December 2022 – Physicspupil

December 31, 2022January 7, 2023

202212311431 Problem 1.4

Let $g:\Omega\to\mathbb{R}$ be such that $\displaystyle{\frac{\partial g}{\partial x}}$ and $\displaystyle{\frac{\partial g}{\partial y}}$ exist at $(x_0,y_0)\in\Omega$ , and suppose that one of these partials exists in a neighbourhood of $(x_0,y_0)$ and is continuous at $(x_0,y_0)$ . Show that $g$ is real-differentiable at $(x_0,y_0)$ .

_{Extracted from R. B. Ash & W. P. Novinger. (2004). Complex Variables.}

Roughwork.

Granted that

$\begin{aligned} w=f(z)&=u(x,y)+iv(x,y) \\ \textrm{s.t. }A & =u_0+iv_0 \\ z_0 & =x_0+iy_0 \\ \end{aligned}$

the concepts of existence of limit

$\begin{aligned} &\qquad\enspace \lim_{z\to z_0}f(z) = A \\ & \Longleftrightarrow \begin{cases} \lim_{(x,y)\to (x_0,y_0)}u(x,y)=u_0 \\ \lim_{(x,y)\to (x_0,y_0)}v(x,y)=v_0 \\ \end{cases} \textrm{exist} \end{aligned}$

and continuity at a point

$\begin{aligned} &\qquad\enspace w=f(z)\textrm{ continuous at }z_0 \\ &\Longleftrightarrow \textrm{ both }u(x,y)\textrm{ and }v(x,y)\textrm{ continuous at }z_0 \\ \end{aligned}$

are held by definition.

This problem is not to be attempted.

December 31, 2022December 31, 2022

202212311359 Problem 1.3

Let $z_1$ and $z_2$ be nonzero complex numbers, and let $\theta$ ( $0\leqslant \theta\leqslant \pi$ ) be the angle between them. Show that

(a) $\mathrm{Re}z_1\overline{z}_2=|z_1||z_2|\cos\theta$ , $\mathrm{Im}z_1\overline{z}_2=\pm |z_1||z_2|\sin\theta$ , and consequently
(b) The area of the triangle formed by $z_1$ , $z_2$ , and $z_2-z_1$ is $|\mathrm{Im}z_1\overline{z}_2|/2$ .

_{Extracted from R. B. Ash & W. P. Novinger. (2004). Complex Variables.}

Roughwork.

The area of a triangle $\triangle ABC$ , constructed by any two sides $\mathbf{AB}$ and $\mathbf{AC}$ with an included angle $\theta$ , is

$\begin{aligned} \textrm{Area} & = \frac{1}{2}\mathbf{AB}\times \mathbf{AC} \\ \bigg( & =\frac{1}{2}(AB)(AC)\sin\theta\bigg) \\ \end{aligned}$

i.e., half the area of the parallelogram spanned by these two vectors.

Hence, $\mathtt{(a)} \Longrightarrow \mathtt{(b)}$ .

This problem is not to be attempted.

December 31, 2022January 1, 2023

202212311329 Problem 1.2

Show that $|z_1+z_2|=|z_1|+|z_2|$ iff $z_1$ and $z_2$ lie on a common ray from $0$ iff one of $z_1$ or $z_2$ is a nonnegative multiple of the other.

_{Extracted from R. B. Ash & W. P. Novinger. (2004). Complex Variables.}

Roughwork.

Let $z_1=r_1e^{i\theta_{1}}$ and $z_2=r_2e^{i\theta_{2}}$ , rewrite

$\begin{aligned} |r_1e^{i\theta_{1}}+r_2e^{i\theta_{2}}| & = |r_1e^{i\theta_{1}}|+|r_2e^{i\theta_{2}}| \\ ? = r_3 & = r_1 + r_2 \\ \end{aligned}$

by Lemma.

$\begin{aligned} r_3e^{i\theta_{3}} & = r_1e^{i\theta_{1}}+r_2e^{i\theta_{2}} \\ & = r_1(\cos\theta_1+i\sin\theta_1) + r_2(\cos\theta_2+i\sin\theta_2) \\ & = (r_1\cos\theta_1+r_2\cos\theta_2) + i(r_1\sin\theta_1+r_2\sin\theta_2) \\ r_3^2 & = r_1^2+r_2^2+2r_1r_2\cos (\theta_1-\theta_2) \\ r_3 & = \sqrt{r_1^2+r_2^2+2r_1r_2\cos (\theta_1-\theta_2)} \\ \end{aligned}$

we have

$\begin{aligned} \cos (\theta_1-\theta_2) & = 1 \\ \theta_1 - \theta_2 & = 0 \\ \theta_1 & = \theta_2 \\ \end{aligned}$

The rest is left an exercise for the reader.

