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(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.

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Please see this and this; those two articles are where I picked up my intuition for the third derivative. –  J. M. Dec 19 '10 at 6:37
    
@J.M.: That's great; I think you should make it an answer. –  Jonas Meyer Dec 19 '10 at 6:40
    
Too short for an answer: I like to think about third derivative as telling me how quickly the curvature is changing. Third derivatives also give us "osculating cubics" (for when osculating quadrics just won't do). –  Jesse Madnick May 2 '12 at 9:57
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9 Answers 9

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 $t=2\pi/3$:

cardioid and osculating parabola

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

osculating parabola animation

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.

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Mathematica code for playing around with osculating parabolas and conics will be supplied upon request. –  J. M. May 14 '11 at 8:59
    
Excellent! Thank you very much for this answer. –  Jonas Meyer May 20 '11 at 17:07
    
@J.M. well, the math is beyond me but the animation is brilliant! –  Emmad Kareem Oct 30 '11 at 4:04
    
@JM: would love that mathematica code! I need something to entertain me while I'm "working" :) –  user641 Dec 8 '12 at 1:18
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@Steve, I don't have access to the computer where I stashed my routines for now, but I'll ping you when I get them. –  J. M. Jan 16 '13 at 18:06
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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.

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If I remember correctly, you can perform the "tablecloth trick" by pulling a tablecloth from a table at constant jerk without the tableware moving an inch (assuming friction is linear with regards to speed I think). –  Joel Cohen May 14 '11 at 15:27
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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.

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May I ask what the down-vote was for? –  Arturo Magidin Dec 19 '10 at 19:04
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In support of Arturo's question, I think it's good practice on this site to give a reason for downvoting. –  Mike Spivey Dec 19 '10 at 22:35
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Man, I hate downvote-and-run dudes... –  J. M. Dec 20 '10 at 2:08
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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 :)

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Good point, the torsion first shows up in the third derivative; on the other hand, the torsion could be zero even if the third derivative isn't. If you want examples with nonzero torsion, helices will do, or more generally any regular curve with nonvanishing curvature not contained in a plane. The torsion affects the component of the third derivative orthogonal to the osculating plane. –  Jonas Meyer Dec 19 '10 at 7:46
    
@Jonas : I just wanted to give a simple example, and I observed that when $\dot{\gamma} \times \ddot{\gamma} = \lambda \mathbf{e}_i$, you get $\tau = {\dddot{\gamma}}^i/\lambda$, which is a very direct dependence on the third derivative, so I tried to make such an example. –  Alexei Averchenko Dec 19 '10 at 7:59
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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.

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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

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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.

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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 "equivalence" holds

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

In a plane :

1) Right before take-off, the plane is standing still, no acceleration the derivatives 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 accelerate. As a result of the acceleration, you feel like someone is pushing you toward you 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 you seat but, in addition, it feels like this imaginary person is pushing harder and harder. This is due to the fact that 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 20 km/h to say 60 km/h. The third derivative of the position function is positive.

5) After some time, the plane still accelerate 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 ends up reaching its cruise 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 in, the process is reversed. It feels like someone is pushing you in the back. When it feels like the imaginary guy pushes you stronger and stronger, then the jerk is negative, hence the third derivative of the position function is negative too.

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Sorry, did not read Brady Trainor's answer before writing my answer (which is essentially the same as what he wrote...). Worth keeping my answer? –  Martin Van der Linden Sep 13 '13 at 21:45
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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.

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What is the geometric, physical or other meaning of the velocity being in fact the displacement? –  Jonas Meyer Dec 19 '10 at 7:05
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The movement of a dot on the velocity time diagram plotting the velocity of the displacement of some object. I am starting to think I do not make sense. –  picakhu Dec 19 '10 at 7:06
    
I downvoted this post; This does not answer the question in a meaningful way. Regards. –  user21436 May 5 '12 at 12:06
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