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Function of distance problem

  1. Dec 3, 2005 #1

    daniel_i_l

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    Why does:
    (dV(x)/dx)(dx/dt) = d(V^2(x)/2)/dx ? (V(x) is speed as a function of distance?
    I know that the derivative of V^2(x)/2 if (dV(x)/dx)V(x) but I don't think that V(x) equals (dx/dt), that equal V(t)?

    Thanks in advance!
     
  2. jcsd
  3. Dec 3, 2005 #2

    siddharth

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    Are you sure you wrote the equation down correctly? It should be d(V^2(x)/2)/dt on the RHS shouldn't it?
     
  4. Dec 4, 2005 #3

    HallsofIvy

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    V is, by definition, dx/dt.

    Yes, if you know V(x), and x as a function of t, you could write V as a function of t: V(t) but it would still be V.

    And, as siddharth said, that should be (dV(x))/dx)(dx/dt)= d(V^2(x)/2)/dt
     
  5. Dec 10, 2005 #4

    daniel_i_l

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    Thanks.
    I know that a(x) = F(x) and I have to prove that V^2(x)/2 is the antiderivative of F(x).
    That would mean that I have to prove that:
    d(V^2(x)/2)/dx = F(x) right?

    So if I know that F(x) = dV(X)/dt = dV(X)/dx * dx/dt
    How do I prove that d(V^2(x)/2)/dx = F(x)?
     
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