The ancient lighthouse related rates problem A lighthouse is 7 miles from the shore and sweeps at 10 seconds per revolution. At what rate does the beam move along the shore, in mph, when the beam makes a 30 degree angle with the shore?
I've checked my solution a million times and can't find anything wrong with it except the answer: 3,628,800 mph. That can't possibly be right!
Here's my reasoning:
10 seconds per revolution is 36 degrees per second. So $\frac{d\theta}{dt}=36$ degrees/sec. Then we have
$$s = 7\tan(\theta)$$
$$\frac{ds}{dt} = 7\sec^2{\theta}\frac{d\theta}{dt} = \text{ 3,628,800 mph}$$
where $s$ is the distance along the shore and $\theta = 60^{\circ}$. Where's my mistake?
 A: "Where is my mistake?"
You have to measure $\theta$ in radians, and the time unit is $1$ hour. As the beam makes $360$ full turns per hour one has
$$|\dot\theta|=2\pi\cdot 360\ .$$
You therefore obtain
$$|\dot s|=7\cdot \sec^2(60^\circ)\cdot 2\pi\cdot 360\doteq 63\,335\qquad({\rm mph})\ .$$
A: Put the lighthouse at the origin. Put the shore to be the vertical line $x=7$. Drawing a line segment oriented at angle $\theta$ from the origin to the line, you have ASA so you can solve the triangle. You find the length of the vertical piece is $7 \tan(\theta)$ (assume we're looking at $\theta \geq 0$).
So the speed along the shore is $\frac{d}{dt} 7 \tan(\theta(t)) = 7 \sec(\theta(t))^2 \theta'(t)$. Here $\theta$ changes from $0$ to $2 \pi$ in time $10$ so $\theta'=\frac{\pi}{5}$. The angle with the shore is $\pi/2-\theta$, so we need $\sec(\pi/3)=2$. So the rate in miles per second is $7 \cdot 4 \cdot \frac{\pi}{5} = \frac{28 \pi}{5}$, which is about $18$ miles per second (no calculator, just a rough estimate). Multiplying by 3600 converts to mph. The answer is big, but it's not millions of mph.
A: I used degrees instead of radians, in degrees one doesn't have $\frac{d\tan\theta}{dx} = \sec^2\theta$. When one uses radians one gets 63335 mph.
