       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:= eq = r^2 == 3 + 2*r*Cos[\[Theta]]
>
> In:= eq2 = FullSimplify[ eq /. {r -> Sqrt[x*x + y*y],  \[Theta] -
> > ArcTan[x, y]}]
>
> Out= 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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