Re: Polar plot with a twist.
- To: mathgroup at smc.vnet.net
- Subject: [mg26202] Re: [mg26173] Polar plot with a twist.
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Thu, 30 Nov 2000 22:02:17 -0500 (EST)
- References: <200011300604.BAA08338@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
"Thomas A. Rammer" wrote:
>
> I have a nasty equation that I want to plot:
>
> r^2 * Exp[-r] * (Cos[t])^2 == 1
>
> What I really want to do is plot t as a function of r in polar
> coordinates. Is this possible, and if not, how would you suggest I work
> around this?
>
> Thanks
>
> ~Tom
>
> tomrammer at hotmail.com
I see two problems. First, you will have a much easier time plotting r
as a function of t. To do the reverse you'll need to use very restricted
ranges. If you run the contour plot below you will see why.
eee = r^2*Exp[-r]*Cos[t]^2 - 1;
ContourPlot[eee, {t,-5,5}, {r,-10,0}, Contours->{0},
ContourShading->False, PlotPoints->100];
Next, to get a polar plot there must be nonnegative values of r and real
values of t that satisfy the equation. It is easy to see that this
cannot happen. If instead you use Exp[r] you can satisfy the equation
with nonnegative r. To get a polar plot I simply translated to Cartesian
coordinates and plotted that.
eee2 = (x^2+y^2)*Exp[Sqrt[x^2+y^2]]*Cos[ArcTan[x,y]]^2 - 1;
ContourPlot[eee2, {x,-2,2}, {y,-8,8}, Contours->{0},
ContourShading->False, PlotPoints->100];
Daniel Lichtblau
Wolfram Research
- References:
- Polar plot with a twist.
- From: "Thomas A. Rammer" <tomrammer@hotmail.com>
- Polar plot with a twist.