资源描述
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 Multipliers So far 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 Functions of Several VariablesIn this section we study functions of two or more variables from four points of view:pVerbally (by a description in words)pNumerically (by a table of values) pAlgebraically (by an explicit formula)pVisually (by a graph or level curves) Functions of Two VariablesDefinition A 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 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)xy zf Example 1 Find the domains of the following functions and evaluate f(3,2).(a) (b) Solution (a) (b) 1 1),( x yxyxf )ln(),( 2 xyxyxf 1,01, 2613 123)2,3( xyxyxDf 22, 0)32ln(3)2,3( yxyxDf 2 20 1y xxx y 2 2ln( ) 1 xz y x x y 2 2001 0y xxx y Example Find the domain ofSolution 2 2( ) ( , ) 1 0D f x y x y andy xandx the domain is： yyyyy xooooooo y x 2 2 1x y 000oO o0oo Example 3 2( , ) 2 3f u v u uv v 1 2, ,u v thenx y Let 3 21 2 1 1 2 2( , ) ( ) 2( ) ( ) 3( )f x y x x y y 3 21 4 12x xy y 3 21 2( , ), ( , ) 2 3 .Find f where f x y x xy yx y 3 2( , ) 2 3f x y x xy y Solution Graphs Another way of visualizing 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. 3Rx yzD ( , , ( , )x y f x y( , ,0)x yS 2 2z x y 2 2z R x y Level curvesDefinition The level curves 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 functionExample 2 2( , ) 4f x y x y Functions 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 domain 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. 3R nR 1 2( , , , )nx x x1 2( , , , )nz f x x x 14.2 Limits and ContinuityDefinition Let f be a function of two variables whosedomain D includes 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 y b f x y L 0 0 ( , )f x y L ( , )x y D 2 20 ( ) ( )x a y b Example Find if it exists. 22 2( , ) (0,0) 3limx y x yx y Solution Let 0 Since 2 2 22 23 0 3 3x y y x yx y If we choose 3 ,then 22 23 0 x yx y whenever 2 20 x y Hence 22 2( , ) (0,0) 3lim 0 x y x yx y If 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 y a b f x y1( , )f x y L ( , ) ( , )x y a b 1C 2( , )f x y L( , ) ( , )x y a b 2C 1 2L L2 22 2( , ) (0,0)limx y x yx y ( , ) 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?2 2( , ) (0,0)limx y xyx y Solution 2 2 2 2( , ) (0,0) ( , ) (0,0) 1lim lim 2x y x x y xxy x xx y x x 2 2 2 2( , 2 ) (0,0) ( , ) (0,0) 2 2lim lim 54x y x x y xxy x xx y x x Thus the given limit does not exist. Since Example Find .11lim00 yxyxyx )11( 1)1(lim 200 yxxy yxyx 21111lim00 yxyx .11lim00 yxyxyx Solution Example Find 1Solution 01xy sinxylim .x01xy sinxylim x 01xy sinxylim yxy 2 2 2 2 2001 ( )( )xy cos x ylim x y Example FindSolution 2 2 2r x y , Let 240 4 1r ( cos r )lim r 230 4 2 24r sin r rlim r So ContinuityDefinition A function f of two variables is called 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 y a b f x y f a b 2 3 3 2 2 3 3 2( , ) (1,2)lim ( 3 2 ) 12 12 3 1 2 2 11x y x y x y x y Example where is the following function continuous. 22 23 ( , ) 0( , ) 0 ( , ) 0 x y if x yf x y x y if x y Solution The function f is continuous for ( , ) 0 x y since it is equal to a rational function there. The function f is continuous at (0,0) because 22 2( , ) (0,0) 3lim 0 x y x yx y So the function is continuous in 2.R Example where is the following function discontinuous. 11),( 22 yxyxf 2 2( , ) 1D x y x y Solutionsince f a rational function ,it is continuous on itsdomain, which is the set So the function is discontinuous on the circle . 