Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
    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
    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:
    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
    Upon comparing the coefficient of [itex]dx[/itex], you can see that,
    1=\left(\frac{\partial x}{\partial y}\right)_t \left(\frac{\partial y}{\partial x}\right)_t
    so in other words:
    \frac{1}{\left(\frac{\partial y}{\partial x}\right)_t} = \left(\frac{\partial x}{\partial y}\right)_t
    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:
    \frac{1}{V}\frac{\partial}{\partial t}\int_V f(x,y,z,t)d^3 x
    Last edited: Feb 11, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Partial Derivatives Questions Date
Question about partial derivatives. Oct 6, 2015
Partial Derivatives Question Sep 25, 2015
Question about limit definition of partial derivative May 17, 2014
Partial derivative question Apr 18, 2014
Simple partial derivative question Jun 11, 2013