LinearApproximations,Differentials线性逼近微分

4.5:LinearApproximations,DifferentialsandNewton’sMethodGregKelly,HanfordHighSchool,Richland,WashingtonForanyfunctionf(x),thetangentisacloseapproximationofthefunctionforsomesmalldistancefromthetangentpoint.yx0xafxfaWecalltheequationofthetangentthelinearizationofthefunction.Thelinearizationistheequationofthetangentline,andyoucanusetheoldformulasifyoulike.Startwiththepoint/slopeequation:11yymxx1xa1yfamfayfafaxayfafaxaLxfafaxalinearizationoffatafxLxisthestandardlinearapproximationoffata.Importantlinearizationsforxnearzero:1kx1kxsinxcosxtanxx1x1211112xxx134434415151515133xxxxfxLxThisformulaalsoleadstonon-linearapproximations:Differentials:Whenwefirststartedtotalkaboutderivatives,wesaidthatbecomeswhenthechangeinxandchangeinybecomeverysmall.yxdydxdycanbeconsideredaverysmallchangeiny.dxcanbeconsideredaverysmallchangeinx.Letbeadifferentiablefunction.Thedifferentialisanindependentvariable.Thedifferentialis:yfxdxdydyfxdxExample:Consideracircleofradius10.Iftheradiusincreasesby0.1,approximatelyhowmuchwilltheareachange?2Ar2dArdr2dAdrrdxdxverysmallchangeinAverysmallchangeinr2100.1dA2dA(approximatechangeinarea)2dA(approximatechangeinarea)Comparetoactualchange:Newarea:Oldarea:210.1102.01210100.002.01.012.01ErrorOriginalAreaErrorActualAnswer.00497510.5%0.01%.0001.01100Newton’sMethod2132fxxFindingarootfor:-3-2-1012345-4-3-2-11234WewilluseNewton’sMethodtofindtherootbetween2and3.123Guess:3213331.52f1.5tangent33mf2132fxxfxxz1.51.53z1.53z1.532.53(notdrawntoscale)(newguess)123Guess:2.5212.52.53.1252f1.5tangent2.52.5mf2132fxxfxxz.1252.5z.1252.52.452.5(newguess)123Guess:2.452.45.00125f1.5tangent2.452.45mf2132fxxfxxz.001252.45z.001252.452.449489795922.45(newguess)Guess:2.449489795922.44948979592.00000013016fAmazinglyclosetozero!ThisisNewton’sMethodoffindingroots.Itisanexampleofanalgorithm(aspecificsetofcomputationalsteps.)ItissometimescalledtheNewton-RaphsonmethodThisisarecursivealgorithmbecauseasetofstepsarerepeatedwiththepreviousanswerputinthenextrepetition.Eachrepetitioniscalledaniteration.ThisisNewton’sMethodoffindingroots.Itisanexampleofanalgorithm(aspecificsetofcomputationalsteps.)ItissometimescalledtheNewton-RaphsonmethodGuess:2.449489795922.44948979592.00000013016fAmazinglyclosetozero!Newton’sMethod:1nnnnfxxxfxThisisarecursivealgorithmbecauseasetofstepsarerepeatedwiththepreviousanswerputinthenextrepetition.Eachrepetitioniscalledaniteration.nxnfxnnfx1nnnnfxxxfxFindwherecrosses.3yxx1y31xx301xx31fxxx231fxx0112111.5211.5.8755.75.8751.51.34782615.7521.3478261.10068224.44990551.325200431.32520041.32520041.00205841TherearesomelimitationstoNewton’smethod:WrongrootfoundLookingforthisroot.Badguess.FailuretoconvergeNewton’smethodisbuiltintotheCalculusToolsapplicationontheTI-89.OfcourseifyouhaveaTI-89,youcouldjustusetherootfindertoanswertheproblem.TheonlyreasontousethecalculatorforNewton’sMethodistohelpyourunderstandingortocheckyourwork.Itwouldnotbeallowedinacollegecourse,ontheAPexamorononeofmytests.APPSSelectandpress.CalculusToolsENTERIfyouseethisscreen,press,changethemodesettingsasnecessary,andpressagain.ENTERAPPSNowlet’sdooneontheTI-89:31fxxxApproximatethepositiverootof:Nowlet’sdooneontheTI-89:APPSSelectandpress.CalculusToolsENTERPress(Deriv)F2Press(Newton’sMethod)3Entertheequation.(Youwillhavetounlockthealphamode.)Settheinitialguessto1.Press.ENTER31fxxxApproximatethepositiverootof:Settheiterationsto3.Presstoseethesummaryscreen.ESCPresstoseeeachiteration.ENTERPressandthentoreturnyourcalculatortonormal.ESCHOME