A man is moving away from a tower 41.6 m high at a rate of 2 ~m / s . If the eyelevel of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower is

Step-by-step solution

Let h be the height of the tower, h = 41.6 m. Let h₀ be the height of the man's eye level, h₀ = 1.6 m. The effective height of the tower above the man's eye level is H = h - h₀ = 41.6 - 1.6 = 40 m. Let x be the distance of the man from the foot of the tower. The angle of elevation is given by = H x = 40 x . Differentiating both sides with respect to time t , we get ^2 d dt = - 40 x^2 dx dt . Given dx dt = 2 m/s and x = 30 m. At x = 30 , = 40 30 = 4 3 . Since = 4 3 , we have ^2 = 1 + ^2 = 1 + ( 4 3 )^2 = 1 + 16 9 =