Heart-shape function

StephanP shared this question 8 years ago
Answered

Hi,


I receive an error message when I enter this function:

    x^2 + (y - (x^2)^(1/3))^2 = 1

What am I doing wrong and how should I correct it?


The function is obviously is a mod of a circle (x-square + y-square = 1), where the y-part is manipulated with the the 3-root of x-square.

Is there another way to enter the 3-root of x-square?

(I don't know how to say this in proper English)


Cheers,

Stephan

Comments (6)

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1

Not sure why that one doesn't work - and no time to check, but here's another one:


(x^2+y^2-1)^3=x^2y^3

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1

This is a very nice one for Valentine's day


x=cos(T) abs(tan(T))^abs(1/tan(T))

y=sin(T) abs(tan(T))^abs(1/tan(T))


y in thje range 0,pi

xminmax -1,1

yminmax 0-2

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1

    x^2 + (y - (x^2)^(1/3))^2 = 1

Try writing it as a parametric equation and use the Curve[] command...


Tony


Via Paul, try

f(t)=Curve[cos(t) abs(tan(t))^abs(1/tan(t)), sin(t) abs(tan(t))^abs(1/tan(t)) ,t,0,2 pi]

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1

Also, you could try a rotated cardioid


    g(t)=Curve[sin(t)+sin(2t)/2,-cos(t)-cos(2t)/2,t,0,2 pi]


Cheers

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1

Thanks for all the alternatives.

Nice hearts. They'll be well used today.

I'm still wondering though what is in fact wrong with the (notation?) of the first equation.

Cheers, Stephan

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1

As Noel mentioned, the implicit equation doesn't have only integer exponents. Your equation is effectively


x^2 + (y - x^(2/3))^2 = 1


You can use

    f(x)=sqrt((1-x^2)+x^(2/3)

    g(x)=-sqrt((1-x^2)+x^(2/3)


or try

    x^2 + (y - x^(2/3))^2 - 1

In the 3d view of Geogebra5


Simon

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