# A man 2 m tall is walking from a lampost that is 6 m tall at a rate pf 1.5 m/s. Find the rate of change of the angle of depression when he is 8 m away.

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A man that is 2 m tall is walking away from a lamp post that is 6 m tall at a rate of 1.5 m/s.

The angle of depression is is the angle that a line drawn from the top of the lamp post to the man's head makes with the vertical. Consider the triangle when the man is L m away from the lamp post consisting of the vertical line 4 m long and the horizontal line of length L m. If the angle of depression is X, tan X = L/4.

Differentiating both sides with respect to t, `(dX)/(dt)*sec^2 X = (1/4)*(dL)/(dt)`

The rate at which the man is walking away is 1.5

=> `(dX)/(dt) = (1.5/4)*(1/(sec^2X))`

=> `(dX)/(dt) = (1.5)/4*(1/(1 + tan^2 X))`

=> `(dX)/(dt) = (1.5)/4*(1/(1+L^2/16))`

=> `(dX)/(dt) = (4*1.5)/(16 + L^2)`

At L = 8

`(dX)/(dt) = 6/80`

=> `(dX)/(dt) = 3/40`

**The rate at which the angle of depression is changing when the person is 8 m away is `3/40` radian/s**