Undergrad Partial derivatives in thermodynamics

[anachin6000]

So, I'm now studying thermodynamics and our teacher proved some time ago the following mathematical result:
If f(x,y,z)=0, then (∂x/∂y)_z=1/(∂y/∂x)_z
But today he used this relation for a function of four variables. Does this result still hold, because I'm not really sure how to prove it. If anyone could give me an idea or point out a book that deals with this, I would be grateful.

 

[andrewkirk]

The equation f(x,y,z)=0 defines a 2D surface in 3D space. Fixing z to be constant at some value z' focuses us on the intersection of that surface with the plane z=z'. That intersection will be a 1D curve $\gamma$, which can be parameterised as ($\gamma_x(t), \gamma_y(t), \gamma_z(t))$ where $\gamma_z(t)$ is identically equal to z' and we have $f(\gamma_x(t),\gamma_y(t),z')=0$. By the Implicit Function Theorem, in the vicinity of any point (x,y,z) this defines functions $g,h$ such that $y=g(x)$ and $x=h(y)$. Then $\left(\frac{\partial x}{\partial y}\right)_z = h'$ and
$\left(\frac{\partial y}{\partial x}\right)_z = g'$.
By the inverse function rule of basic calculus, we know that $g'(x) = 1/h'(y)$. So that gives us the desired result.

The result can be generalised. When f is a function of n variables ($n\geq 3$), the equation $f(x_1,...,x_n)=0$ defines an $(n-1)$-dimensional hypersurface in n-dimensional space. Fixing the values of all but two of the variables, say $x_3=x'_3, \ ...,\ x_n=x'_n$ defines a 2D 'plane' in the n-D space. The intersection of that 2D 'plane' with the hypersurface will again be a curve. Why? because we have n variables and n-1 equations, being the n-2 constraints that set the values $x_3,...,x_n$, together with the equation that defines the hypersurface. So we have only one degree of freedom - our solutions have one free dimension and a 1D thing is a curve. The curve can be parameterised as ($\gamma_1(t), \gamma_2(t), \gamma_3(t),...\gamma_n(t))$ where $\gamma_1(t),\gamma_2(t)$ give the x and y coordinates, and for $j>2,\ \gamma_j(t)$ is identically equal to $x'_j$.
By the Implicit Function Theorem again, this defines functions $g,h$ such that $x_2=g(x_1)$ and $x_2=h(x_1)$. The result follows in the same way as above, just replacing $x$ by $x_1$, $y$ by $x_2$ and the subscript $z$ by $x_3,...,x_n$.

 

[anachin6000]

The equation f(x,y,z)=0 defines a 2D surface in 3D space. Fixing z to be constant at some value z' focuses us on the intersection of that surface with the plane z=z'. That intersection will be a 1D curve $\gamma$, which can be parameterised as ($\gamma_x(t), \gamma_y(t), \gamma_z(t))$ where $\gamma_z(t)$ is identically equal to z' and we have $f(\gamma_x(t),\gamma_y(t),z')=0$. By the Implicit Function Theorem, in the vicinity of any point (x,y,z) this defines functions $g,h$ such that $y=g(x)$ and $x=h(y)$. Then $\left(\frac{\partial x}{\partial y}\right)_z = h'$ and
$\left(\frac{\partial y}{\partial x}\right)_z = g'$.
By the inverse function rule of basic calculus, we know that $g'(x) = 1/h'(y)$. So that gives us the desired result.

The result can be generalised. When f is a function of n variables ($n\geq 3$), the equation $f(x_1,...,x_n)=0$ defines an $(n-1)$-dimensional hypersurface in n-dimensional space. Fixing the values of all but two of the variables, say $x_3=x'_3, \ ...,\ x_n=x'_n$ defines a 2D 'plane' in the n-D space. The intersection of that 2D 'plane' with the hypersurface will again be a curve. Why? because we have n variables and n-1 equations, being the n-2 constraints that set the values $x_3,...,x_n$, together with the equation that defines the hypersurface. So we have only one degree of freedom - our solutions have one free dimension and a 1D thing is a curve. The curve can be parameterised as ($\gamma_1(t), \gamma_2(t), \gamma_3(t),...\gamma_n(t))$ where $\gamma_1(t),\gamma_2(t)$ give the x and y coordinates, and for $j>2,\ \gamma_j(t)$ is identically equal to $x'_j$.
By the Implicit Function Theorem again, this defines functions $g,h$ such that $x_2=g(x_1)$ and $x_2=h(x_1)$. The result follows in the same way as above, just replacing $x$ by $x_1$, $y$ by $x_2$ and the subscript $z$ by $x_3,...,x_n$.

Thanks a lot.