Multivariate Calculus – Page 2

June 1, 2022October 24, 2022

202206011159 Problem 5E (Q25)

Test each of the following differentials to see whether they are exact, using two methods for each:

(a) $-y\sin x\,\mathrm{d}x+\cos x\,\mathrm{d}y$ ,

(b) $y\,\mathrm{d}x+x\,\mathrm{d}y$ ,
(c) $yx^3e^x\,\mathrm{d}x+x^3e^x\,\mathrm{d}y$ ,
(d) $(1+x)ye^x\,\mathrm{d}x+xe^x\,\mathrm{d}y$ ,
(e) $4x^3y^{-2}\,\mathrm{d}x-2x^4y^{-3}\,\mathrm{d}y$ .

_{K. S. Stowe. (2007). An Introduction to Thermodynamics and Statistical Mechanics}

Revision. (exact differentials)

The differential of a function is given by Eq. (5.8):

$\mathrm{d}F=\displaystyle{\frac{\partial F}{\partial x}\,\mathrm{d}x+\frac{\partial F}{\partial y}\,\mathrm{d}y}$ .

Therefore, one way to determine whether a differential given by Eq. (5.9):

$\mathrm{d}\Phi =g(x,y)\,\mathrm{d}x+h(x,y)\,\mathrm{d}y$

is exact is to see whether we can find some function $F(x,y)$ such that

$\displaystyle{\frac{\partial F}{\partial x}=g(x,y)}\qquad \textrm{and}\qquad\displaystyle{\frac{\partial F}{\partial y}=h(x,y)}$ .

If we can, the differential is exact, and if we can’t, it is inexact. Alternatively, we can use the identity

$\displaystyle{\frac{\partial^2F}{\partial y\partial x}=\frac{\partial^2F}{\partial x\partial y}}$ .

Combining this with equations (5.8) and (5.9), we can see that for exact differentials, Eq. (5.10):

$\displaystyle{\frac{\partial g}{\partial y}=\frac{\partial h}{\partial x}}$ .

_{Text on pg. 89, Sec. E, Ch. 5}

Roughwork.

(a)

We can see that this is indeed an exact differential of the function

$F=y\cos x+\textrm{constant}$ ,

because

$\displaystyle{\frac{\partial F}{\partial x}=-y\sin x}\qquad\textrm{and}\qquad\displaystyle{\frac{\partial F}{\partial y}=\cos x}$ .

Or we can use Eq. (5.10). For this example $g=-y\sin x$ and $h=\cos x$ , so

$\displaystyle{\frac{\partial g}{\partial y}=-\sin x}\qquad\textrm{and}\qquad\displaystyle{\frac{\partial h}{\partial x}=-\sin x}$ .

The two are the same, so the differential is exact.

Parts (b) to (e) are left to the reader as an exercise. Cf. analytic functions et Cauchy-Riemann equations.

February 4, 2022October 24, 2022

202204021539 Example 3, Chapter 1.3, Methods in Physics II (2015-2016 Lectures)

If $\mathbf{f}(t)=t\,\hat{\mathbf{i}}+t^3\,\hat{\mathbf{j}}$ , $\mathbf{g}(t)=\cos t\,\hat{\mathbf{i}}+\sin t\,\hat{\mathbf{j}}$ , and $\mathbf{v}=2\,\hat{\mathbf{i}}-3\,\hat{\mathbf{j}}$ . Calculate

(a) $(\mathbf{f}+\mathbf{g})'$ ;

(b) $(\mathbf{v}\cdot\mathbf{f})'$ ;

(c) $(\mathbf{f}\cdot\mathbf{g})'$ .

Settings. (Some properties of vector differentiation)

i. If $\mathbf{f}$ and $\mathbf{g}$ are differentiable vector functions,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{f}+\mathbf{g})=\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}+\frac{\mathrm{d}\mathbf{g}}{\mathrm{d}t}}$

ii. If $\mathbf{f}$ is a differentiable vector function and $\alpha$ a constant scalar,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\alpha\mathbf{f})=\alpha\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}$

iii. If $\mathbf{f}(t)$ is a vector function, $\mathbf{v}$ a constant vector, and $\mathbf{v}\cdot\mathbf{f}$ differentiable,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{v}\cdot\mathbf{f})=\mathbf{v}\cdot\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}$

