Graphing $x^{3 / 2} + y^{3/2} =1$

Question

My brother asked me what I thought was a fairly straightforward question, graph the function below over the real numbers:

$$ x^{3/2} + y^{3/2} = 1.$$

Now of course, we can't have any negative values in the square roots, so the graph looks 'similar' to the graph of a standard circle in the +X+Y quadrant of the XY plane. This is all well and good, next however; we simply rearranged the formula to:

$$ y = \left(\left( 1-x^{3/2}\right)^{1/3}\right)^2$$

And this is where the confusion began. Now, it's clear that substituting a large x value would simply produce a large negative value inside the bracket, squaring this would simply make the value positive. So now the graph of the function contains all of the points that made up the original graph, plus another branch of values:

This is unexpected. What step in my algebra permitted these positive values?

Answer 1

It happened when you squared both sides of

$$y^{1/2}=\left(1-x^{3/2}\right)^{1/3}\;.$$

Consider a similar but simpler example: $y^{1/2}=x$; clearly $y=x^2$, but because squaring wipes out the sign, it’s also true that $y=(-x)^2$, even though in the original equation $y^{1/2}=x$ we must have $y\ge 0$.

Answer 2

It is because of the ambiguity: is $y^{1/2}$ positive or negative? where $y^{1/2}$ might be defined to be a solution of $z^2=y$.

If you rearrange the other way, you get a third curve, that is the mirror image of your second curve. $$x=\left(\left(1-y^{3/2}\right)^{1/3}\right)^2$$

Answer 3

The exponent is 3/2 , a cube of square root, so must carry both signs $ \pm $ associated with square root even if not explicitly stated.There is a combination of three curves depending on (+,+),(+,-)(-,+)choices as shown.

$$ \pm\; x^{3/2} \pm y^{3/2} = 1 $$

( None of them are second degree of course even if so by appearances).

$$ x^{3/2} + y^{3/2} = 1 .. (1*) $$

is a slim "circle" and

$$ x^{3/2} - y^{3/2} = 1 ..(2*) $$

is a slim "rectangular hyperbola" that has cusp touching x-axis.

$$ -x^{3/2} + y^{3/2} = 1 ..(3*) $$

is a slim "rectangular hyperbola" that has cusp touching y-axis.

EDIT1:

Every square root has an $ \pm $ implied always in its relational existence. Recognition or choice cannot deny its implied existence.

Although some curves with symmetry are best plotted parametrically, only first quadrant is covered in this particular case due the even number 4 in numerator of exponent:

ParametricPlot[ {Cos[u]^(4/3), Sin[u]^(4/3)}, {u, 0 , 2 Pi}],

but the "hyperbolae" need proper domain choice.

ParametricPlot[ {Sec[u]^(4/3), Tan[u]^(4/3)}, {u, -Pi/3, Pi/3}].

EDIT2:

I added egreg's sextics with my $ \pm $ two curves

$$ ( x^3 \mp y^3 )^2 - 2 ( x^3 \pm y^3)^2 + 1 = 0 ... (4*) $$

in Red and Blue ..showing how everyone's right!

Answer 4

$\sqrt y=f(x)$ is not equivalent to $y=f^2(x)$, simply.

(You introduce an extra solution $\sqrt y=-f(x)$. You must enforce $f(x)\ge 0$, which amounts to $x\le1$ here.)

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Said differently, the domain of the square root is the positive reals, so that the curve is restricted to $x,y\ge0$.

And the range of the main branch of the square root is limited to the positive reals as well, so that (after simple computation) the curve is restricted to $x,y\le1$.

When rearranging the equation, you allowed solutions that invalidate the range condition (square root that yields a negative value).

Answer 5

Should be a comment. If you allow both signs for the square root: $x^{\frac{3}{2}}+y^{\frac{3}{2}}=1$ can be squared twice $x^3=(1-y^{\frac{3}{2}})^2=1-2y^{\frac{3}{2}}+y^3$ so $2y^{\frac{3}{2}}=1+y^3-x^3$ making $4y^3=(1+y^3-x^3)^2$ and this has the plot:

What we see is that $(\sqrt{x})^3+(\sqrt{y})^3=1$ is part of the sextic, where $(-\sqrt{x})^3+(\sqrt{y})^3=1$ and $(\sqrt{x})^3+(-\sqrt{y})^3=1$ are the other branches.