Checking simple differential product rule

  • #1

Homework Statement


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?[/B]


Homework 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[/B]


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)][/B]
 
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Answers and Replies

  • #2
Ray Vickson
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Homework Statement


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
[/B]
In general, the only true equation above is the second one. The first and third are true in case F is constant (that is, does not depend on position or time).

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?



Homework Equations




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)][/B]
 
  • #3
ehild
Homework Helper
15,543
1,909

Homework Statement


Saw a calculation that put differentiation of power in terms of acceleration as follows:
E=Fs[/B]
Valid only for constant force.

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?
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.
 
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  • #4
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
Ray Vickson
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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
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
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
Ray Vickson
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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?
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
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
ehild
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Merry Christmas and happy New Year to you, too.
 

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