November 2022 – Page 2

November 22, 2022December 3, 2022

202211221415 Exercise 29.4.1

Use this fact, that the difference between a vacuum energy density $\rho_v$ and $\rho_m$ or $\rho_r$ is that $\rho_v$ does not change as the universe expands (though it may change for other physical reasons), the basic Friedman equation Eq. (29.13):

$\displaystyle{\dot{a}^2-\frac{8\pi G}{3}(\rho_m+\rho_r+\rho_v)a^2=K}$

with $K=0$ , and the assumption of vacuum dominance ( $\rho_m=\rho_r\approx 0$ ) to show that Eq. (29.9):

$a(t)=a(t_s)\exp \Big(\sqrt{\frac{8}{3}\pi G\rho_v}(t-t_s)\Big)$ .

This form of the equation is valuable because it no longer refers to the present state of the universe.

_{Extracted from Thomas A. Moore. (2013). A General Relativity Workbook.}

Roughwork.

This solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

$\begin{aligned} 0 & = \dot{a}^2-\frac{8\pi G}{3}\rho_va^2\\ \bigg(\frac{\mathrm{d}a}{\mathrm{d}t}\bigg)^2 & = \frac{8\pi G}{3}\rho_va^2 \\ \frac{1}{a}\frac{\mathrm{d}a}{\mathrm{d}t} & = \sqrt{\frac{8\pi G}{3}\rho_v} \\ \ln \big( a(t)\big) & = \bigg(\sqrt{\frac{8\pi G}{3}\rho_v}\bigg)\cdot t + C \\ a(t)& =a(t_s)\exp \bigg(\sqrt{\frac{8}{3}\pi G\rho_v}(t-t_s)\bigg) \\ \end{aligned}$

November 22, 2022November 22, 2022

202211221145 Solution to 1972-CE-AMATH-I-XX

The equation of the ellipse shown in the figure below is given by

$9x^2+ky^2=144$ .

Find the value of $k$ and the coordinates of the point $A$ .

Roughwork.

$\begin{aligned} 9x^2+ky^2-144 & = 0 \\ 18x\,\mathrm{d}x+2ky\,\mathrm{d}y & = 0 \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = -\frac{9x}{ky} \\ \end{aligned}$

The equation of a standard ellipse centred at the origin with width $2a$ and height $2b$ is analytically

$\displaystyle{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$ .

_{Wikipedia on Ellipse}

$\begin{aligned} 9x^2+ky^2 & = 144 \\ \frac{x^2}{(4)^2} + \frac{y^2}{(12/\sqrt{k})^2} & = 1 \\ a & = 4 \\ b & = \frac{12}{\sqrt{k}} \\ \end{aligned}$

This problem is not to be attempted.

November 22, 2022November 22, 2022

202211220908 Solution to 1999-CE-AMATH-I-1

Find

(a) $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}\sin (x^2+1)}$ ,
(b) $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}\bigg[\frac{\sin (x^2+1)}{x}\bigg]}$ .

Roughwork.

Let $g(x)=\sin x$ and $h(x)=x^2+1$ s.t.

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\sin (x^2+1) & = \frac{\mathrm{d}}{\mathrm{d}x}(g\circ h)(x) \\ & = (g\circ h)'(x) \\ & = (g'\circ h)\cdot h'(x) \\ & = \frac{\mathrm{d}g(h(x))}{\mathrm{d}h(x)}\cdot\frac{\mathrm{d}h(x)}{\mathrm{d}x} \\ & = \cos (h(x))\cdot 2x \\ & = 2x\cos (x^2+1) \end{aligned}$

Next, let $f(x)=x^{-1}$ s.t.

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\bigg[\frac{\sin (x^2+1)}{x}\bigg] & = \frac{\mathrm{d}}{\mathrm{d}x}\Big( g(h(x))\cdot f(x)\Big) \end{aligned}$

The remaining are left the reader as an exercise.

November 21, 2022November 21, 2022

202211211714 Solution to 1976-CE-AMATH-I-XX

Express in polar form the following complex numbers:

$z=-\sqrt{3}-\mathrm{i}$ and $w=1-\mathrm{i}$ .

Hence or otherwise express $\displaystyle{\frac{z^3}{w^4}}$ in the form of $a+\mathrm{i}b$ .

