10Mathematics I Study Guide: Total Differentials, Implicit Functions, and Chain Rule

Exercise 1: First and Second Order Total Differentials

The total differential of a function represents the change in the function value as a result of small changes in its input variables. For a function $f(x,y)$, the first-order total differential is defined as: $df(x, y) = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$

The second-order total differential is defined as: $d^2f(x, y) = \frac{\partial^2 f}{\partial x^2} dx^2 + 2 \frac{\partial^2 f}{\partial x \partial y} dx dy + \frac{\partial^2 f}{\partial y^2} dy^2$

a. Function $f(x, y) = x^2 + y^2$

First-order Partial Derivatives:

• $\frac{\partial f}{\partial x} = 2x$

• $\frac{\partial f}{\partial y} = 2y$

Second-order Partial Derivatives:

• $\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(2x) = 2$

• $\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}(2y) = 2$

• $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x}(2y) = 0$

• $\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y}(2x) = 0$

Differentials:

• First order: $df(x, y) = 2x dx + 2y dy$

• Second order: $d^2f(x, y) = 2 dx^2 + 2 dy^2$

b. Function $f(x, y) = e^{xy}$

First-order Partial Derivatives:

• $\frac{\partial f}{\partial x} = e^{xy} \times y$

• $\frac{\partial f}{\partial y} = e^{xy} \times x$

Second-order Partial Derivatives:

• $\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(y e^{xy}) = y^2 e^{xy}$

• $\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}(x e^{xy}) = x^2 e^{xy}$

• $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x}(x e^{xy}) = e^{xy} + x(y e^{xy}) = e^{xy} \times (xy + 1)$

• $\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y}(y e^{xy}) = e^{xy} + y(x e^{xy}) = e^{xy} \times (xy + 1)$

Differentials:

• First order: $df(x, y) = e^{xy} \times (y dx + x dy)$

• Second order: $d^2f(x, y) = e^{xy} \times (y^2 dx^2 + 2(xy + 1) dx dy + x^2 dy^2)$

c. Function $f(x, y) = \sin(xy) \cos(x)$

First-order Partial Derivatives:

• $\frac{\partial f}{\partial x} = \cos(xy) y \cos(x) - \sin(xy) \sin(x)$

• $\frac{\partial f}{\partial y} = \cos(xy) x \cos(x)$

Second-order Partial Derivatives:

• $\frac{\partial^2 f}{\partial x^2} = -2 \sin(xy) y \cos(x) - 2 \cos(xy) y \sin(x) - \sin(xy) \cos(x)$, which simplifies in the text to $(-2 \sin(xy) y \cos(x) - 2 \cos(xy) y \sin(x)) dx^2$

• $\frac{\partial^2 f}{\partial y^2} = -\sin(xy) x^2 \cos(x)$

• Mixed partials ($\frac{\partial^2 f}{\partial y \partial x}$): $\sin(xy) y \cos(x) + \cos(xy) \cos(x) - \sin(xy) \sin(x) x$

Differentials:

• First order: $df(x, y) = (\cos(xy) \cos(x) y - \sin(xy) \sin(x)) dx + \cos(xy) x \cos(x) dy$

• Second order: $d^2f(x, y) = (-2 \sin(xy) y \cos(x) - 2 \cos(xy) y \sin(x)) dx^2 + 2(\cos(xy) \cos(x) - \sin(xy) y \cos(x) - \cos(xy) x \sin(x)) dx dy - \sin(xy) x^2 \cos(x) dy^2$

Exercise 2: Checking for Total Differentials

A differential expression in the form $Q dx + P dy$ is a total (exact) differential if it satisfies Schwarz's Theorem (the equality of mixed partial derivatives), meaning: $\frac{\partial Q}{\partial y} = \frac{\partial P}{\partial x}$

a. $x \times y^2 dx + x^2 \times y dy$

• $Q = xy^2 \rightarrow \frac{\partial Q}{\partial y} = 2xy$

• $P = x^2y \rightarrow \frac{\partial P}{\partial x} = 2xy$

• Result: Since $2xy = 2xy$, this is a total differential.

b. $\sin(xy) dx + \sin(xy) dy$

• $Q = \sin(xy) \rightarrow \frac{\partial Q}{\partial y} = \cos(xy)x$

• $P = \sin(xy) \rightarrow \frac{\partial P}{\partial x} = \cos(xy)y$

• Result: Since $\cos(xy)x \neq \cos(xy)y$, this is not a total differential (it is incomplete).

c. $\frac{x}{y} dx + dy$

• $Q = \frac{x}{y} \rightarrow \frac{\partial Q}{\partial y} = -\frac{x}{y^2}$

• $P = 1 \rightarrow \frac{\partial P}{\partial x} = 0$

• Result: Since $\frac{\partial Q}{\partial y} \neq \frac{\partial P}{\partial x}$, this is not a total differential.

d. $\frac{x}{\sqrt{x^2 + y^2}} dx + \frac{y}{\sqrt{x^2 + y^2}} dy$

• $Q = x(x^2 + y^2)^{-\frac{1}{2}} \rightarrow \frac{\partial Q}{\partial y} = -\frac{xy}{(x^2 + y^2)^{\frac{3}{2}}}$

• $P = y(x^2 + y^2)^{-\frac{1}{2}} \rightarrow \frac{\partial P}{\partial x} = -\frac{xy}{(x^2 + y^2)^{\frac{3}{2}}}$

• Result: The derivatives are equal, so this is a total differential.

