Re: how to interpret formula curvature

From: Virgil (ITSnetNOTcom#virgil_at_COMCAST.com)
Date: 02/11/05 

Date: Fri, 11 Feb 2005 14:19:14 -0700

In article <1108155178.394016.142400@l41g2000cwc.googlegroups.com>,
 "nlscb" <nlscb@yahoo.com> wrote:

> I have a really nasty equation that I needed to use maple to determine
> the curvature where the definition
> of curvature is:
>
> (dy^2/dx^2)/(1+(dy/dx)^2)^(3/2).
>
>
> The final answer is REALLY nasty. I doubt that I could
> solve for it's slope being zero w/o numerical methods.
> It also never equals zero in the range that I am looking
> at (0 to 100). It does have a maximum.
>
> How do I interpret this? I went to wolfram.com, but
> it doesn't really describe what you are supposed to do with
> your results.

If it is continuous and positive and has a single maximum on
0<=x<= 100, then it should look a bit like a section of a parabola
containing the vertex of the parabola, with the maximum curvature
occurring at the "vertex".

If it is in the form y = f(x), then plotting four or more points, with
knowing the general shape, should give you a fair idea of what it looks
like.

If it has more than one maximum, things get more complicated.

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