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Calculus breakdown

  1. May 30, 2010 #1
    quick question, I was watching an MITOCW physics lecture and I want to know where a principle breaks down.

    [tex]\int^t_0 \frac{dp}{dt} dt = \int^{p(t)}_{p(0)} dp [/tex]

    how do I know such a thing is allowed? where does this break down? does it break down if p or t does not have property such as smoothness?

    here is the little section of the lecture where I saw it:
    http://www.youtube.com/watch?v=Lkuo6nZ6nZM#t=12m30s

    I am never comfortable until I know what is being assumed and where something breaks. Thanks!
     
  2. jcsd
  3. May 30, 2010 #2
    Well dp/dt has to exist for a start. p(t) probably needs to be bijective as well (monotonic will do).

    This thing is the chain rule. (It's everywhere, look it up)
    Or integration by substitution (I think there's a proof on Wikipedia that isn't too bad)
     
  4. May 30, 2010 #3
    chain rule has this form:
    092a91734fb9c50d36fba56732186ba2.png

    since the forms are not exactly the same, I am concerned about subtleties.
     
  5. May 30, 2010 #4
    perhaps p(t) has to be differentiable with respect to t, not just piecewise differentiable? otherwise dp/dt would not be defined for certain points?
     
  6. May 30, 2010 #5
    [edit] Oh, let me think about that.
    [edit again] Integration ignores isolated "points", heuristically (sets of zero measure? I don't know any measure theory). Piecewise is fine.


    It's basically the chain rule in the opposite direction.

    meow
    And let f=1 after you get that proof.
    Also, try not to write
    [tex]
    \int^t_0 \frac{dp}{dt} dt
    [/tex]
    Because the 't' inside the integral is like a dummy variable. It's not a big deal usually though.

    Integration is, intuitively, "more likely" to work than differentiation (you can integrate some discontinuous curves, for example...but there's no chance to differentiate on both sides of a discontinuity). As long as derivatives exist, we should be okay. Don't quote me on that though. Forgot all my analysis.

    [edit] We can integrate over piecewise discontinuity just fine. (I hate to say it, but using the 'area under a curve' analogy, we can see there's no problem.[/edit]

    Physical examples are unlikely to involve anything close to pathological functions.
     
  7. May 30, 2010 #6
    yep I know integrals are quite robust

    does not [tex]\frac{dp}{dt}[\tex] on its own imply that p(t) is smooth, or that the derivative is continuous since otherwise the limit could have 2 values at a particular time t (one limit coming from the right (+) and one coming from the left(-))
     
  8. May 30, 2010 #7

    Cyosis

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    We have a chain rule for a differentials as well, [itex]dp=p'(t)dt[/itex]. For example if [itex]p=t^3[/itex] then [itex]dp=3t^2dt[/itex]. Furthermore with the integral given in the OP you know for sure that [itex]p'(t)[/itex] exist and if a function is differentiable it is of course smooth and continuous.
     
  9. May 30, 2010 #8
    Oh, right.
    Okay.
    Well. Usually, yeah.

    But p(t)= -t^2 for t<0 and t^2 for t>0 and p=0 at t=0 isn't smooth, is it?

    When we move to physics though, we don't mind if dp/dt doesn't exist at some particular points.

    If a function is differentiable, it doesn't have to be smooth, but it does have to be continuous.
    A bunch of pathological functions involving sin(1/x) are going to be differentiable once, but not smooth.
     
  10. May 30, 2010 #9

    Cyosis

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    You're right I was sloppy in my use of smooth. Functions are smooth if derivatives of any order exist, not just one. However his question, as I understood it, was how you go from the first integral to the second. If the first integral doesn't exist, because p'(t) does not exist then there is no point in continuing. The question only makes sense if [itex]p(t)[/itex] is differentiable in the first place.
     
    Last edited: May 30, 2010
  11. May 30, 2010 #10
    exactly I am comfortable with where integrals exist. I am trying to explore where
    [tex]\frac{dp}{dt} dt = dp [/tex]
    breaks down. It seems the only way for us to break this is to make
    [tex]\frac{dp}{dt}[/tex]
    not exist. Do you think this is right or could it break some other way?
     
  12. May 30, 2010 #11

    Cyosis

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    Are you aware that this is just the substitution rule for integrals?

    Substitution rule:
    [tex]
    \int g(f(x))f'(x)dx=\int g(y)dy
    [/tex]

    with [itex] y=f(x) [/itex]

    This works if f is differentiable with a continuous derivative, if g is continuous and if the composition [itex] g \circ f[/itex] exists.
     
  13. May 30, 2010 #12

    The Lebesgue-Radon-Nikodym theorem goes into that. You need to know a bit of measure theory, treating the integrals of functions as measures.
     
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