There is a lighthouse 2 miles away from a coastline (which happens to be a straight line). It emits a ray of light that rotates at 6 degrees per second. What is the speed in which this light moves across the coastline when it is currently at a point that is 3 miles away from the closest coastal point to the lighthouse?
Illustrated:
Alright. So basically we want to find out
$$\frac{dy}{dt} = \ ?$$
Since the lighthouse isn't going anywhere, we know that
$$\frac{dx}{dt} = 0$$
And of course, the ray of light is rotating at $6$ degrees per second:
$$\frac{d\theta}{dt} = 6$$
Now then. We need a way to relate $y$ with $x$ and $\theta$. This could work:
$$\tan(\theta) = \frac{y}{x}$$
We have to derive the function. We get that
$$-\sec(\theta)\cdot\frac{d\theta}{dt} = \frac{\frac{dy}{dt} \cdot x - \frac{dx}{dt} \cdot y}{x^2}$$
Before we evaluate this, we ought to know the value of $\sec(\theta)$:
$$\sec(\theta) = \frac{z}{3}$$
The value of $z$ right now is
$$z^2 = 2^2 + 3^2 \implies z = \sqrt{13}$$
So
$$\sec(\theta) = \frac{\sqrt{13}}{3}$$
Now we can plug all our data into the derived function:
$$-\frac{\sqrt{13}}{3}\cdot 6 = \frac{\frac{dy}{dt} \cdot 2 - 0 \cdot 3}{2^2}$$
Solving yields
$$\frac{dy}{dt} = -14.42$$
Sadly, this is wrong. The answer is $$\frac{13\pi}{60}$$
Perhaps it's because I need to transform $14.42$ to radians. That would be like $0.25$ radians (which is still wrong).
What was my mistake?