posted by Tracy .
I also have another question (I'm not doing too good one these realted rates).
A camera, located 2 km from the launch pad, is tracking the rocket that is fired straight up. When the height of the rocket is 20 km, the camera is rotating at the rate of 1/200 radians per second. What is the speed of the rocket at that instant? Give your answer in km/sec.
This is the same sort of thing right? The only thing that is stopping me is the whole radians per second. Would that mean that we would have to get trig into here somewheres. I mean I know its a triangle drawing so it would be possible, but radians really mess me up.
Here you know the angle, and you are given dTheta/dt. You are looking for dh/dt
You know TanTheta=h/2, and you know dtheta/dt when h=2-
start with TanTheta=h/2
take the derivative with respect to time. It will come out rather simply.
sec^2theta * dTheta/dt = 1/2 dh/dt
You know sec from the triangle. (Pyth theorem)