The Physics Language – Page 9

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 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.

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, $y=e^x$ .

January 25, 2022October 24, 2022

202201260719 Problem 1.57

Let $A=\{ x:x^2-3x+2=0\}$ , $B=\{ x:x^2\le 16\}$ . Determine whether or not $A\subset B$ .

_{Extracted from M. R. Spiegel. (1969). Schaum’s Outline of Theory and Problems of Real Variables}

Set $A$ and set $B$ are described according to property method; if described by the roster method:

$\begin{aligned} A & =\{ 1,2\} \\ B & = \{ -4,-3,-2,-1,0,1,2,3,4\} \\ \end{aligned}$

Obviously set $A$ is a subset of set $B$ .

Roughwork.

(set $A$ )

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

(set $B$ )

$\begin{aligned} x^2 & \le 16 \\ x^2 - 16 & \le 0 \\ (x-4)(x+4) & \le 0 \\ -4\le x & \le 4 \\ \end{aligned}$

January 25, 2022October 24, 2022

202201251933 Circumference 001

Draw a circle of radius $R$ around the center $(R,0)$ .

As the center of a circle is equidistant from all points on the circumference, using Pythagorean theorem, we have

$\begin{aligned} (x-R)^2+(y-0)^2 & = R^2 \\ y^2 & = R^2 - (x-R)^2 \\ y & = \pm\sqrt{2xR-x^2} \\ |y| & = \sqrt{2xR-x^2}\\ \end{aligned}$

Attempts. (reinventing the wheel)

Considering the differentials $\mathrm{d}x$ and $\mathrm{d}y$ , we have

$\mathrm{d}y =\displaystyle{\frac{R-x}{\sqrt{2xR-x^2}}\,\mathrm{d}x}$

Suppose I do not know the circumference is $2\pi R$ long. Let its unknown length be $s$ , and let it be partitioned into infinitesimal $\mathrm{d}s$ , such that

$s=\displaystyle{\int\mathrm{d}s}$

Assume we may write

$(\mathrm{d}s)^2=(\mathrm{d}x)^2+(\mathrm{d}y)^2$ .

Expand the right hand side as follows

$\begin{aligned} &\quad \textrm{RHS} \\ & = (\mathrm{d}x)^2+(\mathrm{d}y)^2 \\ & = (\mathrm{d}x)^2 + \bigg( \frac{R-x}{\sqrt{2xR-x^2}}\,\mathrm{d}x\bigg)^2 \\ & = \bigg(\frac{R^2}{2xR-x^2}\bigg)(\mathrm{d}x)^2 \\ \end{aligned}$

Then

$\begin{aligned} \mathrm{d}s & = \sqrt{\bigg( \frac{R^2}{2xR-x^2}\bigg)(\mathrm{d}x)^2} \\ s = \int\mathrm{d}s & = \int_{x=0}^{2R} \frac{R}{\sqrt{2xR-x^2}}\,\mathrm{d}x \\ \end{aligned}$

Suppose I evaluate the integral above $\textrm{\scriptsize{NOT}}$ by direct substitution $\textrm{\scriptsize{BUT}}$ by Riemann sum, so the definite integral due to Riemann is given by

$\displaystyle{\int_{a}^{b}f(x)\,\mathrm{d}x}=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_{i}^{*})\, x$ .

For $i=0,1,2,\dots ,n$ , let $P=\{ x_i\}$ be a regular partition of $[0,2R]$ . Then

$x=\displaystyle{\frac{b-a}{n}=\frac{2R}{n}}$ .

By right-endpoint approximation for Riemann sums, for each interval $[x_{i-1},x_i]$ , we have

$\displaystyle{x_i=x_0+ix=0+i\bigg[\frac{2R}{n}\bigg]=\frac{2Ri}{n}}$ .

