May 2022 – Physicspupil

May 31, 2022October 24, 2022

202205311653 Exercise 1W

When $a=1$ , $b=2$ , $c=5$ , $d=6$ , and $m=0$ , give the value of

1. $ad^2+5bc$ 2. $a^2bc^2-3ad+a^4$ 3. $mabc+3a^2b^2c^2$ 4. $b^2cd-abc^2$ 5. $a^4b^3c+bcd$ 6. $ma^3+4cd-ad$ 7. $\displaystyle{\frac{ab^2}{2}+5c}$ 8. $\displaystyle{\frac{md}{3}+\frac{4cd}{b}}$ 9. $\displaystyle{\frac{ac}{d}+a}$ 10. $\displaystyle{\frac{a^2}{c}+3a^2b}$ 11. $\displaystyle{\frac{ab^2}{a}-\frac{a^5d}{b}}$ 12. $\displaystyle{\frac{c+d}{d}+\frac{d-c}{a}}$ 13. $\displaystyle{\frac{d^2-a^2b^2}{a^2b^2}}$ 14. $\displaystyle{\frac{mcd+3abc}{d^2-4b^2}}$ 15. $\displaystyle{\frac{a^2d^2-9b^2}{abcd}}$

_{J. R. Lux & R. S. Pieters. (1969). Exercises in Elementary Algebra}

Solution.

1.

$\begin{aligned} ad^2+5bc & = (1)(6)^2+5(2)(5) \\ & = 36 + 50 \\ & = 86 \\ \end{aligned}$

2.

$\begin{aligned} a^2bc^2-3ad+a^4 & = (1)^2(2)(5)^2-3(1)(6)+(1)^4 \\ & = 50-18+1 \\ & = 33 \\ \end{aligned}$

3.

$\begin{aligned} mabc+3a^2b^2c^2 & = (0)(1)(2)(5)+3(1)^2(2)^2(5)^2 \\ & = 0 + 3(4)(25) \\ & = 300 \\ \end{aligned}$

4.

$\begin{aligned} b^2cd-abc^2 & = (2)^2(5)(6)-(1)(2)(5)^2 \\ & = (4)(5)(6)-(2)(25) \\ & = 120-50 \\ & = 70 \\ \end{aligned}$

5.

$\begin{aligned} a^4b^3c+bcd & = (1)^4(2)^3(5)+(2)(5)(6) \\ & = 8(5) + 60 \\ & = 40 + 60 \\ & = 100 \\ \end{aligned}$

6.

$\begin{aligned} ma^3+4cd-ad & = (0)(1)^3+4(5)(6)-(1)(6) \\ & = 0 + 120 - 6 \\ & = 114 \\ \end{aligned}$

7.

$\begin{aligned} \frac{ab^2}{2} + 5c & = \frac{(1)(2)^2}{2}+5(5) \\ & = \frac{4}{2} + 25 \\ & = 2 + 25 \\ & = 27 \\ \end{aligned}$

8.

$\begin{aligned} \frac{md}{3}+\frac{4cd}{b} & = \frac{(0)(6)}{3}+\frac{4(5)(6)}{(2)} \\ & = 0 + \frac{120}{2} \\ & = 60 \\ \end{aligned}$

9.

$\begin{aligned} \frac{ac}{d}+a & = \frac{(1)(5)}{(6)} + (1) \\ & = \frac{5}{6} + 1 \\ & = 1\frac{5}{6} \\ \end{aligned}$

10.

$\begin{aligned} \frac{a^2}{c}+3a^2b & = \frac{(1)^2}{(5)}+3(1)^2(2) \\ & = \frac{1}{5}+6 \\ & = 6\frac{1}{5} \\ \end{aligned}$

11.

$\begin{aligned} \frac{ab^2}{a}-\frac{a^5d}{b} & = \frac{(1)(2)^2}{(1)}-\frac{(1)^5(6)}{(2)} \\ & = 4 - 3 \\ & = 1 \\ \end{aligned}$

12.

$\begin{aligned} \frac{c+d}{d}+\frac{d-c}{a} & = \frac{(5)+(6)}{(6)} + \frac{(6)-(5)}{(1)} \\ & = \frac{11}{6} + \frac{1}{1} \\ & = 1\frac{5}{6} + 1 \\ & = 2\frac{5}{6} \\ \end{aligned}$

13.

$\begin{aligned} \frac{d^2-a^2b^2}{a^2b^2} & = \frac{(6)^2-(1)^2(2)^2}{(1)^2(2)^2} \\ & = \frac{36-(1)(4)}{(1)(4)} \\ & = \frac{32}{4} \\ & = 8 \\ \end{aligned}$

14.