2 2 1x y 00( , ) ( ) ( , )z f x y g x f x yy y 00( , ) ( ) ( , )z f x y h y f x yx x yxz0 x yTo xT 0y0M 机 动 目 录 上 页 下 页 返 回 结 束 14.3 Partial Derivatives If f is a function of two variables , its partial derivativesare the functions and defined by Notations for partial derivatives If , we write xf yf0 ( , ) ( , )( , ) limx h f x h y f x yf x y h 0 ( , ) ( , )( , ) limy h f x y h f x yf x y h 1 1( , ) ( , )x x xf zf x y f f x y f D f D fx x x 2 2( , ) ( , )y y yf zf x y f f x y f D f D fy y y ( , )z f x y Rule for finding partial derivatives of 1.To find regard as a constant and differentiate with respect to 2.To find regard as a constant and differentiate with respect toExample If , find and .Solution ( , )z f x yyfxf xy ( , )f x y( , )f x y xy3 2 3 2( , ) 2f x y x x y y 2 32 3( , ) 3 2(2,1) 3 2 2 3 1 16xxf x y x xyf 2 22 2( , ) 3 4(2,1) 3 2 1 4 1 8yyf x y x y yf (2,1)xf (2,1)yf Solution1: xz)2,1(xzSolution2: )2,1(xz )2,1(yz ,32 yx yz yx 23 ,82312 )2,1(yz 72213 462 xx 1)62( xx 81xz 231 yy 2)23( yy 72yz 机 动 目 录 上 页 下 页 返 回 结 束 Example If find and22 3 yyxxz (1,2)xf (1,2).yf Example If zyzxxzyx 2ln1 proof: xz yzxxzyx ln1 yy xx yz,1yxy xxy ln z2 机 动 目 录 上 页 下 页 返 回 结 束 ，）and 1,0( xxxz yprove that Solution: 机 动 目 录 上 页 下 页 返 回 结 束 Example If find and ,)32( sin2 2 xyyxz zx zy Interpretations of partial derivatives00 ),(dd00 xxyxfxxf xx yy 0 ),(yy yxfz 00 ),(dd00 yyyxfyyf xx yy 0 ),(xx yxfzcan be interpreted geometrically as the slope of tangent through of curve yxz 0 x yTo xT 0y0M 机 动 目 录 上 页 下 页 返 回 结 束 o yM T 0M 0Mo xM Tcan be interpreted geometrically as the slope of tangent through of curve ),()y,x(, ),()y,x(,yx xy)y,x(f 000 0022Solution),(fx 00 0 0000 x ),(f),x(flimx xxxlimx 00020 ,0),(fy 00 0 000 0 y ),(f)y,(flimy yyylimy 000 20 ,0does not exist2200 yx xylim ),()y,x( (0,0).at )y,x(flim ),()y,x( 00),(fx 00 ),(fy 00and both 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 defind as0 ( , , ) ( , , )( , , ) limx h f x h y z f x y zf x y z h Example If 222 zyxr Solution: yr 2222 zyx rxrzzr ,ry find rx ry .rzandxr x2 Higher derivativesIf then the second partial derivatives of f ( , )z f x y 2 211 2 22 212 2 221 2 222 2 2( )( )( )( )x x xxx y xyy x yxy y yy f f zf f f x x x xf f zf f f y x y x y xf f zf f f x y x y x yf f zf f f y y y y If then the partial derivatives of order 3( , )z f x y 2 3 3111 2 3 32 3 3 112 2 2 2( )( )xx x xxxxx y xxy f f zf f f x x x xf f zf f f y x y x y 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 3 2 3 2( , ) 2f x y x x y y 2 32 3 32 3 2( , ) 3 2( , ) 3 2 6 2( , ) 3 2 6xxx xyf x y x xyf x y x xy x yxf x y x xy xyy 2 2 2 2 22 2 2( , ) 3 4( , ) 3 4 6( , ) 3 4 6 4yyxyyf x y x y yf x y x y y xyxf x y x y y x yy xyfyxf ( , ) ( , )xy yxf a b f a bSolution yzSolution ,xxyyx 23 922z zx y x y ,yyx 196 22 3 2 2 ( )z zx y xx y ,xy12 Example find 3 2 33 1,If z x y xy xy 32zx y 22 yxlnz .yz 022 proof ),yxln(z 2221 xz yz 22xz 22yz ,yx x 22 ,yx y 22 222 22 2)yx( xx)yx( ,)yx( xy 222 22222 22 2)yx( yy)yx( .)yx( yx 222 2222xz 22yz 222 22 )yx( xy 222 22 )yx( yx .0 Example Show that the function 22xzis a solution of the equation 2 3 2tan .x z z zIf z findy x y x y (1 ) . y z zIf z xy find x y Linear Approximations and Differentials( ) ,( , ( )( ) ( )( )( ) ( ) ( )( ) ( )f x is a differentiable function the tangentline of f at a f a isy f a f a x aThe approximationf x f a f a x ais called the linear approximation or