# Related Rates

• Mar 12th 2009, 08:02 AM
Hockey_Guy14
Related Rates
At a marathon there is loud music at a water station to cheer the runners on. The marathon route turns a right-angled corner 200m past the water station. The runner's speed is 4m/s. How fast is the distance between the runner and the water station increasing when the runner is 100m past the corner?

Im pretty sure I have to use the Pythagoras theorem, to solve, but I cant figure out how to set it up properly. Any help would be greatly appreciated. If possible show all steps to getting the answer. Thanks again!
• Mar 12th 2009, 08:25 AM
sinewave85
Quote:

Originally Posted by Hockey_Guy14
At a marathon there is loud music at a water station to cheer the runners on. The marathon route turns a right-angled corner 200m past the water station. The runner's speed is 4m/s. How fast is the distance between the runner and the water station increasing when the runner is 100m past the corner?

Im pretty sure I have to use the Pythagoras theorem, to solve, but I cant figure out how to set it up properly. Any help would be greatly appreciated. If possible show all steps to getting the answer. Thanks again!

Ok imagine (or even better draw out) a right triangle in standard position. The water station is +200 m from the origin along the x-axis. The runner is at a variable point on the +y-axis. The distance between the water station and the runner forms the hypotenuse of the right triangle. So:

$\displaystyle x^2 +y^2 = D^2$

$\displaystyle 200^2 + y^2 = D^2$

$\displaystyle \frac{d}{dt}(200^2 + y^2) = \frac{d}{dt}(D^2)$

$\displaystyle 2y\frac{dy}{dt} = 2D\frac{dD}{dt}$

And for the specific instant given:

$\displaystyle y = 100, \frac{dy}{dt} = 4, D = \sqrt{100^2 + 200^2}\approx223.6$

From there you solve for $\displaystyle \frac{dD}{dt}$. I hope that helps!
• Mar 12th 2009, 08:31 AM
u2_wa
Quote:

Originally Posted by Hockey_Guy14
At a marathon there is loud music at a water station to cheer the runners on. The marathon route turns a right-angled corner 200m past the water station. The runner's speed is 4m/s. How fast is the distance between the runner and the water station increasing when the runner is 100m past the corner?

Im pretty sure I have to use the Pythagoras theorem, to solve, but I cant figure out how to set it up properly. Any help would be greatly appreciated. If possible show all steps to getting the answer. Thanks again!

I think $\displaystyle \frac{\Delta distance}{sec}=$ $\displaystyle \sqrt{(100+4)^2+200^2}$$\displaystyle -\sqrt{100^2+200^2}$
• Mar 12th 2009, 08:37 AM
Hockey_Guy14
Alright that makes more sense now, thanks alot for the help! :)