What is the meaning of the third derivative of a function at a point

Question

(Originally asked on MO by AJAY.)

What is the geometric, physical, or other meaning of the third derivative of a function at a point?

If you have interesting things to say about the meaning of the first and second derivatives, please do so.

Answer 1

I've found the time, so I deleted one of my original comments in the OP and decided to expand it into a full answer.

One way to geometrically interpret the third derivative is in the notion of the osculating parabola. In much the same way that the first derivative enters into the defining equation for the tangent line (the line that best approximates your curve in the vicinity of a given point), and that the second derivative is involved in the expression for the osculating circle (the circle that best approximates your curve in the vicinity of a given point), the third derivative is required for expressing the osculating parabola, which is the parabola that best approximates... oh, you catch on quick. ;)

More specifically, if you remember the fact that four points uniquely determine a parabola, you can think of the osculating parabola as the limiting case of the parabola through four neighboring points of a given curve when those four points coalesce, or come together. The so-called aberrancy (a translation of the French "déviation") is the tangent of the angle the axis of the osculating parabola makes with the normal line, and is given by the formula

$$\tan\,\delta=\frac1{3\varrho}\frac{\mathrm d\varrho}{\mathrm d\phi}=\frac{\mathrm d y}{\mathrm d x}-\frac{1+\left(\frac{\mathrm d y}{\mathrm d x}\right)^2}{3\left(\frac{\mathrm d^2 y}{\mathrm d x^2}\right)^2}\frac{\mathrm d^3 y}{\mathrm d x^3}$$

where $\varrho$ is the radius of curvature and $\phi$ is the tangential angle.

From these considerations, one could derive an expression for the osculating parabola: given a curve represented parametrically as $(f(t)\quad g(t))^T$, the parametric equations for the osculating parabola of the curve at $t=t_0$ are

$$\begin{pmatrix}f(t_0)\\g(t_0)\end{pmatrix}+\frac{\varrho\;\cos^4\delta}{2}\begin{pmatrix}\cos\,\phi&-\sin\,\phi\\\sin\,\phi&\cos\,\phi\end{pmatrix}\cdot\begin{pmatrix}(u^2-2)\tan\,\delta-2u-\tan^3\,\delta\\(u+\tan\,\delta)^2\end{pmatrix}$$

Here for instance is the cardioid $(2\cos\,t+\cos\,2t\quad 2\sin\,t+\sin\,2t)^T$ and its osculating parabola at $t=2\pi/3$:

and an animation of the various osculating parabolas for the curve $(3\cos\,t-2\cos\,3t\quad 3\sin\,t-2\sin\,3t)^T$:

Further, one could also give a geometric interpretation for the fourth derivative; what one now considers is the osculating conic (the limiting conic through five neighboring points of a curve when those five points coalesce), and one could classify points of a plane curve as elliptic, parabolic or hyperbolic depending on the nature of the osculating conic. In this respect, the discriminant of the osculating conic depends on the first four derivatives.

A lot more information is in these two articles by Steven Schot (who also wrote a nice article on the "jerk"), and the references therein.

Answer 2

For the position function $p=f(t)$, you probably know that the first derivative $f'(t)$ gives the instantaneous velocity, and that the second derivative $f''(t)$ gives the instantaneous acceleration.

The rate of change of the acceleration is called the "jerk" (also known as the "surge", the "jolt", or the "lurch"). So the third derivative $f'''(t)$ would give the instantaneous jerk.

Answer 3

The rate of change of acceleration is studied in various situations in physics, mechanics and engineering design.

From wikipedia:

In physics, jerk, also known as jolt (especially in British English), surge and lurch, is the rate of change of acceleration; that is, the derivative of acceleration with respect to time, the second derivative of velocity, or the third derivative of position. Jerk is defined by any of the following equivalent expressions:

$$ \vec j=\frac {\mathrm{d} \vec a} {\mathrm{d}t}=\frac {\mathrm{d}^2 \vec v} {\mathrm{d}t^2}=\frac {\mathrm{d}^3 \vec s} {\mathrm{d}t^3}$$ where

$\vec a$ is acceleration, $\vec v$ is velocity, $\vec s$ is position and $\mathit{t}$ is time.

Jerk is a vector, and there is no generally used term to describe its scalar magnitude (e.g. "speed" as the scalar magnitude for velocity).

Think of roller coaster designs.

Indeed, in mechanics, the fourth derivative is also studied. It is called Jounce or Snap. From wikipedia:

In physics, jounce or snap is the fourth derivative of the position vector with respect to time, with the first, second, and third derivatives being velocity, acceleration, and jerk, respectively; in other words, the jounce is the rate of change of the jerk with respect to time.

$$\vec s =\frac {d \vec j} {dt}=\frac {d^2 \vec a} {dt^2}=\frac {d^3 \vec v} {dt^3}=\frac {d^4 \vec r} {dt^4}$$

Sometimes even the higher derivatives are nontrivial and come into play. Think of a sudden impact, an earthquake, a shock, or a lightning effect in electrical systems. Constructs like Dirac Delta function are very convenient for dealing with such situations. For instance, for a lightning, there is approximately a very high surge of current for a very brief instant and for any smooth approximation all the higher derivatives are nonzero. So you take the limit and manipulate it as if everything were concentrated at a single point.

