RE: Polar plot with a twist.
- To: mathgroup at smc.vnet.net
- Subject: [mg26200] RE: [mg26173] Polar plot with a twist.
- From: "David Park" <djmp at earthlink.net>
- Date: Thu, 30 Nov 2000 22:02:15 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Thomas,
If you want to use PolarPlot to make a Plot of your equation, you can do it
this way.
Needs["Graphics`Graphics`"]
eqn = r^2 * Exp[-r] * (Cos[t])^2 == 1;
Mathematica, with a few warning messages, will solve your equation for r or
t.
Solve[eqn, r]
{{r -> -2*ProductLog[-(Sec[t]/2)]}, {r -> -2*ProductLog[Sec[t]/2]}}
PolarPlot[{-2*ProductLog[Sec[t]/2], -2*ProductLog[-(Sec[t]/2)]}, {t, 0,
2*Pi},
AspectRatio -> Automatic, Frame -> True, FrameLabel -> {x, y},
PlotPoints -> 50, PlotRange -> {{-1, 1}, {-5, 5}}, ImageSize -> 450];
But if you literally wish to plot t as a function of r:
tsol = Solve[eqn, t]
{t -> -ArcCos[-(E^(r/2)/r)]}, {t -> ArcCos[-(E^(r/2)/r)]},
{t -> -ArcCos[E^(r/2)/r]}, {t -> ArcCos[E^(r/2)/r]}}
Off[Plot::plnr];
Plot[{-ArcCos[-(E^(r/2)/r)], ArcCos[-(E^(r/2)/r)], -ArcCos[E^(r/2)/r],
ArcCos[E^(r/2)/r]}, {r, -10, 0}, Frame -> True, FrameLabel -> {r, t},
ImageSize -> 450];
David Park
djmp at earthlink.net
http://home.earthlink.net/~djmp/
> -----Original Message-----
> From: Thomas A. Rammer [mailto:tomrammer at hotmail.com]
To: mathgroup at smc.vnet.net
>
> 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
>
>
>