# Checking simple differential product rule

• Matthew Travers
I am not a staff member, I am a volunteer moderator. The staff are the people with the blue or red badge. In summary, the conversation discusses the calculation of differentiation of power in terms of acceleration, where the equations E=Fs, dE/dt=Fv=P, and dP/dt=Fa=ma^2 are mentioned. It is pointed out that these equations are only valid for constant force, and when the force depends on time or position, the product rule must be applied. The concept of kinetic energy and its relationship to velocity is also discussed, and the meaning of the term a(dm/dt) in a variable-mass problem is explained.
Matthew Travers

## 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 shouldn't 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]

Last edited:
Matthew Travers said:

## 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 shouldn't the product rule be applied here?

## 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]

Matthew Travers said:

## Homework Statement

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

Matthew Travers said:
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 shouldn't 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.

Last edited:
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 don't 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?

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.

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.

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?

Matthew Travers said:
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.

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.

Merry Christmas and happy New Year to you, too.

## What is the simple differential product rule and when is it used?

The simple differential product rule, also known as the product rule, is a rule in calculus that is used to find the derivative of a product of two functions. It is used when finding the slope or rate of change of a function that is a product of two other functions.

## How do you apply the simple differential product rule?

To apply the simple differential product rule, you must first identify the two functions within the product. Then, you will need to find the derivative of each individual function. Finally, you will use the product rule formula, which states that the derivative of a product of two functions is equal to the first function times the derivative of the second function, plus the second function times the derivative of the first function.

## What are some common mistakes when using the simple differential product rule?

One common mistake when using the simple differential product rule is forgetting to use parentheses when finding the derivatives of the individual functions. Another mistake is forgetting to use the product rule formula and instead trying to find the derivative of the product as if it were a single function.

## Can the simple differential product rule be applied to more than two functions?

Yes, the simple differential product rule can be extended to find the derivative of a product of any number of functions. This is known as the generalized product rule and follows a similar formula to the product rule, but with additional terms for each additional function.

## Why is the simple differential product rule important in science?

The simple differential product rule is important in science because it allows us to find the rate of change of a quantity that is a product of two or more variables. This is especially useful in fields such as physics and chemistry, where many quantities are related to each other through multiplication. It also allows us to find the slope of a curve that represents a relationship between two variables.

• Differential Geometry
Replies
2
Views
587
• Calculus and Beyond Homework Help
Replies
10
Views
937
• Classical Physics
Replies
15
Views
529
• Calculus and Beyond Homework Help
Replies
12
Views
4K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
3K
• Calculus and Beyond Homework Help
Replies
40
Views
3K
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Classical Physics
Replies
31
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
2K