Answer
A rider is rising at a rate of $~~25.1~m/min$
Work Step by Step
Let $x$ be the horizontal distance of the rider from the center of the wheel.
Let $y$ be the vertical distance of the rider above the center of the wheel.
Let $r$ be the radius of the wheel.
We can find $y$:
$y^2 = r^2-x^2$
$y^2 = 10^2-6^2$
$y^2 = 64$
$y = 8$
Let $\theta$ be the angle at the center of the wheel between a horizontal line and the line connecting the rider and the center of the wheel.
Then: $~~\frac{d\theta}{dt} = \frac{2\pi~rad}{2~min} = (\pi)~rad/min$
We can find $\frac{dy}{dt}$:
$\frac{y}{r} = sin~\theta$
$\frac{1}{r}~\frac{dy}{dt}-\frac{y}{r^2}~\frac{dr}{dt} = (cos~\theta)~\frac{d\theta}{dt}$
$\frac{1}{10}~\frac{dy}{dt}-\frac{y}{r^2}~(0) = (\frac{8}{10})~(\pi)$
$\frac{dy}{dt} = (10)(\frac{8}{10})~(\pi)$
$\frac{dy}{dt} = (8\pi)~m/min$
$\frac{dy}{dt} = 25.1~m/min$
A rider is rising at a rate of $~~25.1~m/min$
