Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 40

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}
$