Re: Polar Plot

*To*: mathgroup at smc.vnet.net*Subject*: [mg82757] Re: [mg82709] Polar Plot*From*: "W. Craig Carter" <ccarter at mit.edu>*Date*: Tue, 30 Oct 2007 03:37:31 -0500 (EST)*References*: <200710291036.FAA06379@smc.vnet.net>

Dear Miguel, You will probably get some answers about Solve and which root to plot, but this may be an instructive example for the positive root: ContourPlot[(r^2 - 3 + 2*r*Cos[theta]) /. {r -> Sqrt[x^2 + y^2], theta -> ArcTan[y/x]}, {x, -2, 2}, {y, -2, 2}, Contours -> {0}] Or better yet: simpler = Simplify[(r^2 - 3 + 2*r*Cos[theta]) /. {r -> Sqrt[x^2 + y^2], theta -> ArcTan[y/x]}, {x \[Element] Reals && y \[Element] Reals}] ContourPlot[simpler, {x, -2, 2}, {y, -2, 2}, Contours -> {0}] W. Craig Carter On Mon, 29 Oct 2007, Miguel wrote: > Date: Mon, 29 Oct 2007 05:36:28 -0500 (EST) > From: Miguel <misvrne at gmail.com> > To: mathgroup at smc.vnet.net > Subject: [mg82709] Polar Plot > > How can I to plot a polar expresion in implicit form. For example, > r^2==3+2*r*Cos[teta] > > >

**References**:**Polar Plot***From:*Miguel <misvrne@gmail.com>