CHALLENGING 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 two friends changing when the distance between them is 200 m?
Solution
Let C be the center of the circular track. Let A be the location of the runner on the track and
let B be the location of his friend standing exactly 200 feet away from C outside the track.
We denote the distance AB between the two friends as l and the arc length of the circle that
the runner traces on the circular track as L . Lastly, we denote the angle ACB as  .
The situation is then depicted as follows:
We know:
dL
 7m / s
dt
(constant speed of runner)
We seek:
dl
dt
(how fast the distance between the two friends is changing when
l  200 m
The distance between them is 200 meters)
Applying the Law of Cosines in triangle ACB, we have:
l 2  2002  1002  2 100  200  cos   50,000  40,000cos  (*)
15
1
50, 000  l 2
1
This implies that cos  
. So for l  200 , cos   and sin   1    
.
4
4
40, 000
4
2
Differentiating (*) with respect to t , we have:
2l
dl
d
 40, 000sin 
.
dt
dt
dl
This implies that

dt
20, 000sin 
l
d
dt .
Moreover, the arc length L is equal to 100 (in radians) based on the arc length formula.
This implies that
dL
dL
d
d
7
 100
, or that
.
 dt 
dt
dt
dt 100 100
Therefore, when l  200 , we have:
dl
dt
l  200 m
 15   7 
20, 000 


4   100  7 15



 6.78m / s
200
4