A revolving light, located 5 ${\rm km}$ from a straight shoreline, turns at constant angular speed of 3 ${\rm rad}/\! {\rm min}$. With what speed is the spot of light moving along the shore when the beam makes an angle of $60^\circ$ with the shoreline?

Solution Figure 6 illustrates the triangle that describes the problem. $$\begin{eqnarray*} x &=& \hbox{the distance (in kilometers) of the beam of light from the point}\ B \\ \theta &=& \hbox{the angle (in radians) the beam of light makes with } AB \end{eqnarray*}$$

Both variables $x$ and $\theta$ change with time $t$ (in minutes). The rates of change are $$\begin{eqnarray*} \dfrac{\textit{dx}}{\textit{dt}} &=& \hbox{the speed of the spot of light along the shore (in kilometers per minute)} \\[4pt] \dfrac{d\theta }{\textit {dt}} &=& \hbox{the angular speed of the beam of light (in radians per minute)} \end{eqnarray*}$$

We are given $\dfrac{d\theta }{dt}=3 \,{\rm rad}/{\rm min}$ and we seek $\dfrac{\textit{dx}}{\textit{dt}}$ when the angle $\textit{AOB}=60{{}^{\circ}}$.

From Figure 6, $$\tan \theta =\frac{x}{5}\qquad\hbox{ so }x=5\tan \theta$$

Then $$\frac{\textit{dx}}{\textit{dt}}=5\sec ^{2}\!\theta \frac{d\theta }{dt}$$

When $\textit{AOB}=60 {{}^{\circ}} ,$ angle $\theta =30 {{}^{\circ}} =\dfrac{\pi }{6}{\rm rad}$, and $$\frac{\textit{dx}}{\textit{dt}}=\frac{5}{\cos ^{2}\theta }\frac{d\theta }{dt}=\frac{5(3)}{\left( {\cos \,}\dfrac{{\pi }}{6}\right) ^{2}}=\frac{15}{\dfrac{3}{4}}=20 \text{ }$$

When $\theta =30 {{}^\circ}$, the light is moving along the shore at a speed of $20 \,{\rm km}/{\rm min}$.