數(shù)學分析簡明教程答案

第十六章偏導數(shù)與全微§1導數(shù)與全微分的（1）ux2ln(x2y2)u(xy)uarctanyxuxyxyuxyesin(xy)uxyyxu

x解

2xln(

y)

x2y

y)

x2y2]u

2x2

2x2．x2y2ucos(xyxy)(sinxyycos(xyy(xysinxyxyycos(xyx(xysinxyu 1

(y)x

(y)x

；x2y

u

1(y)x

1 x2y

uy1 uxx uyesin(xy)xyesin(xy)cos(xyyy(1xycos(xy))esin(xy)xyyx(1xy

sin(xy)uyxy1yxlny

yxlnx 設

f(x,f(x,y)

1x2y

,x2y20,x2y2函數(shù)在(0,0

xf

f(x,0)f

0

0，即fx(0,00，而

yf

f(0,y)f

lim

不存在，

fy(0,0x2yx2y證明函數(shù)u x2yx2y(x)2(x)2lim lim

x0x2y不存在，由對稱性limyu(0,0)不存在，因而ux2y x2y2x2y2zuxeyzexyx2y2x2y2z2x2y2z

d(x2y2z2x2y2zx2y2zx2y2x2y2z2x2y2z2

dy

dzx2y2z2dud(xeyzexy)eyzdxxeyz(zdyydz)x2y2z2(eyzex)dx(xzeyz1)dyxyeyzdzx2x2y

在點(1,0和(0,1（2）uln(xy2在點(0,1和(1,1xzyu xzyxyux(y1)xy

在點(0,1x2yx2y

)

d(x2y2x2yx2yx2y2(x2y2x2

y2dx(x2y2((x2y2(x2y2)x2所以，在點(1,0du0，在點(0,1dudx

xy

(dx2ydy)

xy

dx

2yxy2

dy(0,1，dudx2dy在點(1,1du1dxdy1(1) x(1) 1(yzy2zz

1)z

xy x1

x1

x du

()yz

dx

y2

()y

dyz

()zy

dzy函數(shù)的定義域為{(x,y0xyor0yx}x0dudx

xdy(y1 ydx1 ydx1 y xy2xy2xyx2

(y1)sgn

)dx

x(1y)sgn

2yxy2yxyx2x0

f(x,y)f(0,y)lim(1

y

x

x在(0,y

(0,y

(0,y)

f(0,yy)f(0,y)lim

0 y0xy但在點(0,ydu不存在，因而uxy1xy

在點(0,1)不可微f(x,y在(0,0f(x,f(x,y)0

x2y

,x2y20x2y20解

xf

f(x,0)f

0

0fx(0,0)0fy(0,0)0f(xy在(0,0fxy

f

xf

y1x1x2x2y

更高階的無窮小，為此極

x2y

xx2y

1，x2y x2y

x2y

1(x2y2xx2xx2yx2y

0f(xy在(0,0

x2 f(x,y)x2y2,

x在(0,0

y2x2xx2x2y

x

x，所limf(xy0x2yx2y12

f(0,0)，即f(x,y在點(0,0

xf

f(x,0)f

0limyf(0,0)

f(0,y)f(0,0)0 fx(0,0)fy(0,0)0f(x,y在(0,0f(x,y)f(0,0)[fx(0,0)x

x2fy(0,0)y](x)2(y)x2x2y

更高階的無窮小量，為此極lim

(，(x0x2

x2y2

令yx(xy沿直線yx趨于(0,0

(x2(x2y2(

x2

f(xy在(0,0

f(x,y)

y

) x2y

x2y200

x2y2的偏導數(shù)存在，但偏導數(shù)在(0,0)點不連續(xù)，且在(0,0)點的任何鄰域中，而f在原

證

yf(x,y)2y(sinx2y2x2y2cosx2y2),xy

0,

x2y20,而limfx(x,y不存在，limfy(x,yfx(x,y),fy(x,y在(0,011

10,M0,1

且

MPn

,0)(P,)11

(P,1fx(Pn)fx1f(P)f

,0)1)1

MM 所以，fx(x,y)，fy(x,y)在(0,0)點的任何鄰域中均

f(x,y)f(0,0)[fx(0,0)x

fy(x)(x)2

(x)2

0((x

0,())，f(xy在(0,0可微，且在(0,0df(0,0)0設

2xy f(x,y)x2y2 x

0 證

在(0,0

0

x2y20 2xy 證明f(x2y22,

x y22xyy 2xy22x2y

x2y2

y

0

fx(0,0

fx(x,y在(0,0

亦在(0,0設

1ex(x2y2

,x2y20f(x,y)x2y0f(xy在點(0,0df(0,0

x2y2x證 f(0,0)x

1

x3o(x3

1，f(0,0)lim00yyf(x,y)f(0,0)[fx(0,0)xfy11ex(x2y2

1x(x2y2)ex(x2y2 xx2y

3(x2y2)1x2(x2y2)2(x2(x2y2 3(x2y2) x2(x2y2)2(x2(x2y2)2)0((x,y)(0,0))2f(xy在(0,0df(0,0xdx設

f(x,y)x2y2,x

0

x2y2xx(t),yy(t是通過原點的任意可微曲線（x20y20)0t0x2ty2t)0,x(t),y(t可微f(x(t),y(tf(x,y在(0,0

x3

,t0,證明（1）設(t)

f(x(t),y(t))

