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Re: Plotting Quartic Solutions in Polar Coordinates

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  • Subject: [mg112500] Re: Plotting Quartic Solutions in Polar Coordinates
  • From: "David Park" <djmpark at>
  • Date: Fri, 17 Sep 2010 06:43:29 -0400 (EDT)


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:

    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:

     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  

From: Ed Frank [mailto:satchelp at] 

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

Ed Frank
satchelp at

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