Multivariate Calculus – Physicspupil

August 22, 2024

202408221017 Exercise 5.68

A particle moves along the space curve

$\mathbf{r}=e^{-t}\cos t\,\hat{\mathbf{i}}+e^{-t}\sin t\,\hat{\mathbf{j}}+e^{-t}\,\hat{\mathbf{k}}$ .

Find the magnitude of the (a) velocity and (b) acceleration at any time $t$ .

_{Extracted from Spiegel, M. R. (1971). Schaum’s Outline of Theory and Problems of Advanced Mathematics for Engineers and Scientists.}

Roughwork.

Let displacement

$\mathbf{r}=r_x\,\hat{\mathbf{i}}+r_y\,\hat{\mathbf{j}}+r_z\,\hat{\mathbf{k}}$

where

$\begin{aligned} r_x(t) & = e^{-t}\cos t \\ r_y(t) & = e^{-t}\sin t \\ r_z(t) & = e^{-t} \\ \end{aligned}$

then velocity

$\mathbf{v}=v_x\,\hat{\mathbf{i}}+v_y\,\hat{\mathbf{j}}+v_z\,\hat{\mathbf{k}}$

where

$\begin{aligned} v_x & = \frac{\mathrm{d}r_x}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}\cos t) \\ v_x(t)& = -e^{-t}\sin t-e^{-t}\cos t \\ v_y & = \frac{\mathrm{d}r_y}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}\sin t) \\ v_y(t)& = e^{-t}\cos t-e^{-t}\sin t \\ v_z & = \frac{\mathrm{d}r_z}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}) \\ v_z(t)& = -e^{-t} \\ \end{aligned}$

then acceleration

$\mathbf{a}=a_x\,\hat{\mathbf{i}}+a_y\,\hat{\mathbf{j}}+a_z\,\hat{\mathbf{k}}$

where

$\begin{aligned} a_x & = \frac{\mathrm{d}v_x}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(-e^{-t}\sin t-e^{-t}\cos t) \\ a_x(t)& = 2e^{-t}\sin t\\ a_y & = \frac{\mathrm{d}v_y}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}\cos t-e^{-t}\sin t) \\ a_y(t)& = -2e^{-t}\cos t\\ a_z & = \frac{\mathrm{d}v_z}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(-e^{-t}) \\ a_z(t)& = e^{-t} \\ \end{aligned}$

Cheat.

$\mathbf{r}(t) = \underbrace{e^{-t}}_{\textrm{(1)}}\underbrace{(\cos t,\sin t, 1)}_{\textrm{(2)}}$

where

$\begin{aligned} \textrm{(1):}&\enspace \textrm{radial centripetal} \\ \textrm{(2):}&\enspace \textrm{circumferential anticlockwise} \\ \end{aligned}$

_{By courtesy of WolframAlpha}

In a cylindrical coordinate system, the position of a particle can be written as

$\mathbf{r}=\rho\,\hat{\boldsymbol{\rho}}+z\,\hat{\mathbf{z}}$ .

The velocity of the particle is the time derivative of its position,

$\mathbf{v}=\displaystyle{\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}}=\dot{\rho}\,\hat{\boldsymbol{\rho}}+\rho\dot{\varphi}\,\hat{\boldsymbol{\varphi}}+\dot{z}\,\hat{\mathbf{z}}$ ,

where the term $\rho\dot{\varphi}\,\hat{\boldsymbol{\varphi}}$ comes from the Poisson formula

$\displaystyle{\frac{\mathrm{d}\hat{\boldsymbol{\varphi}}}{\mathrm{d}t}}=\dot{\varphi}\,\hat{\mathbf{z}}\times\hat{\boldsymbol{\rho}}$ .

Its acceleration is

$\mathbf{a}=\displaystyle{\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}}=(\ddot{\rho}-\rho\dot{\varphi}^2)\,\hat{\boldsymbol{\rho}}+(2\dot{\rho}\dot{\varphi}+\rho\ddot{\varphi})\,\hat{\boldsymbol{\varphi}}+\ddot{z}\,\hat{\mathbf{z}}$ .

_{Kinematics in Wikipedia on Cylindrical coordinate system}

Stop wandering! You are facing up to

$\begin{aligned} v=|\mathbf{v}|=\sqrt{v_x^2+v_y^2+v_z^2} & = \cdots \\ a=|\mathbf{a}|=\sqrt{a_x^2+a_y^2+a_z^2} & = \cdots \\ \end{aligned}$

But then, I’m fond of opting out.

