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Re: Curvature for a circle

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  • Subject: [mg83109] Re: [mg83082] Curvature for a circle
  • From: Andrzej Kozlowski <akoz at>
  • Date: Sun, 11 Nov 2007 02:59:15 -0500 (EST)
  • References: <>

On 10 Nov 2007, at 17:38, dpdoughe at wrote:

> It can easily be shown that the curvature for a circle of radius r is
> 1/r.  How can I get Mathematica to show me this?
> I have defined a function to calculate the curvature for me assuming a
> vector-valued function.
> Curvature =
>   Function[{V, x}, Norm[D[D[V, x]/Norm[D[V, x]], x]/Norm[D[V, x]]]];
> S = {2*Cos[theta], 2*Sin[theta]}
> ListPlot[Table[{t, Curvature[S, theta] /. theta -> t}, {t, 0, 2*Pi}]]
> I don't get anything for a plot.  If I look at what my curvature
> formula calculates for me there are some questions as to if
> Abs'  (Derivative of absolute value??) being left unevaluated.
> Any thoughts?  It should be a straight-forward calculation....

One way is replace the buil in Norm by "real" norm:


you could also use

norm[v_] := Simplify[Norm[v], Element[_, Reals]]

but the first approach should be faster. Then with

S = {2*Cos[t], 2*Sin[t]}

Plot[Evaluate[Curvature[S, t]], {t, 0, 2 Pi}]

will give you the rather trivial plot you were expecting. In fact, in  
this case:

Simplify[Curvature[S, t]]

Andrzej Kozlowski

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