A revolving searchlight, which is $100$ m from the nearest point on a straight highway, casts a horizontal beam along a highway. The beam leaves the spotlight at an angle of $\frac{π}{16}$ rad and revolves at a rate of $\frac{π}{6}$ rad/s. Let $w$ be the width of the beam as it sweeps along the highway and $θ$ be the angle that the center of the beam makes with the perpendicular to the highway. What is the rate of change of $w$ when $θ = \frac{π}{3}?$ Neglect the height of the lighthouse.
100% directly from the textbook. Where do I begin?
Here’s a bit to get you started. As Ross said, you start with a diagram:
Here $S$ is the spotlight, the heavy horizontal line is the highway, the line $\overline{SC}$ marks the centre of the spotlight’s beam, and the lines $\overline{SA}$ and $\overline{SD}$ mark the edges of the beam. You’re told that the angle $\angle BSD=\frac{\pi}{16}$, so $\angle BSC=\angle CSD=\frac{\pi}{32}$. The angle $\theta$ is the angle that the centre of the beam makes with the perpendicular $\overline{SA}$ from $S$ to the highway, so $\theta=\angle ASC$. Finally, $w$ is the width of the beam measured along the highway, so $w=|BD|$.
You know how fast $\theta$ is changing, since you’re given the rate of rotation of the spotlight. To solve the problem, therefore, you need to express $w$ in terms of $\theta$. This can be done with a little trigonometry involving tangents and the following data:
$$\begin{align*} &\angle ASB=\theta-\frac{\pi}{32}\\ &\angle ASC=\theta\\ &\angle ASD=\theta+\frac{\pi}{32}\\ &|SA|=100 \end{align*}$$