tangent lineapproximation of f at a x is near a a( , ( )a f a ( )f x( )L xx 0 ( ) ( )( ) ( )( )lim 0 x y f x x f xdy f x x f x dxy dyy f x x x ( , ( )x f xx x x y dy( ) ( ) ( )( )f x f a f a x a Differentials Let y=f(x) be a differentiable function with independent 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 tangent plane to the surface at the point is . ( , ) ( , )( ) ( , )( ) x yz f a b f a b x a f a b y b ( , )z f x y ( , , ( , )p a b f a b ( , ) ( , ) ( , )( ) ( , )( )x yf x y f a b f a b x a f a b y b The approximation is called the linear approximation.( , ) ( , ) ( , )( ) ( , )( )x yf x y f a b f a b x a f a b y b ( , ) ( , ) ( , ) ( , ) x yf a x b y f a b f a b x f a b y ( , ) ( , )x yz f a b x f a b y x x a y y 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).( , )z f x yz 1 2( , ) ( , )x yz f a b x f a b y x y 1 2 0 ( , ) (0,0)x y xf yf DifferentialsFor 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( , )z f x y dx dydz( , ) ( , ) x y z zdz f x y dx f x y dy dx dyx y If we take then the differential of is ,dx x x a dy y y b z( , )( ) ( , )( )x ydz f a b x a f a b y b ( , ) ( , ) ( , )( ) ( , )( )( , ) x yf x y f a b f a b x a f a b y bf a b dz the linear approximation can be written as ),()y,x(, ),()y,x(,yx xy)y,x(f 000 0022Solution),(fx 00 0 0000 x ),(f),x(flimx xxxlimx 00020 ,0),(fy 00 0 000 0 y ),(f)y,(flimy yyylimy 000 20 ,0(0,0).at),(fx 00 ),(fy 00andbut is not continuousboth exist),(fx 00 ),(fy 00and (0,0) (0,0) 0 x ydz f x f y 1( , ) (0,0) 2z f x y f We have But at all points on the lineSo f is not differentiable at (0,0).y x For a differentiable function of three variables, , we define the differentials , and to be independent variables; that is, they can be given any values. Then the total differential du, is defined by( , , )u f x y z dx dydz u u udu dx dy dzx y y Example Find the differential of the function: )1,0().3(,2sin).2(,).1( 22 xxxueyxuyyxz yzyzSolution(1). dyyzdxxzdz xydx2 dyyx )2( 2 dzzudyyudxxudu ).2( dx dyzey yz)2cos21( dzyeyzdzzudyyudxxudu ).3( dxyzxyz 1 xdylnzxyz xdzlnyxyz 14.5 The Chain RuleExample If 2 ,x yz e dtdz 3sin ,x t and y t whereFindSolution 3sin 2t tz e 3 3sin 2 3 2 sin 2(sin 2 ) (cos 6 )t t t tdz e t t t t edt Because So z yx t The 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 ordz f dx f dydt x dt y dt dz z dx z dydt x dt y dt ( , )z f x y ( )x g t( )y h t x yz t tz yx t proof A change of in produces changes of in and in .these ,in turn .produce a change of in .since f is differentiable ,sot xyz t xyz 1 2f fz x y x yx y 1 2 0 ( , ) (0,0)x y where and asDividing both sides of this equation by ,t we get 1 2z f x f y x yt x t y t t t 1 20 0lim lim( )t tdz z f x f y x ydt t x t y t t tf dx f dyx dt y dt Thus dtdz dtdyyzdtdxxz tcose yx 2 22 32 t)(e yx ).tt(cose ttsin 22 63 z yx tExample 2 ,x yz e dtdz 3sin ,x t and y t whereFindSolution Applying Case 1 of the Chain Rule,we get Let , , , 0u f t v g t u g t vthen z=f t =uSo dz z du z dvdt u dt v dt 1 lnv vvu f t u ug t 1ln g tg tg t f t f tf t f t g t , 0,.g td f t where f t f t and g tdtare differentiable Find Example Let z = excosy, x = sint and y = t2. FindSolution We find that .dtdz sin 2 2cos (cos ) ( sin )(2 )(cos cos 2 sin ).x xtdz z dx z dydt x dt y dte y t e y te t t t t The Chain Rule (Case 2) Suppose that is a differentiable function of and , where and are differentiable functions of and .Then ( , )z f x y ( , )x g u v( , )y h u v x y u vz z x z