Re: PolarPlot
- To: mathgroup at smc.vnet.net
- Subject: [mg87538] Re: [mg87484] PolarPlot
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sat, 12 Apr 2008 07:01:10 -0400 (EDT)
- Reply-to: hanlonr at cox.net
PolarPlot[Flatten[Table[
{2 a Sqrt[Cos@t], -2 a Sqrt[Cos@t]},
{a, 7}]], {t, 0, 2 \[Pi]},
AspectRatio -> 1,
Frame -> True, Axes -> False]
Cos[ArcTan[y/x]]
1/Sqrt[y^2/x^2 + 1]
ContourPlot[Evaluate[Table[
(x^2 + y^2) == 4 a^2 /Sqrt[y^2/x^2 + 1],
{a, 7}]], {x, -15, 15}, {y, -10, 10},
AspectRatio -> 1,
ContourLabels -> (Text[Sqrt[Numerator[#3[[2]]]/4], {#1, #2}] &)]
Put the arrow over one of the lines in the ContourPlot.
Bob Hanlon
---- 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]
>