This problem is not to be attempted.

May 10, 2024June 13, 2024

202405101842 Exercise 16.67

The cross product of a vector with itself is

A. the vector $\mathbf{i}^2+\mathbf{j}^2+\mathbf{k}^2$ .
B. the vector $\mathbf{i}+\mathbf{j}+\mathbf{k}$ .
C. the zero vector.
D. the scalar quantity $1$ .
E. the scalar quantity $0$ .

_{Extracted from Stan Gibilisco. (2006). Technical Math Demystified.}

Erratum.

Option A should be a scalar.

Roughwork.

Most commonly, it is the three-dimensional Euclidean space $\mathbf{E}^3$ (or $\mathbb{E}^3$ ) that models physical space $\mathbf{R}^3$ (or $\mathbb{R}^3$ ).

_{Wikipedia on Three-dimensional space}

The Euclidean metric is given by

$\mathbf{G}=(g_{\alpha\beta})\equiv \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\\end{bmatrix}$

and so line element $\mathrm{d}s$

$\begin{aligned} \mathrm{d}s^2 & = g_{\alpha\beta}\,\mathrm{d}x^\alpha\,\mathrm{d}x^\beta \\ & = (\mathrm{d}x^1)^2 + (\mathrm{d}x^2)^2 + (\mathrm{d}x^3)^2 \\ & = \mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2 \\ \end{aligned}$

or distance

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

is equivalent to applying twice the Pythagorean equation

$a^2+b^2=c^2$

to two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ with respect to the standard basis $(\mathbf{e}_x,\mathbf{e}_y,\mathbf{e}_z)$ :

$\begin{cases} \hat{\mathbf{i}}=\mathbf{e}_x=(1,0,0) \\ \hat{\mathbf{j}}=\mathbf{e}_y=(0,1,0) \\ \hat{\mathbf{k}}=\mathbf{e}_z=(0,0,1) \\ \end{cases}$

of a Cartesian coordinate system. To carry a point from one position to another, a vector $\mathbf{v}$ has both magnitude $v=|\mathbf{v}|$ and direction $\hat{\mathbf{v}}=\frac{\mathbf{v}}{|\mathbf{v}|}$ . Scalar multiplication $c\mathbf{v}$ will scale its magnitude by some factor $c$ (the scalar) without change in its direction. I.e.,

$\begin{aligned} \mathbf{u} & \stackrel{\mathrm{def}}{=}c\mathbf{v} \\ u & = |\mathbf{u}| = |c\mathbf{v}| = c|\mathbf{v}|\\ & = cv \\ \hat{\mathbf{u}} & = \frac{\mathbf{u}}{|\mathbf{u}|} = \frac{c\mathbf{v}}{c|\mathbf{v}|} = \frac{\mathbf{v}}{|\mathbf{v}|} \\ & = \hat{\mathbf{v}} \\ \end{aligned}$

Vector multiplication is for whose vectors be the products scalar or vector. For instance, we have

Dot product (aka scalar product) of two vectors $\mathbf{u}=(u_x,u_y,u_z)$ and $\mathbf{v}=(v_x,v_y,v_z)$

$\begin{aligned} \mathbf{u}\cdot\mathbf{v} & = \sum_{x,y,z}u_iv_i \\ & = u_xv_x+u_yv_y+u_zv_z \\ & = |\mathbf{u}| |\mathbf{v}|\cos\theta \\ \end{aligned}$

as a scalar quantity; but their cross product (aka vector product)

$\begin{aligned} \mathbf{u}\times\mathbf{v} & = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \\ \end{vmatrix} \\ & = (u_yv_z-u_zv_y, u_zv_x-u_xv_z, u_xv_y-u_yv_x) \\ & = |\mathbf{u}||\mathbf{v}|\sin\theta\,\hat{\mathbf{n}} \\ \end{aligned}$

a vector quantity. Added on are tensor product $\mathbf{u}\otimes\mathbf{v}$ , wedge product $\mathbf{u}\wedge\mathbf{v}$ , and more.

This problem is not to be attempted.

