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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



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