A television camera is positioned 2500 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 that its speed is 800 ft/s when it has risen 4000 ft.
How fast is the distance from the camera to the rocket changing in ft/s at that moment?
If the televison camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing in radians/s at that moment?
Calculus - Steve, Tuesday, July 2, 2013 at 12:12am
The distance y when the rocket has reached height h is given by
y^2 = h^2 + 2500^2
So, when h=4000, y = 500√89
2y dy/dt = 2h dh/dt
2(500√89) dy/dt = 2(4000)(800)
dy/dt = 6400/√89 = 678.4 ft/s