May 7, 2024May 7, 2024

202405071349 Exercise 13.2.1

A right pyramid, $12\,\mathrm{cm}$ high, stands on a rectangular base $6\,\mathrm{cm}$ by $10\,\mathrm{cm}$ . Calculate (a) the length of an edge of the pyramid; (b) the angles the triangular faces made with the base; (c) the volume of the pyramid.

_{Extracted from A. Godman & J. F. Talbert. (1973). Additional Mathematics Pure and Applied in SI Units.}

Roughwork.

Commit my visualization to drawing.

A right pyramid is a pyramid where the base is circumscribed about the circle and the altitude of the pyramid meets at the circle’s center.

_{Wikipedia on Pyramid (geometry)}

The pyramid above has a polygonal base, here the rectangle $QRST$ , and an apex $P$ , here the common vertex of triangles $\triangle PTQ$ , $\triangle PQR$ , $\triangle PRS$ , and $\triangle PST$ . The altitude is based on the origin $O$ . To suit our coordinates to this problem, we write

$\begin{aligned} P & =P(0,0,12) \\ Q & =Q(3,5,0) \\ R & =R(-3,5,0) \\ S & =S(-3,-5,0) \\ T & =T(3,-5,0) \\ \end{aligned}$

such that

$\begin{aligned} a & = c = 10 \\ b & = d = 6 \\ e & = f = g = h \\ & = \surd \big\{ (12)^2 + \big( \sqrt{(6)^2+(10)^2}/2\big)^2 \big\} \\ & = \sqrt{178} \\ & = 13.3417\quad\textrm{(4 d.p.)} \\ \end{aligned}$

For the edges of its base, write

$\begin{aligned} a=\overline{TQ} & :\begin{cases} x=3 \\ |y|\leqslant 5 \\ z=0 \\ \end{cases} \\ b=\overline{QR} & :\begin{cases} |x|\leqslant 3 \\ y=5 \\ z=0 \\ \end{cases} \\ c=\overline{RS} & :\begin{cases} x=-3 \\ |y|\leqslant 5 \\ z=0 \\ \end{cases} \\ d=\overline{ST}& :\begin{cases} |x|\leqslant 3 \\ y=-5 \\ z=0 \\ \end{cases} \\ \end{aligned}$

and for, the lateral, edge $e=\overline{PQ}$ :

$\begin{aligned} \frac{x-3}{0-3} & = \frac{y-5}{0-5} = \frac{z-0}{12-0} \\ \end{aligned}$

edge $f=\overline{PR}$ :

$\begin{aligned} \frac{x-(-3)}{0-(-3)} & = \frac{y-5}{0-5} = \frac{z-0}{12-0} \\ \end{aligned}$

edge $g=\overline{PS}$ :

$\begin{aligned} \frac{x-(-3)}{0-(-3)} & = \frac{y-(-5)}{0-(-5)} = \frac{z-0}{12-0} \\ \end{aligned}$

and edge $h=\overline{PT}$ :

$\begin{aligned} \frac{x-3}{0-3} & = \frac{y-(-5)}{0-(-5)} = \frac{z-0}{12-0} \\ \end{aligned}$

For lateral surface $A$ enclosed by edges $a$ , $h$ , and $e$ , write

$A(x,y,z):\begin{cases} (x,y,z)\in [0,3]\times [-5,5]\times [0,12] \\ \textrm{s.t. }\displaystyle{\frac{|y|}{5}\leqslant \frac{x}{3}=1-\frac{z}{12}} \\ \end{cases}$

for lateral surface $B$ by edges $b$ , $e$ , and $f$ , write:

$B(x,y,z):\begin{cases} (x,y,z)\in [-3,3]\times [0,5]\times [0,12] \\ \textrm{s.t. }\displaystyle{\frac{|x|}{3}\leqslant \frac{y}{5}=1-\frac{z}{12}} \\ \end{cases}$

for lateral surface $C$ by edges $c$ , $f$ , and $g$ , write:

$C(x,y,z):\begin{cases} (x,y,z)\in [-3,0]\times [-5,5]\times [0,12] \\ \textrm{s.t. }\displaystyle{\frac{|y|}{5}\leqslant -\frac{x}{3}=1-\frac{z}{12}} \\ \end{cases}$

for lateral surface $D$ by edges $d$ , $g$ , and $h$ , write:

$D(x,y,z):\begin{cases} (x,y,z)\in [-3,3]\times [-5,0]\times [0,12] \\ \textrm{s.t. }\displaystyle{\frac{|x|}{3}\leqslant -\frac{y}{5}=1-\frac{z}{12}} \\ \end{cases}$

and for base $E$ by edges $a$ , $b$ , $c$ , and $d$ :

write

$E(x,y,z):\begin{cases} |x|\leqslant 3 \\ |y|\leqslant 5 \\ z=0 \\ \end{cases}$

This problem is not to be attempted.

