Asymptotic Curves and Lines of Curvature of Helicoid

Kevin, note that your computation of $e$ is wrong: You should have $e=0$. From this we get the fact that $X_u$ and $X_v$ are both asymptotic directions. And the differential equation you have for lines of curvature will simplify immensely, as well.

To double-check what's going on, you should note that the helicoid is a minimal surface ($k_1+k_2=0$) and the principal directions therefore bisect the asymptotic directions. Therefore, since the asymptotic directions are orthogonal (!), the principal directions will be at angle $\pm \pi/4$ from the asymptotic directions.

We try to generalize all the cases in between catenoid and helicoid.

Let $\, \boldsymbol {x} (u,v)= \cos t \left( \begin{array}{c} \cos u \cosh v \\ \sin u \cosh v \\ v \end{array} \right)+ \sin t \left( \begin{array}{c} \sin u \sinh v \\ -\cos u \sinh v \\ u \end{array} \right)= \Re \left[ e^{it}\left( \begin{array}{c} \cos (u+vi) \\ \sin (u+vi) \\ -i(u+vi) \end{array} \right) \right] $

It's a catenoid for $t=0, \pi$ and helicoid for $t=\frac{\pi}{2}, \frac{3\pi}{2}$.

Now $\, \left| \begin{array}{ccc} E & F & G \\ e & f & g \\ dv^{2} & -du \, dv & du^{2} \\ \end{array} \right| = \left| \begin{array}{ccc} \cosh^{2} v & 0 & \cosh^{2} v \\ -\cos t & \sin t & \cos t \\ dv^{2} & -du \, dv & du^{2} \\ \end{array} \right| = 0 $.

We have $\left(\sin t \, du^{2}+2\cos t \, du \, dv-\sin t \, dv^{2}\right) \cosh^{2} v=0, \,$ thus

$ \left \{ \begin{array}{rcl} \cos \frac{t}{2} du-\sin \frac{t}{2} dv & = & 0 \\ \sin \frac{t}{2} du+\cos \frac{t}{2} dv & = & 0 \\ \end{array} \right. \implies \left \{ \begin{array}{rcl} u\cos \frac{t}{2}-v\sin \frac{t}{2} & = & \alpha \\ u\sin \frac{t}{2}+v\cos \frac{t}{2} & = & \beta \\ \end{array} \right. \; $ or equivalently,

$ u+vi=\exp\left( \frac{1}{2}it \right) (\alpha+\beta i) \,$ where $\alpha, \beta$ are integration constants and hence the parameters for the lines of curvature.

Note that $t=0$ is trivial for catenoid.