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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
>
>
> 



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