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I tried $\;r=-1-\sin\theta,\,$ for $\,\theta$ between $[-2\pi,2\pi]$. But it's not as accurate as expected.

Any one to help me?

Thank you in advance.

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4 Answers 4

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You can use the union of $$y = |x| + \sqrt{1 - x^2} \quad \cup \quad y = |x| - \sqrt{1 - x^2}$$ to obtain the following (compliments of Wolfram):

enter image description here

Then perhaps tweak the "domain" to obtain the desired figure.

Search Wolfram Alpha by typing heart curve in the input field: you'll find some heart curves to select from in a drop down menu.

For example, using $$(x^2 + y^2 - 1)^3 = x^2y^3$$ You'll obtain the following graph:

enter image description here


I couldn't resist also adding this:

If you want the "real thing": an anatomical heart: See this link for a very complex function that will yield the following graph!:

enter image description here

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  • $\begingroup$ The $(x^2+y^2−1)^3 = x^2y^3$ that you mentioned from WolframAlpha looks nice as a curve, but it has some discontinuities (?) at $y=0$ that make it less useful for anything involving derivatives. $\endgroup$
    – LarsH
    Feb 15, 2018 at 14:14
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I love this one which is from one of my old works in Maple. There is another function in it:

> with(plots);
> plot([16*sin(t)^3, 13*cos(t)-5*cos(2*t)-2*cos(3*t)-cos(4*t), t = 0 .. 6.3], thickness = 3);
> implicitplot3d((x^2+9*y^2*(1/4)+z^2-1)^3-x^2*z^3-9*y^2*z^3*(1/80) = 0, x = -2 .. 2, y = -2 .. 2, z = -2 .. 2, axes = boxed, numpoints = 100000, color = red);

enter image description here

enter image description here

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How about $y=|x|\pm \sqrt{1-x^2}$

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A variation on the equations given by Michael and amWhy:

$1.2y = \sqrt{|x| + 0.2} \pm\sqrt{1 - x^2}$

heart curve variation

Tweak the constants to taste. Increasing the $+ 0.2$ makes the cusps less pointy.

Credit to IQ and Dave_Hoskins.

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