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
>
>
>