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Re: Curvature in Cartesian Plane
Posted:
Nov 14, 2012 11:50 PM
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Thank you for your replies.
"Mike Terry" <news.dead.person.stones@darjeeling.plus.com> wrote in message
news:<-aSdnXxKRNVDZj7NnZ2dnUVZ7rWdnZ2d@brightview.co.uk>...
> "dy/dx" <dydx-1@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 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?
> > >
> > > 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 V-shaped pair of line segments joining the
> > three points, then rounding the corner of the V so it's
> > twice-differentiable (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 high-frequency "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 self-intersect.)
>
> ..or it is easy to think of example curves where the curvature stays
> arbitrarily low (e.g. sort of resembling a large 3-leafed 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 3-leafed 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(1-x^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
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