# Checking simple differential product rule

1. Dec 20, 2014

### Matthew Travers

1. The problem statement, all variables and given/known data
Saw a calculation that put differentiation of power in terms of acceleration as follows:
E=Fs
dE/dt=Fv=P
dP/dt=Fa=ma^2

It doesn't make sense to me because if power was changing, acceleration must change. Correct me if I'm wrong, but shouldnt the product rule be applied here?

2. Relevant equations
If z=xy then
dz/dt=x(dy/dt)+y(dx/dt)
F=ma
E=Fs
P=F(ds/dt) with s(dF/dt)=0 so it is discarded

3. The attempt at a solution
Instantaneously
P=Fs/t=F(ds/dt) provided ds/dt is smaller than c. Now first application of product rule with second diff of position defined as first diff of velocity
dP/dt=F(dv/dt)+(ds/dt)(dF/dt) since F=ma, subst in and apply product rule again
dP/dt=ma^2+(ds/dt)[m(da/dt)+a(dm/dt)]

Last edited: Dec 20, 2014
2. Dec 20, 2014

### Ray Vickson

3. Dec 20, 2014

### ehild

Valid only for constant force.

dP/dt = Fa is only valid if the force is constant. It can be pulled out from the differentiation then and you get dP/dt=F dv/dt = F a. In case of constant force, the acceleration is also constant, but the power changes with time.
When the force depends on time or position, apply the product rule.

Last edited: Dec 21, 2014
4. Dec 21, 2014

### Matthew Travers

First of all, thank you for the feedback I appreciate the distinction for the first equation and pointing it out for me.
I must be a bit thick. To me it seems that if acceleration is constant, then energy is being delivered at a constant rate, ie power stays constant and therefore rate of power change should be zero.
I dont understand how a constant force causing a constant acceleration causes a change in the rate of energy applied. If I drop a ball, force and acceleration and rate of increase of kinetic energy are all constants.

What am I missing here?

5. Dec 21, 2014

### Ray Vickson

You are missing the fact that the kinetic energy involves v^2, so its derivative is proportional to v, even when a = dv/dt is constant.

6. Dec 22, 2014

### Matthew Travers

Thanks mate. I had missed that....so energy rises parabolically, power rises linearly, and change in power is a constant when force and acceleration are constant.

7. Dec 22, 2014

### Matthew Travers

One last question please? When I applied the product rule above, I ended up with a a(dm/dt) term. Does this have a physical meaning?

8. Dec 22, 2014

### Ray Vickson

It would in a variable-mass problem (eg., rockets). However, in a constant-mass problem you would have dm/dt = 0, so that term would not make a contribution.

9. Dec 22, 2014

### Matthew Travers

Thanks Ray and Ehild for your indulgence and patience. Merry Christmas and a happy new year to you and all the staff at physics forum. It's a pleasure becoming that little less ignorant.

10. Dec 22, 2014

### ehild

Merry Christmas and happy New Year to you, too.