Exercise 3: Implicit Functions and the Van-der-Waals Gas

The van-der-Waals equation of state is written as an implicit function $F(p, V, T)$, where pressure $p$, volume $V$, and temperature $T$ are variables, and $a$, $b$, and $R$ are constants: $F(p, V, T) = (p + \frac{a}{V^2})(V - b) - RT = 0$

a. Temperature Dependence of Volume at Constant Pressure

To find $\left( \frac{\partial V}{\partial T} \right)_p$, we use the theorem of implicit functions: $\left( \frac{\partial V}{\partial T} \right)_p = - \frac{\frac{\partial F}{\partial T}}{\frac{\partial F}{\partial V}}$

Partial Derivatives:

• $\frac{\partial F}{\partial T} = -R$

• $\frac{\partial F}{\partial V} = \frac{\partial}{\partial V} [(p + a V^{-2})(V-b)] = (p + a V^{-2}) \times (1) + (V-b) \times (-2a V^{-3})$

• $\frac{\partial F}{\partial V} = p + \frac{a}{V^2} - \frac{2a(V-b)}{V^3}$

Calculation: $\left( \frac{\partial V}{\partial T} \right)_p = - \frac{-R}{p + \frac{a}{V^2} - \frac{2a(V-b)}{V^3}} = \frac{R}{p - \frac{a}{V^2} + \frac{2ab}{V^3}}$

b. Temperature Dependence of Pressure at Constant Volume

To find $\left( \frac{\partial p}{\partial T} \right)_V$, we use: $\left( \frac{\partial p}{\partial T} \right)_V = - \frac{\frac{\partial F}{\partial T}}{\frac{\partial F}{\partial p}}$

Partial Derivatives:

• $\frac{\partial F}{\partial T} = -R$

• $\frac{\partial F}{\partial p} = V - b$

Calculation: $\left( \frac{\partial p}{\partial T} \right)_V = - \frac{-R}{V - b} = \frac{R}{V - b}$

c. Difference of Molar Heat Capacities

The expression to calculate is $\Delta C = T \times \left( \frac{\partial p}{\partial T} \right)_V \times \left( \frac{\partial V}{\partial T} \right)_p$.

Substitution Step 1: $\Delta C = T \times \frac{R}{V - b} \times \frac{R}{p + \frac{a}{V^2} - \frac{2a(V-b)}{V^3}}$

Substitution Step 2 (Eliminating $p$): From the implicit equation, $p = \frac{RT}{V-b} - \frac{a}{V^2}$. Substituting this into the denominator for the volume derivative: $\text{Denominator} = (\frac{RT}{V-b} - \frac{a}{V^2}) + \frac{a}{V^2} - \frac{2a(V-b)}{V^3} = \frac{RT}{V-b} - \frac{2a(V-b)}{V^3}$

Final Result: $\Delta C = T \times \frac{R}{V-b} \times \frac{R}{\frac{RT}{V-b} - \frac{2a(V-b)}{V^3}}$ Multiplying the fractions results in: $\Delta C = \frac{T R^2}{RT - \frac{2a(V-b)^2}{V^3}} = \frac{R}{1 - \frac{2a(V-b)^2}{RT V^3}}$

Exercise 4: Variable Substitution and Chain Rule

Given the equation: $y \frac{\partial z}{\partial x} - x \frac{\partial z}{\partial y} = 0$, where $z = z(x,y)$.

Substitution

Replace $x$ and $y$ with new variables $u = x$ and $v = x^2 + y^2$. We now treat $z$ as a function of $u$ and $v$ ($z(u, v)$).

Chain Rule Application

For $z(u(x,y), v(x,y))$: $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x} = \frac{\partial z}{\partial u} (1) + \frac{\partial z}{\partial v} (2x)$ $\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} = \frac{\partial z}{\partial u} (0) + \frac{\partial z}{\partial v} (2y)$

Transforming the Equation

Substitute the partial derivatives back into the original equation: $y(\frac{\partial z}{\partial u} + 2x \frac{\partial z}{\partial v}) - x(2y \frac{\partial z}{\partial v}) = 0$ $y \frac{\partial z}{\partial u} + 2xy \frac{\partial z}{\partial v} - 2xy \frac{\partial z}{\partial v} = 0$ $y \frac{\partial z}{\partial u} = 0$

Interpretation

Since the equation must hold for arbitrary values of $y$, it implies that $\frac{\partial z}{\partial u} = 0$. This means function $z$ does not depend on $u$; it depends only on $v$. Therefore, $z = z(v) = z(x^2 + y^2)$. This shows that the function is constant on any circle where $x^2 + y^2$ is constant (e.g., rotational symmetry). Examples include $z = x^2 + y^2$ or $z = e^{x^2+y^2}$.