Deriving The Work-Energy Theorem

  • #1

Homework Statement



Trying to derive the work-energy theorem, without manipulating differentials.


Homework Equations



a=[itex]\frac{dv}{dt}[/itex] v=[itex]\frac{dx}{dt}[/itex]
W=[itex]\int[/itex] F dx =ΔKE=[itex]\frac{1}{2}[/itex]mvf[itex]^{2}[/itex]-mvi[itex]^{2}[/itex]

The Attempt at a Solution



F=ma

[itex]\int[/itex] F dx=m[itex]\int[/itex]a dx

=m[itex]\int[/itex][itex]\frac{dv}{dt}[/itex]dx <-- I cannot continue, unless I start using the differentials as fractions. Can you move forward without thinking of them as fractions? Or if you choose to use them as fractions could you justify the act?
 
Last edited:

Answers and Replies

  • #2
Andrew Mason
Science Advisor
Homework Helper
7,656
383

Homework Statement



Trying to derive the work-energy theorem, without manipulating differentials.


Homework Equations



a=[itex]\frac{dv}{dt}[/itex] v=[itex]\frac{dx}{dt}[/itex]
W=[itex]\int[/itex] F dx =ΔKE=[itex]\frac{1}{2}[/itex]mvf[itex]^{2}[/itex]-mvi[itex]^{2}[/itex]

The Attempt at a Solution



F=ma

[itex]\int[/itex] F dx=m[itex]\int[/itex]a dx

=m[itex]\int[/itex][itex]\frac{dv}{dt}[/itex]dx <-- I cannot continue, unless I start using the differentials as fractions. Can you move forward without thinking of them as fractions? Or if you choose to use them as fractions could you justify the act?
dv is just the arbitrarily small change in velocity occurring in the arbitrarily small time interval dt. So long as dx is the arbitrarily small change in displacement in that same time interval, and you can effectively treat them as fractions. So, (dv/dt)dx = (dx/dt)dv

Does that help?

AM
 
Last edited:
  • #3
vanhees71
Science Advisor
Insights Author
Gold Member
17,077
8,179
Keep in mind that velocity and force are vectors and energy is a scalar. Write down Newton's equation of motion and think about, how to derive a scalar equation and then integrate. Just another hint: Think about, how work is defined in terms of the force and the trajectory of the particle as a function of time!
 
  • #4
UltrafastPED
Science Advisor
Gold Member
1,912
216
I always use F=dp/dt in vector form. But then I manipulate the differentials - this is permitted by the chain rule.

So perhaps you should go to the definitions, and work it out from that perspective. Clearly the problem is intended to be instructional, and working with definitions is always that.
 
  • #5
dv is just the arbitrarily small change in velocity occurring in the arbitrarily small time interval dt. So long as dx is the arbitrarily small change in displacement in that same time interval, and you can effectively treat them as fractions. So, (dv/dt)dx = (dx/dt)dv

Does that help?

AM
Yes, this is how I was taught to derive this theorem. I still hope for something more, since I have not looked at non-standard analysis and have not defined differentials in a rigorous way.
 
  • #7
Keep in mind that velocity and force are vectors and energy is a scalar. Write down Newton's equation of motion and think about, how to derive a scalar equation and then integrate. Just another hint: Think about, how work is defined in terms of the force and the trajectory of the particle as a function of time!
My equations are 1 dimensional, force is acting in the positive or negative x-axis. Isn't that a scalar equation from newtons's equation of motion? Work is defined as the sum of all contributions of force in the direction of motion multiplied by their respective distances? I'm sorry I don't see how these hints help.
 
  • #8
I always use F=dp/dt in vector form. But then I manipulate the differentials - this is permitted by the chain rule.

So perhaps you should go to the definitions, and work it out from that perspective. Clearly the problem is intended to be instructional, and working with definitions is always that.
That is interesting. How does the chain rule allow you to manipulate differentials as fractions.
 
