Physicspupil – Page 27

November 22, 2021October 24, 2022

202111221624 Exercises 9.1.A (Q2)

For Exercises 1-8, determine if the given sequence is convergent. If so then find its limit.

2. $\displaystyle{\bigg\{ \frac{n^2}{3n^2+7n-2}\bigg\}_{n=1}^{\infty} }$

Solution.

By L’Hôpital’s Rule, treating an integer $n \ge 1$ as a real-valued variable $x$ ,

$\begin{aligned} &\quad \lim_{n\to\infty}\frac{n^2}{3n^2+7n-2} \\ & = \lim_{n\to\infty}\frac{2n}{6n+7} \\ & = \frac{2}{6} \\ & = \frac{1}{3}\\ \end{aligned}$

Thus the sequence is convergent and its limit is $1/3$ .

November 22, 2021October 24, 2022

202111221615 Exercises 9.1.A (Q1)

For Exercises 1-8, determine if the given sequence is convergent. If so then find its limit.

$\{ ne^{-n}\}_{n=0}^{\infty}$

Solution.

By L’Hôpital’s Rule, treating an integer $n \ge 0$ as a real-valued variable $x$ ,

$\begin{aligned} & \quad \lim_{n\to\infty} \frac{n}{e^n} \\ & = \lim_{n\to\infty} \frac{0}{e^n} \\ & = 0 \\ \end{aligned}$

Thus the sequence is convergent and its limit is $0$ .

November 22, 2021October 24, 2022

202111221509 Solution to 1971-CE-AMATH-1

When the functions $x^2-bx+3$ and $2b-x$ are divided by $(x-a)$ , the remainders are 1 and 4 respectively. Find $a$ and $b$ .

Solution.

Translating mathematically, we have

$\begin{aligned} x^2-bx+3 & = (x-a)F(x)+1 \\ 2b-x & = (x-a)G(x)+4\\ \end{aligned}$

Plugging in $x=a$ , we have

$\begin{aligned} (a)^2-b(a)+3 & = \big( (a)-a\big) F(a)+1 \\ 2b-(a) & = \big( (a)-a\big) G(a)+4 \\ \end{aligned}$

After simplifying it, we have

$\begin{aligned} a^2-ab & = -2\\ -a+2b & = 4\\ \end{aligned}$

Solving two equations with two unknowns,

$\begin{cases} & a =2\\ & b =3 \\ \end{cases}$

is the solution.

October 25, 2021October 24, 2022

202110251056 Solution to 1970-CE-AMATH-1

Given that $F(x)=(b-c)x^3-(a+2b-3c)x^2-(b+2c-3a)x-2(a-b)$ . By using the remainder theorem, prove that $(x-1)(x-2)$ is a factor of $F(x)$ . Find also the remaining factor of $F(x)$ .

Solution.

$\begin{aligned} F(1) & = (b-c)(1)^3 - (a+2b-3c)(1)^2 - (b+2c-3a)(1) - 2(a-b) = 0 \\ F(2) & = (b-c)(2)^3 - (a+2b-3c)(2)^2 - (b+2c-3a)(2) - 2(a-b) = 0 \\ \end{aligned}$

$\therefore$ $(x-1)(x-2)$ is a factor of $F(x)$ .

By trial-and-error,

$\begin{aligned} F(3) & = (b-c)(3)^3-(a+2b-3c)(3)^2-(b+2c-3a)(3)-2(a-b) \\ & = -2a+8b-6c \\ & \neq 0 \\ F(4) & = (b-c)(4)^3-(a+2b-3c)(4)^2-(b+2c-3a)(4)-2(a-b) \\ & = -6a+30b-24c \\ & \neq 0 \\ F(5) & = (b-c)(5)^3-(a+2b-3c)(5)^2-(b+2c-3a)(5)-2(a-b) \\ & = -12a+112b-60c \\ & \neq 0 \\ F(6) & = (b-c)(6)^3-(a+2b-3c)(6)^2-(b+2c-3a)(6)-2(a-b) \\ & = -20a+140b-120c \\ & \neq 0 \\ F(7) & = (b-c)(7)^3-(a+2b-3c)(7)^2-(b+2c-3a)(7)-2(a-b) \\ & = -30a+240b-210c \\ & \neq 0 \\ F(8) & = (b-c)(8)^3-(a+2b-3c)(8)^2-(b+2c-3a)(8)-2(a-b) \\ & = -42a+378b-336c \\ & \neq 0 \\ F(9) & = (b-c)(9)^3-(a+2b-3c)(9)^2-(b+2c-3a)(9)-2(a-b) \\ & = -56a + 560b - 504c \\ & \neq 0 \end{aligned}$

