Thomas Calculus 13th [Solutions]

Chapter 16 Practice Exercises 1225
2 1 2
F 5
1 dr1 2 t 1 dt Work 1
0
2 t 1 dt
3
; r2
i j tk , 0 t 1 x 1, y 1, z t and
1
dr 5 8
2 k dt F2 2i j k F2 dr2 dt Work2 dt 1 Work Work
0
1 Work2 1
3 3
36. Over Path 1: r ti tj tk , 0 t 1 x t, y t,
z t and
2 1 2
F dr
3t 1 dt Work 3t 1 dt 2;
0
2 2
dr ( i j k) dt F 2t i t j k
Over Path 2: Since f is conservative, F dr 0 around any simply closed curve C. Thus consider
C
F ,
curve
d r F
C
d r F
C
d r where C 1 is the path from (0, 0, 0) to (1, 1, 0) to (1,1, 1) and C 2 is the
1 2
path from (1, 1, 1) to (0, 0, 0) . Now, from Path 1 above,
F dr
C
1
2
F dr C
2 0 F dr F dr
curve C
( 2)
2 1
t t t t
37. (a) r e cos t i e sin t j x e cos t,
y e sin t from (1, 0) to
e
2 ,0 0 t 2
dr t t t t
xi
yj
e cost i e sin t j
e cos t e sin t i e sin t e cos t j and F
dt
2 2
3/2
2 2 2 2
3/2
t
t
x y e cos t e sin t
cos t sin t 2 sin cos 2
cos t t sin sin t cos t t t
2
i j F dr
t t
e Work e dt 1 e
2t 2t t t t t
e e dt e e e e
0
xi
yj
f 2 2
1/2
x
f y g
x y y
x y x y x y
(b) F
f ( x, y, z) x y g( y, z)
2 2
3/2
2 2
3/2
2 2
3/2
y
2 2 3/2
( x y )
C
F
2 2 1/2
g( y, z) C f ( x, y, z) ( x y ) is a potential function for F
2 2
d f e ,0 f (1,0) 1 e
r
t
t
2
38. (a)
(b)
2 y
F x ze F is conservative F dr 0 for any closed path C
C
C
(1, 0, 2 ) 2 y 2 y 2 y
d x ze d x ze x ze
(1, 0, 0)
(1, 0, 2 ) (1, 0, 0)
F r r
2 0 2
39.
i j k
2i 6j 3k
F
2 yk ; unit normal to the plane is n 2 i 6 j 3 k
x y z
4 36 9 7 7 7
2 2
y y 3z
F n 6
7 y;
p k and | f |
f ( x, y, z) 2x 6y 3 z | f p| 3 d dA 7 dA
| f p| 3
F
C
6 6 7
2 1 2
dr
yd y dA 2y dA 2r sin rdrd 2 sin d 0
7 7 3 0 0 0 3
R R R
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