iv. If $h(t)$ is a scalar function, $\mathbf{f}(t)$ a vector function and $h\mathbf{f}$ differentiable,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(h\mathbf{f})=h\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}+\mathbf{f}\frac{\mathrm{d}h}{\mathrm{d}t}}$

v. If $\mathbf{f}$ and $\mathbf{g}$ are vector functions and $\mathbf{f}\cdot\mathbf{g}$ differentiable,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{f}\cdot\mathbf{g})=\mathbf{f}\cdot\frac{\mathrm{d}\mathbf{g}}{\mathrm{d}t}+\mathbf{g}\cdot\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}$

vi. If $\mathbf{f}$ and $\mathbf{g}$ are vector functions and $\mathbf{f}\times\mathbf{g}$ differentiable,

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}}(\mathbf{f}\times\mathbf{g})=\displaystyle{\bigg( \frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}\times\mathbf{g}\bigg) + \bigg( \mathbf{f}\times \frac{\mathrm{d}\mathbf{g}}{\mathrm{d}t}\bigg)}$ .

Solution.

(a)

$\begin{aligned} \mathbf{f}+\mathbf{g} & = (t+\cos t)\,\hat{\mathbf{i}}+(t^3+\sin t)\,\hat{\mathbf{j}} \\ (\mathbf{f}+\mathbf{g})' & = (t+\cos t)'\,\hat{\mathbf{i}}+(t^3+\sin t)'\,\hat{\mathbf{j}} \\ & = (1-\sin t)\,\hat{\mathbf{i}}+(3t^2+\cos t)\,\hat{\mathbf{j}} \\ \end{aligned}$

(b)

$\begin{aligned} \mathbf{v}\cdot\mathbf{f} & = (2,-3)\cdot (t,t^3) \\ & = 2t-3t^3 \\ (\mathbf{v}\cdot\mathbf{f})' & = (2t-3t^3)' \\ & = 2-9t^2 \\ \end{aligned}$

(c)

$\begin{aligned} \mathbf{f}\cdot\mathbf{g} & = (t,t^3) \cdot (\cos t,\sin t) \\ & = t\cos t+t^3\sin t \\ (\mathbf{f}\cdot\mathbf{g})' & = \cos t-t\sin t+t^3\cos t + 3t^2\sin t\\ \end{aligned}$

February 4, 2022October 24, 2022

202204021524 Example 1, Chapter 1.3, Methods in Physics II (2015-2016 Lectures)

If $\mathbf{f}(\theta)=\cos\theta \,\hat{\mathbf{i}}+e^{2\theta}\,\hat{\mathbf{j}}$ , find $\mathbf{f}'(\theta)$ .

Setting.

If $\mathbf{f}(t)=f_1(t)\,\hat{\mathbf{i}}+f_2(t)\,\hat{\mathbf{j}}+f_3(t)\,\hat{\mathbf{k}}$ is a vector function and $f_1$ , $f_2$ , and $f_3$ are differentiable, then the derivative of $f(t)$ is defined as

$\displaystyle{\frac{\mathrm{d}\mathbf{f}}{\mathrm{d}t}}=\mathbf{f}'(t)=f_1'(t)\,\hat{\mathbf{i}}+f_2'(t)\,\hat{\mathbf{j}}+f_3'(t)\,\hat{\mathbf{k}}$ .

Solution.

From $\mathbf{f}(\theta)=\cos\theta \,\hat{\mathbf{i}}+e^{2\theta}\,\hat{\mathbf{j}}$ , identify that

$\begin{aligned} f_1(\theta ) & =\cos\theta \\ f_2(\theta ) & = e^{2\theta} \\ f_1'(\theta ) & = -\sin\theta \\ f_2'(\theta ) & = 2e^{2\theta }\\ \end{aligned}$

Thus, $\mathbf{f}'(\theta )= -\sin\theta\,\hat{\mathbf{i}}+2e^{2\theta}\,\hat{\mathbf{j}}$ .

February 4, 2022October 24, 2022

202202041514 Example 1, Chapter 1.2, Methods in Physics II (2015-2016 Lectures)

Find the Cartesian equation of the locus represented by the parametric equation:

$\mathbf{f}(t)=t\,\hat{\mathbf{i}}+e^t\,\hat{\mathbf{j}}$ .

Solution.

$\begin{aligned} x(t) & =t \\ y(t) & = e^t\\ \end{aligned}$

Hence, $y=e^x$ .

