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Topic: Curvature on a curve and circle
Replies: 6   Last Post: Mar 11, 2013 3:54 PM

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Brad Cooper

Posts: 171
Registered: 12/8/04
Re: Curvature on a curve and circle
Posted: Mar 5, 2013 6:11 PM
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Hi James, William and 1treePetrifiedForestLane

"James Waldby" <not@valid.invalid> wrote in message
> On Mon, 04 Mar 2013 13:16:22 +1000, Brad Cooper wrote:
>> The "well behaved, smooth" function f(x) has endpoints f(0) = f(h) = 0.
>> The
>> curve of the function has length s1.
>> An arc of a circle passing through (0, 0) and (0, h) has fixed curvature
>> k
>> and its arc length is also s1.
>> It is required to show that a point must exist on f(x) where curvature is
>> also k.

> ...
>> I am not making much headway. Any ideas appreciated.
> ...
> Presumably "well behaved, smooth" f has a continuous curvature function,
> say
> K(f,x) for the curvature of f at x. By continuity, unless a point x
> exists
> where K(f,x) = k, either K(f,x) < k for all x, or K(f,x) > k for all x.
> From <> which says "the total
> curvature of an immersed plane curve is the integral of curvature along
> a curve taken with respect to arclength" and that "The total curvature of
> a closed curve is always an integer multiple of 2?" (that last symbol is
> pi), and with a bit of additional reasoning about the integral of
> curvature
> across a couple of discontinuities in curvature, you can conclude that the
> integral from 0 to h of the curvature of f equals the integral from 0 to h
> of the curvature of the arc. Etc.

I took on board you idea regarding total curvature and investigated for the
specific case of the first cycle of the cycloid:

x := t - sin(t)
y := 1 - cos(t) t = 0..2*PI

I chose this because it is tractable.

I found:

Total curvature of the first cycle of the cycloid is 3.141592654 (PI)
Total curvature of the corresponding arc of the circle is 2.349234368

These are not equal, but there are two places where curvature of the cycle
of the cycloid equals that of the circular arc i.e. where s = 1.901550418
and s = 6.098449582

I am still trying to prove the original proposition. I agree that Rolle's
Theorem looks like the way to go, but how?

The work I did:


In terms of arc length s, the cycloid can be written as:

x := 4*arcsin(sqrt(s/8)) + (s-4)/8*sqrt(s*(8-s))
y := s*(8-s)/8 s = 0..8

from this:

k(s) := 1/sqrt(s*(8-s))

Total Curvature = int( k(s), s = 0..8 ) = 3.141592654 (PI)


With some simple geometry, the radius of the circle is given by:

R = abs(numeric::solve( PI/R = sin(4/R), R))

R = 3.40536479

k = 1/R = 0.293654296

Total Curvature = numeric::int(k, s = 0..8) = 2.349234368


> --
> jiw

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