Thomas Calculus 13th [Solutions]

970 Chapter 13 Vector-Valued Functions and Motion in Space
(b) ur (cos ) i (sin ) j v ur x cos y sin r cos r sin (cos ) r sin r cos (sin )
by part (a), v u r r ; therefore, r x cos y sin ; u (sin ) i (cos ) j
v u xsin y cos r cos r sin ( sin ) r sin r cos (cos ) by part (a)
v u r ; therefore, r xsin y cos
6.
2 2
2
r f ( ) dr f ( ) d d r f ( ) d f ( ) d ;
dt dt 2 dt
2
dt
dt
v dr u d cos dr sin d sin dr cos d
dt r r u r i r j
dt dt dt dt dt
2 2
1 2
2
1 2
2 2
v dr r d f f d ; v ×a x y y x ,
dt dt dt
where x r cos and y r sin . Then dx r sin d cos dr
dt dt dt
2 2 2
d x
2
( 2sin ) d dr ( cos ) d ( sin ) d (cos ) d r ;
2 dt dt r dt
r dy ( r cos ) d (sin ) dr
2 2
dt dt dt dt dt dt
2
d y 2
2 2
(2cos ) d dr ( sin ) d ( cos ) d (sin ) d r .
2 dt dt r dt
r Then, after much algebra v a
2 2
dt dt dt
3 2 2
2 3 2
2
2 2
r d r d dr r d d r 2 d dr d f f f 2 f
dt dt dt dt dt dt dt dt
2 2 3 2
2
v a
v
f f f f
f
2 2
f
2
2 2
7. (a) Let r 2 t and 3t dr 1 and d 3 d r d 0. The halfway point is (1, 3) t 1;
dt
dt 2 2
dt dt
2 2
2
v dr u d
r r u v(1) ur
3 u ; a d r r d u d 2 dr d
dt dt
2 dt r r
u
2
dt
dt dt dt
a(1) 9ur
6u
(b) It takes the beetle 2 min to crawl to the origin
the rod has revolved 6 radians
6 2 2 6 2 2 6
2
L f ( ) f ( ) d 2 1 d 4 4 1 d
0 0 3 3 0 3 9 9
6 37 12 1
6 2 1 ( 6) 2 1
2
d ( 6) 1 d
( 6) 1 ln 6 ( 6) 1
0 9 3 0
3 2 2
0
2 6
37 1 ln 37 6 6.5 in.
6
8. (a) x r cos dx cos dr r sin d ; y r sin dy sin dr r cos d ; thus
2 2 2 2 2 2
dx cos dr 2r sin cos dr d r sin d and
2 2 2 2 2 2 2 2 2 2 2 2 2 2
dy sin dr 2r sin cos dr d r cos d ds dx dy dz dr r d dz
(c) r e dr e d
ln8 2 2 2 2
L dr r d dz
0
ln8 2 2 2
e e e d
0
ln8
ln 8
3e d 3e
0 0
8 3 3 7 3
(b)
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