I know there is a parameterization of a hyperboloid $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$ in terms of $\cosh$ and $\sinh$, but I don't see how these equations are derived. I would appreciate it if either someone could explain to me how such a parameterization is derived or recommend a reference.
Deriving parameterization for hyperboloid
-
0The derivation is just writing down the relevant identities, $\cos^2 u+\sin^2 u=1$ and $\cosh^2 v-sinh^2 v=1$. We also need to check we do obtain the whole hyperboloid, which involves the range of $\sinh$. – 2012-06-14
-
0I know they work, using these identities. I guess I'd like a geometric reason. I can see easily how to derive spherical coordinates, which I find similar to these. Can they be derived by considering the hyperboloid as a surface of revolution? – 2012-06-14
-
1Surface of revolution works fine, $\cosh and \sinh$ are to the hyperbola $x^2-y^2=1$ as $\cos$ and $\sin$ are to the circle $x^2+y^2=1$. That gets you to the parametrization of the hyperboloid of revolution $x^2+y^2-z^2=1$. The multiplications by $a$, $b$, $c$ are at the end, they are scalings in the $x$, $y$, and $z$ directions. – 2012-06-14
-
0Oh! Thank you. I just realized this whole time I was misreading the cos and sin terms in the parameterizations that come from surface of revolution as a second pair of cosh and sinh. If you leave your comment as an answer I will accept. – 2012-06-14
-
1There was a partial answer by @Argon, later deleted. Argon can revive and amplify her/his answer. Or, perhaps better, now that you have figured it out you can answer, and, after some time has passed, accept. – 2012-06-14
2 Answers
This derivation has been done by André Nicolas!
The parametrization of the hyperbola is
$$x(t)=\cosh t$$ $$y(t)=\pm \sinh t$$
A circle of radius $r$ is parametrized as:
$$x(t)=\cos t$$ $$y(t)=\sin t$$
Rotating the hyperbola above around a circle of radius $\cosh$ (distance of a regular hyperbola from y axis):
$$x(u, v)=\cosh v \cos u$$ $$y(u, v)=\cosh v \sin u$$ $$z(u, v)=\sinh v$$
It is easy to imaging the hyperboloid from two ways - from the top and from the side. This helped me understand the derivation.
By virtue of user bondesan, here's a picture:
The $(P_u, P_z)$ might confuse, so I'd rewrite it as follows.
$\color{green}{\text{On the $uv$-plane, any hyperbola is given by: } u = \cosh(s) \text{ and } z = c\sinh(s) \, \,\forall -1 \leq s \leq 1}$.
Then by definition of polar coordinates, $x = (a\cos \theta) \color{green}{u} \text{ and } y = (b\sin \theta) \color{green}{u} \text{ and } \color{green}{z = z}$.
Altogether, $x = (a\cos \theta) \color{green}{\cosh(s)} \text{ and } y = (b\sin \theta) \color{green}{\cosh(s)} \text{ and } \color{green}{z = c\sinh(s)}$.