(t)

t

(t)

x2(t)[x2(t)x(t)3y2(t)x(t)2x(t)y(t)y(t)][x2(t)y2(t)]2

,t0而在t

t

(t)(0)t

t0t(x

x3(t)y

0 t

(x(t))2(y(t))

[x(0)]2[

0

x3t(x3(t)y2x3t(x3(t)y2t

x3

t

t0

(t)

，即

0f(x(t),y(tf

(0,0)

f(x,0)f(0,0)f(0,y)fyf(0,0)y

0f(x,y在(0,0f(x,y)f(0,0)(fx(0,0)xfy x2

x

，x2x2x2y

更高階的無窮小，為此極lim1

)

xy，設

y033

3

(2x2)

22該極限不存在，因而

xy )不是 更高階的無窮小量，因此f(x,y)

x,y（1）(1x)m(1y)nx 1 解（1）f(x,y)m(1x)m11y)nf(x,y)n(1x)m(1y)n1 fx(0,0)m，fy(0,0)n (1x)m(1y)nf(0,0)f(0,0)xf(0,0)y()1mxny x,y(1x)m(1y)n1mxny

x2y2)fx(x,y)

1

xy

1y(1

1y (1xy)2(xy)211xfy(x,y)(1xy)2(xy)2f

(0,0)1

f(0,0)f(0,0arctan00，故arctanx 1

xy()，x,yarctanx1

xyuf(xyaxbcyd內(nèi)可微，且全微分du恒為零，問f(x,y f(x,y)在該矩形內(nèi)應取常數(shù)值．證明如下由于uf(x,y在矩形內(nèi)可微，故(x,ya,bcddu

fx(x,y)dxfy(x,y)dy0

fx(x,y)0，fy(x,y)0P0x0y0f(x,y)f(x0,y0)[f(x,y)f(x0,y)][f(x0,y)f(x0,y0fx(x01(xx0),y)(xx0)fy(x0,y02(yy0)(yy0 (011,021)f(xy)

f(x0y0)Cf(x,y取常數(shù)值C

f(x0,y0)設

在

,

在

,

f(x,y在

,

證 f(x0x,y0y)f(x0,y0f(x0x,y0y)f(x0x,y0)f(x0x,y0)f(x0,y0)y在(x0y0P0(x0y0Lagrange01f(x0x,y0y)f(x0x,y0)fy(x0x,y0y)yf在

,

fy

x,

y)f

,

fy(x0x,y0y)

fy(x0y0lim0x而對f(x0xy0f(x0y0，設(xf(xy0f(x0x,y0)f(x0,y0)(x0x)(x0)，由于(x0)fx(x0,y0，故(xx0f(x0x,y0)f(x0,y0)(x0x)(x0(x0)x其中o(xx0

fx(x0,y0)xf(x0x,y0y)f(x0,y0)fy(x0,y0)yfx(x0,y0)xyy

0(0f(x,y在(x0,y0x2y（1）x2y（2）uxyyxuxsin(xy)ycos(xy)uexy解u1ln(x2y2)，u

，由對稱性

2x2

x2

x2y2ux

y2x(x2y2)2

(x2y2)

(x2y2)2

2u

x2y2(x2y2)2（2）uyy，ux1 x 2

2u

1 x

0usin(xy)xcos(xy)ysin(xy)yxcos(xy)cos(xy)ysin(xy)

2cos(xy)xsin(xy)ycos(xy)cos(xy)xsin(xy)sin(xy)ycos(x

cos(x

xsin(x

y)

y)

ycos(x

y)y)2u

xsin(x

y)

2sin(x

y)

ycos(x

y)

2

2u

y

xye2u

x2exyu

siny

sinx

；x3yxu 1u

y

)，求

x3

；yu

xyz

pqr；xryqzu

xyx

(xy

；xmynuln(axby

．xmyn解

u

3x

sinyy

cosx

6xsiny

sinx

6sinyy

cosx，

6cosy

cosxx3y

6siny

6ycosx，

6cosy6cosx

u

xy

1y

1y(1xy)2(x

1x1 1xyu

1y

，x

，

2(1y2

，

03u2(3x2

3u2(3y2

3u

，

00

u

2xcos(x

y2

2

)

y2)

12xsin(

)

y2)

y

)8y

cos(x

y2)（4）uyzexyzxyzexyz(x1)yzexyz

xy

(x1)

xy

(x2)

xyz

(xp)

xyzxp

(x

xy

(xp)

xy

(xp)(y

xyz

pquxp

(xp)(y

xyzxpyq

(xp)(y

xy

(xp)(y

xy(xp)(yq)(z1)exyz

pqrxpyqz

(xp)(yq)(z

xyzu 2 mu2(1)mm!

