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?


enter image description here

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

$$z^2 = 2^2 + 3^2 \implies z = \sqrt{13}$$


$$\sec(\theta) = \frac{\sqrt{13}}{3}$$

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?


1 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


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

So we have

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

  • $\begingroup$ Ah, I see my problem. Thanks. However the answer is $$\frac{13\pi}{60}$$ $\endgroup$ Nov 9, 2015 at 6:53
  • $\begingroup$ Woops! Forgot the 2. $\endgroup$
    – Ben Longo
    Nov 9, 2015 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.