1. ## Derivative/rate problem

A lighthouse 1 km from a beach shore revolves at 10 $\pi$ radians/minute. How fast is the light sweeping across the shore 2 km from the lighthouse?
I drew a diagram here to try and help:

I think it will have something to do with a triangle. So just to find the other side, I'll use the Pythagorean theorem and get $\sqrt{3}$. I already knew this because I recognized the 30-60-90 triangle, so I know the angles too, but I don't know if we need them yet.

What I don't understand is what the question is asking. I thought the light was going around at a constant speed of 10 $\pi$ radians/minute? What speed do they want?

2. ## Re: Derivative/rate problem

$\frac{d\theta}{dt} = 10\pi \, rad/min$

let the horizontal distance along the beach between the ray and the perpendicular be $x$ ... you are looking for $\frac{dx}{dt}$

using the right triangle, the following relationship between $\theta$ and $x$ exists (why?) ...

$\theta = \arctan(x)$

or

$\tan{\theta} = x$

take the derivative of either of the above equations w/r to time, sub in your given/calculated values and determine the desired rate of change.

3. ## Re: Derivative/rate problem

$\frac{dx}{dt} = (tan\theta)' \\\\\frac{dx}{dt} = \sec^{2}\theta \cdot \frac{d\theta}{dt} \\\\\frac{dx}{dt} = \sec^{2}\theta \cdot 10\pi \\\\\frac{dx}{dt} = \sec^{2}(60) \cdot 10\pi \\\\\frac{dx}{dt} = \frac{1}{\cos^{2}(60)} \cdot 10\pi \\\\\frac{dx}{dt} = \frac{10\pi}{(\frac{1}{2})^{2}} \\\\\frac{dx}{dt} = \frac{10\pi}{\frac{1}{4}} \\\\\frac{dx}{dt} = 40\pi$km./min.

Wow, I wish I could figure out relationship that quickly. I worked on this for an hour and couldn't come up with the information you provided..

Except I still don't get why you need to take the derivative of the trigonometric function with respects to time to arrive at the solution..

4. ## Re: Derivative/rate problem

Except I still don't get why you need to take the derivative of the trigonometric function with respects to time to arrive at the solution.
the mathematical relationship between $x$ and $\theta$ is

$x = \tan{\theta}$

their respective rates of change are also related (that is why these problems are called "related rates" problems) ... you have to take the derivative of the original relation with respect to time to see how those respective rates of change are related ...