January 4, 2024January 4, 2024

202401041119 Exercise 8.2.541

Evaluate the integral

$\displaystyle{\underset{\varGamma}{\int}(x\,\mathrm{d}y+y\,\mathrm{d}x)}$ ,

where (a) $\varGamma$ is an arc of the parabola $y=\sqrt{x}$ , (b) $\varGamma$ is a segment of a straight line, (c) $\varGamma$ is the arc of the parabola $y=x^2$ connecting the points $(0,0)$ and $(1,1)$ in the direction indicated by the arrows (see Figure below).

_{Extracted from Yakov Stepanovič Bugrov. (1984). A collection of problems.}

Answer. (a) 1; (b) 1; (c) 1.

Roughwork.

(a)

When a point moves along the curve $\varGamma :y=\sqrt{x}$ in the indicated direction,

$\begin{aligned} &\quad \underset{\varGamma}{\int}(x\,\mathrm{d}y+y\,\mathrm{d}x) \\ & = \int_{(0,0)}^{(1,1)}(y^2\,\mathrm{d}y+\sqrt{x}\,\mathrm{d}x) \\ & = \bigg[\frac{y^3}{3} + \frac{2x^{3/2}}{3}\bigg]\bigg|_{(0,0)}^{(1,1)}\\ & = \bigg(\frac{(1)^3}{3} + \frac{2(1)^{3/2}}{3}\bigg) - \bigg(\frac{(0)^3}{3} + \frac{2(0)^{3/2}}{3}\bigg) \\ & = \bigg(\frac{1}{3}+\frac{2}{3}\bigg) - (0+0) \\ & = 1 \\ \end{aligned}$

(b)

When a point moves along the curve $\varGamma :y=x$ in the indicated direction,

$\begin{aligned} &\quad \underset{\varGamma}{\int}(x\,\mathrm{d}y+y\,\mathrm{d}x) \\ & = \int_{(0,0)}^{(1,1)}(y\,\mathrm{d}y+x\,\mathrm{d}x) \\ & = \bigg[\frac{y^2}{2} + \frac{x^2}{2}\bigg]\bigg|_{(0,0)}^{(1,1)} \\ & = \bigg(\frac{(1)^2}{2} + \frac{(1)^2}{2}\bigg) - \bigg(\frac{(0)^2}{2} + \frac{(0)^2}{2}\bigg)\\ & = \bigg( \frac{1}{2} + \frac{1}{2}\bigg) - (0+0) \\ & = 1 \\ \end{aligned}$

(c)

When a point moves along the curve $\varGamma :y=x^2$ in the indicated direction,

$\begin{aligned} &\quad \underset{\varGamma}{\int}(x\,\mathrm{d}y+y\,\mathrm{d}x) \\ & = \int_{(0,0)}^{(1,1)}(\sqrt{y}\,\mathrm{d}y+x^2\,\mathrm{d}x) \\ & = \bigg[\frac{2y^{3/2}}{3} + \frac{x^3}{3}\bigg]\bigg|_{(0,0)}^{(1,1)} \\ & = \bigg(\frac{2(1)^{3/2}}{3} + \frac{(1)^3}{3}\bigg) - \bigg(\frac{2(0)^{3/2}}{3} + \frac{(0)^3}{3}\bigg) \\ & = \bigg(\frac{2}{3}+\frac{1}{3}\bigg) - (0+0) \\ & = 1 \\ \end{aligned}$

Afterword.

What am I doing?

$\displaystyle{\int (x\,\mathrm{d}y+y\,\mathrm{d}x) = \int x\,\mathrm{d}y + \int y\,\mathrm{d}x}$

Note the first term on RHS:

and the second term on RHS:

Many simple formulae in physics, such as the definition of work as $W=\mathbf{F}\cdot\mathbf{s}$ , have natural continuous analogues in terms of line integrals, in this case $\displaystyle{W=\underset{L}{\int}\mathbf{F}(\mathbf{s})\,\mathrm{d}\mathbf{s}}$ , which computes the work done on an object moving through an electric or gravitational field $\mathbf{F}$ along a path $L$ .

