December 2, 2021 – Physicspupil

December 2, 2021October 24, 2022

202112021444 Solution to 1979-CE-AMATH-I-1

Let $\displaystyle{y=x+\frac{1}{x^2}}$ . Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ from first principles.

Solution.

From first principles,

$\begin{aligned} y:=f(x) & =x+\frac{1}{x^2} \\ f(x+\Delta x) & =(x+\Delta x) + \frac{1}{(x+\Delta x)^2} \\ f(x+\Delta x)-f(x) & = \Delta x + \frac{1}{(x+\Delta x)^2} - \frac{1}{x^2} \\ & = \Delta x + \frac{x^2-(x+\Delta x)^2}{x^2(x+\Delta x)^2} \\ & = \Delta x - \frac{2x\Delta x+(\Delta x)^2}{x^2(x+\Delta x)^2} \\ \frac{f(x+\Delta x)-f(x)}{\Delta x} & = 1 - \frac{2x + \Delta x}{x^2(x+\Delta x)^2} \\ \lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} & = 1 - \frac{2x + 0}{x^2(x+0)^2} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = 1 - \frac{2}{x^3} \\ \end{aligned}$

Playground.

To make fun of calculus, do let

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

where $f(x,y)$ is a degree $3$ polynomial in two variables $x$ and $y$ .

The set of solutions to $f(x,y)=0$ is

$\begin{cases} \enspace x \neq 0 \\ \enspace y = \frac{x^3+1}{x^2} \end{cases}$

Computing $\partial_xf$ and $\partial_yf$ as follow:

$\begin{aligned} \partial_xf & = \frac{\partial}{\partial x}f(x,y) \\ & = 3x^2-2xy \\ \end{aligned}$

and

$\begin{aligned} \partial_yf & = \frac{\partial}{\partial y}f(x,y) \\ & = -x^2 \\ \end{aligned}$ ,

we oversee the relation

$\mathrm{d}f = (\partial_xf)(\Delta x)+(\partial_yf)(\Delta y)$ .

Definition. (Singularity and Smoothness)

A point $p=(a,b)$ on a curve $\mathcal{C}=\{ (x,y)\in\mathbb{C}^2:f(x,y)=0\}$ is said to be singular if

$\begin{aligned} \frac{\partial f}{\partial x}(a,b) & = 0 \\ \frac{\partial f}{\partial y}(a,b) & = 0 \\ \end{aligned}$ .

A point that is not singular is called smooth. If there is at least one singular point on $\mathcal{C}$ , then curve $\mathcal{C}$ is called a singular curve. If there are no singular points on $\mathcal{C}$ , the curve $\mathcal{C}$ is called a smooth curve.

_{C.f. Definition 1.9.1, T. Carrity, et al., Algebraic Geometry: A Problem Solving Approach}

That said, in our scenario

$f(x,y)=x^3-x^2y+1$ ,

the set of candidates for singularity

$\{ (a,b)\in\mathbb{C}^2\enspace \mathrm{s.t.}\enspace f(a,b)=0\textrm{ and }\partial_xf(a,b)=\partial_yf(a,b)=0\}$ .

is empty.

$\therefore$ The curve $\mathcal{C}=\{ (x,y)\in\mathbb{C}^2:x^3-x^2y+1=0\}$ is smooth, thus everywhere differentiable.

Wait, the function in question should be the curve

$F(x)=\displaystyle{x+\frac{1}{x^2}}$ ;

still, for better or worse, $F(x)$ is $\textrm{\scriptsize{NOT}}$ a differentiable function because the derivative does $\textrm{\scriptsize{NOT}}$ exist at $x=0$ , for that point is a discontinuity.

December 2, 2021October 24, 2022

202112021138 SQL Table 004

Background. (Equilibrium)

Equilibrium is established by the exchange of energy, volume, or particle number between different systems or subsystems.

_{Section 2.2. Arovas. (2019). Thermodynamics and Statistical Mechanics}

CREATE TABLE SQL_Table_004
    (ID int,
    Object_of_exchange varchar(255),
    Constancy varchar(255)
    Subject_of_equilibrium varchar(255)
    );
INSERT INTO SQL_Table_004
    (Object_of_exchange, Constancy, Subject_of_equilibrium)
    VALUES
    ('energy', 'temperature (T)', 'thermal')
    ('volume', 'momentum/temperature (p/T)', 'mechanical')
    ('particle', 'chemical potential/temperature (\mu /T)', 'chemical')

CREATE TABLE SQL_Table_004

(ID int,

Object_of_exchange varchar(255),

Constancy varchar(255)

Subject_of_equilibrium varchar(255)

);

INSERT INTO SQL_Table_004

(Object_of_exchange, Constancy, Subject_of_equilibrium)

VALUES

('energy', 'temperature (T)', 'thermal')

('volume', 'momentum/temperature (p/T)', 'mechanical')

('particle', 'chemical potential/temperature (\mu /T)', 'chemical')

Output.
+------+---------------------+-------------------------------+--------------------------+
|  ID  | Object_of_exchange  |  Constancy                    |  Subject_of_equilibrium  |
+------+---------------------+-------------------------------+--------------------------+
|  1   |  energy             |  temperature (T)              |  thermal                 |
|  2   |  volume             |  momentum/temperature (p/T)   |  mechanical              |
|  3   |  particle           |  chemical potential/          |  chemical                |
|      |                     |    temperature (\mu /T)       |                          |
+------+---------------------+-------------------------------+--------------------------+

Output.

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

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

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

December 2, 2021December 12, 2022

202112021037 Exercise 1.4 Lagrangian actions

For a free particle an appropriate Lagrangian is

Eq. (1.8):

$\mathcal{L}(t,x,v)=\displaystyle{\frac{1}{2}mv^2}$ .

Suppose that $x$ is the constant-velocity straight-line path of a free particle, such that $x_a=x(t_a)$ and $x_b=x(t_b)$ . Show that the action on the solution path is

Eq. (1.9):

$\displaystyle{\frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a}}$ .

_{Extracted from Structure and Interpretation of Classical Mechanics, SICP, 2e}

Background. (Lagrangians and Lagrangian actions)

The function $\mathcal{L}$ is called a Lagrangian for the system, and the resulting action,

Eq. (1.4):

$S[q](t_1,t_2)=\displaystyle{\int_{t_1}^{t_2}\mathcal{L}\circ\Gamma [q]}$ ,

is called the Lagrangian action. For Lagrangians that depend only on time, positions, and velocities the action can also be written

Eq. (1.5):

$S[q](t_1,t_2)=\displaystyle{\int_{t_1}^{t_2}\mathcal{L}(t,q(t),\mathrm{D}q(t))\,\mathrm{d}t}$ .

_{(Section 1.3 The Principle of Stationary Action)}

Working. (roughly)

$\begin{aligned} S & = \int_{t_a}^{t_b}\frac{1}{2}mv^2\,\mathrm{d}t \\ & = \frac{m}{2}\int_{t_a}^{t_b}(\mathbf{v}\cdot\mathbf{v})\,\mathrm{d}t \\ & = \frac{m}{2}\int_{t_a}^{t_b}\bigg( \frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}\cdot\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}\bigg) \,\mathrm{d}t \\ & = \frac{m}{2}\int_{x_a,t_a}^{x_b,t_b}\frac{(\mathrm{d}x)^2}{(\mathrm{d}t)} \\ & = \frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a} \\ \end{aligned}$