《高数双语》课件section 9.5.pptx

1、Section 9.5FermatJacobi,Jakob 2Differentiation of Implicit Functions隐函数隐函数For an equation 1212,(,)0,nnF xxxxxx if there exists a function of n variables 12(x),x(,)R,nnyx xx such that when we substitute it into the equation,the equationbecomes an identity 12(,)0,nF xxxy is called an implicit function

2、 determined by the equation.(x)y then Differentiation of Implicit FunctionsTheorem(Existence of implicit functions 隐函数存在定理隐函数存在定理)(2)Both partial derivatives of the function F are continuous in asatisfies the conditions:Suppose that the function(,)F x y00(,)0.yFxy(3)Then(1)00(,)0;F xy neighborhood o

3、f the point 00(,);xyand ,()0.F x f x in a()yf x (1)There exists one and only one function has a continuous derivative in the()yf x(2)The function 00(,),xy00()yf x such that neighbourhood of the point 00(,)U xy00(,),U xyandneighbourhood.xyFdydxF 34Differentiation of Implicit FunctionsThe proof of thi

4、s theorem is beyond the scope of this book.We justwith a continuous derivative.determines a implicit function()yf x deduce the formula under the assumption that the equation(,)0F x y ),0.(f xF x()yf x we obtain Substituting into(,)0F x y Differentiate both sides of the above identity using the chain

5、 rule,we have0.xydyFFdxthere exists a neighbourhood00(,)0,yFxy Since yFis continuous and 00(,);U xythus 0yF such that 00(,)U xyin.xyFdydxF 5Differentiation of Implicit Functionsis determined by(,)zf x y Similarly,if the function of two variables ,(,)0.F x y f x y in three variables,then we havean eq

6、uation(,)0F x y z Differentiate both sides of the above identity,we have0.yzzFFy 0,xzzFFx so that,0.yxzzzFFzzFxFyF Suppose that the function z=z(x,y)is determined bythe equationFind the total differential dz2222.xyzxyzat the point(1,0,-1).6Differentiation of Implicit FunctionsExampleSolution(I)Using

7、 the formula for derivation2220.xyzxyzx 2220.zxzzxyzxyxxyz (1,0,1)1.zx 222(1,0,1)0.zxzzxyzxyxxyz 7Differentiation of Implicit FunctionsSolution(I)Using the formula for derivation2220.xyzxyzy 2220.zyzzyxzxyyxyz (1,0,1)2.zy 222(1,0,1)0.zyzzyxzxyyxyz Therefore,(1,0,1)2.dzdxdy 8Differentiation of Implic

8、it FunctionsSolution(II)Using the invariance of the total differential form 2222.d xyzxyzd2222220.2xdxydyzdzyzdxxzdyxydzxyz222(1,0,1)0.xdxydyzdzyzdxxzdyxydzxyz(1,0,1)0.2dxdzdy (1,0,1)2.dzdxdy Finish.9Differentiation of Implicit FunctionsSolution Obviously,all the first order,.ucxaz vcybzLet has cont

9、inuous first Example Suppose that the function(,)u v is determined by order partial derivatives and the function(,)zz x y constants.Find.xyazbz where a,b and c are allthe equation(,)0,cxaz cybz(,)cxaz cybz exist and partial derivatives of the composite function we haveare continuous.221212,yyzFcczFa

10、bab 111212,xxzFcczFabab so.xyazbzc(,)(,),F x y zcxaz cybz Let10Differentiation of Implicit Functions Defined By More Than One EquationConsider the system of two equations of functions(,)0,(,)0.F x y u vG x y u vIn general,two of theThese two equations contain four variables.If they can determine two

11、variables can be determined by the others.functions of two variables with continuous partial derivatives(,),(,),uu x yvv x yso thatDifferentiate both sides of above identities,we have ,(,),(,)0.G x y u x y v x y ,(,),(,)0,F x y u x y v x y 0.xuvuvGGGxx0,xuvuvFFFxx11Differentiation of Implicit Functi

12、ons Defined By More Than One EquationThese two equations form a linear algebraic system for the two unknownand.vx quantitiesux If the determinant of coefficients(,)0,(,)uvuvFFF GJGGu v then,by the Cramers rule,we obtain(,)(,).F Gvu yyJ (,)(,),F Guy vyJ By the same way,we have(,)(,).F Gvu xxJ (,)(,),

