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Math Help - Need Calculus Word Problem Help Really Bad!

  1. #1
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    Need Calculus Word Problem Help Really Bad!

    A ferris wheel 30 meters high turns at a uniform rate of 2.2 revolutions/min. How far above the bottom of the Ferris wheel is a passenger when his or her height is changing at a rate of 2 m/second?

    This is my last one and it is due today!
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  2. #2
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    First, \frac{22}{10}\frac{rev}{min}*2*\pi*15\frac{m}{rev}  *\frac{1 min}{60 sec}\;=\;\frac{11}{10}\pi\frac{m}{sec}

    At the top, the the height is changing (dv/dt) at 0 m/sec with the horizontal displacement changing (dh/dt) at 11 \pi/10 m/sec. Vamp 1/4 turn to the far right. At this point, the situation is reversed, dh/dt = 0 m/sec and dv/dt = 11 \pi/10 m/sec. This should suggest to you some ambiguity in the problem statement. Does it mean +2 m/sec or does it mean 2 m/sec in either direction? I think this is unclear. I would find all four solutions and decide later if the problem statement didn't mean some of them.

    In any case, since we're talking about a circle, you should see that:

    \left(\frac{dv}{dt}\right)^{2}+\left(\frac{dh}{dt}  \right)^{2}\;=\;\left(\frac{11}{10}\pi\frac{m}{sec  }\right)^{2}

    and

    \frac{dv}{dt}\;=\;\frac{11}{10}\pi\frac{m}{sec}*co  s(\alpha)

    Where \alpha is the angular location of the passenger with reference to the center of the wheel, \pi/2 being the vertical position and 0 being the full forward position.

    Simply solving, then,

    2 m/sec\;=\;\frac{11}{10}\pi\frac{m}{sec}*cos(\alpha)

    and

    -2 m/sec\;=\;\frac{11}{10}\pi\frac{m}{sec}*cos(\alpha)

    Produces 0.954, 2.188, 4.095, and 5.329. It's up to you to decide which are applicable, if any.

    Clear as mud?
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