微积分教学-chapter14课件.ppt

1、Chapter 14 Partial Derivatives 14.1 Functions of Several Variables 14.2 Limits and Continuity 14.3 Partial Derivatives 14.4 Tangent Planes and Linear Approximations 14.5 The Chain Rule 14.6 Directional Derivatives and the Gradient Vector 14.7 Maximum and Minimum Values 14.8 Lagrange MultipliersSo fa

2、r we have dealt with the calculus of functions of a single variable.But,in the real world,physical quantities often dependon two or more variables,so in this chapter we turn our attention to functions of several variables and extend the basic ideas of differential calculus to such functions.14.1 Fun

3、ctions of Several VariablesIn this section we study functions of two or more variables from four points of view:nVerbally (by a description in words)nNumerically (by a table of values)nAlgebraically (by an explicit formula)nVisually (by a graph or level curves)Functions of Two VariablesDefinition A

4、function of two variables is a rule that assigns to each ordered pair of real numbers(x,y)in a set D a unique real number denoted by f(x,y).The set D is the domain of f and its range is the set of values that f takes on,that is,f(x,y)|(x,y)D.We often write z=f(x,y)to make explicit the value taken on

5、 by f at the general point(x,y).The variables x and y are independent variablesand z is the dependent variable.（x,y）f(x,y)xyzfExample 1 Find the domains of the following functions and evaluate f(3,2).(a)(b)Solution(a)(b)11),(xyxyxf)ln(),(2xyxyxf1,01,2613123)2,3(xyxyxDf22,0)32ln(3)2,3(yxyxDf2201yxxxy

6、22ln()1xzyxxy220010yxxxyExample Find the domain ofSolution 22()(,)10D fx y xyandyxandxthe domain is：yyyyyxooooo ooyx221xy000oO o0ooExample32(,)23f u vuuvv12,uvthenxyLet 321 21122(,)()2()()3()fx yxxyy321412xxyy3212(,),(,)23.Find fwhere f x yxxyyxy32(,)23f x yxxyySolutionGraphs Another way of visualiz

7、ing the behavior of a function of two variables is to consider its graph.Definition If f is a function of two variables with domain D,then the graph of f is the set of all points(x,y,z)in suchthat z=f(x,y)and(x,y)is in D.3RxyzD(,(,)x y f x y(,0)x yS 22zxy22zRxyLevel curvesDefinition The level curves

8、 of a function of two variables are the curves with equations f(x,y)=kwhere k is a constant(in the range of f).Sketch some level curves of the functionExample22(,)4f x yxyFunctions of Three or More Variables A function of three variables,f,is a rule that assigns to each ordered triple(x,y,z)in a dom

9、ain D a unique real number denoted by f(x,y,z).A function of n variables,f,is a rule that assigns a number to an n-tuple of real number.We denoted by the set of all such n-tuples.3RnR12(,)nx xx12(,)nzf x xx14.2 Limits and ContinuityDefinition Let f be a function of two variables whosedomain D includ

10、es points arbitrarily close to(a,b).Then we say that the limit of f(x,y)as(x,y)approaches(a,b)is L and we write if for every number there is a corresponding number such that whenever and(,)(,)lim(,)x yabf x yL00(,)f x yL(,)x yD220()()xaybExample Find if it exists.222(,)(0,0)3limx yx yxySolutionLet 0

11、Since 222223033x yyxyxyIf we choose 3,then22230 x yxywhenever220 xyHence 222(,)(0,0)3lim0 x yx yxyIf as along a path and as along a path ,where ,then does not exist.Example does not exist.Solution as along the x-axis as along the y-axis(,)(,)lim(,)x ya bf x y1(,)f x yL(,)(,)x ya b1C2(,)f x yL(,)(,)x

12、 ya b2C12LL2222(,)(0,0)limx yxyxy(,)1f x y(,)(0,0)x y(,)1f x y (,)(0,0)x y Thus the given limit does not exist.Example does not exist?22(,)(0,0)limx yxyxySolution2222(,)(0,0)(,)(0,0)1limlim2x y xx y xxyx xxyxx2222(,2)(0,0)(,)(0,0)22limlim54x yxx y xxyxxxyxxThus the given limit does not exist.Since E

