Page 144 - How To Solve Word Problems In Calculus

By the law of cosines,

2
2
2
x = 5 + 10 − 2 · 5 · 10 · cos θ
2
x = 125 − 100 cos θ
Differentiating with respect to t,
dx dθ
2x = 0 − 100(−sin θ)
dt dt
dx dθ
x = 50 sin θ
dt dt
π
We need the value of x when θ = · Since
3
π
2
x = 125 − 100 cos θ, when θ = ,
3
2
x = 125 − 50 = 75 √
π 3
√ √ sin =
x = 75 = 5 3 3 π 1 2
cos =
Substituting, 3 2
√
√ dx 3 π
5 3 = 50 · ·
dt 2 36
√
√ dx 25π 3
5 3 =
dt 36
dx 5π
= in/min
dt 36
EXAMPLE 2
A kite is flying 100 ft above the ground, moving in a strictly
horizontal direction at a rate of 10 ft/sec. How fast is the angle
between the string and the horizontal changing when there is
300 ft of string out?

Solution
x
θ

100
z

131