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Odd integration problem

  1. Oct 6, 2013 #1
    1. The problem statement, all variables and given/known data

    In the process of solving a physics problem, I've run across an integration that I'm not sure I've ever had to perform before. It's of the form $$\int \frac{1}{a - (\frac{dy}{dx})(x)} dx$$ where ##a## is a constant.

    2. Relevant equations

    Table integral?

    3. The attempt at a solution

    I honestly have no clue where to even begin with this. I do not have a functional form for ##y##, it's simply a quantity that changes with respect to ##x##. Any hints on where to begin?
     
  2. jcsd
  3. Oct 6, 2013 #2

    Ray Vickson

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    Your notation is confusing: is the denominator supposed to be ## 1 - a x y'(x)## or ##1-a y'(x)##, where ##y'(x) = dy(x)/dx.## That is, is the (x) an argument of f', or does it multuply f'(x)?

    As far as I can see you can write more-or-less ANY integrand h(x) in that form: just put
    [tex] h(x) = \frac{1}{a-y'(x)} \Longrightarrow y'(x) = \frac{h(x)-1}{a h(x)} [/tex] if your integrand is 1/(1-ay'), or put
    [tex] h(x) = \frac{1}{a - x y'(x)} \Longrightarrow y'(x) = \frac{1-h(x)}{a x h(x)}[/tex]
    if your integrand is 1/(1-a*x*y').

    So, to summarize: your integrand can be anything at all, so there is no way to do the integral.
     
  4. Oct 6, 2013 #3
    I'm sorry, ##x## is being multiplied in the denominator. Well, um, that's certainly disappointing. I guess it's back to the drawing board. Thank you for your help!
     
  5. Oct 6, 2013 #4

    Dick

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    If it's a physics problem you probably know something about the relation between y' and x. What kind of problem is it?
     
  6. Oct 6, 2013 #5
    Momentum transport. In this case, ##x## is time, and ##y'(x)## is the change in mass with respect to time. So really the integral is $$\int_0^t \bigg(\frac{1}{M + m - (t')(\frac{dm}{dt'})}\bigg) dt'$$ where ##t'## is a dummy variable. ##M## and ##m## are the masses of a truck and of an original mass of sand, respectively, so they're constants, and ##(t)(\frac{dm}{dt})## is the amount of sand that falls out the truck after a time ##t##.
     
  7. Oct 6, 2013 #6

    Dick

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    Since you are multiplying t' times dm/dt', I think it's pretty likely you are assuming the rate of mass change dm/dt' is constant. Are you?
     
  8. Oct 6, 2013 #7
    Oh no, you're right, I certainly am! So I should be able to manipulate that into something of the form $$\int \frac{1}{1 + x} dx$$ right? Which is just ##ln(1 + x)##?
     
    Last edited: Oct 6, 2013
  9. Oct 6, 2013 #8

    Dick

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    Sure, so the integral isn't a real problem. Now I'd worry that the integral is computing something like the average value of 1/mass over a portion of the trip, which doesn't sound very useful for momentum problems, but then I don't know exactly what you are doing.
     
  10. Oct 6, 2013 #9
    I'm actually calculating the impulse and dividing by the mass to get velocity, but the force in this particular case happens to be time independent, so I just pulled it out of the integral.
     
  11. Oct 6, 2013 #10

    Dick

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    Ok, carry on.
     
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