blima 01 a x x x2+ y2dydx = limb 01 blima 0 1 1+ y2 a y2+ a2 dy = limb 01 bdy 1+ y2= 1 0dy 1+ y2= 4 ThisshowsthatthelimitoftheRiemannsumasafunctionof two variables x and y doesnotexistbecauseitdependsonapathalong whichthelimitpointisapproached(thefunctionisnotintegrable).106.3.MultipleIntegralsOverUnboundedRegions.Thetreatmentof multipleintegralsoverunboundedregionsfollowsthesamestepsintroducedwhendiscussingtheintegrabilityofunboundedfunctions. Definition 14.19 Let E beanunboundedregionandlet Ekbean exhaustionof E whereeach Ekisbounded.Supposethat f isintegrable oneach Ek.Then EfdV =limk EkfdV ifthelimitexistsandisindependentofthechoiceof Ek. Doubleintegralsoverunboundedregionsaredenedinthesame way.Theorems14.16,14.17,and14.18holdforunboundedregions. Thefollowingpracticalapproachmaybeusedtoevaluateimproper integralsoverunboundedregions.Let DRbetheintersectionof D with adiskofradius R centeredattheoriginandlet ERbetheintersectionof E withaballofradius R centeredattheorigin.Let f beacontinuous functionthatisabsolutelyintegrableon E .Theintegralof f over D (or E )isevaluatedbytherule Df ( r ) dA =limR DRf ( r ) dA or Ef ( r ) dV =limR ERf ( r ) dV Theabsoluteintegrabilityof f meansthattheselimitsexistandare nitefortheabsolutevalue | f | .Theasymptoticbehaviorofafunction sucientforabsoluteintegrabilityonanunboundedregionisstatedin thefollowingtheorem,whichisananalogofTheorem14.19.

PAGE 405

106.IMPROPERMULTIPLEINTEGRALS399 Theorem 14.20 Suppose f isacontinuousfunctiononanunboundedregion D ofaEuclideanspacesuchthat | f ( r ) | M r 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 1 a R= R1 1 +lima a1 1 if =1.Thelimitisniteif > 1.When =1,theintegraldiverges asln a .Let D Rbethepartof D thatliesoutsidetheball BRofradius R andlet B Bbethepartofthespaceoutside BR(seeFigure14.36, leftpanel).Notethat B Rincludes D R.Inthetwo-variablecase,the useofthepolarcoordinatesgives D R| f | dA B R| f | dA B RMdA r = M 2 0d Rrdr r=2 M Rdr r 1, whichisnite,provided 1 > 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 M R2d =4 M Rd 2, whichconvergesif > 3. Example 14.32 Evaluatethedoubleintegralof f ( x,y ) =exp( x2 y2) overtheentireplane. Solution: Inpolarcoordinates, | f | = e r2.So,as r | f | decreasesfasterthananyinversepower r n, n> 0,andbyvirtueof Theorem14.20, f isabsolutelyintegrableontheplane.Bymakinguse

PAGE 406

40014.MULTIPLEINTEGRALS BRO R DRD D RB Rx y z = singularpoints = Figure14.36.Left :Anunboundedregion D issplitinto twoparts: DRliesinsidetheball BRofradius R ,and D Ris thepartof D thatliesoutsidetheball BR.Theregion B Ris theentirespacewiththeball BRremoved.Theregion D Riscontainedin B R. Right :Aregularizationprocedurefor theintegralinStudyProblem14.9.Theintegrationregion E containssingularpointsalongthe z axis.Theintegralis regularizedbyremovingtheball < andthesolidcone < from E .Aftertheevaluationoftheintegral,thelimit 0istaken.ofthepolarcoordinates, De x2 y2dA =limR 2 0R 0e r2rdrd = limR R20e udu = limR (1 e R2)= wherethesubstitution u = r2hasbeenmade. Itisinterestingtoobservethefollowing.Asthefunctionisabsolutelyintegrable,thedoubleintegralcanalsobeevaluatedbyFubinis theoreminrectangularcoordinates: De x2 y2dA = e x2dx e y2dy = I2, I = e x2dx =

PAGE 407

106.IMPROPERMULTIPLEINTEGRALS401 because I2= bythevalueofthedoubleintegral.Adirectevaluation of I bymeansofthefundamentaltheoremofcalculusisproblematicas anantiderivativeof e x2cannotbeexpressedinelementaryfunctions.106.4.StudyProblem.Problem14.9. Evaluatethetripleintegralof f ( x,y,z )= ( x2+ y2) 1 / 2( x2+ y2+ z2) 1 / 2over E ,whichisboundedbythecone z = x2+ y2andthesphere x2+ y2+ z2=1 ifitexists. Solution: Thefunctionissingularatallpointsonthe z axis.Consider Eobtainedfrom E byeliminatingfromthelatterasolidcone < andaball < ,where and aresphericalcoordinates.To investigatetheintegrability,consider | f | dV = fdV inthespherical coordinates: fdV =( 2sin ) 12sin ddd = ddd ,whichis regular.Sothefunction f isintegrableastheimage Eof E inthe sphericalcoordinatesisarectangle(i.e.,itisbounded).Hence, lim 0EfdV =lim 0E ddd = 2 0d / 4 0d 1 0d = 2 2 SotheJacobiancancelsoutallthesingularitiesofthefunction. 106.5.Exercises.(1) Letthefunction g ( x,y )beboundedsothat0 0, q> 0,where D isdened bythecondition | x | + | y | 1 (iii) Dg ( x,y )(1 x2 y2) pdA ,where D isdenedbythecondition x2+ y2 1 (iv) Dg ( x,y ) | x y | pdA ,where D isthesquare[0 ,a ] [0 ,a ] (v) De ( x + y )dA ,where D isdenedby0 x y (2) Letthefunction g ( x,y,z )beboundedsothat0
PAGE 408

40214.MULTIPLEINTEGRALS (iii) Eg ( x,y,z )( | x |p+ | y |q+ | z |s) 1dV ,where p q ,and s are positivenumbersand E isdenedby | x | + | y | + | z | 1 (iv) Eg ( x,y,z ) | x + y z | dV ,where E =[ 1 1] [ 1 1] [ 1 1] (3) Evaluatetheimproperintegralifitexists.Useappropriatecoordinateswhenneeded. (i) E( x2+ y2+ z2) 1 / 2( x2+ y2) 1 / 2dV ,where E istheregionin therstoctantboundedfromabovebythesphere x2+ y2+ z2= 2 z andfrombelowbythecone z = 3 x2+ y2(ii) Ez ( x2+ y2) 1 / 2dV ,where E isintherstoctantandbounded fromabovebythecone z =2 x2+ y2andfrombelowby theparaboloid z = x2+ y2(iii) Exy ( x2+ y2) 1( x2+ y2+ z2) 1dV ,where E istheportion oftheball x2+ y2+ z2 a2abovetheplane z =0 (iv) Ee x2 y2 z2( x2+ y2+ z2) 1 / 2dV ,where E istheentirespace (v) D( x2+ y2) 1 / 2dA ,where D liesbetweenthetwocircles x2+ y2=4and( x 1)2+ y2=1intherstquadrant, x,y 0 (vi) Dln( x2+ y2) dA ,where D isthedisk x2+ y2 a2(vii) E( x2+ y2+ z2)ln( x2+ y2+ z2) dV ,where E istheball x2+ y2+ z2 a2and isreal.Doestheintegralexistforall ? (viii) D( x2+ y2)ln( x2+ y2) dA ,where D isdenedby x2+ y2 a2> 0and isreal.Doestheintegralexistforall ? (ix) E( x2+ y2+ z2)ln( x2+ y2+ z2) dV ,where E isdenedby x2+ y2+ z2 a2> 0and isreal.Doestheintegralexistfor all ? (x) D[( a x )( x y )] 1 / 2dA ,where D isthetriangleboundedby thelines y =0, y = x ,and x = a (xi) Dlnsin( x y ) dA ,where D isboundedbythelines y =0, y = x ,and x = (xii) D( x2+ y2) 1dA ,where D isdenedby x2+ y2 x (xiii) Ex py qz sdV ,where E =[0 1] [0 1] [0 1] (xiv) E( x2+ y2+ z2) 3dV ,where E isdenedby x2+ y2+ z2 1 (xv) E(1 x2 y2 z2) dV ,where E isdenedby x2+ y2+ z2 1 (xvi) Ee x2 y2 z2dV ,where E istheentirespace (xvii) De x2 y2sin( x2+ y2) dA ,where D istheentireplane (xviii) De ( x/a )2 ( y/b )2dA ,where D istheentireplane (xix) Deax2+2 bxy + cy2dA ,where a< 0, ac b2> 0,and D isthe entireplane( Hint: Findarotationthattransforms x and y so

PAGE 409

107.LINEINTEGRALS403 thatinthenewvariablesthebilinearterm xy isabsentin theexponential.) (xx) Ee ( x/a )2 ( y/b )2 ( z/c )2+ x + y + zdV ,where E istheentirespace (4) Let n beaninteger.Showthat limn Dnsin( x2+ y2) dA = ,Dn: | x | n, | y | n, limn Dnsin( x2+ y2) dA =0 ,Dn: x2+ y2 2 n. Notethatineachcase Dncoverstheentireplaneas n .What canbesaidabouttheconvergenceoftheintegralovertheentireplane? (5) Showthattheintegral D( x2 y2)( x2+ y2) 2dA ,where D isdenedby x 1, y 1,diverges,whereastheiteratedintegralsinboth ordersconverge. (6) Showthatthefollowingimproperintegralsconverge.Usethegeometricseriestoshowthattheirvaluesaregivenbythespeciedconvergentseries: (i)lima 1Da(1 xy ) 1dA = n =1 1 n2,where Da=[0 ,a ] [0 ,a ] (ii)lima 1Ea(1 xyz ) 1dV = n =1 1 n3,where Ea=[0 ,a ] [0 ,a ] [0 ,a ] 107.LineIntegrals Considerawiremadeofanonhomogeneousmaterial.Theinhomogeneitymeansthatifonetakesasmallpieceofthewireoflength 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

PAGE 410

40414.MULTIPLEINTEGRALS aRiemannsumandleadstotheconceptofa lineintegral of alonga curve C .107.1.LineIntegralofaFunction.Let f beaboundedfunctionin E andlet C beasmooth(orpiecewise-smooth)curvein E .Suppose C hasanitearclength.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 upperandlowersumsaredenedby U ( f,N )= N p =1Mp spand L ( f,N )= N p =1mp sp. Definition 14.20 (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 ),andbythesqueeze principlethelimitoftheRiemannsumis independent ofthechoiceof samplepoints r p(seetheleftpanelofFigure14.37). Itisalsointerestingtoestablisharelationofthelineintegralwith atriple(ordouble)integral.Supposethat f isintegrableonaregion thatcontainsasmoothcurve C .Let Eabea neighborhood of C thatis denedasthesetofpointswhosedistance(inthesenseofDenition 11.14)to C cannotexceed a> 0.So,for a smallenough,thecross sectionof Eabyaplanenormalto C isadiskofradius a whosearea is A = a2(seetherightpanelofFigure14.37).Then,inthelimit a 0, (14.31) 1 a2Eaf ( r ) dV Cf ( r ) ds.

PAGE 411

107.LINEINTEGRALS405 SpC rpr prp +1 A Ea spC Vp= A spFigure14.37.Left :Apartitionofasmoothcurve C by segmentsofarclength spusedinthedenitionoftheline integralanditsRiemannsum. Right :Theregion Eais aneighborhoodofasmoothcurve C .Itconsistsofpoints whosedistanceto C cannotexceed a> 0(recallDenition 11.14).For a and spsmallenough,planesnormalto C throughthepoints rppartition Eaintoelementswhosevolumeis Vp= A sp,where A = a2istheareaofthe crosssectionof Ea.Thispartitionisusedtoestablishthe relation(14.31)betweenthetripleandlineintegrals.Inotherwords,lineintegralscanbeviewedasthelimitingcaseof tripleintegralswhentwodimensionsoftheintegrationregionbecome innitesimallysmall.Thisfollowsfrom(14.13)bytakingapartition of Eabyvolumeelements Vp= A spandsamplepointsalongthe curve C in Ea.IntheRiemannsumfortheleftsideof(14.31),the factor A = a2in Vpcancelsthesamefactorinthedenominator sothattheRiemannsumbecomesaRiemannsumforthelineintegral ontherightsideof(14.31).Inparticular,itcanbeconcludedthat thelineintegralexistsforany f thatiscontinuousorhasonlyanite numberofboundedjumpdiscontinuitiesalong C .Also, thelineintegral inheritsallthepropertiesofmultipleintegrals Theevaluationofalineintegralisbasedonthefollowingtheorem. Theorem 14.21 (EvaluationofaLineIntegral) Supposethat f iscontinuousinaregionthatcontainsasmoothcurve C .Letavectorfunction r ( t ) t [ a,b ] ,traceoutthecurve C justonce. Then (14.32) 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 a ) /N .Itinducesapartitionof C bypieces Cpso that r ( t )tracesout Cpwhen t [ tp 1,tp], p =1 2 ,...,N .Thearc

PAGE 412

40614.MULTIPLEINTEGRALS lengthof Cpis tptp 1 r( t ) dt = sp.Since C issmooth,thetangent vector r( t )isacontinuousfunctionandsoisitslength r( t ) .Bythe integralmeanvaluetheorem,thereis t p [ tp 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 ( t )) 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 hasanitenumber ofboundedjumpdiscontinuitiesand C ispiecewisesmooth.Thelatterimpliesthatthetangentvectormayonlyhaveanitenumberof discontinuitiesandsodoes r( t ) .Therefore, F ( t )hasonlyanite numberofboundedjumpdiscontinuitiesandhenceisintegrable.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 ) Step4 .Substitute x = x ( t ), y = y ( t ),and z = z ( t )into f ( x,y,z )and evaluatetheintegral(14.32). Remark. Acurve C maybetracedoutbydierentvectorfunctions. Thevalueofthelineintegralis independent ofthechoiceofparametricequationsbecauseitsdenitionisgivenonlyinparameterizationinvariantterms(thearclengthandvaluesofthefunctiononthecurve). Theintegrals(14.32)writtenfortwodierentparameterizationsof C arerelatedbyachangeoftheintegrationvariable(recalltheconcept ofreparameterizationofaspatialcurve). Example 14.33 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 axis countedcounterclockwise.Theequationofthecircleinquestionis

PAGE 413

107.LINEINTEGRALS407 x2+( y a )2= R2.So,byanalogy,onecanput x = R cos t and y a = R sin t (byshiftingtheorigintothepoint(0 ,a )).Theparametric equationofthecirclecanbetakenintheform r ( t )=( R cos t,a + R sin t ).Therangeof t mustberestrictedtotheinterval t [0 2 ] sothat r ( t )tracesthecircleonlyonce.Then r( t )=( 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 14.34 Evaluatethelineintegralof f ( x,y,z )= 3 x2+3 y2 z2overthecurveofintersectionofthecylinder x2+ y2= 1 andtheplane x + y + z =0 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 )= cos t sin t .Theintervalof t is[0 2 ]asthecurvewinds aboutthecylinder.Therefore, r( t )=( sin t, cos t, sin t cos t )and r( t ) = 2 2sin t cos t = 2 sin(2 t ).Thevaluesofthefunction alongthecurveare f = 3 (cos t +sin t )2= 2 sin(2 t ).Note thatthefunctionisdenedonlyintheregion3( x2+ y2) z2(outside thedoublecone).Ithappensthatthecurve C liesinthedomainof f becauseitsvaluesalong C arewelldenedas2 > sin(2 t )forany t Hence, Cfds = 2 0 2 sin(2 t ) 2 sin(2 t ) dt = 2 0(2 sin(2 t )) dt =4 107.3.Exercises.(1) Evaluatethelineintegral: (i) Cxy2ds ,where C istherighthalfofthecircle x2+ y2=4 (ii) Cx sin yds ,where C isthelinesegmentfrom(0 ,a )to( b, 0) (iii) Cxyzds ,where C isthehelix x =2cos t y = t z = 2sin t 0 t (1)(iv)] C(2 x +9 z ) ds ,where C isthecurve x = t y = t2, z = t3from(0 0 0)to(1 1 1)

PAGE 414

40814.MULTIPLEINTEGRALS (v) Czds ,where C istheintersectionoftheparaboloid z = x2+ y2andtheplane z =4 (vi) Cyds ,where C isthepartofthegraph y = exfor0 x 1 (vii) C( x + y ) ds ,where C isthetrianglewithvertices(0 0),(1 0), and(0 1) (viii) Cy2ds ,where C isanarcofthecycloid x = R ( t sin t ), y = R (1 cos t )from(0 0)to(2 R, 0) (ix) Cxyds ,where C isanarcofthehyperbola x = a cosh t y = a cosh t for0 t T (x) C( x4 / 3+ y4 / 3) ds ,where C istheastroid x2 / 3+ y2 / 3= a2 / 3(xi) Cxds ,where C isthepartofthespiral r = aethatliesin thedisk r a ;here( r, )arepolarcoordinates (xii) C x2+ y2ds ,where C isthecircle x2+ y2= ax (xiii) Cy 2ds ,where C is y = a cosh( x/a ) (xiv) C( x2+ y2+ z2) ds ,where C isoneturnofthehelix x = R cos t y = R sin t z = ht (0 t 2 ) (xv) Cy2ds ,where C isthecircle x2+ y2+ z2=1, x + y + z =0 (xvi) Czds ,where C istheconichelix x = t cos t y = t sin t z = t (0 t T ) (xvii) Czds ,where C isthecurveofintersectionofthesurfaces x2+ y2= z2and y2= ax fromtheorigintothepoint( a,a,a 2) (2) Findthemassofanarcoftheparabola y2=2 ax ,0 x a/ 2,if itslinearmassdensityis ( x,y )= | y | (3) Findthemassofthecurve x = at y = at2/ 2, z = at2/ 3,where 0 t 1,ifitslinearmassdensityis ( x,y,z )= 2 y/a 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 x2 y2and z = 1 x2 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.Supposethat f hascontinuouspartialderivatives,andhence itslinearizationatapointin D denesatangentplanetothegraph.

PAGE 415

108.SURFACEINTEGRALS409 Then Sijcanbeapproximatedbytheareaoftheparallelogramthat liesabove Rijinthetangentplanetothegraphthroughthepoint ( x i,y j,z ij),where z ij= f ( x i,y j)and( x i,y j) Rijisanysample point.Recallthatthedierentiabilityof f meansthatthedeviation of f fromitslinearizationtendsto0fasterthan ( x )2+( y )2as ( x, y ) (0 0).Therefore,inthislimit, x and y canbeviewed asthedierentials dx and dy ,and,whencalculating Sij,onlyterms linearin dx and dy mustberetained.Therefore,theareas Sijand A = x y havetobeproportional: Sij= Jij A. Thecoecient Jijisfoundbycomparingtheareaoftheparallelogram inthetangentplaneabove Rijwiththearea A of Rij.Thinkofthe roofofabuildingofshape z = f ( x,y )coveredbyshinglesofarea Sij. Theequationofthetangentplaneis z = z ij+ f x( x i,y j)( x x i)+ f y( x i,y j)( y 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(seetheleftpanel ofFigure14.38).If O A ,and B arethepointsinthetangentplane above O, A,and B,respectively,thentheadjacentsidesoftheparallelograminquestionare a = OA and b = OB and Sij= a b 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) L ( xi,yj)= f x( x i,y j) x and a =( x, 0 ,f x( x i,y j) x ).Similarly, b =(0 y,f y( x i,y j) y ).Hence, a b =( f x, 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 .

PAGE 416

41014.MULTIPLEINTEGRALS A z = L ( x,y ) a Pz = f ( x,y ) O b B A x P O y Bx y z 2 4 2 S DFigure14.38.Left :Therectanglewithadjacentsides OAand OBisanelementofarectangularpartitionof D and P isasamplepoint.Thepoint Pisthepointon thegraph z = f ( x,y )for( x,y )= P .Thelinearizationof f at Pdenesthetangentplane z = L ( x,y )tothegraph through P.Thesurfaceoftheportionofthegraphabove thepartitionrectangleisapproximatedbytheareaofthe portionofthetangentplaneabovethepartitionrectangle, whichistheareaoftheparallelogramwithadjacentsides OA and OB .Itequals a b Right :Anillustrationto Example14.36.Thepartoftheparaboloidwhoseareaisto beevaluatedisobtainedbyrestricting( x,y )tothepart D ofthediskofradius2thatliesintherstquadrant.Definition 14.21 (SurfaceArea) Supposethat f ( x,y ) hascontinuousrst-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 14.35 Showthatthesurfaceareaofasphereofradius R is 4 R2. Solution: Thehemisphereisthegraph z = f ( x,y )= R2 x2 y2onthedisk x2+ y2 R2ofradius R .Theareaofthesphereistwice

PAGE 417

108.SURFACEINTEGRALS411 theareaofthisgraph.Onehas f x= x/f and f y= y/f .Therefore, J =(1+ x2/f2+ y2/f2)1 / 2=( f2 x2 y2)1 / 2/f = R/f .Hence, A ( S )=2 R DdA R2 x2 y2=2 R 2 0d R 0rdr R2 r2=4 R R 0rdr R2 r2=2 R R20du u =4 R2, wherethedoubleintegralhasbeenconvertedtopolarcoordinates andthesubstitution u = R2 r2hasbeenusedtoevaluatethelast integral. Example 14.36 Findtheareaofthepartoftheparaboloid z = x2+ y2intherstoctantandbelowtheplane z =4 Solution: Thesurfaceinquestionisthegraph z = f ( x,y )= x2+ y2. Next,theregion D mustbespecied(itdeterminesthepartofthe graphwhoseareaistobefound).Onecanview D asthevertical projectionofthesurfaceontothe xy plane.Theplane z =4intersects theparaboloidalongthecircle4= x2+ y2ofradius2.Sincethesurface alsoliesintherstoctant, 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 / 2 1) wherethedoubleintegralhasbeenconvertedtopolarcoordinates andthesubstitution u =1+4 r2hasbeenusedtoevaluatethelast integral. 108.2.SurfaceIntegralofaFunction.Anintuitiveideaoftheconcept ofthesurfaceintegralofafunctioncanbeunderstoodfromthefollowingexample.Supposeonewantstondthetotalhumanpopulation ontheglobe.Thedataaboutthepopulationareusuallysuppliedas thepopulation density (i.e.,thenumberofpeopleperunitarea).The populationdensityisnotaconstantfunctionontheglobe.Itishigh incitiesandlowindesertsandjungles.Therefore,thesurfaceofthe globemustbepartitionedbysurfaceelementsofarea Sp.If ( r )is thepopulationdensityasafunctionofposition r ontheglobe,then thepopulationoneachpartitionelementisapproximately ( r p) Sp,

PAGE 418

41214.MULTIPLEINTEGRALS where r pisasamplepointinthepartitionelement.Theapproximationneglectsvariationsof withineachpartitionelement.Thetotal populationisapproximatelytheRiemannsum p ( r p) Sp.Toget anexactvalue,thepartitionhastoberenedsothatthesizeofeach partitionelementbecomessmaller.Thelimitisthesurfaceintegral of overthesurfaceoftheglobe,whichisthetotalpopulation.In general,onecanthinkofsomequantitydistributedoverasurfacewith somedensity(theamountofthisquantityperunitareaasafunction ofpositiononthesurface).Thetotalamountisthesurfaceintegralof thedensityoverthesurface. Let f beaboundedfunctioninanopenregion E andlet S be asurfacein E thathasanitesurfacearea.Considerapartition of S by N pieces Sp, p =1 2 ,...,N ,whichhavesurfacearea Sp. Supposethat S isdenedasalevelsurface g ( x,y,z )= k ofafunction g thathascontinuouspartialderivativeson E andwhosegradientdoes notvanish.ItwasshowninSection93.2thatgivenapoint P on S thereisafunctionoftwovariableswhosegraphcoincideswith S in aneighborhoodof P .Sothesurfacearea Spofasucientlysmall partitionelement SpcanbefoundbyDenition14.21.Put mp= infSpf and Mp=supSpf ;thatis, mpisthelargestlowerboundof valuesof f forall r Spand Mpisthesmallestupperboundonthe valuesof f forall r Sp.Theupperandlowersumsaredenedby U ( f,N )= N p =1Mp Spand L ( f,N )= N p =1mp Sp.Let Rpbethe radiusofthesmallestballthatcontains SpandmaxpRp= RN.A partitionof S issaidtoberenedif RNN .Inother words,undertherenement,thesizes Rpofpartitionelementsbecome uniformly smaller. Definition 14.22 (SurfaceIntegralofaFunction) Thesurfaceintegralofaboundedfunction f overasurface S is Sf ( r ) dS =limN U ( f,N )=limN L ( f,N ) providedthelimitsoftheupperandlowersumsexistandcoincide.The limitisunderstoodinthesense RN 0 as N ThesurfaceintegralcanalsoberepresentedbythelimitofaRiemannsum: (14.33) 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 ),andbythe

PAGE 419

108.SURFACEINTEGRALS413 r p SpS E Spa Vp= a SpFigure14.39.Left :Apartitionofasurface S byelementswithsurfacearea Sp.Itisusedinthedenitionof thesurfaceintegralandalsotoconstructitsRiemannsums. Right :Aneighborhood Eaofasmoothsurface S denedas thesetofpointswhosedistanceto S cannotexceed a> 0. Forsucientlynepartitionof S andsmall a ,theregion Eaispartitionedbyelementsofvolume Vp= a Sp.squeezeprinciplethelimitoftheRiemannsumis independent ofthe choiceofsamplepoints r p.Riemannsumscanbeusedinnumerical approximationsofthesurfaceintegral. Similartolineintegrals,surfaceintegralsarerelatedtotripleintegrals.Consideraneighborhood Eaofasmoothsurface S thatis denedasthesetofpointswhosedistanceto S cannotexceed a/ 2 > 0 (inthesenseofDenition11.14).Theregion Ealookslikeashellwith thickness a (seetherightpanelofFigure14.39).Supposethat f is integrableon Ea.Then,inthelimit a 0, (14.34) 1 a Eaf ( r ) dV Sf ( r ) dS. ThisrelationcanbeunderstoodbyconsideringtheRiemannsum(14.13) forthetripleintegralin(14.34)inwhich Eaispartitionedbyvolume elements Vp= a Spwithsamplepointstakenon S .Partitionelementsarecylindersofheight a alongthenormaltothesurfaceand withtheareaofthecrosssection Spdenedbythetangentplane. Thefactor1 /a ontherightsideof(14.34)cancelsthecommonfactor a in Vp,andtheRiemannsumsturnsintoaRiemannsumforasurface integral.Hence, thesurfaceintegralexistsforany f thatiscontinuous orhasboundedjumpdiscontinuitiesalonganitenumberofsmooth curveson S ,anditinheritsallthepropertiesofmultipleintegrals .

PAGE 420

41414.MULTIPLEINTEGRALS 108.3.EvaluationofaSurfaceIntegral.Theorem 14.22 (EvaluationofaSurfaceIntegral) Supposethat f iscontinuousinaregionthatcontainsasurface S denedbythegraph z = g ( x,y ) on D .Supposethat g hascontinuous rst-orderpartialderivativesonanopenregionthatcontains D .Then (14.35) 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 (14.33),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 toDenition14.22,andhencethelimitoftheRiemannsumexistsand isindependentofthechoiceofsamplepoints.Itcanbeprovedthat underthehypothesisofthetheoremthesurfaceintegralexists. Theevaluationofthesurfaceintegralinvolvesthefollowingsteps: Step1 .Represent S asagraph z = g ( x,y )(i.e.,ndthefunction g usingageometricaldescriptionof S ). Step2 .Findtheregion D thatdenesthepartofthegraphthat coincideswith S (if S isnottheentiregraph). Step3 .Calculatethederivatives g xand g yandtheareatransformationfunction J dS = JdA Step4 .Evaluatethedoubleintegral(14.35). Example 14.37 Evaluatetheintegralof f ( x,y,z )= z overthe partofthesaddlesurface z = xy thatliesinsidethecylinder x2+ y2=1 intherstoctant. Solution: Thesurfaceisapartofthegraph z = g ( x,y )= xy .Since itlieswithinthecylinder,itsprojectionontothe xy planeisbounded bythecircleofunitradius, x2+ y2=1.Thus, D isthequarterofthe

PAGE 421

108.SURFACEINTEGRALS415 disk x2+ y2 1intherstquadrant.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 1) udu = 1 2 u5 / 2 5 u3 / 2 3 2 1= 2(4 2+1) 15 wherethedoubleintegralhasbeenconvertedtopolarcoordinatesand thelastintegralisevaluatedbythesubstitution u =1+ r2. 108.4.ParametricEquationsofaSurface.Thegraph z = g ( x,y )of acontinuousfunction g ,where( x,y ) D ,denesasurface S in space.Considerthevectors r ( u,v )=( u,v,g ( u,v ))wherethepair ofparameters( u,v )spanstheregion D .Foreverypair( u,v ),the rule r ( u,v )=( u,v,g ( u,v ))denesavectorinspace,whichisthepositionvectorofapointonthesurface.Onecanmakeacontinuous transformationfrom D to Dbychangingvariables( u,v ) ( u,v). Thenthecomponentsofpositionvectorsofpointsof S becomegeneralcontinuousfunctionsofthenewvariables( u,v).Thisobservationsuggeststhatasurfaceinspacecanbedenedbyspecifying threecontinuousfunctionsoftwovariables, x ( u,v ), y ( u,v ),and z ( u,v ), onaregion D thatareviewedascomponentsofthepositionvector r ( u,v )=( x ( u,v ) ,y ( u,v ) ,z ( u,v )).Amappingof D intospacedened bythisruleiscalleda vectorfunction on D .Therangeofthismapping iscalleda parametricsurfaceinspace ,andtheequations x = x ( u,v ), y = y ( u,v ),and z = z ( u,v )arecalled parametricequations ofthe surface. Forexample,theequations (14.36) x = R cos v sin u,y = R sin v sin u,z = R cos u areparametricequationsofasphereofradius R .Indeed,bycomparing theseequationswiththesphericalcoordinates,onendsthat( ,, )= ( R,u,v );thatis,when( u,v )rangeovertherectangle[0 ] [0 2 ], thevector( x,y,z )= r ( u,v )tracesoutthesphere = R .Anapparentadvantageofusingparametricequationsofasurfaceisthat thesurfacenolongerneedstoberepresentedastheunionofgraphs. Forexample,thewholesphereisdescribedbythesinglevector-valued

PAGE 422

41614.MULTIPLEINTEGRALS function(14.36)oftwovariablesinsteadoftheunionoftwographs z = R2 x2 y2. Definition 14.23 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 visnot0. Ananalogycanbemadewithparametricequationsofacurvein space.Acurveinspaceisamappingofan interval [ a,b ]intospace denedbyavectorfunctionof 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 14.38 Findtheparametricequationsofthedoublecone z2= x2+ y2. Solution: Suppose z =0.Then( x/z )2+( y/z )2=1.Asolution 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 ) ( )forthewholedoublecone.Ofcourse, therearemanydierentparameterizationsofthesamesurface.They arerelatedbyachangeofvariables( u,v ) D ( s,t ) D,where s = s ( u,v )and t = t ( u,v ). Example 14.39 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 a )2= R2.Let( x0, 0 ,z0)beasolutiontothisequation.Thepoint( x0, 0 ,z0) tracesoutthecircleofradius x0upontherotationaboutthe z axis. Allsuchpointsare( x0cos v,x0sin v,z0),where v [0 2 ].Sinceall points( x0, 0 ,z0)areonthecircle z2+( x a )2= R2,theycanbe parameterizedas x0 a = R cos u z0= R sin u ,where u [0 2 ]. Thus,theparametricequationsofatorusare (14.37) x =( R + a cos u )cos v,y =( R + a cos u )sin v,z = R sin u,