It may be feasible to work in this manner, however endlessly, to find a zero. But it is high time for me to work in another way round.

By long division, the remaining factor is

$(b-c)x+(b-a)$ .

One should check that

$\begin{aligned} &\quad (x-1)(x-2)\big( (b-c)x+(b-a)\big) \\ & = (x^2-3x+2)\big( (b-c)x+(b-a)\big) \\ & = (b-c)x^3 - (a+2b-3c)x^2 - (b+2c-3a)x - 2(a-b) \\ & = F(x) \\ \end{aligned}$

Remark.

This remaining zero, i.e., $\displaystyle{x=\frac{b-a}{b-c}}$ , is hard to find by trial-and-error, so long division is necessary.

Roughwork.

$\begin{aligned} & (b-c)x & +(-a+b) & & \\\cline{2-5} x^2-3x+2\quad \Big)& (b-c)x^3 & +(-a-2b+3c)x^2 & +(3a-b-2c)x & +(-2a+2b) \\ & (b-c)x^3 & +(-3b+3c)x^2 & + (2b-2c)x & \\\cline{2-5} & & (-a+b)x^2 & +(3a-3b)x & + (-2a+2b) \\ & & (-a+b)x^2 & +(3a-3b)x & + (-2a+2b) \\\cline{3-5} \end{aligned}$

October 25, 2021October 24, 2022

202110251042 Solution to 1967-CE-AMATH-1

When $14x^4-27x^3+ax^2+12x+9$ is divided by $(x-2)$ the remainder is $5$ . What is the value of $a$ ?

Solution.

$\begin{aligned} f(x) & \stackrel{\textrm{def}}{=} 14x^4-27x^3+ax^2+12x+9 \\ f(x) & = (x-2)g(x) + 5 \\ f(2) & = 5 \\ 5 & = 14(2)^4-27(2)^3+a(2)^2+12(2)+9 \\ a & = 9 \\ \end{aligned}$

$\therefore$ The value of $a$ is $9$ .

October 15, 2021October 24, 2022

202110150945 Exercises 8.1.A (Q1)

For Exercises 1-6, find the area of the region bounded by the given curves.

$y=x^2$ and $y=2x+3$

Roughwork.

$\begin{aligned} y& = x^2 = 2x+3 \\ 0 &= x^2-2x-3 \\ 0 &= (x-3)(x+1) \\ x &=3,-1 \end{aligned}$

$\therefore$ $\{(3,9),(-1,1)\}$ are the points of intersection.

The area $A$ of the shaded region is

$\begin{aligned} A & = \int_{-1}^{3}|(2x+3)-(x^2)|\,\mathrm{d}x \\ & = \bigg[ x^2+3x-\frac{x^3}{3}\bigg]\bigg|_{-1}^{3} \\ & = \bigg( (3)^2+3(3)-\frac{(3)^3}{3}\bigg) - \bigg( (-1)^2+3(-1)-\frac{(-1)^3}{3}\bigg) \\ & = (9) - \bigg(-\frac{5}{3}\bigg) \\ & = \frac{32}{3}\quad \textrm{(squared units)} \\ \end{aligned}$

(alternatively)

