Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 40
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At what rate is the height of the rider increasing when his seat is 16m above ground level?
Let the center of the ferris wheel be the origin.
Using sine function,
$ \begin{equation} \begin{aligned} \sin \theta &= \frac{y}{10}\\ \\ y &= 10 \sin \theta \end{aligned} \end{equation} $
Taking the derivative with respect to time,
$\displaystyle \frac{dy}{dt} = 10 \cos \theta \frac{d \theta}{dt} \qquad \Longleftarrow \text{ Equation 1}$
When the rider is 16m above ground level,
$ y = 16 - 10 = 6$m
Also,
$\displaystyle \frac{d \theta}{dt} = \frac{1 \text{rev}}{2\text{mins}} = 0.5 \frac{\cancel{\text{rev}}}{\text{min}} \left( \frac{2 \pi \text{rad}}{\cancel{\text{rev}}}\right) = \pi \frac{\text{rad}}{\text{min}}$
Now, plugging all these values in Equation 1 to get,
$ \begin{equation} \begin{aligned} \frac{dy}{dt} &= 10 \cos (36.8699)\left( \pi \frac{\text{rad}}{\text{min}}\right)\\ \\ \frac{dy}{dt} &= 8 \pi \frac{m}{\text{min}} \end{aligned} \end{equation} $
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