PAGE 423

108.SURFACEINTEGRALS417 x y z ( x,y,z ) ( x0, 0 ,z0) R u a v a R uFigure14.40.Atorus.Consideracircleofradius R in the zx planewhosecenterispositionedonthepositive x axisatadistance a>R .Anypoint( x0, 0 ,z0)onthecircle isobtainedfromthepoint( a + R, 0 0)byrotationaboutthe centerofthecirclethroughanangle0 u 2 sothat x0= a + R cos u and z0= R sin u .Atorusisasurface sweptbythecirclewhenthe xz planeisrotatedaboutthe z axis.Agenericpoint( x,y,z )onthetorusisobtainedfrom ( x0, 0 ,z0)byrotatingthelatteraboutthe z axisthroughan angle0 v 2 .Underthisrotation, z0doesnotchange and z = z0,whilethepair( x0, 0)inthe xy planechangesto ( x,y )=( x0cos v,x0sin v ).Parametricequationsofatorus are x =( a + R cos u )cos v y =( a + R cos u )sin v z = R sin u where( u,v )rangesovertherectangle[0 2 ] [0 2 ].where( u,v ) [0 2 ] [0 2 ].Analternative(geometrical)derivation oftheseparametricequationsisgiveninthecaptionofFigure14.40. ATangentPlanetoaParametricSurface.Theline v = v0in D is mappedontothecurve r = r ( u,v0)in S (seeFigure14.41).Thederivative r u( u,v0)istangenttothecurve.Similarly,theline u = u0in D ismappedtothecurve r = r ( u0,v )in S ,andthederivative r v( u0,v ) istangenttoit.Ifthecrossproduct r u r vdoesnotvanishin D thenonecandeneaplanenormaltothecrossproductatanypoint of S .Furthermore,if r u r v = 0 inaneighborhoodof( u0,v0),then,

PAGE 424

41814.MULTIPLEINTEGRALS withoutlossofgenerality,onecanassumethat,say,the z component ofthecrossproductisnot0: x uy v x vy u= ( x,y ) / ( u,v ) =0.This showsthatthe transformation x = x ( u,v ) y = y ( u,v ) withcontinuouspartialderivativeshasanonvanishingJacobian .Bytheinverse functiontheorem(Theorem14.10),thereexistsaninversetransformation u = u ( x,y ), v = v ( x,y )thatalsohascontinuouspartialderivatives.Sothevectorfunction r ( u,v )canbewritteninthenewvariables ( x,y )as R ( x,y )= r ( u ( x,y ) ,v ( x,y ))=( x,y,z ( u ( x,y ) ,v ( x,y ))=( x,y,g ( x,y )) whichisavectorfunctionthattracesoutthegraph z = g ( x,y ).Thus, asmoothparametricsurfacenearanyofitspointscanalwaysberepresentedasthegraphofafunctionoftwovariables .Bythechainrule,the function g hascontinuouspartialderivatives.Therefore,itslinearizationnear( x0,y0)=( x ( u0,v0) ,y ( u0,v0))denesthe tangentplane tothe graphandhencetotheparametricsurfaceatthepoint r0= r ( u0,v0). Inparticular,thevectors r vand r umustlieinthisplaneastheyare tangenttotwocurvesinthegraph.Thus,thevector r u r visnormal tothetangentplane.SoDenition14.23ofasmooth parametric surfaceagreeswiththenotionofasmoothsurfaceintroducedinSection 103.1andthefollowingtheoremholds. Theorem 14.23 (NormaltoaSmoothParametricSurface) Let r = r ( u,v ) beasmoothparametricsurface.Thenthevector n = r u r visnormaltothesurface.AreaofaSmoothParametricSurface.Owingtothedenitionofthe surfaceareaelementofthegraphandtheestablishedrelationbetween graphsandsmoothparametricsurfaces,theareaasmoothsurfacecan befoundusingthetangentplanestoit(seeFigure14.38,leftpanel). Letaregion D spannedbytheparameters( u,v )bepartitionedby rectanglesofarea A = u v .Thenthevectorfunction r ( u,v ) denesapartitionofthesurface(apartitionelementofthesurfaceis theimageofapartitionrectanglein D ).Considerarectangle[ u0,u0+ 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

PAGE 425

108.SURFACEINTEGRALS419 v u D v0Ou0r ( u,v ) S O r ur vFigure14.41.Thelines u = u0and v = v0in D are mappedontothecurvesin S thataretracedoutbythe vectorfunctions r = r ( u0,v )and r = r ( u,v0),respectively. Thecurvesintersectatthepoint O withthepositionvector r ( u0,v0).Thederivatives r v( u0,v0)and r u( u0,v0)aretangentialtothecurves.Iftheydonotvanishandarenotparallel,thentheircrossproductisnormaltotheplanethrough O thatcontains r uand r v.Iftheparametricsurfaceissmooth, thisplaneistangenttoit.approximatedbytheareaoftheparallelogram a b withadjacent sides: a = OA = ra r0= r ( u0+ u,v0) r ( u0,v0)= r u( u0,v0) u, b = OB = rb r0= r ( u0,v0+ v ) r ( u0,v0)= r v( u0,v0) v. Thelastequalitiesareobtainedbythelinearizationofthecomponents of r ( u,v )near( u0,v0),whichisjustiedbecausethesurfacehasatangentplaneatanypoint.Theareatransformationlawisnoweasyto nd: S = a b = r u r v A .Thus,thesurfaceareaofasmooth parametricsurfaceisgivenbythedoubleintegral A ( S )= D r u r v dA. Accordingly,thesurfaceintegralofafunction f ( r )overasmoothparametricsurfaceis Sf ( r ) dS = Df ( r ( u,v )) r u r v dA. Example 14.40 Findthesurfaceareaofthetorus(14.37).

PAGE 426

42014.MULTIPLEINTEGRALS Solution: Toshortenthenotation,put w = R + a cos u .Onehas r u=( a sin u cos v, a sin u sin v,R cos u ) r v=( ( R + a cos u )sin v, ( R + a cos u )cos v, 0) = w ( sin v, cos v, 0) n = r u r v= w ( a cos v cos u, a cos v cos u, a sin u ) J = r u r v = aw = a ( R + a cos u ) Thesurfaceareais A ( S )= DJ ( u,v ) dA = 2 02 0a ( R + a cos u ) dvdu =4 2Ra. Example 14.41 Evaluatethesurfaceintegralof f ( x,y,z )= z2( x2+ y2) overasphereofradius R centeredattheorigin. Solution: Usingtheparametricequations(14.36),onends r u=( R cos v cos u,R sin v cos u, R sin u ) r v=( R sin v sin u,R cos v sin u, 0) = R sin u ( 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 = R sin u r ( u,v ) = R2sin u, f ( r ( u,v ))=( R cos u )2R2sin2u = R4cos2u (1 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 cos2u )sin udu =2 R61 1w2(1 w2) dw = 8 15 R6, wherethesubstitution w =cos u hasbeenmadetoevaluatethelast integral.

PAGE 427

108.SURFACEINTEGRALS421 108.5.Exercises.(1) Findthesurfaceareaofthespeciedsurface: (i)Thepartoftheplaneintherstoctantthatintersectsthe coordinateaxesat( a, 0 0),(0 ,b, 0),and(0 0 ,c ),where a b and c arepositivenumbers (ii)Thepartoftheplane3 x +2 y + z =1thatliesinsidethe cylinder x2+ y2=4 (iii)Thepartofthehyperboloid z = y2 x2thatliesbetweenthe cylinders x2+ y2=1and x2+ y2=4 (iv)Thepartoftheparaboloid z = x2+ y2thatliesbetweentwo planes z =1and z =9 (v)Thepartofthesurface y =4 x + z2thatliesbetweentheplanes x =0, x =1, z =0,and z =1 (2) Evaluatetheintegraloverthespeciedsurface: (i) SyzdS ,where S isthepartoftheplane x + y + z =1that liesintherstoctant (ii) Sx2z2dS ,where S isthepartofthecone z2= x2+ y2that liesbetweentheplanes z =1and z =2 (iii) SxzdS ,where S istheboundaryofthesolidregionenclosed bythecylinder y2+ z2=1andtheplanes x =0and x + y =3 ( Hint: Usetheadditivityofthesurfaceintegral.) (iv) SzdS ,where S isthepartofthesphere x2+ y2+ z2=2 thatliesabovetheplane z =1 (v) Sz (sin( x2) sin( y2)) dS ,where S isthepartoftheparaboloid z =1 x2 y2thatliesintherstoctant( Hint: Usethe symmetry.) (3) Supposethat f ( r )= g ( r ),where r =( x,y,z ).If g ( a )=2,use thegeometricalinterpretationofthesurfaceintegraltond SfdS where S isthesphereofradius a centeredattheorigin. (4) Identifyandsketchtheparametricsurface: (i) r ( u,v )=( u + v, 2 v, 2 2 v +3 u ) (ii) r ( u,v )=( a cos u,b sin u,v ) (5) Forthegivenparametricsurface,sketchthecurves r ( u,v0)for severalxedvalues v = v0andthecurves r ( u0,v )forseveralxed values u = u0.Usethemtovisualizetheparametricsurfaceif (i) r ( u,v )=(sin v,u sin v, sin u sin(2 v )) (ii) r ( u,v )=( u cos v sin ,u sin u sin ,u cos ),where0 / 2 isaparameter (6) Findaparametricrepresentationforthefollowingsurfaces

PAGE 428

42214.MULTIPLEINTEGRALS (i)Theplanethrough r0thatcontainstwononzeroandnonparallelvectors a and b (ii)Theellipticcylinder y2/a2+ z2/b2=1 (iii)Thepartofthesphere x2+ y2+ z2= a2thatliesbelowthe cone z = x2+ y2(iv)Theellipsoid x2/a2+ y2/b2+ z2/c2=1 (7) Findanequationofthetangentplanetothegivenparametric surfaceatthespeciedpoint P : (i) r ( u,v )=( u2,u v,u + v )at P =(1 1 3) (ii) r ( u,v )=(sin v,u sin v, sin u sin(2 v ))at P =(1 ,/ 2 0) (8) Evaluatethesurfaceintegraloverthespeciedparametricsurface: (i) Sz2dS ,where S isthetorus(14.37)with R =1and a =2 (ii) S(1+ x2+ y2)1 / 2dS ,where S isthe helicoid withparametric equations r ( u,v )=( u cos v,u sin v,v )and( u,v ) [0 1] [0 ] (iii) SzdS ,where S isthepartofthehelicoid r ( u,v )=( u cos v,u sin v,v ), ( u,v ) [0 ,a ] [0 2 ] (iv) Sz2dS ,where S isthepartofthecone x = u cos v sin y = u sin u sin z = u cos ,and( u,v ) [0 ,a ] [0 2 ],and 0 / 2isaparameter (9) Evaluatethesurfaceintegral.Ifnecessary,usesuitableparametric equationsofthesurface: (i) S( x2+ y2+ z2) dS ,where S isthesphere x2+ y2+ z2= R2(ii) S( x2+ y2+ z2) dS ,where S isthesurface | x | + | y | + | z | = R ; comparetheresultwiththepreviousexercise (iii) S( x2+ y2) dS ,where S istheboundaryofthesolid x2+ y2 z 1 (iv) S(1+ x + y ) 2dS ,where S istheboundaryofthetetrahedron boundedbythecoordinateplanesandbytheplane x + y + z =1 (v) S| xyz | dS ,where S isthepartoftheparaboloid z = x2+ y2belowtheplane z =1 (vi) S(1 /h ) dS ,where S isanellipsoid( x/a )2+( y/b )2+( z/c )2=1 and h isthedistancefromtheorigintotheplanetangentto theellipsoidatthepointwherethesurfaceareaelement dS is taken (vii) S( xy + yz + zx ) dS ,where S isthepartofthecone z = x2+ y2cutoutbythecylinder x2+ y2=2 ax

PAGE 429

109.MOMENTSOFINERTIAANDCENTEROFMASS423 (10) Provethe Poissonformula Sf ( ax + by + cz ) dS =2 1 1f ( u a2+ b2+ c2) du, where S isthesphere x2+ y2+ z2=1. (11) Evaluate F ( a,b,c,t )= Sf ( x,y,z ) dS ,where S isthesphere ( x a )2+( y b )2+( z c )2= t2, f ( x,y,z )=1if x2+ y2+ z2R> 0. 109.MomentsofInertiaandCenterofMass Animportantapplicationofmultipleintegralsisndingthe center ofmass and momentsofinertia ofanextendedobject.Thelawsof mechanicssaythatthecenterofmassofanextendedobjectonwhich noexternalforceactsmovesalongastraightlinewithaconstantspeed. Inotherwords,thecenterofmassisaparticularpointofanextended objectthatdenesthetrajectoryoftheobjectasawhole.Themotion ofanextendedobjectcanbeviewedasacombinationofthemotionof itscenterofmassandrotationaboutitscenterofmass.Thekinetic energyoftheobjectis K = Mv2 2 + Krot, where M isthetotalmassoftheobject, v isthespeedofitscenter ofmass,and Krotisthekineticenergyofrotationoftheobjectabout itscenterofmass; Krotisdeterminedby momentsofinertia discussed later.Forexample,whendockingaspacecrafttoaspacestation,one needstoknowexactlyhowlongtheengineshouldberedtoachieve therequiredpositionofitscenterofmassandtheorientationofthe craftrelativetoit,thatis,howexactlyitskineticenergyhastobe changedbyringtheengines.Soitscenterofmassandmomentsof inertiamustbeknowntoaccomplishthetask.109.1.CenterofMass.Considerapointmass m xedatanendpoint ofarodthatcanrotateaboutitsotherend.Iftherodhaslength L andthegravitationalforceisnormaltotherod,thenthequantity gmL iscalledthe rotationalmoment ofthegravitationalforce mg ,where g isthefree-fallacceleration.Iftherotationisclockwise(themassisat therightendpoint),themomentisassumedtobepositive,anditis negative, gmL ,foracounterclockwiserotation(themassisattheleft endpoint).Moregenerally,ifthemasshasaposition x onthe x axis, thenitsrotationmomentaboutapoint xcis M =( x xc) m (omitting theconstant g ).Itisnegativeif xxc.

PAGE 430

42414.MULTIPLEINTEGRALS m1m1g m2m2g x1xcx2x m1m2m3C r1r2r3Figure14.42.Left :Twomassesconnectedbyarigid masslessrod(oritsmassismuchsmallerthanthemasses m1and m2)arepositionedat x1and x2.Thegravitationalforce isperpendiculartotherod.Thecenterofmass xcisdeterminedbytheconditionthatthesystemdoesnotrotateabout xcunderthegravitationalforces. Right :Anextendedobjectconsistingofpointmasseswithxeddistancesbetween them.Ifthepositionvectorsofthemassesrelativetothe centerofmass C are ri,then m1r1+ m2r2+ + mNrN= 0 .Thecenterofmassisunderstoodthroughtheconceptofrotational moments. Thesimplestextendedobjectconsistsoftwopointmasses m1and m2connectedbyamasslessrod.ItisshownintheleftpanelofFigure14.42.Supposethatonepointoftherodisxedsothatitcanonly rotateaboutthatpoint.Thecenterofmassisthepointontherod suchthattheobjectwouldnotrotateaboutitunderauniformgravitationalforceappliedalongthedirectionperpendiculartotherod. Evidently,thepositionofthecenterofmassisdeterminedbytheconditionthatthetotalrotationalmomentaboutitvanishes.Suppose thattherodliesonthe x axissothatthemasseshavethecoordinates x1and x2.Thetotalrotationalmomentoftheobjectaboutthepoint xcis M = M1+ M2=( x1 xc) m1+( x2 xc) m2.If xcissuchthat M =0,then m1( x1 xc)+ m2( x2 xc)=0= xc= m1x1+ m2x2 m1+ m2. Thecenterofmass( xc,yc)ofpointmasses mi, i =1 2 ,...,N ,positioned onaplaneat( xi,yi)canbeunderstoodasfollows.Thinkoftheplane asaplateonwhichthemassesarepositioned.Thegravitationalforce isnormaltotheplane.Ifarod(aline)isputunderneaththeplane, thenduetoanunevendistributionofmasses,theplanecanrotate abouttherod.Whentherodisalignedalongeithertheline x = xcor theline y = yc,theplanewithdistributedmassesonitdoesnotrotate

PAGE 431

109.MOMENTSOFINERTIAANDCENTEROFMASS425 underthegravitationalpull.Inotherwords,therotationalmoments aboutthelines x = xcand y = ycvanish.Therotationalmoment abouttheline x = xcor y = ycisdeterminedbythedistancesofthe massesfromthisline:Ni =1( xi xc) mi=0= xc= 1 mNi =1mixi= My m ,m =Ni =1mi,Ni =1( yi yc) mi=0= yc= 1 mNi =1miyi= Mx m where m isthetotalmass.Thequantity Myisthemomentaboutthe y axis(theline x =0),whereas Mxisthemomentaboutthe x axis (theline y =0). Consideranextendedobjectthatisacollectionofpointmasses shownintherightpanelofFigure14.42.Itscenterofmassisdened similarlybyassumingthatthetotalmomentsaboutanyoftheplanes x = xc,or y = yc,or z = zcvanish.Thus,if rcisthepositionvectorof thecenterofmass,itsatisesthecondition: imi( ri rc)= 0 wherethevectors ri rcarepositionvectorsofmasses relativetothe centerofmass Definition 14.24 (CenterofMass) Supposethatanextendedobjectconsistsof N pointmasses mi, i = 1 2 ,...,N ,whosepositionvectorsare ri.Thenitscenterofmassisa pointwiththepositionvector (14.38) rc= 1 mNi =1miri,m =Ni =1mi, where m isthetotalmassoftheobject.Thequantities Myz=Ni =1mixi,Mxz=Ni =1miyi,Mxy=Ni =1miziarecalledthemomentsaboutthecoordinateplanes. Ifanextendedobjectcontainscontinuouslydistributedmasses, thentheobjectcanbepartitionedinto N smallpieces.Let Bibe thesmallestballofradius Riwithinwhichthe i thpartitionpiecelies. Althoughallthepartitionpiecesaresmall,theystillhavenitesizes Ri,andthedenition(14.38)cannotbeusedbecausethepoint ricould

PAGE 432

42614.MULTIPLEINTEGRALS 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,thetripleintegralissimplied toeitherasurface(ordouble)integralforshell-like E ,accordingto (14.34),ortoalineintegral,accordingto(14.31).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. Theconceptofrotationalmomentsisalsousefulforndingthe centerofmassusingthesymmetriesofthemassdistributionofan extendedobject.Forexample,thecenterofmassofadiskwitha uniformmassdistributionapparentlycoincideswiththediskcenter (thediskwouldnotrotateaboutitsdiameterunderthegravitational pull). Example 14.42 Findthecenterofmassofthehalf-disk x2+ y2 R2, y 0 ,ifthemassdensityatanypointisproportionaltothe distanceofthatpointfromthe x axis. Solution: Themassisdistributedevenlytotheleftandrightfrom the y axisbecausethemassdensityisindependentof x ( x,y )= ky ( k isaconstant).Sotherotationalmomentaboutthe y axisvanishes;

PAGE 433

109.MOMENTSOFINERTIAANDCENTEROFMASS427 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 14.43 Findthecenterofmassofthesolidthatliesbetweenspheresofradii a
PAGE 434

42814.MULTIPLEINTEGRALS R v m E R( r i) r iFigure14.43.Left :Themomentofinertiaofapoint massaboutanaxis .Apointmassrotatesaboutanaxis withtherate ,calledtheangularvelocity.Itslinearvelocityis v = R ,where R isthedistancefrom .Sothekineticenergyoftherotationalmotionis mv2/ 2= mR22/ 2= I2/ 2,where I= mR2isthemomentofinertia. Right : Themomentofinertiaofanextendedobject E aboutan axis isdenedasthesumofmomentsofinertiaofpartitionelementsof E : Ii= m ( r i) R2 ( r i),where R( r i)is thedistancetotheaxis and m ( r i)isthemassofthe partitionelement.109.2.MomentsofInertia.Considerapointmass m rotatingabout anaxis ataconstantrateof rad/s(calledthe angularvelocity ). ThesystemisshowninFigure14.43(leftpanel).Iftheradiusofthe circulartrajectoryis R ,thenthelinearvelocityoftheobjectis v = R Theobjecthasthekineticenergy Krot= mv2 2 = mR22 2 = I2 2 Theconstant Iiscalledthe momentofinertia ofthepointmass m abouttheaxis .Similarly,consideranextendedsolidobjectconsisting of N pointmasses.Thedistancesbetweenthemassesdonotchange whentheobjectmoves(theobjectis solid ).So,iftheobjectrotates aboutanaxis ataconstantrate ,theneachpointmassrotatesat thesamerateandhencehaskineticenergy miR2 i2/ 2,where Riisthe distancefromthemass mitotheaxis .Thetotalkineticenergyis Krot= I2/ 2,wheretheconstant I=Ni =1miR2 i

PAGE 435

109.MOMENTSOFINERTIAANDCENTEROFMASS429 iscalledthe momentofinertiaoftheobjectabouttheaxis .Itis independentofthemotionitselfanddeterminedsolelybythemass distributionanddistancesofthemassesfromtherotationaxis. Supposethatthemassiscontinuouslydistributedinaregion E withthemassdensity ( r )(seetherightpanelofFigure14.43).Let R( r )bethedistancefromapoint r E toanaxis(line) .Consider apartitionof E bysmallelements Eiofvolume Vi.Themassofeach partitionelementis mi= ( r i) Viforsomesamplepoint r i Eiin thelimitwhenallthesizesofpartitionelementstendto0uniformly. 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 axesis, respectively, Rx= y2+ z2, Ry= x2+ z2,and Rz= x2+ y2.So themomentsofinertiaaboutthecoordinateaxesare Ix= E( y2+ z2) dV,Iy= E( x2+ z2) dV, Iz= E( x2+ y2) dV. Ingeneral,iftheaxis goesthroughtheoriginparalleltoaunitvector u ,thenbythedistanceformulabetweenapoint r andtheline, R2 ( r )= u r 2=( u r ) ( u r )= u ( r ( u r )) = r2 ( u r )2, (14.39) wherethe bac cab rule(11.9)hasbeenusedtotransformthedouble crossproduct. Ifoneortwodimensionsoftheobjectaresmallrelativetothe other,thetripleintegralisreducedtoeitherasurfaceintegralora lineintegral,respectively,inaccordancewith(14.34)or(14.31);that is,fortwo-orone-dimensionalobjects,themomentofinertiabecomes, respectively, I= SR2 ( r ) ( r ) dS,I= CR2 ( r ) ( r ) ds, where iseitherthesurfaceorlinearmassdensity. Example 14.44 Arockettipismadeofthinplateswithaconstant surfacemassdensity = k .Ithasacircularconicshapewithbase

PAGE 436

43014.MULTIPLEINTEGRALS h x y z a cR m Rcr u u r rcrcFigure14.44.Left :AnillustrationtoExample14.44. Right :Anillustrationtotheproofoftheparallelaxistheoremformomentsofinertia(StudyProblem14.11).Theaxis cisparallelto andgoesthroughthecenterofmasswith thepositionvector rc.Thevectors r and r rcareposition vectorsofapartitionelementofmass m relativetothe originandthecenterofmass,respectively.diameter 2 a anddistance h fromthetiptothebase.Findthemoment ofinertiaofthetipaboutitsaxisofsymmetry. Solution: Setupthecoordinatesystemsothatthetipisatthe originandthebaseliesintheplane z = h ;thatis,thesymmetry 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) 1 / 2and g y=( hy/a )( x2+ y2) 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.

PAGE 437

109.MOMENTSOFINERTIAANDCENTEROFMASS431 Example 14.45 Findthemomentofinertiaofahomogeneousball ofradius a andmass m aboutitsdiameter. Solution: Setupthecoordinatesystemsothattheoriginisatthe centeroftheball.Thenthemomentofinertiaaboutthe z axishas tobeevaluated.Sincetheballishomogeneous,itsmassdensityis 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 1(1 u2) du = 2 5 ma2, wherethesubstitution u =cos hasbeenmadetoevaluatetheintegral.Itisnoteworthythattheproblemadmitsasmartersolutionby notingthat Iz= Ix= Iyowingtotherotationalsymmetryofthemass distribution.Bytheidentity Iz=( Ix+ Iy+ Iz) / 3,thetripleintegral canbesimplied: Iz= 1 3 E2( x2+ y2+ z2) dV = 1 3 8 a 04d = 2 5 ma2. Example 14.46 Findthecenterofmassandthemomentofinertia ofahomogeneousrodofmass m bentintoahalf-circleofradius R about thelinethroughtheendpointsoftherod. Solution: Setupthecoordinatesystemsothehalf-circleliesabove the x axis: x2+ y2= R2, y 0.Thelinearmassdensityisconstant = m/ ( R ),where R isthelengthoftherod.Bythesymmetryof themassdistribution,thecenterofmassliesonthe y axis, xc=0, and yc=(1 /m ) Cyds .Toevaluatethelineintegral,choosethe followingparametricequationofthehalf-circle r ( t )=( R cos t,R sin t ), 0 t .Then r( t )=( R sin t,R cos t )and ds = r( t ) dt = Rdt Therefore, yc= 1 m Cyds = 1 R 0R sin tRdt = 2 R If Risthedistancefromthelineconnectingtheendpointoftherod toitsparticularpoint,then,inthechosencoordinatesystem, R= y Therefore,themomentofinertiainquestionis I= CR2 ds = m R Cy2ds = mR2 0sin2tdt = mR2 2

PAGE 438

43214.MULTIPLEINTEGRALS 109.3.StudyProblems.Problem14.10. Findthecenterofmassoftheshelldescribedin Example14.44. Solution: Bythesymmetryofthemassdistributionabouttheaxis oftheconicshell,thecenterofmassmustbeonthataxis.Usingthe algebraicdescriptionofashellgiveninExample14.44,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. Problem14.11.(ParallelAxisTheorem) Let Ibethemomentofinertiaofanextendedobjectaboutanaxis andlet cbeaparallelaxisthroughthecenterofmassoftheobject. Provethat I= Ic+ mR2 c, where Rcisthedistancebetweentheaxis andthecenterofmassand m isthetotalmass. Solution: Choosethecoordinatesystemsothattheaxis goes throughtheorigin(seetherightpanelofFigure14.44).Letitbe paralleltoaunitvector u .Thedierence I Icistobeinvestigated. If rcisthepositionvectorofthecenterofmass,thentheaxis cis obtainedfrom byparalleltransportofthelatteralongthevector rc.Therefore,thedistance R2 c( r )isobtainedfrom R2 ( r )(see(14.39)) bychangingthepositionvector r inthelattertothepositionvector relativetothecenterofmass, r rc.Inparticular, R2 ( rc)= R2 cbythe denitionofthefunction R( r ).Hence,by(14.39), R2 ( r ) R2 c( r )= R2 ( r ) R2 ( r rc) =2 rc r r2 c ( u rc)(2 u r u rc) = r2 c ( u rc)2+2 rc ( r rc) 2( u rc) u ( r rc) = R2 c 2 a ( r rc) ,

PAGE 439

109.MOMENTSOFINERTIAANDCENTEROFMASS433 where a = rc ( u rc) u .Therefore, I Ic= E R2 ( r ) R2 c( r ) ( r ) dV = R2 cE ( r ) dV 2 a E( r rc) ( r ) dV = R2 cm, wherethesecondintegralvanishesbythedenitionofthecenterof mass. Problem14.12. Findthemomentofinertiaofahomogeneousball ofradius a andmass m aboutanaxisthatisatadistance R fromthe ballcenter. Solution: Thecenterofmassoftheballcoincideswithitscenter becausethemassdistributionisinvariantunderrotationsaboutthe center.Themomentofinertiaoftheballaboutitsdiameteris Ic= (2 / 5) ma2byExample14.45.Bytheparallelaxistheorem,forany axis atadistance R fromthecenterofmass, I= Ic+ mR2= m ( R2+2 a2/ 5). 109.4.Exercises.(1) Findthecenterofmassofthespeciedextendedobject: (i)Ahomogeneousthinrodoflength L (ii)Ahomogeneousthinwirethatoccupiesthepartofacircleof radius R thatliesintherstquadrant (iii)Ahomogeneousthinwirebentintooneturnofthehelixof radius R thatrisesbythedistance h pereachturn (iv)Ahomogeneousthinshellthatoccupiesahemisphereofradius R (v)Ahomogeneousthindiskofradius R thathasacircularhole ofradius a
PAGE 440

43414.MULTIPLEINTEGRALS (x)Thepartofthesolidenclosedbytheparaboloid z =2 x2 y2andthecone z = x2+ y2thatliesintherstoctantand whosemassdensityatanypointisproportionaltoitsdistance fromthe z axis (xi)Ahomogeneoussurfacecutfromthecone z = x2+ y2by thecylinder x2+ y2= ax (xii)Thepartofahomogeneousspheredenedby z = a2 x2 y2, x 0, y 0, x + y a (xiii)Thearcofthehomogeneouscycloid x = a ( t sin t ), y = a (1 cos t ),0 t (xiv)Thearcofthehomogeneouscurve y = a cosh( x/a )fromthe point(0 ,a )tothepoint( b,h ) (xv)Thearcofthehomogeneousastroid x2 / 3+ y2 / 3= a2 / 3inthe rstquadrant (xvi)Thehomogeneouslaminaboundedbythecurves x + y = a x =0, y =0 (xvii)Thepartofthehomogeneouslaminaboundedbythecurve x2 / 3+ y2 / 3= a2 / 3intherstquadrant (xviii)Thehomogeneoussolidboundedbythesurfaces x2+ y2=2 z x + y = z (xix)Thehomogeneoussolidboundedbythesurfaces z = x2+ y2, 2 z = x2+ y2, x + y = 1, x y = 1 (2) Showthatthecentroidofatriangleisthepointofintersection ofitsmedians(thelinesjoiningeachvertexwiththemidpointofthe oppositeside). (3) Showthatthecentroidofapyramidislocatedonthelinesegment thatconnectstheapextothecentroidofthebaseandis1/4thedistancefromthebasetotheapex. (4) Findthespeciedmomentofinertiaofthegivenextendedobject: (i)Thesmallerwedgecutoutfromaballofradius R bytwo planesthatintersectalongthediameteroftheballatanangle 0 <0 .Thewedgeishomogeneousandhasmass m .Find themomentofinertiaaboutthediameter. (ii)Themomentofinertiaaboutthe z axisofthesolidthatis enclosedbythecylinder x2+ y2 1andtheplanes z =0, y + z =5andhasamassdensityof ( x,y,z )=10 2 z (iii)Athinhomogeneousshellintheshapeofthetoruswithradii R and a>R thathasmass m .Themomentofinertiaabout thesymmetryaxisofthetorus.

PAGE 441

109.MOMENTSOFINERTIAANDCENTEROFMASS435 (iv)Themomentsofinertia Ixand Iyofthepartofthediskof radius a thatliesintherstquadrantandwhosemassdensity atanypointisproportionaltoitsdistancefromthe y axis. (v)Thesolidhomogeneousconewithheight h andradiusofthe base a .Themomentsofinertiaaboutitssymmetryaxis,the axisthroughitsvertexandperpendiculartothesymmetry axis,andanaxisthatcontainsadiameterofthebase. (vi)Thepartofthehomogeneousplane x + y + z = a ;themoments ofinertiaaboutthecoordinateaxes. (vii)Thehomogeneoustriangleofmass m whoseverticesinpolar coordinatesare( r, )=( a, 0),( a, 2 / 3),( a, 4 / 3);the polar momentofinertia I0= Ix+ Iy. (viii)Thehomogeneoussolidcylinder x2+ y2 a2, h z h of mass m ;themomentofinertiaaboutthelineparalleltothe z axisthroughthepoint( a, 0 0); (ix)Thehomogeneoussolidofmassdensity 0boundedbythe surface( x2+ y2+ z2)2= a2( x2+ y2);thesumofthemoments ofinertia Ix+ Iy+ Iz. (x)Thelaminawithaconstantmassdensity 0boundedbythe circle( x a )2+( y a )2= a2andbythesegments0 y a 0 x a ;themomentsofinertiaaboutthecoordinateaxes. (xi)Thelaminawithaconstantmassdensity 0boundedbythe curves xy = a2, xy =2 a2, x =2 y ,2 x = y ;themomentsof inertiaaboutthecoordinateaxes. (xii)Thesolidthathasaconstantmassdensity 0andisbounded bytheellipsoid( x/a )2+( y/b )2+( z/c )2=1;momentsofinertia aboutthecoordinateaxes. (xiii)Thesphericalhomogeneousshellofmass m andradius R ;the momentofinertialaboutadiameterofthesphere.

