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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