$\begin{array}{cl} {A & = \displaystyle\int_{1}^{9}\bigg( \sqrt{y} - \frac{y-3}{2}\bigg)\,\mathrm{d}y + \int_{0}^{1}\Big( (\sqrt{y})-(-\sqrt{y})\Big)\,\mathrm{d}y }\\ & = \displaystyle\bigg( \frac{2y^{3/2}}{3}-\frac{y^2}{4}+\frac{3y}{2}\bigg)\bigg|_{1}^{9} + \bigg( \frac{4y^{3/2}}{3}\bigg)\bigg|_{0}^{1}} \\ & = \displaystyle\Bigg\{ \bigg[ \frac{2(9)^{3/2}}{3}-\frac{(9)^2}{4}+\frac{3(9)}{2}\bigg] -\bigg[\frac{2(1)^{3/2}}{3}-\frac{(1)}{4}+\frac{3(1)}{2}\bigg]\Bigg\} }\\ &\displaystyle\qquad \quad + \Bigg\{\bigg[\frac{4(1)^{3/2}}{3}\bigg] - \bigg[ \frac{4(0)^{3/2}}{3} \bigg]\Bigg\} }\\ & =\displaystyle \bigg( 18-\frac{81}{4}+\frac{27}{2}\bigg) - \bigg( \frac{2}{3}-\frac{1}{4}+\frac{3}{2} \bigg) + \bigg(\frac{4}{3}\bigg) - (0)} \\ & = \displaystyle\frac{32}{3}\quad \textrm{(squared units)}} \end{array}$

October 12, 2021October 24, 2022

202110121413 Exercise 8.5.B (Q21)

This exercise is related to Einstein’s famous law $E=mc^2$ . The relativistic momentum $p$ of a particle of mass $m$ moving at a speed $v$ along a straight line (say, the $x$ -axis) is

$\displaystyle{p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}}$ ,

where $c$ is the speed of light. The relativistic force on the particle along that line is

$\displaystyle{F=\frac{\mathrm{d}p}{\mathrm{d}t}}$ ,

which is the same formula as Newton’s Second Law of motion in classical mechanics. Assume that the particle starts at rest at position $x_1$ and ends at position $x_2$ along the $x$ -axis. The work done by the force $F$ on the particle is:

$\displaystyle{W=\int_{x_1}^{x_2}F\,\mathrm{d}x=\int_{x_1}^{x_2}\frac{\mathrm{d}p}{\mathrm{d}t}\,\mathrm{d}x}$

(a) Show that

$\displaystyle{\frac{\mathrm{d}p}{\mathrm{d}v}=\frac{m}{\bigg( \displaystyle{1-\frac{v^2}{c^2}}\bigg)^{3/2}}}$ .

(b) Use the Chain Rule formula

$\displaystyle{\frac{\mathrm{d}p}{\mathrm{d}t}=\frac{\mathrm{d}p}{\mathrm{d}v}\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}}$

to show that

$\displaystyle{F\,\mathrm{d}x=v\frac{\mathrm{d}p}{\mathrm{d}v}\,\mathrm{d}v}$ .

(c) Use parts (a) and (b) to show that

$\displaystyle{W=\int_{0}^{v}\frac{\mathrm{d}p}{\mathrm{d}v}\, v\,\mathrm{d}v=\int_{0}^{v}\frac{mv}{\bigg(\displaystyle{ 1-\frac{v^2}{c^2}}\bigg)^{3/2}}\,\mathrm{d}v}$ .

(d) Use part (c) to show that

$\displaystyle{W=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}-mc^2}$ .

(e) Define the relativistic kinetic energy $K$ of the particle to be $K=W$ , and define the total energy $E$ to be

$\displaystyle{E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}}$ .

So by part (d), $K=E-mc^2$ . Show that

$E^2=p^2c^2+(mc^2)^2$ .

(Hint: Expand the right side of that equation.)

(f) What is $E$ when the particle is at rest?

Solution.

