A television camera at ground level away from the launching pad of a space rocket

Question

A television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically, as seen in the following figure. The angle of elevation of the camera can be found by θ=tan^−1 (x/2000) where x is the height of the rocket.

Find the rate of change of the angle of elevation after launch when the camera and the rocket are 5367 feet apart.

find θ' in radians per foot.

Answer 1

tanθ = x/2000; then we differentiate by x: sec²θdθ/dx = 1/2000;

FDrom here dθ/dx = 1/2000cos²θ = 1/2000(2000/5367)² = 6.944·10^-5 rad/foot