1. ## Related Rates Problem

A runner sprints around a circular track of radius 100 m at a constant speed of 7 m/s. The runner's friend is standing at a distance 200 m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200 m.

I attached a sketch of the problem so you can visualize what I'm doing.

$\displaystyle l=r\theta$

$\displaystyle \frac{dl}{dt}=r\frac{d\theta}{dt}$

$\displaystyle 7=r\frac{d\theta}{dt}$

$\displaystyle \frac{d\theta}{dt}=\frac{7}{100}$

I decided to use the law of cosines, $\displaystyle s$ is the distance between them:

$\displaystyle s^2=(100)^2+(200)^2-2(200)(100)\cos(\theta)$

Differentiating with respect to $\displaystyle t$

$\displaystyle 2s\frac{ds}{dt}=40,000\sin(\theta)\frac{d\theta}{d t}$

$\displaystyle \frac{ds}{dt}=\frac{20,000\sin(\theta)}{2s}\frac{d \theta}{dt}$

When the $\displaystyle s=200$

$\displaystyle 40,000=10,000+40,000-40,000\cos(\theta)$

$\displaystyle cos(\theta)=\frac{1}{4}\rightarrow \sin(\theta)=\frac{\sqrt{15}}{4}$

So no I plug everything into the equation for the time derivative of s:

$\displaystyle \frac{ds}{dt}=\frac{(20,000)(\sqrt{15})(7)}{4(100) }$

$\displaystyle =\frac{7\sqrt{15}}{8}$

This is very close to the answer in my book which is $\displaystyle \frac{7}{4}\sqrt{15}$

I can't seem to find where I want wrong.

2. It's where you go from $\displaystyle 2s\frac{ds}{dt}=,$ to $\displaystyle \frac{ds}{dt}=.$ You've missed a cancellation by 2.