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