Super Hard Challenge Calculus Problem. PLEASE HELP ME!!?

Question

So the question goes like this:
A TV camera at ground level is filming the liftoff of a space shuttle that is rising vertically according to the position eqn: s(t)=50t^2, where s is measured in feet and t is measured in seconds. The camera is 2000 feet from the launch pad.

Find the velocity of the space shuttle when it is 5000ft high.
[[I did this already, and I got 1000 ft/s. I know that it is right.]]

Find the rate of change in the angle of elevation of the camera when the space shuttle is 5000 ft. high.
[[I can't figure this one out for the life of me. I think it has something to do with inverse tangent, and with the previous anwser (1000 ft/sec).]]

If you can solve this, you're my hero and a freaking genius!!

Answer 1

tan ( angle ) = rise / run = 50t² / 2000 = t² / 40.


So the angle is, as you said, inv-tan ( t² / 40 ).
The rate of change of the angle is the first derivative.

The derivative of arctan ( x ) = 1 / ( 1 + x² ) so

Rate of change of angle = ( 1 / ( 1 + t^4 / 40 ) ) ( t / 20 )

Find the time corresponding to a height of 5000 ft, then plug that time into the last equation to get the rate of change of the angle.