  • #9
vanhees71
Science Advisor
Insights Author
Gold Member
17,077
8,179
UltrafastPED (unfortunately) posted already the answer in his link :-(.
 
  • #11
6,054
391
## dv = \frac {dv} {dt} dt ## holds pretty much by definition, so whether you interpret that as if the derivative were a ratio of differentials is a matter of style more than anything else.
 
  • #12
## dv = \frac {dv} {dt} dt ## holds pretty much by definition, so whether you interpret that as if the derivative were a ratio of differentials is a matter of style more than anything else.
I do not think this is a matter style. The derivative is based on a limit, and is given the notation[itex]\frac{dy}{dx}[/itex]. For you to treat them as if they were fractions is quite strange to me.

How does that hold by definition? How can you isolate dv like that? It seems intuitively to be correct, but still not enough to be considered justified.
 
  • #14
1,540
135
  • Like
Likes 1 person
  • #15
6,054
391
I do not think this is a matter style. The derivative is based on a limit, and is given the notation[itex]\frac{dy}{dx}[/itex]. For you to treat them as if they were fractions is quite strange to me.
In fact, historically it was exactly how they were treated and this is why the notation is in terns of fractions.

How does that hold by definition? How can you isolate dv like that? It seems intuitively to be correct, but still not enough to be considered justified.
Give us your definitions of the differential and the derivative and we can discuss that.

One definition of the differential is that if ## f(x) ## is given, and if there exist functions ## df(x, z), \ R(x, z) ## such that ##df(x, z)## is linear in the z-argument and ## f(x + z) - f(x) = df(x, z) + R(x, z) ## and moreover ## \lim_{z \to 0} R(x, z)/z = 0 ##, then ## df ## is the differential of the function.

It is then trivial to show that if the derivative ## f'(x) ## exists, then ## df(x, z) = f'(x)z ##. Now if we use ## dx ## in the stead of ## z ##, and the symbol ## \frac {df} {dx} ## for ## f'(x) ##, we obtain ## df = \frac {df} {dx} dx ##.
 
  • #16
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
38
There is no need to split up the derivative (in Leibniz notation) as though it were a fraction. For your integrand, you have, by the chain rule:[tex]\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}[/tex] Where the last equality follows because dx/dt = v by definition. Now, if you examine the rightmost expression, you can see that it looks like something that has been differentiated w.r.t. x using the chain rule. If you "reverse" this application of the chain rule, you can turn the expression into something from which the kinetic energy will follow.
 
  • Like
Likes 1 person
  • #17
In fact, historically it was exactly how they were treated and this is why the notation is in terns of fractions.



Give us your definitions of the differential and the derivative and we can discuss that.

One definition of the differential is that if ## f(x) ## is given, and if there exist functions ## df(x, z), \ R(x, z) ## such that ##df(x, z)## is linear in the z-argument and ## f(x + z) - f(x) = df(x, z) + R(x, z) ## and moreover ## \lim_{z \to 0} R(x, z)/z = 0 ##, then ## df ## is the differential of the function.

It is then trivial to show that if the derivative ## f'(x) ## exists, then ## df(x, z) = f'(x)z ##. Now if we use ## dx ## in the stead of ## z ##, and the symbol ## \frac {df} {dx} ## for ## f'(x) ##, we obtain ## df = \frac {df} {dx} dx ##.


Thanks for this, I can sort of follow it. R(x,z) is some sort of remainder function?
I don't have a definition of the differential, which is my problem.
The definition of the derivative which I have been using is

lim h->0 [itex]\frac{f(x+h)-f(x)}{h}[/itex]

Could you recommend some problems to look at? Or some sources to read to get a better grasp of this idea. I have been hung up about this for a while now.
 