PAGE 443

CHAPTER15 VectorCalculus 110.LineIntegralsofaVectorField110.1.VectorFields.Consideranairowintheatmosphere.Theair velocityvariesfrompointtopoint.Inordertodescribethemotionof theair,theairvelocitymustbedenedasafunctionofposition,which meansthatavelocity vector hastobeassignedtoeverypointinspace. Inotherwords,incontrasttoordinaryfunctions,theairvelocityisa vector-valued functionofthepositionvectorinspace. Definition 15.1 (VectorField) Let E beasubsetinspace.Avectoreldon 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 Avectoreldis continuous ifitscomponentsarecontinuous.Avectoreldis dierentiable ifitscomponentsaredierentiable.Asimple exampleofavectoreldisthegradientofafunction, F ( r )= f ( r ). Thecomponentsofthisvectoreldaretherst-orderpartialderivatives: F ( r )= f ( r ) F1( r )= f x( r ) ,F2( r )= f y( r ) ,F3( r )= f z( r ) Manyphysicalquantitiesaredescribedbyvectorelds.Electricand magneticeldsarevectorelds.Allmoderncommunicationdevices (radio,TV,cellphones,etc.)useelectromagneticwaves.Visible lightisalsoelectromagneticwaves.Thepropagationofelectromagneticwavesinspaceisdescribedbydierentialequationsthatrelate electromagneticeldsateachpointinspaceandeachmomentoftime toadistributionofelectricchargesandcurrents(e.g.,antennas).The gravitationalforcelooksconstantnearthesurfaceoftheEarth,buton thescaleofthesolarsystemthisisnotso.Ifonethinksaboutaplanet asahomogeneousballofmass M ,thenthegravitationalforceexerted byitonapointmass m dependsonthepositionofthepointmass relativetotheplanetscenteraccordingtoNewtonslawofgravity: F ( r )= GMm r3r = GMm x r3, GMm y r3, GMm z r3 ,437

PAGE 444

43815.VECTORCALCULUS F x yFigure15.1.Left :Flowlinesofavectoreld F arecurves towhichthevectoreldistangential.Theowlinesare orientedbythedirectionofthevectoreld. Right :Flow linesofthevectoreld F =( y,x, 0)inExample15.1are concentriccirclesorientedcounterclockwise.Themagnitude F = x2+ y2isconstantalongtheowlinesandlinearly increaseswithincreasingdistancefromtheorigin.where G isNewtonsgravitationalconstant, r isthepositionvector relativetotheplanetscenter,and r = r isitslength.Theforce isproportionaltothepositionvectorandhenceparalleltoitateach point.Theminussignindicatesthat F isdirectedoppositeto r ,that is,theforceis attractive ;thegravitationalforcepullstowarditssource (theplanet).Themagnitude F = GMmr 2decreaseswithincreasingdistance r .Sothegravitationalvectoreldcanbevisualizedby plottingvectorsoflength F ateachpointinspacepointingtoward theorigin.Themagnitudesofthesevectorsbecomesmallerforpoints fartherawayfromtheorigin.Thisobservationleadstotheconceptof owlines ofavectoreld.110.2.FlowLinesofaVectorField.Definition 15.2 (FlowLinesofaVectorField) Theowlineofavectoreld F isacurveinspacesuchthat,atany point r ,thevectoreld F ( r ) istangenttoit. Thedirectionof F denesthe orientation ofowlines.Thedirectionofatangentvector F isshownbyarrowsontheowlinesas depictedintheleftpanelofFigure15.1.Forexample,theowlines oftheplanetsgravitationaleldarestraightlinesorientedtowardthe centeroftheplanet.Flowlinesofagradientvectoreld F = f = 0 arenormaltolevelsurfacesofthefunction f andorientedinthedirectioninwhich f increasesmostrapidly(Theorem13.16).Theyarethe

PAGE 445

110.LINEINTEGRALSOFAVECTORFIELD439 curvesofsteepestascentofthefunction f .Flowlinesofthevelocity vectoreldoftheairareoftenshowninweatherforecaststoindicate thewinddirectionoverlargeareas.Forexample,owlinesoftheair velocityinahurricanewouldlooklikeclosedloopsaroundtheeyeof thehurricane. Thequalitativebehaviorofowlinesmaybeunderstoodbyplotting vectors F atseveralpoints riandsketchingcurvesthroughthemso thatthevectors Fi= F ( ri)aretangenttothecurves.Findingthe exactshapeoftheowlinesrequiressolvingdierentialequations.If r = r ( t )isaparametricequationofaowline,then r( t )isparallel to F ( r ( t )).Sothederivative r( t )mustbeproportionalto F ( r ( t )), whichdenes asystemofdierentialequations forthecomponentsof thevectorfunction r ( t ),forexample, r( t )= F ( r ( t )).Tondaow linethroughaparticularpoint r0,thedierentialequationsmustbe supplementedby initial conditions,forexample, r ( t0)= r0.Ifthe equationshaveauniquesolution,thentheowthrough r0existsand isgivenbythesolution. Example 15.1 Analyzeowlinesoftheplanarvectoreld F = ( y,x, 0) Solution: Bynotingthat F r =0,itisconcludedthat,atany point, F isperpendiculartothepositionvector r =( x,y, 0)inthe plane.Soowlinesarecurveswhosetangentvectorisperpendicular tothepositionvector.If r = r ( t )isaparametricequationofsucha curve,then r ( t ) r( t )=0or( d/dt ) r2( t )=0andhence r2( t )=const, whichisacirclecenteredattheorigin.Soowlinesareconcentric circles.Atthepoint(1 0 0),thevectoreldisdirectedalongthe y axis: F (1 0 0)=(0 1 0)= e2.Therefore,theowlinesareoriented counterclockwise.Themagnitude F = x2+ y2remainsconstant oneachcircleandincreaseswithincreasingcircleradius.Theowlines areshownintherightpanelofFigure15.1. 110.3.LineIntegralofaVectorField.Theworkdonebyaconstant force F inmovinganobjectalongastraightlineisgivenby W = F d where d isthedisplacementvector(Section73.5).Supposethatthe forcevariesinspaceandthedisplacementtrajectoryisnolongera straightline.Whatistheworkdonebytheforce?Thisquestionis evidentlyofgreatpracticalsignicance.Toanswerit,theconceptof thelineintegralofavectoreldwasdeveloped.

PAGE 446

44015.VECTORCALCULUS C Cirir iF ( r i) T ( r i) ri +1rarbh R C1C2Figure15.2.Left :Tocalculatetheworkdonebyacontinuousforce F ( r )inmovingapointobjectalongasmooth curve C ,thelatterispartitionedintosegments Ciwitharc length s .Theworkdonebytheforcealongapartitionsegmentis F ( r i) di,wherethedisplacementvectorisapproximatedbytheorientedsegmentoflength s thatistangent tothecurveatasamplepoint r i,thatis, di= T ( r i) s where T istheunittangentvectoralongthecurve. Right : AnillustrationtoExample15.2.Theclosedcontourofintegrationinthelineintegralconsistsoftwosmoothpieces,one turnofthehelix C1andthestraightlinesegment C2.The lineintegralisthesumoflineintegralsalong C1and C2.Let C beasmoothcurvethatgoesfromapoint ratoapoint rband haslength L .Considerapartitionof C bysegments Ci, i =1 2 ,...,N oflength s = L/N .Followingthediscussionofsmoothcurvesin Sections80.3and84.3,eachsegmentcanbeapproximatedbyastraight linesegmentoflength s orientedalongtheunittangentvector T ( r i) atasamplepoint r i Ci(seetheleftpanelofFigure15.2).The workalongthesegment Cicanthereforebeapproximatedby Wi= F ( r i) T ( r i) s sothatthetotalworkisapproximatelythesum W = W1+ W2+ + WN.Theactualworkshouldnotdependonthe choiceofsamplepoints.Thisproblemisresolvedbytheusualtrick ofintegralcalculusbyreningapartition,ndingthelowandupper sums,andtakingtheirlimits.Iftheselimitsexistandcoincide,the limitingvalueshouldnotdependonthechoiceofsamplepointsand isthesought-afterwork.Thetechnicalitiesinvolvedmaybesparedby notingthat Wi= f ( r i) s ,where f ( r )= F ( r ) T ( r )and T ( r )denotes theunittangentvectoratapoint r C .Theapproximatetotalwork appearstobeaRiemannsumof f along C .So, ifthefunction f

PAGE 447

110.LINEINTEGRALSOFAVECTORFIELD441 isintegrableonthecurve C ,thentheworkisthelineintegralofthe tangentialcomponent F T oftheforce Definition 15.3 (LineIntegralofaVectorField) Thelineintegralofavectoreld F alongasmoothcurve C is CF d r = CF T ds, where T istheunittangentvectorto C ,providedthetangentialcomponent F T ofthevectoreldisintegrableon C Theintegrabilityof F T isdenedinthesenseoflineintegrals forordinaryfunctions(seeDenition14.20).Inparticular,theline integralofa continuous vectoreldoverasmoothcurveofanite lengthalwaysexists.110.4.EvaluationofLineIntegralsofVectorFields.Thelineintegralof avectoreldisevaluatedinmuchthesamewayasthelineintegralof afunction. Theorem 15.1 (EvaluationofLineIntegrals) Let F =( F1,F2,F3) beacontinuousvectoreldon 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 onlyoncesothat 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. (15.1) Proof. Theunittangentvectorreads T = r/ r and ds = r dt Therefore, T ds = r( t ) dt .Forasmoothcurve, r( t )iscontinuouson [ a,b ].Therefore,bythecontinuityofthevectoreld,thefunction f ( t )= F ( r ( t )) r( t )iscontinuouson[ a,b ],andtheconclusionofthe theoremfollowsfromTheorem14.21. Equation(15.1)alsoholdsif C ispiecewisesmoothand F hasa nitenumberofboundedjumpdiscontinuitiesalong C muchlikein thecaseofthelineintegralofordinaryfunctions.Owingtotherepresentation(15.1)andtherelations dx = xdt dy = ydt ,and dz = zdt thelineintegralisoftenwrittenintheform:

PAGE 448

44215.VECTORCALCULUS (15.2) CF d r = CF1dx + F2dy + F3dz. Forasmoothcurvetraversedbyavectorfunction r ( t ),thedierential d r ( t )istangenttothecurve. Incontrasttothelineintegralofordinaryfunctions,thelineintegral ofavectorelddependsontheorientationof C .Theorientationof C isxedbytheconditions 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 C .Thelineintegralchanges itssignwhentheorientationofthecurveisreversed: (15.3) CF d r = CF d r becausethedirectionofthederivative r( t )isreversedforall t .If C is piecewisesmooth(e.g.,theunionofsmoothcurves C1and C2),then theadditivityoftheintegralshouldbeused: CF d r = C1F d r + C2F d r .LineIntegralAlongaParametricCurve.Aparametriccurveisdened byavectorfunction r ( t )on[ a,b ](recallDenition12.4).Thevector function r ( t )maytraceitsrange(asapointsetinspace)orsomeparts ofitseveraltimesas t changesfrom a to b .Furthermore,two dierent vectorfunctions r1( t )and r2( t )on[ a,b ]mayhavethesamerange.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.Thelineintegraloveraparametriccurveis dened bytherelation(15.1).Aparametriccurveismuchlikethetrajectory ofaparticlethatcanpassthroughthesamepointsmultipletimes. Sotherelation(15.1)denestheworkdonebyanonconstantforce F alongaparticlestrajectory r = r ( t ). Theevaluationofalineintegralincludesthefollowingsteps: Step1 .Ifthecurve C isdenedasapointsetinspacebysome geometricalmeans,thennditsparametricequations r = r ( t )that agreewiththeorientationof C .Hereitisusefultorememberthat,if r ( t )correspondstotheorientationoppositetotherequiredone,then itcanstillbeusedaccordingto(15.3). Step2 .Restricttherangeof t toaninterval[ a,b ]sothat C istraced outonlyonceby r ( t ).

PAGE 449

110.LINEINTEGRALSOFAVECTORFIELD443 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(15.1). Example 15.2 Evaluatethelineintegralof F =( y,x,z2) along aclosedcontour C thatconsistsoftwoparts.Therstpartisoneturn ofahelixofradius R ,whichwindsaboutthe z axiscounterclockwiseas viewedfromthetopofthe z axis,startingfromthepoint ra=( R, 0 0) andendingatthepoint rb=( R, 0 2 h ) .Thesecondpartisastraight linesegmentfrom rbto ra. Solution: Let C1beoneturnofthehelixandlet C2bethestraight linesegment.Twolineintegralshavetobeevaluated.Theparametric equationsofthehelixare r ( t )=( R cos t,R sin t,ht )sothat r (0)= ( R, 0 0)and r (2 )=( R, 0 ,h )asrequiredbytheorientationof C1.Note thepositivesignsatcos t andsin t intheparametricequationsthatare necessarytomakethehelixwindingaboutthe z axiscounterclockwise (seeStudyProblem12.1).Therangeof t hastoberestrictedto[0 2 ]. Then r( t )=( R sin t,R cos t,h ).Therefore, F ( r ( t )) r( t )=( R sin t,R cos t,h2t2) ( 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 Theparametricequationsofthelinethroughtwopoints raand rbare r ( t )= ra+ v t ,where v = rb raisthevectorparalleltothe line,or,inthecomponents, r =( R, 0 0)+ t (0 0 2 h )=( R, 0 2 ht ). Then r (0)= raand r (1)= rbsothattheorientationisreversedif t [0 1].Theseparametricequationsdescribethecurve C2.One has 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 = C2F d r = (2 h )31 0t2dt = (2 h )3 3 Thelineintegralalong C isthesumoftheseintegrals,whichisequal to2 R2.

PAGE 450

44415.VECTORCALCULUS 110.5.StudyProblem.Problem15.1. Findtheworkdonebytheforce F =(2 x, 3 y2, 4 z3) alonganysmoothcurveoriginatingfromthepoint (0 0 0) andending atthepont (1 1 1) Solution: Foranyinnitesimalpartofthecurve,theworkis F d r =2 xdx +3 y2dy +4 z3dz = d x2+ y3+ z4 If r ( t )=( x ( t ) ,y ( t ) ,z ( t ))isaparametricequationofasmoothcurve, a t b ,suchthat r ( a )=(0 0 0)and r ( b )=(1 1 1),thenthetotal workis CF d r = b ad x2( t )+ y3( t )+ z4( t ) =( x2( t )+ y3( t )+ z4( t )) b a=3 110.6.Exercises.(1) Sketchowlinesofthegivenplanarvectoreld: (i) F =( ax,by ),where a and b arepositive (ii) F =( ay,bx ),where a and b arepositive (iii) F =( ay,bx ),where a and b havedierentsigns (iv) F = u u =tan 1( y/x ) (v) F = u u =ln[( x2+ y2) 1 / 2] (vi) F = u u =ln[( x a )2+( y b )2] (2) Sketchowlinesofthegivenvectoreldinspace: (i) F =( ax,by,cz ),where a b ,and c arepositive (ii) F =( ax,by,cz ),where a and b arepositive,while c isnegative (iii) F =( y, x,a ),where a isaconstant (iv) F = r r =( x,y,z ) (v) F = r 1, r =( x,y,z ) (vi) F = u u =( x/a )2+( y/b )2+( z/c )2; (vii) F = u u = x2+ y2+( z + c )2+ x2+ y2+( z c )2, where c ispositive (viii) F = a r ,where a isaconstantvectorand r =( x,y,z ) (ix) F = u u = z/ x2+ y2+ z2(3) Aballrotatesataconstantrate aboutitsdiameterparalleltoa unitvector n .Iftheoriginofthecoordinatesystemissetatthecenter oftheball,ndthevelocityvectoreldasafunctionoftheposition vector r ofapointoftheball. (4) Evaluatethelineintegral CF d r forthegivenvectoreld F and thespeciedcurve C :

PAGE 451

110.LINEINTEGRALSOFAVECTORFIELD445 (i) F =( y,xy, 0)and C istheparametriccurve r ( t )=( t2,t3, 0) for t [0 1] (ii) F =( z,yx,zy )and C istheellipse x2/a2+ y2/b2=1oriented clockwise (iii) F =( z,yx,zy )and C istheparametriccurve r ( t )=(2 t,t + t2, 1+ t3)fromthepoint( 2 0 0)tothepoint(2 2 2) (iv) F =( y,x,z )and C istheboundaryofthepartoftheparaboloid z = a2 x2 y2thatliesintherstoctant; C isoriented counterclockwiseasviewedfromthetopofthe z axis (v) F =( z, 0 ,x )and C istheboundaryofthepartofthesphere x2+ y2+ z2= a2thatliesintherstoctant; C isoriented clockwiseasviewedfromthetopofthe z axis (vi) F = a r ,where a isaconstantvectorand r =( x,y,z ); C is thestraightlinesegmentfrom r1to r2(vii) F =( y sin z,z sin x,x sin y )and C istheparametriccurve r =(cos t, sin t, sin(5 t ))for t [0 ] (viii) F =( y, xz,y ( x2+ x2))and C istheintersectionofthe cylinder x2+ z2=1withtheplane x + y + z =1thatis orientedcounterclockwiseasviewedfromthetopofthe y axis (ix) F =( y sin( z2) ,x cos( z2) ,exyz)and C istheintersectionof thecone z = x2+ y2andthesphere x2+ y2+ z2=2; C is orientedcounterclockwiseasviewedfromthetopofthe z axis (x) F =( e y,ex, 0)and C istheparabolainthe xy planefrom theorigintothepoint(1 1) (xi) F =( x,y,z )and C isanelliptichelix r ( t )=( a cos t,b sin t,ct ), 0 t 2 (xii) F =( y 1,z 1,x 1)and C isthestraightlinesegmentfrom thepoint(1 1 1)tothepoint(2 4 8) (xiii) F =( ey z,ez x,ex y)and C isthestraightlinesegmentfrom theorigintothepoint(1 3 5) (xiv) F =( y + z, 2+ x,x + y )and C istheshortestarconthesphere x2+ y2+ z2=25fromthepoint(3 4 0)tothepoint(0 0 5) (5) Findtheworkdonebytheconstantforce F inmovingapoint objectalongasmoothpathfromapoint ratoapoint rb. (6) Findtheworkdonebytheforce F = f( r ) r /r inmovingapoint objectalongasmoothpathfromapoint ratoapoint rb,wherethe derivative fof f isacontinuousfunctionof r = r (7) Findtheworkdonebytheforce F =( y,x,c ),where c isa constant,inmovingapointobjectalong: (i)Thecircle x2+ y2=1, z =0 (ii)Thecircle( x 2)+ y2=1, z =0

PAGE 452

44615.VECTORCALCULUS (8) Theforceactingonachargedparticlethatmovesinamagnetic eld B andanelectriceld E is F = e E +( e/c ) v B ,where v isthe velocityoftheparticle, e isitselectriccharge,and c isthespeedof lightinavacuum.Findtheworkdonebytheforcealongatrajectory originatingfromapoint raandendingatthepoint rbif (i)Theelectricandmagneticeldsareconstant. (ii)Theelectriceldvanishes,butthemagneticeldisacontinuousfunctionofthepositionvector, B = B ( r ). 111.FundamentalTheoremforLineIntegrals Recallthefundamentaltheoremofcalculus,whichassertsthat,if thederivative f( x )iscontinuousonaninterval[ a,b ],then b af( x ) dx = f ( b ) f ( a ) Itappearsthatthereisananalogofthistheoremforlineintegrals.111.1.ConservativeVectorFields.Definition 15.4 (ConservativeVectorFieldandItsPotential) Avectoreld F inaregion E issaidtobe conservative ifthereisa function f ,calleda potential of F ,suchthat F = f in E Conservativevectoreldsplayasignicantroleinmanypractical applications.Ithasbeenprovedearlier(seeStudyProblem13.14) thatifaparticlemovesalongatrajectory r = r ( t )undertheforce F = U ,thenitsenergy E = m v2/ 2+ U ( r ),where v = risthe velocity,isconservedalongthetrajectory, dE/dt =0.Inparticular, Newtonsgravitationalforceisconservative, F = U ,where U ( r )= GMm r 1.Astaticelectriceld(theCoulombeld)createdbya distributionofstaticelectricchargesisalsoconservative.Continuous conservativevectoreldshavearemarkableproperty. Theorem 15.2 (FundamentalTheoremforLineIntegrals) Let C beasmoothcurveinaregion E withinitialandterminalpoints raand rb,respectively.Let f beafunctionon E whosegradient f is continuouson C .Then (15.4) C f d r = f ( rb) f ( ra) .

PAGE 453

111.FUNDAMENTALTHEOREMFORLINEINTEGRALS447 Proof. Let r = r ( t ), t [ a,b ],betheparametricequationsof C such that r ( a )= raand r ( b )= rb.Then,by(15.1)andthechainrule, C f d r = b a( f xx+ f yy+ f zz) dt = b ad dt f ( r ( t )) dt = f ( rb) f ( ra) Thelatterequalityholdsbythefundamentaltheoremofcalculusand thecontinuityofthepartialderivativesof f and r( t )forasmooth curve. 111.2.PathIndependenceofLineIntegrals.Definition 15.5 (PathIndependenceofLineIntegrals) Acontinuousvectoreld F haspath-independentlineintegralsif C1F d r = C2F d r foranytwosimple,piecewise-smoothcurvesinthedomainof F with thesameendpoints. Recallthatacurveissimpleifitdoesnotintersectitself(seeSection79.3).Animportantconsequenceofthefundamentaltheoremfor lineintegralsisthattheworkdonebyacontinuousconservativeforce, F = f ,is path-independent .Soacriterionforavectoreldtobeconservativewouldbeadvantageousforevaluatinglineintegralsbecause foraconservativevectoreldacurvemaybedeformedatconvenience withoutchangingthevalueoftheintegral. Theorem 15.3 (Path-IndependentProperty) Let F beacontinuousvectoreldonanopenregion 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 curve. Proof. Pickapoint r0in E andconsideranysmoothcurve C from r0toapoint r =( x,y,z ) E .Theideaistoprovethatthefunction (15.5) f ( r )= CF d r isapotentialof F ,thatis,toprovethat f = F underthecondition thatthelineintegralof F vanishesforeveryclosedcurvein E .This

PAGE 454

44815.VECTORCALCULUS guessfor f ismotivatedbythefundamentaltheoremforlineintegrals (15.4),where rbisreplacedbyagenericpoint r E .Thepotential isdeneduptoanadditiveconstant( ( f +const)= f )sothe choiceofaxedpoint r0isirrelevant.First,notethatthevalueof f isindependentofthechoiceof C .Considertwosuchcurves C1and C2.Thentheunionof C1and C2(thecurve C2whoseorientation isreversed)isaclosedcurve,andthelineintegralalongitvanishes bythehypothesis.Ontheotherhand,thislineintegralisthesum oflineintegralsalong C1and C2.Bytheproperty(15.3),theline integralsalong C1and C2coincide.Tocalculatethederivative f x( r )= limh 0( f ( r + h e1) f ( r )) /h ,where e1=(1 0 0),letusexpressthe dierence f ( r + h e1) 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) 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 a 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. Althoughthepathindependencepropertydoesprovideanecessaryandsucientconditionforavectoreldtobeconservative,itis ratherimpracticaltoverify(onecannotevaluatelineintegralsalong everyclosedcurve!).Amorefeasibleandpracticalcriterionisneeded, whichisestablishednext.Itisworthnotingthat(15.5)givesapracticalmethodofndingapotentialifthevectoreldisfoundtobe conservative(seeStudyProblem15.3).111.3.TheCurlofaVectorField.Accordingtotherulesofvectoralgebra,theproductofavector a =( a1,a2,a3)andanumber s isdened

PAGE 455

111.FUNDAMENTALTHEOREMFORLINEINTEGRALS449 by s a =( sa1,sa2,sa3).Byanalogy,thegradient f canbeviewedas theproductofthevector =( /x,/y,/z )andascalar f : f = x y z f = f x f y f z Thecomponentsof arenotordinarynumbers,butrathertheyare operators (i.e.,symbolsstandingforaspeciedoperationthathasto 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) Thecurlofadierentiablevectoreld F is curl F = F Thecurlofavectoreldisavectoreldwhosecomponentscanbe computedaccordingtothedenitionofthecrossproduct: F =det e1 e2 e3 x y z F1F2F3 # = F3 y F2 z e1+ F1 z F3 x e2+ F2 x F1 y e3. Whencalculatingthecomponentsofthecurl,theproductofacomponentof andacomponentof F meansthatthecomponentof operatesonthecomponentof F ,producingthecorrespondingpartial derivative. Example 15.3 Findthecurlofthevectoreld F =( yz,xyz,x2) Solution: F =det e1 e2 e3 x y z yzxyzx2 # = ( x2) y ( xyz ) z, ( x2) x+( yz ) z, ( xyz ) x ( yz ) y =( xy,y 2 x,yz z )

PAGE 456

45015.VECTORCALCULUS Thegeometricalsignicanceofthecurlofavectoreldwillbe discussedinSection114.4.Herethecurlisusedtoformulatesucient conditionsforavectoreldtobeconservative.OntheUseoftheOperator.Therulesofvectoralgebraareusefulto simplifyalgebraicoperationsinvolvingtheoperator .Forexample, curl f = ( f )=( ) f = 0 becausethecrossproductofavectorwithitselfvanishes.However, thisformalalgebraicmanipulationshouldbeadoptedwithprecaution becauseitcontainsatacitassumptionthattheactionofthecomponentsof on f vanishes.Thelatterimposesconditionsonthe classoffunctionsforwhichsuchformalalgebraicmanipulationsare justied.Indeed,accordingtothedenition, f =det e1 e2 e3 x y z f xf yf z # =( f zy f yz,f zx f xz,f xy f yx) Thisvectorvanishes,providedtheorderofdierentiationdoesnot matter(i.e.,Clairautstheoremholdsfor f ).Thus, therulesofvector algebracanbeusedtosimplifytheactionofanoperatorinvolving ifthepartialderivativesofafunctiononwhichthisoperatoractsare continuousuptotheorderdeterminedbythataction .111.4.TestforaVectorFieldtoBeConservative.Aconservativevector eldwithcontinuouspartialderivativesinaregion E hasbeenshown tohavethevanishingcurl: F = f = curl F = 0 Unfortunately,theconverseis not trueingeneral.Inotherwords,the vanishingofthecurlofavectorelddoes not guaranteethatthevector eldisconservative.Theconverseistrueonlyiftheregioninwhich thecurlvanishesbelongstoaspecialclass.Aregion E issaidtobe connected ifanytwopointsinitcanbeconnectedbyapaththatlies in E .Inotherwords,aconnectedregioncannotberepresentedasthe unionoftwoormorenonintersecting(disjoint)regions. Definition 15.7 (SimplyConnectedRegion) Aconnectedregion E is simplyconnected ifeverysimpleclosedcurve in E canbecontinuouslyshrunktoapointin E whileremainingin E throughoutthedeformation.

PAGE 457

111.FUNDAMENTALTHEOREMFORLINEINTEGRALS451 Figure15.3.Fromlefttoright:Aplanarconnectedregion(anytwopointsinitcanbeconnectedbyacontinuous curvethatliesintheregion);aplanardisconnectedregion (therearepointsinitthatcannotbeconnectedbyacontinuouscurvethatliesintheregion);aplanarsimplyconnected region(everysimpleclosedcurveinitcanbecontinuously shrunktoapointinitwhileremainingintheregionthroughoutthedeformation);aplanarregionthatisnotsimplyconnected(ithasholes).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 cannothaveholesinit. Theorem 15.4 (TestforaVectorFieldtoBeConservative) Suppose F isavectoreldwhosecomponentshavecontinuouspartial derivativesonasimplyconnectedopenregion E .Then F isconservativein E ifandonlyifitscurlvanishesforallpointsof E : curl F = 0 onsimplyconnected E F = f on E. ThistheoremfollowsfromStokestheoremdiscussedlaterinSection114andhastwousefulconsequences.First, thetestforthepath independenceoflineintegrals : curl F = 0 onsimplyconnected E C1F d r = C2F d r foranytwopaths C1and C2in E originatingfromapoint ra E and terminatingatanotherpoint rb E .ItfollowsfromTheorem15.3

PAGE 458

45215.VECTORCALCULUS forthecurve C thatistheunionof C1and C2.Second,thetestfor vanishinglineintegralsalongclosedpaths: 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 StudyProblem15.2. Newtonsgravitationalforcecanbewrittenasthegradient F = U ,where U ( r )= GMm r 1everywhereexcepttheorigin.Therefore,itscurlvanishesin E ,whichistheentirespacewithonepoint removed;itissimplyconnected.Hence,theworkdonebythegravitationalforceis independent ofthepathtraveledbytheobjectanddeterminedbythedierenceinvaluesofitspotential U (alsocalled potential energy )attheinitialandterminalpointsofthepath.Moregenerally, sincetheworkdonebyaforceequalsthechangeinkineticenergy(see Section73.5),themotionunderaconservativeforce F = U hasthe fundamentalpropertythat thesumofkineticandpotentialenergies, m v2/ 2+ U ( r ) ,isconservedalongatrajectoryofthemotion (recall StudyProblem13.14). Example 15.4 Evaluatethelineintegralofthevectoreld F = ( F1,F2,F3)=( yz,xz + z +2 y,xy + y +2 z ) alongthepath C thatconsists 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 arepolynomialsandhencehavecontinuouspartialderivativesintheentirespace. Therefore,ifitscurlvanishes,then F isconservativeintheentirespace byTheorem15.4astheentirespaceissimplyconnected: F =det e1 e2 e3 x y z F1F2F3 # =det e1 e2 e3 x y z yzxz + z +2 yxy + y +2 z # = ( F3) y ( F2) z, ( F3) x+( F1) z, ( F2) x ( F1) y =( x +1 ( x +1) y + y,z z )=(0 0 0) .

