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
The same question 
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
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
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
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
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
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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