_{Cited from Wikipedia on Line integral}

$\begin{aligned} &\quad \int (x\,\mathrm{d}y+y\,\mathrm{d}x) \\ & = \int \begin{bmatrix} y & x \end{bmatrix} \begin{bmatrix} \mathrm{d}x \\ \mathrm{d}y \end{bmatrix} \\ & = \int \mathbf{F}(x,y)\,\mathrm{d}\mathbf{s} \\ \cdots & \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ & \mathbf{F}(x,y) = y\,\hat{\mathbf{i}} + x\,\hat{\mathbf{j}} \\ \cdots & \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \end{aligned}$

As an aside, work done $W$ is the scalar (/dot) product of force $\mathbf{F}$ and displacement $\mathbf{s}$ . A conservative vector field in physics is analogous to a path-independent vector field in mathematics. That is, over any paths, the line integral depends only upon the starting point and the end point.

This problem is not to be attempted.

November 7, 2023November 30, 2023

202311070929 Exercise 21.1.2

Find (by drawing or calculation) the resultant of the following forces, in magnitude and direction.

(a)

(b)

(c)

_{Extracted from A. Godman & J. F. Talbert. (1973). Additonal Mathematics Pure and Applied in SI units.}

(a)

So that the two forces are labelled as vectors $\mathbf{OA}$ and $\mathbf{OB}$ , let their initial points for both be $O$ and the terminal points of each $A$ and $B$ .

In Cartesian coordinates,

$\begin{aligned} \mathbf{OA} & = |\mathbf{OA}|\,\hat{\mathbf{i}} \\ & = OA\,\hat{\mathbf{i}} \\ & = 12\,\mathrm{N}\,\hat{\mathbf{i}} \\ \mathbf{OB} & = |\mathbf{OB}|\,\hat{\mathbf{j}} \\ & = OB\,\hat{\mathbf{j}} \\ & = 10\,\mathrm{N}\,\hat{\mathbf{j}} \\ \end{aligned}$

With, on purpose, no diagrams being provided, let also the resultant force $\mathbf{OC}$ be directed from the same initial point $O$ to some terminal point we let be $C$ . Then,

$\begin{aligned} \mathbf{OC} & = \mathbf{OA} + \mathbf{OB} \\ & = 12\,\mathrm{N}\,\hat{\mathbf{i}} + 10\,\mathrm{N}\,\hat{\mathbf{j}} \\ OC & = |\mathbf{OC}| \\ & = \sqrt{12^2+10^2} \\ & = 15.6\,\mathrm{N}\qquad\textrm{(3 s.f.)} \\ \end{aligned}$

Trying in polar coordinates,

$\begin{aligned} \mathbf{OA} & = |\mathbf{OA}|\,\hat{\mathbf{r}} \\ & = 12\,\mathrm{N}\,\hat{\mathbf{r}} \\ \mathbf{OB} & = |\mathbf{OB}|\,\hat{\mathbf{r}} +\frac{\pi}{2}\,\hat{\boldsymbol{\theta}}\\ & = \bigg( 10\,\hat{\mathbf{r}} + \frac{\pi}{2}\,\hat{\boldsymbol{\theta}}\bigg) \,\mathrm{N} \\ \end{aligned}$

Recall the formulae for addition of vectors

$\begin{aligned} r_3 & = \sqrt{r_1^2+2r_1r_2\cos(\theta_2-\theta_1)+r_2^2} \\ \theta_3 & = \theta_1+\arctan \bigg(\frac{r_2\sin (\theta_2-\theta_1)}{r_1+r_2\cos (\theta_2-\theta_1)}\bigg) \\ \end{aligned}$

beside the rectangular form I am stupid enough to do the polar. Quit.

(b)

By tail-to-tip method, draw

$\begin{aligned} OC & = \sqrt{OB^2+BC^2} \\ & = \sqrt{OB^2+OA^2} \\ & = \sqrt{40^2+30^2} \\ & = 50\,\mathrm{N} \\ \tan\theta & = \frac{30}{40} \\ \theta & = \tan^{-1} \bigg(\frac{30}{40}\bigg) \\ & = 36.9^\circ\qquad\textrm{(3 s.f.)} \\ \end{aligned}$

Or, by parallelogram method, draw

which makes no matter how you visualise, only are the calculations the matters.

