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Re: Curvature in Cartesian Plane
Posted:
Nov 15, 2012 8:53 AM
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Thanks for the replies.
--
We know that there is intelligent life in the universe because they have
never attempted to contact us.
Lily Tomlin
"William Elliot" <marsh@panix.com> wrote in message
news:Pine.NEB.4.64.1211150156450.16641@panix3.panix.com...
> On Thu, 15 Nov 2012, Brad Cooper wrote:
>> "Mike Terry" <news.dead.person.stones@darjeeling.plus.com> wrote in
>> message
>> 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''|
>>
> No, rho = (1 + y'^2)^(3/2) / |y"|
>
>> I tested this with the simple hemisphere y = sqrt(1-x^2).
>
> That's a semicircle.
Oops - why did I say hemisphere??!
Thanks for the correct formula. There is a typo in Spiegel's book.
I wrote a short program in the MuPAD CAS which does the following...
1. Generate a random polynomial y(x) of order between 5 and 10.
2. Generate 3 random points in the domain [-10, 10] which lie on the curve
of the polynomial.
3. Calculate the radius of the circle through the 3 points.
4. Calculate rho = (1 + y'^2)^(3/2) / |y"| and solve the equation rho =
radius for x in [-10, 10].
Every time I execute this (without exception) I obtain at least one value of
x and usually several.
The original problem is...
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?
The findings from MuPAD support the proposition for a simple curve which
does not cross itself. It looks to be true for a simple curve. Hmm! Now for
a way forward towards a proof.
Cheers,
Brad
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