December 30, 2022December 31, 2022

202212301726 Solution to 1987-CE-AMATH-I-3

For any complex number $z$ , let $\overline{z}$ , $|z|$ , and $\mathrm{Re}(z)$ be its conjugate, modulus, and real part respectively. Show that

$z+\overline{z}=2\mathrm{Re}(z)$ and $|z|\geqslant \mathrm{Re}(z)$ .

Hence, or otherwise, show that for any complex numbers $z_1$ and $z_2$ ,

$z_1z_2+\overline{z_1z_2}\leqslant 2|z_1||z_2|$ .

Roughwork.

In the field $\mathbb{C}=\{ x+iy:x,y\in\mathbb{R}\}$ of complex numbers, $x$ is called the real part of $z$ and $y$ the imaginary part, i.e.,

$z=x+iy=\mathrm{Re}(z)+i\,\mathrm{Im}(z)$ ,

its trigonometric form being $z=r(\cos\theta +i\sin\theta )$ with $r$ the modulus and $\theta$ the argument, and its exponential form, $z=re^{i\theta}$ .

This problem is not to be attempted.

December 30, 2022December 31, 2022

202212301149 Excerpt Article 001

Reading Comprehension

The three states of matter

_{(Excerpted passages from Ch. II Sec. I by S. E. Coleman in A Text-Book of Physics, 1911.)}

Matter exists in three physical states or conditions, called the solid, the liquid, and the gaseous states. Much of physics depends upon the characteristic properties which distinguish the states of matter from one another. Thus we have the mechanics of solids, the mechanics of liquids, and the mechanics of gases. These properties, therefore, require some attention at the outset.

Liquids are distinguished from solids by the fact that they tend to flow, and hence must be contained in vessels. Every solid, on the contrary, has a shape of its own, which it tends to preserve. Some solids, e.g. stone and iron, offer great resistance to a change of shape; others, such as wet clay and putty, can readily be molded into any form. But even the small amount of resistance offered by soft solids distinguishes them from liquids.

Many of the physical properties of gases may be learned from a study of the air, which is a mixture of several gases, principally nitrogen and oxygen. Although the air is everywhere about us, we are ordinarily unconscious of its existence unless it is in motion. When it is in motion, we recognize it as a current of air, a breeze, or a wind. We commonly call a vessel “empty” when it is full of air; and seldom stop to think that when the so-called empty vessel is being filled with a liquid or a solid, the air in it is being pushed out.

It will help toward clear thinking on this point to push an inverted tumbler into a vessel of water. The water does not rise to fill the tumbler, being prevented from doing so by the confined air; but when the tumbler is slowly inclined, the air escapes in a succession of bubbles, and the water enters at the same time to take its place.

This simple experiment shows that a body of air confined in any space tends to keep other bodies out of that space; and the same is true of all gases. But we know that, after a bicycle or an automobile tire is fully inflated, much air must still be pumped in to make it hard. Now air can be forced into the fully inflated tire only by compressing the air already in it into a smaller space; and experience teaches that the compression of the confined air can be carried as far as the strength of the tire or of the operator will permit. The great compressibility of air can be shown simply by pushing in the piston of a bicycle pump or other compression pump, while the outlet is closed with the finger. A vigorous push will compress the air perhaps to one half or even to one third of its original volume. When the piston is released, the air expands and drives it back.

All gases are highly compressible and expansible, like air. When any quantity of gas, however small, is admitted into an otherwise empty space it instantly expands so as to fill the space completely.

If the above experiment is repeated with the compression pump filled with water, it will be found that the water is as unyielding as a board, for the piston cannot be pushed in at all.

All liquids and most solids are only very slightly compressible. Even under very great pressure their change of volume is commonly so slight as to escape notice; and, for all practical purposes, they are regarded as incompressible. Hence great compressibility and expansibility are distinguishing properties of the gaseous state.

December 28, 2022December 29, 2022

202212281204 Solution to 2015-DSE-PHY-IA-5

A constant net force acting on an object of mass $m_1$ produces an acceleration $a_1$ while the same force acting on another object of mass $m_2$ produces an acceleration $a_2$ . If this net force acts on an object of mass $(m_1+m_2)$ , what would be the acceleration produced?

