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...
2 Answers
$\begingroup$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.
Besides that, all you really need to do is to find the time when the person on the ferris wheel fits your initial conditions.
$\endgroup$ 4 $\begingroup$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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