Question

A runner sprints around a circular track of radius \(100\) m at a constant speed of \(7 {\rm{m/sec}}\). 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?

Short answer

The distance between the friend's changes when the distance between them is \(200\) m is about \(6.78\;{\rm{m/sec}}\).

Step-by-step solution

Write the formula of Law of Cosines and trigonometric identities

Law of Cosines:\({c^2} = {a^2} + {b^2} - 2ab\cos \theta \);\({\sin ^2}\theta + {\cos ^2}\theta = 1\)

Construct the diagram according to the given conditions

Step 3 Find the distance between the friends changing when the distance between them is \(200\) m

Assume the distance between the runner and the friend as \(l\). Apply the Law of Cosines and solve as shown below:

\(\begin{aligned}{l^2} &= {200^2} + {100^2} - 2\left( {200} \right)\left( {100} \right)\cos \theta \\ &= 50000 - 40000\cos \theta .............\left( 1 \right)\end{aligned}\)

Differentiate the above equation w.r.t \(t\).

\(\begin{aligned}\frac{d}{{dt}}\left( {{l^2}} \right) &= \frac{d}{{dt}}\left( {50000 - 40000\cos \theta } \right)\\2l\frac{d}{{dt}} &= - 40000\left( { - \sin \theta } \right)\frac{{d\theta }}{{dt}}............\left( 2 \right)\end{aligned}\)

Apply the formula of the length of the arc on the circle we have,

\(\begin{aligned}D &= 100\theta \\\theta &= \frac{1}{{100}}D\end{aligned}\)

Differentiate the above equation w.r.t \(t\).

\(\begin{aligned}\frac{{d\theta }}{{dt}} &= \frac{1}{{100}}\frac{{dD}}{{dt}}\\ &= \frac{1}{{100}}\left( 7 \right)\\ &= \frac{7}{{100}}\end{aligned}\)

Now find \(\sin \theta \) when \(l = 200\) to put it into the equation \(\left( 1 \right)\).

\(\begin{aligned}{200^2} &= 50000 - 40000\cos \theta \\400 &= 50000 - 40000\cos \theta \\\cos \theta &= \frac{1}{4}\end{aligned}\)

On applying trigonometric identity we get,

\(\begin{aligned}{\sin ^2}\theta + {\cos ^2}\theta & = 1\\{\sin ^2}\theta + \left( {\frac{1}{4}} \right) &= 1\\\sin \theta &= \sqrt {1 - \frac{1}{4}} \\\sin \theta &= \frac{{\sqrt {15} }}{4}\end{aligned}\)

Put \(\sin \theta = \frac{{\sqrt {15} }}{4}\), \(l = 200\)and \(\frac{{d\theta }}{{dt}} = \frac{7}{{100}}\) in equation \(\left( 2 \right)\).

\(\begin{aligned}2\left( {200} \right)\frac{{dl}}{{dt}} &= \left( {40000} \right)\frac{{\sqrt {15} }}{4}\left( {\frac{7}{{100}}} \right)\\\frac{{dl}}{{dt}} &= \frac{{7\sqrt {15} }}{4}\\ \approx 6.78\;{\rm{m/sec}}\end{aligned}\)

Thus, the distance between the friends changes when the distance between them is \(200\) m is about \(6.78\;{\rm{m/sec}}\).