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