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

**References**:**Curvature for a circle***From:*dpdoughe@dialup4less.com