Page 158 - How To Solve Word Problems In Calculus
P. 158
It follows that
29 dθ
500 = 5000 · ·
25 dt
dθ
500 = 5800
dt
dθ 500 5
= = rad/sec
dt 5800 58
dθ 5 180
In terms of degrees = · ≈ 4.94 degrees/sec
dt 58 π
4.
x
θ P
L 2
dθ dx
Given: = 6π rad/min Find: when x = 1
dt dt
(3 revolutions = 6π radians)
x
= tan θ
2
x = 2 tan θ
dx dθ
2
= 2 sec θ
dt dt
1
2
2
When x = 1, tan θ = . Since sec θ = tan θ + 1, it follows that
2
5
2
sec θ = .
4
dx 5
= 2 · · 6π
dt 4
= 15π km/min
145
