(2)
?z?z?2xyf1?x2f2, ?y2f1?2xyf2,?y?x?f2?2z?22?f1?yf?2xyf?y?2yf?2xy 1222?x?x?x?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.
???2yf1?y2(f11?2xy?f12?x2)?2xf2?2xy(f21?2xy?f22?x2)?2yf1?2xf2?2xyf11?2xyf22?5xyf123322
?f1?2z?22?f2 ?2xyf?xf?2xf?2xy?x1212?y?y?y?y???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?u?u?u)?()2?()2?()2. ?s?t?x?y?(x?y?z)?(x?y?z)?(x?y?z), Fx?1?e,Fy?1?e,
?(6 (1) 设F(x,y,z)?x?y?z?eFz?1?e?(x?y?z),
FyF?z?z??x??1,????1 ?xFz?yFz(2)设F(x,y,z)?z?x2?y2tanFx??xx2?y2tanzx2?y2zx?y22,?1(?)(x2?y2)22xz2x2?y2?x2?y2sec2z3
6
=?xx?yyx?y2222tanzx?yzx?y2222?xz2secx2?y2222zx?yz22,
3 Fy?tan?x?ysec?1(?)(x2?y2)2(?2yz) 2x2?y2=
yx?y22tanzx?y22?yzsec222x?yzzx?y22,
Fz?1?x2?y2sec212x?y2x?y22=?tan22zx?yzx?yzx?y222,
F?zxzxzcot?2csc2??x??2Fz?xx2?y2x2?y2x?yFy?z????yFzyx?y2222,
cotzx?y22?yzcsc222x?y.
xy, z(3) 设F(x,y,z)?x?2y?z?2xyz,Fx?1?yzxzFx?1? Fy?2?xy.
F?z??x=
Fz?xFyyz?xyz?z??,=
Fzxyz?xy?yxz?2xyzxyz?xy(4) 设F(x,y,z)?xzx11x1?ln??lnz?lny,Fx?,Fy?Fz??2?, zyzzyzzFyFxz?z?zz2???,, ???Fzx?z?y?xFzy(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),
?
Fy2F1?z?z??x?,???,
Fz3?y?xFz3?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?z?z????c. ??,?, ? a?bFza?1?b?2?y?y?x?xFza?1?b?2
7
9. (1)方程两边同时对x求导得
x(6z?1)?dydy?dz??,?2x?2y,??dx?dx2y(3z?1)dx解之得? ?dydzdyx???2x?4y?6z?0,?dxdx??dx3z?1(2) 方程两边同时对z求导得
?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?ucosv,u???xe(sinv?cosv)?1???y?y?y ?解之得? u?u?u?v?vsinv?e?1?eu???cosv?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
22（1）u?x?2y?3z,P0?1,1,0?,l??1,?1,2?
解：
?u?xP0?2xP0?2 ，
?u?yP0?4yP0?4，
?u?zP0?6z0P0?0，l?(16,?16,26)
??u?lP0?2*16?4*(?16)??26.
（2）u?(),P0(1,1,1),l?(2,1,?1); 解：
yxz?u?x?u?zyz?1y?z()(?2)P0xxyzy?()ln()P0xxP0??1，
?u?yyz?11?z()()P0xxP0?1,
0P0?0, l?(26,16,?16)
8
??u?lP0?(?1)*26?1*16??16.
22（3）u?ln(x?y),P0(1,1),l与ox轴夹角为
?3;
解：
?u?x?u?yP0?2xx2?y22yx2?y2P0?1,
P0?P0?1,
由题意知??0?3,则???6,
l?(cos,cos)?(,??3613) 22 ??u?lP0?1*131?3?1*?. 222（4）u?xyz,P0(5,1,2),P1(9,4,14),l?P0P1.
?u?x?u?yP0?yz?xzP0?2,
?10,
P0P0
?u?zP0?xyP0?5,
4312l?(4,3,12),?l0?(,,),
131313?u431298??2*?10*?5*?. P0?l13131313
2． 求下列函数的梯度gradf
（1）f(x,y)?sin(xy)?(cos(xy); 解：
22?f?cos(x2y)*(2xy)?sin(xy2)*y2, ?x?f?cos(x2y)*x2?sin(xy2)*(2xy), ?y222222
(),xcos(xy)?2xysin(xy)) f(2xycosx(y)?ysinxy ?grad? 9
yy(2)f(x,y)?e.
xx?fyy11y解：?(?2)ey?ey?ey(1?),
?xxyxxx?f1yyyx11?e?e(?2)?ey(?), ?yxxxyy1y11。 ?gradf?(ey(1?)，ey(?)）
xxxy3． 一个登山者在山坡上点(?xxxxxxxx33,?1,)处，山坡的高度z由公式z?5?x2?2y2近似，其24中x和y是水平直角坐标，他决定按最陡的道路上登，问应当沿什么方向上登。 解：
?z?x?z?y33(?,?1,)24??2x33(?,?1,)24?3,
33(?,?1,)24??4y33(?,?1,)24?4,
?按最陡的道路上登，应当沿（3，4）方向上登。
4． 解：
?T?T?y(1?y)(1?2x),?x(1?x)(1?2y) ?x?y11(,)43沿方向?gradT11?(?,?)
9165． 解：设路径为y?f(x)，在点(x,y)处gradT?(?2x,?8y)
y?f(x)在(x,y)点的切向量为??(1,dy) dx?,? ?grad平行于切向量T 因为过(1,2),?y??2x
4dxdy?y?cx4 ??2x?8y习题1-5
tt?1,y?,z?t2在对应于t?1点处的切线及法平面方程。 1?tt1解：当t?1时，x(1)?,y(1)?2,z(1)?1，
21、求曲线x?
10