Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

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:
  • Prev by Date: Re: Re: Show and scaling
  • Next by Date: Re: Bug of Integrate
  • Previous by thread: Re: Polar Plot
  • Next by thread: Re: Polar Plot