PAGE 459

111.FUNDAMENTALTHEOREMFORLINEINTEGRALS453 Thus, F isconservative.Nowtherearetwooptionstonishtheproblem. Option1 .Onecanusethepathindependenceofthelineintegral, whichmeansthatonecanpickanyotherpath C1connectingtheinitialpoint A andtheterminalpoint D toevaluatethelineintegralin question.Forexample,astraightlinesegmentconnecting A and D is simpleenoughtoevaluatethelineintegral.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 .TheprocedureofSection89.1maybeusedtondapotential f of F (seealsothestudyproblemsattheendofthissection foranalternativeprocedure).Thelineintegralisthenfoundbythe fundamentaltheoremforlineintegrals.Put f = F .Thentheproblemisreducedtonding f fromitsrst-orderpartialderivatives(the existenceof f hasalreadybeenestablished).Followingtheprocedure ofSection89.1, f x= F1= yz = f ( x,y,z )= xyz + g ( y,z ) 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) f (0 0 0)=4 bythefundamentaltheoremforlineintegrals. 111.5.StudyProblems.Problem15.2. Verifythat F = f = y x2+ y2, x x2+ y2, 2 z ,f ( x,y,z )=tan 1( y/x )+ z2,

PAGE 460

45415.VECTORCALCULUS x y z a c x = y =0 =0 x y z C1C2C3( x,y,z ) ( x0,y0,z0)Figure15.4.Left :AnillustrationtoStudyProblem15.2. Right :AnillustrationtoStudyProblem15.3.Tondthe potentialofaconservativevectoreld,onecanevaluateits lineintegralfromanypoint( x0,y0,z0)toagenericpoint ( x,y,z )alongtherectangularcontour C thatisistheunion ofthestraightlinesegments C1, C2,and C3paralleltothe coordinateaxes.and curl F = 0 inthedomainof F .Evaluatethelineintegralof F alongthecircularpath C : x2+ y2= R2intheplane z = a .Thepathis orientedcounterclockwiseasviewedfromthetopofthe z axis.Doesthe resultcontradictthefundamentaltheoremforlineintegrals?Explain. Solution: Astraightforwarddierentiationof f showsthatindeed f = F ,andthereforecurl F = 0 everywhereexcepttheline x = y =0,where F isnotdened.Thepath C istracedoutby r ( t )= ( R cos t,R sin t,a ),where t [0 2 ].Then F ( r ( t ))=( R 1sin t, R 1cos t, 2 a )and r( t )=( R sin t,R cos t, 0).Therefore, F ( r ( t )) r( t )=1and 'CF d r = 2 0dt =2 Sotheintegralovertheclosedcontourdoesnotvanishdespitethe factthat F = f ,whichseemstobeinconictwiththefundamental 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

PAGE 461

111.FUNDAMENTALTHEOREMFORLINEINTEGRALS455 ittakes two values,0and2 ,atthesamepointonthepositive x axis. Theonlywaytomake f afunctionistoremovethehalf-plane =0 fromthedomainof f .Thinkofacutinspacealongthehalf-plane. But,inthiscase,anyclosedpaththatintersectsthehalf-planebecomes nonclosedasithastwo distinct endpointsontheoppositeedgesofthe cut.Ifthefundamentaltheoremforlineintegralsisappliedtosucha path,thennocontradictionarisesbecausethevaluesof f ontheedges ofthecutdierexactlyby2 infullaccordancewiththeconclusionof thetheorem. Alternatively,theissuecanbeanalyzedbystudyingwhether F is conservativeinitsdomain E .Thevectoreldisdenedeverywherein spaceexcepttheline x = y =0(the z axis).So E isnotsimplyconnected.Therefore,theconditioncurl F = 0 isnotsucienttoclaim thatthevectoreldisconservativeonitsdomain.Indeed,theevaluatedlineintegralalongtheclosedpath(whichcannotbecontinuously contracted,stayingwithin E ,toapointin E )showsthatthevector eldcannotbeconservativeon E .Ifthehalf-plane =0isremoved from E ,then F isconservativeonthisreducedregionbecausethe latterissimplyconnected.Naturally,thelineintegralalonganyclosed paththatdoesnotcrossthehalf-plane =0(i.e.,itlieswithinthe reduceddomain)vanishes. Problem15.3. 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(15.5),assume C consistsofthreestraightlinesegments, ( x0,y0,z0) ( x,y0,z0) ( x,y,z0) ( x,y,z ),asdepictedintheright panelofFigure15.4.Theparametricequationoftherstline 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).Sothelineintegralof F along C1gives thersttermintheaboveexpressionfor f .Similarly,thesecondterm isthelineintegralof F alongthesecondline r ( t )=( x,t,z0),where y0 t y ,sothat r( t )=(0 1 0).Thethirdtermisthelineintegralof 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 )

PAGE 462

45615.VECTORCALCULUS isconservative.Forsimplicity,choose( 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. Problem15.4.(Operator inCurvilinearCoordinates) Letthetransformation ( u,v,w ) ( x,y,z ) beachangeofvariables.If eu, ev,and ewareunitvectorsnormaltothecoordinatesurfaces(see exercise14inSection105.4),showthat = u eu u + v ev v + w ew v Inparticular,ndthe operatorinthecylindricalandsphericalcoordinates. Solution: Bythechainrule, x = u x u + v x v + w x w andsimilarlyfor /y and /z .Then = e1 x + e2 y + e3 z = u x e1+ u y e2+ u z e3 u + v x e1+ v y e2+ v z e3 v + w x e1+ w y e2+ w z e3 w = u u + v v + w w = u eu u + v ev v + w ew v wheretheunitvectorsaredenedin(14.23).Makinguseof(14.24), (14.25),(14.26),and(14.27),theoperator isobtainedinthecylindricalandsphericalcoordinates: = er r + 1 r e + e3 z = e + 1 e + 1 sin e

PAGE 463

111.FUNDAMENTALTHEOREMFORLINEINTEGRALS457 111.6.Exercises.(1) Calculatethecurlofthegivenvectoreld: (i) F =( xyz, y2x, 0) (ii) F =(cos( xz ) sin( yz ) 2) (iii) F =( h ( x ) ,g ( y ) ,h ( z )),wherethefunctions h g ,and h are dierentiable (iv) F =(ln( xyz ) ln( yz ) ln z ) (v) F = a r ,where a isaconstantvectorand r =( x,y,z ) (2) Supposethatavectoreld F ( r )andafunction f ( r )aredierentiable.Showthat ( f F )= f ( F )+ f F (3) Find ( c r f ( r )),where r = r f isdierentiable,and c isa constantvector. (4) Auid,llingtheentirespace,rotatesataconstantrate about anaxisparalleltoaunitvector n .Findthecurlofthevelocityvector eldatagenericpoint r .Assumethatthepositionvector r originates fromapointontheaxisofrotation. (5) Determinewhetherthevectoreldisconservativeand,ifitis,nd itspotential: (i) F =(2 xy,x2+2 yz3, 3 z2y2+1) (ii) F =( yz,xz +2 y cos z,xy y2sin z ) (iii) F =( ey,xey z2, 2 yz ) (iv) F =(6 xy + z4y, 3 x2+ z4x, 4 z3xy ) (v) F =( yz (2 x + y + z ) ,xz ( x +2 y + z ) ,xy ( x + y +2 z )) (vi) F =( y ( x2+ y2) 1+ z,x ( x2+ y2) 1,x ) (vii) F =( y cos( xy ) ,x cos( xy ) ,z + y ) (viii) F =( yz/x2,z/x,y/x ) (6) Determinerstwhetherthevectoreldisconservativeandthen evaluatethelineintegral CF d r : (i) F ( x,y,z )=( y2z2+2 x +2 y, 2 xyz2+2 x, 2 xy2z +1)and C consistsofthreelinesegments:(1 1 1) ( a,b,c ) (1 2 3) (ii) F =( zx,yz,z2)and C isthepartofthehelix r ( t )=(2sin t, 2cos t,t ) thatliesinsidetheellipsoid x2+ y2+2 z2=6 (iii) F =( y z2,x +sin z,y cos z 2 xz )and C isoneturnofa helixofradius a from( a, 0 0)to( a, 0 ,b ). (iv) F = g ( r2) r ,where r =( x,y,z ), r = r g iscontinuous,and C isasmoothcurvefromapointonthesphere x2+ y2+ z2= a2toapointonthesphere x2+ y2+ z2= b2.Whatisthework donebytheforce F if g = 1 /r3? (v) F = 2( y + z )1 / 2, x ( y + z )3 / 2, x ( y + z ) 3 / 2 and C isa smoothcurvefromthepoint(1 1 3)and(2 4 5)

PAGE 464

45815.VECTORCALCULUS (7) Supposethat F and G arecontinuouson E .Showthat (CF d r = (CG d r foranysmoothclosedcurve C in E ifthereisafunction f withcontinuouspartialderivativesin E suchthat F G = f (8) Usethepropertiesofthegradienttoshowthatthevectors er= (cos sin )and e=( sin cos )areunitvectorsorthogonalto thecoordinatecurves r ( x,y )=constand ( x,y )=constofpolar coordinates.Givenaplanarvectoreld,put F = Fr er+ F e.Usethe chainruletoexpressthecurlofaplanarvectoreld F ( r, )inpolar coordinates( r, )asalinearcombinationof erand e. (9) Evaluatethepairwisecrossproductsoftheunitvectors(14.27) andthepairwisecrossproductsoftheunitvectors(14.26).Usethe obtainedrelationsandtheresultofStudyProblem15.4toexpressthe curlofavectoreldinsphericalandcylindricalcoordinates: F = 1 sin (sin F) F e+ 1 1 sin F ( F) e+ 1 ( F) F e, F = 1 r Fz F z er+ Fr z Fz r e+ 1 r ( rF) r Fr ez, wheretheeld F isdecomposedoverthebases(14.27)and(14.26): F = F e+ F e+ F eand F = Fr er+ F e+ Fz ez. Hint: Show e/ = e, e/ =sin e,andsimilarrelationsforthepartial derivativesofotherunitvectors. 112.Green'sTheorem Greenstheoremshouldberegardedasthecounterpartofthefundamentaltheoremofcalculusforthedoubleintegral. Definition 15.8 (OrientationofPlanarClosedCurves). Asimpleclosedcurve C inaplanewhosesingletraversaliscounterclockwise(clockwise)issaidtobepositively(negatively)oriented. Asimpleclosedcurvedividestheplaneintotwoconnectedregions. Ifaplanarregion D isboundedbyasimpleclosedcurve,thenthe positivelyorientedboundaryof D isdenotedbythesymbol D (see theleftpanelofFigure15.5). Recallthatasimpleclosedcurvecanberegardedasacontinuous vectorfunction r ( t )=( x ( t ) ,y ( t ))on[ a,b ]suchthat r ( a )= r ( b )and,

PAGE 465

112.GREEN'STHEOREM459 x y D D x y a b C1C2C3C4y = ytop( x ) y = ybot( x )Figure15.5.Left :Asimpleclosedplanarcurveencloses aconnectedregion D intheplane.Thepositiveorientation oftheboundaryof D meansthattheboundarycurve D istraversedcounterclockwise. Right :Asimpleregion D is boundedbyfoursmoothcurves:twographs C1and C3and twoverticallines C2( x = b )and C4( x = a ).Theboundary D istheunionofthesecurvesorientedcounterclockwise.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, sin t )onthe interval[0 2 ]denethepositivelyandnegativelyorientedcirclesof unitradius,respectively. Theorem 15.5 (GreensTheorem) Let C beapositivelyoriented,piecewise-smooth,simple,closedcurvein theplaneandlet D betheregionboundedby C = D .Ifthefunctions F1and F2havecontinuouspartialderivativesinanopenregionthat contains D ,then D F2 x F1 y dA = 'DF1dx + F2dy. Justlikethefundamentaltheoremofcalculus,Greenstheoremrelatesthederivativesof F1and F2intheintegrandtothevaluesof F1and F2ontheboundaryoftheintegrationregion.AproofofGreens theoremisratherinvolved.Hereitislimitedtothecasewhenthe region D issimple. Proof(forsimpleregions) .Asimpleregion D admitstwoequivalentalgebraicdescriptions: D = { ( x,y ) | ybot( x ) y ytop( x ) ,x [ a,b ] } (15.6) D = { ( x,y ) | xbot( y ) x xtop( y ) ,y [ c,d ] } (15.7)

PAGE 466

46015.VECTORCALCULUS Theideaoftheproofistoestablishtheequalities (15.8) 'DF1dx = DF1 y dA, 'DF2dy = DF2 x dA using,respectively,(15.6)and(15.7).Theconclusionofthetheoremis thenobtainedbyaddingtheseequations. Thelineintegralistransformedintoanordinaryintegralrst.The boundary D containsfourcurves,denoted C1, C2, C3,and C4(see therightpanelofFigure15.5).Thecurve C1isthegraph y = ybot( x ) whoseparametricequationsare r =( t,ybot( t )),where t [ a,b ].So C1istracedoutfromlefttorightasrequiredbythepositiveorientation of D .Thecurve C3isthetopboundary y = ytop( x ),and,similarly, itsparametricequations r ( t )=( t,ytop( t )),where t [ a,b ].Thisvector functiontraverses C3fromlefttoright.Sotheorientationof C3must bereversedtoobtainthecorrespondingpartof D .Theboundary curves C2and C4(thesidesof D )aresegmentsoftheverticallines x = b (orientedupward)and x = a (orienteddownward),whichmay collapsetoasinglepointifthegraphs y = ybot( x )and y = ytop( x ) intersectat x = a or x = b orboth.Thelineintegralsalong C2and C4donotcontributetothelineintegralwithrespectto x along D because dx =0along C2and C4.Byconstruction, x = t and dx = dt forthecurves C1and C2.Hence, 'DF1dx = C1F1dx + C2F1dx = b a F ( x,ybot( x )) F ( x,ytop( x )) dx, wheretheproperty(15.3)hasbeenused.Next,thedoubleintegralis transformedintoanordinaryintegralbyconvertingittoaniterated integral: DF1 y dA = b aytop( x ) ybot( x )F1 y dydx = b a F ( x,ytop( x )) F ( x,ybot( x )) dx, wherethelatterequalityfollowsfromthefundamentaltheoremof calculusandthecontinuityof F1onanopenintervalthatcontains [ ybot( x ) ,ytop( x )]forany x [ a,b ](thehypothesisofGreenstheorem). Comparingtheexpressionofthelineanddoubleintegralsviaordinary integrals,thevalidityoftherstrelationin(15.8)isestablished.The

PAGE 467

112.GREEN'STHEOREM461 x D C C x y C2C2D D D D C1C1Figure15.6.Left :Aregion D issplitintotworegions byacurve C .Iftheboundaryoftheupperpartof D has positiveorientation,thenthepositivelyorientedboundaryof thelowerpartof D hasthecurve C Right :Greenstheoremholdsfornonsimplyconnectedregions.Theorientation oftheboundariesofholesin D isobtainedbymakingcuts alongcurves C1and C2sothat D becomessimplyconnected. Thepositiveorientationoftheouterboundaryof D induces theorientationoftheboundariesoftheholes.secondequalityin(15.8)isprovedanalogouslybyusing(15.7).The detailsareomitted. Supposethatasmooth,orientedcurve C dividesaregion D into two simple regions D1and D2(seetheleftpanelofFigure15.6).If theboundary D1contains C (i.e.,theorientationof C coincideswith thepositiveorientationof D1),then D2mustcontainthecurve C andviceversa.Usingtheconventionalnotation F1dx + F2dy = F d r where F =( F1,F2),oneinfersthat 'DF d r = 'D1F d r + 'D2F d r = D1 F2 x F1 y dA + D2 F2 x F1 y dA = D F2 x F1 y dA. Therstequalityholdsbecauseofthecancellationofthelineintegralsalong C and C accordingto(15.3).Thevalidityofthesecond equalityfollowsfromtheproofofGreenstheoremforsimpleregions. Finally,theequalityisestablishedbytheadditivitypropertyofdoubleintegrals.Bymakinguseofsimilararguments,theproofcanbe extendedtoaregion D thatcanberepresentedastheunionofanite numberofsimpleregions.

PAGE 468

46215.VECTORCALCULUS Green'sTheoremforNonsimplyConnectedRegions.Lettheregions D1and D2beboundedbysimple,piecewise-smooth,closedcurvesandlet D2lieintheinteriorof D1(seetherightpanelofFigure15.6).Consider theregion D thatwasobtainedfrom D1byremoving D2(theregion D hasaholeoftheshape D2).MakinguseofGreenstheorem,one nds D F2 x F1 y dA = D1 F2 x F1 y dA D2 F2 x F1 y dA = 'D1F d r 'D2F d r = 'D1F d r + ' D2F d r = 'DF d r (15.9) ThisestablishesthevalidityofGreenstheoremfornotsimplyconnectedregions.Theboundary D consistsof D1and D2;thatis, theouterboundaryhasapositiveorientation,whiletheinnerboundary isnegativelyoriented.AsimilarlineofreasoningleadstotheconclusionthatGreenstheoremholdsforanynumberofholesin D :allinner boundariesof D mustbenegativelyoriented.Suchorientationofthe boundariescanalsobeunderstoodasfollows.Letacurve C connecta pointoftheouterboundarywithapointoftheinnerboundary.Letus makeacutoftheregion D along C .Thentheregion D becomessimply connected,and D consistsofa continuous curve(theinnerandouter boundaries,andthecurves C and C ).Theboundary D canalways bepositivelyoriented.Thelatterrequiresthattheouterboundary betracedcounterclockwise,whiletheinnerboundaryistracedclockwise(theorientationof C and C ischosenaccordingly).Byapplying Greenstheoremto D ,onecanseethatthelineintegralsover C and C arecancelledand(15.9)followsfromtheadditivityofthedouble integral.112.1.EvaluatingLineIntegralsviaDoubleIntegrals.Greenstheorem providesatechnicallyconvenienttooltoevaluatelineintegralsalong planarclosedcurves.Itisespeciallybenecialwhenthecurveconsists ofseveralsmoothpiecesthataredenedbydierentvectorfunctions; thatis,thelineintegralmustbesplitintoasumoflineintegralstobe convertedintoordinaryintegrals.Sometimes,thelineintegralturns outtobemuchmorediculttoevaluatethanthedoubleintegral. Example 15.5 Evaluatethelineintegralof F =( y2+ ecos x, 3 xy sin( y4)) alongthecurve C thatistheboundaryofthehalfofthering: 1 x2+ y2 4 and y 0 ; C isorientedclockwise.

PAGE 469

112.GREEN'STHEOREM463 2 1 12 1 2 D Pi +1PnCiCnPiP1Figure15.7.Left :TheintegrationcurveinthelineintegraldiscussedinExample15.5. Right :Ageneralpolygon. ItsareaisevaluatedinExample15.7byrepresentingthearea viaalineintegral.Thecurve C consistsoffoursmoothpieces,thehalf-circlesofradii1 and2andtwostraightlinesegmentsofthe x axis,[ 2 1]and[1 2],as shownintheleftpanelofFigure15.7.Eachcurvecanbeeasilyparameterized,andthelineintegralinquestioncanbetransformedintothe sumoffourordinaryintegrals,whicharethenevaluated.Thereader isadvisedtopursuethisavenuetoappreciatethefollowingalternative basedonGreenstheorem(thisisnotimpossibletoaccomplishifone guresouthowtohandletheintegrationofthefunctions ecos xand sin( y4)whoseantiderivativesarenotexpressibleinelementaryfunctions). Solution: Thecurve C isasimple,piecewise-smooth,closedcurve, andthecomponentsof F havecontinuouspartialderivativeseverywhere.Thus,Greenstheoremappliesif D = C (becausetheorientationof C isnegative)and D isthehalf-ring.Onehas F1/y =2 y and F2/x =3 y .ByGreenstheorem, 'CF d r = 'DF d r = D F2 x F1 y dA = DydA = 02 1r sin rdrd = 0sin d 2 1r2dr = 14 3 wherethedoubleintegralhasbeentransformedtopolarcoordinates. Theregion D istheimageoftherectangle D=[1 2] [0 ]inthe polarplaneunderthetransformation( r, ) ( x,y ). ChangingtheIntegrationCurveinaLineIntegral.Ifaplanarvectoreld isnotconservative,thenitslineintegralalongacurve C originating fromapoint A andterminatingatapoint B dependson C .If Cisanothercurveoutgoingfrom A andterminatingat B ,whatisthe relationbetweenthelineintegralsof F over C and C?Greenstheorem

PAGE 470

46415.VECTORCALCULUS allowsustoestablishsucharelation.Supposethat C and Chaveno self-intersectionsanddonotintersect.Thentheirunionisaboundary ofasimplyconnectedregion D .Letusreversetheorientationofone ofthecurvessothattheirunionisthepositivelyorientedboundary D ,where D istheunionof C and C.Then 'DF d r = CF d r + CF d r = CF d r CF d r ByGreenstheorem, (15.10) CF d r = CF d r + D F2 x F1 y dA, whichestablishestherelationbetweenlineintegralsofanonconservativeplanarvectoreldovertwodierentcurvesthathavecommon endpoints. Example 15.6 Evaluatethelineintegralofthevectoreld F = (2 y +cos( x2) ,x2+ y3) overthecurve C ,whichconsistsoftheline segments (0 0) (1 1) and (1 1) (0 2) Solution: Let Cbethelinesegment(0 0) (0 2).Thentheunion of C and Cistheboundary D (positivelyoriented)ofthetriangular region D withvertices(0 0),(1 1),and(0 2).Therelation(15.10) canbeappliedtoevaluatethelineintegralover C .Theparametric equationsof Care x =0, y = t ,0 t 2.Hence,along C, F d r = F2(0 ,t ) dt = t3dt and CF d r = 2 0t3dt =4 Then F2/x =2 x and F1/y =2.Theregion D admitsanalgebraic descriptionasaverticallysimpleregion: x y 2 x ,0 x 1. Hence, D F2 x F1 y dA = D(2 x 2) dA =2 1 0( x 1) 2 x xdy = 4 1 0( x 1)2dx = 4 3 Therefore,by(15.10), CF d r =4+ 4 3 = 16 3

PAGE 471

112.GREEN'STHEOREM465 112.2.AreaofaPlanarRegionasaLineIntegral.Put F2= x and F1= 0.Then D F2 x F1 y dA = DdA = A ( D ) Thearea A ( D )canalsobeobtainedif F =( y, 0)or F =( y/ 2 ,x/ 2). ByGreenstheorem,theareaof D canbeexpressedbylineintegrals: (15.11) A ( D )= 'Dxdy = 'Dydx = 1 2 'Dxdy ydx, assuming,ofcourse,thattheboundaryof D isasimple,piecewisesmooth,closedcurve(orseveralsuchcurvesif D hasholes).Thereason thevaluesoftheselineintegralscoincideissimple.Thedierence ofanytwovectoreldsinvolvedisthegradientofafunctionwhose lineintegralalongaclosedcurvevanishesowingtothefundamental theoremforlineintegrals.Forexample,for F =(0 ,x )and G ( y, 0), thedierenceis F G =( y,x )= f ,where f ( x,y )= xy ,sothat 'DF d r 'DG d r = 'D( F G ) d r = 'D f d r =0 Therepresentation(15.11)oftheareaofaplanarregionastheline integralalongitsboundaryisquiteusefulwhentheshapeof D istoo complicatedtobecomputedusingadoubleintegral(e.g.,when D isnot simpleand/orarepresentationofboundariesof D bygraphsbecomes technicallydicult). Example 15.7 (AreaofaPolygon) Consideranarbitrarypolygonwhoseverticesincounterclockwiseorder are ( x1,y1) ( x2,y2) ,..., ( xn,yn) .Finditsarea. Solution: Evidently,agenericpolygonisnotasimpleregion(e.g.,it mayhaveastarlikeshape).Sothedoubleintegralisnotatallsuitable forndingthearea.Incontrast,thelineintegralapproachseemsfar morefeasibleastheboundaryofthepolygonconsistsof n straightline segmentsconnectingneighboringverticesasshownintherightpanelof Figure15.7.If Ciissuchasegmentorientedfrom( xi,yi)to( xi +1,yi +1) for i =1 2 ,...,n 1,then Cngoesfrom( xn,yn)to( x1,y1).Avector functionthattracesoutastraightlinesegmentfromapoint ratoa point rbis r ( t )= ra+( rb ra) t ,where0 t 1.Forthesegment Ci, take ra=( xi,yi)and rb=( xi +1,yi +1).Hence, x ( t )= xi ( xi +1 xi) t = xi+ xit and y ( t )= yi+( yi +1 yi) t = yi+ yit .Forthevectoreld F =( y,x )on Ci,onehas F ( r ( t )) r( t )=( y ( t ) ,x ( t )) ( xi, yi)= xi yi yi xi= xiyi +1 yixi +1;

PAGE 472

46615.VECTORCALCULUS thatis,the t dependencecancelsout.Therefore,takingintoaccount that Cngoesfrom( xn,yn)to( x1,y1),theareais A = 1 2 'Dxdy ydx = 1 2ni =1Cixdy ydx = 1 2n 1i =11 0( xiyi +1 yixi +1) dt + 1 2 1 0( xny1 ynx1) dt = 1 2 n 1i =1( xiyi +1 yixi +1)+( xny1 ynx1) SoGreenstheoremoersanelegantwaytondtheareaofageneralpolygonifthecoordinatesofitsverticesareknown.Asimple, piecewise-smooth,closedcurve C inaplanecanalwaysbeapproximatedbyapolygon.Theareaoftheregionenclosedby C cantherefore beapproximatedbytheareaofapolygonwithalargeenoughnumber ofvertices,whichisoftenusedinmanypracticalapplications.112.3.TheTestforPlanarVectorFieldstoBeConservative.GreenstheoremcanbeusedtoproveTheorem15.4forplanarvectorelds.Consideraplanarvectoreld F =( F1( x,y ) ,F2( x,y ) 0).Itscurlhasonly onecomponent: F =det e1 e2 e3 x y z F1( x,y ) F2( x,y )0 # = e3 F2 x F1 y Supposethatthecurlof F vanishesthroughoutasimplyconnected openregion D F = 0 .Bydenition,anysimpleclosedcurve C inasimplyconnectedregion D canbeshrunktoapointof D while remainingin D throughoutthedeformation(i.e.,anysuch C bounds asubregion Dsof D ).ByGreenstheorem,where C = Ds, 'CF d r = Ds F2 x F1 y dA = Ds0 dA =0 foranyclosedsimplecurve C in D .Bythepathindependenceproperty, thevectoreld F isconservativein D .

PAGE 473

112.GREEN'STHEOREM467 x y D D y = x2x = y21 1 x y C Caa DaFigure15.8.Left :AnillustrationtoStudyProblem15.5. Right :AnillustrationtoStudyProblem15.6.Theregion Daisboundedbyacurve C andthecircle Ca.112.4.StudyProblems.Problem15.5. Evaluatethelineintegralof F =( y + ex2, 3 x sin( y2)) alongthecounterclockwise-orientedboundaryof D thatisenclosedby theparabolas y = x2and x = y2. Solution: Onehas F1/y =1and F2/x =3.ByGreens theorem, 'DF d r = D2 dA =2 1 0 x x2dydx =2 1 0( x x2) dx = 2 3 Theintegrationregion D isshownintheleftpanelofFigure5.8. Problem15.6. Provethatthelineintegraloftheplanarvectoreld F = y x2+ y2, x x2+ y2 alonganypositivelyoriented,simple,smooth,closedcurve C thatencirclestheoriginis 2 andthatitvanishesforanysuchcurvethatdoes notencircletheorigin. Solution: Ithasbeenestablished(seeStudyProblem15.2)that thecurlofthisvectoreldvanishesinthedomainthatistheentire planewiththeoriginremoved.If C doesnotencircletheorigin,then F2/x F1/y =0throughouttheregionencircledby C ,andthe lineintegralalong C vanishesbyGreenstheorem.Givenaclosedcurve C thatencirclestheorigin,butdoesnotgothroughit,onecanalways ndadiskofasmallenoughradius a suchthatthecurve C doesnot intersectit.Let Dabetheregionboundedbythecircle Caofradius a andthecurve C .Then F2/x F1/y =0throughout Da.Let C beorientedcounterclockwise,while Caisorientedclockwise.Then

PAGE 474

46815.VECTORCALCULUS Daistheunionof C and Ca.ByGreenstheorem, 'DF d r =0 'CF d r = 'CaF d r = ' CaF d r =2 because Caisthecircleorientedcounterclockwiseandforsuchacircle thelineintegralhasbeenfoundtobe2 (seeStudyProblem15.2). Problem15.7.(VolumeofAxiallySymmetricSolids) Let D bearegionintheupperpartofthe xy plane( y 0 ).Consider thesolid E obtainedbyrotationof D aboutthe x axis.Showthatthe volumeofthesolidisgivenby V ( E )= 'Dy2dx. Solution: Let dA betheareaofapartitionelementof D thatcontains apoint( x,y ).Ifthepartitionelementisrotatedaboutthe x axis,the point( x,y )traversesthecircleofradius y (thedistancefromthepoint ( x,y )tothe x axis).Thelengthofthecircleis2 y .Consequently,the volumeofthesolidringsweptbythepartitionelementis dV =2 ydA Takingthesumoverthepartitionof D ,thevolumeisexpressedvia thedoubleintegralover D : V ( E )=2 DydA. InGreenstheorem,put F1/y =2 y and F2/x =0sothatthe abovedoubleintegralisproportionaltotheleftsideofGreensequation.Inparticular, F1= y2and F2=0satisfytheseconditions.By Greenstheorem, V ( E )= DF1 y dA = 'DF1dx = 'Dy2dx asrequired. 112.5.Exercises.(1) Evaluatethelineintegralbytwomethods:(a)directlyand(b) usingGreenstheorem: (i) (Cxy2dx y2xdy ,where C isthetrianglewithvertices(0 0), (1 0),and(1 2); C isorientedcounterclockwise (ii) (C2 yxdx + x2dy ,where C consistsofthelinesegmentsfrom (0 1)to(0 0)andfrom(0 0)to(1 0)andtheparabola y = 1 x2from(1 0)to(0 1) (2) EvaluatethelineintegralusingGreenstheorem:

PAGE 475

112.GREEN'STHEOREM469 (i) ( x sin( x2) dx +( xy2 x8) dy ,where C isthepositivelyoriented boundaryoftheregionbetweentwocircles x2+ y2=1and x2+ y2=4 (ii) (C( y3dx x3dy ),where C isthepositivelyorientedcircle x2+ y2= a2(iii) (C( x + y3) dx +( x2+ y ) dy ,where C consistsofthearc ofthecurve y =cos x from( / 2 0)to( / 2 0)andtheline segmentfrom( / 2 0)to( / 2 0) (iv) (C( y4 ln( x2+ y2)) dx +2tan 1( y/x ) dy ,where C isthepositivelyorientedcircleofradius a> 0withcenter( x0,y0)such that x0>a and y0>a (v) (C( x + y )2dx ( x2+ y2) dy ,where C isapositivelyoriented trianglewithvertices(1 1),(3 2),and(2 5) (vi) (Cxy2dx x2ydy ,where C isthenegativelyorientedcircle x2+ y2= a2(vii) (C( x + y ) dx ( x y ) dy ,where C isthepositivelyoriented ellipse( x/a )2+( y/b )2=1 (viii) (Cex[(1 cos y ) dx ( y sin y ) dy ],where C isthepositively orientedboundaryoftheregion0 x ,0 y sin x (ix) (Ce x2+ y2[cos(2 xy ) dx sin(2 xy ) dy ],where C isthepositively orientedcircle x2+ y2= a2(3) Usethecontourtransformationlaw(15.10)tosolvethefollowing problems: (i)Let F =(( x + y )2, ( x y )2).Findthedierencebetweenthe lineintegralsof F over C1,whichisthestraightlinesegment (1 1) (2 6),andover C2whichistheparabolawiththe verticalaxisandpassingthrough(1 1),(2 6),and(0 0). (ii) C( exsin y qx ) dx +( excos y q ) dy ,where q isaconstant and C istheupperpartofthecircle x2+ y2= ax oriented from( a, 0)to(0 0). (iii) C[ g ( y ) ex qy ] dx +[ g( y ) ex q ] dy ,where g ( y )and g( y )are continuousfunctionsand C isasmoothcurvefromthepoint P1=( x1,y1)tothepoint P2=( x2,y2)suchthatitandthe straightlinesegment P1P2formaboundaryofaregion D of thearea A ( D ). (4) UseGreenstheoremtondtheworkdonebytheforce F =(3 xy2+ y3,y4)inmovingaparticlealongthecircle x2+ y2= a2from(0 a ) to(0 ,a )counterclockwise. (5) Usearepresentationoftheareaofaplanarregionbytheline integraltondtheareaofthespeciedregion D :

