数学分析简明教程答案

第十六章偏导数与全微§1导数与全微分的（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 设

f(x,f(x,y)

1x2y

,x2y20,x2y2函数在(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证明函数u x2yx2y(x)2(x)2lim lim

x0x2y不存在，由对称性limyu(0,0)不存在，因而ux2y x2y2x2y2zuxeyzexyx2y2x2y2z2x2y2z

d(x2y2z2x2y2zx2y2zx2y2x2y2z2x2y2z2

dy

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

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

在点(0,1x2yx2y

)

d(x2y2x2yx2yx2y2(x2y2x2

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

xy

(dx2ydy)

xy

dx

2yxy2

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

1)z

xy x1

x1

x du

()yz

dx

y2

()y

dyz

()zy

dzy函数的定义域为{(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但在点(0,ydu不存在，因而uxy1xy

在点(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

更高阶的无穷小，为此极

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在点(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

更高阶的无穷小量，为此极lim

(，(x0x2

x2y2

令yx(xy沿直线yx趋于(0,0

(x2(x2y2(

x2

f(xy在(0,0

f(x,y)

y

) x2y

x2y200

x2y2的偏导数存在，但偏导数在(0,0)点不连续，且在(0,0)点的任何邻域中，而f在原

证

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)点的任何邻域中均

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设

2xy f(x,y)x2y2 x

0 证

在(0,0

0

x2y20 2xy 证明f(x2y22,

x y22xyy 2xy22x2y

x2y2

y

0

fx(0,0

fx(x,y在(0,0

亦在(0,0设

1ex(x2y2

,x2y20f(x,y)x2y0f(xy在点(0,0df(0,0

x2y2x证 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设

f(x,y)x2y2,x

0

x2y2xx(t),yy(t是通过原点的任意可微曲线（x20y20)0t0x2ty2t)0,x(t),y(t可微f(x(t),y(tf(x,y在(0,0

x3

,t0,证明（1）设(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

更高阶的无穷小，为此极lim1

)

xy，设

y033

3

(2x2)

22该极限不存在，因而

xy )不是 更高阶的无穷小量，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内可微，且全微分du恒为零，问f(x,y f(x,y)在该矩形内应取常数值．证明如下由于uf(x,y在矩形内可微，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取常数值C

f(x0,y0)设

在

,

在

,

f(x,y在

,

证 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而对f(x0xy0f(x0y0，设(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

，由对称性

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

（使用数学归纳法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设函数u(xyu

u.x yx证明u(xyu(xy)) 2u(x u 2

(y)),

(

(y)u

(x

xu2u

(

(

(y)yx

(

(

(y))u即

u.x yx19．fx,fy在点(x0y0的某邻域内存在且在点(x0,y0fxy(x0,y0)fyx(x0,y0)证明16.4fxy(x0,y0fyx(x0,y0

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

处的可微性，下面证明

f(x

)

f(x,

)

x

51

61二者对充分小的xy同时成立，且当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（当xy0时fy(x0,y0y)fy(x0,y0)fyy(x0,y0)y3y其中30（当y0时W

1{

x,

y)f

,

1{

,y0)

f

,

)x

f

,

)y1x

f(x,

)

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

这正是（*）式．同样，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（当xy0时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合函数与隐函数微分（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对各自变量的二阶混合偏导数与求导次序无关z

f(x2y2

f1zx

1zzy y证 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设v1g(tr)，c为常数，函数x2y2z 2v2v2v1

c2t证明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由函数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为n

x

y

z

nf(x,y,z)证明f(x,ynf(txtytztnf(xyz对t求导

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的函数F(t)

f(tx,ty,tz)，(t0)t它在t0时有定义且是可微的，对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从而当t0F(tc(与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为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设uf(x,yxrcos yrsin

(u)2 r

u(

(u

(u)22u1u1

2u

r r r2

证 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（其中旋转角是常数

yusinvz

z

(z

(z)2这时称(x

(y

证明uxcosysinvxsinycos(z)2(z)2(fcos

sin)2(fsin

cos)2 f

f

u u(x

(y

(x)(y)设函数uf(xyLaplace2u2ux y 0

0（1）x

s2t

y s2txescost（3）x(s,t)

yessinty(st)满足

，

．这组方程称为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

作自变量的变换，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), 作自变量和因变量的变换，取uvww(uvuxy,v

y,wz 2zx

22

2y

0x设u ，x,xzy，变换方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，是可微函数，于是

uv

u，所以，zxwxxyy(xyy(xy(xy，其中、xx设u ,x,xzy,变换方程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证明

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x2xyy21两边对x求导， 2xyx

2y

0

2xyx2

dz

2x2

2xyx2

2(x2y2，x222x2ydy(x2y)2(x2y2)12dy2

dx

dx

(x2y)22(5x38x2y15xy28y3．(x2设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）方程组两边对x求偏导数， 1sincosxcossinx 0sinsinxcoscosxzcos

cos，

sinzcoscot y 0sincosycossiny 1sinsinycoscosyzcos

，代入第三个方程 sincot 1xx

xvu 02ux2vxz3u2u3v2v

xvuz3uv3(yx2)

y 0yy

y2(vu) 12uy2vyz3u2u3v2v

y2(vu)z3(uv)3x

§3何应xasin2tybsintcostzccos2t,在点t42x23y2z29z23x2y2，在点(1,1,2x2y2z26xyz0，在点(1,2,1xtcosty3sin2tz1cost，在点t2解

t4

，对应的点为(y(y(

c)2x()2asint444

t

a，b(cos2tsin24

t

0z )2ccostsin4

4t4

c

t444

对应的点(2

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

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 法平面方程为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对应点为(,4,1x(1sin

222y()2sint2

0

z()3sin

t23t t

x

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

y

zc

1在点(a,b

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

法线方程为 1a

1b

31cn（3）F(xyz)2x24y2zn

的切平面方程为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，法线方程为：xu0cosv0asinv0

yu0sinacos

zav0xaetcostyaetsintzaetx2y2z2的母线相证明x2y2z2P(xyz因此，母线的方向向量为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求导，有3

3 3

3 0 x在曲线上任一点(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为所求切平面方程．F(xazybz)0的切平面与某一定直线平行，其中ab证 G(x,y,z)F(xaz,ybz)，则曲面为G(x,y,z)0，曲面上任意一P0(x0,y0z0n(Gx,Gy,Gz0

nab,10nab,1，故曲面过P0点的切平面平行于方向向量为ab,1F(xazybz0的切平面与一方向向量为axzxeyx00证明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)的交线在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的交线在Oxy x

y (F,(z,(F,(z, (y,0§4向导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求函数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轴正向的夹角为600 （2）uxexy(xy1,1l与向量(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设函数f(xy)

,

)可微，单位向量l1

),

)75f(x0,y0)1,f(x0,y0)0，确定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的方向导数是1，指向原点的方向导数是3，试回答：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，高于二阶偏导数均为0在(1,2)点，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