Thomas Calculus 13th [Solutions]

Section 14.3 Partial Derivatives 995
59.
f ( 2 h, 3) f ( 2, 3) 2( 2 h) 9 1 4 9 1 2h 4 2 2h 4 2 2h
4 2
fx ( 2,3) lim lim lim lim
h 0
h
h 0
h
h 0
h
h 0
h 2h
4 2
lim 2 1
0 2 4 2 2
,
f ( 2, 3 h) f ( 2, 3) 4 3(3 h) 1 4 9 1 3h
4 2
f ( 2, 3) lim lim lim
h h
y
h 0
h
h 0
h
h 0
h
3h
4 2 3h
4 2
lim
lim 3 3
h 0
h 3h 4 2 h 0 3h
4 2 4
60.
sin h
3
0
0
2
3
h 0
3
f (0 h, 0) f (0, 0) sin h
fx (0, 0) lim lim lim 1,
h 0
h
h 0
h
h 0 h
sin 0 h
4
(0, 0 ) (0, 0) 0 h
2
0
4 4
sin sin
h
3 4
f h f h h
f y (0, 0) lim lim lim lim 0.1 0
h 0
h
h 0
h
h 0 h h 0 h
61. (a) In the plane x 2 f y ( x, y) 3 f y (2, 1) 3 m 3
(b) In the plane y 1 fx
( x, y) 2 f y (2, 1) 2 m 2
2 2
62. (a) In the plane x 1 f y ( x, y) 3 y f y ( 1, 1) 3(1) 3 m 3
(b) In the plane y 1 fx
( x, y) 2 x f y ( 1,1) 2( 1) 2 m 2
f ( x , y , z h) f ( x , y , z )
fz
( x0 , y0, z0) lim ;
h 0
h
63.
0 0 0 0 0 0
2 2
f (1, 2, 3 h) f (1, 2, 3) 2(3 h) 2(9)
f (1, 2, 3) lim lim lim 12h 2h
z lim (12 2 h) 12
h 0
h
h 0
h
h 0
h
h 0
f ( x , y , h, z ) f ( x , y , z )
f y ( x0 , y0, z0
) lim ;
h 0
h
64.
0 0 0 0 0 0
f ( 1, h, 3) f ( 1, 0, 3) 2h
9h
0
f y ( 1, 0, 3) lim lim lim (2h
9) 9
h 0
h
h 0
h
h 0
2
65.
y 2 3 2 3
3 z z 2 z 0 3 2 z
x x z y x xz y x
y z at (1,1, 1) we have (3 2) z
x
1 1
or z 2
x
66. x y x y
2 x 0 2 x
z z x x z x z z x x z
x at (1, 1, 3) we have ( 3 1 2) x
z
1 or x 1
z 6
67.
2 2 2
a b c 2bc cos A 2 a (2bc sin A ) A A a ; also 0 2b 2c cos A (2bc sin A) A
a a bcsin
A
b
c cos A b
2c cos A 2 b (2bc sin A)
A A
b b bc sin A
a
68. a b
(sin A) a cos A
A
0 (sin A) a a cos A 0 a a cos A ; also 1
sin A sin B 2
sin A
x A sin A sin A
a b csc B cot B sin A
B
a
B
b( csc B cot B)
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