# Partial Derivatives Questions

1. Feb 7, 2013

### Mandelbroth

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, $(\frac{\partial x(t)}{\partial t}) = \frac{1}{y}$, where y is a function, not necessarily of t.

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

In general, does $(\frac{\partial x(t)}{\partial t})^{-1} \neq \frac{\partial t}{\partial x(t)}$? 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 $g = \frac{\partial f}{\partial t}$, 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. Feb 11, 2013

### antibrane

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:
\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}
where the, for example, $\left(\frac{\partial y}{\partial x}\right)_t$ means $t$ means is held constant. Then you can eliminate $dy$ in the first relation:
\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}
Upon comparing the coefficient of $dx$, 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 $f(x,y,z,t)$ with respect to time, at a specific spatial coordinate $(x,y,z)$ (the points you are holding constant). In order to find a change in average $f$, 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