Re: : Polar Plot

*To*: mathgroup at smc.vnet.net*Subject*: [mg82808] Re: : Polar Plot*From*: "David Park" <djmpark at comcast.net>*Date*: Wed, 31 Oct 2007 06:17:51 -0500 (EST)*References*: <fg71qq$jp4$1@smc.vnet.net>

Another possible method is to use the DrawingTransform command in DrawGraphics to convert from polar coordinates to xy-coordinates. Needs["DrawGraphics6`DrawingMaster`"] Draw2D[ {ContourDraw[ r^2 == 3 + 2*r*Cos[theta], {r, 0, 3}, {theta, 0, 2 \[Pi]}] /. DrawingTransform[#1 Cos[#2] &, #1 Sin[#2] &]}, Frame -> True] -- David Park djmpark at comcast.net http://home.comcast.net/~djmpark/ "Tom Burton" <Tom.Burton at DigiLore.com> wrote in message news:fg71qq$jp4$1 at smc.vnet.net... >I think you must convert to Cartesian coordinates first: > > In[41]:= eq = r^2 == 3 + 2*r*Cos[\[Theta]] > > In[44]:= eq2 = FullSimplify[ eq /. {r -> Sqrt[x*x + y*y], \[Theta] - > > ArcTan[x, y]}] > > Out[44]= y^2 + (x - 2)*x == 3 > > This seems to describe a circle of radius 2 centered at {1,0}, which > you can verify in version 6 with > > ContourPlot[Evaluate[eq2],{x,-1,3},{y,-2,2}] > > Use ImplicitPlot in version 5. > > Tom Burton > > >