(x

m1u

(xm

（使用數(shù)學歸納法x

m!(xy)m2m2xmy

m!(m1)(2xmy)(xy)xmy

m!(m1)(m2)(3xmy)(x

(mn

nx．xmyn

(xy)x(ba)（6）xx(ba)

(a0)xmu(1)m1(m1)!(1)m1(m1)!am(1)m1(m1)!a

(xa

(ax

bm(y

ab

xmyn

(1)m1(m1)!amm(m1)(mnbm(yab(1)mn1(mn．(ax

2u2ux y 0（1）uln(x2y2)（2）ux2y2uexcosyuarctanyx

2u2(y2x2

2u2(x2y2

2u2u1（1

x

(x

22y2 22

(x2

)2

x

y 0u u 2x 2 2y

所 0

x

y

2ux2u2u

cosy，x2

cosy，y

sinyy ecosy

0u

2u

1

yx

)x x2y

，

(x2y2)2u 1

2u

1

y)2x

x2y2，

(x2y2)22u2u

0設函數(shù)u(xyu

u.x yx證明u(xyu(xy)) 2u(x u 2

(y)),

(

(y)u

(x

xu2u

(

(

(y)yx

(

(

(y))u即

u.x yx19．fx,fy在點(x0y0的某鄰域內(nèi)存在且在點(x0,y0fxy(x0,y0)fyx(x0,y0)證明16.4fxy(x0,y0fyx(x0,y0

f(x0x,y0y)f(x0,y0y)f(x0x,y0)f(x0,y0f,f在(x

處的可微性，下面證明

f(x

)

f(x,

)

x

51

61二者對充分小的xy同時成立，且當x0y0i0i16011．于是令xy0 （*（**）(y)f(x0x,y)f(x0,y)W式11

W

y)(

)]

1[

x,

y)f

(x0,

y)]

0

1fy在(x0y0fy(x0x,y0y)

fy(x0,y0)fyx(x0,y0)xfyy(x0,y01x2y其中1,20（當xy0時fy(x0,y0y)fy(x0,y0)fyy(x0,y0)y3y其中30（當y0時W

1{

x,

y)f

,

1{

,y0)

f

,

)x

f

,

)y1x

f(x,

)

fy(x0,y0)fyy(x0,y0)yyy

這正是（*）式．同樣，令(xf(x,y0yf(x,y0W

x)

1

1y{fx(x01x,y0y)fx(x01x,y0)}1

011fx在(x0y0fx(x01x,y0y)

fx(x0,y0)fxx(x0,y0)1xfxy(x0,y041x5y其中4,50（當xy0時fx(x01x,y0)fx(x0,y0)fxx(x0,y0)1x61xW

1{

1x,

y)f

1x,y01{f(x,

)

(x,y)x

(x,

)yx

4 fx(x0,y0)fxx(x0,y0)1x fxy(x0y041y561y，§2合函數(shù)與隱函數(shù)微分（1）u

f(ax,by)（2）uf(xy,xy)（3）uf(xy2,x2y)（4）u

f(xy

y)z（5）uf(x2y2z2)；（6）u

f(xy,

x)y解（1）uf(axbyaaf(axbyu

(ax,by)b

(ax,by) 2u

(ax,by)

af11(ax,

2u

abf21(ax,by)，

f22(ax,by)（2）u

f1(xy,xy)f

(xy,xy)yf1(xy,xy)f2(xy,xy)

(xy,xy)

(xy,xy)f

(xy,xy)f

(xy,xf11(xy,xy)2f12(xy,xy)f22(xy,xy)，

(xy,xy)

(xy,xy)

f

(xy,xy)f

(xy,xf11(xy,xy)f22(xy,xy)

(xy,xy)

(xy,xy)f

(xy,xy)f

(xy,xf11(xy,xy)f22(xy,xy)2u

(xy,xy)

(xy,xy)

(xy,xy)

(xy,x2f11(xy,xy)2f12(xy,xy)f22(xy,xy)2（3）uy22 2

(xy2,x2y)

(xy2,x2y)u2xyf(xy

,x

y)x

f2

,x

y)；2u

y2[y

f11

,x

y)2xyf12

,x

y)]2yf2

,x2 2222xy[y2f(xy2,x2y)2xyf(xy2,x2 222 2222y4f(xy2,x2y)4xy3f(xy2,x2y)4x2y2f(xy2,x2y)2yf(xy2,x 2222

2yf1

,x

y)

