1. ## Differentiation word problem

A runner sprints around a circular track of radius 100m at a constant speed of 7 m/s. The runner's friend is standing at a distance of 200m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200m ?

2. Originally Posted by cityismine
A runner sprints around a circular track of radius 100m at a constant speed of 7 m/s. The runner's friend is standing at a distance of 200m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200m ?
Set the geometry up so that the runners position at time $\displaystyle t$ is:

$\displaystyle p(t)=(100 \cos(0.07 t), 100 \sin (0.07 t))$

That is we take their position at time $\displaystyle t=0$ to be $\displaystyle p(0)=(100,0)$, the circle is centred at the origin, and we assume the runner is running anti-clockwise.

We may place the friend at any convienient point $\displaystyle 200$ m from the centre of the circle, so lets put them at $\displaystyle (200,0)$.

So now your problem is to find:

$\displaystyle \frac{d}{dt} \sqrt{(p_1(t)-200)^2+p_2(t)^2}$

when $\displaystyle \sqrt{(p_1(t)-200)^2+p_2(t)^2}=200$

RonL

3. Originally Posted by cityismine
A runner sprints around a circular track of radius 100m at a constant speed of 7 m/s. The runner's friend is standing at a distance of 200m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200m ?
If D = distance in meters between the two friends at any time t in seconds, then we are looking for
dD/dt = rate of change of the D, in m/sec.

We draw the figure based on the givens. (I do not know how to draw in computers; neither do I know how to use Latex for easy reading of math expressions.)

It is a triangle whose 3 sides are 100m, 200m and D. The center of the circle is at the intersection of the 100m and 200m sides.
Let us call the central angle as theta.
The D side of the triangle is opposite the angle theta.

Connecting the givens and the D, we will use the Law of Cosines,
D^2 = 100^2 +200^2 -2(100)(200)cos(theta)
D^2 = 50,000 -40,000cos(theta) ------------------(1)

Since we are looking for dD/dt, we differentiate (1) with respect to time t,
2D*dD/dt = 40,000sin(theta) *dtheta/dt ------------(2)

In this Eq.2, when D=200m, we don't know theta and dtheta/dt, so we solve for them first.

For theta, use the Eq.1,
200^2 = 50,000 -40,000cos(theta)
theta = arccos(10,000 /40,000) = 75.5224878 degrees ----**

For dtheta/dt, use the rotational speed of the central angle theta,
w = dtheta/dt = (tangential velocity)/(radius) = (7m/sec)/100m = 0.07 rad/sec -------------------**

Hence, substituting all those in (2),
2(200)*dD/dt = 40,000sin(75.5224878 deg) *(0.07)
400dD/dt = 2711.088
dD/dt = 2711.088 /400 = 6.778 m/sec

Therefore, at the instant when the distance between the two friends is 200 meters, they are separating at the rate of 6.778 m/sec. -----answer.