PAGE 476

47015.VECTORCALCULUS (i) D isboundedbyanellipse x = a cos t y = b sin t ,0 t 2 (ii) D isunderonearcofthecycloid x = a ( t sin t ), y = a (1 cos t ) (iii) D isthe astroid enclosedbythecurve x = a cos3t y = a sin3t (iv) D isboundedbythecurve x ( t )= a cos2t y ( t )= b sin(2 t )for t [0 ] (v) D isboundedbytheparabola( x + y )2= ax andbythe x axis, a> 0 (vi) D isboundedbyoneloopofthecurve x3+ y3=3 axy a> 0 ( Hint: Put y = tx .) (vii) D isboundedbythecurve( x2+ y2)2= a2( x2 y2)( Hint: Put y = x tan t .) (viii) D isboundedby( x/a )n+( y/b )n=1, n> 0( Hint: x = a cosn/ 2t y = b sinn/ 2t .) (6) Letacurve C havexedendpoints.Underwhatconditiononthe function g ( x,y )isthelineintegral Cg ( x,y )( ydx + xdy )independent of C ? (7) Let D beaplanarregionboundedbyasimpleclosedcurve.If A istheareaof D ,showthatthecoordinates( xc,yc)ofthecentroidof D are xc= 1 2 A 'Dx2dy,yc= 1 2 A 'Dy2dx. Hint: Useanapproachsimilartothederivationof(15.11). (8) Letalaminawithaconstantsurfacemassdensity occupya planarregion D enclosedbyasimplepiecewisesmoothcurve.Show thatitsmomentsofinertiaaboutthe x and y axesare Ix= 3 'Dy3dx,Iy= 3 'Dx3dy. Hint: Useanapproachsimilartothederivationof(15.11). 113.FluxofaVectorField Theideaofauxofavectoreldstemsfromanengineeringproblemofmasstransferacrossasurface.Supposethereisaowofauid orgaswithaconstantvelocity v andaconstantmassdensity (mass perunitvolume).Let A beaplanarareaelementplacedintothe ow.Atwhatrateistheuidorgascarriedbytheowacrossthearea A ?Inotherwords,whatisthemassofuidtransferredacross A perunittime?Thisquantityiscalleda ux ofthemassowacross thearea A Supposerstthatthemassowisnormaltotheareaelement. Considerthecylinderwithanaxisparallelto v withcross-sectional

PAGE 477

113.FLUXOFAVECTORFIELD471 h A n vnv F ( r i) S Si nir iFigure15.9.Left :Amasstransferredbyahomogeneous massowwithaconstantvelocity v acrossanareaelement A intime t is m = V ,where V = h A isthevolumeofthecylinderwithcross-sectionalarea A andheight h = tvn; vnisthescalarprojectionof v ontothenormal n Right :Apartitionofasmoothsurface S byelements Si. If r iisasamplepointin Si, niistheunitnormalto S at r i, and Siisthesurfaceofthepartitionelement,thentheux ofacontinuousvectoreld F ( r )across Siisapproximated byi= F ( r i) ni Si.area A andheight h = v t ,where v = v istheowspeedand t is atimeinterval.Thevolumeofthecylinderis V = h A = v t A Intime t ,allthemassstoredinthiscylinderistransferredbythe owacross A .Thismassis m = V = v t A ,andtheuxis = m t = v A. Theuxdependsontheorientationofanareaelementrelativeto theow.Iftheowisparalleltotheareaelement,thennomassis transferredacrossit.Thevelocityvectorcanbeviewedasthesumof avectornormaltotheareaelementandavectortangentialtoit.Only thenormalcomponentoftheowcontributestotheux.If n isthe unitnormalvectortotheareaelementand istheanglebetween v and n ,thenthenormalcomponentofthevelocityis vn= v cos = v n and(seetheleftpanelofFigure15.9) (15.12)= vn A = v n A = F n A = Fn A, wherethevector F = v characterizesthemassow(howmuch ( )andhowfast( v ))and Fnisitscomponentnormaltothearea element. Ifthemassowisnotconstant(i.e., F becomesavectoreld),then itsuxacrossasurface S canbedenedbypartitioning S intosmall surfaceareaelements Si, i =1 2 ,...,N ,whosesurfaceareasare Si

PAGE 478

47215.VECTORCALCULUS asshownintherightpanelofFigure15.9.Let r ibeasamplepoint in Siandlet nibetheunitvectornormalto Siat r i.Ifthesize(the radiusofthesmallestballcontaining Si)issmall,then,byneglecting variationsof F andthenormal n within Si,theuxacross Sicanbe approximatedby(15.12),i F ( r i) ni Si.Theapproximation becomesbetterwhen N sothatthesizesof Sidecreaseto0 uniformlyandhencethetotaluxis =limN Ni =1i=limN Ni =1F ( r i) ni Si=limN Ni =1Fn( r i) Si. ThesuminthisequationisnothingbuttheRiemannsumofthefunction Fn( r )overapartitionofthesurface S .Naturally,itslimitisthe surfaceintegralof Fn( r )over S .Thus, theuxofavectoreldacross asurfaceisthesurfaceintegralofthenormalcomponentofthevector eld .113.1.OrientableSurfaces.Theabovedenitionoftheuxsounds ratherplausible.However,itcontainsatacitassumptionthatthe normalcomponentofavectoreldcanalwaysbe uniquely denedas acontinuousfunctiononasmoothsurface.Itappearsthatthereare smoothsurfacesforwhichthiscannotbedone! Thenormal n = n ( r )dependsonthepointofasurface.Soit isavectoreldon S .Inorderforthenormalcomponent Fntobe uniquelydened,therule n = n ( r )shouldassignjustone n forevery pointof S .Furthermore, n ( r )shouldbecontinuouson S andhence alongeveryclosedcurve C inasmoothsurface S .Inotherwords,if n istransportedalongaclosedcurve C in S ,theinitial n mustcoincide withthenal n asillustratedintheleftpanelofFigure15.10.Since, ateverypointof S ,thereareonlytwopossibilitiestodirecttheunit normalvector,bycontinuitythedirectionof n ( r )denesonesideof S whilethedirectionof n ( r )denestheotherside.Thus,thenormal componentofavectoreldiswelldenedfortwo-sidedsurfaces.For example,theoutwardnormalofasphereiscontinuousalonganyclosed curveonthesphere(itremainsoutwardalonganyclosedcurve)and hencedenestheoutersideofthesphere.Ifthenormalonthesphere ischosentobeinward,thenitisalsocontinuousanddenestheinner sideofthesphere. Arethereone-sidedsurfaces?Ifsuchasurfaceexists,itshouldhave quiteremarkableproperties.Takeapointonit.Inaneighborhoodof thispoint,onealwaysthinksabouttwosides(asurfaceissmooth). Onesideisdenedbyanormal n (face-uppatch),whiletheother

PAGE 479

113.FLUXOFAVECTORFIELD473 n C S n S S nFigure15.10.Left :Ifthereisacontinuousunitnormal vector n onasurface S ,thenwhen n istransportedalong aclosedcurvein S ,itsinitialdirectionshouldmatchthe naldirection. Right :Asmallpatch S ofasurface S can beorientedintwodierentwaysaccordingtotwopossible choicesoftheunitnormalvector n or n .Ifthereisa one-sidedsurface,theface-uppatchcanbetransported alongaclosedcurvein S totheface-downpatchatthesame positionon S .If S hasaboundary,thentheclosedcurveis notallowedtocrosstheboundary.hasthesameshapebutitsnormalis n (face-downpatch)asshown intherightpanelofFigure15.10.Foraone-sidedsurface,thefaceupandface-downpatchesmustbeonthesamesideofthesurface. Thisimpliesthatthereshouldexistaclosedcurveonthesurfacethat startsatapointononesideandcanreachtheverysamepointbut fromtheotherside without crossingthesurfaceboundaries(ifany)or piercingthesurface.Bymovingtheface-uppatchalongsuchacurve, itbecomestheface-downpatch.Thus,thenormalcannotbeuniquely denedonaone-sided S .ExamplesofOne-SidedSurfaces.One-sidedsurfacesdoexist.Toconstructanexample,takearectangularpieceofpaper.Putupwardarrowsonitsverticalsidesandgluethesesidessothatthearrowsremain parallel.Indoingso,acylinderisobtained,whichisatwo-sidedsurface(thereisnocurvethattraversesfromonesidetotheotherwithout crossingtheboundarycirclesformedbythehorizontalsidesoftherectangle).Thegluingcanbedonedierently.Beforegluingthevertical sides,twisttherectanglesothatthearrowsonthembecomeopposite andthengluethem.TheprocedureisshowninFigure15.11.The resultingsurfaceisthefamous M¨ obiusstrip (namedaftertheGerman mathematicianAugustM¨ obius).Itisone-sided.Allcurveswinding

PAGE 480

47415.VECTORCALCULUS n n gluing n n gluing nnFigure15.11.Constructionofatwo-sidedsurface(aportionofacylinder)fromabandbygluingitsedges(left). Constructionofaone-sidedsurface(aM¨ obiusstrip)froma bandbygluingitsedgesaftertwistingtheband(right).aboutittraversebothsidesofthegluedrectanglewithoutcrossingits boundaries(thehorizontaledges).Aface-uppatchcantransported intoaface-uppatchatthesamepositionalongaclosedcurveinthe band. Thereareone-sidedsurfaceswithoutboundaries(likeasphere). Themostfamousoneisa Kleinbottle .Takeabottle.Drillaholeon thesidesurfaceandinthebottomofthebottle.Supposetheneck ofthebottleisexible(arubberbottle).Benditsneckandpullit throughtheholeonthebottlessidesurface(sothatnecktstightly intothehole).Finally,attachtheedgeofthebottlesnecktothe edgeoftheholeinthebottlebottom.Theresultisasurfacewithout boundariesanditisone-sided.Abugcancrawlalongthissurfaceand getinandoutofthebottle.FluxandOne-SidedSurfaces.Theuxmakessenseonlyfortwo-sided surfaces.Indeed,theuxmeansthatsomethingisbeingtransferred fromonesidetotheothersideofthesurface(i.e., across it)atacertain rate.Ifthesurfaceisone-sided,thenonecangetacrossitbymerely slidingalongit!Forexample,amassow tangential toaone-sided surfacecantransfermassacrossthesurface.

PAGE 481

113.FLUXOFAVECTORFIELD475 x y z n S DFigure15.12.Left :AKleinbottleisanexampleofa one-sidedclosedsurface(ithasnoboundaries). Right :An illustrationtoExample15.8.Definition 15.9 (OrientableSurface) Asmoothsurfaceiscalled orientable ifthereisnoclosedcurveinit suchthatthenormalvectorisreversedwhenmovedaroundthiscurve. Soorientablesurfacesaretwo-sidedsurfaces.Theuxofavector eldcanonlybedenedacrossanorientablesurface.113.2.FluxasaSurfaceIntegral.Definition 15.10 (FluxofaVectorField) Let S beanorientablesmoothsurfaceandlet n betheunitnormal vectoron S .Theuxofavectoreld F across S isthesurfaceintegral = SF n dS, providedthenormalcomponent F n ofthevectoreldisintegrable on S Theintegrabilityofthenormalcomponent Fn( r )= F n isdened inthesenseofsurfaceintegralsofordinaryfunctions(seeDenition 14.22).Inparticular,theuxofacontinuousvectoreldacrossa smoothorientablesurfaceexists.113.3.EvaluationoftheFluxofaVectorField.Supposethatasurface S isagraph z = g ( x,y )and g hascontinuouspartialderivativesin aregion( x,y ) D .Therearetwopossibleorientationsof S .The normalvectortothetangentplaneatapointof S is n =( g x, g y, 1)

PAGE 482

47615.VECTORCALCULUS (seeSection90).Its z componentispositive.Forthisreason,the graphissaidtobe orientedupward .Alternatively,onecantakethe normalvectorintheoppositedirection, n =( g x,g y, 1).Inthiscase, thegraphissaidtobe orienteddownward .Accordingly,the upward ( downward )ux,denoted(),ofavectoreldisassociatedwith theupward(downward)orientationofthegraph.Whentheorientation ofasurfaceisreversed,theuxchangesitssign: = Considertheupward-orientedgraph z = g ( x,y ).Theunitnormal vectorreads n = 1 n n = 1 J ( g x,g y, 1) ,J = 1+( g x)2+( g y)2. InSection108.1,itwasestablishedthattheareaoftheportionofthe graphaboveaplanarregionofarea dA is dS = JdA .Therefore,in theinnitesimaluxacrossthesurfacearea, dS canbewritteninthe form F n dS = F n 1 J JdA = F n dA, wherethevectoreldmustbeevaluatedon S ,thatis, F = F ( x,y,g ( x,y ))(thevariable z isreplacedby g ( x,y )because z = g ( x,y ) foranypoint( x,y,z ) S ).Ifthevectoreld F iscontinuous,thenthe dotproduct F n isacontinuousfunctionon D sothattheuxexists andisgivenbythedoubleintegralover D .Thefollowingtheoremhas beenproved. Theorem 15.6 (EvaluationoftheFluxAcrossaGraph) Supposethat S isagraph z = g ( x,y ) ofafunction g whoserst-order partialderivativesarecontinuouson D .Let S beorientedupwardby thenormalvector n =( g x, 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 )= g xF1( x,y,g ) g yF2( x,y,g )+ F3( x,y,g ) Theevaluationofthesurfaceintegralinvolvesthefollowing steps: Step1 .Represent S asagraph z = g ( x,y )(i.e.,ndthefunction g usingageometricaldescriptionof S ).If S cannotberepresentedasa graphofasinglefunction,thenithastobesplitintopiecessothat eachpiececanbedescribedasagraph.Bytheadditivityproperty,the

PAGE 483

113.FLUXOFAVECTORFIELD477 surfaceintegralover S isthesumofintegralsovereachpiece. Step2 .Findtheregion D thatdenesthepartofthegraphthat coincideswith S (if S isnotthegraphonthewholedomainof g ).One canthinkof D astheverticalprojectionof S ontothe xy plane. Step3 .Determinetheorientationof S (upwardordownward)from theproblemdescription.Thesignoftheuxisdeterminedbytheorientation.Calculatethenormalcomponent Fn( x,y )ofthevectoreld asafunctionon D Step4 .Evaluatethedoubleintegralof Fnover D Example 15.8 Evaluatethedownwarduxofthevectoreld F = ( xz,yz,z ) acrossthepartoftheparaboloid z =1 x2 y2intherst octant. Solution: Thesurfaceisthepartofthegraph z = g ( x,y )=1 x2 y2intherstoctant.Theparaboloidintersectsthe xy plane( z =0)along thecircle x2+ y2=1.Therefore,theregion D isthequarterofthe diskboundedbythiscircleintherstquadrant( x,y 0).Since S is orienteddownward, n =( g x,g y, 1)=( 2 x, 2 y, 1)andthenormal componentof F is Fn( x,y )=( xg,yg,g ) ( 2 x, 2 y, 1)= (1 x2 y2)(1+2 x2+2 y2) Convertingthedoubleintegralof Fntopolarcoordinates, = DFn( x,y ) dA = / 2 01 0(1+ r2)(1+2 r2) rdrd = 19 24 Thenegativevalueofthedownwarduxmeansthattheactualtransfer ofaquantity(likeamass),whoseowisdescribedbythevectoreld F ,occursintheupwarddirectionacross S 113.4.ParametricSurfaces.Ifthesurface S intheuxintegralisdenedbytheparametricequations r = r ( u,v ),where( u,v ) D ,then, byTheorem14.23,thenormalvectorto S is n = r u r v(or n ;the signischosenaccordingtothegeometricaldescriptionoftheorientationof S ).Since n = J ,where J determinestheareatransformation law dS = JdA ( dA = dudv ),theuxofavectoreld F acrossthe surfacearea 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, andtheuxisgivenbythedoubleintegral = FF n dS = DF ( r ( u,v )) ( r u r v) dA = DFn( u,v ) dA.

PAGE 484

47815.VECTORCALCULUS 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 15.9 Evaluatetheoutwarduxofthevectoreld F = ( z2x,z2y,z3) acrossthesphereofunitradiuscenteredattheorigin. Solution: Theparametricequationsofthesphereofradius R =1are givenin(14.36),andthenormalvectoriscomputedinExample14.41: 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.Theoutwarduxreads = SF n dS = Dcos2u sin udA = 2 0dv 0cos2u sin udu = 4 3 NonorientableParametricSurfaces.Nonorientablesmoothsurfacescan bedescribedbytheparametricequations r = r ( u,v )orbyanalgebraic equation F ( x,y,z )=0(asalevelsurfaceofafunction).Forexample, aM¨ obiusstripofwidth2 h withmidcircleofradius R andheight z =0 isdenedbytheparametricequations (15.13) r ( u,v )= [ R + u cos( v/ 2)]cos v, [ R + u cos( v/ 2)]sin v,u sin( v/ 2) where( u,v ) D =[ h,h ] [0 2 ].ItalsofollowsfromtheseparametricequationsthattheM¨ obiusstripisdenedbya cubic surface: R2y + x2y + y3 2 Rxz 2 x2z 2 y2z + yz3=0 Thisisveriedbysubstitutingtheparametricequationsintothisalgebraicequationandshowingthattheleftsidevanishesforall( u,v ) D Letusprovethatthesurfacedenedbytheparametricequations (15.13)isnotorientable.Todoso,oneshouldanalyzethebehaviorof anormalvectorwhenthelatterismovedaroundaclosedcurveinthe

PAGE 485

113.FLUXOFAVECTORFIELD479 surface.Considerthecircleinthe xy planedenedbythecondition 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 )=( R sin v,R cos v, 0) When r (0 ,v )returnstotheinitialpoint,thatis, r (0 ,v +2 )= r (0 ,v ), thenormalvectorisreversed.Indeed, r u(0 ,v +2 )= 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 )= r u(0 ,v ) r v(0 ,v ) = n (0 ,v ); thatis,thesurfacedenedbytheseparametricequationsis not orientablebecausethenormalvectorisreversedwhenmovedarounda closedcurve. So,ifasurface S isdenedbyparametricoralgebraicequations, onestillhastoverifythatitisorientable(i.e.,itistwo-sided!)when evaluatingtheuxacrossit;otherwise,theuxmakesnosense.113.5.Exercises.(1) Findtheuxofaconstantvectoreld F =( a,b,c )acrossthe speciedsurface S : (i) S isarectangleofarea A ineachofthecoordinateplanes orientedalongthecoordinateaxisorthogonaltotherectangle (ii) S isthepartoftheplane( x/a )+( y/b )+( z/c )=1inthe positiveoctantorientedoutwardfromtheoriginand a b ,and c arepositive (iii) S istheboundaryofthepyramidwhosebaseisthesquare [ q,q ] [ q,q ]inthe xy planeandwhosevertexis(0 0 ,h ) (iv) S isthecylinder x2+ y2= R2,0 z h (v) S isthesurfaceofarectangularboxorientedoutward (vi) S isthesphere x2+ y2+ z2= R2orientedoutward (vii) S isatorusorientedinward (2) Findtheuxofthevectoreld F acrossthespeciedoriented surface S : (i) F =( xy,zx,xy )and S isthepartoftheparaboloid z = 1 x2 y2thatliesabovethesquare[0 1] [0 1]andis orientedupward (ii) F =( y, x,z2)and S isthepartoftheparaboloid z =1 x2 y2thatliesabovethe xy planeandisorienteddownward

PAGE 486

48015.VECTORCALCULUS (iii) F =( xz,zy,z2)and S isthepartofthecone z = x2+ y2beneaththeplane z =2intherstoctantandisoriented upward (iv) F =( x, z,y )and S thepartofthesphereintherstoctant andisorientedtowardtheorigin (v) F = a r ,where a isaconstantvectorand S isthesphereof radius R orientedoutwardandcenteredattheorigin (vi) F =(2 y + x,y +2 z x,z y ),where S istheboundaryof thecubewithvertices( 1 1 1)andisorientedoutward (vii) F = c r / r 3,where r =( x,y,z ), c isaconstant,and S isthe sphereofradius a thatiscenteredattheoriginandoriented inward (viii) F =(2 y,x, z )and S isthepartoftheparaboloid y =1 x2 z2intherstoctantandisorientedupward (ix) F =( xy,zy,z )and S isthepartoftheplane2 x 2 y z =3 thatliesinsidethecylinder x2+ y2=1andisorientedupward (x) F =( x,y,z )and S isthepartofthecylinder x = z2+ y2that liesbetweentheplanes x =0and x =1 (xi) F =( x,y,z )and S isthesphere x2+ y2+ z2= R2oriented outward (xii) F =( f ( x ) ,g ( y ) ,h ( z )),where f g ,and h arecontinuousfunctionsand S istheboundaryoftherectangularbox[0 ,a ] [0 ,b ] [0 ,c ]orientedoutward (xiii) F =( y z,z x,x y )and S isthepartofthecone x2+ y2= z2,0 z h ,orientedawayfromthe z axis (3) Useparametricequationsofthespeciedsurface S toevaluatethe uxofthevectoreldacrossit: (i) F =( x, y,z2)and S isthepartofthedoublecone z2= x2+ y2betweentheplanes z = 1and z =1 (ii) F =( z2+ y2,x2+ z2,x2+ y2)and S istheboundaryofthe solidenclosedbythecylinder x2+ z2=1andtheplanes y =0 and y =1; S isorientedoutward (iii) F =( y,x,z )and S isthepartofthesphere x2+ y2+ z2=4 thatliesoutsidethedoublecone z2=3( x2+ y2)andisoriented towardtheorigin (iv) F =( y,x,z )and S isthetoruswithradii R and a thatis orientedoutward (v) F =( x 1,y 1,z 1)and S istheellipsoid( x/a )2+( y/b )2+ ( z/c )2=1orientedoutward (vi) F =( x2,y2,z2)and S isthesphere( x a )2+( y b )2+( z c )2= R2orientedoutward

PAGE 487

114.STOKES'THEOREM481 114.Stokes'Theorem114.1.VectorFormofGreen'sTheorem.ItwasshowninSection112.3 thatthecurlofaplanarvectoreld F ( x,y )=( F1( x,y ) ,F2( x,y ) 0)is paralleltothe z axis, F =( F2/ F1/x ) e3.Thisobservation allowsustoreformulateGreenstheoreminthefollowingvectorform: 'DF d r = D(curl F ) e3dA. Thus, thelineintegralofavectoreldalongaclosedsimplecurveis determinedbytheuxofthecurlofthevectoreldacrossthesurface boundedbythiscurve .Itturnsoutthatthisstatementholdsnotonly inaplanebutalsoinspace.Itisknownas Stokestheorem .114.2.Positive(Induced)OrientationofaClosedCurve.Suppose S is asmoothsurfaceorientedbyitsnormalvector n andboundedbya closedsimplecurve.Consideratangentplaneatapoint r0of S .Any circleinthetangentplanecenteredat r0canalwaysbeorientedcounterclockwiseasviewedfromthetopofthenormalvector n = n0at r0. Thiscircleissaidtobe positivelyorientedrelativetotheorientationof S .Sincethesurfaceissmooth,acircleofasucientlysmallradiuscan alwaysbeprojectedontoaclosedsimplecurve C in S bymovingeach pointofthecircleparallelto n0.Thiscurveisalso positivelyoriented relativeto n0.Itcanthenbecontinuously(i.e.,withoutbreaking) deformedalong S sothatitspartliesontheboundaryof S afterthe deformationandtheorientationsoftheboundaryof S and C canbe compared.Theboundaryof S is positivelyoriented ifithasthesame orientationas C .Thepositivelyorientedboundaryof S isdenotedby S .Theproceduretodeneapositiveorientationoftheboundaryof anorientedsurface S isillustratedinFigure15.13(leftpanel). Inotherwords,thepositive(orinduced)orientationof C means thatifonewalksinthepositivedirectionalong C withoneshead pointinginthedirectionof n ,thenthesurfacewillalwaysbeonones left.Let S beagraph z = g ( x,y )over D orientedupward.Then S isobtainedfrom D (apositivelyorientedboundaryof D )byliftingpointsof D to S paralleltothe z axis(seetherightpanelof Figure15.13). Theorem 15.7 (StokesTheorem) Let S beanoriented,piecewise-smoothsurfacethatisboundedbya simple,closed,piecewise-smoothcurve C withpositiveorientation C = S .Letthecomponentsofavectoreld F havecontinuouspartial

PAGE 488

48215.VECTORCALCULUS n S S C x y z n S S D DFigure15.13.Left :Thepositive(orinduced)orientation oftheboundaryofanorientedsurface S .Thesurface S is orientedbyitsnormalvector n .Takeaclosedcurve C in S thathascounterclockwiseorientationasviewedfromthe topof n .Deformthiscurvetowardtheboundaryof S .The boundaryof S haspositiveorientationifitcoincideswith theorientationof C Right :Thesurface S isthegraphof afunctionon D .If S hasupwardorientation,thenthepositivelyorientedboundary S isobtainedfromthepositively (counterclockwise)orientedboundary D .derivativesonanopenspatialregionthatcontains S .Then 'SF d r = Scurl F n dS, where n istheunitnormalvectoron S Stokestheoremisdiculttoproveingeneral.Hereitisproved foraparticularcasewhen S isagraphofafunction. Proof(for S beingagraph) .Let S betheupward-orientedgraph z = g ( x,y ),( x,y ) D ,where g hascontinuoussecond-orderpartial derivativeson D and D isasimpleplanarregionwhoseboundary D correspondstotheboundary S .Inthiscase,thenormalvector n = ( g x, g y, 1)andtheupwarduxofcurl F across S canbeevaluated accordingtoTheorem15.6inwhich F isreplacedby F : Scurl F n dS = D(curl F )ndA, (curl F )n= F3 y F2 z z x F1 z F3 z z y + F2 x F1 y ,

PAGE 489

114.STOKES'THEOREM483 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.Bythechainrule, 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 byGreenstheorem: 'SF d r = D x F2+ F3z y y F1+ F3z x dA = D(curl F )ndA = Scurl F n dS, (15.14) wherethemiddleequalityisveriedbythedirectevaluationofthe partialderivativesusingthechainrule.Forexample, x F2( x,y,g ( x,y ))= F2 x + F2 z g x x F3g y = F3 x + F3 z g x g y + F32g xy Thetermscontainingthemixedderivatives g xy= g yxarecancelledout owingtoClairautstheorem,whiletheothertermscanbearrangedto coincidewiththeexpressionforthenormalcomponent(curl F )nfound above.Thelastequalityin(15.14)holdsbyTheorem14.22( dS = JdA and n = J n ). 114.3.UseofStokes'Theorem.Stokestheoremisveryhelpfulfor evaluatinglineintegralsalongclosedcurvesofcomplicatedshapeswhen adirectuseofTheorem15.1istechnicallytooinvolved.Theprocedure includesafewbasicsteps. Step1 .Givenaclosedsimplecurve C ,choose any smoothorientable surface S whoseboundaryis C .Notethat,accordingtoStokestheorem,thevalueofthelineintegralisindependentofthechoiceof S .

PAGE 490

48415.VECTORCALCULUS S1S2C = S1= S2C = S S n DFigure15.14.Left :Giventhecurve C = S ,thesurface S mayhaveanydesiredshapeinStokestheorem.Thesurfaces S1and S2havethesameboundaries S1= S2= C Thelineintegralover C canbetransformedintotheux integraleitheracross S1or S2. Right :Anillustrationto Example15.10.Theintegrationcontour C istheintersectionofthecylinderandaplane.WhenapplyingStokes theorem,thesimplestchoiceofasurface,whoseboundaryis C ,isthepartoftheplanethatliesinsidethecylinder.Thisfreedomshouldbeusedtomake S assimpleaspossible(seethe leftpanelofFigure15.14). Step2 .Findtheorientationof S (thedirectionofthenormalvector) sothattheorientationof C ispositiverelativetothenormalof S ,that is, C = S Step3 .Evaluate B =curl F andcalculatetheuxof B across S Example 15.10 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 x y ,where( x,y ) D and D isthedisk x2+ y2 1(asshownintherightpanelofFigure15.14).Since C is orientedclockwiseasviewedfromabove,theorientationof S mustbe downwardtomaketheorientationpositiverelativetothenormalon S ,thatis, n =( g x,g y, 1)=( 1 1 1).Next, B = F =det e1 e2 e3 x y z xyyzxz # =( y, z, x ) .