13、xample Find.11lim00yxyxyx)11(1)1(lim200yxxyyxyx21111lim00yxyx.11lim00yxyxyxSolutionExample Find1Solution01xysinxylim.x01xysinxylimx01xysinxylimyxy22222001()()xycos xylimxyExample FindSolution222rxy,Let2404 1r(cos r)limr2304224rsin rrlimrSoContinuityDefinition A function f of two variables is called

14、continuous at(a,b)if We say f is continuous on D if f is continuous at Every point(a,b)in D.Example (,)(,)lim(,)(,)x ya bf x yf a b23322 33 2(,)(1,2)lim(32)12123 1 2 2 11x yx yx yxy Example where is the following function continuous.2223(,)0(,)0(,)0 x yifx yf x yxyifx ySolutionThe function f is cont

15、inuous for(,)0 x y since it is equal to a rational function there.The function f is continuous at(0,0)because 222(,)(0,0)3lim0 x yx yxySo the function is continuous in 2.RExample where is the following function discontinuous.11),(22yxyxf22(,)1Dx y xySolutionsince f a rational function,it is continuo

16、us on itsdomain,which is the set So the function is discontinuous on the circle.221xy00(,)()(,)zf x yg xf x yyy00(,)()(,)zf x yh yf xyxxyxz0 xyToxT0y0M机动 目录 上页 下页 返回 结束 14.3 Partial DerivativesIf f is a function of two variables,its partial derivativesare the functions and defined by Notations for p

17、artial derivatives If ,we writexfyf0(,)(,)(,)limxhf x h yf x yf x yh0(,)(,)(,)limyhf x y hf x yf x yh11(,)(,)xxxfzfx yff x yfD fD fxxx22(,)(,)yyyfzfx yff x yfD fD fyyy(,)zf x yRule for finding partial derivatives of 1.To find regard as a constant and differentiate with respect to 2.To find regard as

18、 a constant and differentiate with respect toExample If ,find and .Solution(,)zf x yyfxfxy(,)f x y(,)f x yxy3232(,)2f x yxx yy2323(,)32(2,1)3 22 3 116xxfx yxxyf 2222(,)34(2,1)3 2 14 18yyfx yx yyf (2,1)xf(2,1)yfSolutionSolution1:xz)2,1(xzSolutionSolution2:)2,1(xz)2,1(yz,32yx yzyx23,82312)2,1(yz722134

19、62xx1)62(xx81xz231yy 2)23(yy72yz机动 目录 上页 下页 返回 结束 Example If find and223yyxxz(1,2)xf(1,2).yfExample Ifzyzxxzyx2ln1 proof:xzyzxxzyxln1 yyxx yz,1yxyxxylnz2机动 目录 上页 下页 返回 结束，）and1,0(xxxzyprove thatSolutionSolution:机动 目录 上页 下页 返回 结束 Example If find and,)32(sin22xyyxzzxzyInterpretations of partial deriva

20、tives00),(dd00 xxyxfxxfxxyy0),(yyyxfz00),(dd00yyyxfyyfxxyy0),(xxyxfzcan be interpreted geometrically as the slope of tangent through of curveyxz0 xyToxT0y0M机动 目录 上页 下页 返回 结束 o yM T0M0MoxM Tcan be interpreted geometrically as the slope of tangent through of curve ),()y,x(,),()y,x(,yxxy)y,x(f0000022So

21、lution),(fx00 00000 x),(f),x(flimxxxxlimx00020 ,0),(fy00 00000 y),(f)y,(flimyyyylimy00020 ,0does not exist2200yxxylim),()y,x(0,0).at)y,x(flim),()y,x(00),(fx00),(fy00 andboth exist but f is not continuousbecause If f is a function of three variables x,y,z,its partial derivatives with respect to x is