(a)

$\begin{aligned} \frac{\mathrm{d}p}{\mathrm{d}v} & = \frac{\mathrm{d}}{\mathrm{d}v}\bigg( \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\bigg) \\ & = \frac{\bigg(\sqrt{1-\displaystyle{\frac{v^2}{c^2}}}\bigg)(m)-(mv)\bigg(\displaystyle{\frac{1}{2}}\Big( 1-\displaystyle{\frac{v^2}{c^2}}\Big)^{-1/2}\Big( -\displaystyle{\frac{2v}{c^2}}\Big)\bigg)}{\bigg(\sqrt{1-\displaystyle{\frac{v^2}{c^2}}}\bigg)^2} \\ & = \frac{\bigg( 1-\displaystyle{\frac{v^2}{c^2}}\bigg) (m)-(mv)\bigg( \displaystyle{-\frac{v}{c^2}}\bigg)}{\bigg( 1-\displaystyle{\frac{v^2}{c^2}}\bigg)^{3/2}} \\ & = \frac{m}{\bigg( 1-\displaystyle{\frac{v^2}{c^2}}\bigg)^{3/2}} \end{aligned}$

(b)

$\begin{aligned} \int_{x_1}^{x_2}F\,\mathrm{d}x & = \int_{x_1}^{x_2}\frac{\mathrm{d}p}{\mathrm{d}t}\,\mathrm{d}x \\ \int_{x_1}^{x_2}F\,\mathrm{d}x & = \int_{x_1}^{x_2}\frac{\mathrm{d}p}{\mathrm{d}v}\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}x \\ F\,\mathrm{d}x & = \frac{\mathrm{d}p}{\mathrm{d}v}\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}x \\ & = v\,\frac{\mathrm{d}p}{\mathrm{d}v}\,\mathrm{d}v\\ \end{aligned}$

(c)

$\begin{aligned} W & =\int_{0}^{v}\frac{\mathrm{d}p}{\mathrm{d}v}\, v\,\mathrm{d}v \\ & = \int_{0}^{v}\Bigg( \frac{m}{\big( 1-\frac{v^2}{c^2}\big)^{3/2}}\Bigg) (v)\,\mathrm{d}v\\ & =\int_{0}^{v}\frac{mv}{\bigg(\displaystyle{ 1-\frac{v^2}{c^2}}\bigg)^{3/2}}\,\mathrm{d}v \\ \end{aligned}$ .

(d)

$\begin{aligned} W & = \int_{0}^{v}\frac{mv}{\bigg(\displaystyle{1-\frac{v^2}{c^2}}\bigg)^{3/2}}\,\mathrm{d}v \\ \dots\enspace & \textrm{let }u=\frac{v}{c}\enspace\dots \\ \dots\enspace & \textrm{then }\mathrm{d}v=c\,\mathrm{d}u\enspace\dots \\ & = \int_{0}^{\frac{v}{c}}\frac{mc^2u}{(1-u^2)^{\frac{3}{2}}}\,\mathrm{d}u \\ \dots\enspace & \textrm{let }u=\sin\theta\enspace\dots \\ \dots\enspace & \textrm{then }\mathrm{d}u=\cos\theta\,\mathrm{d}\theta\enspace\dots \\ & = \int\frac{mc^2\sin\theta}{\cos^3\theta}\cdot\cos\theta\,\mathrm{d}\theta \\ & = \int\frac{mc^2\sin\theta}{\cos^2\theta}\,\mathrm{d}\theta \\ \dots\enspace & \textrm{let }w=\cos\theta\enspace\dots \\ \dots\enspace & \textrm{then }\mathrm{d}w=-\sin\theta\,\mathrm{d}\theta\enspace\dots \\ & \quad -\int \frac{mc^2}{w^2}\,\mathrm{d}w \\ & = \frac{mc^2}{w} \\ & = \frac{mc^2}{\cos\theta} \\ & = mc^2\sec\theta \\ & = \frac{mc^2}{\sqrt{1-u^2}} \\ & = \enspace\dots \\ & = \bigg[\frac{mc^2}{\sqrt{1-u^2}}\bigg]\bigg|^{\frac{v}{c}}_{0} \\ & = \frac{mc^2}{\sqrt{1-\displaystyle{\frac{v^2}{c^2}}}}-mc^2 \\ \end{aligned}$

Parts (e) and (f) are left to the readers.