PAGE 491

114.STOKES'THEOREM485 Therefore, Bn( x,y )= B n =( y, g, x ) ( 1 1 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 15.11 Evaluatethelineintegralof F =( z2y, z2x,z ) alongthecurve C thatistheboundaryofthepartoftheparaboloid z = 1 x2 y2intherstoctant.Thecurve C isorientedcounterclockwise asviewedfromabove. Solution: Choose S tobethespeciedpartoftheparaboloid z = g ( x,y )=1 x2 y2,where( x,y ) D and D isthepartofthedisk x2+ y2 1intherstquadrant( D istheverticalprojectionofthe saidpartoftheparaboloidontothe xy plane).Theparaboloidmustbe orientedupwardsothatthegivenorientationof C ispositiverelativeto thenormalon S .Therefore,thenormalvectoris n =( g x, g y, 1)= (2 x, 2 y, 1).Next, B = F =det e1 e2 e3 x y z z2y z2xz # =(2 zx, 2 zy, 2 z2) sothat Bn( x,y )= B n =(2 gx, 2 gy, 2 g2) (2 x, 2 y, 1)=4 g ( x2+ y2) 2 g2=4 g (1 g ) 2 g2=4 g 6 g2.Thus, CF d r = SB n dS = DBn( x,y ) dA = / 2 01 0[4(1 r2) 6(1 r2)2] rdrd = 7 15 wherethedoubleintegralhasbeenconvertedtopolarcoordinates, g ( x,y )=1 r2. 114.4.GeometricalSignicanceoftheCurl.Stokestheoremreveals thegeometricalsignicanceofthecurlofavectoreld.Thelineintegralofavectoreldalongaclosedcurve C isoftencalledthe circulation ofavectoreldalong C .Let B = F andlet B0= B ( r0)atsome point r0.Consideraplanethrough r0normaltoaunitvector n .Let Sabeasimpleregionintheplanesuchthat r0isaninteriorpointof Sa.Let a betheradiusofthesmallestdiskcenteredat r0thatcontains

PAGE 492

48615.VECTORCALCULUS v n Sa n S2S1E nFigure15.15.Left :Anillustrationtothemechanicalinterpretationofthecurl.Asmallpaddlewheelwhoseaxis ofrotationisparallelto n isplacedintoauidow.The workdonebythepressureforcealongtheloop Sathrough thepaddlescausesrotationofthewheel.Itisdeterminedby thelineintegralofthevelocityvectoreld v .Theworkis maximal(thefastestrotationofthewheel)when n isaligned parallelwith v Right :AnillustrationtoCorollary15.2. Theuxofavectoreldacrossthesurface S1canberelated totheuxacross S2bythedivergencetheoremif S1and S2haveacommonboundaryandtheirunionenclosesasolid region E .Sa.If Saistheareaof Sa,consider thecirculationofavectoreld perunitarea atapoint r0denedastheratio (SaF d r / Sainthe limit a 0,thatis,inthelimitwhentheregion Sashrinkstothe point r0.Then,byvirtueofStokestheoremandtheintegralmean valuetheorem, lima 01 Sa'SaF d r =lima 01 SaSaB n dS = B0 n =(curl F )0 n Indeed,sincethefunction f ( r )= B n iscontinuouson Sa,thereisa point ra Sasuchthatthesurfaceintegralof f equals Saf ( ra).As a 0, ra r0and,bythecontinuityof f f ( ra) f ( r0).Thus, the circulationofavectoreldperunitareaismaximalifthenormalto theareaelementisinthesamedirectionasthecurlofthevectoreld, andthemaximalcirculationequalsthemagnitudeofthecurl ThisobservationhasthefollowingmechanicalinterpretationillustratedintheleftpanelofFigure15.15.Let F describeauidow F = v ,where v istheuidvelocityvectoreld.Imagineatinypaddle wheelintheuidatapoint r0whoseaxisofrotationisdirectedalong

PAGE 493

114.STOKES'THEOREM487 n .Theuidexertspressureonthepaddles,causingthepaddlewheel torotate.Theworkdonebythepressureforceisdeterminedbytheline integralalongtheloop Sathroughthepaddles.Themoreworkdone bythepressureforce,thefasterthewheelrotates.Thewheelrotates fastest(maximalwork)whenitsaxis n isparalleltocurl v because,in thiscase,thenormalcomponentofthecurl,( v ) n = v ismaximal.Forthisreason,thecurlisoftencalledthe rotation ofa vectoreldandalsodenotedasrot F = F Definition 15.11 (RotationalVectorField) Avectoreld F thatcanberepresentedasthecurlofanothervector eld A ,thatis, F = A ,iscalleda rotationalvectoreld Thefollowingtheoremholds(theproofisomitted). Theorem 15.8 (HelmholtzsTheorem) Let F beavectoreldonaboundeddomain E whosecomponentshave continuoussecond-orderpartialderivatives.Then F canbedecomposed intothesumofconservativeandarotationalvectorelds;thatis,there isafunction f andavectoreld A suchthat F = f + A Thevectoreld A iscalleda vectorpotential oftheeld F .The vectorpotentialisnotunique.Itcanbechangedbyaddingthegradient ofafunction, A A + g ,because ( A + g )= A + ( g )= A forany g thathascontinuoussecond-orderpartialderivatives.Electromagneticwavesarerotationalcomponentsofelectromagneticelds, whiletheCoulombeldcreatedbystaticchargesisconservative.The velocityvectoreldofanincompressibleuid(likewater)isarotationalvectoreld.114.5.TestforaVectorFieldtoBeConservative.Thetestforavector eldtobeconservative(Theorem15.4)followsfromStokestheorem. Indeed,inasimplyconnectedregion E ,anysimpleclosedcurvecanbe shrunktoapointwhileremainingin E throughoutthedeformation. Therefore,foranysuchcurve C ,onecanalwaysndasurface S in E suchthat S = C (e.g., C canbeshrunktoapointalongsuch S ).If curl F = 0 throughout E ,then,byStokestheorem, 'CF d r = Scurl F n dS =0

PAGE 494

48815.VECTORCALCULUS foranysimpleclosedcurve C in E .Bythepathindependenceproperty, F isconservative.Thehypothesisthat E issimplyconnectediscrucial. Forexample,if E istheentirespacewiththe z axisremoved(seeStudy Problem15.2),thenthe z axisalwayspiercesthroughanysurface S boundedbyaclosedsimplecurveencirclingthe z axis,andonecannot claimthatthecurlvanisheseverywhereon S .114.6.StudyProblem.Problem15.8. Provethattheuxofacontinuousrotationalvectoreld F vanishesacrossanysmooth,closed,andorientablesurface. Whatcanbesaidaboutauxinaowofanincompressibleuid? Solution: Acontinuousrotationalvectoreldcanbewrittenasthe curlofavectoreld A whosecomponentshavecontinuouspartial derivatives, F = A .Considerasmoothclosedsimplecontour C inasurface S .Itcuts S intotwopieces S1and S2.Supposethat S isorientedoutward.Thentheinducedorientationsoftheboundaries S1and S2areopposite: S1= S2.Thelatteralsoholdsif S is orientedinward.ByvirtueofStokestheorem, S( A ) n dS = S1( A ) n dS + S2( A ) n dS = 'S1A d r + 'S2A d r = 'S1A d r + ' S1A d r =0 Recallthatthelineintegralchangesitssignwhentheorientationofthe curveisreversed.Sincetheowofanincompressibleuidisdescribed byarotationalvectoreld,theuxacrossaclosedsurfacealways vanishesinsuchaow. 114.7.Exercises.(1) VerifyStokestheoremforthegivenvectoreld F andsurface S by calculatingthecirculationof F along S andtheuxof F across S : (i) F =( y, x,z )and S isthepartofthesphere x2+ y2+ z2=2 thatliesabovetheplane z =1 (ii) F =( x,y,xyz )and S isthepartoftheplane2 x + y + z =4 intherstoctant (iii) F =( y,z,x )and S isthepartoftheplane x + y + z =0inside thesphere x2+ y2+ z2= a2

PAGE 495

114.STOKES'THEOREM489 (2) UseStokestheoremtoevaluatethelineintegralofthevectoreld F alongthespeciedclosedcontour C : (i) F =( x + y2,y + z2,z + x2)and C isthetriangletraversedas (1 0 0) (0 1 0) (0 0 1) (1 0 0) (ii) F =( yz, 2 xz,exy)and C istheintersectionofthecylinder x2+ y2=1andtheplane z =3orientedclockwiseasviewed fromabove (iii) F =( xy, 3 z, 3 y )and C istheintersectionoftheplane x + y = 1andthecylinder y2+ z2=1 (iv) F =( z,y2, 2 x )and C istheintersectionoftheplane x + y + z = 5andthecylinder x2+ y2=1;thecontour C isoriented counterclockwiseasviewedfromthetopofthe z axis (v) F =( yz,xz, 0)and C istheintersectionofthehyperbolic paraboloid z = y2 x2andthecylinder x2+ y2=1; C is orientedclockwiseasviewedfromthetopofthe z axis (vi) F =( z2y/ 2 z2x/ 2 0)and C istheboundaryofthepartof thecone z =1 x2+ y2thatliesintherstquadrant; C isorientedcounterclockwiseasviewedfromthetopofthe z axis. (vii) F =( y z, x,x )and C istheintersectionofthecylinder x2+ y2=1andtheparaboloid z = x2+( y 1)2; C isoriented counterclockwiseasviewedfromthetopofthe z axis (viii) F =( y z,z x,x y )and C istheellipse x2+ y2= a2, ( x/a )+( z/b )=1, a> 0, b> 0,orientedpositivelywhen viewedfromthetopofthe z axis (ix) F =( y + z,z + x,x + y )and C istheellipse x = a sin2t y =2 a sin t cos t z = a cos2t ,0 t ,orientedinthe directionofincreasing t (x) F =( y2 z2,z2 x2,x2 y2)and C istheintersectionofthe surfaceofthecube[0 ,a ] [0 ,a ] [0 ,a ]bytheplane x + y + z = 3 a/ 2,orientedcounterclockwisewhenviewedfromthetopof the x axis (xi) F =( y2z2,x2z2,x2y2),where C istheclosedcurvetracedout bythevectorfunction r ( t )=( a cos t,a cos(2 t ) ,a cos(3 t ))in thedirectionofincreasing t (3) Let C beaclosedcurveintheplane n r = d thatboundsaregion ofarea A .Find 'C( n r ) d r (4) UseStokestheoremtondtheworkdonebytheforce F inmoving aparticlealongthespeciedclosedpath C :

PAGE 496

49015.VECTORCALCULUS (i) F =( yz,zx,yx )and C isthetriangle(0 0 6) (2 0 0) (0 3 0) (0 0 6) (ii) F =( yz,xz,z2)and C istheboundaryofthepartofthe paraboloid z =1 x2 y2intherstoctantthatistraversed clockwiseasviewedfromthetopofthe z axis (iii) F =( y +sin x,z2+cos y,x3)and C istraversedby r ( t )= (sin t, cos t, sin(2 t ))for0 t 4 ( Hint: Observethat C lies inthesurface z =2 xy .) (5) Findthelineintegralof F =( ex2 yz,ey2 xz,z2 xy )along C whichisthehelix x = a cos t y = a sin t z = ht/ (2 )fromthepoint ( a, 0 0)tothepoint( a, 0 ,h ). Hint: Supplement C bythestraightline segment BA tomakeaclosedcurveandthenuseStokestheorem. (6) Supposethatasurface S satisesthehypothesesofStokestheoremandthefunctions f and g havecontinuouspartialderivatives. Showthat 'S( f g ) d r = S( f g ) n dS. Usetheresulttoshowthatthecirculationofthevectoreldsofthe form F = f f and F = f g + g f vanishesalong S (7) Considerarotationallysymmetricsolid.Letthesolidberotating aboutthesymmetryaxisataconstantrate (angularvelocity).Let w bethevectorparalleltothesymmetryaxissuchthat w = and therotationiscounterclockwiseasviewedfromthetopof w .Ifthe originisonthesymmetryaxis,showthatthelinearvelocityvector eldinthesolidisgivenby v = w r ,where r isthepositionvectorof apointinthesolid.Next,showthat v =2 w .Thisgivesanother relationbetweenthecurlofavectoreldandrotations. 115.Gauss-Ostrogradsky(Divergence)Theorem115.1.DivergenceofaVectorField.Definition 15.12 (DivergenceofaVectorField) Supposethatavectoreld F =( F1,F2,F3) isdierentiable.Thenthe scalarfunction div F = F = F1 x + F2 y + F3 z iscalledthe divergenceofavectoreld Example 15.12 Findthedivergenceofthevectoreld F = ( x3+cos( yz ) ,y +sin( x2z ) ,xyz ) .

PAGE 497

115.GAUSS-OSTROGRADSKY(DIVERGENCE)THEOREM491 Solution: Onehas div F =( x3+cos( yz )) x+( y +sin( x2z )) y+( xyz ) z=3 x2+1+ yx. Corollary 15.1 Arotationalvectoreldwhosecomponentshave continuouspartialderivativesisdivergencefree, divcurl A =0 Proof. Bydenition,arotationalvectoreldhastheform F = curl A = A ,wherethecomponentsof A havecontinuoussecondorderpartialderivativesbecause,bythehypothesis,thecomponents of F havecontinuousrst-orderpartialderivatives.Therefore, div F =divcurl A = curl A = ( A )=0 bytherulesofvectoralgebra(thetripleproductvanishesifanytwo vectorsinitcoincide).Theserulesareapplicablebecausethecomponentsof A havecontinuoussecond-orderpartialderivatives(Clairauts theoremholdsforitscomponents;seeSection111.4). LaplaceOperator.Let F = f .Thendiv F = f = f xx+ f yy+ f zz. Theoperator = 2iscalledthe Laplaceoperator .115.2.AnotherVectorFormofGreen'sTheorem.Greenstheoremrelatesalineintegralalongaclosedcurveofthe tangential componentof aplanarvectoreldtotheuxofthecurlacrosstheregionbounded 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 ) x( t )) T n =0 aretheunittangentvectorandtheoutwardunitnormalvectortothe curve C ,respectively.Considerthelineintegral (CF n ds ofthenormal componentofaplanarvectoreldalong C .Onehas ds = r( t ) dt andhence F n ds = F1ydt F2xdt = F1dy F2dx = G d r where G =( F2,F1).ByGreenstheoremappliedtothelineintegral ofthevectoreld G 'CF n ds = 'CG d r = D G2 x G1 y dA = D F1 x + F2 y dA.

PAGE 498

49215.VECTORCALCULUS Theintegrandinthedoubleintegralisthedivergenceof F .Thus, anothervectorformofGreenstheoremhasbeenobtained: 'DF n ds = Ddiv F dA. Foraplanarvectoreld(thinkofamassowonaplane),theline integralontheleftsidecanbeviewedastheoutwarduxof F across theboundaryofaregion D (e.g.,themasstransferbyaplanarow acrosstheboundaryof D ).AnextensionofthisformofGreenstheoremtothree-dimensionalvectoreldsisknownasthe divergence,or Gauss-Ostrogradsky,theorem .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.Ifcomponentsofavector eld F havecontinuouspartialderivativesinanopenregionthatcontains E ,then EF n dS = Ediv F dV. Thedivergencetheoremstatesthattheoutwarduxofavector eldacrossaclosedsurface S isgivenbythetripleintegralofthe divergenceofthevectoreldoverthesolidregionboundedby S .It providesaconvenienttechnicaltooltoevaluatetheuxofavector eldacrossaclosedsurface. Remark. Itshouldbenotedthattheboundary E maycontain severaldisjointpieces.Forexample,let E beasolidregionwitha cavity.Then E consistsoftwopieces,theouterboundaryandthe cavityboundary.Bothpiecesareorientedoutwardinthedivergence theorem. Example 15.13 Evaluatetheuxofthevectoreld F = (4 xy2z + ez, 4 yx2z,z4+sin( xy )) acrosstheclosedsurfaceorientedoutwardthatistheboundaryofthepartoftheball x2+ y2+ z2 R2in therstoctant( x,y,z 0 ). Solution: Thedivergenceofthevectoreldis div F =(4 xy2z + ez) x+(4 yx2z ) y+( z4+sin( xy )) z=4 z ( x2+ y2+ z2) .

PAGE 499

115.GAUSS-OSTROGRADSKY(DIVERGENCE)THEOREM493 Bythedivergencetheorem, SF n dS = E4 z ( x2+ y2+ z2) dV = / 2 0/ 2 0R 04 3cos 2sin ddd = R6 24 wherethetripleintegralhasbeenconvertedtosphericalcoordinates. Thereaderisadvisedtoevaluatetheux without usingthedivergence theoremtoappreciatethepowerofthelatter! Thedivergencetheoremcanbeusedtochange(simplify)thesurface intheuxintegral. Corollary 15.2 Lettheboundary E ofasolidregion E bethe unionoftwosurfaces S1and S2.Supposethatallthehypothesesofthe divergencetheoremhold.Then S2F n dS = Ediv F dV S1F n dS. Thisestablishesarelationbetweentheuxacross S1andtheux across S2withacommonboundarycurve(seeFigure15.15,right panel).Indeed,since E istheunionoftwodisjointpieces S1and S2,thesurfaceintegralover E isthesumoftheintegralsover S1and S2.Ontheotherhand,theintegralover E canbeexpressed asatripleintegralbythedivergencetheorem,whichestablishesthe statedrelationbetweentheuxesacross S1and S2.Notethat S1and S2mustbeorientedsothattheirunionis E ;thatis,ithasoutward orientation. Example 15.14 Evaluatetheupwarduxofthevectoreld F = ( z2tan 1( y2+1) ,z4ln( x2+1) ,z ) acrossthepartoftheparaboloid z = 2 x2 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 x2 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, 1)=(0 0 1),and

PAGE 500

49415.VECTORCALCULUS x y z 2 1 1 1 n E S2S1n D F Eadiv F > 0 F Eadiv F < 0Figure15.16.Left :AnillustrationtoExample15.14. Thesolidregion E isenclosedbytheparaboloid z =2 x2 y2andtheplane z =1.ByCorollary15.2,theuxofavector eldacrossthepart S2oftheparaboloidthathasupward orientationcanbeconvertedtotheuxacrossthepart S1oftheplanethathasdownwardorientation.Theunionof S1and S2hasoutwardorientation. Right :Thedivergence ofavectoreld F determinesthedensityofsourcesof F .If div F > 0atapoint,thentheuxof F acrossasurfacethat enclosesasmallregion Eacontainingthepointispositive(a faucet).Ifdiv F < 0atapoint,thentheuxof F acrossa surfacethatenclosesasmallregion Eacontainingthepoint isnegative(asink).hence Fn= F n = F3( x,y,g )= 1on S1and S1F n dS = DFn( x,y ) dA = DdA = A ( D )= Next,thedivergenceof F is div F =( z2tan 1( y2+1)) x+( z4ln( x2+1)) y+( z ) z=0+0+1=1 Hence, Ediv F dV = EdV = 2 01 02 r21rdzdrd =2 1 0(1 r2) rdr = 2 wherethetripleintegralhasbeentransformedintocylindricalcoordinatesfor E = { ( x,y,z ) | zbot=1 z 2 x2 y2= ztop, ( x,y ) D } .

PAGE 501

115.GAUSS-OSTROGRADSKY(DIVERGENCE)THEOREM495 Theupwarduxof F acrosstheparaboloidisnoweasytondby Corollary15.2: S2F n dS = Ediv F dV S1F n dS = 2 + = 3 2 Thereaderisagainadvisedtotrytoevaluatetheuxdirectlyvia thesurfaceintegraltoappreciatethepowerofthedivergencetheorem! Corollary 15.3 Theuxofarotationalvectoreld,whosecomponentshavecontinuouspartialderivatives,acrossanorientable,closed, piecewise-smoothsurface S vanishes: Scurl A n dS =0 Proof. Thehypothesesofthedivergencetheoremaresatised.Therefore, Scurl A n dS = Edivcurl A dV =0 byCorollary15.1. ByHelmholtzstheorem,avectoreldcanalwaysbedecomposed intothesumofconservativeandrotationalvectorelds.Itfollowsthen thatonlytheconservativecomponentofthevectoreldcontributesto theuxacrossaclosedsurface: div( f + A )= 2f + ( A )= 2f. Sothedivergenceofavectoreldisdeterminedbytheactionofthe Laplaceoperatorofthescalarpotential f ofthevectoreld.This observationisfurtherelucidatedwiththehelpoftheconceptofvector eldsources.115.4.SourcesofaVectorField.Considerasimpleregion Eaofvolume Vaandaninteriorpoint r0of Ea.Let a betheradiusofthe smallestballthatcontains Eaandiscenteredat r0.Letuscalculatetheoutwardux perunitvolume ofavectoreld F acrossthe boundary Ea,whichisdenedbytheratio EaF n dS/ Vainthe limit a 0,thatis,when Eashrinksto r0.Supposethatcomponents of F havecontinuouspartialderivatives.Byvirtueofthedivergence theoremandtheintegralmeanvaluetheorem, lima 01 VaEaF n dS =lima 01 VaEadiv F dV =div F ( r0) .

PAGE 502

49615.VECTORCALCULUS Indeed,bythecontinuityofdiv F ,andtheintegralmeanvaluetheorem,thereisapoint ra Easuchthatthetripleintegralequals Vadiv F ( ra).Inthelimit a 0, ra r0anddiv F ( ra) div F ( r0). Thus,ifthedivergenceispositivediv F ( r0) > 0,theuxofthevectoreldacrossanysmallsurfacearound r0ispositive.This,inturn, meansthattheowlinesof F areoutgoingfrom r0asifthereisa source creatingaowat r0.Followingtheanalogywithwaterow,sucha sourceiscalleda faucet .Ifdiv F ( r0) < 0,theowlinesdisappearat r0(theinwardowispositive).Suchasourceiscalleda sink .Thus, thedivergenceofavectorelddeterminesthedensityofthesourcesof avectoreld .Forexample,owlinesofastaticelectriceldoriginate frompositiveelectricchargesandendonnegativeelectriccharges.The divergenceoftheelectricelddeterminestheelectricchargedensityin space. Thedivergencetheoremstatesthattheoutwarduxofavector eldacrossaclosedsurfaceisdeterminedbythetotalsourceofthe vectoreldintheregionboundedbythesurface.Inparticular,the uxoftheelectriceld E acrossaclosedsurface S isdeterminedby thetotalelectricchargeintheregionenclosedby S .Incontrast,the magneticeld B isarotationalvectoreldandhenceisdivergencefree. Sothereareno magneticcharges alsoknownas magneticmonopoles Thesetwolawsofphysicsarestatedintheform: div E =4 div B =0 where isthedensityofelectriccharges.Flowlinesofthemagnetic eldareclosed,whileowlinesoftheelectriceldendatpointswhere electricchargesarelocated(asindicatedbythearrowsintheright panelofFigure15.16).115.5.StudyProblem.Problem15.9.(VolumeofaSolidastheSurfaceIntegral). Let E beboundedbyapiecewisesmoothsurface S = E orientedby anoutwardunitnormalvector n .Provethatthevolumeof E is V ( E )= 1 3 E n r dS. Solution: Considerthreevectorelds F1=( x, 0 0), F2=(0 ,y, 0), and F3=(0 0 ,z ).Then div F1=1 div F2=1 div F3=1 .

PAGE 503

115.GAUSS-OSTROGRADSKY(DIVERGENCE)THEOREM497 Then,byvirtueoftheequality r = F1+ F2+ F3andbythedivergence theorem, E n r dS = E n F1dS + E n F2dS + E n F3dS = E(div F1+div F2+div F3) dV =3 EdV =3 V ( E ) andtherequiredresultfollows. 115.6.Exercises.(1) Findthedivergenceofthespeciedvectoreld: (i) F = f ,where f = x2+ y2+ z2(ii) F = r /r ,where r = r (iii) F = a f ( r ),where r = r and a isaconstantvector (iv) F = r f ( r ),where r = r .Whendoesthedivergencevanish? (v) F = a g ,where a isaconstantvectorand g isadierentiable function.Whendoesthedivergencevanish? (vi) F = a r ,where a isaconstantvector (vii) F = a g ,where a isaconstantvector.Whendoesthe divergencevanish? (viii) F = a G ,where a isaconstantvector.Whendoesthe divergencevanish? (2) Provethefollowingidentities,assumingthattheappropriatepartialderivativesofvectoreldsandfunctionsexistandarecontinuous: (i)div( f F )= f div F + F f (ii)div( F G )= G curl F F curl G (iii)div( f g ) (iv)curlcurl F = (div F ) 2F (3) Let a beaxedvectorandlet n betheunitnormaltoaplanar closedcurve C directedoutwardfromtheregionboundedby C .Show that (Ca n ds =0. (4) Let C beasimpleclosedcurveinthe xy planeandlet n bethe unitnormalto C directedoutwardfromtheregion D boundedby C If A ( D )istheareaof D ,nd (Cr n ds (5) Verifythedivergencetheoremforthegivenvectoreld F onthe region E : (i) F =(3 x,yz, 3 xz )and E istherectangularbox[0 ,a ] [0 ,b ] [0 ,c ]

PAGE 504

49815.VECTORCALCULUS (ii) F =(3 x, 2 y,z )and E isthesolidboundedbytheparaboloid z = a2 x2 y2andtheplane z =0 (6) Let a beaconstantvectorandlet S beaclosedsmoothsurface orientedoutwardbytheunitnormalvector n .Provethat Sa n dS =0 (7) Evaluatetheuxofthegivenvectoreldacrossthespeciedclosed surface S .Ineachcase,determinethekindofsourceof F intheregion enclosedby S (sinkorfaucet): (i) F =( x2,y2,z2)and S istheboundaryoftherectangularbox [0 ,a ] [0 ,b ] [0 ,c ]orientedoutward (ii) F =( x3,y3,z3)and S isthesphere x2+ y2+ z2= R2oriented inward (iii) F =( xy,y2+sin( xz ) cos( yx ))and S isboundedbytheparaboliccylinder z =1 x2andtheplanes z =0, y =0, y + z =2; S isorientedoutward (iv) F =( xy2, yz2,zx2)and S isthesphere x2+ y2+ z2=1 withinwardorientation (v) F =( xy,z2y,zx )and S istheboundaryofthesolidregion insidethecylinder x2+ y2=4andbetweentheplanes z = 2; S isorientedoutward (vi) F =( xz2,y3/ 3 ,zy2+ xy )and S istheboundaryofthepart oftheball x2+ y2+ z2 1intherstoctant; S isoriented inward (vii) F =( yz,z2x + y,z xy )and S istheboundaryofthesolid enclosedbythecone z = x2+ y2andthesphere x2+ y2+ z2=1; S isorientedoutward (viii) F =( x +tan( yz ) cos( xz ) y, sin( xy )+ z )and S istheboundaryofthesolidregionbetweenthesphere x2+ y2+ z2=2 z andthecone z = x2+ y2(ix) F =(tan( yz ) ln(1+ z2x2) ,z2+ eyx)and S istheboundary ofthesmallerpartoftheball x2+ y2+ z2 a2betweentwo half-planes y = x/ 3and y = 3 x x 0; S isoriented inward (x) F =( xy2,xz,zx2)and S istheboundaryofthesolidbounded bytwoparaboloids z = x2+ y2and z =1+ x2+ y2andthe cylinder x2+ y2=4; S isorientedoutward (xi) F =( x,y,z )and S istheboundaryofthesolidobtainedfrom thebox[0 2 a ] [0 2 b ] [0 2 c ]byremovingthesmallerbox [0 ,a ] [0 ,b ] [0 ,c ]; S isorientedinward

PAGE 505

115.GAUSS-OSTROGRADSKY(DIVERGENCE)THEOREM499 (xii) F =( x y + z,y z + x,z x + y )and S isthesurface | x y + z | + | y z + x | + | z x + y | =1orientedoutward (xiii) F =( x3,y3,z3)and S isthesphere x2+ y2+ z2= x oriented outward (8) Let S1and S2betwosmoothorientablesurfacesthathavethe sameboundary.Supposethat F =curl A andthecomponentsof F havecontinuouspartialderivatives.Comparetheuxesof F across S1and S2. (9) Usethedivergencetheoremtondtheuxofthegivenvectoreld F acrossthespeciedsurface S byanappropriatedeformationof S : (i) F =( xy2,yz2,zy2+ x2)and S isthetophalfofthesphere x2+ y2+ z2=4orientedtowardtheorigin (ii) F =( z cos( y2) ,z2ln(1+ x2) ,z )and S isthepartoftheparaboloid z =2 x2 y2abovetheplane z =1; S isoriented upward (iii) F =( yz,xz,xy )and S isthecylinder x2+ y2= a2,0 z b orientedoutwardfromitsaxisofsymmetry (iv) F =( yz + x3,x2z3,xy )and S isthepartofthecone z = 1 x2+ y2orientedupward (10) Theelectriceld E andthechargedensity arerelatedby theGausslawdiv E =4 .Supposethechargedensityisconstant, = k> 0,insidethesphere x2+ y2+ z2= R2and0otherwise.Findtheoutwarduxoftheelectriceldacrosstheellipsoid x2/a2+ y2/b2+ z2/c2=1inthetwofollowingcases:rst,when R is greaterthananyof a,b,c ;second,when R islessthananyof a,b,c (11) Let F beavectoreldsuchthatdiv F = 0=constinasolid boundedregion E anddiv F =0otherwise.Let S beaclosedsmooth surfaceorientedoutward.Considerallpossiblerelativepositionsof E and S inspace(thesolidregionboundedby S mayormaynothavean overlapwith E ).If V isthevolumeof E ,whatareallpossiblevalues oftheuxof E across S ? (12) UsethevectorformofGreenstheoremtoproveGreensrstand secondidentities: Df 2gdA = 'D( f g ) n ds Dg 2fdA, D( f 2g f 2g ) dA = 'D( f g g f ) n ds, where D satisesthehypothesesofGreenstheoremandtheappropriatepartialderivativesof f and g existandarecontinuous.

PAGE 506

50015.VECTORCALCULUS (13) UsetheresultofStudyProblem15.9tondthevolumeofasolid boundedbythespeciedsurfaces: (i)Theplanes z = c andtheparametricsurface x = a cos u cos v + b sin u sin v y = a cos u sin v b sin u cos v z = c sin u (ii)Theplanes x =0and z =0andtheparametricsurface x = u cos v y = u sin v z = u + a cos v ,where u 0and a> 0 (iii)Thetorus x =( R + a cos u )sin v y =( R + a cos u )sin v z = a sin u (14) UsetheresultsofStudyProblem15.4toexpressthedivergenceofavectoreldincylindricalandsphericalcoordinates: F = 1 r ( rFr) r + 1 r F + Fz z F = 1 2 ( 2F) + 1 sin (sin F) + F Hint: Show e/ = e, e/ =sin e,andsimilarrelationsfor thepartialderivativesofotherunitvectors. (15) UsetheresultsofStudyProblem15.4toexpresstheLaplace operatorincylindricalandsphericalcoordinates: 2f = 1 r r r f r + 1 r22f 2 + 2f z2, 2f = 1 2 2f + 1 2sin sin f + 1 2sin2 2f 2. Hint: Show e/ = e, e/ =sin e,andsimilarrelationsfor thepartialderivativesofotherunitvectors.

PAGE 507

Acknowledgments TheauthorwouldliketothankhiscolleaguesDr.DavidGroisser andDr.ThomasWalshfortheirusefulsuggestionsandcommentsthat werehelpfultoimprovethetextbook.501

PAGE 1

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

PAGE 3

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

PAGE 4

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.Nothinginthislicenseimpairsorrestrictstheauthorsmoral 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

PAGE 5

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 .

PAGE 6

viCONTENTS Chapter14.MultipleIntegrals20197.DoubleIntegrals20198.PropertiesoftheDoubleIntegral20799.IteratedIntegrals211100.DoubleIntegralsOverGeneralRegions215101.DoubleIntegralsinPolarCoordinates222102.ChangeofVariablesinDoubleIntegrals227103.TripleIntegrals234104.TripleIntegralsinCylindricalandSphericalCoordinates243105.ChangeofVariablesinTripleIntegrals250106.ImproperMultipleIntegrals254107.LineIntegrals261108.SurfaceIntegrals265109.MomentsofInertiaandCenterofMass273Chapter15.VectorCalculus283110.LineIntegralsofaVectorField283111.FundamentalTheoremforLineIntegrals288112.GreensTheorem297113.FluxofaVectorField303114.StokesTheorem310115.Gauss-Ostrogradsky(Divergence)Theorem315

PAGE 7

CHAPTER11 VectorsandtheSpaceGeometry Ourspacemaybeviewedasacollectionofpoints.Everygeometricalgure,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,butthishasyettobeveriedbyexperiments.71.RectangularCoordinatesinSpace Theelementaryobjectinspaceisapoint.Sothediscussionshould beginwiththequestion:Howcanonedescribeapointinspacebyreal numbers?Thefollowingprocedurecanbeadopted.Selectaparticular pointinspacecalledthe origin andusuallydenoted O .Setupthree mutuallyperpendicularlinesthroughtheorigin.Arealnumberis associatedwitheverypointoneachlineinthefollowingway.The origincorrespondsto0.Distancestopointsononesideoftheline1

PAGE 8

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 speciedasan orderedtripleofreal numbers relativetoagivenrectangularcoordinatesystem.Consider arectanglewhosetwooppositevertices(theendpointsofthelargest diagonal)aretheoriginandapoint P ,whileitssidesthatareadjacent attheoriginlieontheaxesofthecoordinatesystem.Foreverypoint P thereisonlyonesuchrectangle.Therectangleisuniquelydetermined byitsthreesidesadjacentattheorigin.Letthenumber x marksthe positionofonesuchsidethatliesontherstaxis,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.Toreecttheorderin( x,y,z ),the axesofthecoordinatesystemwillbemarkedas x y ,and z axes.Thus, tondapointinspacewithrectangularcoordinates(1 2 3),onehas toconstructarectanglewithavertexattheoriginsuchthatitssides adjacentattheoriginoccupytheintervals[0 1],[0 2],and[ 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 denedsimilarly.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 coordinateisxedtoaparticular value z = z0(e.g., z =1).Theyformaplaneparalleltothe xy plane thatlies | z0| unitsoflengthaboveitif z0> 0orbelowitif z0< 0.

