# Basic kinematics confusion all derivatives 0?

1. Dec 10, 2011

1. The problem statement, all variables and given/known data

Let's say an object is moving with x(t)= e^(-1/t^2)
it's motion is continuous everywhere and differentiable because its exponential
so v(t)= 2e^(-1/t^2)/t^3
a(t)= e^(-1/t^2) *(4-6t^2)/t^6

I have asked to a calculus teacher and he said all the derivatives will be 0 at t=0
yet if all the derivatives are 0 then one would not expect the object to move in time but for any t>0 x(t) >0 but there is no velocity to move the object ,no acceleration to increase the velocity and no jerk to make acceleration etc
so whats causing them motion?
I am talking about only at t=0
i know the velocity isnt 0 other where but i am saying
at t=0the velocity is 0 and it shoulda stayed 0 so is acceleration etc because all the derivatives that may increase it is 0 i asked this before and got a reply y=x^4 has derivative 0 at t=0 but still increase though d^4(y)/dx^4 = 4 which is a positive number that increase d the third derivative to increase the second derivative to increase the first derivative to increase the function value so please dont give me the same unsatisfactory reply :X
2. Relevant equations
v=dx/dt
a=dv/dt

3. The attempt at a solution

2. Dec 10, 2011

should i ask this in the math section cause even though its a physics question i think the answer will be better by a calculus expert?

3. Dec 10, 2011

### Delphi51

Most interesting! I'm stuck at your statement that ALL the derivatives are zero at t=0. Were you given any evidence of that? It is getting more complex with each differentiation and the balance between two infinitesimals may not hold out forever! It would take forever to check! I would definitely suggest posting this question over in the math forum as well.

4. Dec 11, 2011

I wasn't given an evidence of it but I did ask a calculus professor and he said that he had encountered this question before in term of infinite series and said that all the derivatives are equal to 0 and this function is not equal to it's maclurian series representation, though I don't know if this helps.

5. Dec 11, 2011

bump :X

6. Dec 11, 2011

### ehild

Neither the function nor its derivatives are defined at t=0. There can not be assigned any motion or process that starts at t=0 to this function.

ehild

7. Jun 19, 2012

### utesfan100

The limit of the function on both sides is 0, so the discontinuity is removable.

A more interesting function is

x(t) = 0, |t|>=1
x(t) = e^(1-(1-t^2)^-2), |t|<1

This displacement starts at rest, at -1 begins to increase, reaches a maximum of 1 at 0, then decreases and flattens out at 1, after which it is at rest.

This function is infinitely differentiable at all points, starts and stops at rest in finite time, yet is not 0 everywhere. Such a function is called a bump function. Like tortoises, the derivatives are bumps, all the way down.

The real confusion here appears to be that velocities don't cause displacement, accelerations don't cause changes in velocity and jerks don't cause acceleration. Rather, they measure the magnitude of these. In fact, these must also be bump functions.

Such a displacement function would still be caused by a force proportional to the acceleration, which would also be a bump function.