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So I have this question here...

''You have designed a Ferris wheel of diameter 20 m that rotates at a rate of 1 revolution per minute. How fast is a rider rising or falling when he/she is 6 m horizontally away from the vertical line passing through the center of the wheel?''

Can someone help me set this up? I think I have to use the arc length formula $s=r\theta$ somehow to relate the revolutions per minute somehow but I am not sure...

Here is work...

enter image description here

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  • $\begingroup$ You want to find the height of the rider, and differentiate it in terms of t. $\endgroup$ – Kaynex Nov 4 '16 at 23:36
  • $\begingroup$ Yep, that looks good! $\endgroup$ – Kaynex Nov 5 '16 at 3:51
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Hint: The height of the person at any time $t$ is based on $\sin t$, which can be derived from the parametric definition of the circle. Change the parameters of $\sin t$ to fit your specific problem.

Sine and Cosine

Besides that, all you really need to do is to find the time when the person on the ferris wheel fits your initial conditions.

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  • $\begingroup$ Is there a way to do this without parametric equations? Because I didn't learn it in class so I can't technically use it. I know about parametric equations though. $\endgroup$ – Future Math person Nov 4 '16 at 23:39
  • $\begingroup$ While it is using parametric equations, y = sin(t) and x = cos(t) is the definition of trigonometry, and something you've definitely learned. $\endgroup$ – Kaynex Nov 4 '16 at 23:52
  • $\begingroup$ Okay so now my idea is to use the Pythagorean for $x^2 + y^2=r^2$ and then substitute for $y=rsin(\theta)$ and $x=rcos(\theta)$ and then differentiate that. Is that okay so far? $\endgroup$ – Future Math person Nov 5 '16 at 0:20
  • $\begingroup$ I edited my question to show my work. Is this okay? $\endgroup$ – Future Math person Nov 5 '16 at 0:42
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I don't think that you have to use related rates here. At any point on the wheel, the rider's speed is described by $2\pi R\over T$, where $R$ is the radius of the Ferris wheel (10 m) and $T$ is the period (1 rev/min). The vertical component of the speed is the total speed multiplied by the sine of the angle between the vertical axis and the line connecting the center of the wheel to the rider. After drawing the triangle, it is clear that this sine is $6\over 10$. Thus, the vertical component of the speed is $\frac{2*10*\pi}{1}*\frac{6}{10}=12\pi$, without using any calculus at all.

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  • $\begingroup$ The radius is 10m actually since it's given the diameter is 20m. This helps. I'll try and solve this with calculus if I can though. Thank you. $\endgroup$ – Future Math person Nov 5 '16 at 0:29
  • $\begingroup$ Whoops, sorry about that. Glad it helps! $\endgroup$ – Andrew Stelzer Nov 5 '16 at 0:32

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