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Re: Polar plot with a twist.

  • To: mathgroup at
  • Subject: [mg26202] Re: [mg26173] Polar plot with a twist.
  • From: Daniel Lichtblau <danl at>
  • Date: Thu, 30 Nov 2000 22:02:17 -0500 (EST)
  • References: <>
  • Sender: owner-wri-mathgroup at

"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

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

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