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Partial Derivatives Questions

  1. Feb 7, 2013 #1
    In a thermodynamics question, I was recently perplexed slightly by some partial derivative questions, both on notation and on physical meaning.

    I believe my questions are best posed as examples. Suppose we have an equation, [itex](\frac{\partial x(t)}{\partial t}) = \frac{1}{y}[/itex], where y is a function, not necessarily of t.

    When we solve for y, does this become [itex]y = (\frac{\partial x(t)}{\partial t})^{-1}[/itex]?

    In general, does [itex](\frac{\partial x(t)}{\partial t})^{-1} \neq \frac{\partial t}{\partial x(t)}[/itex]? I was under the impression that they are, in general, not equal.

    Suppose we had a function that is dependent on both position in space (x,y,z) and time (t). I shall denote it f(x,y,z,t). If we take [itex]g = \frac{\partial f}{\partial t}[/itex], the others are said to be held constant. Can g be seen as the change in f with respect to time at a specific point, or is it simply talking about the average change in f over an infinitesimal change in time? I thought that the former of the two (change at a point) was true, but now I'm not sure.
     
  2. jcsd
  3. Feb 11, 2013 #2
    Yes, you are allowed to invert both sides of an equation.

    In general, if the same variables are held constant, you can do this. I'll show you what I mean; take the total derivatives of your functions, and assume that they are all functions of each other for generality:
    [tex]
    \begin{align}
    dx &= \left(\frac{\partial x}{\partial t}\right)_y dt + \left(\frac{\partial x}{\partial y}\right)_t dy\\
    dy &= \left(\frac{\partial y}{\partial t}\right)_x dt + \left(\frac{\partial y}{\partial x}\right)_t dx
    \end{align}
    [/tex]
    where the, for example, [itex]\left(\frac{\partial y}{\partial x}\right)_t[/itex] means [itex]t[/itex] means is held constant. Then you can eliminate [itex]dy[/itex] in the first relation:
    [tex]
    \begin{align}
    dx &=\left(\frac{\partial x}{\partial t}\right)_y dt + \left(\frac{\partial x}{\partial y}\right)_t \left[\left(\frac{\partial y}{\partial t}\right)_x dt +\left(\frac{\partial y}{\partial x}\right)_t dx\right]\\
    &=\left[\left(\frac{\partial x}{\partial t}\right)_y + \left(\frac{\partial x}{\partial y}\right)_t\left(\frac{\partial y}{\partial t}\right)_x\right]dt +\left(\frac{\partial x}{\partial y}\right)_t \left(\frac{\partial y}{\partial x}\right)_t dx
    \end{align}
    [/tex]
    Upon comparing the coefficient of [itex]dx[/itex], you can see that,
    [tex]
    1=\left(\frac{\partial x}{\partial y}\right)_t \left(\frac{\partial y}{\partial x}\right)_t
    [/tex]
    so in other words:
    [tex]
    \frac{1}{\left(\frac{\partial y}{\partial x}\right)_t} = \left(\frac{\partial x}{\partial y}\right)_t
    [/tex]
    So, like I said earlier, this is true only if the same variable is held constant in each derivative.
    This is the change in [itex]f(x,y,z,t)[/itex] with respect to time, at a specific spatial coordinate [itex](x,y,z)[/itex] (the points you are holding constant). In order to find a change in average [itex]f[/itex], you would need to use the average value of the function, like:
    [tex]
    \frac{1}{V}\frac{\partial}{\partial t}\int_V f(x,y,z,t)d^3 x
    [/tex]
     
    Last edited: Feb 11, 2013
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