Roughwork.

$\begin{aligned} \mathrm{mod}(z) & = |z| \\ & = |-\sqrt{3}-\mathrm{i}|\\ & = \sqrt{(-\sqrt{3})^2+(-1)^2} \\ & = 2 \\ \mathrm{arg}(z) & = \arctan\bigg(\frac{-1}{-\sqrt{3}}\bigg) \\ & = \frac{4\pi}{3} \\ \mathrm{mod}(w) & = |w| \\ & = |1-\mathrm{i}| \\ & = \sqrt{(1)^2+(-1)^2} \\ & = \sqrt{2} \\ \mathrm{arg}(w) & = \arctan\bigg(\frac{-1}{1}\bigg) \\ & = \frac{3\pi}{4} \\ \end{aligned}$

$\begin{cases} z=2e^{\mathrm{i}\frac{4\pi}{3}} \\ w=\sqrt{2}e^{\mathrm{i}\frac{3\pi}{4}} \\ \end{cases}$

$\begin{aligned} \frac{z^3}{w^4} & = z^3w^{-4} \\ & = \big( 2e^{\mathrm{i}\frac{4\pi}{3}}\big)^3\big(\sqrt{2}e^{\mathrm{i}\frac{3\pi}{4}}\big)^{-4} \\ & = \dots \\ \end{aligned}$

This problem is not to be attempted.

November 21, 2022November 21, 2022

202211211635 Solution to 1976-CE-AMATH-I-XX

Let the complex number $z_1=-1+\mathrm{i}\sqrt{3}$ .

i. Represent $z_1$ and its conjugate $\overline{z_1}$ on the Argand diagram.
ii. Calculate the modulus and the argument of $z_1$ .
iii. If $z_2=(z_1)^2$ , express $z_2$ in the form of $a+\mathrm{i}b$ .
iv. Show that both $z_1$ and $\displaystyle{\frac{1}{2}z_2}$ are roots of the equation $z^3-8=0$ .

Roughwork.

$\begin{aligned} \textrm{modulus }\mathrm{mod}(z_1) & = |z_1| \\ & = |-1+\mathrm{i}\sqrt{3}| \\ & = \sqrt{(-1)^2+(\sqrt{3})^2} \\ & = 2 \\ \textrm{argument }\mathrm{arg}(z_1) & = \arctan\bigg(\frac{\sqrt{3}}{-1}\bigg) \\ & = \frac{2\pi}{3}\\ \end{aligned}$

Thus, $z_1=2e^{\mathrm{i}\frac{2\pi}{3}}$ .

As is given,

$\begin{aligned} z_2 & = (z_1)^2 \\ & = \big( 2e^{\mathrm{i}\frac{2\pi}{3}}\big)^2 \\ & = 4e^{\mathrm{i}\frac{4\pi}{3}} \\ \mathrm{mod}(z_2) & = 4 \\ \mathrm{arg}(z_2) & = \frac{4\pi}{3} \\ \end{aligned}$

or,

$\begin{aligned} z_2 & = (z_1)^2 \\ & = (-1+\mathrm{i}\sqrt{3})^2 \\ & = (-1)^2 +2(-1)(\mathrm{i}\sqrt{3}) + (\mathrm{i}\sqrt{3})^2 \\ & = 1 -\mathrm{i}(2\sqrt{3}) -3 \\ & = -2-\mathrm{i}(2\sqrt{3}) \\ \end{aligned}$

This problem is not to be attempted.

November 8, 2022November 8, 2022

202211081129 Python Formula 002

Coulomb’s law states that $F$ the magnitude of the electrostatic force of attraction or repulsion between two point charges $Q_1$ and $Q_2$ is directly proportional to $Q_1Q_2$ the product of the magnitudes of charges and inversely proportional to $r^2$ the square of the distance between them:

$\displaystyle{F=\frac{Q_1Q_2}{4\pi\epsilon_0r^2}}$

_{Wikipedia on Coulomb’s law}

Roughwork.

pi = 3.1415926535

epsilon_0 = 8.8541878128e-12

charge_1 = float(input("Charge of Q_1: "))

charge_2 = float(input("Charge of Q_2: "))

distance = float(input("Separating distance: "))

force = charge_1 * charge_2 / (4 * pi * epsilon_0 * distance ** 2)

print("Electrostatic force (in Newton) is:", force)

The nucleus of a helium atom contains two protons $\sim 3.8\times 10^{-15}\,\mathrm{m}$ apart each of electric charge $+e=+1.6\times 10^{-19}\,\mathrm{C}$ . The electrostatic force of repulsion between them is

Charge of Q_1: 1.6e-19

Charge of Q_2: 1.6e-19

Separating distance: 3.8e-15

Electrostatic force (in Newton) is: 15.933609826070786

November 8, 2022November 8, 2022

202211081044 Python Formula 001

Converting between temperature scales.

