A man is moving away from a tower 85 m high at a speed of 4 m/s. Find the rate at which his angle of elevation of the top of the tower is changing when he is at a distance of 60 m from the foot of the tower.
In the figure
OT = tower of height h = 85 m
M = position of the man at any instant of time
x = distance of the man from the tower
$\alpha =$ angle of elevation of the top of the tower as seen by the man
Now, speed at which the man is moving from the tower $= \dfrac{dx}{dt} = 4 \; m/s$
From the figure
$\tan \left(\alpha\right) = \dfrac{h}{x}$ $\;\; \cdots$ (1)
Differentiating equation (1) w.r.t time t gives
$\sec^2 \left(\alpha\right) \dfrac{d\alpha}{dt} = - \dfrac{h}{x^2} \dfrac{dx}{dt}$
$\begin{aligned}
\text{i.e. } \dfrac{d \alpha}{dt} & = -\dfrac{h}{x^2 \sec^2 \left(\alpha\right)} \dfrac{dx}{dt} \\
& = - \dfrac{h}{x^2 \left[1+ \tan^2 \left(\alpha\right)\right]} \dfrac{dx}{dt} \\
& = - \dfrac{h}{x^2 \left[1+ \left(\dfrac{h}{x}\right)^2\right]} \dfrac{dx}{dt} \;\; \left[\text{In view of equation (1)}\right] \\
& = \dfrac{-h}{x^2 + h^2} \dfrac{dx}{dt}
\end{aligned}$
Now, when $h = 85 \; m$, $x=60 \; m$ and $\dfrac{dx}{dt} = 4 \; m/s$,
$\dfrac{d\alpha}{dt} = \dfrac{-85}{60^2 + 85^2} \times 4 = -0.0314 \; \text{rad/s}$
$\therefore$ The angle of elevation of the top of the tower is decreasing at the rate of $-0.0314 \; \text{rad/s}$.