Roughwork.

Write

$\begin{aligned} F_\textrm{net} & = m_1a_1 = m_2a_2 \\ & = (m_1+m_2)a_3 \\ \end{aligned}$

Provided are four options. Let’s check them one by one.

A. $a_3\stackrel{?}{=}a_1+a_2$

$\begin{aligned} (m_1+m_2)a_3 & = (m_1+m_2)(a_1+a_2) \\ & = (m_1a_1) + (m_2a_2) +m_1a_2+m_2a_1 \\ & = F_\textrm{net} + F_\textrm{net} +m_1a_2+m_2a_1 \\ & \gneq 2F_\textrm{net} \\ \therefore\enspace a_3 & \neq a_1+a_2 \\ \end{aligned}$

B. $a_3\stackrel{?}{=}\displaystyle{\frac{a_1+a_2}{2}}$

$\begin{aligned} (m_1+m_2)a_3 & = (m_1+m_2)\bigg(\frac{a_1+a_2}{2}\bigg) \\ & \stackrel{\textrm{(A)}}{=} F_\textrm{net} + \frac{m_1a_2+m_2a_1}{2} \\ & \gneq F_\textrm{net} \\ \therefore\enspace a_3 & \neq \frac{a_1+a_2}{2} \\ \end{aligned}$

C. $a_3\stackrel{?}{=}\displaystyle{\frac{a_1a_2}{a_1+a_2}}$

$\begin{aligned} (m_1+m_2)a_3 & = (m_1+m_2)\bigg(\frac{a_1a_2}{a_1+a_2}\bigg) \\ & = \frac{(m_1a_1)a_2+(m_2a_2)a_1}{a_1+a_2} \\ & = \frac{(F_\textrm{net})(a_1+a_2)}{a_1+a_2} \\ & = F_\textrm{net} \\ \therefore\enspace a_3 & = \frac{a_1a_2}{a_1+a_2} \\ \end{aligned}$

D. $a_3\stackrel{?}{=}\displaystyle{\frac{2a_1a_2}{a_1+a_2}}$ is so not to check.

Try-and-err was slower if steadier paced than fright-but-fight from head start,

$\begin{aligned} a_3 & = \frac{F_\textrm{net}}{m_1+m_2} \\ & = \bigg(\frac{m_1}{F_\textrm{net}}+\frac{m_2}{F_\textrm{net}}\bigg)^{-1} \\ & = \bigg(\frac{1}{a_1}+\frac{1}{a_2}\bigg)^{-1} \\ & = \bigg(\frac{a_1+a_2}{a_1a_2}\bigg)^{-1} \\ & = \frac{a_1a_2}{a_1+a_2} \\ \end{aligned}$

And the answer is C.

This problem is not to be attempted.

December 28, 2022January 6, 2023

202212281153 Solution to 1999-CE-AMATH-II-2

Find $\displaystyle{\int x(x+2)^{99}\,\mathrm{d}x}$ .

Roughwork.

$\begin{aligned} &\quad\enspace \int x(x+2)^{99}\,\mathrm{d}x \\ \dots\, &\textrm{let }u=x+2\textrm{ s.t. }\mathrm{d}u=\mathrm{d}x\,\dots \\ & = \int (u-2)(u)^{99}\,\mathrm{d}u \\ & = \int \big( u^{100}-2u^{99}\big)\,\mathrm{d}u \\ & = \frac{u^{101}}{101} - \frac{u^{100}}{50} + C\textrm{ for some constant} \\ & = \frac{(x+2)^{101}}{101} - \frac{(x+2)^{100}}{50} + C \\ \end{aligned}$

This problem is not to be attempted.

December 28, 2022December 28, 2022

202212281117 Problem 1.18

Show that an equation for the circle $C(z_0,r)$ is

$z\overline{z}-\overline{z}_0z-z_0\overline{z}+z_0\overline{z}_0=r^2$ .

_{Extracted from R. B. Ash & W. P. Novinger. (2004). Complex Variables.}

Roughwork.

