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):

Prolate cycloid animation

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 ☺

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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.

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