Related rates: using angle to find rate of change of opposite side

Question

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?

Answer 1

We are given $\frac{d\theta}{dt}=6$deg/s$=\frac{\pi}{30}$rad/s and we know the lighthouse is always 2 miles away from the coast so $\tan\theta=y/2$. We are interested in $\frac{dy}{dt}$.

Differentiating implicitly we have have

$$\frac{dy}{dt}=2\sec^2(\theta)\frac{d\theta}{dt}$$

When $x=3$, $\sec(\theta)=\frac{\sqrt{13}}{2}$ (not $\frac{\sqrt{13}}{3}$).

So we have

$$\frac{dy}{dt}=2\left ( \frac{\sqrt{13}}{2}\right )^2\frac{\pi}{30}$$ $$=\frac{13\pi}{60}$$