Find the parametrization for the hyperboloid of two sheets${(x,y,z) \in \mathbb{R}^3}; -x^2-y^2+z^2=1$.
Ok so I saw two answers for this question: $x(u,v)=(\sinh u \cos v, \sinh u \sin v, \cosh u)$ and $x=(u,v)= (\cosh u \sinh v, \sinh v, \cosh u \cosh v)$. I'm pretty sure the first one is the correct one. But I'm confused on how to parametrize.
So what I think is,
$$x^2+y^2-z^2=-1$$
$$x^2+y^2=z^2-1$$
$$r^2=z^2-1$$
$$r=\sqrt{z^2-1}$$
So, $x=r\cos v$=$\sqrt{z^2-1}\cos v$, $y=r\sin v= \sqrt{z^2-1}\sin v$, and $z= \sqrt{z^2-1}$.
Clearly I know this isn't right but I'm not sure how to go from here. I know we need to get the partial derivative and I guess somehow that gets us the missing part. Can someone help please.
Best Answer
There are many parametrizations, so no such thing as the parametrization.
You have $x^2 + y^2 + 1 = z^2$. If you write $x^2 + y^2 = r^2$ (so you can take $x = r \cos(v)$, $y = r\sin(v)$), then $r^2 + 1 = z^2$. Now since $\cosh^2(u) - \sinh^2(u) = 1$, you could take $z = \cosh(u)$, $r = \sinh(u)$, and thus $$ \eqalign{x &= \sinh(u) \cos(v)\cr y &= \sinh(u) \sin(v)\cr z &= \cosh(u)\cr}$$
However,this is not a good parametrization at $u=0$ (corresponding to the single point $(x,y,z) = (0,0,1)$.
Now let's try a different way. Write the equation as $t^2 - y^2 = 1$ where $t^2 = z^2 - x^2$. We can take $y = \sinh(v)$ (note that this ranges over $\mathbb R$ as $v$ does), and then $t^2 = \cosh^2(v)$. If we choose the positive $t$ (and we can), we have $t = \cosh(v)$. Now $(z/t)^2 - (x/t)^2 = 1$. For the sheet with $z > 0$ we can take $z/t = \cosh(u)$ and $x/t = \sinh(u)$. As $u$ ranges over $\mathbb R$, we get all possible $(x,z)$ pairs with $z > 0$. So we have
$$ \eqalign{ x &= \sinh(u) \cosh(v)\cr y &= \sinh(v)\cr z &= \cosh(u) \cosh(v)\cr} $$
This parametrization is better because it maps ${\mathbb R}^2$ one-to-one onto one sheet.