Differential Geometry and Its Application - Exercise 2.1.12

Problem:

Find a patch for the catenoid obtained by revolving the catenary $y = \cosh(x)$ about the x-axis.

Solution:

$(u. \cosh u \cos v, \cosh u \sin v)$, $u \in (-\infty, +\infty)$, $v \in [-\pi, \pi]$.

We will show the patch is one-to-one by giving its inverse. Given a point $(x, y, z)$ on the catenoid, we know $u = x$, now we know $y = \cosh x \cos v$, so $v = \cos^{-1}(\frac{y}{\cosh x})$.

Next, we will show the patch is regular. We have

$x_u = (1, \sinh u \cos v, \sinh u \sin v)$
$x_v = (0, -\cosh u \sin v, \cosh u \cos v)$.t

$x_u \times x_v = \left|\begin{array}{ccc} i & j & k \\ 1 & \sinh u \cos v & \sinh u \sin v \\ 0 & -\cosh u \sin v & \cosh u \cos v \end{array}\right| = (\sinh u \cosh u \cos^2 v + \sinh u \cosh u \sin^2 v, -\cosh u \cos v, -\cosh u \sin v)$

Now $\cosh u$ is never 0, $\cos v$ and $\sin v$ is never simultaneously 0, so the vector is never zero, the patch is regular.