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Re: Curvature in Cartesian Plane
Posted:
Nov 15, 2012 8:53 AM


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(1x^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 xy 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 xy 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



