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Could somebody give me an intuitive physical interpretation of higher order derivatives (from 2 and so on), that is not related to position - velocity - acceleration - jerk - etc?

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I think it's going to be hard to give an intuitive description if we're not allowed to use intuitively obvious starting points like position, velocity and acceleration. –  John Rennie Nov 30 '12 at 11:43
    
I would like to find an orthogonal explanation to the one that is used always? I can't seem to figure out one by myself, so I thought that the hive might be of help. –  user680111 Nov 30 '12 at 11:46
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Curvature of a surface i.e. the derivatives of height with position? –  John Rennie Nov 30 '12 at 11:56
    
Excellent! Thanks –  user680111 Nov 30 '12 at 12:07
    
Beyond the second derivates comes the performance, i.e. the magtitude needed to describe the good or not quality of accelerating cars. –  Mario Enrique Nov 30 '12 at 13:15
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7 Answers

Take a curve in ther $xy$ plane, described by $y=f(x)$, and consider some point $(x_0,y_0)$ on the curve.

The function value $f(x_0)=y(0)$ gives the possition of the point on a vertical line through $x_0$ on the $x$-axis.

The derivative $f'(x_0)=y_0'$ gives the slope of the tangent (linear approximation) at that point, $f(x_0+h) \approx y_0+y_0'h$.

The second derivative $f''(x_0)=y_0''$ measures the deviation from linearity close to the point, in the sense that $f(x_0+h) \approx y_0+y_0'h+y_0''h^2/2$ is a more accurate parabolic approximation close to $x_0$.

The third derivative $f'''(x_0)=y_0'''$ measures the deviation from the optimal parabolic approximation close to the point, in the sense that $f(x_0+h) \approx y_0+y_0'h+y_0''h^2/2+y_0'''h^3/6$ is a more accurate cubic approximation close to $x_0$.

etc. The series $y_0+y_0'h+y_0''h^2/2+y_0'''h^3/6+...$ is the Taylor series expansion, and gives more and more accurate approximations (but sometimes in smaller and smaller neigborhoods).

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While this is a good explanation of derivatives and Taylor series, it does not give the physical interpretation which is what the OP is asking. –  Sklivvz Dec 9 '12 at 14:19
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+1 to reverse Skliwz's unjustified downvote. –  Dilaton Dec 9 '12 at 16:19
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@Nemo The only votes that are unjustified are those which are given to cancel out other people's votes. (Well, also other kinds of votes that are not based on the content of the post) Not to mention, you don't know who voted for this. –  David Z Dec 9 '12 at 20:37
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Beyond derivatives that yield position, velocity and acceleration the higher order derivatives are for approximating and are used in engineering rather than in physics or maths. If you look into the Taylor series you will find the intuition you're seeking.

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I don’t know about your physics or maths, but I tend to use higher order ($n \geq 2$) derivatives all the time in my physics or maths, mostly for approximation purposes where the first few orders are not enough (for example, because the first order gives zero due to some term-cancelling or the like). –  Claudius Dec 9 '12 at 12:47
    
Derivatives don't yield position, really :-) –  Sklivvz Dec 9 '12 at 14:20
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Sure they do, derivative of the integral of position is position. Acceleration is position in acceleration phase-space –  Hobo2 Dec 9 '12 at 14:39
    
@claudius I dont really associate approximation with pure mathematics its more like a real life application than a theoretical one. I'm aware of the extreme examples of high order derivatives in physics like jounce. The third derivative, for example, jolt, is an example of a high order derivative used in engineering. Beyond the 6th derivative all you can use differentiation for is approximations. But youre right I didn't say what I mean. –  Borab Dec 9 '12 at 17:15
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The smoothness of acceleration and thus applied and reactive forces is related to the higher derivatives (i.e. jerk, snap, crackle and pop).

So a physical interpretation would be if you are standing in a bus, and the driver brakes with constant deceleration up to the point the bus stops at which point you feel like you are thrown backwards. Drivers to avoid this problem typically ease off the brakes before stopping to achieve a better transition. Ideally you want a 2nd order curve for acceleration, or constant snap to transition nicely into a stop.

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I've always liked the discrete explanation.

In physics, you can almost always approximate a function $f: \mathbb{R} \to \mathbb{R}$ by saying what the value of $f$ is at a discrete set of points $\{x_i\} \subset \mathbb{R}$.

