Revolving Beacon confusing calculus problem This is not homework. I have the solutions, I just can't understand why mine are wrong.
The problem states: A revolving beacon $3600$ feet off a straight shore makes $2$ revolutions per minute. How fast does its beam sweep along the shore, (a) at the point on the shore nearest the beacon? (b) at the point on the shore $4800$ feet away from the beacon?
My solutions (second one is wrong): I didn't feel I needed calculus to solve (a). Basically the beacon is travelling 2 times the circumference of a circle of radius 3600 feet at (assumedly) constant speed in 1 minute, so $$C = 2 \pi r = 7200 \pi$$ and therefore $v = 14400 \frac{feet}{minute}$, which is going to be the instantaneous speed at any point on the circumference of that circle, and this agrees with the solution in the book.
For (b), I figured that since the point is $4800$ feet away from the beacon, I could just again take the constant speed of the beacon along the circumference of a circle of radius 4800 this time and have that be the instantaneous speed at any point along the circumference of that circle, and therefore at the point on the shore $4800$ feet away from the beacon.
Which gives me $$C = 2 \pi r = 9600 \pi$$ and $v = 19200 \frac{feet}{minute}$.
However, the book disagrees with that result, and states:
Measuring the distance, $x$, along the shore from the foot of the perpendicular from the beacon to the shore and taking $A$ as the angle between this perpendicular and the beam, we have $x = 3600 \tan{A} $. Hence, $\frac{dx}{dt}=\frac{14400 \pi}{\cos^2{A}}$, making the answer to (b) $25600 \pi \frac{feet}{minute}$.
I don't understand why my answer to (b) is wrong. If constant speed made sense as instantaneous speed at a point given a radius of $3600$, why doesn't it work for a radius of $4800$. I fail to see how the two questions are fundamentally different.
 A: The difference is the angle the shore makes with the light beam at each spot.
Note that the question asks "How fast does its beam sweep along the shore," which implies that the direction is parallel to the shore. Your method calculates the speed along (tangent to) the circle swept out by the light beam.
In your part (a) the shoreline is tangent to the swept-out circle, so your direction is along the shore and your answer agrees with the book's. In part (b) the circle intersects the shoreline at an angle, which your answer did not take into account.

(As pointed out by @Michael in the comments, the following analysis is incorrect. I'll leave it here so the comments make sense.)
It is possible to get from your answer to the book's answer. Your answer is the component of the desired velocity vector that is perpendicular to the ray from the beacon to the intersection point of circle and shore. You can use trigonometry to find the angle of the desired velocity vector to your component and then the magnitude of the desired velocity vector. We do that kind of problem in my 12th-grade Physics class, but I won't bother doing it here. Ask if you want those details, but I think my first three paragraphs answer your stated question.
A: The position on shore is related to the angle by
$x(t) = d \tan \theta(t)$.

Differentiating gives
$\dot{x}(t) = d { 1\over \cos^2 \theta(t) } \dot{\theta}(t)$.
In all cases $\dot{\theta}(t) = 4 \pi$ $\text{min}^{-1}$, $d=3,600$ $\text{ft}$.
For Part (a), if $\theta(t) = 0$, then we have
$\dot{x}(t) = d \dot{\theta}(t) = 14,400 \pi $ $\text{ft } \text{min}^{-1}$.
For Part (b), we have $\cos \theta(t) = {3600 \over 4800 }= {3 \over 4}$ so
$\dot{x}(t) = d ({ 4 \over 3})^2 \dot{\theta}(t)  = 25,600 \pi$ $\text{ft } \text{min}^{-1}$.
