The problem comes from MIT's calculus exercise book (2E-2), and goes as follows:
A beacon light 4 miles offshore (measured perpendicularly from a straight shoreline) is rotating at 3 revolutions per minute. How fast is the spot of light on the shoreline moving when the beam makes an angle of 60* with the shoreline?
The answer from the answer book is:
$tan\theta = x/4$ and $ d\theta/dt=3(2\pi)=6\pi$ with $t$ is measured in minutes and $\theta$ measured in radians. The light makes an angle of $\pi/3$ with the shore when $\theta$ is $pi/6$. Differentiate with respect to $t$ to get $(sec^2 \theta)(d\theta/dt) = (1/4)(dx/dt)$. Since $sec^2 (\theta/6) = 4/3$, we get $dx/dt = 32 \pi$ miles per minute.
My solution is different than theirs:
When the beam rotates by an angle $\theta$, then it travels a distance equal to $l = x\theta$ (x is the radius of the circle whose center is the lighthouse). Differentiating with respect to $t$, we get $dl/dt = x (d\theta/dt)$. With $x = 4 tan(\pi/6)$.
Of course the answers are different. My understanding of the problem is the beam rotates in a circular fashion so the radius of that circle is a constant (which the variable $x$ I used). With that understanding it's only a matter of converting from angular velocity to distance velocity.
However, from the the answer book the distance $x$ is a variable ($x = 4tan\theta$). My struggle is visualising how does this beam of light operate.