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Thread: Related Rates

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    Related Rates

    A Ferris wheel with a radius of 10 m is rotating at a rate of one revolution every 2 minutes. How fast is a rider rising when his seat is 16 m above ground level?

    Can someone help me figure out what equation I should use? I want to use the circumference of a circle forumla but I'm not sure how that would work with the revolutions.

    Thanks!
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    Re: Related Rates

    Quote Originally Posted by calcmaster View Post
    A Ferris wheel with a radius of 10 m is rotating at a rate of one revolution every 2 minutes. How fast is a rider rising when his seat is 16 m above ground level?
    Please see the attached picture below (click it for the full version).

    We are told that $\displaystyle \frac{d\theta}{dt} = \frac{2\pi\ \mathrm{rad}}{2\ \mathrm{min}} = \pi\ \mathrm{rad}/\mathrm{min}.$ We are asked to find $\displaystyle \frac{dy}{dt}$ when $\displaystyle y = 6.$ We know that

    $\displaystyle y = 10\sin\theta,$

    so we differentiate,

    $\displaystyle \frac{dy}{dt} = 10\cos\theta\cdot\frac{d\theta}{dt}.$

    When $\displaystyle y = 6$ we have

    $\displaystyle \sin\theta = \frac35$

    and since $\displaystyle \theta$ is an acute angle, this gives

    $\displaystyle \theta = \arcsin\frac35.$

    Now substitute the given quantities:

    $\displaystyle \frac{dy}{dt} = 10\cos\theta\cdot\frac{d\theta}{dt}$

    $\displaystyle \Rightarrow\frac{dy}{dt} = 10\cos\left(\arcsin\frac35\right)\cdot\pi$

    $\displaystyle \Rightarrow\frac{dy}{dt} = 10\pi\cdot\frac45 = 8\pi\ \mathrm m/\mathrm{min} \approx 25.13\ \mathrm m/\mathrm{min}.$
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