Thomas Calculus 13th [Solutions]

Section 16.3 Path Independence, Potential Functions, and Conservative Fields 1177
(b) Since
f ( x, y)
3 2
(1,1)
x y is a potential function for F, F dr
f (1, 1) f ( 1,1) 2
( 1,1)
P N M P N M
y z z x x y
32. 0 , 0 , 2x sin y F is conservative there exists an f so that F f ;
f 2 f 2 g 2
g
x y y y
(a)
(b)
(c)
(d)
2x cos y f ( x, y, z) x cos y g( y, z) x sin y x sin y 0 g( y, z) h( z)
2 f
2 2
F
z
f ( x, y, z) x cos y h( z) h ( z) 0 h( z) C f ( x, y, z) x cos y C x cos y
C
C
C
C
2 2
(0,1)
2x cos y dx x sin y dy x cos y 0 1 1
2 2
(1, 0)
(1, 0)
2x cos y dx x sin y dy x cos y 1 ( 1) 2
2 2
( 1, )
(1, 0)
2x cos y dx x sin y dy x cos y 1 1 0
2 2
( 1, 0)
(1, 0)
2x cos y dx x sin y dy x cos y 1 1 0
(1, 0)
P N M P
y z z x
33. (a) If the differential form is exact, then 2ay c y for all y 2 a c, 2cx 2cx for all
(b)
N
x
M
y
x, and by 2ay for all y b 2a and c 2a
F f the differential form with a 1 in part (a) is exact b 2 and c 2
34.
( x, y, z) ( x, y, z)
g f g f
F f g x y z F dr f dr
f x y z f
(0, 0, 0) (0, 0, 0)
x x y y
g f
z z
( , , ) ( , , ) (0, 0, 0) 0, 0, and
0 g f F , as claimed
35. The path will not matter; the work along any path will be the same because the field is conservative.
36. The field is not conservative, for otherwise the work would be the same along C 1 and C 2 .
37. Let the coordinates of points A and B be xA, yA,
z A and xB , yB, z B , respectively. The force
F ai bj ck is conservative because all the partial derivatives of M, N, and P are zero. Therefore, the
potential function is f ( x, y, z) ax by cz C , and the work done by the force in moving a particle along
any path from A to B is f ( B) f ( A) f xB, yB , zB f xA, yA,
zA
axB byB czB C axA byA czA C a xB xA b yB yA c zB zA
F BA
38. (a) Let GmM C F C
i j k
3/2 3/2 3/2
x
2 2 2 2 2 2 2 2 2
x y z x y z x y z
P 3yzC
N M 3xzC P N 3xyC
M
y 2 2 2
5/2 z z 2 2 2
5/2 x x 2 2 2
5/2 y
x y z x y z x y z
y
, , F f for some f;
f xC
f yC g
f ( x, y, z) C g( y, z)
x 2 2 2
3/2
2 2 2
1/2 y 2 2 2
3/2 y
x y z x y z x y z
z
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