Let be $X(u,v) = (\cosh v \cos u,\cosh v \sin u, v)$ the parametrization of catenoid. Determine the asymptotic curves of the catenoid.
$\textbf{My attempt:}$
$X_u(u,v) = (-\cosh v \sin u,\cosh v \cos u, 0)$
$X_v(u,v) = (\sinh v \cos u,\sinh v \sin u, 1)$
$X_{uu}(u,v) = (-\cosh v \cos u,- \cosh v \sin u, 0)$
$X_{uv}(u,v) = (-\sinh v \sin u,\sinh v \cos u, 0)$
$X_{vv}(u,v) = (\cosh v \cos u,\cosh v \sin u, 0)$
$X_u \times X_v = (\cosh v \cos u, – \cosh v \sin u, – \sinh v \cosh v)$
$||X_u \times X_v|| = \cosh^2 v$
$N = \frac{X_u \times X_v}{||X_u \times X_v||}$
$e = \langle X_{uu}, N \rangle = – \cos (2u)$
$f = \langle X_{uv}, N \rangle = \frac{-2 \sinh v \cosh v \sin u \cos u}{\cosh ^2 v} = – \frac{\sinh v \sin (2u)}{\cosh v}$
$g = \langle X_{vv}, N \rangle = \cos (2u)$
Since we want the asymptotic curve of the catenoid, we look for curves $\alpha: I \longrightarrow S$ such that
$II(\alpha'(t)) = 0$ for each $t \in I$, i.e., $(u'(t))^2 e + 2u'(t)v'(t)f + (v'(t))^2 g = 0$ for each $t \in I$, then we need $e = f = g = 0$, the condition is satisfied when
$\cos(2u) = 0$ and $\sinh v = 0$, i.e., when $u = \frac{\pi}{4} + \frac{k \pi}{2}$ for each $k \in \mathbb{Z}$ and $v = 0$.
This question is from Do Carmo's Differential Geometry of Curves and Surfaces and there he states the asymptotic curves of the catenoid for this parametrization is such that $u + v = c$ or $u – v = c$, where $c$ is a constant. Did I wrong something? Someone can explain me how can I get that the asymptotic curve is such that $u + v = c$ or $u – v = c$?
Thanks in advance!
Best Answer
It seems your computation was mistaken.
Xu$\times$ Xv =$(\cosh v\cos u,\cosh v \sin u,-\cosh v\sinh v)$
Hence, $f=0,e=-1,g=1$.
The differential equation of the asymptotic curve is \begin{equation} -(u')^2+(v')^2=-(u'-v')(u'+v')=0 \end{equation}
Solve the differential equation, the asymptotic curve is $u+v=c$ and $u-v=c$
In addition, $f$ of the surface of revolution is always zero.