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Topic: Curvature in Cartesian Plane
Replies: 6   Last Post: Nov 15, 2012 8:53 AM

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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>,

> 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,

> > >
> > > 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
>
> 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 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(1-x^2).

In my lexicon, "y = sqrt(1-x^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(1-x^2)" is certainly 1.
>
> Cheers,

--

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