[2xyf11

,x

y)x

f12

,x

2xf2(xy2,x2y)2xy[2xyf21(xy2,x2y)x2f22(xy2,x2 2xy3f(xy2,x2y)5x2y2f(xy2,xy)2x3yf(xy2, 2yf(xy2,x2y)2xf(xy2,x2y)

2yf1

2,x2

y)2xy[y

2f112

,x

y)2xyf12

2,x2

2xf(xy2,x2y)x2[y2f(xy2,x2y)2xyf(xy2,x 2xy3f(xy2,x2y)5x2y2f(xy2,xy)2x3yf(xy2, 2yf(xy2,x2y)2xf(xy2,x2y) 2u

2xf1

2,x2

y)2xy[2xyf11

2,x2

x

f12

2,x2

x2[2xyf(xy2,x2y)x2f(xy2,x2 4x2y2

(xy2,x2y)4x3

(xy2,x2y)x4

(xy2,x2y)2xf(xy2,x2y)．1 x 1

x x

x（4）x

f ) y

)y

f2 ) y z

f2 )y2ux y

f11(y

y)z

x

f ) y y

f11 )

f12 zy

f1(y

y)

f11(y

y)

f12(y

y)z2u x

x y y

y(2)f12 )

f12 )y2u

x

1 x

f ) y y

f11 )

yy2u

f1(yx

y) y2x2

f11(yx

y)zx

yzf121

x,y)yx

y3f1(y,z)

[y

f11(y,z)

f12(y,z1[x

f21(y

y)

f22(y

yz 3f1

y)

f11

y)

2f12 2

y)

2f22 2

y)24 y 242u x

y y

y yx x

f ) ) )f )y y y

) y x

f12(y

y)

f2(y

y)

f22(y

y)z2u

y

y x

z

z

f12 )z y

x

x x

z2f2(y

) z

f21(y,z)

f22(y,zz2

f2(y

y)z

f12(y

y)

f22(y

y)z2u2 x x

x y x

z3f2(y,z)z

f22(y

)

)y

f2(y,z)z

f22(y,z)2u2xf(x2y2z22

2yf(x

u2zf(x2y2z2)，，

2

(x

y

z

)4x

f(x

y

z2)，

2u

4xyf(x

z)，xzzx4xzf(x

z) 2

(x

y

z

)4y

f(x

y

z2)

4

(x

y

z2)

2

(x

y

z

)4zx

f(x

y

z2) （6）xf1(xy,xy,y)yf2(xy,xy,y)yf3(xy,xy,y)

yf1(xy,xy,y)xf2(xy,xy,y)y2f3(xy,xy,y)2u

f11(xy,xy,y)yf129xy,xy,y)

f13(xy,xy,yy[

(xy,xy,x)

(xy,xy,x)1

(x

y

y, y1[

(xy,xy,x)

(xy,xy,x)1

(x

y

y, yf(xy,xy,x)2

(xy,xy,x)2

(x

y

y, yy2

(xy,xy,x)2

(xy,xy,x)

(x

xy, )y

2u

f11(xy,xy,y)xf12(xy,xy,y)y

f13(xy,xy,yf

(xy,xy,x)y[

(xy,xy,x)

(x

y, yx

(x y

y,xy,y)](y2)f3(xy,xy,y)y[f31(xy,xy,y

(xy,xy,x) y

(x

y, yf

(x

y, )xyx

f3(x

y, y

(x

y, y(xy)

(x y(x

(x y

y, y xyf22(xy,xy,y)y3f33(xy,xy,y)2u

f11(xy,xy,y)xf12(xy,xy,y)y

f13(xy,xy,yx[

(x

xy, y

(x

y, )xyx

(x

xy, y2xf(x y3

y,xy,y)y2[f31(xy,xy,y)xf32(xy,xy,yxy

(x

y, y2xf(x y3

y,xy,y)f11(xy,xy,y)2xf12(xy,xy,yy

f(xy,xy,x)x

f(xy,xy,x)

2xy

f(x y,xy,f(xxy

f(x y,f(x （f對各自變量的二階混合偏導數(shù)與求導次序無關z

f(x2y2

f1zx

1zzy y證 z

f2(x2y2

f(x

y

)2x

2xyf(x2y2，f2(x2y2z

f2(x2y2

f(x2y2)(2y)

f(x2y22y2f(x2y2)f2(x2y2

f(x2y2)1z1z2yf(x2y2)2yf(x2y2) x y f2(x2y2 f2(x2y2 yf(x2y2 y2f(x2y2 yx2y2z設v1g(tr)，c為常數(shù)，函數(shù)x2y2z 2v2v2v1

c2t證明v

g(t

r)