  • #18
There is no need to split up the derivative (in Leibniz notation) as though it were a fraction. For your integrand, you have, by the chain rule:[tex]\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}[/tex] Where the last equality follows because dx/dt = v by definition. Now, if you examine the rightmost expression, you can see that it looks like something that has been differentiated w.r.t. x using the chain rule. If you "reverse" this application of the chain rule, you can turn the expression into something from which the kinetic energy will follow.
m[itex]\int[/itex][itex]\frac{dv}{dt}[/itex]dx

=m[itex]\int[/itex][itex]\frac{dv}{dx}[/itex][itex]\frac{dx}{dt}[/itex]dx

=m[itex]\int[/itex]v[itex]\frac{dv}{dx}[/itex]dx

[itex]\frac{d(1/2v^2)}{dx}[/itex] =1/2*(2v)*[itex]\frac{dv}{dx}[/itex]=v*[itex]\frac{dv}
{dx}[/itex]

∴m[itex]\int[/itex]v[itex]\frac{dv}{dx}[/itex]dx=m[itex]\int[/itex][itex]\frac{d(1/2v^2)}{dx}[/itex]dx

=[itex]\frac{1}{2}[/itex]mv2!

Thank you everyone for your help :).
 
  • #19
6,054
391
I would recommend Courant's Differential and Integral Calculus. It is freely available online: http://archive.org/details/DifferentialIntegralCalculusVolI

Section II.3.9 (p. 109) explains the connection between the derivative and the differential; II.3.10 discusses why in physics and natural science it is perfectly valid to treat the derivative as a quotient.

More modern books tend to omit the entire discussion of differentials in single-variable calculus (which they can get away with). When it comes to multi-variate calculus, they have to define the differential and discuss its properties; even then, however, they may not use the name "differential" - but you should be able to recognize one via the definition I gave earlier. The whole idea of the differential, in single and multi-variate calculus, is that it is a linear approximation to a function at a point; and the derivative is the slope of the approximation. It is because of the linearity that we can use the quotient trickery.
 
  • Like
Likes 1 person
  • #20
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
38
m[itex]\int[/itex][itex]\frac{dv}{dt}[/itex]dx

=m[itex]\int[/itex][itex]\frac{dv}{dx}[/itex][itex]\frac{dx}{dt}[/itex]dx

=m[itex]\int[/itex]v[itex]\frac{dv}{dx}[/itex]dx

[itex]\frac{d(1/2v^2)}{dx}[/itex] =1/2*(2v)*[itex]\frac{dv}{dx}[/itex]=v*[itex]\frac{dv}
{dx}[/itex]

∴m[itex]\int[/itex]v[itex]\frac{dv}{dx}[/itex]dx=m[itex]\int[/itex][itex]\frac{d(1/2v^2)}{dx}[/itex]dx

=[itex]\frac{1}{2}[/itex]mv2!

Thank you everyone for your help :).
Glad to be of help and glad you worked it out. By the way, I wasn't trying to negate what others in the thread were saying about how you will get the right answer (in most situations) by using "differentials" or treating dy/dx as a fraction. That is true. In fact, you can think of my post as answering your question in post #8 (why does the chain rule allow you to treat derivatives as fractions) by showing that, at the very least, the two methods are mutually consistent.
 
Last edited:
  • #21
I would recommend Courant's Differential and Integral Calculus. It is freely available online: http://archive.org/details/DifferentialIntegralCalculusVolI

Section II.3.9 (p. 109) explains the connection between the derivative and the differential; II.3.10 discusses why in physics and natural science it is perfectly valid to treat the derivative as a quotient.

More modern books tend to omit the entire discussion of differentials in single-variable calculus (which they can get away with). When it comes to multi-variate calculus, they have to define the differential and discuss its properties; even then, however, they may not use the name "differential" - but you should be able to recognize one via the definition I gave earlier. The whole idea of the differential, in single and multi-variate calculus, is that it is a linear approximation to a function at a point; and the derivative is the slope of the approximation. It is because of the linearity that we can use the quotient trickery.
I will check it out, thanks!
 

Related Threads on Deriving The Work-Energy Theorem

  • Last Post
Replies
2
Views
10K
  • Last Post
Replies
1
Views
3K
Replies
2
Views
839
Replies
6
Views
1K
Replies
3
Views
2K
Top