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Help with (dx^2/dy^2)

  1. Oct 8, 2008 #1
    is this correct?

    ∫(dx²/dy²) dx = dx/dy
     
  2. jcsd
  3. Oct 8, 2008 #2

    CompuChip

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    What do you mean by (dx^2/dy^2) ?
    Is it the second derivative of x with respect to y? Is it the first derivative of x with respect to y, squared? Is x a function of y (this is a bit unconventional notation, then)
     
  4. Oct 8, 2008 #3
    i am sorry i placed the square on x but it is.. ∫ (d²x/dy²) dx ...
     
  5. Oct 9, 2008 #4
    Suppose [tex]\frac{dx}{dy} = x + 1[/tex].

    Then [tex]\frac{d^2x}{dy^2} = 1[/tex]

    However, it is *not* true that [tex]\int 1 dx = x + 1[/tex]. As your calc teacher will say it, [tex]\int 1 dx = x + C[/tex] for some abuse of notation/constant C.

    The problem with your statement is that the integral isn't the inverse of the derivative. Integrating a derivative of a function doesn't give you back the original function.

    However, the reverse *is* true. Taking the derivative of an integral of a function DOES give you back the original function. This is the fundamental theorem of calculus: [tex]\frac{d}{dx}(\int f(x) dx) = f(x)[/tex]

    (Note that I sort of disregarded your use of the second derivative, since it is just an added layer of confusion to this particular problem).
     
  6. Oct 9, 2008 #5
    so u r saying this is not correct..∫ (d²x/dy²) dx not= dx/dy..

    my basic question is i want to integrate acceleration.so
    ∫ a dx = ∫ (dv/dt) dx since [ a=dv/dt]
    so to solve the above integration i thought
    ∫ a dx = ∫ (dv/dt) dx = ∫ (d²x/dt²) dx since [a=d²x/dt² or v=dx/dt]
    = dx/dt ( i guess this is answer)
    if its not,so what is the answer for this?
     
  7. Oct 9, 2008 #6

    CompuChip

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    Actually,
    ∫ a dt = dx/dt =: v
    by precisely that the theorem mentioned by Tac-Tics: a = d/dt( dx/dt ) so integrating it gives the velocity dx/dt.
    You might be able to rework the integral over x to an integral over t, by a variable substitution:
    ∫ a(x(t)) dx = ∫ a(t) v(t) dt = ∫ v'(t) v(t) dt
    (try to work that out for yourself).

    I am thinking that partial integration may help you forward in this problem, but I'm not quite sure... play around a bit
     
  8. Oct 9, 2008 #7
    I understand the first part ∫ a dt = dx/dt =: v
    but i am confused with rest...
    i want answer for ∫ a dx = ∫ (dv/dt) dx= ?
    dont know how to continue..
     
  9. Oct 9, 2008 #8
    I tried using integration by parts..i dont know is this correct..
    ∫ a dx = ∫ a v dt

    u = v du = dv
    dv= a dt v=v

    ∫ u dv = uv - ∫ vdu

    = v.v - ∫ v dv
    = v² - v²/2 or = v² - x = (dx/dt)² - x
    = v²/2
    = (dx/dt)²/2

    is this correct...please someone help me...
     
  10. Oct 9, 2008 #9
    Maybe my explanation was a little much for what you're trying to do (though it is important to understand it eventually too).

    If you integrate acceleration, the function you get is something that looks a LOT like the function for velocity. In fact, if you know the velocity at any point in time, you can reconstruct the function for velocity entirely.
     
  11. Oct 9, 2008 #10
    Now i am getting this value...

    ∫ a dx = ∫ a v dt

    u = a du = da
    dv= v dt v= a

    ∫ u dv = uv - ∫ vdu

    = a.a - ∫ a da
    = a² - a²/2
    = a²/2
    = (d²x/dt²)²/2

    is this correct...
     
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