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.

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.
