## An unbounded curve with a positive lower bound for curvature

Yesterday Zorn Leung conjectured that if a plane curve has curvature ${k\geq c>0}$, then it is contained in a circle (of some radius). I will give an example to show that this is actually false. ${\underset{\bigtriangledown}{\underline{ \ }{ \ \ }\underline{ \ }}}$

The example is the following curve:

$\displaystyle \alpha(t)= (t+2\cos t, 2\sin t).$

This curve is called the prolate cycloid (I’ve just learned this name today). Geometrically it is the path traced out by a fixed point at a radius ${b>a}$ (here ${b=2, a=1}$) , where ${a}$ is the radius of a rolling circle (the animation is not of the same scale as my example because I am too lazy to make the animation myself, I just download it from the net):

It is easy to calculate that

$\displaystyle k=\frac{\det(\alpha', \alpha'')}{|\alpha'|^3}= \frac{4-2\sin t}{ ( 5-4 \sin t)^{3/2}},$

noting that ${|\alpha'|^2=5-4\sin t\geq 1.}$

It is not necessary to calculate exactly the minimum value of ${k(t)}$, just observe that ${k(t)}$ is periodic and that ${k>0}$ for all ${t}$, this implies ${k\geq c>0}$ for some constant ${c}$.

But it is easy to see that ${\alpha}$ is unbounded, and in particular it cannot be contained in any circle, no matter how large we choose the radius.

Remark 1 The conjecture should be true if the curve is simple closed and the radius of the circle can be chosen to be ${1/\min_\alpha{k}}$. I haven’t written down the full details of the proof, though. I think it would be a nice exercise for the geometry students ☺

This entry was posted in Geometry. Bookmark the permalink.

### 3 Responses to An unbounded curve with a positive lower bound for curvature

1. Anonymous says:

If the curve is closed, then it is compact. Therefore, we do not need the curvature condition. Perhaps, the interesting case is that the curve is simple (no self-intersection) and has a positive lower bound of curvature. It should be a consequence of Fundamental Theorem of DG.

2. KKK says:

Of course a closed curve must be contained in a large circle, since it’s compact as you described (and in the remark I’ve actually forgotten to state that it is simple), the emphasis is that the radius can be $1/\min _\alpha k$. It would be nice if someone (perhaps… you?) can post a proof of it here.
And by the way, forgive my ignorance, what is the fundamental theorem of differential geometry?

3. Hon Leung says:

The curvature function determines a plane curve up to rigid motion.