What does the second derivative represent in a curve?

Question

I have a very simple question

If I have a curve represented by four coefficients

y= c0+c1x+c2x^2+c3x^3

and I take the derivative of dy/dx in a point, I believe that represents the tangent slope, am I wrong?

Anyway, my question is what does the d2y/dx2 represents geometrically?

Answer 1

For a function $y = f(x)$, the curvature of the graph of the function, $\kappa(x)$, can be written as

\begin{equation*} \kappa(x) = \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\left(1+\mathcal{O}\left[\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}\right]\right). \end{equation*}

In that sense, the second derivative represents an approximation to the curvature of the graph of the function (with the goodness of the approximation dependent on the square of the magnitude of $\mathrm{d}y/\mathrm{d}x$). Therefore, for graphs of functions with small first derivative, the second derivative is sometimes used in certain applications in place of the curvature.

More precisely, though, the value of the second derivative of a function at a point represents how concave or convex the graph of the function is at that point.

Answer 2

There's a few interpretations:

One is to link the second derivative to the curvature. Denoting the curvature at $x$ by $k(x)$ we have $$ k(x) = \frac{y''(x)}{(1+y'(x)^2)^{3/2}} $$ so the sign of the curvature is always the same as that of the second derivative.

Another interpretation I like is to interpret the second derivative as the one-dimensional Laplace-operator. The Laplacian intuitively tells you how much the function's value at some point deviates from the mean of the function around that point.

Lastly there's the physics-y interpretation: if your parameter is time, then the second derivative is the instantaneous acceleration of a particle traveling along the graph.