Intuition behind gradient VS curvature In Newton's method, one computes the gradient of a cost function, (the 'slope') as well as its hessian matrix, (ie, second derivative of the cost function, or 'curvature'). I understand the intuition, that the less 'curved' the cost landscape is at some specific weight, the bigger the step we want to take on the landscape terrain. (This is why it is somewhat superior to simply gradient ascent/descent). 
Here is the equation:
$$
\mathbf {w_{n+1} = w_n - \alpha\frac{\delta J(w)}{\delta w}\begin{bmatrix}\frac{\delta^2J(w)}{\delta w^2}\end{bmatrix}^{-1}}
$$
What I am having a hard time visualizing, is what is the intuition behind 'curvature'? I get that is the 'curviness of a function', but isnt that the slope? Obviously it is not, that is what the gradient measures, so what is the intuition behind curvature? 
Thanks!

The gradient and the radius of curvature at a point x are both normal to the equipotential surface through x.
The radius of curvature is the radius of the largest sphere that is tangent to the surface at that point.  It has nothing to do with the slope, but only with the equipotential surface.  Curvature is the reciprocal of the radius of curvature.  It is used rather than the radius of curvature because the radius of curvature is infinite when curvature is 0.  0 is more convenient to use in computations than infinity.
The gradient is a vector that points "up hill" and has magnitude equal to the slope.
