微积分 第一章习题解答(下)(2)

2018-11-21 22:31

4x3arcsin1?x2?y2x?y44?xln(x4?y4)(1?x?y)(x?y)2222

??z?f???f??=???y???y???y1??24y3arcsin1?x2?y2x4?y43. (1)

4y3?arcsin??4=422x?y1?x?y?y?yln(x4?y4)(1?x?y)(x?y)2222.

?u?u=2xf1?yexyf2, =?2yf1?xexyf2. ?x?yyx1?u1?u?u=?f1, =?2?f1?f2,=?2?f2. ?xy?z?yzzy(2)

(3)

?u?u?u=f1?yf2?yzf3,=xf2?xzf3,=xyf3.

?z?x?y?u?u?u=2xf1?yf2?f3=2yf1?xf2?f3,=f3. ?x?z?y?z?z?yf1,?xf1?f2, ?x?y(4)

4 .(1)

?f1?2z2???y?yf?y?yf11, 112?x?x?f?2z???yf1??f1?y1?f1?y(f11?x?f12)?f1?xyf11?yf12, ?x?y?y?y?f1?f2?2z?2???xf?f?x??x(f?x?f)?f?x?f?xf11?2xf12?f2212111221222?y?y?y?y(2)

?z?z?y2f1?2xyf2,?2xyf1?x2f2, ?x?y?f2?2z?22?f1?yf?2xyf?y?2yf?2xy 122?x?x?x2?x???y2(f11?y2?f12?2xy)?2yf2?2xy(f21?y2?f22?2xy)?2yf2?yf11?4xyf12?4xyf22?f?f?2z?2?yf1?2xyf2?2yf1?y21?2xf2?2xy2 ?x?y?y?y?y4322.

?? 6

?2yf1?y2(f11?2xy?f12?x2)?2xf2?2xy(f21?2xy?f22?x2)?2yf1?2xf2?2xyf11?2xyf22?5xyf12?f1?2z?22?f2 ?2xyf?xf?2xf?2xy?x1212?y?y?y?y3322

???2xf1?2xy(f11?2xy?f12?x2)?x2(f21?2xy?f22?x2)?2xf1?4xyf11?4xyf12?xf225 ?2234

?u?u?x?u?y1?u3?u?u?u?x?u?y3?u1?u, ????,??????s?x?s?y?s2?x2?y?t?x?t?y?t2?x2?y(?u21?u23?u?u3?u2?u23?u23?u?u1?u2)?()??(),()?()??(), ?s4?x2?x?y4?y?t4?x2?x?y4?y?u2?u2?u?u)?()?()2?()2. ?s?t?x?y?(x?y?z)?(6 (1) 设F(x,y,z)?x?y?z?e?(x?y?z), Fx?1?e?(x?y?z),Fy?1?e,

Fz?1?e?(x?y?z),

FyFx?z?z????1,????1 ?xFz?yFz(2)设F(x,y,z)?z?x2?y2tanFx??xx?y22zx?y22,1(?)(x2?y2)2x2?y2z3?2tanzx?y22

?x2?y2sec22xz=?xx?yyx2?y222tanzx?yzx2?y222?xzsec222x?yzx?yz22,

3 Fy?tan?x2?y2sec2?122(?)(x?y)2(?2yz) 2x2?y2=

yx?y22tanzx?y22?yz2secx2?y2z12zx?y22,

Fz?1?x2?y2sec2x?y2x?y22=?tan22zx?y2,

7

F?zxzxz??x??cot?2csc22?xFzx2?y2x2?y2x?yFy?z???Fz?yyx?y22zx?yzx?y2222,

cotzx?y22?yz2cscx2?y2.

(3) 设F(x,y,z)?x?2y?z?2xyz,Fx?1?xyyzxz Fy?2?, Fx?1?xzy.

F?z??x=?xFzFyyz?xyz?z,=??Fzxyz?xy?yxz?2xyzxyz?xy(4) 设F(x,y,z)?x1xzx11?ln??lnz?lny,Fx?,Fy?Fz??2?,

zzyzzyzFyFx?zz?zz2??,, ?????xFzx?z?yFzy(x?z)7.设F(x,y,z)?x?2y?3z?2sin(x?2y?3z),Fx?1?2cos(x?2y?3z),

?Fy?2?4cos(x?2y?3z),Fz??3?6cos(x?2y?3z),

?

Fy2F?z1?z??x?,???, ?xFz3?yFz3?z?z??1. ?x?y?

8.设F(x,y,z)??(cx?az,cy?bz),Fx?c?1,Fy?c?2,Fz??a?1?b?2,

FyFxc?1c?2?z?z??,????, ?x?yFza?1?b?2Fza?1?b?2? a?z?z?b?c. ?x?y9. (1)方程两边同时对x求导得

x(6z?1)?dydy?dz??,??dx?2x?2ydx,?dx2y(3z?1)

解之得??dydz?dy?x?2x?4y?6z?0,?dxdx??dx3z?1(2) 方程两边同时对z求导得

8

?dxdy?dz?dz?1?0, ?解之得

dydx?2x?2y?2z?0dz?dz (3) 方程两边同时对x求偏导得

?dx???dz?dy????dzy?z,x?y z?x.x?y?u1?e? ??0?eu?sinv??u?u?u?v?,?sinv?ucosv,u??xe(sinv?cosv)?1?x?x?x解之得? ?u?u?u?v?vcosv?e???cosv?usinv,.u?x?x?x???xu[e(sinv?cosv)?1]同理方程两边同时对y求偏导得

?cosv??u?u?v?u?u?,0?e?sinv?ucovs,u?????xe(sinv?cosv)?1?y?y?y ?解之得? u?u?u?v?vsinv?e?1?eu???covs?usinv,.u??y?y?y????xu[e(sinv?cosv)?1]

习题1-4

1. 求下列函数的方向导数

?u?lPo

(1)u?x2?2y?3z2,P0?1,1,0?,l??1,?1,2? 解:

?u?xP0?2xP0?2

?u?y?u?zP0?4y?6zP0?4 ?0

P0P0l0?(?u?l16,?16,26)

1626?P0?2*yxz16?4*(?)??.

(2)u?(),P0(1,1,1),l?(2,1,?1); 解:

?u?x?u?yyz?1y?z()(?)P0xx2yz?11?z()()P0xxP0??1,

P0?1,

9

?u?z0yzy?()ln()P0xx26,16,?1P0?0,

l?(626)

1616 ??u?lP0?(?1)*?1*??.

(3)u?ln(x2?y2),P0(1,1),l与ox轴夹角为解:

?; 3?u?x?u?yP0?2xx2?y22yx2?y2P0?1,

P0?P0?1,

由题意知???3,则???6,

l?(cos,cos)?(,0??3613) 22 ??u?lP0131?3?1*?1*?.

222(4)u?xyz,P,2),P0(5,11(9,4,14),l?P0P1.

?u?xP0?yzP0?2,

?u?y?u?zP0?xz?xyP0?10, ?5,

P0P0l?(4,3,12),?l0?(?

2. 求下列函数的梯度gradf

?u?lP04312,,), 131313431298?2*?10*?5*?.

13131313(1)f(x,y)?sin(xy)?(cos(xy);

22 10


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