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

"dy/dx" <dydx-1@gmail.invalid> wrote in messagenews: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 thatpasses through these three points has a constant curvature of k.> >> > If we  have a doubly differentiable curve in the x-y plane that passesthrough these points, is there always some point on the curve which hascurvature 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" onthis> 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 staysarbitrarily low (e.g. sort of resembling a large 3-leafed clover).Mike.
```