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Understanding derivation of -du/dx = F

  1. Oct 2, 2016 #1
    1. The problem statement, all variables and given/known data

    Where U is potential energy, show that F = -du/dx

    2. Relevant equations

    ΔU = U(x) - U(x_0)

    $$ΔU = U(x) - U(x_0) = \int_{x_0}^{x}Fdx \qquad (1)$$

    3. The attempt at a solution

    I am confused why
    $$\frac{d}{dx}ΔU = \frac{du}{dx} \qquad (2)$$

    Doesn't
    $$ΔU = du \qquad (3)$$

    I recognize that (3) is simple misunderstanding of calculus but for some reason I can't make sense of why it isn't true in this case.
     
  2. jcsd
  3. Oct 2, 2016 #2

    TSny

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    I'm a little confused with the notation. Is ##u## the same as ##U##?

    Note that ##\frac{d}{dx}ΔU = \frac{d}{dx} \left[U(x) - U(x_0) \right]##.

    Simplify this.
    Hint: What is ##\frac{d}{dx} U(x_0)##?

    Why do you think it is true in this case?
     
  4. Oct 3, 2016 #3
    Yes it is meant to, my mistake for lack of clarity. From now on I will use ##dU## instead of ##du## to represent the derivative of ##U##
    Does it ##=F+dU/dx##?

    Ok, I can see why ##ΔU \neq dU## in this case. After all, ##ΔU## represents a real change in potential, not an infinitesimally small one. But I am still having trouble understanding this algebraically and intuitively. How can you take the derivative of ##ΔU##? In a situation where all potential energy is provided by gravity, ##ΔU## does not represent a single point where gravity has a single instantaneous force, like ##U(x)## and ##U(x_0)## do. It is rather, a range. Does taking the derivative of ##ΔU## then yield an average force over that range?


    I think my grasp of Leibniz notation is poor.
     
  5. Oct 3, 2016 #4

    TSny

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    You have the relation $$ΔU = U(x) - U(x_0) = \int_{x_0}^{x}Fdx $$ I believe there should be a negative sign in this relation, so that it should read $$ΔU = U(x) - U(x_0) = - \int_{x_0}^{x}Fdx $$ Think of ##x_0## as having a fixed value; but consider ##x## to be a variable. So, ##U(x_0)## is just some number (i.e., a constant), but ##U(x)## is a function of the variable ##x##.

    Use rules of calculus to simplify ##\frac{d}{dx}ΔU = \frac{d}{dx}\left[ U(x) - U(x_0) \right] = -\frac{d}{dx}\int_{x_0}^{x}Fdx##
     
  6. Oct 3, 2016 #5
    I get it! Thank you :)
     
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