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.