长沙理工大学高数下册练习册答案(2)

的方向向量s1 (1,1,2),点B( 1,3,0)在已知直线上, 112

因为所求直线与已知直线相交,所以三向量,s1，s共面,故有

已知直线

1 ( 1)3 00 4

1

m1n2 0, p

即 10m 4n 3p 0. (2)

联立(1)、(2)式可得

161928 . mnp

因此所求直线方程为

x 1yz 4 . 161928

5.解:设柱面上任一点为M(x,y,z),准线上相应点为M0(x0,y0,0),M0与M在同一 条母线上,

M0M (x x0,y y0,z)//(1,2,1),

代入y0 x02,得y 2z (x z).

2

x x0y y0z

,x0 x z,y0 y 2z, 121

yzx2y2z2a2yz

0 2 2 2 2 ( ) 0 1, 三、解: M(x,y,z) L x a,

bcabcabc

x2y2z2x2y2z2

M 曲面2 2 2 1 L 曲面2 2 2 1,即L在曲面上.

abcabc

七、设旋转曲面上任一点为M(x,y,z),直线L上相应点为M0(x0,y0,z0),M0与M在同一垂

直于z轴的圆周上,z0 z,x02 y02 x2 y2.又由

2

2

2

2

2

2

x0 1y0z0

,得 011

2

2

2

2

x0 1,y0 z0,代入x0 y0 x y得x y 1 z,或x y z 1.

习题9.1 （A）

4224

1.(1)x 2xy 2y；(2)曲线y 2x上所有点.

2

2.(1) x,y 0 x y,x2 y2 1；(2) x,yx2 y2 z2,z 0. 3.(1)解:

2

(x,y) (1,0)

lim

ln(x ey)x y

2

2

ln(1 e0) 0

2

ln2.

(2)解:

(x,y) (0,0)

lim

xy(xy 1 1)xy

lim lim(xy 1 1) 2. (x,y) (0,0)(x,y) (0,0)(xy 1) 1xy 1 1

(3)解:

sin(xy)sin(xy) lim x 1 2 2.

(x,y) (2,0)(x,y) (2,0)yxylim

(4)解:

(x,y) (0,0)

lim(1 2xy)

1

tan(xy)

(x,y) (0,0)

lim(1 2xy)

12xy

2xytan(xy)

e2.

4.解:

(x,y) (0,0)

lim

x2yxy

f(x,y) lim lim y,

(x,y) (0,0)x2 y2(x,y) (0,0)x2 y2

(x,y) (0,0)

limy 0,

xy

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

(x,y) (0,0)x2 y2

则f(x,y)在点(0,0)处连续.

(B)

1.(1)

1 y2

x；(2)直线x m或y n(m,n Z)上所有点. 1 y

2.(1) x,yy2 4x,0 x2 y2 1；(2) x,y y2 x y2,0 y 1.

x2 y2

2sin

1 cos(x2 y2) lim3.(1)解:lim 2222222(x,y) (0,0)xy(x y)(x,y) (0,0)xy(x y2)

2

x2 y2sin

x2 y211 lim 1 lim( ) . 222222(x,y) (0,0)(x,y) (0,0)x y22xy2y2x

()

2

2

t22t2

(2)解： limt limt limt 0，

t et et e

lim(x y)e

x y

2

2

(x y)

x21y21

lim(x y y x) 0 0 0 0 0. x eeeey

2

4.证:当(x,y)沿曲线y kx趋于(0,0)时,有

x2yx2 kx2k

lim, lim lim

(x,y) (0,0)x4 y2(x,y) (0,0)x4 (kx2)2(x,y) (0,0)1 k2

显然它是随着k的值不同而改变的,故所给极限不存在.

