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.
Best Answer
The rotating beam moves by say $\theta$ but it isn't a constant radius circle that is relevant. Rather, the beam hits a spot on the shore which at different times will be at different distances from the source of the beam. For example when the rotating beam is perpendicular to the shore the distance will be shortest, and as it rotates away from there the distance increases. Draw several pictures and see the relation to $\tan \theta.$