What does second-order derivative really tell you geometrically?

Question

I understand the first-order derivative (or partial derivative) tells you the slope of the tangent line at that point. If this is the case, what does second-order derivative tell you about?

Thank you

Answer 1

Second derivative is related to rate of change and slope of first derivative that is convexity/concavity of the function, notably if

• $f''(x)\ge 0\implies f(x)$ is convex
• $f''(x)\le 0 \implies f(x)$ is concave

Answer 2

I always find it a bit underwhelming that the second derivative only tells you a binary piece of information: whether the function is convex or concave. After all, the derivative tells you more than just whether the function is increasing or decreasing; it also tells you how quickly the function is increasing or decreasing.

The second-order analogue of this is how tightly the function's graph is curving. While the derivative measures the slope of a tangent line, the second-order analogue is the curvature of an osculating circle, the largest tangent circle that fits (near the point of tangency) inside the curve of the graph. Numerically, the curvature of the circle is the reciprocal of its radius (because larger circles curve less tightly); and a straight line (thought of as a degenerate circle with infinite radius) has curvature zero. Then just as we can say that a nonlinear graph has a slope (equal to the slope of the tangent line) at a given point, we can say that a non-circular non-linear graph has a curvature (equal to the curvature of its osculating circle) at a given point. And while a circle by itself can only be said to have an unsigned curvature, when it comes to the curvature of the graph of a function, we can count it as positive when the graph curves upwards (convex) and negative when the graph curves downwards (concave).

One reason that they don't bring this up as much, is that the second derivative alone is not enough to calculate the curvature! So while the slope of the graph of $ f $ at the point $ ( x , f ( x ) ) $ is simply $ D f ( x ) = f ' ( x ) $, the curvature there is $$ K f ( x ) = \frac { f ' ' ( x ) } { \bigl ( 1 + f ' ( x ) ^ 2 \bigr ) ^ { 3 / 2 } } \text . $$ Since the denominator is always positive, the sign of the second derivative is always enough to tell you whether the sign of the curvature is positive or negative, that is whether the function is convex or concave. But the full value of the curvature is more complicated.

For a simple example, take $ f ( x ) = x ^ 2 $, so that $ f ' ( x ) = 2 x $ and $ f ' ' ( x ) = 2 $. Since the second derivative is always positive, this function is everywhere convex. But although the second derivative is constant, a glance at the graph shows that the tightest curvature occurs at the vertex, when $ x = 0 $. When we write the full formula for the curvature, $ K f ( x ) = 2 / ( 1 + 4 x ^ 2 ) ^ { 3 / 2 } $, we see that indeed this is $ 2 $ when $ x = 0 $ but less than $ 2 $ when $ x \ne 0 $. (If you draw in a circle with radius $ 1 / K f ( 0 ) = 1 / 2 $, the you can see a good example of what an osculating circle looks like. This circle is tangent to the graph and fits inside the curve of the graph, but any larger tangent circle would fall, at least near the point $ ( 0 , 0 ) $ of tangency, outside the curve of the graph.)