$\begin{aligned} & x\,^\circ\mathrm{F}\triangleq (x-32)\times \frac{5}{9}\,^\circ\mathrm{C} & \qquad & x\,^\circ\mathrm{C}\triangleq (x\times \frac{9}{5}+32)\,^\circ\mathrm{F} \\ & x\,^\circ\mathrm{F}\triangleq (x+459.67)\times \frac{5}{9}\,\mathrm{K} & \qquad & x\,\mathrm{K}\triangleq (x\times \frac{9}{5}-459.67)\,^\circ\mathrm{F} \\ & x\,^\circ\mathrm{F} \triangleq (x+459.67) ^\circ\mathrm{R} & \qquad & x\,^\circ\mathrm{R}\triangleq (x-459.67)\,^\circ\mathrm{F} \\ \end{aligned}$

Roughwork.

degree_Celsius = float(input("Input degrees Celsius: "))

degree_Fahrenheit = degree_Celsius * (9 / 5) + 32

print(degree_Fahrenheit)

One has a slight fever when one’s body temperature is $37.8\,^\circ\mathrm{C}$ , or in degrees Fahrenheit,

1 2	Input degrees Celsius: 37.8 100.03999999999999

The rest of conversion of temperature scales, in Celsius, Fahrenheit, Kelvin, Rankine, etc., are left for the reader to code.

November 4, 2022November 4, 2022

202211041452 Solution to 1972-CE-AMATH-I-10

A point $P$ moves such that its distance from the point $(1,5)$ is equal to its distance from the line $y=x$ . Write down and simplify the equation of the locus of $P$ .

Roughwork.

Let the locus of $P$ be $t$ -parametrized as $(x_P(t),y_P(t))=(t,f(t))$ , and let the line $L:y=x$ be $s$ -parametrized as $(x_L(s),y_L(s))=(s,s)$ .

For the shortest line joining any point $P$ on the locus to its perpendicular projection on $L$ , its slope is:

$\textrm{Slope}_\alpha = \displaystyle{\frac{-1}{y_L'(s)} = \frac{-1}{1} = -1}$

with angle of inclination $\displaystyle{\alpha =\frac{3\pi}{4}}$ , and for the line joining any point $P$ on the locus to point $(1,5)$ , its slope is

$\textrm{Slope}_\beta =\displaystyle{\frac{y_P(t_0)-5}{x_P(t_0)-1}}$ ,

with angle of inclination $\displaystyle{\beta =\tan^{-1}\bigg(\frac{y_P-5}{x_P-1}\bigg)}$ , whereas the slope of tangent at any point $(x_P(t_0),y_P(t_0))$ on the locus is

$\textrm{Slope}_\gamma =f'(t_0)=\displaystyle{\frac{\mathrm{d}f(t)}{\mathrm{d}t}\bigg|_{t=t_0}}$ .

with angle of inclination $\gamma =\tan^{-1}(f'(t_0))$ .

Have a picture in mind

Recall

$\displaystyle{\tan (A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}}$ .

Hence,

$\begin{aligned} \tan (\gamma -\alpha ) & = \tan (\gamma -\beta ) \\ \frac{\tan\gamma -\tan\alpha}{1+\tan\gamma\tan\alpha} & = \frac{\tan\gamma -\tan\beta}{1+\tan\gamma\tan\beta} \\ \frac{(f'(t_0))-(-1)}{1+(f'(t_0))(-1)} & = \frac{(f'(t_0))-\bigg(\displaystyle{\frac{y(t_0)-5}{x(t_0)-1}}\bigg)}{1+(f'(t_0))\bigg(\displaystyle{\frac{y(t_0)-5}{x(t_0)-1}}\bigg)} \\ \end{aligned}$