October 12, 2021October 24, 2022

202110121137 Exercise 1.1.1

(a) Show that $x^2\in\langle x-y^2,xy\rangle$ in $k[x,y]$ ( $k$ any field).

(b) Show that $\langle x-y^2,xy,y^2\rangle = \langle x,y^2\rangle$ .

(c) Is $\langle x-y^2,xy\rangle = \langle x^2,xy\rangle$ ? Why or why not?

Definition.

Let $f_1,\dots ,f_s\in k[x_1,\dots ,x_n]$ . We let $\langle f_1,\dots ,f_s\rangle$ denote the collection $\langle f_1,\dots ,f_s\rangle = \{ p_1f_1+\cdots +p_sf_s:p_i\in k[x_1,\dots ,x_n]\}$ for $i= 1,\dots ,s$ .

(a)

$\begin{aligned} x^2 & = (x)(x-y^2) + (y)(xy) \\ & \in \langle x-y^2,xy\rangle \\ \end{aligned}$

(b)

First, we want to show

$\langle x-y^2,xy,y^2\rangle \in \langle x,y^2\rangle$ .

For any $p_1,p_2,p_3\in k[x,y]$ ,

$\begin{aligned} & \quad \langle x-y^2,xy,y^2\rangle \\ & = (p_1)(x-y^2)+(p_2)(xy)+(p_3)(y^2) \\ & = p_1x-p_1y^2+p_2xy+p_3y^2 \\ & = (p_1+p_2y)(x)+(p_3-p_1)(y^2) \\ & \in \langle x,y^2\rangle \\ \end{aligned}$

Next, we want to show

$\langle x,y^2\rangle\in\langle x-y^2,xy,y^2\rangle$ .

For any $p_1,p_2\in k[x,y]$ ,

$\begin{aligned} &\quad \langle x,y^2\rangle \\ & = (p_1)(x)+(p_2)(y^2) \\ & = (p_1)(x-y^2)+(0)(xy)+(p_1+p_2)(y^2) \\ & \in \langle x-y^2,xy,y^2\rangle \\ \end{aligned}$

All in all,

$\langle x-y^2,xy,y^2\rangle = \langle x,y^2\rangle$ .

(c) I guess $\textrm{\scriptsize{NOT}}$ .

October 12, 2021October 24, 2022

202110120936 Exercises 13.3.6 (Q43)

Graph each function by algebraically determining its key features. Then state the domain and range of the function.

$f(x)=x^2-7x+12$

Solution.

$\begin{aligned} f(x) & = x^2-7x+12 \\ & = (x-3)(x-4) \end{aligned}$

We see that $y=f(x)=0$ when $x=\{ 3,4\}$ .

i. $\therefore$ The $x$ -intercepts of $f(x)$ are thus $(3,0)$ and $(4,0)$ .

Plugging in $x=0$ will give the $y$ -intercept:

$\begin{aligned} y\textrm{-intercept}& = f(0) \\ & = (0)^2-7(0)+12 \\ & = 12 \end{aligned}$

ii. $\therefore$ The $y$ -intercept is thus $(0,12)$ .

Differentiating $y=f(x)$ with respect to $x$ ,

$\begin{aligned} & \quad f'(x) \\ & = \frac{\mathrm{d}}{\mathrm{d}x}\big( f(x)\big) \\ & = \frac{\mathrm{d}}{\mathrm{d}x}(x^2-7x+12) \\ & = 2x-7 \\ \end{aligned}$

When $x=3.5$ and $f(3.5)=-0.25$ , the slope of $f(x)$ is zero, i.e., $f'(x)=0$ .

iii. $\therefore$ $(\frac{7}{2},-\frac{1}{4})$ is a turning point (/extreme point/vertex).

$\begin{aligned} y & =f(x) \\ & = x^2-7x+12 \\ & = x^2-7x+\bigg( \frac{7}{2}\bigg)^2 - \bigg( \frac{7}{2}\bigg)^2 + 12 \\ & = \bigg( x-\frac{7}{2}\bigg)^2 - \frac{1}{4}\\ \end{aligned}$

iv. $\therefore$ The axis of symmetry of the graph is $x=\frac{7}{2}$ .

