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,

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