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I keep hearing that when one rolls a parabola on a straight line, the focus traces a catenary. I kept trying to find a proof on the Internet, but no dice. How does one prove this to be true?

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By the way, Timmy, you can suggest edits, instead of flagging for moderators to make an edit :) Thank you for your help though! –  Zev Chonoles Sep 9 '11 at 5:10
    
I googled "rolling parabola on straight line" and found this Word document as the top hit. At first glance, it looks like it contains a proof. –  Rahul Sep 9 '11 at 5:14
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1 Answer 1

up vote 8 down vote accepted

Let's roll. :)

Deriving the parametric equations of a straight line roulette isn't terribly complicated. As I mentioned in this previous answer, rolling is best decomposed as a rotation and a translation. For this case, I'll take the straight line to be the horizontal axis.

Let's start again with a convenient parabola parametrization:

$$\begin{pmatrix}2at\\at^2\end{pmatrix}$$

where $a$ is the focal length (the distance from vertex to focus). The focus of this parabola is at the point $(0,a)$.

We also require the arclength function for this parametrization of the parabola: $s(t)=a(t\sqrt{1+t^2}+\mathrm{arsinh}(t))$.

The trick to rolling a parabola is to consider the transformations necessary for a point on the parabola to touch an appropriate point on the straight line it is rolling on. The parametrization I have chosen is particularly convenient, in that the vertex of the parabola already touches the horizontal axis at the origin.

Going forward, we can translate the parabola to be rolled so that the intended contact point coincides with the origin. We then perform a rotation such that the parabola is now tangent to the horizontal axis, and then horizontally translate by an amount equal to the parabola's arclength. (A similar derivation is done for the cycloid.)

Mathematically, we perform this sequence of transformations on the point $(0,a)$; here is the translation:

$$\begin{pmatrix}0\\a\end{pmatrix}-\begin{pmatrix}2at\\at^2\end{pmatrix}$$

The rotation then needed is given by the tangential angle rotation matrix. I derived the expression for the parabola in my previous answer, so I won't repeat it here. The only difference from the previous answer is that to go forward, we require a clockwise rotation, and thus we must transpose the tangential angle rotation matrix. This now gives us

$$\begin{pmatrix}\frac1{\sqrt{1+t^2}}&\frac{t}{\sqrt{1+t^2}}\\-\frac{t}{\sqrt{1+t^2}}&\frac1{\sqrt{1+t^2}}\end{pmatrix}\cdot\left(\begin{pmatrix}0\\a\end{pmatrix}-\begin{pmatrix}2at\\at^2\end{pmatrix}\right)$$

Finally, we translate horizontally with the arclength expression given earlier:

$$\begin{pmatrix}a(t\sqrt{1+t^2}+\mathrm{arsinh}(t))\\0\end{pmatrix}+\begin{pmatrix}\frac1{\sqrt{1+t^2}}&\frac{t}{\sqrt{1+t^2}}\\-\frac{t}{\sqrt{1+t^2}}&\frac1{\sqrt{1+t^2}}\end{pmatrix}\cdot\left(\begin{pmatrix}0\\a\end{pmatrix}-\begin{pmatrix}2at\\at^2\end{pmatrix}\right)$$

The parametric equations for the roulette surprisingly simplify to

$$\begin{align*}x&=a\operatorname{arsinh}(t)\\y&=a\sqrt{1+t^2}\end{align*}$$

Eliminating the parameter $t$ yields the usual equation for the catenary, $y=a\cosh\frac{x}{a}$.

Here's a picture I previously did:

rolling parabola

A similar derivation can be done to show that the directrix of the same rolling parabola envelopes a reflection about the horizontal axis of the catenary being traced by the focus.

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