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:
- Polar Plot
- From: Miguel <misvrne@gmail.com>
- Polar Plot