Asymptotic Curves and Lines of Curvature of Helicoid

Question

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.

Answer 1

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.

Answer 2

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.