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Help understanding math on inclined plane example

  1. Sep 8, 2009 #1
    I'm trying to follow a very simple example example in a text (Marion/Thornton example 2.1) and I think my rusty calculus is tripping me up and I'm just being stupid.

    I understand how to derive the acceleration down the plane:

    [tex]\ddot{x}=g\sin\Theta[/tex]

    but next they say 'we can find the velocity of the block after it moves from rest a distance x0 down the plane by multiplying by 2x' and integrating.

    [tex]2\dot{x}\ddot{x}=2\dot{x}g\sin\Theta[/tex]

    the next steps confuse me...

    [tex]\frac{d}{dt}(\dot{x}^2)=2g\sin\Theta\frac{dx}{dt}[/tex]

    What happens to the 2 on the left? What am I doing wrong here?
    [tex]2\dot{x}\ddot{x}=2\dot{x}\frac{d}{dt}\dot{x}=2\frac{d}{dt}(\dot{x}^2)[/tex]

    Next, the limits of integration are chosen:

    [tex]\int_{0}^{v^2_0}d(\dot{x}^2)=2g\sin\Theta\int_{0}^{x_0}dx[/tex]

    I have no idea what is happening on the left... where do they pull v^2 0 from? What happens to time? What is going on here again.. last time I had calculus was years ago. Thanks.
     
  2. jcsd
  3. Sep 8, 2009 #2
    2VdV/dt = dV^2/dt
     
  4. Sep 9, 2009 #3

    Redbelly98

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    Staff Emeritus
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    They are using the chain rule from calculus.

    du/dt = (du/dv) (dv/dt) , where u is a function of v -- i.e. u(v)

    Letting u=v2, note that (du/dv) = 2v. So applying the chain rule:

    d(v2)/dt = (du/dv) (dv/dt)
    = (2v) (dv/dt)
    = 2 v dv/dt​

    .

    As for the limits on the integral: note that "v2" is the variable that is being integrated. The limits must be in terms of that variable. And since v2 varies from 0 to vo2, those are the limits of integration.

    Hope that helps. And welcome to Physics Forums!
     
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