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Re: Plotting Quartic Solutions in Polar Coordinates
*To*: mathgroup at smc.vnet.net
*Subject*: [mg112500] Re: Plotting Quartic Solutions in Polar Coordinates
*From*: "David Park" <djmpark at comcast.net>
*Date*: Fri, 17 Sep 2010 06:43:29 -0400 (EDT)
Ed,
I take it that you want to plot the curve and not the roots of the equation
(which I think could also be done as a function of the "a" parameter).
This is easily done with the Presentations package. We use a ContourPlot,
which plots in the r, phi plane, and then transform to Cartesian coordinates
using DrawingTransform. DrawingTransform contains pure functions for x and y
in terms of r (#1) and phi (#2).
<< Presentations`
quartic[a_][r_, \[Phi]_] := r^4 - (2 Cos[2 \[Phi]]) r^2 - (a - 1)
A specific case:
Draw2D[
{ContourDraw[
quartic[2][r, \[Phi]] == 0, {r, 0, 2}, {\[Phi], 0, 2 \[Pi]}] /.
DrawingTransform[#1 Cos[#2] &, #1 Sin[#2] &]},
Frame -> True]
Within a Manipulate statement:
Manipulate[
Draw2D[
{ContourDraw[
quartic[a][r, \[Phi]] == 0, {r, 0, 2}, {\[Phi], 0, 2 \[Pi]},
PerformanceGoal -> "Quality"] /.
DrawingTransform[#1 Cos[#2] &, #1 Sin[#2] &]},
PlotRange -> 2,
Frame -> True],
{a, 0.1, 5, Appearance -> "Labeled"}]
David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/
From: Ed Frank [mailto:satchelp at earthlink.net]
I have used "Solve" in (M7HE) to provide solutions to a Quartic equation in
(r,Phi), and have attempted to plot the results in Polar coordinates without
much success.
Can I send you what I have done - in an email attachment perhaps - and ask
for your comments?
The equation involves the 2 variables (r, Phi) and a single, constant
(though adjustable) parameter:
r^4 - (2 Cos[2*[Phi]]) r^2 - (a - 1) == 0
These are "Cassinian Ovals" for different vales of "a".
I think part of the problem may be in the self-defined formulas I created to
consolidate the solution(s).
PolarPlot seems to plot a Test constant value OK, but won't plot my solution
formulas.
Thanks,
Ed Frank
satchelp at earthlink.net
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