Principal and normal curvature I am having trouble understanding the principal curvature and normal curvature. What are the basic intuition or ideas between the two. How are they related? And if given a problem, how can you compute the normal and principal curvature? 
For example, how can I compute the curvature at $(0,0,0)\in \sum_g$ such that $g(x,y)=x^2+y^2$ and $\sum_g = \text{graph of g} = \{(x,y,g(x,y)\}$? I know the prinicpal curvature for this problem is $2$ and the normal curvature is $2$ as well, but I don't know how to get it. 
Another a similar problem is $f(x,y) = y^2-x^2$ but here there are two principal curvature $2$ and $-2$. 
 A: If $\alpha(t)$ is a curve on a regular surface $S$ with a unit speed then we have curvatue $k$, $\alpha''(t) = k{\bf n}$ where $|{\bf n}|=1$. Hence we have normal curvature $$ k_n = k\ N\cdot {\bf n}$$
where $N$ is unit normal to $S$. 
Here principal curvatures are maximum and minimum of normal curvatures at $(0,0,0)$ ;
There exists well-known facts. There exists an orthonormal basis $\{ e_1, e_2\}$ on $T_pS$ such that $$dN_p (e_1)=-k_1e_1,\ dN_p(e_2)=-k_2e_2$$
Here $k_i$ are principal curvatures. 
(1) In fact in case of $(x,y, x^2+y^2)$ by symmetry three curvatures are same :
 $$N = \frac{1}{\sqrt{4x^2+4y^2+1}}(-2x,-2y,1),\ N'(0)=(-2x'(0),-2y'(0),0)$$ So $dN$ has any vector $(x'(0),y'(0),0)$ as eigenvector wrt eigenvalue $-2$. So normal curvature is $2$. (For details, see (2)). 
(2) In second example, $$ {\bf x}(u,v) = (u,v,v^2-u^2),\ N = \frac{{\bf x}_u\times {\bf x}_v}{ |{\bf x}_u\times {\bf x}_v|} = \frac{1}{\sqrt{u^2+v^2+1/4}}(u,-v,1/2) $$
If $a(t)={\bf x}(u(t),v(t)),\ i.e., a'(0)=(u'(0),v'(0),0)$,
 is a curve through $(0,0,0)$ then $$ dN (a'(0)=(u'(0),v'(0),0))
 =\frac{d}{dt}|_0 N(a(t))= (2u'(0),-2v'(0),0) $$
Hence we conclude that $dN$ has eigenvector $(1,0,0)$ wrt eigenvalue 
$2$ and eigenvector $(0,1,0)$ wrt eigenvalue $-2$. So principal curvatures are $-2$ and $2$