PAGE 9

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.Thisisachievedbytranslatingtheoriginrstalongthe x axisbythedistance x0(asshowninthegure),thenalongthe y axisbythe distance y0,andnallyalongthe z axisbythedistance z0.Asaresult,apoint P thathadcoordinates( x,y,z ) intheoldsystemwillhavethecoordinates x= x x0, y= y y0,and z= z 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

PAGE 10

411.VECTORSANDTHESPACEGEOMETRY coordinates( x,y,z ).Consideranewcoordinatesystemwhoseaxesare paralleltothecorrespondingaxesoftheoldcoordinatesystem,but whoseoriginisshiftedtothepoint Owithcoordinates( x0, 0 0).It isstraightforwardtoseethatthepoint P wouldhavethecoordinates ( x x0,y,z )relativetothenewcoordinatesystem(Figure11.1,right panel).Similarly,iftheoriginisshiftedtoapoint Owithcoordinates ( x0,y0,z0),whiletheaxesremainparalleltothecorrespondingaxesof theoldcoordinatesystem,thenthecoordinatesof P aretransformedas (11.1)( x,y,z ) ( x x0,y y0,z 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.Thequestionthereforeishowtondthelengthofthis diagonal. Considerthreesidesofthisrectanglethatareadjacent,say,atthe vertex P1.Thesideparalleltothe x axisliesbetweenthecoordinate planes x = x1and x = x2andisperpendiculartothem.Sothe lengthofthissideis | x2 x1| .Theabsolutevalueisnecessaryasthe

PAGE 11

71.RECTANGULARCOORDINATESINSPACE5 dierence x2 x1maybenegative,dependingonthevaluesof x1and x2,whereasthedistancemustbenonnegative.Similarargumentslead totheconclusionthatthelengthsoftheothertwoadjacentsidesare | y2 y1| and | z2 z1| .Ifarectanglehasadjacentsidesoflength a b and c ,thenthelength d ofitslargestdiagonalsatisestheequation 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 = | x1 x2| b = | y1 y2| ,and c = | z1 z2| .Thelengthofthediagonal d isobtainedbythedoubleuseofthePythagorean theoremineachoftheindicatedrectangles: d2= c2+ f2(topright)and f2= a2+ b2(bottomright).

PAGE 12

611.VECTORSANDTHESPACEGEOMETRY Put a = | x2 x1| b = | y2 y1| ,and c = | z2 z1| .Then d = | P1P2| is thedistancebetween P1and P2.Thedistanceformulaisimmediately found: (11.2) | P1P2| = ( x2 x1)2+( y2 y1)2+( z2 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( x1 x0,y1 y0,z1 z0)and( x2 x0,y2 y0,z2 z0),respectively.Thenumericalvalueofthedistancedoesnotchangebecause thecoordinatedierences,( x2 x0) ( x1 x0)= x2 x1(similarly forthe y and z coordinates ),donotchange. RotationsinSpace. Theorigincanalwaysbetranslatedto P1so thatinthenewcoordinatesystem P1is(0 0 0)and P2is( x2 x1,y2 y1,z2 z1).Sincethedistanceshouldnotdependontheorientationof thecoordinateaxes,anyrotationcannowbedescribedalgebraically as alineartransformationofanorderedtriple ( x,y,z ) underwhich thecombination x2+ y2+ z2remainsinvariant .Alineartransformationmeansthatthenewcoordinatesarelinearcombinationsofthe oldones.Itshouldbenotedthatreectionsofthecoordinateaxes, x x (similarlyfor y and z ),arelinearandalsopreservethedistance.However,acoordinatesystemobtainedbyanoddnumberof reectionsofthecoordinateaxescannotbeobtainedbyanyrotation oftheoriginalcoordinatesystem.So,intheabovealgebraicdenition ofarotation,thereectionsshouldbeexcluded.71.5.SpheresinSpace.Inthiscourse,relationsbetweentwoequivalent descriptionsofobjectsinspacethegeometricalandthealgebraic willalwaysbeemphasized.Oneofthecourseobjectivesistolearn howtointerpretanalgebraicequationbygeometricalmeansandhow todescribegeometricalobjectsinspacealgebraically.Thesimplest exampleofthiskindisasphere. GeometricalDescriptionofaSphere .Asphereisasetof pointsinspacethatareequidistantfromaxedpoint.Thexedpoint iscalledthe spherecenter .Thedistancefromthespherecentertoany pointofthesphereiscalledthe sphereradius AlgebraicDescriptionofaSphere .Analgebraicdescriptionof asphereimpliesndinganalgebraicconditiononcoordinates( x,y,z ) ofpointsinspacethatbelongtothesphere.Soletthecenterofthe

PAGE 13

71.RECTANGULARCOORDINATESINSPACE7 Figure11.3. Left :Asphereisdenedasapointset inspace.Eachpoint P ofthesethasaxeddistance R fromaxedpoint P0.Thepoint P0isthecenterofthe sphere,and R istheradiusofthesphere. Right :Transformationofcoordinatesunderarotation ofthecoordinatesysteminaplane. spherebeapoint P0withcoordinates( x0,y0,z0)(denedrelativeto 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 x0)2+( y y0)2+( z 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 x0)2+( y y0)2+( z z0)2= R2 Thisrelationmeansthattheset S isacollectionofallpoints( x,y,z ) suchthat(theverticalbar)theirrectangularcoordinatessatisfy(11.3).

PAGE 14

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 liesintherstoctant.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 x2 ( x1+ x2) / 2=( x2 x1) / 2and ( x1+ x2) / 2 x1=( x2 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

PAGE 15

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 = Therefore, x= r cos =cos( )= r cos cos + r sin sin = x cos + y sin y= r sin = r sin( )= r sin cos r cos sin = y cos x sin Problem11.3. Giveageometricaldescriptionoftheset S = ( x,y,z ) x2+ y2+ z2 4 z =0 Solution: Theconditiononthecoordinatesofpointsthatbelong tothesetcontainsthesumofsquaresofthecoordinatesjustlikethe equationofasphere.Thedierenceisthat(11.3)containsthesum ofperfectsquares.Sothesquaresmustbecompletedintheabove equationandtheresultingexpressioncomparedwith(11.3).Onehas z2 4 z =( z 2)2 4sothattheconditionbecomes x2+ y2+( z 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+ y2 2 x 4 y =4 Solution: Asinthepreviousproblem,theconditioncanbewritten asthesumofperfectsquares( x 1)2+( y 2)2=9bymeansthe ofrelations x2 2 x =( x 1)2 1and y2 4 y =( y 2)2 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 x )=0 .

PAGE 16

1011.VECTORSANDTHESPACEGEOMETRY Solution: Theconditionissatisedifeither 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) Findthedistancebetweenthefollowingspecied points: (i) A (1 2 3)and B ( 1 0 2) (ii) A ( 1 3 2)and B ( 1 2 1) (2) Lettheset S consistofpoints( t, 2 t, 3 t )where 0 (iv) x2+ y2 4 y< 0, z> 0 (v)4 x2+ y2+ z2 9 (vi) x2+ y2 1, x2+ y2+ z2 4 (vii) x2+ y2+ z2 2 z< 0, z> 1 (viii) x2+ y2+ z2 2 z =0, z =1 (ix)( x a )( y b )( z c )=0 (4) Sketcheachofthefollowingsetsandgivetheiralgebraicdescription: (i)Aspherewhosediameteristhestraightlinesegment AB where A =(1 2 3)and B =(3 2 1). (ii)Aspherecenteredat(1 2 3)thatliesintherstoctantand touchesoneofthecoordinateplanes. (iii)Thelargestsolidcubethatiscontainedinaballofradius R centeredattheorigin.Solvethesameproblemiftheballis notcenteredattheorigin. (iv)Thesolidregionthatisaballofradius R thathasacylindrical holeofradius R/ 2whoseaxisisatadistanceof R/ 2fromthe centeroftheball.Chooseaconvenientcoordinatesystem.

PAGE 17

72.VECTORSINSPACE11 (v)Theportionofaballofradius R thatliesbetweentwoparallel planeseachofwhichisatsdistanceof a
PAGE 18

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 algebraicdenitionofavector. 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

PAGE 19

72.VECTORSINSPACE13 endpointoftheorientedsegmentthatrepresents a iftheinitialpoint coincideswiththeorigin. Definition 11.2 (EqualityofTwoVectors) Twovectors a and b areequalorcoincideiftheircorrespondingcomponentsareequal: a = b a1= b1,a2= b2,a3= b3. Thisdenitionagreeswiththegeometricaldenitionofavector asaclassofallorientedsegmentsthatareparallelandhavethesame length.Indeed,iftwoorientedsegmentsrepresentthesamevector, then,afterparalleltransportsuchthattheirinitialpointscoincide withtheorigin,theirnalpointscoincidetooandhencehavethesame 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= x2 x1,y2 y1,z2 z1 accordingtothecoordinatetransformationlaw(11.1),where P0= P1. Thus,inordertondthecomponentsofthevector P1P2fromthe coordinatesofitspoints,onehastosubtractthecoordinatesofthe initialpoint P1fromthecorrespondingcomponentsofthenalpoint 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 .

PAGE 20

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 .Recallthatanif andonlyifstatementactuallyimpliestwostatements.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 .

PAGE 21

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 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

PAGE 22

1611.VECTORSANDTHESPACEGEOMETRY applications,thecomponentsofavectoroftenhavedimensions.For instance,thecomponentsofadisplacementvectoraremeasuredin unitsoflength(meters,inches,etc.),thecomponentsofavelocity vectoraremeasuredin,forexample,meterspersecond,andsoon. Themagnitudeofavector a hasthesamedimensionasitscomponents. Therefore,thecorrespondingunitvector a isdimensionless.Itspecies 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.Theshipsvelocityvectorrelative 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 withtheshipsvelocity u because B travelsthedistance u parallelto u .Thissuggestsasimplegeometricalrulefornding ABasshownin Figure11.6.Takethevector AB = v ,placethevector u sothatits initialpointcoincideswith B ,andmaketheorientedsegmentwiththe initialpointof v andthenalpointof u inthisdiagram.Theresulting vectoristhedisplacementvectorofthepersonrelativetothesurface ofthewaterin1secondandhencedenesthevelocityoftheperson 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 ;

PAGE 23

72.VECTORSINSPACE17 Figure11.6. Left :Parallelogramruleforaddingtwo vectors.Iftwovectorsformadjacentsidesofaparallelogramatavertex A ,thenthesumofthevectorsisa vectorthatcoincideswiththediagonaloftheparallelogramandoriginatesatthevertex A Right :Addingseveralvectorsbyusingtheparallelogramrule.Giventherstvectorinthesum,allother vectorsaretransportedparallelsothattheinitialpoint ofthenextvectorinthesumcoincideswiththeterminalpointofthepreviousone.Thesumisthevector thatoriginatesfromtheinitialpointoftherstvectorandterminatesattheterminalpointofthelast vector.Itdoesnotdependontheorderofvectorsin thesum. thatis,theorderinwhichthevectorsareaddeddoesnotmatter.To addseveralvectors(e.g., a + b + c ),onecanrstnd a + b bythe parallelogramruleandthenadd c tothevector a + b .Alternatively, thevectors b and c canbeaddedrst,andthenthevector a canbe addedto b + c .Accordingtotheparallelogramrule,theresulting vectoristhesame: ( a + b )+ c = a +( b + c ) Thismeansthattheadditionofvectorsis associative .Soseveralvectorscanbeaddedinanyorder.Taketherstvector,thenmovethe secondvectorparalleltoitselfsothatitsinitialpointcoincideswith thenalpointoftherstvector.Thethirdvectorismovedparallelso thatitsinitialpointcoincideswiththenalpointofthesecondvector, andsoon.Finally,makeavectorwhoseinitialpointcoincideswith theinitialpointoftherstvectorandwhosenalpointcoincideswith thenalpointofthelastvectorinthesum.Tovisualizethisprocess, imagineamanwalkingalongtherstvector,thengoingparallelto thesecondvector,thenparalleltothethirdvector,andsoon.The endpointofhiswalkisindependentoftheorderinwhichhechooses thevectors.

PAGE 24

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 Thisdenitionisequivalenttothegeometricaldenitionofadding 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 = c1 a1,c2 a2,c3 a3 ,where therelationbetweenthecomponentsofavectorandthecoordinates ofitsendpointshasbeenused.Theequalityoftwovectorsmeans theequalityofthecorrespondingcomponents,thatis, b1= c1 a1, b2= c2 a2,and b3= c3 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 Thedierenceoftwovectorscanbedenedas a b = a +( 1) b Intheparallelogramwithadjacentsides a and b ,thesumofvectors a and( 1) b representsthevectorthatoriginatesfromtheendpoint of b andendsattheendpointof a because b +[ a +( 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

PAGE 25

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

PAGE 26

2011.VECTORSANDTHESPACEGEOMETRY Solution: Let a = 1 2 3 b = 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 = 3 3 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 a )= 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 donotdependonthecoordinatesystemrelativetowhichthecomponentsofallthevectorsaredened.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 = 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 ,onends3= t s and3 3= s 3,or s =3 and t =6.Hence, c =6 a +3 b Problem11.10. Supposethethreecoordinateplanesareallmirrored. Alightraystrikesthemirrors.Determinethedirectioninwhichthe reectedraywillgo. Solution: Let u beavectorparalleltotheincidentray.Under areectionfromaplanemirror,thecomponentof u perpendicularto theplanechangesitssign.Therefore,afterthreeconsecutivereections fromeachcoordinateplane,allthreecomponentsof u changetheir signs,andthereectedraywillgoparalleltotheincidentraybutin theexactoppositedirection.Forexample,supposetherayisreected rstbythe xz plane,thenbythe yz plane,andnallybythe xy plane. Inthiscase, u = u1,u2,u3 u1, u2,u3 u1, u2,u3 u1, u2, u3 = u Remark. Thisprincipleisusedtodesignreectorslikethecatseyesonbicyclesandthosethatmarktheborderlinesofaroad.No matterfromwhichdirectionsuchareectorisilluminated(e.g.,bythe headlightsofacar),itreectsthelightintheoppositedirection(so thatitwillalwaysbeseenbythedriver).72.5.Exercises.(1) Findthecomponentsofeachofthefollowing vectorsandtheirnorms: (i)Thevectorhasendpoints A (1 2 3)and B ( 1 5 1)andisdirectedfrom A to B .

PAGE 27

73.THEDOTPRODUCT21 (ii)Thevectorhasendpoints A (1 2 3)and B ( 1 5 1)andisdirectedfrom B to A (iii)Thevectorhastheinitialpoint A (1 2 3)andthenalpoint C thatisthemidpointofthelinesegment AB ,where B = ( 1 5 1). (iv)Thepositionvectorisofapoint P obtainedfromthepoint A ( 1 2 1)bytransportingthelatteralongthevector u = 2 2 1 3unitsoflengthandthenalongthevector w = 3 0 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 Usevectoralgebratond a + b + c (3) Let uk, k =1 2 ,...,n ,beunitvectorsintheplanesuchthat thesmallestanglebetweenthetwovectors ukand uk +1is2 /n .What canbesaidaboutthesum u1+ u2+ + un?Whathappenswhen n ? (4) Aplaneiesataspeedof 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 ( 1 0 1),and C (1 1 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.

PAGE 28

2211.VECTORSANDTHESPACEGEOMETRY Itfollowsfromthisdenitionthatthedotproducthasthefollowing 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 .Therst propertystatesthattheorderinwhichtwovectorsaremultipliedinthe dotproductdoesnotmatter;thatis,thedotproductis commutative Thesecondpropertymeansthattheresultofthedotproductdoesnot dependonwhetherthevector a isscaledrstandthenmultipliedby b orthedotproduct a b iscomputedrstandtheresultmultiplied by s .Thethirdrelationshowsthatthedotproductis distributive .73.1.GeometricalSignicanceoftheDotProduct.Asitstands,thedot productisanalgebraicruleforcalculatinganumberoutofsixgiven numbersthatarecomponentsofthetwovectorsinvolved.Thecomponentsofavectordependonthechoiceofthecoordinatesystem. Naturally,oneshouldaskwhetherthenumericalvalueofthedotproductdependsonthecoordinatesystemrelativetowhichthecomponents ofthevectorsaredetermined.Itturnsoutthatitdoesnot.Therefore, itrepresentsanintrinsicgeometricalquantityassociatedwithtwovectorsinvolvedintheproduct.Toelucidatethegeometricalsignicance ofthedotproduct,notersttherelationbetweenthedotproductand thenorm(length)ofavector: a a = a2 1+ a2 2+ a2 3= a 2or a = a a Thus,if a = b inthedotproduct,thenthelatterdoesnotdepend onthecoordinatesystemwithrespecttowhichthecomponentsof a aredened.Next,considerthetrianglewhoseadjacentsidesarethe vectors a and b asdepictedinFigure11.8(leftpanel). Thentheothersideofthetrianglecanberepresentedbythedifference c = b a .Thesquaredlengthofthislattersideis (11.4) c c =( b a ) ( b a )= b b + a a 2 a b wherethealgebraicpropertiesofthedotproducthavebeenused. Therefore,thedotproductcanbeexpressedviathegeometricalinvariants,namely,thelengthsofthesidesofthetriangle: (11.5) a b = 1 2 c 2 b 2 a 2 .

PAGE 29

73.THEDOTPRODUCT23 Figure11.8. Left :Independenceofthedotproduct fromthechoiceofacoordinatesystem.Thedotproduct oftwovectorsthatareadjacentsidesofatrianglecanbe expressedviathelengthsofthetrianglesidesasshown in(11.5). Right :Geometricalsignicanceofthedotproduct. 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 .Bydenition,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)revealsthegeometricalsignicanceofthedotproduct. Itdeterminestheanglebetweentwoorientedsegmentsinspace.It providesasimplealgebraicmethodtoestablishamutualorientation oftwostraightlinesegmentsinspace.Thefollowingtheoremisuseful inpracticalapplications.

PAGE 30

2411.VECTORSANDTHESPACEGEOMETRY Theorem 11.2 (GeometricalSignicanceoftheDotProduct) Twononzerovectorsareperpendicularifandonlyiftheirdotproduct vanishes: a b a b =0 Inparticular,foratrianglewithsides a b ,and c andanangle betweensides a and b ,itfollowsfromtherelation(11.4)that c2= a2+ b2 2 ab cos Foraright-angledtriangle,thePythagoreantheoremisrecovered: c2= a2+ b2. Example 11.2 Consideratrianglewhoseverticesare A (1 1 1) B ( 1 2 3) ,and C (1 4 3) .Findalltheanglesofthetriangle. Solution: Lettheanglesatthevertices A B ,and C be ,and respectively.Then + + =180.Soitissucienttondanytwo angles.Tondtheangle ,denethevectors a = AB = 2 1 2 and b = AC = 0 3 4 .Theinitialpointofthesevectorsis A ,and hencetheanglebetweenthevectorscoincideswith .Since a =3 and b =5,bythegeometricalpropertyofthedotproduct, cos = a b a b = 0+3 8 15 = 1 3 = =cos 1( 1 / 3) 109 5. Tondtheangle ,denethevectors a = BA = 2 1 2 and b = BC = 2 2 6 withtheinitialpointatthevertex B .Thenthe anglebetweenthesevectorscoincideswith .Since a =3, b = 2 11,and a b =4 2+12=14,onendscos =14 / (6 11)and =cos 1(7 / (3 11)) 45 3.Therefore, 180 109 5 45 3= 25 2.Notethattherangeofthefunctioncos 1mustbetakenfrom 0to180inaccordancewiththedenitionoftheanglebetweentwo vectors. Theorem 11.3 (Cauchy-SchwarzInequality) Foranytwovectors a and b | a b | a b wheretheequalityisreachedonlyifthevectorsareparallel. Thisinequalityisadirectconsequenceoftherstrelationin(11.6) andtheinequality | cos | 1.Theequalityisreachedonlywhen =0 or = ,thatis,when a and b areparallel.

PAGE 31

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 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). Denetheangle thatiscountedfromthepositivedirectionofthe x axistowardthevector a .Inotherwords,theangle istheangle between e1and a .Similarly,theangles and are,bydenition,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 .

PAGE 32

2611.VECTORSANDTHESPACEGEOMETRY Figure11.9. Left :Directionanglesofavectoraredenedastheanglesbetweenthevectorandthreecoordinatesaxes.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.),vectorsareoftenspeciedby theirmagnitude a = a anddirectionangles.Thecomponentsare thenfoundby a1= a cos a2= a cos ,and a3= a cos .

PAGE 33

73.THEDOTPRODUCT27 73.3.PracticalApplications. 73.3.1.StaticProblems.AccordingtoNewtonsmechanics,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 tworopessothatthesmallestanglebetweentherstropeandtheceiling is 1andtheangle 2isdenedsimilarlyforthesecondrope.Findthe magnitudesofthetensionforcesintheropes. Solution: Setthecoordinatesystemsothatthe x axisishorizontal andorientedfromtherstropetothesecondropesasdepictedin Figure11.10(leftpanel).Theropesareinthe xy plane,whilethe gravitationalforceisinthedirectionoppositetothe y axis.Let T1and T2bethemagnitudesofthetensionforces.Theninthiscoordinate systemtheforcesactingontheballare T1= T1cos 1,T1sin 1, 0 T2= T2cos 2,T2sin 2, 0 G = 0 mg, 0 where G isthegravitationalforceand g istheaccelerationofthe freefall( g 9 8m/s2);thatis, mg istheweightoftheball.The equilibriumcondition T1+ T2+ G = 0 leadstotwoequationsforthecomponents(thethirdcomponentsofall vectorsareidentically0): T1cos 1+ T2cos 2=0 ,T1sin 1+ T2sin 2 mg =0 whichcanbesolvedfor T1and T2.Bymultiplyingtherstequation bysin 1andthesecondbycos 1andthenaddingthem,onegets T2= mg cos 1/ sin( 1+ 2).Substituting T2intotherstequation, thetension T1isobtained.

PAGE 34

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 .Alawofphysicsstatesthatachangeinanobjectskinetic energyisequaltothework W donebythisforce: K2 K1= F P1P2= W, where K1and K2arethekineticenergiesattheinitialandnalpoints ofthemotion,respectively. Example 11.4 Letanobjectslideonaninclinedplanewithout frictionunderthegravitationalforce.Findthenals peed v ofthe objectiftherelativeheightoftheinitialandnalpointsis h andthe objectwasinitiallyatrest. Solution: Choosethecoordinatesystemsothatthedisplacement vector P1P2andthegravitationalforceareinthe xy plane.Letthe y axisbeverticalsothatthegravitationalforceis F = 0 mg, 0 where m isthemassand g istheaccelerationofthefreefall.Theinitial pointischosentohavethecoordinates(0 ,h, 0)whilethenalpoint is( L, 0 0),where L isthedistancetheobjecttravelsinthehorizontal

PAGE 35

73.THEDOTPRODUCT29 directionwhilesliding.Thedisplacementvectoris P1P2= L, 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 = 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 t, 1 and b = t,t, 3+ t areorthogonal. Solution: Bythegeometricalpropertyofthedotproduct,twovectorsareorthogonalifandonlyiftheirdotproductvanishes.Therefore, a b =2 t2+ t (3 t ) (3+ t )=( t +1)2 4=0.Thesolutionsofthis equationare t =1and t = 3. Problem11.13. Describethesetofpointsinspacewhoseposition vector r satisesthecondition ( r a ) ( r b )=0 .Hint: Notethat thepositionvectorsatisfyingthecondition r c = R describesa sphereofradius R whosecenterhasthepositionvector c Solution: Theequationofaspherecanalsobewrittenintheform r c 2=( r c ) ( r c )= R2.Theequation( r a ) ( r b )=0can

PAGE 36

3011.VECTORSANDTHESPACEGEOMETRY betransformedintothesphereequationbycompletingthesquares. Usingthealgebraicpropertiesofthedotproduct, ( r a ) ( r b )= r r r ( a + b )+ a b =( r c ) ( r c ) c c + a b c =1 2( a + b ) c c a b = R R R =1 2( a b ) Hence,thesetisasphereofradius R = R ,anditscenterispositioned at c 73.5.Exercises.(1) Findthedotproduct a b if (i) a = 1 2 3 and b = 1 2 0 (ii) a = e1+3 e2 e3and b =3 e1 2 e2+ e3(2) Forwhatvaluesof b arethevectors 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 ( 2 4 3),and C (1 2 1). (5) Findtheunitvectorparallelto a = 2 1 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 b areorthogonal. (12) Consideraparallelogramwithadjacentsidesoflength a and b .If d1and d2arethelengthsofthediagonals,provetheparallelogram law: d2 1+ d2 2=2( a2+ b2).

PAGE 37

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 = a11a22 a12a21, thatis,theproductofthediagonalelementsminustheproductofthe o-diagonalelements. Definition 11.10 Thedeterminantofa 3 3 matrix A isthe numberobtainedbythefollowingrule: det a11a12a13a21a22a23a31a32a33 = a11det A11 a12det A12+ a13det A13=3k =1( 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( 1)k + mamkdet Amkforany m =1 2 3 det A =3m =1( 1)k + mamkdet Amkforany k =1 2 3 ,

PAGE 38

3211.VECTORSANDTHESPACEGEOMETRY wherethematrix Amkisobtainedfrom A byremovingtherowand columncontaining amk.Thisdenitionofthedeterminantisextended 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 = 123 013 121 Solution: Expandingthedeterminantovertherstrowyields det A =1(1 1 2 3) 2(0 1 ( 1) 3)+3(0 2 ( 1) 1)= 8 Alternatively,expandingthedeterminantoverthesecondrowyields thesameresult: det A = 0(2 1 3 2)+1(1 1 ( 1) 3) 3(1 2 ( 1) 2)= 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 e2det a1a3b1b3 + e3det a1a2b1b2 = a2b3 a3b2,a3b1 a1b3,a1b2 a2b1 (11.8)

PAGE 39

74.THECROSSPRODUCT33 Notethattherstrowofthematrixconsistsoftheunitvectors paralleltothecoordinateaxesratherthannumbers.Forthisreason,it isreferredastoa formal matrix.Theuseofthedeterminantismerely acompactwaytowritethealgebraicruletocomputethecomponents ofthecrossproduct. Thecrossproducthasthefollowingpropertiesthatfollowfromits denition: a b = b a ( a + c ) b = a b + c b ( s a ) b = s ( a b ) Therstpropertyisobtainedbyswappingthecomponentsof 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 computedrstandthenscaledby s (change aito saiin(11.8)andthenfactorout s ).74.3.GeometricalSignicanceoftheCrossProduct.Theabovealgebraicdenitionofthecrossproductusesaparticularcoordinatesystemrelativetowhichthecomponentsofthevectorsaredened.Does thecrossproductdependonthechoiceofthecoordinatesystem?To answerthisquestion,oneshouldinvestigatewhetherbothits direction andits magnitude dependonthechoiceofthecoordinatesystem.Let usrstinvestigatethemutualorientationoftheorientedsegments a b ,and a b .Asimplealgebraiccalculationleadstothefollowing result: a ( a b )= a1( a2b3 a3b2)+ a2( a3b1 a1b3)+ a3( a1b2 a2b1)=0 Bytheskewsymmetryofthecrossproduct,itisalsoconcludedthat b ( a b )= 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 thechoiceofthecoordinatesystemmodulothereection a b a b So,byasuitablerotation,thecoordinatesystemcanbeorientedsothat thecrossproductisparalleltothe z axis.Thenthevectors a and b

PAGE 40

3411.VECTORSANDTHESPACEGEOMETRY Figure11.11. Left :Geometricalinterpretationofthe crossproductoftwovectors.Thecrossproductisavectorthatisperpendiculartobothvectorsintheproduct. Itslengthequalstheareaoftheparallelogramwhoseadjacentsidesarethevectorsintheproduct.Ifthengers oftherighthandcurlinthedirectionofarotationfrom thersttosecondvectorthroughthesmallestanglebetweenthem,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 b sin acos b)= e3 a b sin( b a) Twoimportantconclusionscanbededucedfromthisexpression.74.3.1.TheRight-HandRule(theCross-ProductDirection).Bydenition,theangles aand brangeovertheinterval[0 2 ).Let0 bethe(smallest)anglebetweenthevectors a and b (justasdenedby theirdotproduct).Thelengths a and b andtheangledierence b aareindependentoftheorientationofthecoordinateaxesinthe planeandsomustbethecrossproduct.Inparticular,onecanchoose the x axisparalleltothevector a (or a=0).Then b= if b and b=2 if
PAGE 41

74.THECROSSPRODUCT35 knownasthe right-handrule : Ifthengersoftherighthandcurlin thedirectionofarotationfrom a toward b throughthesmallestangle betweenthem,thenthethumbpointsinthedirectionof a b Thus,thecrossproductisalwaysperpendiculartotheplanecontaining a and b andoriented(upordownrelativetotheplane) accordingtotheright-handrule. Remark. Thetransformationinwhichthecoordinateaxeschange theirdirectiontotheoppositeiscalledthe paritytransformation .Evidently,undertheparitytransformation,coordinatesofeverypoint changetheirsign,andhenceeveryvector(denedasanorderedtriple ofnumbers)changesitsdirection, a = a1,a2,a3 a1, a2, a3 = a .However,thecrossproductoftwovectorsdoesnotchangeunder theparitytransformation: a b ( a ) ( b )= a b .Forthis reason,thecrossproductissometimesreferredtoasa pseudovector or an axialvector .Thecoordinatesystemsrelatedbytheparitytransformationcannotbeobtainedfromoneanotherbyrotations,justlike leftandrightareswappedinamirrorreection.Thereareforces innaturethatareaxialvectors.Sotheworldanditsmirrorimagecan bedistinguishedbystudyingtheresultsoftheactionsofsuchforces. Physicalexperimentsrevealthattheparitysymmetryisindeedbroken inourUniverse!74.3.2.TheAreaofaParallelogram(theCross-ProductMagnitude).By thedenitionoftheangle ,sin 0.Therefore,themagnitudeofthe crossproductisexpressedviathegeometricalinvariantsthelength 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 (GeometricalSignicanceoftheCrossProduct) 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.

PAGE 42

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 calculationsoftheareasofplanarguresinspace. 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 1 3) ,and C =( 1 3 1) Findtheareaofthetriangle ABC andavectornormaltotheplane thatcontainsthetriangle. Solution: Accordingtothegeometricalpropertiesofthecrossproduct,inordertondavectornormaltoaplane,oneshouldtake thecrossproductofanytwononparallelvectorsintheplane.For example, a = AB = 1 2 2 and b = AC = 2 2 0 .Then a b = 4 4 6 isnormaltotheplane.Notethatthecross productofanyotherpairofvectorscorrespondingtothesidesofthe trianglecanonlybeascaledvector s 4 4 6 becauseanytwonor-malvectorsofagivenplanemustbeparallelandhenceproportional. Since 4 4 6 =2 2 2 3 =2 17,theareaofthetriangle ABC is 17byCorollary11.2.Theunitsherearesquaredunitsof lengthusedtomeasurethecoordinatesofthetrianglevertices(e.g., m2ifthecoordinatesaremeasuredinmeters). 74.4.StudyProblems.Problem11.14. Findthemostgeneralvector r thatsatisestheequations a r =0 and b r =0 ,where a and b arenonzero,nonparallel vectors.

PAGE 43

74.THECROSSPRODUCT37 Solution: Theconditionsimposedon r holdifandonlyifthevector r isorthogonaltobothvectors a and b .Therefore,itmustbeparallel totheircrossproduct.Thus, r = t ( a b )foranyreal t Problem11.15. Usegeometricalmeanstondthecrossproductsof 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= e3). Sinceacyclicpermutationofthreeindices { ijk }{ kij } (andsoon) consistsoftwopermutationsofanytwoindices,therelationbetween theunitvectorscanbecastintheform ei= ej ek, { ijk } = { 123 } andcyclicpermutations Problem11.16. Provethebac cabrule: d = a ( b c )= b ( a c ) 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 = p ( a b )forsomereal p .Sincethemagnitudeofthecrossproduct isindependentofthechoiceofthecoordinatesystem,thenumber p canbexedbycomputing 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 = e3b2 c andtherefore,fora generic a = a1,a2,a3 a ( b c )= e1 c a2b2+ e2b2 c a1= c a2b2+( b b1 e1) c a1= b c a1 c ( a1b1+ a2b2)= b ( c a ) c ( a b ) thatis, p =1.Ofcourse,thestatementcanalsobeprovedbya directuseofthealgebraicdenitionofthecrossproduct(abrute-force method).

