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.