Person A sprints around a circular track of radius 100m at a constant speed of 7 m/s. Person B is standing...?

Question

1. Person A sprints around a circular track of radius 100m at a constant speed of 7m/s. Person B is standing at a distance 200m from the center of the track. How fast is the distance between the 2 changing when the distance between them is 200m?

2. Show that the volume of the sphere is decreasing at rate proportional to its surface area then the radius of the sphere is shrinking at a constant rate.

michael · Accepted Answer

1.  This means person A is at the point on the track nearest person B and person A's direction is perpendicular to the line from person A to person B.
So a right triangle with a=100, b=0, c=100,  satisfies a^2+b^2=c^2 and has db/dt=7.

2a da/dt +2b db/dt = 2c dc/dt is the equation to use.  
Substituting you get 200*0+0*7 = 200 dc/dt.  
This tells you 0 = dc/dt which may seem surprising at first but it is correct.

As the runner approaches the point 200 m from person B the distance between them is decreasing and once the runner passes that point the distance starts increasing again, so the rate of change at that critical point is 0.

This is similar to the velocity of an object thrown straight up being 0 when it reaches its apex.

Person A sprints around a circular track of radius 100m at a constant speed of 7 m/s. Person B is standing...?

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