Date: Nov 14, 2012 11:50 PM Author: Brad Cooper Subject: Re: Curvature in Cartesian Plane Thank you for your replies.

"Mike Terry" <news.dead.person.stones@darjeeling.plus.com> wrote in message

news:<-aSdnXxKRNVDZj7NnZ2dnUVZ7rWdnZ2d@brightview.co.uk>...

> "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.

I had not considered that the curve could cross back over itself and so I

never added this restriction.

The helical telephone cord projected down to 2D and the 3-leafed clover both

cross over themselves. I would like to just consider a simple curve which

doesn't cross itself.

While investigating this I came across this in Spiegel's Vector Analysis:

The radius of curvature rho of a plane curve with equation y = f(x), i.e. a

curve in the xy plane is given by

rho = sqrt(1+(y')^2) / |y''|

I tested this with the simple hemisphere y = sqrt(1-x^2).

At x = 0, it correctly gives rho as 1, but at x = 0.4 it incorrectly gives

rho = 0.84

For any value other than x = 0, rho is incorrect.

I was trying to use the above formula for rho to help with my investigation

(amongst a lot of other things). Now I seem to be further away!!

Isn't the radius of curvature of the hemisphere 1?

Cheers,

Brad