
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 rightangled 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.

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 ycoordinate is always 200.
The distance formula gives:
$\displaystyle (distance)^2=d^2=x^2+y^2$
Then we take the derivative with respect to time since we want the time rate of change of distance:
$\displaystyle 2d\dot d = 2x \dot x + 2y \dot y$, then
$\displaystyle \dot d = \frac{x \dot x + y \dot y}{d} = \frac{x \dot x + y \dot y}{\sqrt[]{x^2+y^2}}$
In our case, the problem says:
$\displaystyle x = 100, \dot x = 4, y = 200, \dot y = 0$
So that gives us:
$\displaystyle \dot d = \frac{400 + 0}{\sqrt[]{100^2+200^2}} \approx 1.788$ meters/second