Let $f(x)\stackrel{\textrm{def}}{=}\displaystyle{\frac{R}{\sqrt{2xR-x^2}}}$ . Thus,

$\begin{aligned} f(x_i) & = \frac{R}{\sqrt{(2)\displaystyle{\bigg(\frac{2Ri}{n}\bigg)}(R)-\displaystyle{\bigg(\frac{2Ri}{n}\bigg)^2}}} \\ & = \dots \\ & = \frac{n}{2\sqrt{in-i^2}} \\ \end{aligned}$

Writing the Riemann sum in the form

$\begin{aligned} \sum_{i=1}^{n}f(x_i)\, x & = \sum_{i=1}^{n}\frac{n}{2\sqrt{in-i^2}}\bigg(\frac{2Ri}{n}\bigg) \\ & = R\cdot \sum_{i=1}^{n}\frac{1}{\sqrt{\frac{n}{i}-1}} \\ & = R\cdot g(n) \\ \end{aligned}$

Inspecting $R\cdot g(n)$ where

$g(n) = \displaystyle{ \sum_{i=1}^{n}\frac{1}{\sqrt{\frac{n}{i}-1}}}$

I guess, under correction, that $g(n)=2\pi$ .

(to be continued)

Solution. (arc-length parametrization)

Referring to the equation of locus on the very first line:

$(x-R)^2+(y-0)^2=R^2$ ,

then parameterizing $x(\theta )$ , $y(\theta )$ by $\theta$ ,

$\begin{aligned} x & = R+R\cos\theta \\ y & = R\sin\theta \\ \end{aligned}$

and computing the derivatives wrt $\theta$ :

$\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}\theta} & = -R\sin\theta \\ \frac{\mathrm{d}y}{\mathrm{d}\theta} & = R\cos\theta \\ \end{aligned}$

Note that

$\displaystyle{s=\int\mathrm{d}s}$

where

$\mathrm{d}s=\sqrt{\displaystyle{\bigg(\frac{\mathrm{d}x}{\mathrm{d}\theta}\bigg)^2+\bigg(\frac{\mathrm{d}y}{\mathrm{d}\theta}\bigg)^2}}\,\mathrm{d}\theta$ .

Then,

$\begin{aligned} s & = \int \mathrm{d}s \\ & = \int \sqrt{\bigg( \frac{\mathrm{d}x}{\mathrm{d}\theta}\bigg)^2+\bigg( \frac{\mathrm{d}y}{\mathrm{d}\theta}\bigg)^2}\,\mathrm{d}\theta \\ & = \int \sqrt{(-R\sin\theta )^2+(R\cos\theta )^2}\,\mathrm{d}\theta \\ & = \int \sqrt{R^2}\,\mathrm{d}\theta \\ & = \int_{\theta =0}^{2\pi} R\,\mathrm{d}\theta \\ & = 2\pi R \\ \end{aligned}$

(to be continued)

January 13, 2022July 25, 2022

202201131704 Practice in the Making

Let $x=2$ , $y=3$ , and $z=5$ . Then,

$\begin{aligned} &\quad 2x+3y+5z \\ &= 2(2)+3(3)+5(5) \\ &= 38 \\ & \\ &\quad 3x+5y+2z \\ & = 3(2)+5(3)+2(5) \\ & = 31 \\ & \\ & \quad 5x+2y+3z \\ & = 5(2)+2(3)+3(5) \\ & = 31 \\ \end{aligned}$

That is to say, I have made a set of three equations:

$\begin{cases} 2x+3y+5z =38 \\ 3x+5y+2z =31 \\ 5x+2y+3z =31 \\ \end{cases}$

in matrix form,

$\begin{bmatrix} 2 & 3 & 5 \\ 3 & 5 & 2 \\ 5 & 2 & 3 \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ z\\ \end{bmatrix} = \begin{bmatrix} 38 \\ 31 \\ 31 \\ \end{bmatrix}$

complicating it,

$\begin{bmatrix} 2 & 0 & 3 & 0 & 5 \\ 0 & 0 & 0 & 0 & 0 \\ 3 & 0 & 5 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 \\ 5 & 0 & 2 & 0 & 3 \\ \end{bmatrix}\begin{bmatrix} x \\ 0 \\ y \\ 0 \\ z\\ \end{bmatrix} = \begin{bmatrix} 38 \\ 0 \\ 31 \\ 0 \\ 31 \\ \end{bmatrix}$