Asymptotic Curves and Lines of Curvature of Helicoid I have a question that asks me to find the asymptotic curves and lines of curvature of the helicoid given by: $x = v \cos u$, $y = v \sin u$, $z = cu$, for some fixed real $c$. Can you show me how best to do this every step of the way, from finding the coefficients of the fundamental forms to solving the differential equations for $u$ and $v$?
$X = (x,y,z)$.
$X_{u} = (-v \sin u, v \cos u, c)$,
$X_{v} = (\cos u, \sin u, 0)$.
First fundamental form:
$E = X_{u} • X_{u} = v^{2} + c^{2},$
$F = X_{u} • X_{v} = 0$,
$G = X_{v} • X_{v} = 1$.
Information that follows swiftly from it:
$X_{uu} = (-v \cos u, - v \sin u, 0)$
$X_{uv} = (-\sin u, \cos u, 0)$,
$X_{vv} = (0,0,0)$.
$N = \frac{1}{\sqrt{(EG-F^2)}}\cdot(-c \sin u, c \cos u, -v).$
Second Fundamental Form:
$e = \langle N, X_{uu}\rangle = \frac{2c v \sin u\cos u}{\sqrt{EG-F^2}},$
$f = \langle N, X_{uv}\rangle = \frac{c \sin^2 u - c \cos^2 u}{\sqrt{EG-F^2}},$
$g =\langle N, X_{vv}\rangle = 0.$
These can be simplified somewhat.
Asymptotic curves must satisfy:
$e(u')^{2} - 2f u' v' + g (v')^{2} = 0.$
I can plug these values in, but cannot really proceed.
The differential equation for the lines of curvature is given by:
$$\begin{vmatrix}(v')^2& -u'v'& (u')^2\\ E& F& G\\ e& f& g\end{vmatrix}= 0.$$
Again, I can plug them in, but how to proceed eludes me.
 A: 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.
A: 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.
