Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 50

Suppose that a lighthouse is located on a small island, 3 km away from the nearest point $P$ on a straight shore line, and its light makes four revolutions per minute. How fast is the beam of light moving along the shore line when it is 1km from $P$





$
\begin{equation}
\begin{aligned}
\tan \theta &= \frac{y}{3}\\
\\
\theta &= \tan^{-1} \left( \frac{y}{3} \right)
\end{aligned}
\end{equation}
$


By taking the derivative with respect to time,

$
\begin{equation}
\begin{aligned}
\frac{d \theta}{dt} &= \frac{\frac{d}{dy}\left( \frac{y}{3} \right) \frac{dy}{dt}}{1 + \left( \frac{y}{3} \right)^2}\\
\\
\frac{d \theta}{dt} &= \frac{\left( \frac{1}{3} \right) \frac{dy}{dt}}{1 + \frac{y^2}{9}}\\
\\
\frac{d \theta}{dt} &= \frac{\left( \frac{1}{3} \right) \frac{dy}{dt} }{\frac{9+y^2}{9}}\\
\\
\frac{d \theta}{dt} &= \frac{3\frac{dy}{dt}}{9 + y^2}\\
\\
\frac{d \theta}{dt} &= \frac{9+y^2}{3} \frac{d \theta}{dt}
\end{aligned}
\end{equation}
$

Since, $\displaystyle 1 \text{km } \text{ and } \frac{d \theta}{dt} = 4 \frac{\cancel{\text{rev}}}{\text{min}} \left( \frac{2\pi\text{rad}}{1\cancel{\text{rev}}} \right) = 8 \pi \frac{\text{rad}}{\text{min}}$

So,

$
\begin{equation}
\begin{aligned}
\frac{dy}{dt} &= \frac{9+ (1)^2}{3} \left( 8 \pi \frac{\text{rad}}{\text{min}} \right)\\
\\
\frac{dy}{dt} &= \frac{80 \pi}{3} \frac{\text{km}}{\text{min}}
\end{aligned}
\end{equation}
$