22、defind as0(,)(,)(,)limxhf x h y zf x y zf x y zhExample If 222zyxrSolution:yr2222zyxrxrzzr,ryfindrxry.rzandxrx2Higher derivativesIf then the second partial derivatives of f(,)zf x y22112222122221222222()()()()xxxxxyxyyxyxyyyyffzfffxxxxffzfffyxy xy xffzfffxyx yx yffzfffyyyy If then the partial deriva

23、tives of order 3(,)zf x y233111233233112222()()xxxxxxxxyxxyffzfffxxxxffzfffyxy xy x Example find the second partial derivatives of Clairauts Theorem Suppose f is defined on a disk D that contains the point(a,b).If the functions and are both continuous on D,then 3232(,)2f x yxx yy23233232(,)32(,)3262

24、(,)326xxxxyf x yxxyfx yxxyxyxfx yxxyxyy22222222(,)34(,)346(,)3464yyxyyf x yx yyfx yx yyxyxfx yx yyx yyxyfyxf(,)(,)xyyxfa bfa bSolutionyz Solution,xxyyx 23922zzx yxy,yyx19622 322()zzxy xxy,xy12 Example find32331,Ifzx yxyxy32zxy22yxlnz .yz022 proof),yxln(z2221 xz yz 22xz 22yz ,yxx22 ,yxy22 222222)yx(x

25、x)yx(,)yx(xy22222 222222)yx(yy)yx(.)yx(yx22222 22xz 22yz 22222)yx(xy 22222)yx(yx .0 Example Show that the function 22xz is a solution of the equation232tan.xzzzIf zfindyxyxy(1).yzzIf zxyfindxyLinear Approximations and Differentials(),(,()()()()()()()()()f x is a differentiable function thetangentlin

26、e of f at a f aisyf afa xaThe approximationf xf afa xais called thelinear approximation or tangent lineapproximation of f at a x is near aa(,()a f a()f x()L xx0()()()()()lim0 xyf xxf xdyfxxfx dxydyyfxxx (,()x f xxxxydy()()()()f xf afa xaDifferentials Let y=f(x)be a differentiable function with indep

27、endent variable x.Then the differential,dx,of the independent variable x is an arbitrary increment of x;that is,dx=x;the differential,dy,of dependent variable y at the point x is dy=f(x)dx.14.4 Tangent Planes and Linear ApproximationsSuppose f has continuous partial derivatives.An equation of the ta

28、ngent plane to the surface at the point is .(,)(,)()(,)()xyzf a bf a b x af a b y b(,)zf x y(,(,)p a b f a b(,)(,)(,)()(,)()xyf x yf a bf a b x af a b y bThe approximation is called the linear approximation.(,)(,)(,)()(,)()xyf x yf a bf a b x af a b y b(,)(,)(,)(,)xyf ax byf a bf a b xf a b y(,)(,)x

29、yzf a b xf a b y xx ayy b Definition If ,then f is differentiable at(a,b),if can be expressed in the form where and as .Theorem If the partial derivatives and existnear(a,b)and are continuous at(a,b),then f isdifferentiable at(a,b).(,)zf x yz12(,)(,)xyzf a b xf a b yxy 120(,)(0,0)x y xfyfDifferentia

30、lsFor a differentiable function of two variables,we define the differentials and to be independent variables;that is,they can be given any values.Then the differential ,also called the total differential,is defined by(,)zf x ydxdydz(,)(,)xyzzdzfx y dxfx y dydxdyxyIf we take then the differential of

31、is ,dxxxa dyyyb z(,)()(,)()xydzf a b x af a b y b(,)(,)(,)()(,)()(,)xyf x yf a bf a b x af a b y bf a bdzthe linear approximation can be written as ),()y,x(,),()y,x(,yxxy)y,x(f0000022Solution),(fx00 00000 x),(f),x(flimxxxxlimx00020 ,0),(fy00 00000 y),(f)y,(flimyyyylimy00020 ,0(0,0).at),(fx00),(fy00

32、andbut is not continuousboth exist),(fx00),(fy00 and(0,0)(0,0)0 xydzfxfy 1(,)(0,0)2zf x yf We have But at all points on the lineSo f is not differentiable at(0,0).yxFor a differentiable function of three variables,we define the differentials ,and to be independent variables;that is,they can be given

