Thomas Calculus 13th [Solutions]

1014 Chapter 14 Partial Derivatives
37. The directional derivative is the scalar component. With f evaluated at P 0 , the scalar component of
the direction of u is f u ( Du
f ) P .
0
f in
38. Di f f i fxi f y j fzk i fx;
similarly, Dj f f j f y and Dk
f f k fz
39. If ( x, y ) is a point on the line, then T( x, y) ( x x0 ) i ( y y0)
j is a vector parallel to the line
T N 0 A( x x0 ) B( y y0
) 0, as claimed.
40. (a)
(b)
( kf ) ( kf ) ( kf ) f f f f f f
( kf ) i j k k i k j k k k i j k k f
x y z x y z x y z
( f g) ( f g) ( f g)
f g f g f g
( f g)
i j k i j k
x y z x x y y z z
f g f g f g f f f g g g
i i j j k k i j k i j k
x x y y z z x y z x y z
(c) ( f g)
f g (Substitute g for g in part (b) above)
(d)
( fg) ( fg) ( fg)
f g f g f g
( fg)
i j k g f i g f j g f k
x y z x x y y z z
f g f g f g g g g f f f
g i f i g j f j g k f k f i j k g i j k
x x y y z z x y z x y z
f g g f
f f f f g f g
f g
f g g g g f g f
x x y y
g f
z z
(e)
2 2 2
i j k i j k
g x y z g g g
f f f g g g
f f f g g g
i j k i j k g i j k f i j k
x y z x y z
x y z x y z
2 2 2 2 2 2 2
g g g f f f g f f g g f f g
g g g g g g g
f
g
14.6 TANGENT PLANES AND DIFFERENTIALS
1. (a) f 2xi 2yj 2 zk f (1, 1, 1) 2i 2j 2k
Tangent plane: 2( x 1) 2( y 1) 2( z 1) 0 x y z 3;
(b) Normal line: x 1 2 t, y 1 2 t, z 1 2t
2. (a) f 2xi 2yj 2 zk f (3, 5, 4) 6i 10j 8k
Tangent plane: 6( x 3) 10( y 5) 8( z 4) 0 3x 5y 4z
18;
(b) Normal line: x 3 6 t, y 5 10 t, z 4 8t
3. (a) f 2xi 2 k f (2, 0, 2) 4i 2k
Tangent plane: 4( x 2) 2( z 2) 0 4x 2z 4 0 2x z 2 0;
(b) Normal line: x 2 4 t, y 0, z 2 2t
4. (a) f (2x 2 y) i (2x 2 y) j 2 zk f (1, 1, 3) 4j 6k
Tangent plane: 4( y 1) 6( z 3) 0 2y 3z
7;
(b) Normal line: x 1, y 1 4 t, z 3 6t
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