Problem source

Find the ratio, over one revolution, of the distance moved by a wheel rolling on a flat surface to the distance traced out by a point on its circumference.

Solution source

The point on the circumference will have position $(a\cos t, a \sin t )$ relative to the circumference where $t \in [0, 2\pi]$. the wheel will travel $2\pi a$, therefore the position is $(a\cos t + at, a \sin t )$.

The total distance travelled can be computed using the arc length:

\begin{align*}
&& s &= \int_0^{2\pi} \sqrt{\left ( \frac{\d y}{\d t} \right)^2 +\left (  \frac{\d x}{\d t} \right)^2} \d t \\
&&&= \int_0^{2\pi} \sqrt{(a - a\sin  t)^2 +(a \cos t)^2 } \d t \\
&&&= a \int_0^{2\pi} \sqrt{2 - 2 \sin t } \d t \\
&&&= \sqrt{2}a \int_0^{2 \pi} \sqrt{1 - \sin t} \d t \\
&&&= \sqrt{2}a \int_0^{2 \pi} \frac{|\cos t|}{\sqrt{1 + \sin t}} \d t \\ 
&&&= 2\sqrt{2} a \int_{-\pi/2}^{\pi/2} \frac{\cos t}{\sqrt{1+\sin t}} \d t \\
&&&=  2\sqrt{2} a  \left [ 2\sqrt{1+\sin t} \right]_{-\pi/2}^{\pi/2} \\
&&& = 2\sqrt{2} a 2\sqrt{2} \\
&&&= 8a
\end{align*}

Therefore the ratio is $\frac{4}{\pi}$