Principal directions Consider a catenoid $C$ parametrized by 
$$r(u,v)= (u, \cosh u \cos v, \cosh u \sin v), u\in \mathbb{R}, v\in(-\pi, \pi)$$
I am required to show that the principal directions are the same as the coordiante curves, u= constant, v=constant.
The formula I want to use is $$dN_p (v)=\lambda v$$ for $v$ in the tangent space to $C$ at $p$. 
I have found the the differential of the Gauss map($N$) by using 
$$N= \frac{r_u\times r_v}{|r_u\times r_v|}$$
But once I start calculating the differential of $N$, things get messy. Am I on the right track? Is there a better way to find the principal directions?
Thanks in advance.
 A: For the catenoid $C$ parametrized by $r(u,v)= (u, \cosh u \cos v, \cosh u \sin v)$, we have
$$r_u(u,v)= (1, \sinh u \cos v, \sinh u \sin v),$$ 
$$r_v(u,v)= (0, -\cosh u \sin v, \cosh u \cos v).$$
This implies 
$$r_u\times r_v=(\sinh u\cosh u,-\cosh u\cos v,-\cosh u \sin v).$$
This gives $|r_u\times r_v|=\cosh u\sqrt{(\sinh u)^2+1}=(\cosh u)^2$ and 
$$N=\frac{r_u\times r_v}{|r_u\times r_v|}=\frac{1}{\cosh u}(\sinh u,-\cos v,-\sin v).$$
I think this is what got for the unit normal.
Now, fixed a point  $p\in C$ and assume $p=r(u_0,v_0)$. Then the coordinate curves at $p$ are given by $\alpha:(-\epsilon,\epsilon)\rightarrow C$ and $\beta:(-\epsilon,\epsilon)\rightarrow C$ such that $\alpha(t)=r(u_0+t,v_0)$ and $\alpha(t)=r(u_0,v_0+t)$ are coordinate curves at $p$, because $\alpha(0)=\beta(0)=r(u_0,v_0)=p$. Then 
$$\alpha'(0)=\frac{d}{dt}r(u_0,v_0+t)\Big|_{t=0}=r_u(u_0,v_0)\in T_pC,$$
$$\beta'(0)=\frac{d}{dt}r(u_0+t,v_0)\Big|_{t=0}=r_v(u_0,v_0)\in T_pC.$$
Differentiate $N(\alpha(t))$ with respect to $t$ and using $(2)$, we have
$$\frac{d}{dt}N(\alpha(t))\Big|_{t=0}=\frac{d}{dt}\left(\frac{1}{\cosh(u_0+t)}(\sinh(u_0+t),-\cos v_0,-\sin v_0)\right)\Big|_{t=0}$$
$$=\left(-\frac{\sinh(u_0+t)}{\cosh^2(u_0+t)}(\sinh(u_0+t),-\cos v_0,-\sin v_0)
\right)\Big|_{t=0}$$
$$+\left(\frac{1}{\cosh(u_0+t)}(\cosh(u_0+t),0,0)\right)\Big|_{t=0}$$
$$=\frac{1}{\cosh^2(u_0)}(1,\sinh u_0\cos v_0,\sinh u_0\sin v_0).$$
Using chain rule, the left hand side is $dN_{\alpha(0)}(\alpha'(0))=dN_{p}(\alpha'(0))$. On the other hand, the right hand side is equal to $\displaystyle\frac{1}{\cosh^2(u_0)}\alpha'(0)$ by $(1)$ and $(3)$. Therefore, we have shown that $$dN_{p}(\alpha'(0))=\frac{1}{\cosh^2(u_0)}\alpha'(0),$$
i.e. the coordinate curve $\alpha$ is principal direction with principal curvature $\cosh^2(u_0)$. 
Similarly you can show that the coordinate curve $\beta$ is principal direction. I will let you do it. 