yu x u y u z z x z yv x v y v yxz uv Example Let z = xy, x = 3u2 + v2, and y = 4u +2v. Find and Solution zu .zv 1 2 2 4 2 1 2 2 4 2 2 21 2 2 4 2 1 2 2 4 2 2 2(6 ) 4( ln )6 (4 2 )(3 ) 4(3 ) ln(3 ),(2 ) 2( ln )2 (4 2 )(3 ) 2(3 ) ln(3 ).y y u v u vy y u v u vz z x z yu x u y uyx u x xu u v u v u v u vz z x z yv x v y vyx v x xv u v u v u v u v Applying Case 2 of the Chain Rule,we getyxz uv ),(),( mtsvmtsu sztz ),( vufz ),(),( mtsmtsfz vuz stsvvz suuz tvvz tuuz m mz muuz muuz Let Then Example ,z zx y (1 )yz xy If FindSolution 1 2 1ln(1 ) ln(1 )(1 ) (1 )( ) ln(1 )(1 ) ln(1 ) 1y yy xy y xyyz y xy y y xyxz e e y xyy y yxxy xy y xy Example ,z zx y (1 )yz xy If FindSolution 1 2 111 1 ln 0 (1 )ln 1(1 ) ln(1 ) 1v v v yv vyLet u xy v ythen z u u xy v yz z u z v vu y u u y xyx u x v xz z u z v vu x u uy u y v y xxy xy y xy vuz xy Example z zx y z xyx y ( )z xy xF u If and F is differentiable, show thatSolution 2 1( ) ( ) ( ) ( )( ) ( ) ( )z uy F u xF u y F u xF ux x yz u xx xF u x xF uy y yz zx y z xyx y where xu y Example ( , )z f x yIf findSolution 2 2z z x z y z zr sr x r y r x y where 2 2, 2 ,x r s y rs .2 rs z Let f),xyz,zyx(fw has continuous second ., 2 xz wxw Let ,zyxu ,xyzv then ),( vufwxw 1f 2fyz 2wz x 21 fyzfz zf 1 2fy zfyz 2 w xyzvu1f 2f xvvfxuuf )xyz,zyx(ff 11 )xyz,zyx(ff 22-order partial derivatives, find zf 1 zvvfzuuf 11 11f zf 2 zvvfzuuf 22 21f 12fxy 22fxy xz w2 11f 12fxy 2fy 21f(yz )fxy 22 1211 f)zx(yf 222 fzxy 2fy ,uff 1 ,vff 2 ,uff 111,vff 112 ,uff 221 ,vff 222 Example 2u x yz If findSolution where cos , sin ,x pr y pr z p r ., urupu Implicit 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 y x( )y f x ( , ( ) 0Fx f x x0 xyF d x F d yx d x y d xF Fd y xFd x Fy 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 z xy ( , )z f x y ( , , ( , ) 0F x y f x y ( , )x y00F x F zx x z xF y F zy y z y xzyzF Fz xFx FzF Fz yFy Fz Example 1 Find if Solution Let then Example 2 Find and if Solution Let then y sn cos sin cosx y x y ( , ) sin cos sin cosF x y x y x y cos cos cos .sin cos sinxyFdy x x yy dx F y x y zx zy 3 3 3 6 1x y z xyz 3 3 3( , , ) 6 1F x y z x y z xyz 2222 22 .22xzyzFz x yzx F z xyFz y xzy F z xy Example 1 Find if Solution Let then Example 2 Find and if Solution Let then y 3 3 6x y xy 3 3( , ) 6 0F x y x y xy 22 2 .2xyFdy x yy dx F y x zx zy 3 3 3 6 1x y z xyz 3 3 3( , , ) 6 1F x y z x y z xyz 2222 22 .22xzyzFz x yzx F z xyFz y xzy F z xy 0 xyzez),( zyxF xyzez xz zxFF xye yzz xye yzz xy z2 )xz(y )xye yz(y z 2)xye( )xyze(yz)xye)(yzyz( z zz 3 222 )xye( )yxxyzee(z z zz yz zyFF xye xzz Example 1 Find and if yzSolution xy z2 Example Find and if Solution zx zy( ), ( ) ( )xyz f u u u p t dt 14.6 Directional Derivatives and the Gradient VectorDefinition The directional derivative of f at in the direction of a unit vector is ),( 00 yx ba,uh yxfhbyhaxfyxf h ),(),(lim)(D 000000,0u if this limit exists.The partial derivative of with respect to and are special cases of the directional derivative.f x y of at the point in the direction off ),( 00 yxp)(D 0,0u yxf is the rate of change.,u ba Theorem if is a differentiable function of and ,then has a directional derivative in the direction of any unit vector andf x yf ba,u byxfayxfyxfD yxu ),(),(),( ProofIf we define a function of the single variableg hby ),()( 00 hbyhaxfhg then ),(),(),(lim )0()(lim)0( 0000000 0 yxfDh yxfhbyhaxf h ghgg uh h On the other hand,we can write ).,(),()( 00 yxfhbyhaxfhg By the Chain Rule,we have