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....