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**calcmaster** · July 11th 2012, 01:45 PM

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!

**Reckoner** · July 11th 2012, 02:49 PM

Originally Posted by calcmaster

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

$y = 10\sin\theta,$

so we differentiate,

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

When $y = 6$ we have

$\sin\theta = \frac35$

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

$\theta = \arcsin\frac35.$

Now substitute the given quantities:

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

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

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

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