If you do this, you approximate the derivative by looking at how $f$ changes when you move from a point to one of its nearest neighbors. So if your mesh is a lattice of points $a\mathbb{Z} = \{an | n \in \mathbb{Z}\}$, you only get to step a distance $a$. Second order derivatives are constructed from nearest neighbor differences of derivatives, so you can travel a distance $2a$ when constructing 2nd order derivatives. Likewise, for $n$-th order derivatives, you can see information that is distance $na$ from you in the lattice.

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If you want a simple intuitive explanation, you can get a lot from vehicles.

In a car traveling at a constant speed, suppose there is a white dot painted on the top of the steering wheel. If that dot is in the center, you are traveling in a straight line. If you turn it some angle to the left, say 90 degrees, then the car is traveling in a circular arc at a constant lateral acceleration. That is the second derivative of lateral position.

If you turn the steering at a constant rate from 0 degrees to 90 degrees, then the rate of lateral acceleration is changing constantly while you are turning the wheel. That is jerk, and it is constant because you are turning the steering wheel at constant speed. It is the third derivative of lateral position.

(While you are doing this, the path traced by the car is a spiral of linearly increasing curvature. Highway and railway curves are built using these spirals to connect segments of constant curvature. Such a spiral gives a place to gradually bank the roadway, and without it drivers tend to cross lanes, and trains actually jerk when starting or ending a curve.)

However, if you don't turn the steering wheel at a constant rate, but rather accelerate it leftward from 0 degrees until you are turning it quickly at 45 degrees, and then decelerate it until it reaches 90 degrees, then you are giving it a doublet of snap, first positive, then negative, and that is the fourth derivative of lateral acceleration.

Another vehicle to illustrate it is a submarine having bow planes to control depth. Suppose there is a motor that rotates the bow planes at a constant speed, up, down, or 0. The angle of the bow planes determines the pitch rate of the submarine. The pitch angle of the submarine determines the rate of change of depth.

So the pitch angle is proportional to the first derivative of depth, the bow plane angle determines the second derivative, and the speed of the bow plane motor is the third derivative.

Also, take a rocket with gimballed engines. If the thrust line of a rocket engine does not go through the rocket's center of mass, then it produces angular acceleration, or the second derivative of directional orientation. There are motors that move the engine gimbals, and the rate at which they move them determines the third derivative of direction.

I'm sure you can think of other examples.

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Its easy, f' tells you how fast f is changing, f'' tells you how fast the change is changing etc...

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Hmm, this really doesn't answer the question--it gives no "physical intuition". Could you edit this post to a more substantial answer giving physical intuition? –  Manishearth Dec 10 '12 at 9:32
    
@Manishearth You must know exactly the definition of "physical intuition" then, could you enlighten us? –  Hobo2 Dec 10 '12 at 14:53
    
Basically using our intuitive image of the real world to analyse something. The OP already knows that f' is the rate of change of f, etc. Arnold's answer above has the same issue. Though you're right, this question isn't really that clear. –  Manishearth Dec 10 '12 at 14:59
    
@Manishearth How do you know what the OP knows? What if someone besides the OP is reading this question? Some of the other thousands of users this site is intended for? –  Hobo2 Dec 11 '12 at 9:25
    
@Hobo2: Definitely Manish doesn't know what the OP really knows. But, his question does indicate that "he knows" the rate of change of (rate of change of...) and how many other orders you continue inserting..! :-) –  Waffle's Crazy Peanut Dec 11 '12 at 9:37
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Higher derivatives of position are related to "generalized curvatures". In 3D, for instance, the derivative of acceleration is secretly related to the torsion of a curve. The hint is the Frenet-Serret (binormal, normal, tangent) triplet or the so-called repere mobile (a la Cartan). A higher dimensional extension of this moving reference frame does exist for higher derivatives. So, higher order derivatives can be mathematically imagined to be different classes of "classical curvatures". For instance, the sign of f''(0) is related to the maximum/minimum character of f'(0)=0. Usually, people work with theories in which only "curvature" matters and torsion is usually neglected. It happens with general relativity, but some theories include torsion terms. Fluid dynamics, indeed, relates torsion to vorticity/vortex lines.

With respect to the classical meaning of some variables, we have some interesting interpretations of derivatives of position (in the physical sense, not in the mathematical sense I have spoken here as related to curvature/torsion). See my link below.

Moreover, another physical intuition for higher derivatives in modern physics comes from Cosmology. I recall that the so-called statefinders of the distance-scale factor in General Relativity relates Hubble constant to "an acceleration", and higher derivatives are also related to cosmological observables.

My post about derivatives of position and "their meaning":

derivatives of position explained

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