1

(t

r) r2x2r2x2y2z

g(tr)1g(tr)(

r3

cx2y2cx2y2zc

g(tr)cr3x3r22v

g(tr)

r2x2rr

g(tr

rxg(tr)r

x)

g(tr)(x 3x2rr

g(t

r)c

3x2r

g(t

r)c

xc2r

g(t

r)c2由函數(shù)vxyz22v3y2r

g(t

r)

3y2r

g(t

r)

g(t

r)

r3z2rr

g(t

r)c

3z2r

g(t

r)c

c2rzc2r

g(tr)c2v2v2v3(x2y2z2)3r

r 3(x2y2z2)3r

g(t

r)c

x2y2z

g(trcc2r

g(t

r)

．f(xyztf(tx,ty,tz)tnf(x,y,z)，f(xyznf(xyzf(x,y為n

x

y

z

nf(x,y,z)證明f(x,ynf(txtytztnf(xyz對t求導

xf(tx,ty,tz)yf(tx,ty,tz)zf(tx,ty,tz)ntn1f(x,y, 令txtytzf(,,)

f2(,,)

f3(,,)

ft

,,)

t

f(,,)再把x,y,zx

y

z

nf(x,y,z)f(x,yzx

y

z

nf(x,yz(x,y,z)，下面的t的函數(shù)F(t)

f(tx,ty,tz)，(t0)t它在t0時有定義且是可微的，對tF(t)

tn

(tx,ty,tz)yf

(tx,ty,tz)zf

(tx,ty,tz)}

f(tx,ty,t

(tx,ty,tz)tyf

(tx,ty,tz)tzf

(tx,ty,tz)nf(tx,ty,0從而當t0F(tc(與tF(t的等式中令t1，得cF(1)

f(x,y,z)

F(t)

f(tx,ty,tz)f(x,y,z)tf(txty,tztnf(x,yzf(x,yz為n

（1）u

yyxxy0 （2）u

y)yxxyy（3）ux(xyy(xy

2ux

y

0 2

2（4）ux()()

0

解（1）u(x2y22x2x(x2y2u2y(x2y2 yuxu2xy(x2y2)2xy(x2y2)0 （2）u2xy(x2y2u(x2y22y2(x2y2 yuxu2xy2(x2y2)x(x2y2)2xy2(x2y2)xu （3）u(xy)x(xy)y(xy)ux(xy)(xy)y(xy) 2u2(xy)x(xy)y

y)

(xy)x(xy)(x

y)

y(x

y)2ux(xy)2

y) 2

2u(2(xy)x(xy)y

2((xy)x(xy)(xy)y(x(x(xy)2(xy)y(x0

u

y)x(

yx

y)x

y)x

y)x

yx

y)x

yx

y)xu

1y( ( 2u

yx

y)x

yx

y)x

y2x32

y)x

2y

(y)x

y(y2x 2

y)x

2y

y)x

y22

y)x

yx

y)x

1x

y)x

yx3

y)x

1x

y)x

1

y)x xx2

yy2

y2x2

y)x

(y)x

y(y2x 22yx

y)x

2y

(y)x

2y2x

(y)x

y2x2

y)x

y2x2

y)0x設uf(x,yxrcos yrsin

(u)2 r

u(

(u

(u)22u1u1

2u

r r r2

證 rxcosysin x(rsin)y(rcos)xrsinrcosy(u)2

1(u)2

cos

sin)2

1(

rsinrcosf)2 r2 r (f)2cos2(f)2sin22f

sin x(f)2sin22ffsincos(f)2cos2 x (f)2(f)2(u)2(u)2 2ur

2(

cos

2

sin)cos

2

cos

2

sin)2

2

sincos

2

sin2

2[x

(rsin)

2

rcos](rsin)

f(rcos2[

(rsin)

2

rcos]rcos

f(rsin2

r2sin

2

r2sincos

2

r2cos2 xrcosyrsin

2u1u1 r r22

2

sincos

2

sin2

1fr

cos

1fr

2

sin

2

sincos

2

cos2

1fr

cos

1fr

2

2

2u

zf(xyxucosvsin下（其中旋轉角是常數(shù)

yusinvz

z

(z

(z)2這時稱(x

(y

證明uxcosysinvxsinycos(z)2(z)2(fcos

sin)2(fsin

cos)2 f

f

u u(x

(y

(x)(y)設函數(shù)uf(xyLaplace2u2ux y 0

0（1）x

s2t

y s2txescost（3）x(s,t)

yessinty(st)滿足

，

．這組方程稱為u

t2s

u

s2t解

x(s2t2

y(s2t2)2，

x(s2t2

y(s2t2)22u

[

t2

2u

t2]

u2s(s23t2

x2(t2s2

(s2t2

(s2t2

(s2t2[

t2s

u2t(3s2t2yx(t2s2

(s2t2

(s2t2

y(s2t22u(t2s2)2

4st(t2s2)

