1. ## Optimization problem

4. [5 marks] At a marathon there is loud music at a water station to cheer the
runners on. The marathon route turns a right-angled corner 200 m past the
water station. The runner’s speed is 4 m/s. How fast is the distance between the runner and the water station increasing when the runner is 100 m past the corner? Include a sentence answer with units.

Help would be appreciated. Thank you.

2. This is less optimization, more practice in implicit differentiation.

Set your origin at the water station. I assume he runs north from the water station and then turns the corner going east. Then after the runner has turned the corner his y-coordinate is always 200.

The distance formula gives:

$(distance)^2=d^2=x^2+y^2$

Then we take the derivative with respect to time since we want the time rate of change of distance:

$2d\dot d = 2x \dot x + 2y \dot y$, then
$\dot d = \frac{x \dot x + y \dot y}{d} = \frac{x \dot x + y \dot y}{\sqrt[]{x^2+y^2}}$

In our case, the problem says:

$x = 100, \dot x = 4, y = 200, \dot y = 0$

So that gives us:

$\dot d = \frac{400 + 0}{\sqrt[]{100^2+200^2}} \approx 1.788$ meters/second