Differentiating $y=f(x)$ twice with respect to $x$ ,

$\begin{aligned} &\quad f''(x)\\ & = \frac{\mathrm{d}}{\mathrm{d}x}\big( f'(x)\big) \\ & = \frac{\mathrm{d}}{\mathrm{d}x}(2x-7) \\ & = 2 \quad (>0)\\ \end{aligned}$

We see that the slope is increasing with $x$ .

v. $\therefore$ The graph of $f(x)$ is concave upward (/convex downward).

vi. $\therefore$ The domain is $\mathbb{R}$ and the range $\{ y\in\mathbb{R}:y\ge -\frac{1}{4}\}$ .

October 11, 2021October 25, 2022

202110111120 Solution to 2012-DSE-PHY-1A-5

There are two forces $\mathbf{F_1}$ and $\mathbf{F_2}$ , of constant magnitudes (i.e., $F_1=|\mathbf{F_1}|=\textrm{Const.}$ ; $F_2=|\mathbf{F_2}|=\textrm{Const.}$ ), acting at the same point. The angle $\theta$ between $\mathbf{F_1}$ and $\mathbf{F_2}$ increases from $0^\circ$ to $180^\circ$ .Apparently from the figure, $F_1>F_2$ .

Let the direction of $\mathbf{F_2}$ be fixed due east.

Then,

$\begin{aligned} \mathbf{F_1} & = F_1\cos\theta\,\hat{\mathbf{i}}+F_1\sin\theta\,\hat{\mathbf{j}} \\ \mathbf{F_2} & = F_2\,\hat{\mathbf{i}} + 0\,\hat{\mathbf{j}} \\ \mathbf{F_3} & = \mathbf{F_1} + \mathbf{F_2} \\ & = ( F_1\cos\theta + F_2 )\,\hat{\mathbf{i}} + F_1\sin\theta\,\hat{\mathbf{j}} \\ \end{aligned}$

$\begin{aligned} F_3 & = |\mathbf{F_3}| \\ & = \sqrt{(F_1\cos\theta + F_2)^2+(F_1\sin\theta )^2} \\ & = \sqrt{(F_1)^2\cos^2\theta + 2F_1F_2\cos\theta +(F_2)^2+(F_1)^2\sin^2\theta} \\ & = \sqrt{(F_1)^2+(F_2)^2+2F_1F_2\cos\theta} \\ \end{aligned}$

If $\theta =0^\circ$ , then $\cos\theta =1$ and $F_3=F_1+F_2$ ;

if $\theta =90^\circ$ , then $\cos\theta =0$ and $F_3=\sqrt{(F_1)^2+(F_2)^2}$ ;

if $\theta =180^\circ$ , then $\cos\theta =-1$ and $F_3=F_1-F_2$ .

By the triangle inequality,

$F_1+F_2>F_3=\sqrt{(F_1)^2+(F_2)^2}$ .

So the magnitude $F_3$ of the resultant force $\mathbf{F_3}$ decreases throughout.

(Countercheck)

Differentiating $F_3$ w.r.t. $\theta$ ,

$\begin{aligned} \quad \frac{\mathrm{d}}{\mathrm{d}\theta}(F_3) & = \frac{\mathrm{d}}{\mathrm{d}\theta} \sqrt{(F_1)^2+(F_2)^2+2F_1F_2\cos\theta} \\ & = \frac{-F_1F_2\sin\theta}{\sqrt{(F_1)^2+(F_2)^2+2F_1F_2\cos\theta}}\\ \end{aligned}$

As $\sin\theta \ge 0$ for $0\le \theta \le \pi$ , and $\sqrt{[\cdots ]}\ge 0$ , we have

$\displaystyle{\frac{\mathrm{d}F_3}{\mathrm{d}\theta}\le 0}$ .

And the answer is A.