Runner and his friends (Implicit Differentiation)

Question

A runner sprints around a circular track of radius 100m at a constant speed of 7m/s.The runner's friend is standing at a distance 200m from the center of track.

How fast is the distance between the friends changing when the distance between them is 200m?

The distance between the friends and distance between the runner's friend and the track center is constant, which is both 200m, but even though we have the rate of change of the runner, how can i find the rate of change of the other two?

Answer 1

The law of cosines can be used as the input to the process of implicit differentiation.   Let D be the distance between the friends and θ be the angle between (the line joining the center of the circle and the standing friend ) and (the radius line joining the center of the circle and the instantaneous position of the runner).  The law of cosines then is:

 

    D² =  100² + 200² - 2 100 200 cos(θ)

 

Differentiating this equation with respect to time gives

 

     2 D D'  = 2 100 200 sin(θ) θ'    

 

where D' is the derivative of D with respect to time and θ' is the derivative of θ with respect to time.

  θ'  = velocity / radius =  7/100 = 0.07

   The point of interest is where D = 200.    From the law of cosines equation, it can be worked out that

   cos(θ) at this point is equal to 1/4 .  This means that sin(θ) at this point is sqrt(15/16) =  .968

 

  Plugging into the the second equation gives

 

    400 D'   =  2 100 200 .968 0.07    Solving for D;  gives

 

  D'  =  6.776