Question

A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and its speed is 700 ft/s when it has risen 3000 ft.

b) If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment

Answer 1

\[\tan \theta = \frac{h}{4000} \]
Take derivative
\[\sec^2 \theta \frac{d \theta}{dt} = \frac{1}{4000} \frac{dh}{dt}\]
\[\frac{d \theta}{dt} = \frac{1}{4000} \frac{dh}{dt}*\cos^2 \theta\]
By using pythagorean thm, obtain ratio for cos
\[\rightarrow \cos^2 \theta = (\frac{4000}{\sqrt{h^2 + 4000^2}})^2\]
\[\frac{d \theta}{dt} = \frac{4000}{h^2 + 4000^2} * \frac{dh}{dt}\]

Answer 2

is it .112? @dumbcow

Answer 3

got it!

Answer 4

yep :)

Answer 5

thank you!

Answer 6

Refer to a solution using Mathematica.