Answer 4

It is a common theme in applied math that you can easily interpret first and second derivative or moment (in case of probability theory), but after that, trouble begins.

That being said, the third derivative is used in calculating the torsion of a curve.

Let's review an example (rather poor one, I admit). Let's work in $\mathbb{R}^3$ with a Cartesian coordinate system $x$, $y$, $z$ and the associated basis $\mathbf{e}_1$, $\mathbf{e}_2$, $\mathbf{e}_3$. Then let $\gamma$ be a unit circle given by $x = \cos \varphi$, $y = \sin \varphi$, $z = z_0$, parametrized naturally (by $\varphi$). Then $\dot{\gamma} \times \ddot{\gamma} = \mathbf{e}_3$, and then $\tau = {\dddot{\gamma}}^3 = 0$. I guess this particular curve is not very instructive, but I can't think of a better one off the top of my head :)

Answer 5

An intuitive complement to Arturo Magidin's answer:

A good way to intuitively grasp the jerk (hence the third derivative of the position function) is to remember the last time you took a plane and realize that the following "equivalences" holds

• No acceleration = constant speed = feels like when sitting in your chair at work = first derivative of the position function is zero.
• Acceleration = speed increases = feels like someone is pushing you toward the back of your seat = second derivative is positive.
• Increasing acceleration = the pace at which your speed increases gets higher and higher = feels like the guy who is pushing you toward the back of your seat is pushing harder and harder = jerk or third derivative is positive.

In a plane :

1) Right before take-off, the plane is still, no acceleration, the derivative of the position function are zero.

2) Now the plane starts moving, you are not still anymore, and the first derivative of the position function is positive.

3) Not only are you moving, but the plane brutally accelerates. As a result of the acceleration, you feel like someone is pushing you toward the back of your seat: the second derivative of the position function is positive.

4) Quickly after the engines are on, not only do you feel like someone is pushing you toward the back or your seat but, in addition, it feels like this imaginary person is pushing harder and harder. This is because you accelerate more and more (the jerk is positive). During the first 2 second you went from 0 km/h to say 20 km/h, and during the 2 following ones, you went from, say, 20 km/h to 60 km/h: The third derivative of the position function is positive.

5) After some time, the plane still accelerates, but at a diminishing rate. It feels like the imaginary guy pushing you toward the back of your seat starts to release the pressure. He is still pushing (you would need a higher effort to stand up from your seat than if you were sitting in your office chair), but less and less intensely. The rate at which you accelerate is diminishing, hence the third derivative is negative. However, you are still accelerating so the second derivative is still positive.

6) Your plane eventually reaches its cruising altitude and maintain a constant speed of say 800 km/h. So now, your are not accelerating at all, the second and third derivative of the position function are zero. Only the first derivative remains positive.

7) When you land, the process is reversed. It feels like someone is pushing you in the back and you need the seatbelt to prevent you from falling forward. When it feels like the imaginary guy pushes you in the back stronger and stronger, then the jerk is negative, hence the third derivative of the position function is negative too.

Answer 6

Suppose $y=f(x)$ is smooth on $\mathbb R$.

The first derivative $y'$ represents the gradient of the curve. If $y'>0$ on $\mathbb R$, $y$ is strictly increasing. If $y'<0$ on $\mathbb R$, $y$ is strictly decreasing.

The second derivative $y''$ represents the rate of change of the gradient. If $y'=0$ and $y''<0$, we have a local maximum. If $y'=0$ and $y''>0$, we have a local minimum.

The third derivative $y'''$ represents the rate of change of gradient change. If $y''=0$ and $y'''\neq 0$, we have a point of inflection.

Answer 7

Since force is a constant scalar multiple of acceleration (at non-relativistic speeds), the third derivative of a position function, jerk, is a constant multiple of the rate of change of force. In other words the jerk of a unit mass object is equal to the rate of change of force, a quantity sometimes called "yank". This is analogous to the relationship between momentum and velocity. Note, force is the time-derivative of momentum.

Then the snap (4'th derivative of position) of a unit mass object is equal to the second derivative of force, called "tug". Apparently, there's a whole heierarchy up to 6'th derivative of position!

http://math.ucr.edu/home/baez/physics/General/jerk.html

Answer 8

when you are in a car, and it is accelerating at a constant rate, the back of your seat is pushing on your back with a constant force, or a constant pressure if you like. if there is jerk, the pressure on your back will change. if the jerk is constant, the pressure on your back will change nicely, perhaps at a linear rate. if the jerk is not constant, the pressure on your back will change more erratically.

Answer 9

By asking the question, you can interpret the 1st and 2nd derivatives in some meaningful ways.

Consider the 3rd as the combination of them. In other words, consider the third derivative as the 'acceleration' if the velocity was in fact the displacement.