Question

The formulas of this problem are useful in thermodynamics. (a) Given \(f(x, y, z)=0,\) find formulas for $$\left(\frac{\partial y}{\partial x}\right)_{z}, \quad\left(\frac{\partial x}{\partial y}\right)_{z}, \quad\left(\frac{\partial y}{\partial z}\right)_{x}, \quad \text { and } \quad\left(\frac{\partial z}{\partial x}\right)_{y}$$ (b) Show that \(\left(\frac{\partial x}{\partial y}\right)_{z}\left(\frac{\partial y}{\partial x}\right)_{z}=1\) and \(\left(\frac{\partial x}{\partial y}\right)_{z}\left(\frac{\partial y}{\partial z}\right)_{z}\left(\frac{\partial z}{\partial x}\right)_{y}=-1\) $$\begin{aligned}&\text { (c) If } x, y, z \text { are each functions of } t, \text { show that }\left(\frac{\partial y}{\partial z}\right)_{x}=\left(\frac{\partial y}{\partial t}\right)_{x} /\left(\frac{\partial z}{\partial t}\right)_{x} \text { and }\\\ &\text { corresponding formulas for }\left(\frac{\partial z}{\partial x}\right)_{y} \text { and }\left(\frac{\partial x}{\partial y}\right)_{z} \end{aligned}$$

Short answer

(\partial y/\partial x)_z = -f_x/f_y, \quad (\partial x/\partial y)_z = -f_y/f_x, \quad (\partial y/\partial z)_x = -f_z/f_y, \quad (\partial z/\partial x)_y = -f_x/f_z(\frac{\text{d}x}{\text{d}y})_z (\frac{\text{d}y}{\text{d}x})_z = 1(\frac{\text{d}x}{\text{d}y})_z (\frac{\text{d}y}{\text{d}z})_x (\frac{\text{d}z}{\text{d}x})_y = -1(\frac{\text{d}y}{\text{d}z})_x = (\frac{\text{d}y}{\text{d}t})_x / (\frac{\text{d}z}{\text{d}t})_x.

Step-by-step solution

- Find Implicit Derivatives

Given the function \( f(x, y, z) = 0 \), we need to find the formulas for partial derivatives. Using implicit differentiation, start by taking the partial derivative of \( f \) with respect to \( x \), keeping \( z \) constant: \[ \frac{\text{d}f}{\text{d}x} = f_x + f_y \left( \frac{\text{d}y}{\text{d}x} \right)_z + f_z \left( \frac{\text{d}z}{\text{d}x} \right) = 0 \] Since \( z \) is constant: \[ \left( \frac{\text{d}z}{\text{d}x} \right)_z = 0 \] Simplifying, we get: \[ \left( \frac{\text{d}y}{\text{d}x} \right)_z = -\frac{f_x}{f_y} \]

- Determine Remaining Derivatives

Repeat the process to find remaining partial derivatives: \( \left( \frac{\text{d}x}{\text{d}y} \right)_z \) is the reciprocal of \( \left( \frac{\text{d}y}{\text{d}x} \right)_z \): \[ \left( \frac{\text{d}x}{\text{d}y} \right)_z = -\frac{f_y}{f_x} \] Find \( \left( \frac{\text{d}y}{\text{d}z} \right)_x \): \[ \left( \frac{\text{d}y}{\text{d}z} \right)_x = -\frac{f_z}{f_y} \] Finally, determine \( \left( \frac{\text{d}z}{\text{d}x} \right)_y \): \[ \left( \frac{\text{d}z}{\text{d}x} \right)_y = -\frac{f_x}{f_z} \]

- Verify Derivative Products

Show that the product of partial derivatives equals 1: \( \left( \frac{\text{d}x}{\text{d}y} \right)_z \left( \frac{\text{d}y}{\text{d}x} \right)_z = \left(-\frac{f_y}{f_x}\right) \left(-\frac{f_x}{f_y}\right) = 1 \) To show: \( \left(\frac{\text{d}x}{\text{d}y}\right)_z \left(\frac{\text{d}y}{\text{d}z}\right)_x \left(\frac{\text{d}z}{\text{d}x}\right)_y \) Using values from the previous step: \[ \left( -\frac{f_y}{f_x} \right) \left( -\frac{f_z}{f_y} \right) \left( -\frac{f_x}{f_z} \right) = -1 \]