September 10, 2021July 25, 2022

202109101713 Exercises 1.2.1

Let $T:\mathbb{R}^2\rightarrow \mathbb{R}^3$ be the function defined by $T(x_1,x_2)=(x_1,x_2,0)$ . Show that $T$ is a linear transformation.

Definition. A function $T:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is called a linear transformation if:
(1) $T(\mathbf{x}+\mathbf{y})=T(\mathbf{x})+T(\mathbf{y})$ ; and
(2) $T(c\mathbf{x})=cT(\mathbf{x})$ .
The conditions must be satisfied for all $\mathbf{x},\mathbf{y}$ in $\mathbb{R}^n$ and all $c$ in $\mathbb{R}$ .

Proof.

Let $\mathbf{x}=(x_1,x_2)$ and $\mathbf{y}=(y_1,y_2)$ .

Then

$\begin{aligned} T(\mathbf{x})& =T(x_1,x_2)=(x_1,x_2,0) \\ T(\mathbf{y})&=T(y_1,y_2)=(y_1,y_2,0) \\ T(\mathbf{x})+T(\mathbf{y}) & = (x_1,x_2,0) + (y_1,y_2,0) \\ \Rightarrow\enspace\textrm{RHS} & = (x_1+y_1,x_2+y_2,0)\\ \mathbf{x}+\mathbf{y} & = (x_1,x_2) + (y_1,y_2) \\ & = (x_1+y_1,x_2+y_2) \\ T(\mathbf{x}+\mathbf{y}) & = T(x_1+y_1,x_2+y_2) \\ \Rightarrow\enspace\textrm{LHS} & = (x_1+y_1,x_2+y_2,0) \\ \therefore\enspace \textrm{LHS} & = \textrm{RHS} \\ \end{aligned}$

Condition (1) is thus satisfied.

$\forall\, c\in\mathbb{R}$ and $\forall\, \mathbf{x}=(x_1,x_2)\in\mathbb{R}^2$ ,

$\begin{aligned} cT(\mathbf{x}) & = cT(x_1,x_2) \\ & = c(x_1,x_2,0) \\ \Rightarrow\enspace\textrm{RHS} & = (cx_1,cx_2,0) \\ T(c\mathbf{x}) & = T(c(x_1,x_2)) \\ & = T(cx_1,cx_2) \\ \Rightarrow \enspace\textrm{LHS} & = (cx_1,cx_2,0) \\ & = \textrm{RHS} \\ \end{aligned}$

Condition (2) is also satisfied.

In conclusion, $T$ is a linear transformation.

May 24, 2021October 24, 2022

202105241204 Homework 1 (Q3)

Prove

(a) $\nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A})-A\times (\nabla f)$

(b) $\nabla \times (\mathbf{A}\times \mathbf{B})=(\mathbf{B}\cdot\nabla )\mathbf{A}+(\nabla\cdot\mathbf{B})\mathbf{A}-(\mathbf{A}\cdot\nabla )\mathbf{B}-(\nabla\cdot\mathbf{A})\mathbf{B}$

Attempts. (brute force)

(a)

$\begin{aligned} & \quad \nabla\times (f\mathbf{A}) \\ & = \nabla \times (fA_x,fA_y,fA_z) \\ & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ fA_x & fA_y & fA_z \end{vmatrix} \\ & = \bigg( \frac{\partial}{\partial y}(fA_z)-\frac{\partial}{\partial z}(fA_y),\, \frac{\partial}{\partial z}(fA_x) - \frac{\partial}{\partial x}(fA_z),\, \frac{\partial}{\partial x}(fA_y) - \frac{\partial}{\partial y}(fA_x) \bigg) \\ & = \Bigg( \bigg( f\frac{\partial A_z}{\partial y} + \frac{\partial f}{\partial y}A_z - f\frac{\partial A_y}{\partial z} - \frac{\partial f}{\partial z}A_y \bigg) , \\ & \quad \qquad \bigg( f\frac{\partial A_x}{\partial z} + \frac{\partial f}{\partial z}A_x - f\frac{\partial A_z}{\partial x} - \frac{\partial f}{\partial x}A_z \bigg) , \\ & \qquad \qquad \bigg( f\frac{\partial A_y}{\partial x}-\frac{\partial f}{\partial x} - \frac{\partial f}{\partial y}A_x - f\frac{\partial A_x}{\partial y} \bigg) \Bigg) \\ & = f\Bigg( \bigg( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \bigg) ,\, \bigg( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial z} \bigg),\, \bigg( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\bigg) \Bigg) \\ & \quad \qquad + \bigg( \frac{\partial f}{\partial y}A_z - \frac{\partial f}{\partial z}A_y,\, \frac{\partial f}{\partial z}A_x - \frac{\partial f}{\partial x}A_z,\, \frac{\partial f}{\partial x}A_y - \frac{\partial f}{\partial y}A_x \bigg) \\ & = f(\nabla \times \mathbf{A}) + \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\ A_x & A_y & A_z \end{vmatrix} \\ & = f(\nabla \times \mathbf{A}) + (\nabla f)\times\mathbf{A} \\ & = f(\nabla \times \mathbf{A}) - \mathbf{A}\times (\nabla f) \\ \end{aligned}$

