A man is moving away from a tower 41.6 m high at the rate of 2 m/sec. Find the rate at which the angle of elevation of the top of tower is changing, when he is at a distance of 30m from the foot of the tower. Assume that the eye level of the man is 1.6m from the ground.

Explanation

`tan theta=frac{41.6-1.6}{{x}}`



Or


`tan theta=frac{40}{x}` Or `x=40 cot theta`


Differentiating with respect to time, we get



`{v}_{{x}}=-40 {cosec}^{2} theta cdot frac{{d} theta}{{dt}}`



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