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

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Brad Cooper

Posts: 171
Registered: 12/8/04
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" <> wrote in message
> On Thu, 15 Nov 2012, Brad Cooper wrote:
>> "Mike Terry" <> 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.


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