(b)

$\begin{aligned} \textrm{LHS}\enspace & = \nabla \times (\mathbf{A}\times \mathbf{B}) \\ & = \nabla \times (A_yB_z-A_zB_y,\, -A_xB_z+A_zB_x,\, A_xB_y-A_yB_x) \\ & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_yB_z-A_zB_y & -A_xB_z + A_zB_x & A_xB_y - A_yB_x \end{vmatrix} \\ & = \Bigg( \bigg( \frac{\partial}{\partial y}(A_xB_y) - \frac{\partial}{\partial y}(A_yB_x) - \frac{\partial}{\partial z}(A_xB_z) + \frac{\partial}{\partial z}(A_zB_x) \bigg) ,\, \\ & \quad \qquad \bigg( -\frac{\partial}{\partial x}(A_xB_y) + \frac{\partial}{\partial x}(A_yB_x) + \frac{\partial}{\partial z}(A_yB_z) - \frac{\partial}{\partial z}(A_zB_y) \bigg) ,\, \\ & \qquad \qquad \bigg( \frac{\partial}{\partial x}(-A_xB_z) - \frac{\partial}{\partial x}(A_zB_x) - \frac{\partial}{\partial y}(A_yB_z) + \frac{\partial}{\partial y}(A_zB_y) \bigg) \Bigg) \\ & = \Bigg( \bigg( A_x\frac{\partial B_y}{\partial y} + \frac{\partial A_x}{\partial y}B_y - A_y\frac{\partial B_x}{\partial y} - \frac{\partial A_y}{\partial y}B_x - A_x\frac{\partial B_z}{\partial z} - \frac{\partial A_x}{\partial z}B_z + A_z\frac{\partial B_x}{\partial z} + \frac{\partial A_z}{\partial z}B_x \bigg) ,\, \\ & \quad \qquad \bigg( -A_x\frac{\partial B_y}{\partial x} - \frac{\partial A_x}{\partial x}B_y + A_y\frac{\partial B_x}{\partial x}+\frac{\partial A_y}{\partial x}B_x + A_y\frac{\partial B_z}{\partial z} + \frac{\partial A_y}{\partial z}B_z - A_z\frac{\partial B_y}{\partial z} - \frac{\partial A_z}{\partial z}B_y \bigg) ,\, \\ & \qquad \qquad \bigg( A_x\frac{\partial B_z}{\partial x} + \frac{\partial A_x}{\partial x}B_z - A_z\frac{\partial B_x}{\partial x} - \frac{\partial A_z}{\partial x}B_x - A_y\frac{\partial B_z}{\partial y} - \frac{\partial A_y}{\partial y}B_z + \frac{\partial A_z}{\partial y}B_y + A_z\frac{\partial B_y}{\partial y} \bigg) \Bigg) \\ \end{aligned}$

$\textrm{RHS}=(\mathbf{B}\cdot\nabla )\mathbf{A}+(\nabla\cdot\mathbf{B})\mathbf{A}-(\mathbf{A}\cdot\nabla )\mathbf{B}-(\nabla\cdot\mathbf{A})\mathbf{B}$

Inspect these four terms on the right hand side by expanding one after the other.

