# Related Rates Problem involving two runners on a circular path

Problem:

There was a typo in the original statement. I fixed it now!!

Two runners start running (from the same point) in opposite directions along a circular path of radius $100\ m$ at a speed of $5\ m/s$. At what rate is the shortest distance between these two runners growing after $10$ seconds?

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Here is my attempt:

Assume that both runners start running from the point $A$. After $10$ seconds, the first runner reaches the point $B$ and the second runner reaches the point $C$. Since the speed of both runners is the same $5\ m/s$, the arc $AB$ is equal to the arc $AC$ (actually the length of the arc $AB=AC=D=5*10=50\ meter$. Let's call $d$ the length of the shortest distance between $B$ and $C$ (which is the segment $[BC]$).

Let's call $\theta$ the angle between $[OA]$ and $[OB]$ where $O$ is the center of the circle (i.e, $\theta$ is the angle swapped when the first runner runs from $A$ to $B$).

By using the law of cosines on the triangle $OBC$, we get: $d^2=100^2+100^2-2(100)(100)cos(2\theta)$ .....................(**)

By differentiating the above equation with respect to $t$, we get:\

$d \cdot \frac{\partial d }{\partial t}=2\cdot10^4 \cdot sin(2\theta) \cdot \frac{\partial \theta }{\partial t}$ .................(1)

Let's call $D$ the distance traversed along the path by each runner and $\theta$ the angle traversed (in radians). We have: $D=100\cdot \theta$. Then $\frac{\partial D }{\partial t}=5\ m/s=\ 100 \frac{\partial \theta}{\partial t}$ $\Rightarrow$ $\frac{\partial \theta }{\partial t}= 0.05\ rad/sec$................(2)

When $t=10\ sec$, $D=50\ meters$.

Using the equation $D=100\ theta$, we get $\theta= 50/100=0.5 radians$.............(3)

By replacing the value of $\theta$ in $(**)$, we will get $d=100\sqrt{2-2cos(1)}$...........(4)

By pluggin-in $(2)$, $(3)$ and $(4)$ in $(1)$, we get $\frac{\partial d}{\partial t}=\frac{10sin(1)}{\sqrt{2-2cos(1)}}$.

Please let me know if there is any mistake in my proof, and whether there is an easier way to do this problem. Thanks!

I think you're overthinking things. Yes; it's a circle, but you're looking at the arclength which can almost be treated as a "straight" line. So $\frac{dx}{dt} = -\frac{dy}{dt} = 5m/s$ Then $\frac{d(x-y)}{dt} = \frac{dx}{dt} - \frac{dy}{dt} = 10m/s$.

** Edit: Nevermind. I'm underthinking the problem. Although you might be able to convert easily enough. Let me check your work again and try to wake up a bit.

It might be a little difficult to explain, but draw a triangle between your two positions on the circle and the center. Then the shortest distance D between the two runners has the following relationship: $2sin(\frac{\theta}{2}) = \frac{D}{r}$. So $cos(\frac{\theta}{2})\frac{d\theta}{dt} = \frac{1}{r}\frac{dD}{dt}$ or $\frac{dD}{dt} = rcos(\frac{\theta}{2})\frac{d\theta}{dt}$.

We know $S = r\theta \rightarrow \frac{dS}{dt} = r\frac{d\theta}{dt}$ and $\frac{dS}{dt} = 10m/s$. The rest is basically plug and chug.

$\frac{d\theta}{dt} = 0.1s^{-1} \rightarrow \theta = 0.2$ and $\frac{dD}{dt} = 100m\cdot cos(\frac{0.2}{2})\cdot 0.1s^{-1} = 9.95m/s$

• Didn't see the edit for 10s rather than 2s, but the process should hold. Want me to edit it for t=10s? – Daniel Goldman Oct 29 '14 at 3:53

The (signed) distance between the runners is

$$s = 2 \ r \sin {\theta \over 2}$$

$\ldots$ where

• $\theta$ is the angle between the runners
• $r$ is the radius of the circle - a constant.

Its rate of change is

$$\dot s = r \ \dot \theta \cos {\theta \over 2}$$

In this case, the runners separate at rate

$$r \ \dot \theta = 10 \ m \ of \ arc / sec$$

So

$$\dot \theta = {10 \over 100} / sec = 0.1 / sec$$

$\ldots$ and, since $s = 0$ when $t = 0$

$$\theta = 0.1 \ t$$

After 10 seconds,

$$\theta = 1$$

$\ldots$ when the rate of change of the distance between the runners

$$\dot s = {100 \times 0.1} \times \cos {1 \over 2} \ m / sec$$

$$\approx \ 8.78 \ m / sec$$.

Sanity check

• When $\theta = 0$, $\dot s = r \ \dot \theta$.
• When $\theta = \pi$, $\dot s = 0$.