# A Thermodynamics. Partial derivative tricks.

1. May 7, 2017

### LagrangeEuler

If we consider function $z=z(x,y)$ then $dz=(\frac{\partial z}{\partial x})_ydx+(\frac{\partial z}{\partial y})_xdy$. If $z=const$ then $dz=0$. So,
$$(\frac{\partial z}{\partial x})_ydx+(\frac{\partial z}{\partial y})_xdy=0$$
and from that
$$\frac{dx}{dy}=-\frac{(\frac{\partial z}{\partial y})_x}{(\frac{\partial z}{\partial x})_y}$$
or
$$(\frac{\partial x}{\partial y})_z=-\frac{(\frac{\partial z}{\partial y})_x}{(\frac{\partial z}{\partial x})_y}$$
My question is is $z=const$ isn't then also $(\frac{\partial z}{\partial x})_y=0$ and $(\frac{\partial z}{\partial y})_x=0$? This is confusing to me. Could you please answer me that.

2. May 7, 2017

### vanhees71

Everything you wrote is correct. Your confusion is due to a unfortunate formulation of what's meant by the mathematical operations. What you study is the following:

You define a function $x=x(x)$ (let me be a bit sloppy; pure mathematicians would kill me for this formulation already ;-)) in an implicit way by using a function $z=z(x,y)$ by setting
$$z(x,y)=c=\text{const}.$$
So $z$ is always a function of two independent variables $x$ and $y$. Setting this function to a constant leads to an equation for $x$ to be solved given the value of the variable $y$.

Now taking the partial derivatives of $z$ on both sides of the defining equation makes sense. The total differential on the left-hand side indeed is
$$\mathrm{d} z=\mathrm{d} x \partial_x z+\mathrm{d} y \partial_{y} z.$$
Now since you restrict yourself along the line, where $z(x,y)=c=\text{const}$ this means that $\mathrm{d} z=0$. On the other hand this line can (at least locally) be described by a function $x=x(y)$, and then you get your formula
$$\frac{\mathrm{d} x}{\mathrm{d} y}=-\frac{\partial z}{\partial y} \left (\frac{\partial z}{\partial x} \right)^{-1}=-\frac{\partial_y z}{\partial_x z}.$$
Let's give a very simple example. Let's set
$$z(x,y)=x y^2.$$
Now we set
$$z(x,y)=1=\text{const} \; \Rightarrow\; x y^2=1 \; \Rightarrow x=\frac{1}{y^2}.$$
Obviously you get
$$\partial_x z=y^2, \quad \partial_y z=2xy.$$
Now
$$\frac{\partial_y z}{\partial_x z}=\frac{2xy}{y^2}=\frac{2x}{y}.$$
Along the implicitly defined curve you have
$$\left (\frac{\partial_y z}{\partial_x z} \right)_{x=1/y^2}=\frac{2}{y^3}.$$
Indeed, calculating the function $x=1/y^2$ directly you get
$$\frac{\mathrm{d} x}{\mathrm{d} y}=-\frac{2}{y^3}=-\left (\frac{\partial_y z}{\partial_x z} \right )_{x=1/y^2}.$$

3. May 7, 2017

### LagrangeEuler

Thanks a lot. Very useful answer. Could you just explain in this case why you say "at least locally".

4. May 7, 2017

### Staff: Mentor

What you are doing is determining is how x and y have to vary in tandem in order for z to remain constant.

5. May 7, 2017

### vanhees71

Take as an example
$$z(x,y)=x^2+y^2.$$
Obviously the implicitly defined curve
$$z(x,y)=R^2=\text{const}$$
is a circle of radius $R$ around the origin in the $xy$ plane, which cannot be described by a function $y=y(x)$ or $x=x(y)$, but you can of course do that for two parts of the circle, e.g., one semi-circle in the upper part,
$$y=\sqrt{R^2-x^2}$$
or in the lower part
$$y=-\sqrt{R^2-x^2}.$$
That's what I meant with "local".