(Originally published on September 26, 2010)

What is the shape of the curve traced by one tip of a pencil as you roll it up a cylinder (the pencil being tangent to the cylinder at all times)? The pencil starts just touching the cylinder. As the pencil moves closer to the cylinder, the tip first moves away, then quickly moves back, eventually to stop as the pencil becomes vertical.

Below is the diagram of the pencil’s start point, the end point, and an arbitrary point.

The pencil rolling on a cylinder. The tip tracing a curve is marked.

It’ll be easier to parameterize the curve: determine the coordinates of each point as a function of the distance of the bottom tip of the pencil to the bottom of the cylinder (where it is tangent with the table). Initially the pencil’s tip is just touching the cylinder and the distance \(t\) is equal to \(l\), the length of the pencil. At the end, as the pencil is vertical, \(t=r\) the radius of the cylinder.

We have, from the arbitrary point,

\[\begin{aligned}
\frac{y}{x+t}&=tan 2\alpha = \frac{2tan \alpha}{1-tan^2\alpha} = \frac{2r/t}{1-r^2/t^2} = \frac{2rt}{t^2-r^2}\\ 
\frac{x+t}{l}&=\cos 2\alpha = 2\cos^2\alpha-1 = \frac{2t^2}{r^2+t^2}-1 = \frac{t^2-r^2}{t^2+r^2}
\end{aligned}\] 

Therefore,

\[\begin{aligned}
x &= \frac{l(t^2-r^2)}{t^2+r^2}-t\\
y &= \frac{2rt}{t^2-r^2}\cdot (x+t) = \frac{2rt}{t^2-r^2}\cdot\frac{l(t^2-r^2)}{t^2+r^2} = \frac{2rtl}{t^2+r^2}
\end{aligned}\]

We can plot the curve as a function of \(t\):