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Thread: another related rates

  1. #1
    Junior Member winterwyrm's Avatar
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    another related rates

    A ferris wheel with a radius of 5 m is rotating at a rate of one revolution every 2 minutes. How fast is a rider rising when the rider is 9 m above ground level?

    Sorry I keep posting these, I just really have trouble with finding the proper equation to start it off with. Thanks in advance.
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  2. #2
    Eater of Worlds
    galactus's Avatar
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    We are assuming the wheel is tangent to the ground. Unrealistic, but let's go with it.

    The y coordinate is given by $\displaystyle y=5-5cos({\theta})$...[1]

    The wheel rotates $\displaystyle \frac{\pi}{60}$ rad/sec.

    When the wheel is 9 feet off the ground, then it has swept out an angle of

    $\displaystyle sin^{-1}(4/5)+\frac{\pi}{2}\approx{2.49809} \;\ radians$

    From [1], differentiating, we get

    $\displaystyle y'=5sin({\theta})\frac{d{\theta}}{dt}$

    We know the angular velocity is $\displaystyle \frac{d{\theta}}{dt}=\frac{\pi}{60} \;\ rad/sec$ and $\displaystyle {\theta}=2.49809$

    Entering in the values, we get $\displaystyle \frac{dy}{dt}=\frac{\pi}{20}=0.157 \;\ feet/sec$


    Another way is:

    $\displaystyle y=5-5cos(\frac{\pi}{60}t)$

    $\displaystyle \frac{dy}{dt}=\frac{{\pi}sin(\frac{\pi}{60}t)}{12}$

    Entering in the 47.71 seconds to get to 9 feet, we get:

    0.157 ft/sec. as before.
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