$\begin{aligned} \frac{mcd+3abc}{d^2-4b^2} & = \frac{(0)(5)(6)+3(1)(2)(5)}{(6)^2-4(2)^2} \\ & = \frac{0+30}{36-4(4)} \\ & = \frac{30}{36-16} \\ & = \frac{30}{20} \\ & = \frac{3}{2} \\ & = 1\frac{1}{2} \\ \end{aligned}$

15.

$\begin{aligned} \frac{a^2d^2-9b^2}{abcd} & = \frac{(1)^2(6)^2-9(2)^2}{(1)(2)(5)(6)} \\ & = \frac{(1)(36)-9(4)}{60} \\ & = \frac{36-36}{60} \\ & = \frac{0}{60} \\ & = 0 \\ \end{aligned}$

May 31, 2022October 24, 2022

202205311107 Problem 4.23

Boxes $A$ and $B$ are in contact on a horizontal, frictionless (i.e. $f=0$ ) surface, as shown in the Figure below. Box $A$ has mass $20.0\,\mathrm{kg}$ and box $B$ has mass $5.0\,\mathrm{kg}$ . A horizontal force of $100\,\mathrm{N}$ is exerted on box $A$ . What is the magnitude of the force that box $A$ exerts on box $B$ ?

_{extracted from Problem 4.23, Sears and Zemansky’s University Physics}

Steps.

Draw the free-body diagrams of $A$ , $B$ , and $A+B$ .

Apply Newton’s $2^\textrm{nd}$ law $\textrm{Net }F=ma$ :

$\begin{aligned} F_A- {}_{B}F_A - f_A & = m_Aa_A \\ {}_{A}F_B - f_B & = m_Ba_B \\ F_A-f_{A+B} & = m_{A+B}a_{A+B} \\ \end{aligned}$

Conditioning the equations of motion:

$\begin{aligned} a_A=a_B & =a_{A+B} \\ f_A=f_B=f_{A+B} & =0 \\ {}_{B}F_A & ={}_{A}F_{B} \\ m_A+m_B & =m_{A+B} \\ \end{aligned}$

and substituting numbers for symbols, write:

$\begin{aligned} 100 - {}_{A}F_{B} & = 20a \\ {}_{A}F_{B} & = 5a \\ 100 & =25a \\ \end{aligned}$

$\therefore$ The magnitude ${}_{A}F_{B}$ of the force that box $A$ exerts on box $B$ is $20\,\mathrm{N}$ .

May 27, 2022July 24, 2022

202205271015 Parametrization 001

equation $y=mx+c$ describes a straight line with slope $m$ and $y$ -intercept $c$ , as shown below:

With some scalar parameter $t$ parametrize the equation, in vector representation of units $\hat{\imath}$ and $\hat{\jmath}$ , by

$\mathbf{s}(t)=t\,\hat{\mathbf{i}} + (mt+c)\,\hat{\mathbf{j}}\quad\textrm{where } t\in (-\infty ,\infty)$

whereas for a quadratic equation $y=ax^2+bx+c$ which describes a parabolic curve, its parametric representation is

$\mathbf{s}(t)=t\,\hat{\mathbf{i}} + (at^2+bt+c)\,\hat{\mathbf{j}}\quad\textrm{where } t\in (-\infty ,\infty)$

and similarly for a circle of radius $r$ centered at the origin $O(0,0)$ , its locus is parametrized as

$\mathbf{s}(t)=r\cos t\,\hat{\mathbf{i}}+r\sin t\,\hat{\mathbf{j}}\quad\textrm{where } t\in [-2\pi ,2\pi ]$ .

(to be continued)

May 26, 2022July 24, 2022

202205261415 Arithmetic in Lisp

Given two operands $a$ and $b$ which are real numbers, define the operators,

$a+b$ , $a-b$ , $a\times b$ , $a\div b$ , $a^{b}$ , $\sqrt[b]{a}$ , $a\bmod b$ ,

with respect to arithmetic operations such as addition, subtraction, multiplication, division, exponentiation, rooting, and Modulo.