byxfayxfhyyfhxxfhg yx ),(),()( .),(),()0( 0000 byxfayxfg yx It follows thatTherefore .),(),(),( 000000 byxfayxfyxfD yxu Example Find the directional derivative of at thegiven point in the direction indicated by the anglef .Solution 3)0,2(sin),( xxyxf byxfayxfyxfD yxu ),(),(),( 3sin,3cossin,cos,u ba The Gradient VectorTheorem if is a function of two variables and , then the gradient of is the vector funtion defined by f x yf fji),(),(),( yfxfyxfyxfyxf yx byxfayxfuyxfyxfD yxu ),(),(),(),( We have greidint Definition The directional derivative of f at in the direction of a unit vector is ),( 000 zyx cba ,u h zyxfhzzhbyhaxfzyxf h ),(),(lim),(D 000000000,0u if this limit exists. h fuhff h )x()x(lim)x(D 0000u If we use vector notation ,thenWhere If n=3. 000 ,x zyx uzyxfzyxfDu ),(),( kji ),(),(),(),( zfyfxf zyxfzyxfzyxfzyxf zyx czyxfbzyxfazyxfzyxfD yyxu ),(),(),(),( We have Theorem Suppose is a differentiable function of two or three variables. The maximum value of the direc-tional derivative is and it occurs when has the same direction as the gradient vector .fu )x(fDu )x(f )x(fproof coscos)x( fufuffDu where is the angle between and )x(f .uThe maximum value of the directional derivative is and it occurs when has the same direction as the gradient vector .)x(f u )x(f Example (a) If find the rate of changeof at the point in the direction from,),( yxeyxf pSolution yyyx xeeyxfyxfyxf ,),(),(),(f )0,2(pto ).2,21(Q(b) In what direction does have the maximum rate of change？ What is this maximum rate of ch-ange? fsince 2,1)0,2(f 54,5325 2,5.1u PQPQThe rate of change of in the direction from tof p)2,21(Q is 154,532,1)0,2()0,2( uffD u The maximum value of the rate of change is.52,1)0,2()x( ffDu（ b） increases fastest in the direction of the gradientvector .2,1)0,2( ff 14.7 Maximum and Minimum ValuesDefinition A function of two variables has a local maximum at (a,b) if when (x,y) is near(a,b). This means that for all points (x,y) in some disk with center (a,b). The number f(a,b) is called a local maximum value. If when (x,y) is near (a,b), then f(a,b) is called a local minimum value. If the inequalities hold for all points(x,y) in the domain of f, then f has an absolute maxi-mum (or absolute minimum) at (a,b). ( , ) ( , )f x y f a b( , ) ( , )f x y f a b ( , ) ( , )f x y f a b Theorem If f has a local maximum or minimumat (a,b) and the first-order partial derivatives of f exist there, then and .( , ) 0 xf a b ( , ) 0yf a b Proof L t If has a local maximum).,()( bxfxg fat ,then ),( ba )(xg has a local maximum at a,so0)( ag but .0),()( bafag xSimilaly, .0),( bafy A point (a,b) is called a critical point (or stationarypoint ) of f if and , or one of partialDerivatives does not exist. Theorem says that if f has a local maximum or mini-mum at (a,b), then (a,b) is a critical point of f . However, not all critical points give rise to maximumor minimum.( , ) 0 xf a b ( , ) 0yf a b Example Let f(x, y) = x2 y2. Show that the origin is the only critical point but that f has no extreme value at the origin.Solution The partial derivatives of f exist at every point in the domain of f, and we have that fx(x, y) = 2x, fy(x, y) = -2y, so that (0, 0) is the only critical point of f. However, f(0, 0) = 0 is no a extreme value of f, because f(x, 0) = x 2 0 for all x0 and f(0, y) = -y2 0. Since A0, we know that f(1, 1) = -1 is a local minimum. For (0, 0) we have A = 0, B = -3, C = 0, and D= -9 0, hence f has no extreme
展开阅读全文