4s2tx2(t2s2

(s2t2)4

(s2t2)2u2s(s23t2)u2t(3s2t22u

[

(s2t2

(s2t2s2t]

u2s(3t2s2t

x2(s2t2

xy(s2t2

(s2t2

(s2t2[

s2t

s2t]

u2t(t23s2yx(s2t2

(s2t2

(s2t2

y(s2t2)2

4s2t

4st(s2t2)

(s2t2

(s2t2

(s2t2)4

(s2t2)2u2s(3t2s2)u2t(t23s2 (s2t2

y(s2t22u2u2u(t2s2)24s2t22u4s2t2(s2t2)2

(t2s2

(s2t2)22u2u 0

u

costu

sint，

u

sintu

cost2us

2u[x2

cost

sin

cost

u

cos

cost

2uy2

sin

sint

u

sin

e2s

t

e2s

sintcost

e2s

sin2uescostuessin eses txs 2u ecost)ecost

sint

escos

sint

essin

e2s

sin

t

e2s

sintcost

e2s

cos2u

costu

sint

2u2u

2u

2s

y2 0 u u

u u x y

xtyt u

u

u

utx

y

sys2u

2u

)

u

(

2u

)

u，s

x2

xy

x

yx

y2

y(2u(

2u

)

u

(

2u

)

u，t

x2

xy

x

yx

y2

y

2u2u

(

)[(

(

)2]0

作自變量的變換，取,, （1）x,

yyxxy0（2）xyxzxuuu0 解（1）zf(xyg(,

g

g

g

g xx

x

x2x，yy

y

2y

0yxxyy(2x)x(2y)yz0z(z(x2y2（2）uf(xyzg(,, u u

u xx

xxuuuuu u u

u zz

zzuuu0u0．故u(,yxzx), 作自變量和因變量的變換，取uvww(uvuxy,v

y,wz 2zx

22

2y

0x設u ，x,xzy，變換方xy2yy

2

2x解（1）w

zwxzxzu uy yw wz zv vy y zwx(wuwv)wxwyw u

v

x w w yx(uyvy)xuv2zwyww

2

2 x2

xyw 2w2wx2

x

v2

x22

2 2

2

y22，

x

x3v2

ww1

2w

2w

1w

2w2w

v

x(u

uvx

x

x

v2x

2wxu2

y2w x)uv

2w，2z

2wx(u2

2wuvx)

2w

2wv2

)

2wu

2w

12w，x

20

22

2

(12y

2w3x3)v232w

w

u，是可微函數(shù)，于是

uv

u，所以，zxwxxyy(xyy(xy(xy，其中、xx設u ,x,xzy,變換方程xy

2

2z wxzyzwyx 1w x

w w z 1

yuyu 1 xu

y2

y2 2

2 12w x 2 x2

y3

y2u2

y2

y3

y4u

y

2

2z y

w y

2w y

w2

x2 ，y3u 2w

w

即

0．解得 v，是任意可微函

u (v)

zwx

1 x

xx

y

x

1xy 其中、z

f(xy)（1）exy2zez0xyzexyzxyzzyzx2y2z22x2y4z50．解（1）xyyexy2zezz0，xexy2zezz0 z

xe

xez2

ez22z

y2exy(ez2)yexyez(ez

y2(e2z4ez4ezxyexy(ez2

2

(exyxyexy)(ez2)yexyez (ez

(1xy)(ez2)2，(ez2)3exy2zx2(e2z4ez4ezxy

exy(ezxy xyz z xyz

z1x 1x，1y

z1z

2z

2

2

02z0

xyyzxyz1z，xzxyz1z 1 1xxy1yxy12yz(xy1)(1yz)2z 2y(yz1)，

(xyz

(xy1)2

z

y(xy1)(1(xy

(xy2z2x(xz

(xyxy2x2zz24z0，2y2zz24z0 z1x，z1y

z z(z2)(1x)2z

(z

，2

(z(1x) y(z

(1x)(1y)(z2)3

(z2，2

(z2)(1y) (z

(z2)2(1．(z．dzzf(xz,zy)F(xy,yz,zx)0f(xyz,x2y2z2)0f(x,y)g(y,z)0解（1）dz

f1(xz,zy)d(xz)f2(xz,zy)d(zf1(xz,zy)(zdxxdz)f2(xz,zy)(dzdzzf1(xzzy)dxf2(xzzy)dy1xf1(xz,zy)f2(xz,z（2）F1(xy,yz,zx)(dxdy)F2(xy,yz,zx)(dyF3(xy,yz,zx)(dzdx)dz

