TITLE: Relation bewteen Hessian Matrix and Curvature
QUESTION [19 upvotes]: According to Hessian matrix,  It describes the local curvature of a function. 
AFAIK, for one-variable function $f(x)$, its local curvature is 
$$\kappa = \frac{|f''|}{(1 + f'^2)^{3/2}},$$
and its Hessian matrix is 
$$\mathcal{Hess}(f) = [f''],$$
right? And here is my problem, I think the local curvature is not just described by its Hessian matrix, because $f'$ also has its role in it, doesn't it?
And furthermore, for 2-variable function $f(x,y)$, its Hessian matrix is 
$$\mathcal{Hess}(f) = \left[ \begin{array}{cc}
f_{xx}'' & f_{xy}'' \\
f_{xy}'' & f_{yy}''
\end{array} \right].$$
How does it relate to the local curvature of $f(x,y)$?

REPLY [13 votes]: Just as a little addendum to Shine's answer who already pinpoints it, here a neat plot for intuitive illustration. The difference between the Hessian of a function $f$ and its actual local curvature at some point $x$ is best illustrated by a simple quadratic function $f=x²$. Its Hessian $f''=2$ is globally constant, while its curvature $\kappa$ of is obviously not.
The determinant of the Hessian equals the curvature only if evaluated at an extremum ($f'=0$).