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Derivatives in an Atwood Machine

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data
    I have the professor's solutions for a homework we handed in. There is a part that is confusing me. We have the following equation:

    $$E = \frac{1}{2}(m_1 + m_2)\dot{x}^2-(m_1-m_2)gx$$


    2. Relevant equations

    We want to find: $$dE/dt = 0$$


    3. The attempt at a solution

    The solution says the correct answer is:

    $$dE/dt = 0 = (m_1 + m_2)\dot{x}_1\ddot{x}_1 - g(m_1-m_2)\dot{x}_1$$

    Why does it contain [itex]\dot{x}\ddot{x}[/itex] instead of just [itex]\ddot{x}[/itex]?

    Is it because of the chain rule?

    Thanks!
     
  2. jcsd
  3. Nov 11, 2012 #2

    Nugatory

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    Staff: Mentor

    For any function U(t), what is [itex]\frac{dU^2}{dt}[/itex]?

    [itex]\dot{x}[/itex] is a function of t.
     
  4. Nov 11, 2012 #3
    So then it's just

    $$ \frac{dU^2}{dU}\frac{dU}{dt} = 2U\dot{U}$$ ?
     
  5. Nov 11, 2012 #4

    Nugatory

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    Staff: Mentor

    That doesn't quite work, because [itex]\dot{U} \equiv \frac{dU}{dt}[/itex]

    Try [itex]\frac{dU^n}{dt} = nU^{n-1}\frac{dU}{dt}[/itex]
     
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