A man is moving away from a 40-m high tower at a speed of 2 m/s. Find the rate at which the angle of elevation of the top of the tower is changing when he is at a distance of 30 metres from the foot of the tower. Assume that the eye level of the man is 1.6 m from the ground.

Answer

0.032 radian/second

Explanation

Let at any time, t the man be at a distance of x metres from tower AB and let `theta` be the angle of elevation at that time
Then, `x = 40 cot theta rArr (dx)/(dt) = -40 "cosec"^(2) theta.(d theta)/(dt)` ltrbgt `:. (d theta)/(dt) = (-2 sin^(2) theta)/(40) = - (1)/(20).((AB)/(BE))^(2) = - (1)/(20).((40)^(2))/((40)^(2) + x^(2))`
Find its value when `x = 30 m`