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Multivariable analog to the total derivative?

  1. Jun 17, 2010 #1
    For a single variable we have

    [tex]\int_{x_1}^{x_2} f(x) dx = F(x_2)-F(x_1)[/tex]

    if f(x) = dF/dx. f(x) is then a total derivative. What is the analog in 3D so that

    [tex]\int_V f(\vec{x}) d^3x[/tex]

    does not depend on the values of f in the interior of V?

    In case there is not a single answer, let me give the context. In the calculus of variations two Lagrangians are equivalent if

    [tex]L_2(q(t),\dot{q}(t),t)=\lambda L_1(q(t),\dot{q}(t),t) + \frac{d}{dt}F(q(t),\dot{q}(t),t)[/tex]

    where lambda is a constant and F is any function. (That is, their actions are extremized for the same function q(t).) What replaces dF/dt in this equivalency if we have a multi-parameter action

    [tex]S=\int L(q(\vec{x}),\partial q(\vec{x}),\vec{x}) d^3x[/tex]

    (where [tex]\partial q[/tex] stands for the various partial derivatives of q)?

    Is it [tex]\nabla \cdot \vec{F}[/tex] for some vector function F? Or is there more to it than that?
    Last edited: Jun 17, 2010
  2. jcsd
  3. Jun 17, 2010 #2


    Staff: Mentor

    I think you have your terminology wrong. In your first example F is an antiderivative of f and f is the derivative of F.

    The total derivative refers to a function of two or more variables, for example f(x, y). The total differential of f in this case is
    [tex]df = \frac{\partial f}{\partial x}~dx + \frac{\partial f}{\partial y}~dy[/tex]

    If it turns out that x and y are differentiable functions of t, then the total derivative of f looks like this:
    [tex]\frac{df}{dt} = \frac{\partial f}{\partial x}~\frac{dx}{dt} + \frac{\partial f}{\partial y}~\frac{dy}{dt}[/tex]
  4. Jun 17, 2010 #3
  5. Jun 18, 2010 #4
    Thanks to both .
  6. Jun 18, 2010 #5
  7. Jun 19, 2010 #6
    Thanks, Studiot
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