-- We know that there is intelligent life in the universe because they have never attempted to contact us. Lily Tomlin "William Elliot" <email@example.com> wrote in message news:Pine.NEB.firstname.lastname@example.org... > On Thu, 15 Nov 2012, Brad Cooper wrote: >> "Mike Terry" <email@example.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.