Thomas Calculus 13th [Solutions]

1016 Chapter 14 Partial Derivatives
15. f 2xi 2j 2k f 1, 1, 1 2i 2j 2k and g j for all points ;
2
i j k
v f g v 2 2 2 2i 2k Tangent line: x 1 2 t, y 1, z 1 2t
2
0 1 0
i j k
16. f i 2 yj k f 1 ,1, 1
2 2
i 2j k and g j for all points; v f g v 1 2 1 i k
0 1 0
Tangent line: x 1 t, y 1, z 1
2 2
t
17.
2 2 2 2
f 3x 6xy 4y i 6x y 3y 4x j 2 zk f (1, 1, 3) 13i 13j 6 k;
i j k
g 2xi 2yj 2 zk g(1, 1, 3) 2i 2j 6 k;
v f g v 13 13 6 90i 90j
Tangent line: x 1 90 t, y 1 90 t, z 3
2 2 6
18. f 2xi 2yj f 2, 2, 4 2 2i 2 2 j;
g 2xi 2yj k g 2, 2, 4 2 2i 2 2 j k;
i j k
v f g v 2 2 2 2 0 2 2i 2 2j Tangent line: x 2 2 2 t, y 2 2 2 t, z 4
2 2 2 2 1
x
y
i j z k f (3, 4, 12) 3 i 4 j 12 k;
x y z x y z x y z
169 169 169
3i 6j 2k
u v
3 i 6 j 2 k f u 9 and df ( f u) ds 9 (0.1) 0.0008
v 2 2 2
3 6 ( 2)
7 7 7 1183
1183
19. f
2 2 2 2 2 2 2 2 2
20. f e x cos yz i ze x sin yz j ye x sin yz k f (0, 0, 0) i;
1 i 1 j 1 k f u 1 and df ( f u) ds 1 (0.1) 0.0577
3 3 3 3
3
u
v
v
2i 2j 2k
2 2 2
2 2 ( 2)
21. g (1 cos z) i (1 sin z) j ( xsin z y cos z) k g(2, 1, 0) 2 i j k;
A P0 P1 2i 2j 2k
2i 2j 2k
u v
1 i 1 j 1 k g u 0 and dg ( g u) ds (0)(0.2) 0
v
2 2 2
( 2) 2 2 3 3 3
22.
2
h y sin ( xy) z i x sin ( xy) j 2 xzk h( 1, 1, 1) ( sin 1) i ( sin ) j 2k
i j k
i 2 k;
v P0 P1
i j k where P
1 1 1
1 (0, 0, 0) u v
i j k
v 2 2 2
1 1 1 3 3 3
h u 3 3 and dh ( h u)
ds 3(0.1) 0.1732
3
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