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Math Help - Related Rates Problem

  1. #1
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    Related Rates Problem

    A runner sprints around a circular track of radius 100 m at a constant speed of 7 m/s. The runner's friend is standing at a distance 200 m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200 m.

    I attached a sketch of the problem so you can visualize what I'm doing.

    l=r\theta

    \frac{dl}{dt}=r\frac{d\theta}{dt}

    7=r\frac{d\theta}{dt}

    \frac{d\theta}{dt}=\frac{7}{100}

    I decided to use the law of cosines, s is the distance between them:

    s^2=(100)^2+(200)^2-2(200)(100)\cos(\theta)

    Differentiating with respect to t

    2s\frac{ds}{dt}=40,000\sin(\theta)\frac{d\theta}{d  t}

    \frac{ds}{dt}=\frac{20,000\sin(\theta)}{2s}\frac{d  \theta}{dt}

    When the s=200

    40,000=10,000+40,000-40,000\cos(\theta)

    cos(\theta)=\frac{1}{4}\rightarrow \sin(\theta)=\frac{\sqrt{15}}{4}

    So no I plug everything into the equation for the time derivative of s:

    \frac{ds}{dt}=\frac{(20,000)(\sqrt{15})(7)}{4(100)  }

    =\frac{7\sqrt{15}}{8}

    This is very close to the answer in my book which is \frac{7}{4}\sqrt{15}

    I can't seem to find where I want wrong.
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  2. #2
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    It's where you go from 2s\frac{ds}{dt}=, to \frac{ds}{dt}=. You've missed a cancellation by 2.
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