Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 38

At what rate is the beam of light moving along the shore line when it is 1km from P






Using the tangent function
$\displaystyle \tan \theta = \frac{y}{3}$ ; when $\displaystyle y = 1 ; \theta = \tan^{-1}\left[ \frac{1}{3} \right] = 18.4349^{\circ}$
Taking the derivative with respect to time we have


$
\begin{equation}
\begin{aligned}
\sec^ \theta \frac{d \theta}{dt} &= \frac{\frac{dy}{dt}}{3}\\
\\
\frac{dy}{dt} &= 3\sec^2 \theta \frac{d \theta}{dt} && \text{ Equation 1}
\end{aligned}
\end{equation}
$


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

Plugging all the values in Equation 1 we have,

$
\begin{equation}
\begin{aligned}
\frac{dy}{dt} &= 3 \sec^2 (18.4349^{\circ}) \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}
$