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Re: Re: PolarPlot

  • To: mathgroup at smc.vnet.net
  • Subject: [mg87510] Re: Re: [mg87484] PolarPlot
  • From: Bruce Colletti <vze269bv at verizon.net>
  • Date: Sat, 12 Apr 2008 06:55:54 -0400 (EDT)

Craig

Thankx for the reply.  The code:

a = 2; PolarPlot[4 a^2 Cos@theta, {theta, 0, 2 Pi}]

only produces half the plot, since only theta values in [0,pi/2] and [3pi/2=
,2pi] can satisfy the function.  The actual polar curve has two parts, each=
 corresponding to a positive and negative value of r.

I don't see how to "easily" apply ParametricPlot (what would the parametric=
 forms be for x and y?), although I may well be missing the obvious. 

It seems that a "PolarContourPlot" command is what's needed, but no such th=
ing exists in Mathematica (I think).

Bruce

=====================
From: W_Craig Carter <ccarter at mit.edu>
Date: 2008/04/11 Fri AM 07:52:22 CDT
To: Bruce Colletti <vze269bv at verizon.net>
Cc: mathgroup at smc.vnet.net
Subject: [mg87510] Re: [mg87484] PolarPlot

Dear Bruce,
It doesn't appear that you are plotting what you say you wish to.
If I understand correctly, you are asking for
a = 2;
PlotPlot[4 a^2 Cos@theta, {theta, 0, 2 Pi}]

In your example, you are plotting two radial functions. Perhaps you were th=
inking of ParametricPlot?

On Fri, Apr 11, 2008 at 1:47 AM, Bruce Colletti <vze269bv at verizon.net> wrot=
e:
Re 6.0.2 under WinXP.

I'm using the command below to plot the polar equation r^2 = 4a^2 Cos@the=
ta. =C2 Is there a better way? =C2 Thankx.

Bruce

PolarPlot[{2 a Sqrt[Cos@t], -2 a Sqrt[Cos@t]}, {t, 0, 2 \[Pi]},
=C2 PlotRange -> 7]




--
W. Craig Carter



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