The first term being

$\begin{aligned} (\mathbf{B}\cdot\nabla )\mathbf{A} & = \bigg( B_x\frac{\partial}{\partial x} + B_y\frac{\partial}{\partial y} + B_z\frac{\partial}{\partial z} \bigg) \mathbf{A} \\ & = \bigg( B_x\frac{\partial A_x}{\partial x} + B_y\frac{\partial A_x}{\partial y} + B_z\frac{\partial A_x}{\partial z},\, \\ & \quad \qquad B_x\frac{\partial A_y}{\partial x} + B_y\frac{\partial A_y}{\partial y} + B_z\frac{\partial A_y}{\partial z},\, \\ & \qquad \qquad B_x\frac{\partial A_z}{\partial x} + B_y\frac{\partial A_z}{\partial y} + B_z\frac{\partial A_z}{\partial z} \bigg)\end{aligned}$

the second term being

$\begin{aligned} (\nabla \cdot \mathbf{B})\mathbf{A} & = \bigg( \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z}\bigg)\mathbf{A} \\ & = \bigg( \frac{\partial B_x}{\partial x}A_x + \frac{\partial B_y}{\partial y}A_x + \frac{\partial B_z}{\partial z}A_x ,\, \\ & \quad \qquad \frac{\partial B_x}{\partial x}A_y + \frac{\partial B_y}{\partial y}A_y + \frac{\partial B_z}{\partial z}A_y , \, \\ & \qquad \qquad \frac{\partial B_x}{\partial x}A_z + \frac{\partial B_y}{\partial y}A_z + \frac{\partial B_z}{\partial z}A_z \bigg) \\ \end{aligned}$

the third term being

$\begin{aligned} -(\mathbf{A}\cdot\nabla )\mathbf{B} & = - \bigg( A_x\frac{\partial}{\partial x} + A_y\frac{\partial}{\partial y} + A_z\frac{\partial}{\partial z} \bigg) \mathbf{B} \\ & = -\bigg( A_x\frac{\partial B_x}{\partial x} + A_y\frac{\partial B_x}{\partial y} + A_z\frac{\partial B_x}{\partial z},\, \\ & \quad\qquad A_x\frac{\partial B_y}{\partial x} + A_y\frac{\partial B_y}{\partial y} + A_z\frac{\partial B_y}{\partial z},\, \\ & \qquad\qquad A_x\frac{\partial B_z}{\partial x} + A_y\frac{\partial B_z}{\partial y} + A_z\frac{\partial B_z}{\partial z} \bigg) \\\end{aligned}$

and the fourth and last term being

$\begin{aligned} -(\nabla \cdot \mathbf{A})\mathbf{B} & = -\bigg( \frac{\partial A_x}{\partial x} +\frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} \bigg)\mathbf{B} \\ & = - \bigg( B_x\frac{\partial A_x}{\partial x} + B_x\frac{\partial A_y}{\partial y} + B_x\frac{\partial A_z}{\partial z} ,\, \\ & \quad\qquad B_y\frac{\partial A_x}{\partial x} + B_y\frac{\partial A_y}{\partial y} + B_y\frac{\partial A_z}{\partial z} ,\, \\ & \qquad \qquad B_z\frac{\partial A_x}{\partial x} + B_z\frac{\partial A_y}{\partial y} + B_z\frac{\partial A_z}{\partial z}\bigg) \\ \end{aligned}$

One can check that $\textrm{LHS}=\textrm{RHS}$ .

Solution. (proof)

(The solution below is based on the manuscript of 2016-2017 PHYS3450 Electromagnetism Homework 1 Solution.)

Using Einstein summation (/notation) and the Levi-Civita symbol $\varepsilon_{ijk}$ ,

(a)

$\begin{aligned} & \quad \nabla \times (f\mathbf{A}) \\ & = \sum_{i,j,k}\hat{\mathbf{e}}_i\frac{\partial}{\partial j}(f\mathbf{A}_k)\cdot\varepsilon_{ijk}\qquad\qquad\qquad i,j,k\in\{ x,y,z\} \\ & = \sum_{i,j,k}\hat{\mathbf{e}}_i\bigg(\frac{\partial}{\partial j}f\bigg)\cdot A_k\cdot\varepsilon_{ijk}+\sum_{i,j,k}\hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}A_k \bigg)\cdot f\cdot \varepsilon_{ijk} \\ & = (\nabla f)\times \mathbf{A} + f\cdot (\nabla\times\mathbf{A}) \\ & = (\nabla f)\times \mathbf{A} - A\times (\nabla f) \end{aligned}$

(b)

