Re: Re: PolarPlot

*To*: mathgroup at smc.vnet.net*Subject*: [mg87512] Re: Re: [mg87484] PolarPlot*From*: "W_Craig Carter" <ccarter at mit.edu>*Date*: Sat, 12 Apr 2008 06:56:16 -0400 (EDT)*References*: <22352342.6066981207920340514.JavaMail.root@vms073.mailsrvcs.net>

Sorry, I missed the (r^2 = ) in your original post. How about: a= 3; PolarPlot[r /. Solve[r^2 == 4 a^2 Cos[theta], r], {theta, 0, 2 Pi}] On Fri, Apr 11, 2008 at 9:25 AM, Bruce Colletti <vze269bv at verizon.net> wrote: > 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 > thing 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: 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 > thinking of ParametricPlot? > > On Fri, Apr 11, 2008 at 1:47 AM, Bruce Colletti <vze269bv at verizon.net> > wrote: > Re 6.0.2 under WinXP. > > I'm using the command below to plot the polar equation r^2 = 4a^2 > Cos@theta. Is there a better way? Thankx. > > Bruce > > PolarPlot[{2 a Sqrt[Cos@t], -2 a Sqrt[Cos@t]}, {t, 0, 2 \[Pi]}, > PlotRange -> 7] > > > > > -- > W. Craig Carter > > -- W. Craig Carter