2
$\begingroup$

First of all, good day and thanks for taking the time to read this question. The problem I'm having, is how i can plot just sections of the Folia of Descartes graph, which is build from the equation:

$x^3 + y^3 - 3xy=0$

I'm trying to do the representations that Morris Tenenbaum and Harry Pollard do, in their Ordinary differential equation book, page 17.

enter image description hereSo far i can do the plot of the Folia of Descartes using:

ContourPlot[x^3 + y^3 - 3 x y == 0, {x, -3, 3}, {y, -3, 3}, Axes -> True, Frame -> False]

enter image description here

And parametric plot:

ParametricPlot[{(3 m)/(1 + m^3), (3 m^2)/(1 + m^3)}, {m, -20, 80},PlotRange -> {-3, 3}]

enter image description here

Now i know that the option RegionFunction of the Function ParametricPlot could help for this, however i have failed in getting the sections.

I'm looking for any pointers or ideas on how to do the sections of the folia of descartes.

PS: My English isn't that good so I hope i explain my self good enough while making this question. Thanks again and have a good day.

$\endgroup$
3
$\begingroup$

As seen on Wikipedia, the folium of Descartes has a polar representation: $$ r={\frac {3a\sin \theta \cos \theta }{\sin ^{3}\theta +\cos ^{3}\theta }}. $$

With this representation it's pretty easy to pick out relevant segments and to visualize them with PolarPlot.

f[min_, max_] := PolarPlot[
  (3 Sin[t] Cos[t])/(Sin[t]^3 + Cos[t]^3), {t, min, max},
  PlotRange -> 2.5, ImageSize -> 300,
  PlotStyle -> Directive[Thickness[0.01], Black],
  Epilog -> {HalfLine[{2^(2/3), 0}, {0, 1}], HalfLine[{0, 0}, {1, 1}]}
  ]

Row[{
  f[0, Pi/4],
  f[Pi/4 - 0.15, Pi/2]
  }]

Mathematica graphics

Row[{f[Pi/2, 2 Pi/3], f[5 Pi/6, Pi]}]

Mathematica graphics

You can combine images with Show. You could also extend f to take several intervals of t.

$\endgroup$
  • $\begingroup$ Awesome thanks, this helps a lot. $\endgroup$ – Leothan Sep 9 '16 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.