Re: Curvature for a circle

*To*: mathgroup at smc.vnet.net*Subject*: [mg83110] Re: Curvature for a circle*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Sun, 11 Nov 2007 02:59:46 -0500 (EST)*Organization*: The Open University, Milton Keynes, UK*References*: <fh3qqc$o9$1@smc.vnet.net>

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.... Using the built-in function *Norm* may be not a good idea when dealing with derivatives. As you witnessed, it might generates absolute values *Abs* and Mathematica does not know how to differentiate such things. What you should do in such cases is witting your own norm function as illustrated in the example below. norm = Function[{V}, Sqrt[V.V]]; 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}]] In doing so, you get rid of the *Abs* function (and its unevaluated derivative) as you can see by evaluating Table[{t, Curvature[S, theta] /. theta -> t}, {t, 0, 2*Pi}] HTH, -- Jean-Marc