Comments on: Waiter, There’s a Derivative in my Cereal

By: Nerd Alert! My mind is gone » Undress Me Robot

Nerd Alert! My mind is gone » Undress Me Robot — Mon, 14 Jul 2008 06:17:45 +0000

[...] Fifth and sixth derivatives – The first and second derivatives (velocity and acceleration, respectively) are easy enough to grasp, but the third and fourth take a bit more thought. Change in position through time? Velocity. Change in velocity through time? Acceleration. Change in acceleration through time? Jerk. But, uh, the fifth and sixth derivatives? Crackle and pop, respectively. [...]

By: Kaleberg

Kaleberg — Fri, 04 Jul 2008 22:23:00 +0000

I was taught that the fourth derivative was called inauguration. A friend of mine who worked for an elevator company confirmed this. Elevator people are very sensitive to higher derivatives, a lot like stomachs. The joke I always heard was that inauguration was change of jerk.

By: Lawrence B. Crowell

Lawrence B. Crowell — Fri, 04 Jul 2008 18:52:19 +0000

A periodic system or a rotation is a case of a C^{infty} function, as e^{-ipx} or cos(px) etc have an infinite number of derivatives. In the case where the angular rotation changes there is a derivative of the angular velocity that is nonzero, where this can be C^k.

In a general setting one can have a Lagrangian of the form

L = L(q, q’, q”, …)

for a large number of primes (time derivatives) on q. Lagrangian systems are a case of Finsler geometry, where for L(q, q’) the one form (dq – vdt), for v = q’, defines the horizonal plus vertical portions of the bundle. For a Lagrangian with higher order derivative on q the bundle one-form extends into various independent vector spaces of dimension n. A related issue is the jet bundle which contains a set or sequence of vector spaces corresponding to higher order differentials.

Lawrence B. Crowell

By: Tom Snyder

Tom Snyder — Fri, 04 Jul 2008 13:35:49 +0000

Mike Webster:

In uniform circular motion, the acceleration vector has a constant magnitude but a changing direction. As the object moves on the circle, the acceleration vector changes its direction so that it always points toward the center of the circle.

So, the time derivative of the acceleration vector (the jerk) is not zero. For uniform circular motion, jerk is a vector with a constant magnitude and a direction always opposite to the velocity. The rate of change of the jerk gives the snap which is nonzero and points radially outward from the center. Crackle, then, would be parallel to the velocity. Etc.

So, according to DanO, the Force would be centripetal, the Yank would be opposite to the velocity, the Tug would be centrifugal, and on and on and on. (Uuhhh, I’m feeling queasy.)

By: Mike Webster

Mike Webster — Fri, 04 Jul 2008 01:15:01 +0000

Hey Claire,

Speed IS a scalar. However, velocity is a vector (where the speed is the magnitude of the vector and the direction is the, umm …, direction of the vector … )

By: DanO

DanO — Thu, 03 Jul 2008 18:36:31 +0000

There are corresponding silly names for derivatives of momentum as well.

Momentum -> Force -> Yank -> Tug -> Snatch -> Shake
Velocity -> Acceleration -> Jerk -> Snap -> Crackle -> Pop

By: Claire C Smith

Claire C Smith — Thu, 03 Jul 2008 02:49:26 +0000

Tom Snyder,

I thought if the rate of change of speed was acceleration and the rate of change of acceleration was the jerk, the constant for the rate of change of acceleration was the jounce.

That’s actually what I meant first time.

Claire

By: Claire C Smith

Claire C Smith — Thu, 03 Jul 2008 02:35:33 +0000

Thought it was, speed a scalar, accleration a vector.

Unless at uniform speed, I would assume that if you accelerate or de-ccelerate on a corner/bend in your car or go in a circle, one of two things, or both will occur,

a) you are going to get pulled up by the coppers, (unless you say your doing a physics experiment in which they will let you off)

b) Your car is wanting to go off on another tangent, direction. Usually outward, to change its position constantly due to the force of acceleration. May the force be with you.

Here I was mainly talking O Level physics stuff. A level goes in to other.

Wierdly I mentioned the jerk, later I meant the jounce (the other pposter said correctly, Kaydubs it was) I actually said the Joust not Jounce by accident. Using the word wrong I loose a point.

The initial thing I meant, thought about the analogy of amusement park rides because of the maximumization of forces to create that, “I feel sick effect”.!

Always been interested in forces, but this is why I like drving too.

Claire

By: Mike Webster

Mike Webster — Wed, 02 Jul 2008 23:02:34 +0000

Tom:
Uniform circular motion, it seems, would have a first order time-dependent linear velocity since the direction is constantly changing (I assume by uniform circular motion you mean a constant circular velocity). But the derivative of that would give a constant acceleration and all the higher derivatives of a constant is 0, no?

By: Tom Snyder

Tom Snyder — Wed, 02 Jul 2008 20:05:02 +0000

Maybe not worth mentioning, but I will anyway. If you’re defining these things as time derivatives of vectors, then a particle in uniform circular motion has velocity, acceleration, jerk, snap, crackle, pop, …(ad inifinitum). (The ratio of the magnitude of any one of these to its predecessor is just the angular velocity of the circular motion.)

Now, if you’re riding on some ride at six flags and going in uniform circular motion, you will *feel* just a steady inertial force. You will not *feel* any jerk (nor any snap, crackle, or pop). That makes me wonder if jerk, etc. would be better defined in terms of rates of change of magnitudes rather than rates of change of vectors. Not sure (and probably nobody cares).