(c)

Draw a labelled diagram below:

$\begin{aligned} \mathbf{OA} & = 4\cos 30^\circ \,\hat{\mathbf{i}} + 4\sin 30^\circ \,\hat{\mathbf{j}} \\ & = 4\bigg(\frac{\sqrt{3}}{2}\bigg)\,\hat{\mathbf{i}} + 4\bigg(\frac{1}{2}\bigg) \,\hat{\mathbf{j}} \\ & = \big(2\sqrt{3}\,\hat{\mathbf{i}} +2\,\hat{\mathbf{j}}\big)\,\mathrm{N} \\ \mathbf{OB} & = \big( 6\,\hat{\mathbf{i}}\big)\,\mathrm{N} \\ \end{aligned}$

Let $\mathbf{OC}$ be the resultant vector,

$\begin{aligned} \mathbf{OC} & = \mathbf{OA} + \mathbf{OB} \\ & = \big( 2\sqrt{3}+6\big) \,\hat{\mathbf{i}} + 2\,\hat{\mathbf{j}} \\ \end{aligned}$

Write $OA$ , $OB$ , and $OC$ the magnitudes of respective vectors $\mathbf{OA}$ , $\mathbf{OB}$ , and $\mathbf{OC}$ ,

$\begin{aligned} OA & = 4 \\ OB & = 6 \\ OC & = |\mathbf{OC}| \\ & = \sqrt{(2\sqrt{3}+6)^2+2^2} \\ & = \sqrt{52+24\sqrt{3}} \\ & = 2\sqrt{13+6\sqrt{3}} \\ \end{aligned}$

With a diagram in mind,

apply cosine law,

$\begin{aligned} \cos \angle COB & = \frac{OC^2+OB^2-OA^2}{2(OC)(OB)} \\ & = \frac{(52+24\sqrt{3})+(36)-(16)}{2(52+24\sqrt{3})(6)} \\ \angle COB & = \cdots \\ \end{aligned}$

This problem is not to be attempted.

March 15, 2023March 15, 2023

202303150916 Exercise 4.6.4

Let $f:\mathbb{R}^2\to\mathbb{R}^2$ be defined by

$f(x,y)=(x^3+y^3,x^3-y^3)$ for $(x,y)\in\mathbb{R}^2$ .

Prove that the Jacobian matrix $J_{f,(0,0)}$ is the zero matrix. Show that nevertheless $f$ is globally invertible on $\mathbb{R}^2$ . [Hint: prove that $f$ is $\textrm{1--1}$ on $\mathbb{R}^2$ .] Show also that the inverse function is not differentiable at $f(0,0)=(0,0)$ .

_{Extracted from P. R. Baxandall. (1986). Vector Calculus.}

Roughwork.

$\begin{aligned} f\bigg(\begin{bmatrix}x\\y\end{bmatrix}\bigg) & = \begin{bmatrix} x^3+y^3 \\ x^3-y^3 \end{bmatrix} \\ J_{f,(x,y)} & = \begin{bmatrix} \partial_x (x^3+y^3) & \partial_y(x^3+y^3) \\ \partial_x(x^3-y^3) & \partial_y(x^3-y^3) \\ \end{bmatrix} \\ & = \begin{bmatrix} 3x^2 & 3y^2 \\ 3x^2 & -3y^2 \\ \end{bmatrix} \\ J_{f,(0,0)} & = \begin{bmatrix} 3(0)^2 & 3(0)^2 \\ 3(0)^2 & -3(0)^2 \end{bmatrix} \\ & = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix} \\ \end{aligned}$

The function $f:\mathbb{R}^n\to\mathbb{R}^n$ is locally invertible at $\mathbf{x}_0$ if there is an $\epsilon >0$ and a function $g:B_{\epsilon}(f(\mathbf{x}_0))\to\mathbb{R}^n$ such that

$\begin{cases} f\circ g(\mathbf{y}) \equiv \mathbf{y}\quad\textrm{for }\mathbf{y}\in B_{\epsilon}(f(\mathbf{x}_0)) \\ g\circ f(\mathbf{x}) \equiv \mathbf{x}\quad \textrm{for }\mathbf{x}\in B_{\epsilon}(\mathbf{x}_0) \\ \end{cases}$

_{Luca Rigotti. (2015). University of Pittsburgh, ECON2001 Lecture 12}

I should have felt no hesitation in applying the inverse function theorem but for some difficulties when interpreting

$\begin{aligned} &\quad\enspace\textrm{ locally invertible everywhere} \\ & \nLeftrightarrow \textrm{ globally invertible} \\ \end{aligned}$

(to be continued)

March 14, 2023March 14, 2023

202303141200 Exercise 7.7.1

Consider the coordinate transformation $(x,y)=G(u,v)$ given by

$G(u,v)=(au,bv)$ for $(u,v)\in\mathbb{R}^2$

where $a$ and $b$ are positive constants.

