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Re: Curvature for a circle

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  • Subject: [mg83102] Re: Curvature for a circle
  • From: Thomas E Burton <tburton at>
  • Date: Sun, 11 Nov 2007 02:55:39 -0500 (EST)

Norm assumes complex-valued arguments and so returns results  
incorporating Abs.  As far as I know, your formula for curvature  
applies only to real-valued vectors. So, without loss of generality  
(and certainly for the case at hand), you can substitute

norm[V_] := Sqrt[V . V]



S = {r*Cos[theta], r*Sin[theta]}

Simplify[curvature[S, theta], r > 0]

> 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....

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