**a cameraman is on the ground 3000 feet away from the vertical launch of a rocket. The rockets speed is constant at 1000 feet per second. Find the rate of change of the angle of elevation of the camera 3 seconds after liftoff.**

**Answer:**

Draw a right triangle whose hypotenuse is the cameraman’s line of sight to the rocket. The base of the triangle is 3000’, the cameraman’s distance from the launch pad. The (changing) altitude of the triangle is the rocket’s vertical distance above the earth.

Theta, the angle of elevation, is given by arctan(altitude/base). Instantaneous altitude is velocity times time in flight or vt.

Theta = arctan[vt/3000].

Differentiate theta with respect to time to get rate of change of angle of elevation.

d{arctan(u)] = du/[1+ u^2]. For u = vt/3000, du = vdt/3000.

Plug stuff in: d(theta)/dt = (v/3000)/[1 + (vt/3000)^2].

Evaluate for v = 1000 ft/s, t = 3 s

d(theta)/dt = (1000/3000)/[1 + (3000/3000)^1]

d(theta)/dt = (1/3)/(1 + 1) = 1/6 degree per second.