# Higher order time derivatives of position

1. Mar 26, 2013

### V0ODO0CH1LD

Newton's laws says $F=ma$. Which, as far as I can see, states that all physical interactions concern the second time derivative of position. And because there is no other way for two bodies to interact in the physical world, the "worst" I can do to a system is change its acceleration, right? My question is: are higher (than three) order derivatives of position with respect to time valid physical concepts? I can clearly see why the third time derivative of displacement, jerk, would be a useful concept. But what I can't see is how defining anything beyond that makes any sense. How can I change the rate at which a body changes its acceleration? I mean, if the only thing I can do is apply more or less force to a body, therefore changing its acceleration. What would I have to do to change anything beyond that?

The other thing is: if I have some equation that relates time and position as $x=t^3$. I have no good reason to stop taking derivatives at the second time around. So the actual "equation of motion" for a particle that behaves in a manner described by that equation would be:
$$x(t)=x_0+\dot{x_0}t+\frac{\ddot{x_0}}{2}t^2+\frac{\dddot{x_0}}{6}t^3$$
by the taylor series approximation.

So is it the point that I could never, in reality, do something to a body so that it behaved in such a way that it's trajectory would obey higher (than three) order polynomial approximations, or is it that we just neglect the higher order derivatives by saying; sure.. the equation says $x=t^n$ but the acceleration is $(n-1)nt^{n-2}$ and thats all we need!

What part of mathematics are Newton's laws constraining?

2. Mar 26, 2013

### AlephZero

I think you are forgetting that $F$ can be a function of time, and have with as many derivatives as you like. You seem to be assuming that $F$ is constant.

Of course $x(t)$ can have more than two time derivatives in a "real world" situation. An obvious example is simple harmonic motion where $x(t) = \sin \omega t$.

3. Mar 27, 2013

### SteamKing

Staff Emeritus
Look up articles on 'jerk' and 'jounce', which are the third and fourth time derivatives.

4. Mar 27, 2013

### cosmik debris

Sometimes the higher derivatives are called jerk, snap, crackle, and pop. Some of these higher orders are used in camshaft design.

5. Mar 27, 2013

### sophiecentaur

I remember being very confused by the equations of motion (SUVAT) that we were taught at School. Not enough emphasis was placed on the fact that they applied for Uniform Acceleration. I had to sort it out in my own mind (some time later) because I was too embarrassed to ask!

6. Mar 27, 2013

### V0ODO0CH1LD

I see! Force concerns acceleration. Which means there is no "type" of force I could apply onto a body that would allow me to interact with any higher time derivatives of position than two. Any force ever results in an acceleration. But the point is that a force may vary with respect to time, which means my acceleration would also vary with respect to time and depending on how that force is related to time, higher and higher time derivatives of position may also be nonzero. What was confusing me is that it's not what force "messes" with that is changing with respect to time, just the force it self.

7. Mar 28, 2013

### sophiecentaur

Most forces that you come across in everyday life are not uniform (with the exception of g, over 'small' distances). Force from your arm, from a car engine and, internally, from the gas expansion in the cylinder plus the effect of the crank, changing reciprocating motion to circular. Then there's the force in an electric motor, from a sky-rocket engine (and the also changes with time), the force from the rubber band on a slingshot etc. etc..

Constant force / acceleration is a theoretical concept rather than a reality.