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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.

Shevliaskovic
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Ruth Gutierrez
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2 Answers2

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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.

Robert Israel
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  • Wow yes this makes more sense. Thank you so much. – Ruth Gutierrez Mar 04 '14 at 05:51
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    the parametrization $\phi_1(x,y)=(x,y,\sqrt{x^2+y^2+1})$ and $\phi_2(x,y)=(x,y,-\sqrt{x^2+y^2+1})$, defined in $\mathbb{R}^2 ,$ is one-to-one onto one sheet, right? @RobertIsrael –  Oct 14 '20 at 21:04
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    Posted date: $\approx \pi$ $\ddot\smile$ – VIVID Jun 17 '21 at 14:30
  • Is it possible to parameterize an n -hyperboloid like this ? Similar to the parameterization for n-sphere in n variables. – NadaBrothers Feb 25 '23 at 04:28
  • what if z is negative how can we generalise the parameterisation for both $z > 0$ and $z < 0$? – R_Squared Nov 10 '23 at 08:36
  • @R_Squared There are two sheets and their union is not connected, so you can't get both in one continuous parametrization. For the other sheet take $z = - \cosh(u) \cosh(v)$ with the same $x$ and $y$. – Robert Israel Nov 10 '23 at 15:16
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If $x^2/a^2-y^2/b^2-z^2/c^2=1$ is the hyperboloid then putting $z=c.\tanɸ$ it reduces to $x^2/a^2-y^2/b^2=\sec^2ɸ$ or,$x^2/(a.\secɸ)^2-y^2/(b.\secɸ)²=1$ which again has parametric representation $x=a.\secɸ.\secɵ$, $y=b.\secɸ.\tanɵ$. So $x=a.\secɸ.\secɵ$,$y=b.\secɸ.\tanɵ$, $z=c.\tanɸ$ is the required parametrization

soap
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amer bnayan
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