- Solve Part c using Chain Rule

Given \( x, y, \) and \( z \) are functions of \( t \), use chain rule to derive: \[ \left( \frac{\text{d}y}{\text{d}z} \right)_x = \frac{\left( \frac{\text{d}y}{\text{d}t} \right)_x}{\left( \frac{\text{d}z}{\text{d}t} \right)_x} \] Similarly: \[ \left(\frac{\text{d}z}{\text{d}x}\right)_y = \frac{\left( \frac{\text{d}z}{\text{d}t} \right)_y}{\left( \frac{\text{d}x}{\text{d}t} \right)_y} \] and \[ \left( \frac{\text{d}x}{\text{d}y} \right)_z = \frac{\left( \frac{\text{d}x}{\text{d}t} \right)_z}{\left( \frac{\text{d}y}{\text{d}t} \right)_z} \]

Key concepts

Partial Derivatives

Partial derivatives are a core concept in multivariable calculus, which is crucial for understanding thermodynamic functions. When a function depends on multiple variables, a partial derivative measures how the function changes as one of the variables changes, while keeping the other variables constant. For instance, if you have a function \( f(x, y, z) \), the partial derivative with respect to \( x \), keeping \( y \) and \( z \) constant, is denoted by \( \left( \frac{\partial f}{\partial x} \right)_y \). This is written as \(\frac{\partial f}{\partial x}\) without subscripts if one assumes context provides constant variables.
Computing partial derivatives involves differentiating the function with respect to one variable while treating other variables as constants. This is essential in thermodynamics to determine how properties like pressure, volume, and temperature interrelated.

Implicit Differentiation

Implicit differentiation is useful when dealing with functions not explicitly solved for one variable. In the given exercise, the function \( f(x, y, z) = 0 \) represents a relationship among three variables. Instead of solving for one variable, we differentiate both sides of the equation with respect to a particular variable, treating other variables as functions of that primary variable.
To find \( \left( \frac{\partial y}{\partial x} \right)_z \), we start with the total derivative of \( f \):
\( \frac{\text{d}f}{\text{d}x} = f_x + f_y \frac{\text{d}y}{\text{d}x} + f_z \frac{\text{d}z}{\text{d}x} = 0 \). Since \( z \) is constant, \( \frac{\text{d}z}{\text{d}x} = 0 \). Simplifying, gives \( \frac{\text{d}y}{\text{d}x} = -\frac{f_x}{f_y} \). Repeat these steps for other derivatives:
\( \frac{\text{d}x}{\text{d}y} = -\frac{f_y}{f_x} \), \( \frac{\text{d}y}{\text{d}z} = -\frac{f_z}{f_y} \), and \( \frac{\text{d}z}{\text{d}x} = -\frac{f_x}{f_z} \).

Chain Rule

The chain rule enables us to find the derivative of a composite function. This is particularly useful when dependent variables are functions of another independent variable. In the exercise, if \( x, y, \) and \( z \) are functions of \( t \), the chain rule helps express derivatives of one variable with respect to another.
To show:
\( \frac{\text{d}y}{\text{d}z} \bigg|_x = \frac{\frac{\text{d}y}{\text{d}t}\big|_x}{\frac{\text{d}z}{\text{d}t}\big|_x} \), we use the chain rule on functions of \( t \). Similarly, apply the chain rule to get:
\(\frac{d\text{z}}{\text{d}x} \bigg|_y = \frac{\frac{\text{d}z}{\text{d}t}\big|_y}{\frac{\text{d}x}{\text{d}t}\big|_y} \) and
\( \frac{\text{d}x}{\text{d}y} \bigg|_z = \frac{\frac{\text{d}x}{\text{d}t}\big|_z}{\frac{\text{d}y}{\text{d}t}\big|_z} \). This expresses the dependency of partial derivatives in a succinct form, making complex thermodynamic relationships easier to handle.

Thermodynamics

Thermodynamics studies the movement and conversion of energy and heat in physical systems. Functions like internal energy, entropy, and enthalpy often depend on multiple variables (e.g., pressure, volume, temperature), requiring the use of partial derivatives.
For example, in deriving thermodynamic relationships, understanding partial derivatives allows us to examine how temperature affects pressure while keeping volume constant, or how entropy changes with temperature at constant pressure.
Given a relationship like \( f(x, y, z)=0 \), the partial derivatives tell us how one thermodynamic variable changes when another is varied. This is essential in equations of state, which describe the state of matter under a given set of physical conditions. When using formulas like
\( \frac{\text{d}y}{\text{d}x} \bigg|_z = -\frac{f_x}{f_y} \), we gain insights into how different thermodynamic properties interrelate.