Can the second derivative of a function be interpreted as the slope of its "concavity lines"? Can the second derivative of a function be interpreted as the slope of its "concavity lines"?
For example consider the following picture:

Does $f''$ for each point $x$ that corresponds to an arrow being drawn (for which there are $9$ in the picature) itself correspond to the slope of the line that would be obtained by extending its concavity arrow into a line?
 A: For regions that are concave up, $f''>0$. Similarly, regions that are concave down are where $f''<0$. For the first three points $f''>0$; however, this does not mean that all the slopes will be positive. For instance, by extending the concavity arrow at the third point x, you notice that the slope is negative. This follows similarly for the second and fifth point(vertical lines), the sixth point is positive, and the ninth point is negative. Therefore, the sign of $f''$ does not correspond to the sign of the slope.
A: The short answer is no.
Each arrow is perpendicular to a tangent line of the curve. A vector pointing in the direction of the tangent line at $x$ is $(1,f'(x))$, so a vector pointing perpendicular to the tangent line is $(-f'(x),1)$. (The dot product of these two vectors is zero.) The sense of the arrows in the picture (up or down) is related to the concavity: pointing up for concave up (convex) and down for concave down (concave). Since $f''(x)\geq 0$ in the former case and $f''(x)\leq 0$ in the latter case, then each arrow is pointing in the direction $$[f''(x)](-f'(x),1).$$ Thus the direction of each arrow depends on both the first and second derivatives (i.e. on both the slope and the concavity).
For example at a local minimum where the function has zero derivative and positive second derivative (convex), the arrow points in the direction $(0,1)$.