习题9.2 (A) 1.(1)

；(2)1. 4

z z yexy 2xy, xexy x2. x y

z z esinx cosx cosy, esinx ( siny). x y

z zxy y(1 xy)y 1 y y2(1 xy)y 1 (1 xy)y[ln(1 xy) ]. x y1 xy

y

y

y

2.(1)解:

(2)解:

(3)解:

uyz 1 u1 uy

(4)解: x, xzlnx , xzlnx ( 2).

xz yz zz zx 2z1yx

第九章解:； ln(x y) ,2 22

xx y xx y(x y)(x y) zx 2zx

； ,2 yx y y(x y)2

2z1xy . 22

x yx y(x y)(x y)

( ) ( ) ( ) z1 z12 z2 zxyxy

第十章证: e 2, e 2, x y 2exy 2z.

xx yy x y

11

11

11

(B)

u 2uxyz

1.解: e yz, exyz xz yz exyz z zexyz(xyz 1),

x x y

3u exyz(xyz 1) zexyz xy(xyz 1) zexyz xy exyz(x2y2z2 3xyz 1),

x y z

2u 3u 2uxyz22

2 e yz,2 (2) exyz xz y2z2 exyz 2yz2 yz2exyz(xyz 2),

x x y y x 3u 2u () zexyz xz(xyz 1) zexyz xz xz2exyz(xyz 2). 2

x y y x y

2.解:

z1y

2f(xy) f'(xy) y '(x y), xxx

2z11y

2f'(xy) x f'(xy) f''(xy) x '(x y) y ''(x y) x yxxx

yf''(xy) '(x y) y ''(x y). 3.解:设u xg(y),v y,则x

uu

,y v,代入即得f(u,v) g(v), g(v)g(v)

f1 2fg (v)

于是. ,

ug(v) u v[g(v)]2

习题9.3

(A)

, 0.2；(2)dx 1.(1) 0.20404

y

1

32

dy. 3

y

y

z zy11

2.(1)解:dz dx dy ex ( 2)dx ex dy 2ex(ydx xdy).

x yxxx

z z1

(2)解:dz dx dy y ( )(x2 y2) 3/2 2xdx

x y2

x2 y2 y

x y

2

2

yx2 y2

dy

xyx2x

2dx dy (ydx xdy).

(x y2)3/2(x2 y2)3/2(x2 y2)3/2

(3)解:dz

z zdx dy x y

1(x y) (x y)1(x y) (x y)

dx dy 22

x y2x y(x y)(x y)1 ()1 ()2x yx y yx ydx xdydx dy .

x2 y2x2 y2x2 y2

u u uy11z1x

dx dy dz ( 2 )dx ( 2)dy ( 2)dz. xy zxzxyyz

1

1

(4)解:du

1xz 111xz 1x

3.解:fx(x,y,z) () ,fy(x,y,z) () ( 2),

zyyzyy

df(1,1,1) fx(1,1,1)dx fy(1,1,1)dy fz(1,1,1)dz dx dy.

(B)

1.解:dz e

arctan

yx

d(x y) (x y)d(e

2

2

2222

yx

)

yx

e

yx

(2xdx 2ydy) (x y)e

[

1y1

( 2dx dy)] yxx1 ()2x

yx

e

yx

(2xdx 2ydy ydx xdy) e

[(2x y)dx (2y x)dy].

y

z zx

(2x y)e 由全微分与偏导数的关系可知,其中dx的系数就是,即, x x

再对y求偏导数,得

2z11y2 xy x2 arctanx

xx

e (2x y)e[ ] e . 22

y x yx y1 ()2xx

y

y

y

2.解:当x y 0时,

22

f x

yx2 y2 xy

x y

2

2

xx2 y2

y3(x2 y2)3

x3(x2 y2)3

,

f y

2

xx2 y2 xy

x2 y2

2

yx2 y2

.

当x y 0时,

x 0

f x

x 0y 0

lim

x 0

( x) 0f(0 x,0) f(0,0)

lim x 0 x x

22

0

0,

0 y

f y

lim

x 0y 0

x 0

0 ( y)f(0,0 y) f(0,0)

lim x 0 y y

22

0

0.

所以

y3x322

,x y 0 f ,x2 y2 0 f

,. (x2 y2)3 (x2 y2)3

x y

22

x y 0x2 y2 0 0, 0,

因为当点(x,y)沿直 …… 此处隐藏：3258字，全部文档内容请下载后查看。喜欢就下载吧 ……