Writing

$\begin{aligned} |z-z_0| & = r \\ |(a+bi)-(a_0+b_0i)| & = r \\ |(a-a_0)+(b-b_0)i| & = r \\ (a-a_0)^2 + (b-b_0)^2 & = r^2 \\ \end{aligned}$

and

$\begin{aligned} &\quad\enspace z\overline{z}-\overline{z}_0z-z_0\overline{z}+z_0\overline{z}_0 \\ & = (a+bi)(a-bi) - (a_0-b_0i)(a+bi) \\ & \qquad\qquad -(a_0+b_0i)(a-bi)+(a_0+b_0i)(a_0-b_0i) \\ & = \cdots \\ \end{aligned}$

is to let ends meet, without use of properties of conjugates:

$\begin{aligned} |\overline{z}| & = |z| \\ \mathrm{arg}\overline{z} & = -\mathrm{arg}z \\ \overline{z_1+z_2} & = \overline{z}_1+\overline{z}_2 \\ \overline{z_1-z_2} & = \overline{z}_1-\overline{z}_2 \\ \overline{z_1z_2} & = \overline{z}_1\overline{z}_2 \\ \mathrm{Re}z & = (z+\overline{z})/2 \\ \mathrm{Im}z & = (z-\overline{z})/2i \\ z\overline{z} & = |z|^2 \\ \end{aligned}$

(to be continued)

December 23, 2022December 29, 2022

202212231640 Solution to 2016-DSE-PHY-IA-2

$0.3\,\mathrm{kg}$ of water at temperature $50\,^\circ\mathrm{C}$ is mixed with $0.2\,\mathrm{kg}$ of ice at temperature $0\,^\circ\mathrm{C}$ in an insulated container of negligible heat capacity. What is the final temperature of the mixture?

Given: specific heat capacity of water $=4200\,\mathrm{J\,kg^{-1}\,^\circ C^{-1}}$ ; specific latent heat of fusion of ice $3.34\times 10^5\,\mathrm{J\, kg^{-1}}$ .

Roughwork.

For all $0.3\,\mathrm{kg}$ liquid water, a temperature drop of $50\,\mathrm{C^\circ}$ will release

$\begin{aligned} \textrm{Sensible heat} & = (0.3)(4200)(50) \\ & = \textrm{63,000}\,\mathrm{J} \\ \end{aligned}$

and a phase transition of freezing will release further latent heat (of fusion)

$\begin{aligned} \textrm{Latent heat}& = ml_f \\ & = (0.3)(3.34\times 10^5) \\ & = \textrm{100,200}\,\mathrm{J} \\ \end{aligned}$

whereas for $0.2\,\mathrm{kg}$ of solid ice to melt all at once, latent heat (of liquidization)

$\begin{aligned} \textrm{Latent heat} & = ml_f \\ & = (0.2)(3.34\times 10^5) \\ & = \textrm{66,800}\,\mathrm{J} \\ \end{aligned}$

is to be absorbed.

Lemma.

Heat always moves from hotter objects to colder objects, unless energy in some form is supplied to reverse the direction of heat flow.

_{Wikipedia on Second law of thermodynamics}

By the inequalities

$0<\textrm{66,800}-\textrm{63,000}<\textrm{100,200}$

one will know.

December 23, 2022December 23, 2022

202212231530 Solution to 1990-CE-AMATH-I-7

Given the curve $C: x^2+4xy+5y^2=1$ , find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ . Hence find the equations of the two tangents to $C$ which are parallel to the line $\displaystyle{y=-\frac{1}{2}x}$ .

Roughwork.

For the curve,

$\begin{aligned} &\quad\enspace x^2+4xy+5y^2 = 1 \\ &\Rightarrow 2x\,\mathrm{d}x + 4(x\,\mathrm{d}y + y\,\mathrm{d}x) + 10y\,\mathrm{d}y = 0 \\ &\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}= -\frac{x+2y}{2x+5y} \\ \end{aligned}$

and for the line,

$\begin{aligned} &\quad\enspace y = -\frac{1}{2}x \\ &\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2} \\ \end{aligned}$

equating them,

$\begin{aligned} &\quad\enspace -\frac{x+2y}{2x+5y} = -\frac{1}{2} \\ &\Rightarrow \begin{pmatrix} x \\ y \end{pmatrix} = \bigg\{\begin{pmatrix} 1 \\ 0 \end{pmatrix} , \begin{pmatrix} -1 \\ 0 \end{pmatrix}\bigg\} \\ \end{aligned}$

are points of contact at which the tangents

$\begin{aligned} &\quad\enspace \frac{y-(\pm 1)}{x-(0)} = -\frac{1}{2} \\ &\Rightarrow \big\{ x + 2y \pm 2 = 0 \big\} \\ \end{aligned}$

are drawn.

This problem is not to be attempted.