(a) Prove that $G$ is $\textrm{1--1}$ on $\mathbb{R}^2$ .
(b) Show that $G$ maps the circular disc

$K=\{(u,v)\in\mathbb{R}^2 : u^2+v^2\leqslant 1\}$

in the $u$ , $v$ plane onto the elliptical disc

$G(K)=\{(x,y)\in\mathbb{R}^2 : x^2/a^2+y^2/b^2\leqslant 1\}$

in the $x$ , $y$ plane.
(c) Calculate $\mathrm{det}\,J_{G,(u,v)}$ .
(d) Calculate, using Theorem 7.7.5, the area of the elliptical disc $G(K)$ . (Assume known that the area of $K$ is $\pi$ .)

_{Extracted from P. R. Baxandall. (1986). Vector Calculus.}

Roughwork.

Symbolically,

$\forall\, a,b\in X,\quad f(a)=f(b)\Rightarrow a=b$ ,

or the contrapositive

$\forall\, a,b\in X,\quad a\neq b\Rightarrow f(a)\neq f(b)$ .

_{Wikipedia on Injective function}

(a)

From $G:\mathbb{R}^2\to\mathbb{R}^2:(u,v)\mapsto (x=au,y=bv)$ ,

$\begin{aligned} & \enspace & (x_1,y_1)=G(u_1,v_1) & = G(u_2,v_2)=(x_2,y_2) \\ \Rightarrow & \enspace & (au_1,bv_1) & = (au_2,bv_2) \\ \Rightarrow & \enspace & (u_1,v_1) & = (u_2,v_2) \\ \end{aligned}$

(b) Obvious.

(c)

$\begin{aligned} J_{G}(u,v) & = \begin{bmatrix} \partial_u(au)&\partial_v(au) \\ \partial_u(bv) & \partial_v(bv) \\ \end{bmatrix} \\ & = \begin{bmatrix} a & 0 \\ 0 & b \\ \end{bmatrix} \\ \mathrm{det}\,J_{G,(u,v)} & = ab \\ \end{aligned}$

(d)

Theorem (Change of variables). Let $G:K\subseteq \mathbb{R}^2\to\mathbb{R}^2$ be a $C^1$ function defined on a compact set $K$ in $\mathbb{R}^2$ , and let $D\subseteq K$ be an open subset of $\mathbb{R}^2$ such that

i. $K\backslash D$ is a null set,
ii. $G$ is $\textrm{1--1}$ on $D$ ,
iii. $\mathrm{det}\,J_{G,(u,v)}\neq 0$ for all $(u,v)\in D$ .

Then, for any bounded function $f:G(K)\subseteq\mathbb{R}^2\to\mathbb{R}$ which is continuous on $G(D)$

$\displaystyle{\iint_{G(K)}f(x,y)\,\mathrm{d}x\,\mathrm{d}y=\iint_{K}f(G(u,v))|\mathrm{det}\,J_{G,(u,v)}|\,\mathrm{d}u\,\mathrm{d}v}$ .

_{See Theorem 7.7.5 on pg. 404.}

The area of the elliptical disc $G(K)$ is $\pi ab$ .

This problem is not to be attempted.

March 14, 2023March 14, 2023

202303141137 Exercise 2.7.1

Calculate the lengths of the following smooth simple arcs in $\mathbb{R}^3$ .

(a) The circular helix parametrized by

$f(t)=(t,\cos t,\sin t)$

where $t\in [a,b]$ ;
(b) the curve parametrized by

$f(t)=(e^t\cos t,e^t\sin t,e^t)$

where $t\in [0,k]$ .

_{Extracted from P. R. Baxandall. (1986). Vector Calculus.}

Roughwork.

