Mathematica can use ContourPlot to draw implicit Cartesian equations, but doesn't seem to have a similar function to plot an implicit polar equation, for example
$\theta ^2=\left(\frac{3 \pi }{4}\right)^2 \cos (r)$
What's the best way to do this?
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$\begingroup$ Well, it's a good question and I see solutions people are coming up with are different. Always interesting to see and compare, no harm in it. $\endgroup$– Vitaliy KaurovJan 23, 2012 at 20:39
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$\begingroup$ @David can I ask why you removed the "storytelling paragraph"? I only wanted to let people know that I planned to answer myself, and would have deleted it myself later. $\endgroup$– one-more-minuteJan 23, 2012 at 20:56
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$\begingroup$ We don't want to discourage people from answering questions. Even if you have one solution, there might be others that are enlightening for yourself or others. $\endgroup$– Brett ChampionJan 23, 2012 at 21:01
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$\begingroup$ Makes sense, I just didn't want to make others spend their time repeating a solution I already have written out. Nevermind, as you said, all contributions are helpful. $\endgroup$– one-more-minuteJan 23, 2012 at 21:09
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$\begingroup$ Related: mathematica.stackexchange.com/q/67261/1871 $\endgroup$– xzczdMar 9, 2019 at 6:22
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5 Answers
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Since ContourPlot[] returns a GraphicsComplex, you could also replace the point list of the plot with g @@@ pointlist where g is the coordinate transformation. For example
f[r_, th_] := th^2 - (3 Pi/4)^2 Cos[r]
g[r_, th_] := {r Cos[th], r Sin[th]}
pl = ContourPlot[f[r, th] == 0, {r, 0, 8 Pi}, {th, 0, 2 Pi}, PlotPoints -> 30];
pl[[1, 1]] = g @@@ pl[[1, 1]];
Show[pl, PlotRange -> All]
which produces
The advantage of this method is that it also works for coordinate transformations for which the inverse transformation is hard to find.
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4$\begingroup$ Very interesting point, applying the coordinate transformation to the finished plot. However, I think this might fail for high frequency functions/coordinate systems etc., since after plotting things like the step distance between points are already set. $\endgroup$– DavidJan 23, 2012 at 20:46
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$\begingroup$ This is basically what I ended up doing, I turned it into a self-contained function so I'll post it up later. +1 for reading my mind. @David I did find that quality was reduced but bumping up MaxRecursions helps without slowing things down too much. $\endgroup$ Jan 23, 2012 at 20:49
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1$\begingroup$ I would include
AspectRatio -> AutomaticinShowso that this "looks nice." $\endgroup$ Jan 24, 2012 at 14:20 -
$\begingroup$ It doesn't work in version 13. It should be
pl[[1, 1, 1]] = g @@@ pl[[1, 1, 1]];$\endgroup$– yodeJan 3 at 19:21
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Does this
ContourPlot[
Evaluate@With[
{r = Sqrt[x^2 + y^2],
θ = ArcTan[x, y]},
θ^2 - Cos[r] == 0
],
{x, 0.1, 4 Pi}, {y, 0, 4 Pi}
]
work?
Plot:
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$\begingroup$ This is a great method, unfortunately it's not as useful if you want to specify the plotting range in polar, rather than cartesian, coordinates. +1 for your
ContourPlotsubstitution, this is something I tried originally but couldn't get to work. $\endgroup$ Jan 23, 2012 at 21:17 -
$\begingroup$ @myk thanks, Heike's is definitely cleverer and also more useful if you can't invert the transformation easily $\endgroup$– aclJan 23, 2012 at 21:20
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If you allow negative radii, there's another entire half of the solution:
PolarPlot[
Evaluate[Flatten[
Table[{-ArcCos[(16 t^2)/(9 Pi^2)], ArcCos[(16 t^2)/(9 Pi^2)]} + k 2 Pi,
{k, -2, 2}]
]],
{t, -Pi, Pi},
PlotStyle -> Table[Directive[Thick, Hue[i/10]], {i, 10}]
]
answered Jan 24, 2012 at 3:55
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You can do something like this:
ContourPlot[ArcTan[x,y]^2 == (3 Pi/4)^2 Cos[Sqrt[x^2 + y^2]],
{x, -23, 23}, {y, -23, 23}, ContourStyle -> Directive[Thick, Orange]]
answered Jan 23, 2012 at 20:31
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$\begingroup$ I think you're still missing a square on the 3Pi/4. $\endgroup$ Jan 23, 2012 at 21:21
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$\begingroup$ @BrettChampion Yes, thanks, have to get my math straight ;-) $\endgroup$ Jan 24, 2012 at 2:15
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All the other three solutions use ContourPlot. Here's a solution using PolarPlot.
PolarPlot[{ArcCos[#2^2/(3 π/4)^2] + 2 π #1,
-ArcCos[#2^2/(3 π/4)^2] + 2 π (#1 + 1)} & @@ QuotientRemainder[Abs@ θ, 2 π],
{θ, -7 π, 8 π}, PlotStyle -> {Thick, Darker@Green}]
This makes use of the fact that the solution to $\theta^2=\displaystyle\left(\frac{3\pi}{4}\right)^2\cos(r)$ is
$$r=\pm\arccos\left(\frac{16\theta^2}{9\pi^2}\right)+2\pi n,\ n\in\mathbb{Z}$$
