For motion that is circular, angular speed \(\omega\) is defined as the rate of change of a central angle \(\theta\) of the circle with respect to time. That is, \(\omega = \dfrac{d \theta}{dt}\), where \(\theta\) is measured in radians.

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 \(20 \,{\rm km}/{\rm min}\).