Let $f:[a,b]\subseteq\mathbb{R}\to\mathbb{R}^n$ be a $C^1$ path in $\mathbb{R}^n$ . The length of $f$ is defined to be

$l(f)=\displaystyle{\int_{a}^{b}\|f'(t)\|\,\mathrm{d}t}$ .

_{See Definition 2.7.2 on pg.60}

(a)

$\begin{aligned} f(t) & = (t,\cos t,\sin t) \\ f'(t) & = (1,-\sin t,\cos t) \\ \|f'(t)\| & = \sqrt{(1)^2+(-\sin t)^2+(\cos t)^2} \\ & = \sqrt{2} \\ l(f) & = \int_{a}^{b} \sqrt{2}\,\mathrm{d}t \\ & = \sqrt{2}(b-a) \\ \end{aligned}$

(b) Not to be attempted.

September 28, 2022October 24, 2022

202209281050 Exercise 22.2 (Q23)

An electric field is given by $\overrightarrow{E}=E_0\,\hat{\jmath}$ , where $E_0$ is a constant. Find the potential as a function of position, taking $V=0$ at $y=0$ .

_{Extracted from R. Wolfson. (2016). Essential University Physics.}

Background.

Electric field vector $\overrightarrow{E}$ is related to electric potential $V$ by:

$\displaystyle{\overrightarrow{E}= -\bigg( \frac{\partial V}{\partial x}\,\hat{\mathbf{i}} + \frac{\partial V}{\partial y}\,\hat{\mathbf{j}} + \frac{\partial V}{\partial z}\,\hat{\mathbf{k}} \bigg)}$

or reversely,

$\displaystyle{V(r)=-\int\overrightarrow{E}\cdot\mathrm{d}\vec{r}}$ .

Write

$\displaystyle{(0,E_0,0) = \bigg( -\frac{\partial V}{\partial x},-\frac{\partial V}{\partial y},-\frac{\partial V}{\partial z}\bigg) }$

that implies $V=V(y)$ is $x$ – and $z$ -independent.

Hence

$\begin{aligned} V(y) & = -\int E_0\,\mathrm{d}y \\ & = -E_0y+C\qquad \textrm{for some constant }C \\ \because\enspace 0 & = V(y=0) = -E_0(0)+C \\ \Rightarrow C& = 0 \\ \therefore\enspace V&=V(y)=-E_0y\\ \end{aligned}$

September 28, 2022September 29, 2022

202209281002 Exercise 14.2 (Q24)

Analysis of waves in shallow water (depth much less than wavelength) yields the following wave equation:

$\displaystyle{\frac{\partial^2y}{\partial x^2}=\frac{1}{gh}\frac{\partial^2y}{\partial t^2}}$

where $h$ is the water depth and $g$ the gravitational acceleration. Give an expression for the wave speed.

_{Extracted from R. Wolfson. (2016). Essential University Physics.}

(top-down approach)

Write a traveling sinusoidal wave in the form

$y(x,t)=A\cos (kx\pm \omega t)$

the wave speed $v$ being such as

$\displaystyle{v=\frac{\lambda}{T}=\frac{2\pi /k}{2\pi /\omega}=\frac{\omega}{k}}$ .

$\begin{aligned} \textrm{LHS} & = \frac{\partial^2y}{\partial x^2} \\ & = \frac{\partial}{\partial x}\bigg(\frac{\partial}{\partial x}\big( A\cos (kx\pm\omega t)\big) \bigg) \\ & = \frac{\partial}{\partial x}\big( kA\sin (kx\pm\omega t)\big) \\ & = -k^2A\cos (kx\pm\omega t) \\ \textrm{RHS} & = \frac{1}{gh}\frac{\partial^2y}{\partial t^2} \\ & = \frac{1}{gh}\frac{\partial}{\partial t}\bigg(\frac{\partial}{\partial t}\big( A\cos (kx\pm\omega t)\big)\bigg) \\ & = \frac{1}{gh}\frac{\partial}{\partial t}\big(\omega A\sin (kx\pm\omega t)\big) \\ & = \frac{1}{gh}\big( -\omega^2A\cos (kx\pm\omega t)\big) \\ \textrm{LHS} & = \textrm{RHS} \\ \Rightarrow v = \frac{\omega}{k} & = \sqrt{gh} \\ \end{aligned}$

$\therefore$ The wave speed $v$ is given by $v=\sqrt{gh}$ .