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Re: Curvature for a circle
*To*: mathgroup at smc.vnet.net
*Subject*: [mg83109] Re: [mg83082] Curvature for a circle
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Sun, 11 Nov 2007 02:59:15 -0500 (EST)
*References*: <200711100838.DAA00716@smc.vnet.net>
On 10 Nov 2007, at 17:38, dpdoughe at dialup4less.com 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:
norm[v_]:=Sqrt[v.v]
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]]
1/2
Andrzej Kozlowski
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