+------+------------+-----+-----+-------------------------+
|  ID  |  operator  |  a  |  b  |  instance for example   |
+------+------------+-----+-----+-------------------------+
|  1   |  +         |  2  |  3  |  a + b  = 2 + 3    = 5  |
|  2   |  -         |  5  |  2  |  a - b  = 5 - 2    = 3  |
|  3   |  ×         |  3  |  2  |  a * b  = 3 * 2    = 6  |
|  4   |  ÷         |  9  |  3  |  a / b  = 9 / 3    = 3  |
|  5   |  ^         |  3  |  2  |  a ^ b  = 3 ^ 2    = 9  |
|  6   |  √         |  3  |  8  |  a √ b  = 3 √ 8    = 2  |
|  7   |  mod       |  7  |  4  |  a mod b = 7 mod 4 = 3  |
+------+------ -----+-----+-----+-------------------------+

+------+------------+-----+-----+-------------------------+

| ID | operator | a | b | instance for example |

+------+------------+-----+-----+-------------------------+

| 1 | + | 2 | 3 | a + b = 2 + 3 = 5 |

| 2 | - | 5 | 2 | a - b = 5 - 2 = 3 |

| 3 | × | 3 | 2 | a * b = 3 * 2 = 6 |

| 4 | ÷ | 9 | 3 | a / b = 9 / 3 = 3 |

| 5 | ^ | 3 | 2 | a ^ b = 3 ^ 2 = 9 |

| 6 | √ | 3 | 8 | a √ b = 3 √ 8 = 2 |

| 7 | mod | 7 | 4 | a mod b = 7 mod 4 = 3 |

+------+------ -----+-----+-----+-------------------------+

Not so accustomed to these, Lisp cannot interpret or compile them until we have defined those seven primitive functions, better known as symbolic computations; as were newborn infants taught human intelligence by grownup adults between homo sapiens.

To Lisp, arithmetic operations are usually $\textrm{\scriptsize{NOT}}$ predefined. We have in our toolkit $\textrm{\scriptsize{ONLY}}$ three mathematical functors, i.e., list, iteration, and recursion.

The roots of quadratic equation are given by the formula:

$\displaystyle{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$

the expected outcome of which is calculated by parts, first the discriminant ‘delta’:

(define (delta a b c) (- (square b) (* 4 a c)))

1	(define (delta a b c) (- (square b) (* 4 a c)))

$\Delta = b^2-4ac$ ;

then the plus-minus ‘pm’:

(define (pm u v)
	(values (cons (+ u v) (- u v))))

1 2	(define (pm u v) (values (cons (+ u v) (- u v))))

$\pm (x)=(+x,-x)$ ;

and hence

(/ (pm (- b) (sqrt (delta a b c))) (* a 2))

1	(/ (pm (- b) (sqrt (delta a b c))) (* a 2))

will solve for real roots, ignoring the complex though.

$\therefore$ Predefining the seven primitive functions is in order for solving any computational problems by Lisp.

(to be continued)

May 25, 2022May 26, 2022

202205251335 SQL Table 005

CREATE TABLE SQL_Table_005
    (ID int,
    Function varchar(255),
    Derivative varchar(255));
INSERT INTO SQL_Table_005
    (Function, Derivative)
    VALUES
    ('position', 'velocity')
    ('velocity', 'acceleration')
    ('momentum', 'force')
    ('work', 'power')
    ('angular momentum', 'torque')
    ('electric charge', 'electric current')
    ('magnetic flux', 'induced voltage')

CREATE TABLE SQL_Table_005

(ID int,

Function varchar(255),

Derivative varchar(255));

INSERT INTO SQL_Table_005

(Function, Derivative)

VALUES

('position', 'velocity')

('velocity', 'acceleration')

('momentum', 'force')

('work', 'power')

('angular momentum', 'torque')

('electric charge', 'electric current')

('magnetic flux', 'induced voltage')

Output.
+------+--------------------+--------------------+
|  ID  |  Function          |  Derivative        |
+------+--------------------+--------------------+
|  1   |  position          |  velocity          |
|  2   |  velocity          |  acceleration      |
|  3   |  momentum          |  force             |
|  4   |  work              |  power             |
|  5   |  angular momentum  |  torque            |
|  6   |  electric charge   |  electric current  |
|  7   |  magnetic flux     |  induced voltage   |
+------+--------------------+--------------------+

Output.

+------+--------------------+--------------------+

| ID | Function | Derivative |

+------+--------------------+--------------------+

| 1 | position | velocity |

| 2 | velocity | acceleration |

| 3 | momentum | force |

| 4 | work | power |

| 5 | angular momentum | torque |

| 6 | electric charge | electric current |

| 7 | magnetic flux | induced voltage |

+------+--------------------+--------------------+

May 25, 2022October 24, 2022

202205250911 Exercise 2.1A (Q1)

Exercises 1-8, show that the given function $y=f(x)$ is one-to-one over the given interval, then find the formulae for the inverse function $f^{-1}$ and its derivative.

