How fast is the distance between the runner and the trainer increasing when the runner has run 1/4 of the way around the track?

Question

A runner and his trainer are standing together on a circular tracks of radius 100 meters.when the trainer gives a signal, the runner starts to run around the track at a speed of 10 m/s. How fast is the distance between the runner and the trainer increasing when the runner has run 1/4 of the way around the track?

Answer 1

This is a related rates problem that can be solved using basic calculus. Draw yourself a picture of the scenario - a circular track with radius 100 m with the runner 1/4 of the way around from the trainer. Draw a line between the runner and the trainer to represent the distance between the runner and trainer and let's call that distance z. We are looking for the rate of change of that distance, or dz/dt.

The distance between the runner and the trainer forms a chord, and the formula for that chord length is:

z = 2rsin (θ/2). Since r is fixed at 100m, we can substitute that in for r, so:

z = 200 sin (θ/2).

The distance between the runner and the trainer depends on the central angle formed between the two. So how can we relate that angle to our known information? We know the arc length and the rate of change of the arc length. Since the runner is running at 10 m/s around the track, that is the rate of change of the arc length, call it s. So ds/dt = 10m/s. We also know the arc length at the moment the runner is a fourth of the way around, but we're not ready to use that information yet.

Arc length is related to the central angle by s = rθ where θ is in radians. Since r = 100, s = 100θ, or

θ = s/100. We can now use this relationship to relate the chord length z (distance between runner and trainer) to the arc length s by substitution. When we substitute into the above formula we get:

z = 200 sin (s/200).

Now we are ready to differentiate both sides with respect to time t. Use the chain rule for the right side - the outer function is 200 sin () and the inner function is s/200.

dz/dt = 200 cos (s/200)• 1/200• ds/dt

dz/dt = cos ( s/200) • ds/dt

Remember we are solving for dz/dt at the point when the runner is 1/4 of the way around the track. That occurs when s = rθ = 100•π/2 = 50π

We also know that ds/dt = 10 m/s. Substitute those values into the above equation:

dz/dt = cos (50π/200) •10

= cos (π/4) •10

= √2/2•10

= 5√2 m/s