33、 any values.Then the total differential du,is defined by(,)uf x y zdxdydzuuududxdydzxyyExample Find the differential of the function:)1,0().3(,2sin).2(,).1(22xxxueyxuyyxzyzyzSolution(1).dyyzdxxzdz xydx2 dyyx)2(2dzzudyyudxxudu).2(dx dyzeyyz)2cos21(dzyeyz dzzudyyudxxudu).3(dxyzxyz 1 xdylnzxyz xdzlnyxy

34、z 14.5 The Chain RuleExample If 2,xyzedtdz3sin,xt and ytwhereFindSolution 3sin2ttze33sin232sin2(sin2)(cos6)ttttdzettttedtBecause So zyxtThe Chain Rule(Case 1)Suppose that is a differentiable function of and ,where and are both differentiable functions of .Then is a differentiable function of and ord

35、zf dxf dydtx dty dtdzz dxz dydtx dty dt(,)zf x y()xg t()yh txyzttzyxtproofA change of in produces changes of in and in .these,in turn.produce a change of in .since f is differentiable,sotxyztxyz12ffzxyxyxy 120(,)(0,0)x y where andasDividing both sides of this equation by,twe get12zfxfyxytxtyttt 1200

36、limlim()ttdzzfxfyxydttxtytttf dxf dyx dty dt Thus dtdzdtdyyzdtdxxz tcoseyx 2 2232t)(eyx ).tt(cosettsin2263 zyxtExample 2,xyzedtdz3sin,xt and ytwhereFindSolution Applying Case 1 of the Chain Rule,we getLet ,0uf tvg tug tvthen z=f t=uSo dzz duz dvdtu dtv dt 1lnvvvuftuug t 1lng tg tg tf tftf tf t gt ,0

37、,.g tdf twhere f tf t and g tdtare differentiableFind Example Let z=excosy,x=sint and y=t2.FindSolution We find that.dtdzsin22cos(cos)(sin)(2)(cos cos2 sin).xxtdzz dxz dydtx dty dteyteytetttt The Chain Rule(Case 2)Suppose that is a differentiable function of and ,where and are differentiable functio

38、ns of and .Then(,)z f x y(,)xg u v(,)yh u vxyuvzzxzyux uy u zzxzyvx vy v yxzuvExample Let z=xy,x=3u2+v2,and y=4u+2v.Find and Solution zu.zv122421224222122421224222(6)4(ln)6(42)(3)4(3)ln(3),(2)2(ln)2(42)(3)2(3)ln(3).yyuvuvyyuvuvzzxzyuxuyuyxuxxuuvuvuvuvzzxzyvxvyvyxvxxvuvuvuvuvApplying Case 2 of the Ch

39、ain Rule,we getyxzuv),(),(mtsvmtsusztz),(vufz),(),(mtsmtsfzvuzstsvvzsuuztvvztuuzmmzmuuzmuuzLet ThenExample ,zzxy(1)yzxyIf FindSolution 121ln(1)ln(1)(1)(1)()ln(1)(1)ln(1)1yyyxyyxyyzyxyyyxyxzeeyxyyyyxxyxyyxyExample ,zzxy(1)yzxyIf FindSolution 121111ln0(1)ln1(1)ln(1)1vvvyvvyLet uxyvythen zuuxyvyzzuzvvu

40、yuuyxyxuxv xzzuzvvuxuuyuyv yxxyxyyxy vuzxyExample zzxyzxyxy()zxyxF uIf and F is differentiable,show thatSolution 21()()()()()()()zuyF uxF uyF uxF uxxyzuxxxF uxxF uyyyzzxyzxyxy where xuyExample (,)zf x yIf findSolution 22zzxzyzzrsrx ry rxy where 22,2,xrsyrs.2rszLetf),xyz,zyx(fw has continuous second.

