Is there a "clean" way of parametrizing the intersection of the sphere and saddle?

Question

I was playing around, trying to find a parametrization for the intersection of the two surfaces,

$$x^2+y^2+z^2=1$$ $$z=x^2-y^2,$$

but I wasn't able to get anything nice-looking. Any suggestions, or is this a hopeless endeavor?

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EDIT: Thanks, Thomas Andrews. With the corrected formulas, $$ a(t) = \sqrt{\frac{2}{1+\sqrt{1+4\cos^2 2t}}} $$ $$ (x(t),y(t),z(t)) = (a(t)\cos t, a(t)\sin t, a(t)^2 \cos2t), $$

we get a very nice picture!

Answer 1

$$x^2+y^2=1-z^2\\ x^2-y^2=z$$

So $x^2=\frac{1+z-z^2}{2}$ and $y^2=\frac{1-z-z^2}{2}$.

Now you just need to pick the range of values for $z$ where both $1-z-z^2$ and $1+z-z^2$ are positive. This is when $1-z^2\geq |z|$ or $1-3z^2+z^4\geq 0$ and $|z|\leq 1$. But $1-3z^2+z^4=\left(z^2-\frac{3}{2}\right)^2 -\frac{5}{4}$, so you need $$z^2\leq \frac{3-\sqrt{5}}{2} \text{ or } z^2\geq\frac{3+\sqrt{5}}{2}$$

The right side is irrelevant, since it is already greater than the known bound $1$. Also: $\frac{3-\sqrt{5}}{2}=\left(\frac{\sqrt{5}-1}{2}\right)^2$. So you get that $|z|\leq\frac{\sqrt{5}-1}{2}$ and:

$$(x,y,z)=\left(\pm\sqrt{\frac{1+z-z^2}{2}},\pm\sqrt{\frac{1-z-z^2}{2}},z\right)$$

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To correct the nice parameterization by Servaes in his wrong but close answer, if $(x,y,z)=(a\cos t,a\sin t, z)$ then $z=a^2(\cos^2 t -\sin^2 t)=a^2\cos 2t$.

Then $1=x^2+y^2+z^2=a^2+a^4\cos^2 2t$ or $$a^2=\frac{-1\pm\sqrt{1+4\cos^2 2t}}{2\cos^2 2t}=\frac{2}{1\pm\sqrt{1+ 4\cos^2 2t}}$$

Since you need $a^2\geq 0$, you can eliminate the $-$ case from $\pm$ and you get:

$$a(t)=\sqrt{\frac{2}{1+\sqrt{1+4\cos^2 2t}}}$$ and $$(x,y,z)=(a(t)\cos(t),a(t)\sin(t),a(t)^2\cos(2t))$$

Answer 2

You could try something like this.

On the curve,

$x^2 = y^2 + z \Rightarrow (y^2 + z) + y^2 + z^2 = 1 \Rightarrow (\sqrt{2}y)^2 + \left( z+\frac{1}{2} \right)^2 = \frac{5}{4}$

so we can write

$$ \sqrt{2}y = \frac{\sqrt{5}}{2}\cos(\phi) \,\,\,,\,\,\, z+ \frac{1}{2} = \frac{\sqrt{5}}{2}\sin(\phi) $$

$$\Rightarrow x = \pm \sqrt{\frac{5}{8}\cos^2(\phi) + \frac{\sqrt{5}}{2}\sin(\phi)-\frac{1}{2}} $$

As Thomas Andrews points out, not every possible value of $y$ and $z$ will be attained and we must restrict $\phi$ to the range where $y^2+z \geq 0$ (the range is something slightly bigger than $[0,\pi]$). To get the entire curve this way it seems like we would have to glue the two pieces, given by the opposing signs of $x$, together.