The Physics Language – Page 4

May 17, 2023May 17, 2023

202305171045 Problem 23.11

In the figure below, $AC$ is a diameter of the circle.

If $AC=1$ , which of the following gives the area of triangle $ABC$ in terms of $\theta$ ?

A. $\displaystyle{\frac{\theta}{2}}$
B. $\displaystyle{\frac{\tan\theta}{2}}$
C. $2\sin\theta$
D. $\displaystyle{\frac{\sin\theta\cos\theta}{2}}$

_{Extracted from Phu Nielson. (2015). SAT Math Advanced Guide and Workbook.}

Remark.

The quantity angle $\theta$ is dimensionless as it is the ratio of arc length to radius, i.e.,

$\frac{[L]}{[L]}=[L]^0$ ;

the sine, the cosine, and the tangent of which, as in a right-angled triangle, are

$\begin{aligned} \sin\theta & = \frac{\textrm{opposite}}{\textrm{hypotenuse}} \\ \cos\theta & = \frac{\textrm{adjacent}}{\textrm{hypotenuse}} \\ \tan\theta & = \frac{\textrm{opposite}}{\textrm{adjacent}} \\ \end{aligned}$

also dimensionless, i.e., $[L]^0$ . Note that the quantity area is $[L]^2$ in dimension. Hence, choices A. to D. are being understood as:

A. $\displaystyle{\frac{\theta}{2}}\enspace\textrm{(sq unit)}$
B. $\displaystyle{\frac{\tan\theta}{2}}\enspace\textrm{(sq unit)}$
C. $2\sin\theta\enspace\textrm{(sq unit)}$
D. $\displaystyle{\frac{\sin\theta\cos\theta}{2}}\enspace\textrm{(sq unit)}$

Warm-up.

Set-up.

where

$\begin{aligned} \mathbf{OA} & = (-1/2,0) \\ \mathbf{OB} & = (x,y) \\ \mathbf{OC} & = (1/2,0) \\ \mathbf{AB} & = \mathbf{OB}-\mathbf{OA} \\ & = (x+1/2,y) \\ AB &= |\mathbf{AB}| \\ & = \sqrt{\bigg( x+\frac{1}{2}\bigg)^2+y^2} \\ \mathbf{BC} & = \mathbf{OC}-\mathbf{OB} \\ & = (1/2-x,-y) \\ BC &= |\mathbf{BC}| \\ & = \sqrt{\bigg(\frac{1}{2}-x\bigg)^2+(-y)^2} \\ \mathbf{AC} & = \mathbf{OC}-\mathbf{OA} \\ & = (1,0) \\ AC & = |\mathbf{AC}| \\ & = \sqrt{(1)^2+(0)^2} =1\\ \sin\theta & = \frac{y}{BC} \\ \cos\theta & = \frac{1/2-x}{BC} \\ \tan\theta & = \frac{y}{1/2-x} \\ \end{aligned}$

Observe that

$\displaystyle{\textrm{Area of }\triangle ABC=\frac{|\mathbf{AB}||\mathbf{BC}|}{2}}$

proceed with

$\begin{aligned} &\quad \frac{AB\cdot BC}{2} \\ & = \frac{\sqrt{(x+0.5)^2+y^2}\sqrt{(0.5-x)^2+(-y)^2}}{2} \\ & = \frac{\sqrt{x^2+x+\frac{1}{4}+y^2}\sqrt{x^2-x+\frac{1}{4}+y^2}}{2} \\ & = \frac{\sqrt{\frac{1}{2}+x}\sqrt{\frac{1}{2}-x}}{2} \\ & = \frac{\sqrt{(\frac{1}{2})^2-x^2}}{2}\\ \end{aligned}$

by noting $OB=|\mathbf{OB}|$ the radius being $\frac{1}{2}$ :

$\displaystyle{x^2+y^2=\bigg(\frac{1}{2}\bigg)^2=\frac{1}{4}}$ .

The function

$\displaystyle{f(x)=\sqrt{\frac{1}{16}-\frac{x^2}{4}}}$

will output the area by inputting $x$ under constraint on $x$ , $y$ :

$g(x,y)=x^2+y^2-\frac{1}{4}=0$ .

(to be refreshed)

Now ready for problem-solving, write simply

$\begin{aligned} \sin\theta & = \frac{AB}{AC}=AB\\ \cos\theta & = \frac{BC}{AC}=BC\\ \tan\theta & = \frac{AB}{BC}\\ \end{aligned}$

and the answer is D.

May 15, 2023May 15, 2023

202305151128 Exercise 1.1

Evaluate $\textrm{\scriptsize{WITHOUT}}$ a calculator.

1. $(-1)^4$ 2. $(-1)^5$ 3. $(-1)^{10}$ 4. $(-1)^{15}$ 5. $(-1)^8$ 6. $-1^8$ 7. $-(-1)^8$ 8. $(-3)^3$ 9. $-3^3$ 10. $-(-3)^3$ 11. $-(-6)^2$ 12. $-(-4)^3$ 13. $2^3\times 3^2\times (-1)^5$ 14. $(-1)^4\times 3^3\times 2^2$ 15. $(-2)^3\times (-3)^4$ 16. $3^0$ 17. $6^{-1}$ 18. $4^{-1}$ 19. $5^0$ 20. $3^2$ 21. $3^{-2}$ 22. $5^3$ 23. $5^{-3}$ 24. $7^2$ 25. $7^{-2}$ 26. $10^3$ 27. $10^{-3}$

_{Extracted from Phu Nielson. (2015). SAT Math Advanced Guide and Workbook.}

Roughwork.