1. $f(x)=x\quad \forall\, x$

2. $f(x)=3x\quad \forall\, x$
3. $f(x)=x^2\quad \forall\, x\ge 0$
4. $f(x)=\sqrt{x}\quad \forall\, x\ge 0$
5. $f(x)=\displaystyle{\frac{1}{x}}\quad \forall\, x>0$
6. $f(x)=\displaystyle{\frac{1}{x}}\quad \forall\, x<0$
7. $f(x)=\displaystyle{\frac{1}{x^2}}\quad \forall\, x>0$
8. $f(x)=x^5\quad \forall\, x$

_{Extracted from Michael Corral. (2020). Elementary Calculus.}

Background. (one-to-one; inverse; derivative of an inverse)

A one-to-one function (aka injection/injective function) is a function $f$ that maps distinct elements to distinct elements; i.e., $f(x_1)=f(x_2)\Rightarrow x_1=x_2$ , or equivalently in the contrapositive that $x_1\neq x_2\Rightarrow f(x_1)\neq f(x_2)$ . Not to be confused with one-to-one correspondence that refers to bijection.

_{Wikipedia on Injective function}

The inverse function of a function $f$ (aka inverse of $f$ ) is a function that undoes the operation of $f$ . The inverse of $f$ exists if and only if $f$ is bijective, and if it exists, is denoted by $f^{-1}$ . For a function $f:X\rightarrow Y$ , its inverse $f^{-1}:Y\rightarrow X$ admits an explicit description: it sends each element $y\in Y$ to the unique element $x\in X$ such that $f(x)=y$ .

_{Wikipedia on Inverse function}

If $y=f(x)$ is differentiable and has an inverse function $x=f^{-1}(y)$ , then $f^{-1}$ is differentiable and its derivative is

$\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}}\quad\textrm{ if }\enspace\frac{\mathrm{d}y}{\mathrm{d}x}\neq 0$ .

_{Text on pg. 38, Sec. 2.1, Ch. 2}

1.

$f(x)=x\quad \forall\, x$ :

$\begin{aligned} \forall\, x\in\mathbb{R},\quad & f(x_1)=x_1 \\ & f(x_2)=x_2 \\ \enspace & f(x_1)=f(x_2) \Rightarrow x_1=x_2 \\ \end{aligned}$

$\therefore$ $y=f(x)=x$ is injective/one-to-one.

In order for an inverse $f^{-1}(x)$ to exist, the original function $f(x)$ must be bijective (et injective et surjective). It remains to check whether or not $y=f(x)=x$ is surjective.

A function $f:X\rightarrow Y$ is said to be surjective if $\forall\, y\in Y,\enspace \exists\, x\in X\textrm{ s.t. }f(x)=y$ . In other words, every element of the function’s codomain is the image of at least one element of its domain.

_{Wikipedia on Surjective function}

Assume $f(x)=x$ is a real-valued function, i.e., $f:\mathbb{R}\rightarrow\mathbb{R}$ given by $x\mapsto x$ . Obviously $f$ is surjective. Thus it is bijective enough to have an inverse $f^{-1}:\mathbb{R}\rightarrow\mathbb{R}$ . And apparently $f^{-1}$ is given by $x\mapsto x$ , such that $f^{-1}\circ f=f\circ f^{-1}=\textrm{id}_{\,\mathbb{R}}$ . Its derivative is $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}\big( f^{-1}(x)\big) =\frac{\mathrm{d}}{\mathrm{d}x}(x)=1}$ .

The remaining questions are left the reader.

May 23, 2022October 24, 2022

202205231502 Exercise 1.3 A (Q1)

Exercises 1-9, let $\mathrm{d}x$ be an infinitesimal and prove the given formula.

1. $(\mathrm{d}x+1)^2=2\,\mathrm{d}x+1$

_{Extracted from Michael Corral. (2020). Elementary Calculus.}

Background. (Infinitesimal)

A number $\delta$ is an infinitesimal if the conditions (a) – (d) hold: (a) $\delta\neq 0$ ; (b) If $\delta >0$ then $\delta$ is smaller than any positive real number; (c) If $\delta <0$ then $\delta$ is larger than any negative real number; (d) $\delta^2=0$ (and hence all higher powers of $\delta$ , such as $\delta^3$ and $\delta^4$ , are also $0$ ) N.b. Any infinitesimal multiplied by a nonzero real number is also an infinitesimal, while $0$ times an infinitesimal is $0$ .

Proof.

Suppose the contrary is true:

$(\mathrm{d}x+1)^2\neq 2\,\mathrm{d}x+1$ .

$\begin{aligned} \textrm{LHS} & = (\mathrm{d}x+1)^2 \\ & = (\mathrm{d}x)^2+2(\mathrm{d}x)(1)+(1)^2 \\ & \stackrel{(\textrm{d})}{=} 0+2\,\mathrm{d}x+1 \\ & = 2\,\mathrm{d}x+1 \\ & = \textrm{RHS} \qquad \perp\\ \end{aligned}$

Thus converse is the case.