F3(xy,yz,zx)F1(xy,yz,zx)dxF3(xy,yz,zx)F2(xy,yz,zx)F1(xy,yz,zx)F2(xy,yz,zx)dyF3(xy,yz,zx)F2(xy,yz,z1f(xyz,x2y2z2)(dxdy12f(xyz,x2y2z2)(2xdx2ydy2zdz)2f(xyz,x2y2z2)2xf(xyz,x2y2z2 dz dx12f(xyz,x2y2z2)2zf(xyz,x2y2z212f(xyz,x2y2z2)2yf(xyz,x2y2z2 dy．12f(xyz,x2y2z2)2zf(xyz,x2y2z212f1(xy)dxf2(xy)dyg1yz)dyg2yz)dz0，所以dz

f1(x,y)dxg2(y,

f2(x,y)g1(y,z)dyg2(y,zz(xy

zxyzyf

2

y(x

z

)x2xyy2xz證明

xy

yf

x,y z1

y

z

zyz2x

yf

，2y

f yf yy

y

y y z z2y f f z

f y

，y

y f y

y (x2y2z2)z2xy

z

z

2y y

f (x2y2z2

z

2xy zf y

f y

4xzy 2xz．zf y zx2y2yd2dx2

f(xx2xyy21 dz=2x2ydy,又在方程x2xyy21兩邊對x求導， 2xyx

2y

0

2xyx2

dz

2x2

2xyx2

2(x2y2，x222x2ydy(x2y)2(x2y2)12dy2

dx

dx

(x2y)22(5x38x2y15xy28y3．(x2設ux2y2z2zf(xyx3y3z33xyz xx2解u2x2zzx3y3z33xyzx 3x2

2

3yz3xyzz

yzx，z2

u

2x

yzxz2

2(xy2x2yyz2x2z2

2z22xzz2xy2yzz2xzx2z(z2

(z22(xz2x2yyz2x2y)2zzy (z22(x5yx3y32x3y2zx4z23x2y2z23xy3z24x2yz33xyz4(z22(y2z42xz5z6．(z2xy)x2y2z2a2

（1）x2y2ax

xu2yv

（2）yv2xu0

u2v3xy

u2v

x2y

uxyz

x2y2z21 x

，y2，xydya2x2x2yy2zz0解（1）2x2yya

2 a 212uuyv0

2vyu

v2u22v

ux

0

x4uvxy，

4uvy2uuvyv0 2v2

2uxv

x

0

4uv

4uv2uuv

u12v1，v32u 4v y

1

1 12uuv

4v

4u 4v

2

y8uv1

y8uv1uyzxyz

z

yz2x22x2z y

0

uxzxyz

uxz22y2z

0

yz， 22

(2yzz2xy)z(yz2x2y)2u

3z2，

z z(z22yzzx2)z(yz2x2y)2u z

y

z4x2z2y2z2x2y，z(2xzz2xy)z(xz2xy2)2u z

y

xy(y23z2．zzxydzdzxcoscosycossin

xuv（2）yu2v2zsin

zu3v3解（1）方程組兩邊對x求偏導數(shù)， 1sincosxcossinx 0sinsinxcoscosxzcos

cos，

sinzcoscot y 0sincosycossiny 1sinsinycoscosyzcos

，代入第三個方程 sincot 1xx

xvu 02ux2vxz3u2u3v2v

xvuz3uv3(yx2)

y 0yy

y2(vu) 12uy2vyz3u2u3v2v

y2(vu)z3(uv)3x

§3何應xasin2tybsintcostzccos2t,在點t42x23y2z29z23x2y2，在點(1,1,2x2y2z26xyz0，在點(1,2,1xtcosty3sin2tz1cost，在點t2解

t4

，對應的點為(y(y(

c)2x()2asint444

t

a，b(cos2tsin24

t

0z )2ccostsin4

4t4

c

t444

對應的點(2

c2x y z 2 2 2 axcz1(a2c22（2）F(xyz)2x23y2z29G(xyz)3x2y2z2為

Fz

6

F(x,y,z)0G(x,y,z)0 由 GG GG

z 6x 2

2z(F,G)6

2z16yz，(F,G)

4x20xz(y,

2

(z,

(F,G)

6y28xy(x,

2(16

x1y1z2 法平面方程為8(x110y17(z2)0，即8x10y7z12F(xyz)x2y2z26G(xyz)xyz

Fz

2

2z由 GG z 1GG(F,G)2

2z2(yz)，(F,G)

2x2(zx)(y, (z, (F,G)

2y2(xy)(x, )x1y2z11 1(x1z1)0xz0t對應點為(,4,1x(1sin

222y()2sint2

0

z()3sin

t23t t

x

y z1 2(x3(z1)0，即2x3z32（1）ye2xz0，在點(1,1,2xa

y

zc

1在點(a,b

c)333z2x24y2在點(2,1,12333xucosv,yusinv,zavp0(u0v0．解（1）F(x,y,z)ye2xz，則法向量(F,F,F

(2e2xz,1,e2xz

z

2(x1y1z2)0即2xyz1x1y1z2

x y z（2）令F(x,y,z) a b2

1c b (F,F,F (,,

2x,, ,,

2(

,1

1))33)