$\begin{aligned} \mathbf{A}\times\mathbf{B} & = \sum_{k,l,m}\hat{\mathbf{e}}_kA_lB_m\varepsilon_{klm}\\ \nabla\times (\mathbf{A}\times\mathbf{B}) & = \sum_{i,j,k}\hat{\mathbf{e}}_i\frac{\partial}{\partial j}(\sum_{l,m}A_lB_m\varepsilon_{klm})\varepsilon_{ijk} \\ & = \sum_{i,j,k,l,m}\hat{\mathbf{e}}_i \bigg[ \bigg( \frac{\partial}{\partial j}A_l \bigg) B_m + A_l\cdot \bigg( \frac{\partial}{\partial j}B_m\bigg) \bigg] \varepsilon_{klm}\varepsilon_{ijk} \\ \textrm{by } & \varepsilon_{klm}\varepsilon_{ijk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} \\ \textrm{Thus, }& = \sum_{i,j,k,l,m}\hat{\mathbf{e}}_i \bigg[ \bigg( \frac{\partial}{\partial j}A_l \bigg) B_m + A_l\cdot \bigg( \frac{\partial}{\partial j}B_m\bigg) \bigg] (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}) \\ & = \sum_{i,j,k,l,m}\bigg[ \hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}A_l \bigg) B_m\delta_{il}\delta_{jm} + \hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}A_l \bigg) B_m (-\delta_{im}\delta_{jl}) \\ & \quad\qquad + \hat{\mathbf{e}}_i\bigg( \frac{\partial}{\partial j}B_m\bigg) A_l\delta_{il}\delta_{jm} + \hat{\mathbf{e}}_i \bigg( \frac{\partial}{\partial j}B_m \bigg) A_l (-\delta_{im}\delta_{jl}) \bigg] \\ & = (\mathbf{B}\cdot\nabla )\mathbf{A}+(\nabla\cdot\mathbf{B})\mathbf{A}-(\mathbf{A}\cdot\nabla )\mathbf{B}-(\nabla\cdot\mathbf{A})\mathbf{B} \end{aligned}$

QED

and the proof is more concise.

May 24, 2021January 19, 2023

202105241144 Homework 1 (Q4)

$\displaystyle{\int \delta (f(x))g(x)\,\mathrm{d}x}=\, ?$

Attempts.

$\begin{aligned} &\quad \int \delta(f(x))g(x)\,\mathrm{d}x \\ & = \int\delta (f(x))g(f^{-1}f(x))\frac{\mathrm{d}f(x)}{f'(x)} \\ & = \frac{g(f^{-1}f(x))}{f'(x)}\bigg|_{f(x)=0} \\ & = \frac{g(f^{-1}(0))}{f'(f^{-1}(0))} \end{aligned}$

Solution.

(The solution below is based on the manuscript of 2016-2017 PHYS3450 Electromagnetism Homework 1 Solution.)

Let the roots to $f(x)=0$ be $a_i$ where $i=1,2,3,\dots$ .
From $y=f(x)$ (or, $x=f^{-1}(y)$ ), we have $\mathrm{d}y=f'(x)\,\mathrm{d}x$ .

$\begin{aligned} &\quad \int \delta (f(x))g(x)\,\mathrm{d}x \\ & = \sum_i \int\delta (f(x))g(x)\,\mathrm{d}x\\ & = \sum_i \int\frac{\delta (y)g(f^{-1}(y))}{f'(f^{-1}(y))}\,\mathrm{d}y \\ & = \sum_i \frac{g(f^{-1}(0))}{f'(f^{-1}(0))} \\ & = \sum_i \frac{g(a_i)}{f'(a_i)} \\ \end{aligned}$

QED

May 24, 2021October 24, 2022

202105241107 Homework 1 (Q2)

Prove that the divergence of a curl is always zero; and the curl of a gradient is always zero.

Solution.

First,

Let $\mathbf{A}=A_x\,\hat{\mathbf{i}} + A_y\,\hat{\mathbf{j}} + A_z\,\hat{\mathbf{k}}$ .

