Date: Nov 15, 2012 8:53 AM
Author: Brad Cooper
Subject: Re: Curvature in Cartesian Plane
Thanks for the replies.

--

We know that there is intelligent life in the universe because they have

never attempted to contact us.

Lily Tomlin

"William Elliot" <marsh@panix.com> wrote in message

news:Pine.NEB.4.64.1211150156450.16641@panix3.panix.com...

> On Thu, 15 Nov 2012, Brad Cooper wrote:

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

>> message

>> 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''|

>>

> No, rho = (1 + y'^2)^(3/2) / |y"|

>

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

>

> That's a semicircle.

Oops - why did I say hemisphere??!

Thanks for the correct formula. There is a typo in Spiegel's book.

I wrote a short program in the MuPAD CAS which does the following...

1. Generate a random polynomial y(x) of order between 5 and 10.

2. Generate 3 random points in the domain [-10, 10] which lie on the curve

of the polynomial.

3. Calculate the radius of the circle through the 3 points.

4. Calculate rho = (1 + y'^2)^(3/2) / |y"| and solve the equation rho =

radius for x in [-10, 10].

Every time I execute this (without exception) I obtain at least one value of

x and usually several.

The original problem is...

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?

The findings from MuPAD support the proposition for a simple curve which

does not cross itself. It looks to be true for a simple curve. Hmm! Now for

a way forward towards a proof.

Cheers,

Brad