13、F Gux vxJ Jacobi,Jakob(1804-1851)German mathematician 12Differentiation of Implicit Functions Defined By More Than One EquationExample Find the partial derivatives of the implicit functions(,)v x ydetermined by the equations(,)u x yand.uvy 222,uvx Solution(I)By formulas of derivatives of the implici

14、t functions,if(,)(,)(,)(,)F GF Gx vu v 22uvyuv 22,uuv Jwe obtain xvuvxvuvFFFFGGGG Similarly,we haveux 22.vuyuv 22,vvxuv anduvuvFFGG 22uvvu 222()uv0,13Differentiation of Implicit Functions Defined By More Than One EquationSolution(II)Find the total differentials to both sides of the equations.2222.vu

15、dxdyuvuv222,.uduvdvdxvduudvdy udxuvvdyvu dvwe have 0,J While 2222,uvdxdyuvuv dxvuvdyuvu duExample Find the partial derivatives of the implicit functions(,)v x ydetermined by the equations(,)u x yand.uvy 222,uvx FermatAppendix15Vector-Valued FunctionsJust as we did for planar curves before,to tract a

16、 particle moving in space,we run a vector r from the origin to the particle and study the xOyzr(),(),()P f tg t h tchanges in r.If the particles position coordinates are twice-differentiable functions of time,then so is r,and wecan find the particles velocity andacceleration vectors at any time by d

17、ifferentiating r.Conversely,if we have enough information about the particles initial velocity andposition,we can find r as a functionof time by integration.16Space CurvesWhen a particle moves through space during a time interval I,we thinkof the particles coordinates as functions defined on I:r()()

18、i()j()ktOPf tg th t make up the curve in space(,)(),(),(),x y zf tg t h ttIThe points The equations and interval in last that we call the particles path.equation parametrize the curve.A curve in space can also be wrote in The vector vector form.(),(),(),.xf tyg tzh ttIdefines r as a vector function

19、of the real variable t on the interval I.More generally,a vector function or vector-valued function on a domainset D is a rule that assigns a vector in space to each element in D.17Limits and ContinuityDefinition Limit and Continuity,then 00limr()r().tttt If r()()i()j()ktf tg th tin its domain if 0t

20、t A vector function r()tis continuous at a point 0000limr()lim()ilim()jlim()k.tttttttttf tg th tThe function is continuous if it is continuous at every point in its domain./4/4/4/4lim r()lim cosilim sinjlimkttttttttExample then r()(cos)i(sin)jk,ttttSuppose that 22ijk.224 18Derivativesr()tis differen

21、tiable if it is differentiable at every point is continuous andThat is if f,g and h have continuous first derivatives that areA vector function Definition Derivative at a Point is differentiable at The vector function 0tt The derivative is the vector 0rr()r()r()limijk.tdtttdfdgdhtdttdtdtdt r()()i()j

22、()ktf tg th tif f,g,and h are differentiable at t0.of its domain.r()t The curve traced by r is smooth if never 0.not simultaneously by 0.19Differentiation RulesLet u and v be differentiable vector functions of t,C a constant vector,c any scalar,and f any differentiable scalar function.1.Constant Fun

23、ction Rule:C0.ddt 6.Chain Rule:u()u()().fdff ftdt uuu.dfffdt2.Scalar Multiple Rules:uu;dccdt 3.Sum and Difference Rules:uvuv.ddt4.Dot Product Rule:u vuvu v.ddt5.Cross Product Rule:uvuvuv.ddt20Integrals of Vector FunctionsDefinition Indefinite IntegralThe indefinite integral of r w.r.t.t is the set o

24、f all antiderivatives of r,If R is any antiderivative of r,then r()R()C.t dtt denoted by r().t dt Example (cos)ij2 k)cosij2kttdttdtdttdt 2123sinijkCCijk.tttCCC 2123sinijktCtCtC21Integrals of Vector FunctionsDefinition Definite Integralare integrable over a,b,r()()i()j()k.bbbbaaaat dtf t dtg t dth t dtthen so is r,and the definite integral of r from a to b isr()()i()()ktf tg t jh tIf the component of Example0000(cos)ij2 k)cosij2kttdttdtdttdt 2200 i0 j0k 2000sinijkttt 2jk.