How can I plot an implicit function?

Question

Say I have a function

How can I plot the following function

Thank you for your help!

I imagine you want the full heart. You can use the real-valued cube root instead of the principal valued one: ContourPlot[x^2 + (y - CubeRoot[x]^2)^2 == 1, {x, -1, 1}, {y, -1, 2}] — Greg Hurst, Feb 10 '17 at 21:18

score 3 · Accepted Answer · answered Feb 10 '17 at 19:36

3

Using the suggestion of @march:

ContourPlot[x^2 + (y - x^(2/3))^2 == 1, {x, -1, 1}, {y, -1, 2}]

Alternatively, if you are more comfortable with the use of Plot:

Plot[Flatten@Solve[x^2 + (y - x^(2/3))^2 == 1, y][[;; , ;; , 2]], {x, -1, 1}]

answered Feb 10 '17 at 19:36

Marchi

Flatten is not necessary in Plot as they formulas aren't to deep, but if you want to see where it transitions from one solution to the other, you can replace it with Evaluate. – rcollyer Feb 10 '17 at 19:46
I think the OP meant something like ContourPlot[{x^2 + (y - x^(2/3))^2 == 1, (-x)^2 + (y - (-x)^(2/3))^2 == 1}, {x, -1, 1}, {y, -1, 2}, ContourStyle -> Red]? (+1) – kglr Feb 10 '17 at 19:47
Thanks for helping me ... excellent answer @Marchi – Emad kareem Feb 10 '17 at 20:10
Thanks for helping me ... excellent answer @rcollyer – Emad kareem Feb 10 '17 at 20:10
Thanks for helping me ... excellent answer @kglr – Emad kareem Feb 10 '17 at 20:11

1 Answers1