I have the following paraboloid $$ S = \{(x,y,z)\in\mathbb{R}^3| x^2+y^2-z=1, z\gneq-1\}$$ which I parametrized by the chart
$$ \phi:\mathbb{R}^2-\{(0,0)\}\longrightarrow\mathbb{R}^3 $$
$$ (u,v)\longrightarrow(u,v,u^2+v^2-1) $$
I was asked to find the principal curvatures, principal directions and asymptotic direction at $p=(1,0,0)=\phi(1,0)$. The second fundamental form in this point is given by
$$ \begin{pmatrix} 2/5 && 0 \\ 0 && 2/5 \end{pmatrix} $$
so that the normal curvature in the direction of $w=(w_1,w_2)\in T_pS$ is
$$ k_n(w) = \begin{pmatrix} w_1 && w_2 \end{pmatrix}\begin{pmatrix} 2/5 && 0 \\ 0 && 2/5 \end{pmatrix}\begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = 2/5$$ as $w_1^2+w_2^2=1 $ because it is an unitary vector in the tangent plane.
So this means that in this point of the paraboloid, the normal curvature is constant. Hence there are no asymptotic directions, and every direction $w$ is principal. Additionally, the principal curvature $k_1$ and $k_2$ are 2/5, for being the maximum and minimum of the normal curvature.
Thing is that I find it difficult to believe that the normal curvature can be constant in the point of the paraboloid. Wouldn't it mean that it curves in the same proportion for every direction? Isn't this weird as it seems that it will curve in a different way in the $z$-axis direction in comparison with the $y$ or $x$ axis directions? A geometrical interpretation of the normal curvature would be useful to check if I'm doing all right.
Best Answer
The only umbilic point on the paraboloid is the vertex $(0,0,-1)$. Your mistake here is in thinking that $w$ is a unit vector. You've forgotten to compute the first fundamental form. Indeed, the lines of curvature of any surface of revolution are the parallels (circles) and meridians (here, parabolas); at the point $(1,0,0)$, the respective principal curvatures are $k_1=2/\sqrt5$ and $k_2=2/(5\sqrt5)$.