41、,2xzwxwLet,zyxu,xyzv then),(vufw xw 1f 2fyz 2wz x 21fyzfz zf 12f y zfyz 2wxyzvu1f 2f xvvfxuuf )xyz,zyx(ff 11)xyz,zyx(ff 22xyz-order partial derivatives,findzf 1zvvfzuuf 1111f zf 2zvvfzuuf 2221f 12fxy 22fxy xzw211f 12fxy 2f y 21f(yz )fxy22 1211f)zx(yf 222f zxy 2f y ,uff 1,vff 2,uff 111,vff 112,uff 22

42、1,vff 222Example 2uxyzIf findSolution where cos,sin,xpryprzpr.,urupuImplicit DifferentiationWe suppose that an equation of the form defines implicitly as a differentiable function of ,that is,where for all in the domain of f.If F is differentiable.Then(,)0F x y yx()yf x(,()0Fx f x x0 xyFd xFd yxd xy

43、d xFFd yxFd xFy We suppose that an equation of the form defines implicitly as a differentiable function of and ,that is,where for all in the domain of f.If F is differentiable.Then (,)0Fxyz zxy(,)zf x y(,(,)0F x y f x y(,)x y00FxFzxxzxFyFzyyzyxzyzFFzxFxFzFFzyFyFzExample 1 Find if Solution Let then E

44、xample 2 Find and if Solution Let then ysincossin cosxyxy(,)sincossin cosF x yxyxycoscos cos.sincos sinxyFdyxxyydxFyxy zxzy33361xyzxyz333(,)61F x y zxyzxyz222222.22xzyzFzxyzxFzxyFzyxzyFzxy Example 1 Find if Solution Let then Example 2 Find and if Solution Let then y336xyxy33(,)60F x yxyxy 222.2xyFdy

45、xyydxFyx zxzy33361xyzxyz333(,)61F x y zxyzxyz222222.22xzyzFzxyzxFzxyFzyxzyFzxy 0 xyzez),(zyxFxyzezxz zxFF xyeyzz xyeyzzxyz2)xz(y )xyeyz(yz 2)xye()xyze(yz)xye)(yzyz(zzz 3222)xye()yxxyzee(zzzz yz zyFF xyexzz Example 1Find and if yzSolutionxyz2Example Find and if Solution zxzy(),()()xyzf u uup t dt14.6

46、 Directional Derivatives and the Gradient VectorDefinition The directional derivative of f at in the direction of a unit vector is ),(00yxba,uhyxfhbyhaxfyxfh),(),(lim)(D000000,0uif this limit exists.The partial derivative of with respect to and are special cases of the directional derivative.fxyof a

47、t the point in the direction off),(00yxp)(D0,0uyxfis the rate of change.,ubaTheorem if is a differentiable function of and ,then has a directional derivative in the direction of any unit vector andfxyfba,ubyxfayxfyxfDyxu),(),(),(ProofIf we define a function of the single variableghby),()(00hbyhaxfhg

48、then),(),(),(lim)0()(lim)0(00000000yxfDhyxfhbyhaxfhghgguhhOn the other hand,we can write).,(),()(00yxfhbyhaxfhgBy the Chain Rule,we havebyxfayxfhyyfhxxfhgyx),(),()(.),(),()0(0000byxfayxfgyxIt follows thatTherefore.),(),(),(000000byxfayxfyxfDyxuExample Find the directional derivative of at thegiven p

49、oint in the direction indicated by the anglef.Solution 3)0,2(sin),(xxyxfbyxfayxfyxfDyxu),(),(),(3sin,3cossin,cos,ubaThe Gradient VectorTheorem if is a function of two variables and ,then the gradient of is the vector funtion defined byfxyffji),(),(),(yfxfyxfyxfyxfyxbyxfayxfuyxfyxfDyxu),(),(),(),(We

50、havegreidint Definition The directional derivative of f at in the direction of a unit vector is ),(000zyxcba,uhzyxfhzzhbyhaxfzyxfh),(),(lim),(D000000000,0uif this limit exists.hfuhffh)x()x(lim)x(D0000uIf we use vector notation,thenWhere If n=3.000,xzyxuzyxfzyxfDu),(),(kji),(),(),(),(zfyfxfzyxfzyxfzy