Date: Nov 14, 2012 3:15 PM
Author: Mike Terry
Subject: Re: Curvature in Cartesian Plane

"dy/dx" <dydx-1@gmail.invalid> wrote in message
news:k80olc$8ee$1@news.mixmin.net...
> On Wed, 14 Nov 2012 09:23:06 +1000, Brad Cooper wrote:
>

> > I expect that this is true...
> >
> > We have three points on a Cartesian x-y plane, and the circle that

passes through these three points has a constant curvature of k.
> >
> > If we have a doubly differentiable curve in the x-y plane that passes

through these points, is there always some point on the curve which has
curvature k?
> >
> > I am finding it tough to prove this. Any help appreciated.
> >
> > Cheers,
> > Brad

>
> If you're having difficulty proving something, it may be worth considering
> the possibility that it's false.
>
> In this case, if I imagine a V-shaped pair of line segments joining the
> three points, then rounding the corner of the V so it's
> twice-differentiable (but widening the V slightly so the curve still goes
> through the middle point), it's clear that the curvature goes from 0 up
> through k to a higher value at the middle point, then down through k to 0
> again.
>
> However, if I then imagine superimposing a high-frequency "coiling" on

this
> curve, like a telephone cord projected down to 2D, arranging that it still
> pass through all three points, it seems it should be possible to keep the
> curvature everywhere higher than some lower bound B > k. (The curve will
> now self-intersect.)


..or it is easy to think of example curves where the curvature stays
arbitrarily low (e.g. sort of resembling a large 3-leafed clover).

Mike.