Thomas Calculus 13th [Solutions]

964 Chapter 13 Vector-Valued Functions and Motion in Space
11.
1 2 1 2 2
y y0 v0 sin t gt y 6.5 (44 ft/sec)(sin 45 )(3sec) 32 ft/sec (3sec) 6.5 66 2 144
2 2
44.16 ft the shot put is on the ground. Now,
2
y 0 6.5 22 2t 16t 0 t 2.13sec (the positive
root) x (44 ft/sec)(cos 45 )(2.13sec) 66.27 ft or about 66 ft, 3 in. from the stopboard
12.
y
2
v0
sin
max y0 2 g
2
(2)(32 ft/sec )
(80 ft/sec)(sin 45 )
7 ft 57 ft
2
1 2
1
13. x v 0 cos t and 1 2
y v0 sin t gt v
2 0 sin g t
2
y v0 sin t gt tan
2 x v cos t v cos
1
0 0 2
2v
sin 2v
cos tan
g
0 0
v cos tan v sin gt t , which is the time when the golf ball hits the
2 0 sin 2 0 cos tan 2 2 2
upward slope. At this time x v v v
2
0 cos 0 sin cos 0 cos tan .
g
g
v v Now
OR x OR 2
cos
g
2 2 2
0 v0
v sin cos cos tan
cos
2 2
0 v0
2v
cos sin cos tan 2 cos sin cos cos sin
g cos cos g
2
cos
0
0
2
0
2v
cos
2
g cos
is maximized:
sin( ) .
dx
d
2
0
2v
g
The distance OR is maximized when x
cos 2 sin 2 tan 0
cos 2 sin 2 tan 0 cot 2 tan 0 cot 2 tan 2
2 2 4
14. (a) x v 0 cos 40 t and 1 2 2
y 6.5 v0 sin 40 t gt 6.5 v
2
0 sin 40 t 16 t ; x 262 5 ft and
12
y 0 ft 262 5 v 262.4167
12 0 cos 40 t or v 0 and
cos 40 t
262.4167
2
0 6.5 sin 40 t 16t
cos 40 t
2
t 14.1684 t 3.764sec. Therefore, 262.4167 v0
cos 40 (3.764sec)
v 262.4167
0 v0
cos 40 (3.764sec)
91 ft/sec
(b)
2
v
(91) sin 40
0 sin
ymax y0 6.5 60 ft
2 g
(2)(32)
2
15.
2
2 2 2
r 2cos t i 2sin t j t k v 2sin t i 2cost j 2tk v 2sin t 2cost 2t
2
2 1 t
4
Length = /4 2 2 2
2 2
0
2 1+t dt t 1 t ln t 1 t
4 1 16 ln 4 1 16
0
16.
3 2 1 2 2 2 1 2
r 3cos t i 3sin t j 2t k v 3sin t i 3cost j 3t k v 3sin t 3cos t 3t
3 3 2
3
9 9t 3 1 t Length = 3 1 t dt 2 1 t 14
0 0
2
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