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!

Image you have a straight line, that has a slope, say 45 degrees, so as an equation y=x
Differentiate it once gives the gradient,a constant dy/dx = 1   differentiate it twice give the curvature,  d2y/dx2 =0  so in this case y=x has no curvature, but it does have a gradient. By the way, if you have a straight line that is horizontal y= 4 say that would have no gradient or curvature. 
Intuitively, I would image holding a straight plastic stick in your hands. When the stick is parallel to the ground, it has no gradient or curvature. If you tilt one end, 45 degrees say, then it has a gradient but still no curvature. If you bend the plastic a bit, it has a gradient and curvature (in this case it has a range of gradients, a different gradient at each point on the stick)
btw - the stick could have a single curvature across the stick whilst having multiple gradients. If you bend the the stick with one bend, then that shape would be a parabola, represented as a quadratic equation ax^2+bx+c=0 . The gradient would be dy/dy = 2ax+b  (different at each x) and the curvature would be d2y/dx2 = 2a  (a constant).
Hope that helps ;)   
