# The product rule applied to the equation for work

• kmarinas86
In summary: This means that power is equal to force times velocity, which is a more intuitive expression for power. The formula you were using before, although mathematically correct, is not physically meaningful as it involves a change in force over time, which would imply that power can exist even when force is zero. This is not possible according to the definition of power, which is the rate at which work is done. Therefore, the term "distance*(change of force/change in time)" is not a physically meaningful component of the formula for power.

#### kmarinas86

This is NOT homework...

If one takes the formula for work, which is force times distance, one can get the formula for power by taking the derivative of it with respect to time. But if we really do this correctly (by using calculus), we get:

power=distance*(change of force/change in time)+force*(change in distance/change in time)
power=distance*(change of force/change in time)+force*velocity

Now we have two terms on the RHS that are not necessarily equal. However, because energy must be conserved, it would seem that that the two terms are somehow separable. In what ways is the first term "physical"? The question refers to the term "distance*(change of force/change in time)" It would suggest that power exists even when force is zero. There is something that does not make sense here...

Imagine that we do the following:

power=distance*(change of force/change in time)+force*velocity

Multiply by distance and divide by velocity:

power*distance/velocity=(distance^2/velocity)*(change of force/change in time)+force*distance

(power*distance/velocity)-((distance^2/velocity)*(change of force/change in time))=force*distance

Now taking the fact that force*distance=work=power*time ...

(power*distance/velocity)-((distance^2/velocity)*(change of force/change in time))=power*time

power-distance*(change of force/change in time)=power

-distance*(change of force/change in time)=0

There is something that really does not make sense here...
.........

Let's put it another way.

Let's assume that power is EQUAL to force*(change in distance/change in time). But in order for this to occur, it cannot equal the time derivative of (force times distance) determined by applying product chain rule (as shown in the first equation of this post). Clearly, we end up trading something that does NOT make sense with another thing that does NOT make sense!

If we integrate this with respect to time, we are left with the problem of choosing a constant.

In light of the definition for power:

time derivative of (force*distance)=distance*(change of force/change in time)+force*(change in distance/change in time)

time derivative of (force*distance)=distance*(change of force/change in time)+time derivative of work

time derivative of work=time derivative of (force*distance)-distance*(change of force/change in time)

This is wacky. There must be something in here that I DON'T understand...

I would use: dW = F dx, thus p = dW/dt = F dx/dt = Fv. (Where F and x are parallel.)

To be honest, it's quite hard to follow because of all the equations have no symbols (velocity = v, etc).

kmarinas86 said:
This is NOT homework...

If one takes the formula for work, which is force times distance, one can get the formula for power by taking the derivative of it with respect to time. But if we really do this correctly (by using calculus), we get:

power=distance*(change of force/change in time)+force*(change in distance/change in time)
power=distance*(change of force/change in time)+force*velocity

The formula for work is better expressed as dW = Fdx, in which F could be taken as a function of x. To evaluate the expression further:
W = ∫Fdx ---- This is what you would derive. What you were deriving was W = Fx, which is valid for when the force is constant along the distance covered; for a constant force, the expression for power you gave would come down to F*dx/dt = F*v considering that dF/dt = 0.

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Just wanted to emphasize the following point:

Gear300 said:
What you were deriving was W = Fx, which is valid for when the force is constant along the distance covered; for a constant force, the expression for power you gave would come down to F*dx/dt = F*v considering that dF/dt = 0.

## 1. What is the product rule applied to the equation for work?

The product rule is a mathematical rule that is used to find the derivative of a product of two functions. In the context of work, it is used to find the derivative of the product of force and displacement, which is the equation for work.

## 2. How is the product rule applied to the equation for work?

The product rule states that the derivative of a product of two functions is equal to the first function multiplied by the derivative of the second function, plus the second function multiplied by the derivative of the first function. In the equation for work, this means that the derivative of force multiplied by displacement is equal to force multiplied by the derivative of displacement, plus displacement multiplied by the derivative of force.

## 3. Why is the product rule important in the context of work?

The product rule is important in the context of work because it allows us to find the rate at which work is being done at any given point. This is useful in understanding and analyzing systems that involve work, such as machines or moving objects.

## 4. Can the product rule be applied to other equations besides the equation for work?

Yes, the product rule can be applied to other equations that involve the product of two functions. It is a fundamental rule in calculus and is used in many different contexts, such as finding the derivatives of exponential and trigonometric functions.

## 5. Are there any limitations or exceptions to the product rule?

While the product rule is a powerful tool in finding derivatives, it does have some limitations and exceptions. These include cases where the functions are not differentiable or when the functions are not continuous. In these situations, alternative methods may need to be used.