October
25, 2006
A
runner runs 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?
Solution. Let the
distance between the runner and the friend be $\ell$.
Then, by the Law of Cosines,
$\ell^2=200^2+100^2-2\cdot 200\cdot
100\cdot\cos\theta = 50,000 - 40,000\cos\theta.$
Differentiating implicitly with respect to $t$, we obtain
$\displaystyle 2\ell\,\frac{d\ell}{dt} =
-40,000(-\sin\theta)\,\frac{d\theta}{dt}$.
Now if $D$ is the distance run when the angle is $\theta$ radians, then
$D=100\theta$, so $\theta = \frac1{100}D$ and
$\displaystyle\frac{d\theta}{dt} =
\frac1{100}\,\frac{dD}{dt} = \frac{7}{100}$.
To substitute into the expression for $d\ell/dt$, we must know
$\sin\theta$ at the time when $\ell = 200$, which we can find from our
first equation:
$200^2 = 50,000 - 40,000\cos\theta
\quad\Longleftrightarrow \quad \cos\theta = \frac14
\quad\Longrightarrow \quad \sin\theta = \sqrt{1-\left( \frac14
\right)^2} = \frac{\sqrt{15}}{4}$.
Substituting, we get
$\displaystyle 2(200)\,\frac{d\ell}{dt} =
40,000\left(\frac{\sqrt{15}}{4} \right)\left( \frac7{100} \right) \quad
\Longrightarrow \quad \frac{d\ell}{dt} = \frac{700\sqrt{15}}{2\cdot
200} = \frac{7\sqrt{15}}{4}\approx 6.78$ m/s.
Whether the distance between them is increasing or decreasing depends
on the direction in which the runner is running.