(a,b,c

ab33333331所求的切平面方程(xa1y3331

b)1(z

c)0，即1x1y1z a3x y3

z

法線方程為 1a

1b

31cn（3）F(xyz)2x24y2zn

的切平面方程為8(x28y1z12)0，即8x8yz128x8yz12x2

y1z128

1 z

cos

sin 0（4）由

z v

usin

ucos a(y,(u,(y,(u,p0

a

asinv0

cos(y,(u,(y,(u,p

acosv0(y,(y,(u,usin

sinvucosp0p

u0n(asinv0,acosv0,u0p0(u0v0(x,y,z)(u0cosv0,u0sinv0,av0)asinv0(xu0cosv0acosv0yu0sinv0u0(zav00，即axsinv0aycosv0u0zau0v0，法線方程為：xu0cosv0asinv0

yu0sinacos

zav0xaetcostyaetsintzaetx2y2z2的母線相證明x2y2z2P(xyz因此，母線的方向向量為1(xyzP2(x(t),y(t),z(t))(aet(costsint),aet(sintcost),aet)(xy,xy,z)2 1212

x(xy)y(xy)z

2z x2x2y2z (xy)2(xy)2z2z 3z6, x3y3a33xy，

(a0)

2

1 解x3y

a

(a0x求導，有3

3 3

3 0 x在曲線上任一點(x0y0kx0

(x00yy0

(xx0)x0x1 1A(x3a30B(0,y3a3 2 24 400dAB(x3a3y3a3)2(a3a3 00ax22y23z221x4y6z000解x22y23z221P00

,

,

n(2x0,4y0,6z0)x4y6z01

44

6z0

z0P(xy

x22y23z221P(1,2,2

1x14y26(z2)0，x4y6z21為所求切平面方程．F(xazybz)0的切平面與某一定直線平行，其中ab證 G(x,y,z)F(xaz,ybz)，則曲面為G(x,y,z)0，曲面上任意一P0(x0,y0z0n(Gx,Gy,Gz0

nab,10nab,1，故曲面過P0點的切平面平行于方向向量為ab,1F(xazybz0的切平面與一方向向量為axzxeyx00證明F(xyz)xeyzP00

,

,

n(Fx,Fy,Fz 0

x0 x20ey0(10)(xx0

)0 y20y2 zxey0 x

x2 x02(10)ey0x0ey0yz02y yF(x,y,z)0，G(x,y,z)的交線在OxyF(x,y,z)0解G(x,y,z)

P0(x0y0z0(F,((F,(y,(F,(z,(F,(x,

y

z F(xyz)0，G(xyz)0的交線在Oxy x

y (F,(z,(F,(z, (y,0§4向導0f(xyzxy2z3fP(1,1,1沿方向l2,2,10 由 1 2y 3z

，故

(1,2,3)

xy0l0l1l1(2,2,1)，所 l03 3求函數(shù)uxyzA(5,1,2B(9,4,14AB解uyzuxzuxy (u,u,u

(2,10,5)，l

AB

Axy A

(2,10,5)

(4,3,12)98

(x,y0 （1）ulnx2y2xy)(1,1lx軸正向的夾角為600 （2）uxexy(xy1,1l與向量(1,1 解

x2y

2x2y

,u

2

x2 x2

(x,y

x2y

(x,y0 000

l(cos600,sin600)

(1 3)2

1 3x,y3

l0(1,1)(

) 2(x,y

0（2）uexy(1xyux2exy

u,y

(1xy),

(2e,e) (x,y02l02

1212121212

(2e,e)

1212

32e2設函數(shù)f(xy)

,

)可微，單位向量l1

),

)75f(x0,y0)1,f(x0,y0)0，確定l使得：f(x0,y0)75解

f(x0,y0) f(x0f(x0,y0) f(x0,y0) 1 f(x,y f(x,y f(x,y) 0 0 0 ) ,22 22解出，f(x0,y0)f(x0,y0) 2 f(x0y0)

，即2

2

7，再由2217575 7575出(45

或((3

，即l5

(45

或l3

5fP0(2,0f(xyP0P12,2的方向導數(shù)是1，指向原點的方向導數(shù)是3，試回答：P33,21f(2,0)0f(2,0)(1)

f(2,0)3

解由 3

f

f

0

f

f

0

f

1f(2,0)

f(2,0) 15255 15255§5Taylor（1）f(xy)2x2xyy26x3y5在(1,2（2）f(xyx2xyy23x2y4在(1,1解

4xy6，

2x2y

24，

2，y

2，高于二階偏導數(shù)均為0在(1,2)點，f(1,2) ，

20

4

2

12y

2

f(x,y)2(x1)2(x1)(y2)(y2)25

2xy3，

x2y2

2

2

2

2

2

1

2

2

2

1，2

2f(x,y)2(x1)(y1)(x1)2(x1)(y1)(y1)2f(xy)

y