$\begin{aligned} & \quad \nabla \cdot (\nabla\times \mathbf{A}) \\ & = \nabla \cdot \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix} \\ & = \nabla \cdot \Bigg( \bigg( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \bigg)\,\hat{\mathbf{i}} - \bigg( \frac{\partial A_z}{\partial x} - \frac{\partial A_x}{\partial z} \bigg)\,\hat{\mathbf{j}} + \bigg( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \bigg)\,\hat{\mathbf{k}} \Bigg) \\ & = \frac{\partial^2 A_z}{\partial x\partial y} - \frac{\partial^2 A_y}{\partial x\partial z}-\frac{\partial^2 A_z}{\partial y\partial x} + \frac{\partial^2A_x}{\partial y\partial z} + \frac{\partial^2A_y}{\partial z\partial x}-\frac{\partial^2A_x}{\partial z\partial y} \\ & = 0 \end{aligned}$

$\therefore$ The divergence of a curl is always zero.

Second,

$\nabla f(x,y,z) = \displaystyle{\bigg( \frac{\partial f(x,y,z)}{\partial x} ,\enspace \frac{\partial f(x,y,z)}{\partial y},\enspace \frac{\partial f(x,y,z)}{\partial z}\bigg)}$
$\mathbf{F}\stackrel{\textrm{def}}{=} \nabla f$

$\begin{aligned} \nabla \times \mathbf{F} & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \end{vmatrix} \\ & =\bigg( \frac{\partial^2f}{\partial y\partial z} - \frac{\partial^2f}{\partial z\partial y},\enspace \frac{\partial^2f}{\partial z\partial x} - \frac{\partial^2f}{\partial x\partial z},\enspace \frac{\partial^2f}{\partial x\partial y}-\frac{\partial^2f}{\partial y\partial x} \bigg) \\ & = 0 \\ \end{aligned}$

$\therefore$ The curl of a gradient is always zero.

April 24, 2021October 24, 2022

202104241713 Homework 1 (Q7)

The height of a mountain is given by $h(x,y)=3000-2x^2-y^2$ , where the $y$ -axis points east, the $x$ -axis points north, and all distances are measured in meters. Suppose a mountain climber is at the point $(30,\, -20,\, 800)$ , will he ascend or descend if he moves in the southwest direction?

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2155 Methods of Physics II Homework Solutions.)

The altitude $h(x,y)$ is given by a function of $x$ and $y$ :

$h(x,y)=3000-2x^2-y^2$ .

Now that the climber moves in the southwest direction

$\begin{aligned} \mathbf{n} & =-1\,\hat{\mathbf{i}}-1\,\hat{\mathbf{j}} \\ \hat{\mathbf{n}} & = \frac{1}{\sqrt{2}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}) \end{aligned}$

$\begin{aligned} h_{\hat{\mathbf{n}}}'(x,y) & = \nabla h(x,y)\cdot \hat{\mathbf{n}} \\ & = (-4x\,\hat{\mathbf{i}}-2y\,\hat{\mathbf{j}}) \cdot \frac{1}{\sqrt{2}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}) \\ & = \frac{1}{\sqrt{2}} (4x+2y) \end{aligned}$

At point $(30,\, -20,\, 800)$ ,

$\begin{aligned} h_{\hat{\mathbf{n}}}'(30,-20) & = \frac{1}{\sqrt{2}}\big( 4(30)+2(-20)\big) \\ & = \frac{1}{\sqrt{2}}\cdot 80 \qquad (>0) \end{aligned}$

he will ascend southwesterly.

April 22, 2021October 24, 2022

202104221623 Homework 2 (Q1)

Find the gradients of the following functions:

(a) $\phi (x,y,z)=x^2+y^2+z^2$ ;

(b) $\phi (x,y,z)=\exp (x+y)\cos z$ .

Solution.

(a)

$\begin{aligned} & \quad \nabla \phi \\ & = \bigg(\hat{\mathbf{i}}\frac{\partial}{\partial x}+\hat{\mathbf{j}}\frac{\partial}{\partial y}+\hat{\mathbf{k}}\frac{\partial}{\partial z}\bigg)\phi (x,y,z) \\ & = \hat{\mathbf{i}} \frac{\partial}{\partial x}(x^2+y^2+z^2) + \hat{\mathbf{j}} \frac{\partial}{\partial y}(x^2+y^2+z^2) + \hat{\mathbf{k}} \frac{\partial}{\partial z}(x^2+y^2+z^2) \\ & = 2x\,\hat{\mathbf{i}} + 2y\,\hat{\mathbf{j}} + 2z\,\hat{\mathbf{k}} \end{aligned}$

(b) It is left as an exercise to the reader.