Calculus
posted by BA on .
A recording camera is located 2500 feet from the launch pad for recording a rocket launch. As the rocket lifts off, the angle of elevation of the camera increases 4 degrees per second while recording the launch. Find the instantaneous velocity and acceleration of the rocket at time t>0.

using ' to denote d/dt, the height h is
h = 2500 tanθ
h' = 2500 sec^2 θ θ'
at t=0, θ'=4*pi/180
h' = 2500 * 4pi/180 = 5.55 ft/s
h'' = 2500 sec^2θ (2tanθ θ' + θ'') 
Thanks, but how do you find the instantaneous velocity from that? I know that acceleration is the derivative from the equation of Instantaneous velocity.. but I don't know how to get the equation for it.

Nevermind, I see what you did. Thank you!