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Math Help - Optimization problem

  1. #1
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    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 runnerís 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.
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  2. #2
    Member Mentia's Avatar
    Joined
    Dec 2008
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    Bellingham, WA
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    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
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