PAGE 44

3811.VECTORSANDTHESPACEGEOMETRY Problem11.17. ProvetheJacobiidentity a ( b c )+ b ( c a )+ c ( a b )= 0 Solution: Notethatthesecondandthirdtermsontheleftsideare obtainedfromtherstbycyclicpermutationsofthevectors.Making useofthe baccab ruleforthersttermandthenaddingtoitits twocyclicpermutations,onecanconvinceoneselfthatthecoecients ateachofthevectors a b ,and c areaddeduptomake0. Remark. NotethattheJacobiidentityimpliesinparticularthat a ( b c ) =( a b ) c ; thatis,themultiplicationlawdenedbythecrossproductdoesnot 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) Denethevectorproductoftwovectors 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=vector)canbedenedinaEuclideanspaceofmorethantwo dimensions. Problem11.19. Let u beavectorrotatinginthe xy planeaboutthe z axis.Givenavector v ,ndthepositionof u suchthatthemagnitude ofthecrossproduct v u ismaximal.

PAGE 45

74.THECROSSPRODUCT39 Solution: Foranytwovectors, v u = v u sin ,where istheanglebetween v and u .Themagnitudeof v isxed,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 = v1/v2.Thisequationhastwosolutionsinthe range0 < 2 : = tan 1( v1/v2)and = tan 1( v1/v2)+ Geometrically,thesesolutionscorrespondtothecasewhen u isparallel totheline y = ( v1/v2) x inthe xy plane. 74.5.Exercises.(1) Findthecrossproduct a b if (i) a = 1 2 3 and b = 1 0 1 (ii) a = e1+3 e2 e3and b =3 e1 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, anditsterminalpointisinrstquadrantofthe 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 =( 1 0 2)beverticesofaparallelogram.Iftheothertwoverticesareobtainedbymoving A and B by 3unitsoflengthalongthevector a = 2 1 2 ,ndtheareaofthe parallelogram. (6) Considerfourpointsinspace.Supposethatthecoordinates ofthepointsareknown.Describeaprocedurebasedonvectoralgebratodeterminewhetherthepointsareinoneplane.Inparticular,arethepoints(1 2 3),( 1 0 1),(1 3 1),and(0 1 2)inone plane? (7) Letthesidesofatrianglehavelengths a b ,and c andletthe anglesattheverticesoppositetothesides a b ,andc be,respectively,

PAGE 46

4011.VECTORSANDTHESPACEGEOMETRY ,and .Provethat sin a = sin b = sin c Hint: Denethesidesasvectorsandexpresstheareaofthetriangle viathevectorsateachvertexofthetriangle. (8) Considerapolygonwithfourvertices A B C ,and D .Ifthe coordinatesoftheverticesarespecied,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 whoseadjacentsidesarediagonalsoftherstparallelogram.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 ) Itfollowsfromthealgebraicdenitionofthecrossproductandthe denitionofthedeterminantofa3 3matrixthat a ( b c )= a1det b2b3c2c3 a2det b1b3c1c3 + a3det b1b2c1c2 =det a1a2a3b1b2b3c1c2c3 Thisprovidesaconvenientwaytocalculatethenumericalvalueof thetripleproduct.Iftworowsofamatrixareswapped,thenits determinantchangessign.Therefore, a ( b c )= b ( a c )= c ( b a ) Thismeans,inparticular,thattheabsolutevalueofthetripleproduct isindependentoftheorderofthevectorsinthetripleproduct.

PAGE 47

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.GeometricalSignicanceoftheTripleProduct.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 (GeometricalSignicanceoftheTripleProduct) Thevolume V ofaparallelepipedwhoseadjacentsidesarethevectors a b ,and c istheabsolutevalueoftheirtripleproduct: V = | a ( b c ) | .

PAGE 48

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 1) ,and D(0 2 3) areinthesameplane. Solution: Considerthevectors a = AB = 1 1 1 b = AC = 2 0 2 ,and c = AD = 1 1 2 .Thepointsinquestionareinthe sameplaneifandonlyifthevectors a b ,and c arecoplanar,or a ( b c )=0.Onends. a ( b c )=det 1 11 202 112 =1(0 2)+1(4+2)+1(2 0)=6 =0 Therefore,thepointsarenotinthesameplane. Thetripleproductcanbeusedtondthedistancesbetweentwosets ofpointsinspace.Let S1and S2betwosetsofpointsinspace.Leta point A1belongto S1,letapoint A2belongto S2,andlet | A1A2| be thedistancebetweenthem.75.2.DistancesBetweenLinesandPlanes.Ifthelinesorplanesin spacearenotintersecting,thenhowcanonendthedistancebetween them?Thisquestioncanbeansweredusingthegeometricalproperties ofthetripleandcrossproducts.

PAGE 49

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 | x1 x2| ,where0 0canbemadesmallerthananypreassigned positivenumberbytaking x1and x2closeenoughto1.Sincethe distance D 0,theonlypossiblevalueis D =0.Intuitively,thesets areseparatedbyasinglepointthatisnotanextendedobject,and hencethedistancebetweenthemshouldvanish.Inotherwords,there aresituationsinwhichtheminimumof | A1A2| isnotattainedforsome A1S1,orsome A2S2,orboth.Nevertheless,thedistancebetween thesetsisstillwelldenedasthelargestnumberthatislessthanor equaltoallnumbers | A1A2| .Suchanumberiscalledthe inmum of thesetofnumbers | A1A2| anddenotedinf | A1A2| .Thus, D =inf | A1A2| ,A1S1,A2S2. Thenotation A1S1standsforapoint A1belongstotheset S1, orsimply A1isanelementof S1.Thedenitionisillustratedin 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

PAGE 50

4411.VECTORSANDTHESPACEGEOMETRY Figure11.13. Left :Distancebetweentwopointsets S1and S2denedasthelargestdistancethatislessthan 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

PAGE 51

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. Denethevectors 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 .

PAGE 52

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 thatsatisestheequation 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.

PAGE 53

75.THETRIPLEPRODUCT47 75.4.Exercises.(1) Determinewhetherthepoints A =(1 2 3), B = (1 0 1), C =( 1 1 2),and D =( 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 e2 e3, b =2 e1 e2+ e3, and c =3 e1+ e2 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 = 1 2 2 (8) Considertwolines.Therstlinepassesthroughthepoints (1 2 3)and(2 1 1),whiletheotherpassesthroughthepoints( 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 1). Hint: Ifthelineisnotparalleltotheplane,thentheyintersectandthe distanceis0.Socheckrstwhetherthelineisparalleltotheplane. Howcanthisbedone? (10) Considerthelinethroughthepoints(1 2 3)and(2 1 2).If asecondlinepassesthroughthepoints(1 ,1 ,s )and(2 1 0),ndall valuesof s ,ifany,atwhichthedistancebetweenthelinesis9 / 2units.

PAGE 54

4811.VECTORSANDTHESPACEGEOMETRY 76.PlanesinSpace76.1.AGeometricalDescriptionofaPlaneinSpace.Letaplane P gothroughapoint P0.Clearly,therearemanyplanesthatcontain aparticularpointinspace.Allsuchplanescanbeobtainedfroma particularplanebyageneralrotationaboutthepoint P0.Toeliminate thisfreedomanddenetheplaneuniquely,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 denedbyapoint P0thatbelongstoitandanormal n .Insome coordinatesystem,thepoint P0hascoordinates( x0,y0,z0)andthe vector n isspeciedbyitscomponents 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 r0= x x0,y y0,z 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 x0)+ n2( y y0)+ n3( z 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

PAGE 55

76.PLANESINSPACE49 Figure11.15. Left :Algebraicdescriptionofaplane. If r0isapositionvectorofaparticularpointintheplane and r isthepositionvectorofagenericpointinthe plane,thenthevector r r0liesintheplaneandis perpendiculartoitsnormal,thatis, n ( r 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 denedbecauseitsmagnitudeisirrelevantforthegeometricaldescriptionoftheplane.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 .

PAGE 56

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 n ).Sotherangeof canalwaysberestrictedtotheinterval[0 ,/ 2].Indeed,if happens tobeintheinterval[ / 2 ](i.e.,cos 0),thentheangle / 2can alsobeviewedastheanglebetweentheplanesbecauseonecanalways reversethedirectionofoneofthenormals n1 n1or n2 n2so thatcos cos Theplanesareperpendiculariftheirnormalsareperpendicular. Forexample,theplanes x + y + z =1and x +2 y 3 z =4are perpendicularbecausetheirnormals n1= 1 1 1 and n2= 1 2 3 areperpendicular: n1 n2=1+2 3=0(i.e., n1 n2). Example 11.8 Findanequationoftheplanethroughthreegiven points A (1 1 1) B (2 3 0) ,and C ( 1 0 3) Solution: Aplaneisspeciedbyaparticularpoint 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 1 and b = AC = 2 1 2 .Then onecantake n = a b = 3 0 3 .Anequationoftheplaneis 3( x 1)+0( y 1)+( 3)( z 1)=0,or x 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

PAGE 57

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 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= r1 r0,and (11.11) D = | n ( r1 r0) | n = | n r1 d | n whichisabitmoreconvenientforcalculatingthedistanceiftheplane isdenedalgebraicallybyanequation n r = d .76.3.1.DistanceBetweenParallelPlanes.Equation(11.11)allowsus toobtainasimpleformulaforthedistancebetweentwoparallelplanes denedbytheequations n r = d1and n r = d2(seeFigure11.15, rightpanel): (11.12) D = | d1 d2| n Indeed,thedistancebetweentwoparallelplanesisthedistancebetween therstplaneandanypoint r0inthesecondplane.By(11.11),this

PAGE 58

5211.VECTORSANDTHESPACEGEOMETRY distanceis D = | n r1 d1| / n = | d2 d1| / n because n r0= d2foranypointinthesecondplane. Example 11.9 Findanequationofaplanethatisparalleltothe plane 2 x y +2 z =2 andatadistanceof3unitsfromit. Solution: Thereareafewwaystosolvethisproblem.Fromthe geometricalpointofview,aplaneisdenedbyaparticularpointinit anditsnormal.Sincetheplanesareparallel,theymusthavethesame normal n = 2 1 2 .Notethatthecoecientsatthevariablesinthe planeequationdenethecomponentsofthenormalvector.Therefore, theproblemisreducedtondingaparticularpoint.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.Tondaparticular pointonthegivenplane,onecansettwocoordinatesto0andnd thevalueofthethirdcoordinatefromtheequationoftheplane.Take, forinstance, P0(1 0 0).Particularpointsontheparallelplanesare r0+ n = 1 0 0 + 2 1 2 = 3 1 2 and,similarly, r0 n = 1 1 2 .Usingthesepointsinthestandardequationofaplane, theequationsoftwoparallelplanesareobtained: 2 x y +2 z =11and2 x y +2 z = 7 Analternativealgebraicsolutionisbasedonthedistanceformula (11.12)forparallelplanes.Anequationofaplaneparalleltothegiven oneshouldhavetheform2 x y +2 z = d .Thenumber d isdetermined bytheconditionthat | d 2 | / n =3or | d 2 | =9,or d = 9+2. 76.4.StudyProblems.Problem11.22. Findanequationoftheplanethatisnormalto astraightlinesegment AB andbisectsitif A =(1 1 1) and B = ( 1 3 5) Solution: Onehastondaparticularpointintheplaneandits normal.Since AB isperpendiculartotheplane, n = AB = 2 2 4 Themidpointofthesegmentliesintheplane.Hence, P0(0 2 3)(the coordinatesofthemidpointsarethehalfsumsofthecorresponding coordinatesoftheendpoints).Theequationreads 2 x +2( y 2)+ 4( z 3)=0or x + y +2 z =8.

PAGE 59

76.PLANESINSPACE53 Problem11.23. Findanequationoftheplanethroughthepoint P0(1 2 3) thatisperpendiculartotheplanes x + y + z =1 and x y + 2 z =1 Solution: Onehastondaparticularpointintheplaneandany vectorperpendiculartoit.Therstpartoftheproblemiseasytosolve: P0isgiven.Let n beanormaloftheplaneinquestion.Then,fromthe geometricaldescriptionofaplane,itfollowsthat n n1= 1 1 1 and n n2= 1 1 2 ,where n1and n2arenormalsofthegivenplanes.So n isavectorperpendiculartotwogivenvectors.Bythegeometrical propertyofthecrossproduct,suchavectorcanbeconstructedas n = n1 n2= 3 1 2 .Hence,theequationreads3( x 1) ( y 2) 2( z 3)=0or3 x y 2 z = 5. Problem11.24. Determinewhethertwoplanes x +2 y 2 z =1 and 2 x +4 y +4 z =10 areparalleland,ifnot,ndtheanglebetweenthem. Solution: Thenormalsare n1= 1 2 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 =cos 1(1 / 9). 76.5.Exercises.(1) Findanequationoftheplanethroughtheorigin andparalleltotheplane2 x 2 y + z =4.Whatisthedistancebetween thetwoplanes? (2) Dotheplanes2 x + y z =1and4 x +2 y 2 z =10intersect? (3) Consideraparallelepipedwithonevertexattheoriginatwhich theadjacentsidesarethevectors a = 1 2 3 b = 2 1 1 ,and c = 1 0 1 .Findequationsoftheplanesthatcontainthefacesofthe parallelepiped. (4) Determinewhethertheplanes2 x + y 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( 1 2, 1).Giveageometricaldescriptionofthe set. (7) Findanequationoftheplanethatisperpendiculartothe plane x + y + z =1andcontainsthelinethroughthepoints(1 2 3) and( 1 1 0). (8) Towhichoftheplanes x + y + z =1and x +2 y z =2isthe point(1 2 3)theclosest?

PAGE 60

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,thereareinnitelymanylinesthrough 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 isuniquelyspeciedbyaparticular point P0of L andanyvector v towhichthelineisparallel, v L Remark. Theverynotionofaline,denedastheshortestpathbetweentwopointsinspace,isdeeplyrootedintheverystructureof spaceitself.Howcanalineberealizedinthespaceinwhichwelive? Onecanuseapieceofrope,asintheancientworld,ortheline ofsight(i.e.,thepathtraveledbylightfromonepointtoanother). Einsteinstheoryofgravitystatesthatstraightlinesdenedastrajectoriestraversedbylightarenotexactlythesameasstraightlines inaEuclideanspace.SoaEuclideanspacemayonlybeviewedasa mathematicalapproximation(ormodel)ofourspace.Agoodanalogy wouldbetocomparetheshortestpathsinaplaneandonthesurfaceof asphere;theyarenotthesameasthelatterarecurved.Theconcept ofcurvatureofapathisdiscussedinthenextchapter.Theshortest

PAGE 61

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.Einsteinstheoryofgeneralrelativityassertsthatabettermodel ofourspaceisa Riemannspace .Asucientlysmallneighborhoodin aRiemannspacelookslikeaportionofaEuclideanspace.77.2.AnAlgebraicDescriptionofaLine.Insomecoordinatesystem,a particularpointofaline L hascoordinates P0( x0,y0,z0),andavectorparallelto L isdenedbyitscomponents, v = v1,v2,v3 .Let r = x,y,z beapositionvectorofagenericpointof L andlet r0= x0,y0,z0 bethepositionvectorof P0.Thevector r r0must beparallelto L becauseitistheorientedstraightlinesegmentconnectingtwopointsof L (seeFigure11.17,leftpanel).Hence,apoint ( x,y,z )belongsto L ifandonlyif r r0 v .Theequivalentalgebraic conditionreads r 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 r0isparalleltothelineandhence mustbeproportionaltoavector v ,thatis, r r0= t v forsomerealnumber t Right :Distancebetweenapoint P1andaline L through apoint P0andparalleltoavector v .Itistheheightof theparallelogramwhoseadjacentsidesarethevectors P0P1and v .

PAGE 62

5611.VECTORSANDTHESPACEGEOMETRY Theorem 11.9 (EquationsofaLine) Thecoordinatesofthepointsoftheline L throughapoint P0and paralleltoavector v satisfythevectorequation (11.13) r = r0+ t v
PAGE 63

77.LINESINSPACE57 77.3.RelativePositionsofLinesinSpace.Twolinesinspacecanbe intersecting,parallel,orskew.Givenanalgebraicdescriptionofthe lines,howcanonendoutwhichoftheabovethreecasesoccurs?If thelinesareintersecting,howcanonendthecoordinatesofthepoint 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.Howcanonendthepointof 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.

PAGE 64

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.Thisdenes threeequationsontwovariables t and s Right :Intersectionpointofaline L andaplane P .The terminalpointofthevector rttraverses L as t ranges overallrealnumbers.Ifthelineintersectstheplane denedbytheequation r n = d ,thenthereshouldexist aparticularvalueof t atwhichthevector rtsatisesthe equationoftheplane: rt n = d Thesystemhasthreeequationsforonlytwovariables.Itisan overdetermined system,whichmayormaynothaveasolution.Fromthe abovegeometricalanalysis,itfollowsthat,ifthelinesareparallel(i.e., v1 v2= 0 ),thenthesystemhasnosolution(thelinesaredistinct), oritmighthaveinnitelymanysolutions(thelinescoincide).For example,put P1= P2and v1=2 v2.Thenthesystemissatisedby anypair( t,s =2 t ),where t isanyreal.Thesystemhasnosolutionif thecriterionfortwolinestobeskewissatised.Finally,thesystem musthavetheonlysolutionifthecriterionfortwononparallellinesto beintersectingissatised.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.

PAGE 65

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

PAGE 66

6011.VECTORSANDTHESPACEGEOMETRY Figure11.19. Left :IllustrationtoStudyProblem 11.25.Thevectors rsand rttraceouttwogivenskewed lines L1and L2,respectively.Thereareparticularvalues of t and s atwhichthedistance rt 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 rs rtisparallelto L ;thatis,thevector rs rtbecomesperpendiculartobothvectors v1and v2.Thecorresponding algebraicconditionsare rs rt v1 ( rs rt) v1=4+4 s 6 t =0 rs rt v1 ( rs rt) v2=5+5 s 4 t =0 Thissystemhasthesolution t =0and s = 1.Thus,thepointswith thepositionvectors rt =0= r1and rs = 1= r2 v2= 2 1 2 belong to L .Sothevector v = rs = 1 rt =0= 1 3 1 isparallelto L .

PAGE 67

77.LINESINSPACE61 Takingaparticularpointof L tobe P1,theparametricequationsread x =1+ t y =1 3 t z =1 t Problem11.26. Consideralinethroughtheoriginthatisparallelto thevector v = 1 1 1 .Findtheportionofthislinethatliesinsidethe sphere x2+ y2+ z2 x 2 y 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 t2 6 t =9or t2 2 t =3andhence t = 1and t =3.The pointsofintersectionare( 1 1 1)and(3 3 3).Thelinesegment connectingthemcanbedescribedbytheparametricequations x = t y = t z = t ,where 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 2 y + z =1. (3) Isthelinethroughthepoints(1 2 3)and(2 1 1)perpendiculartothelinethroughthepoints(0 1 1)and(1 0 2)?Arethelines intersecting?Ifso,ndthepointofintersection. (4) Determinewhetherthelines x =1+2 t y =3 t z =2 t and x +1= y 4=( z 1) / 3areparallel,skew,orintersecting.Ifthey intersect,ndthepointofintersection. (5) Findthevectorequationofthestraightlinesegmentfromthe point(1 2 3)tothepoint( 1 1 2). (6) Let r1and r2bepositionvectorsoftwopointsinspace.Find thevectorequationofthestraightlinesegmentfrom r1to r2. (7) Considertheplane x + y + z =0andapoint P =(1 2 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.

PAGE 68

6211.VECTORSANDTHESPACEGEOMETRY (9) Findparametricequationsofthelinethatisparallelto v = 2 1 2 andgoesthroughthecenterofthesphere x2+ y2+ z2= 2 x +6 z 6.Restricttherangeoftheparametertodescribetheportion ofthelinethatisinsidethesphere. (10) Lettheline L1passthroughthepoint A (1 1 0)parallelto thevector v = 1 1 2 andlettheline L2passthroughthepoint B (2 0 2)paralleltothevector w = 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+ z2 R2=0,thatis, A = B = C =1, p = q = v =0, = = and D = R2,where R istheradiusofthesphere.Theequationthat denesquadricsurfacesisthemostgeneralequation quadratic inallthe coordinates.Thisiswhysurfacesdenedbyitarecalled quadric .The taskhereistoclassifyalltheshapesofquadricsurfaces.Theshape doesnotchangeunderitsrigidrotationsandtranslations.Onthe otherhand,theequationthatdescribestheshapewouldchangeunder rotationsandtranslationsofthecoordinatesystem(seeSection71 andExample11.2).Thefreedominchoosingthecoordinatesystem canbeusedtosimplifytheequationforquadricsurfaceandobtaina classicationofdierentshapesdescribedbyit.78.1.QuadricCylinders.Considerrstasimplerprobleminwhichthe 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 = R2,thecrosssectionofthesurface S byanyhorizontalplaneisacircle x2+ y2= R2.Sothesurface S isacylinderofradius R thatissweptbythecirclewhenthelatter isshiftedupanddownparalleltothe z axis.Similarly,ageneral

PAGE 69

78.QUADRICSURFACES63 cylindricalshapeisobtainedbyshiftingacurveinthe xy planeupand downparalleltothe z axis.Thetaskhereistoclassifyallpossible shapesofquadriccylinders. InExample11.2,itwasestablishedthat,underacounterclockwise rotationofthecoordinatesystemthroughanangle x x cos + y sin and y y cos x sin .Bysubstitutingthetransformedcoordinatesintotheequationfor S ,oneobtainstheequationforthe very same shapeinthenewrotatedcoordinatesystem.Thefreedomof choosingtherotationangle canbeusedtosimplifytheequation.In particular,itisalwayspossibletoadjust sothatinthenewcoordinatesystemtheequationfor S doesnotcontainthemixedterm xy Indeed,afterthesubstitutionofthetransformedcoordinatesintothe equation,thecoecientat xy denesanew p : p 2( A B )cos sin + a (cos2 sin2 )=( A B )sin(2 ) + a cos(2 )= p. Therefore,theterm xy disappearsfromtheequationiftheangle satisesthecondition p=0or (11.15)tan(2 )= q B 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 B )cos(2 ) a sin(2 )]= A, B 1 2[ A + B ( A B )cos(2 )+ a sin(2 )]= B, cos sin = cos + sin = where satises(11.15). Dependingonthevaluesof A B ,and p ,thefollowingthreecases canoccur.First, A= B=0,whichisonlypossibleif A = B = p =0. Inthiscase, S isdenedbytheequation 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 x02+( y y0)=0 ,x0= 2 A,y0= 1 Ax2 0 D ;

PAGE 70

6411.VECTORSANDTHESPACEGEOMETRY hereitisassumedthat =0(otherwise,theequationdoesnotdene acurveinthe xy planeandhence S isnotasurface).Thisequation denesaparabola 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 x0)2+ B( y y0)2= D, where x0= 2 A,y0= 2 B,D= D + 1 2 ( Ax2 0+ By2 0) Finally,afterthetranslationoftheorigintothepoint( x0,y0),the equationbecomes Ax2+ By2= D. If D=0,thenthisequationdenestwostraightlines y = mx ,where m =( A/B) 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, x ),which canbeusedtoreversethesignof A/D.Nowput A/D=1 /a2and B/D= 1 /b2(dependingonwhether B/Dispositiveornegative) sothattheequationbecomes x2 a2 y2 b2=1 Whentheplusistaken,thisequationdenesanellipse.Whenthe minusistaken,thisequationdenesahyperbola. Theaboveresultsaresummarizedinthefollowingtheorem(see Figure11.20). Theorem 11.10 (ClassicationofQuadricCylinders) Ageneralequationforquadriccylinders S = ( x,y,z ) Ax2+ By2+ pxy + x + y + D =0 canbebroughttooneofthefollowingstandardformsbyasuitable rotationandtranslationofthecoordinatesystem: y ax2=0(paraboliccylinder) x2 a2+ y2 b2=1(ellipticcylinder) ,

PAGE 71

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/a2 y2/b2=1. x2 a2 y2 b2=1(hyperboliccylinder) provided A B ,and p donotvanishsimultaneously.If A = B = p =0 then S isaplane.78.2.ClassicationofGeneralQuadricSurfaces.Theclassicationof generalquadricsurfacescanbecarriedoutinthesameway(i.e.,by simplifyingthegeneralquadraticequationbymeansofrotationsand translationsofthecoordinatesystem).First,onecanprovethatthere existsarotationofthecoordinatesystemsuchthatinthenewcoordinatesystemthequadraticequationdoesnothavemixedterms: 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 (ClassicationofQuadricSurfaces) Byrotatingandtranslatingarectangularcoordinatesystem,ageneralequationforquadricsurfacescanbebroughteithertooneofthe

PAGE 72

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 b2 z2 c2=1(hyperboloidofonesheet) x2 a2 y2 b2+ z2 c2=1(hyperboloidoftwosheets) z c = x2 a2+ y2 b2(ellipticparaboloid) z c = x2 a2 y2 b2(hyperbolicparaboloid) Itshouldbenotedagainthattheshapeofthesurfacedoesnot dependonhowthecoordinateaxesarecalled.Sotheshapedoes notchangeunderanypermutationofthecoordinates( x,y,z )inthe standardequations;onlytheorientationoftheshaperelativetothe

PAGE 73

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(asaddle).Across sectionbyahorizontalplaneisahyperbola.Across sectionbyaverticalplaneisaparabola.

PAGE 74

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,anellipsoidisaspherestretched 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 k = y2/b2,where k = x2 0/a2,or z 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/b2 z2/c2= k ,where k = x2 0/a2,or x2/a2 z2/c2= k ,where k = y2 0/b2.So thecharacteristicgeometrical propertyofaconeisthathorizontaltracesareellipses;itsvertical tracesareeitherapairoflines,iftheplanecontainstheconeaxis,

PAGE 75

78.QUADRICSURFACES69 orhyperbolas .If a = b ,theconeiscalleda circularcone .Inthis case,verticaltracesintheplanescontainingtheconeaxisareapair oflineswiththesameslope c/a = c/b .Theangle betweentheaxis oftheconeandanyoftheselinesdenestheconeuniquelybecause 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 hyperbolicparaboloidhasthecharacteristicshapeofasaddle. 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 cc or z = z0< c areellipses. Theuppersheettouchestheplane z = c atthepoint(0 0 ,c ),whilethe lowersheettouchestheplane z = c atthepoint(0 0 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.

PAGE 76

7011.VECTORSANDTHESPACEGEOMETRY For z< 0,theconverseholds.Inotherwords,thehyperboloidoftwo sheetsliesinsidethecone,whilethehyperboloidofonesheetlies outsideit.78.4.StudyProblems.Problem11.27. Classifythequadricsurface 3 x2+3 y3 2 xy =4 Solution: Theequationdoesnotcontainonevariable(the z coordinate).Thesurfaceisacylinder.Todeterminethetypeofcylinder,considerarotationofthecoordinatesysteminthe xy planeand choosetherotationanglesothatthecoecientatthemixedterm vanishes.Accordingto(11.15), A = B =3andhence = / 4.Then A=( A + B 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 x2 2 x + y + z =0 Solution: Bycompletingthesquares,theequationcanbetransformedintotheform( x 1)2+( y 1)+ z =0.Aftershiftingtheorigin tothepoint(1 1 0),theequationbecomes x2+ y z =0.Considerrotationsofthecoordinatesystemaboutthe x axis: y cos y +sin z z cos z sin y .Underthisrotation, y z (cos +sin ) y + (sin cos ) z .Therefore,for = / 4,theequationassumesoneof thestandardforms x2+ 2 y =0,whichcorrespondstoaparabolic cylinder. Problem11.29. Classifythequadricsurface x2+ z2 2 x +2 z y =0 Solution: Bycompletingthesquares,theequationcanbetransformedintotheform( x 1)2+( z +1)2 ( y +2)=0.Thelatter canbebroughtintooneofthestandardformsbyshiftingtheoriginto thepoint(1 2 1): x2+ z2= y ,whichisacircularparaboloid.Its symmetryaxisisparalleltothe y axis(thelineofintersectionofthe planes x =1and z = 1)anditsvertexis(1 2 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 1)2+( y 2)2= z2forall valuesof t and .Thus,thelinesformadoubleconewhoseaxisis paralleltothe z axisandwhosevertexis(1 2 0).Alternatively,one couldnoticethatthevector vrotatesaboutthe z axisas changes.

PAGE 77

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 = x2 z2(iii)4 x2+2 y2+ z2=4 (iv) x2 y2+ z2= 1 (v). y2+4 z2=16 (vi). x2 y2+ z2=1 (2) Reduceeachofthefollowingequationstooneofthestandard form,classifythesurface,andsketchit: (i) x2+ y2+4 z2 2 x +4 y =0 (ii) x2 y2+ z2+2 x 2 y +4 z +2=0 (iii) x2+4 y2 6 x + z =0 (3) Findanequationforthesurfaceobtainedbyrotatingtheline y =2 x aboutthe y axis. (4) Findanequationforthesurfaceconsistingofallpointsthat areequidistantfromthepoint(1 1 1)andtheplane z =2.

PAGE 78

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

PAGE 79

CHAPTER12 VectorFunctions 79.CurvesinSpaceandVectorFunctions Todescribethemotionofapointlikeobjectinspace,itsposition vectorsmustbespeciedateverymomentoftime.Avectorisdened 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. Mostcommonlyusedrulestodeneavectorfunctionarealgebraicrulesthatspecifycomponentsofavectorfunctioninacoordinatesystemasfunctionsofarealvariable: r ( t )= x ( t ) ,y ( t ) ,z ( t ) .For example, r ( t )= 1 t, ln( t ) ,t2 or x ( t )= 1 t,y ( t )=ln( t ) ,z ( t )= t2. Unlessspeciedotherwise,thedomainofthevectorfunctionistheset D ofallvaluesof t atwhichthealgebraicrulemakessense;thatis, allthreecomponentscanbecomputedforany t from D .Intheabove example,thedomainof x ( t )is
PAGE 80

7412.VECTORFUNCTIONS Figure12.1. Left :Theterminalpointofavector r ( t ) whosecomponentsarecontinuousfunctionsof t traces outacurveinspace. Right :Graphingaspacecurve.Drawacurveinthe xy planedenedbytheparametricequations x = x ( t ), y = y ( t ).Itistracedoutbythevector R ( t )= x ( t ) ,y ( t ) 0 Thisplanarcurvedenesacylindricalsurfaceinspace 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.Thefunctionofarealvariableisuniquelydened byitsgraph.Thisisnotsoforvectorfunctions.Supposetheshapeof acurveisdescribedgeometrically,thatis,asapointsetinspace(e.g., alinethroughtwogivenpoints).Onemightaskthequestion:Whatis avectorfunctionthattracesoutagivencurveinspace?Theanswer tothisquestionisnotunique.Forexample,aline L asapointsetin spaceisuniquelydenedbyitsparticularpointandavector v parallel toit.If r1and r2arepositionvectorsoftwoparticularpointsof L thenbothvectorfunctions r1( t )= r1+ t v and r2( t )= r2 2 t v trace out L becausethevector 2 v isalsoparallelto L Thefollowing,moresophisticatedexampleisalsoofinterest.Supposeonewantstondavectorfunctionthattracesoutasemicircle ofradius R .Letthesemicirclebepositionedintheupperpartofthe

PAGE 81

79.CURVESINSPACEANDVECTORFUNCTIONS75 xy plane( y 0).Thefollowingthreevectorfunctionstraceoutthe semicircle: r1( t )= t, R2 t2, 0 R t R, r2( t )= R cos t,R sin t, 0 0 t r3( t )= 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 rangesoverthespeciedinterval.Itcan thereforebeconcludedthattherearemanyvectorfunctionsthattrace outthesamecurveinspacedenedasapointsetinspace. Anotherobservationisthattherearevectorfunctionsthattrace outthesamecurveinoppositedirections.Intheaboveexample,the vectorfunction r2( t )tracesoutthesemicirclecounterclockwise,while thefunctions r1( t )and r3( t )dosoclockwise.Soavectorfunction denesthe 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 determinedbyitspositioninapl