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Re: Curvature in Cartesian Plane
Posted:
Nov 14, 2012 11:50 PM


Thank you for your replies.
"Mike Terry" <news.dead.person.stones@darjeeling.plus.com> wrote in message news:<aSdnXxKRNVDZj7NnZ2dnUVZ7rWdnZ2d@brightview.co.uk>... > "dy/dx" <dydx1@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 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? > > > > > > 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 Vshaped pair of line segments joining the > > three points, then rounding the corner of the V so it's > > twicedifferentiable (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 highfrequency "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 selfintersect.) > > ..or it is easy to think of example curves where the curvature stays > arbitrarily low (e.g. sort of resembling a large 3leafed 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 3leafed 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(1x^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



