A rocket is rising straight up from the ground at a rate of 1000 km/h. An observer 2 km from the launching site is photographing the rocket. How fast is the angle of the camera with the ground changing when the rocket is 1.5 km above the ground. Solution at some time let angle be so tan = y/x x = 2 so differentiating (d/dt)sec^2 = .5(dy/dt) (d/dt)1/cos^2 = .5*1000 = 500 cos = 2/sqrt(1.5^2 + 4) = .8 so (d/dt) = 500*.8^2= 320 d/dt = 320.