The principal curvatures of a surface of revolution The principal curvatures of the surface at a point is defined as the maximal and the minimal curvature among all normal sections. It's claimed (say, on Stillwell's Geometry of Surfaces) that for a pseudosphere (or generally a surface of revolution) the extremal curvatures are obtained when the normal section coincides or is perpendicular to the plane determined by the longitude and the axis of revolution. It's said that the proposition follows clearly from symmetry. According to that book, there's no differential geometric tool (such as the second fundamental form, etc) introduced, so I need an intuitive and elementary explanation for that fact.
After googling, I found an explanation which depends on the knowledge of Dupin indicatrix, which is intuitive and computation-free, though not that elementary. It's really clear that the Dupin indicatrix is reflection-symmetric along the plane determined by the longitude and the axis of revolution, thus one direction of principal curvature is determined. The other one is normal to this.
Any help? Thanks!
 A: Consider a surface $S: z=f(x,y)$ touching the $(x,y)$-plane at the origin $O$. Intersecting $S$ with normal planes at $O$ means looking at the curves
$$\gamma_\phi:\quad z=z(t):=f(t\cos\phi,t\sin\phi)\qquad(-\infty<t<\infty)$$
for various directions $\phi$. The curvature of $\gamma_\phi$ at $O$ is given by
$$\kappa(\phi)=\left.{\ddot z(t)\over\left(1+\dot z^2(t)\right)^{3/2}}\right|_{t=0}=\ddot z(0)\ ,$$
since $\dot z(0)=0$. In this way one obtains 
$$\kappa(\phi)=f_{xx}\cos^2\phi+2f_{xy}\cos\phi\sin\phi+f_{yy}\sin^2\phi\ ,\tag{1}$$
where the partial derivatives of $f$ are taken at $O$.
Now to a surface of revolution: Let $P$ be a point on such a surface and choose $(x,y)$-coordinates on the tangential plane at $P$ such that the $x$-axis goes through the axis of revolution. Then we are in the situation analyzed above, and in addition we conclude from symmetry that   $\kappa(\phi)$ is an even function of $\phi$. It follows that in $(1)$ one has  $f_{xy}=0$, and then it is obvious that $\kappa(\phi)$ takes its extrema at integer multiples of ${\pi\over2}$.
A: The Dupin indicatrix for the pseudosphere is a hyperbola and the principal directions are the axes of the hyperbola (not the asymptotes), which are always orthogonal.  Since the principal directions are the axes, they are necessarily orthogonal, as well.  Therefore the "other" direction is uniquely determined.
A: I suggest please look up Euler's formula:$ k_n = k_1 cos(si)^2 + k_2 sin(si)^2 $. Draw a graph of $k_n$ as a function of $\psi$. It makes things graphically clearer about extremality of curvatures $k_1$ and $k_2$.
Also it is very instructive to draw a Mohr's circle with $k_n$'s on x-axis and geodesic torsion $\tau_g$ =  $(k_1 - k_2)*sin\psi *cos\psi$ on y-axis. Here $\psi$ is the parameter, your reference direction. 
The torsion axis does not touch the Mohr circle for positive, touches for zero and intersects it for negative Gauss curvature.
The two are quite general, need not be restricted to surfaces of revolution only.
