Re: Curvature for a circle
- To: mathgroup at smc.vnet.net
- Subject: [mg83102] Re: Curvature for a circle
- From: Thomas E Burton <tburton at brahea.com>
- 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]
curvature=Function[{V,x},norm[D[D[V,x]/norm[D[V,x]],x]/norm[D[V,x]]]];
Then
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....