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Virgil
Posts:
8,833
Registered:
1/6/11


Re: Curvature in Cartesian Plane
Posted:
Nov 15, 2012 2:05 AM


In article <Gw_os.934$Ow3.101@viwinnwfe02.internal.bigpond.com>, "Brad Cooper" <Brad.Cooper_17@bigpond.com> wrote:
> 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 do not know what Spiegel's says, but the one you present is not correct. The correct one is
rho =[1+(y')^2)]^(3/2) / y'' > > I tested this with the simple hemisphere y = sqrt(1x^2).
In my lexicon, "y = sqrt(1x^2)" represents a semicircle, not a hemisphere. > > 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?
The radius of curvature of semicircle "y = sqrt(1x^2)" is certainly 1. > > Cheers, > Brad 