1.

Since $(x^m)^n=x^{mn}$ ,

$\begin{aligned} (-1)^4 & = ((-1)^2)^2 \\ & = (1)^2\\ & = 1 \\ \end{aligned}$

2.

Since $x^m\cdot x^n=x^{m+n}$ ,

$\begin{aligned} (-1)^5 & = (-1)^{2(2)+1} \\ & = (-1)^{2(2)}\cdot (-1)^1 \\ & = ((-1)^{2})^2\cdot (-1)\\ & = (1)^2\cdot (-1) \\ & = 1\cdot (-1) \\ & = -1 \\ \end{aligned}$

3.

$\begin{aligned} (-1)^{10} & = (-1)^{(2)(5)}\\ & = ((-1)^2)^5 \\ & = (1)^5 \\ & = 1\\ \end{aligned}$

4.

$\begin{aligned} (-1)^{15} & = (-1)^{2(7)+1} \\ & = (-1)^{2(7)}\cdot (-1)^1 \\ & = ((-1)^2)^7\cdot (-1) \\ & = (1)^7\cdot (-1) \\ & = 1\cdot (-1) \\ & = -1 \\ \end{aligned}$

5.

$\begin{aligned} (-1)^8 & = (-1)^{2(4)} \\ & = ((-1)^2)^4 \\ & = (1)^4 \\ & = 1 \\ \end{aligned}$

6.

$\begin{aligned} -1^8 & = -(1^8) \\ & = -(1) \\ & = -1 \\ \end{aligned}$

7.

$\begin{aligned} -(-1)^8 & = -((-1)^8) \\ & = -((-1)^{2(4)}) \\ & = -(((-1)^2)^4) \\ & = -(1^4) \\ & = -(1) \\ & = -1 \\ \end{aligned}$

8.

Since $(xy)^m=x^my^m$ ,

$\begin{aligned} (-3)^3 & = ((-1)(3))^3 \\ & = (-1)^3(3)^3 \\ & = (-1)^{2+1}(3\cdot 3\cdot 3) \\ & = ((-1)^2\cdot (-1))(9\cdot 3) \\ & = (1\cdot (-1))(27) \\ & = (-1)(27) \\ & = -27 \\ \end{aligned}$

9.

$\begin{aligned} -3^3 & = -(3^3) \\ & = -(3\cdot 3\cdot 3) \\ & = -(9\cdot 3) \\ & = -(27) \\ & = -27 \\ \end{aligned}$

10.

$\begin{aligned} -(-3)^3 & = -((-3)^3) \\ & = -(((-1)(3))^3) \\ & = -((-1)^3(3)^3) \\ & = -((-1)^{2+1}(3\cdot 3\cdot 3)) \\ & = -((-1)^2(-1)(9\cdot 3)) \\ & = -((1)(-1)(27)) \\ & = -((-1)(27)) \\ & = -(-27) \\ & = 27 \\ \end{aligned}$

Expressions 11. to 27. are not to be attempted.

May 11, 2023May 11, 2023

202305111603 Exercise 3.1

Solve the following equations:

(a) $13x-4=3x+16$ .

(b) $\displaystyle{\frac{3x-9}{18}+\frac{x}{27}-\frac{2x-5}{4}=\frac{4}{3}-x}$ .

(c) $\displaystyle{\frac{3}{x-1}-\frac{2}{x+4}=\frac{4}{2-2x}}$ .

(d) $\displaystyle{\frac{2x}{x-3}-\frac{4x+1}{2x-1}=\frac{21}{2x^2-7x+3}}$ .

(e) $\displaystyle{\frac{1}{4}(3y-2)-\bigg[ y-\frac{1}{y}(7-3y)\bigg] =-\frac{1}{4}y-7}$ .

_{Extracted from K. L. Nielsen. (1958). College Mathematics.}

Roughwork.

(a)

$\begin{aligned} 13x-4 & = 3x+16 \\ 13x-3x & = 16+4 \\ 10x & = 20 \\ x & = 2 \\ \end{aligned}$

(b)

$\begin{aligned} \frac{3x-9}{18} + \frac{x}{27} - \frac{2x-5}{4} & = \frac{4}{3}-x \\ \frac{3x-9}{2\cdot 3^2} + \frac{x}{3^3} - \frac{2x-5}{2^2} & = \frac{4}{3}-x \\ 6(3x-9) + 4x-27(2x-5) & =36(4)-108x \\ 18x-54+4x-54x+135 & =144-108x \\ 18x+4x-54x+108x & = 144+54-135 \\ 76x & = 63 \\ x & = \frac{63}{76} \\ \end{aligned}$

(c)

$\begin{aligned} \frac{3}{x-1}-\frac{2}{x+4} & = \frac{4}{2-2x} \\ \frac{3}{x-1}-\frac{2}{x+4} & = -\frac{2}{x-1} \\ 3(x+4) - 2(x-1) & = -2(x+4) \\ 3x+12-2x+2 & = -2x-8 \\ 3x & = -22 \\ x & = -\frac{22}{3} \\ \end{aligned}$

(d) Not to be attempted.

(e) Not to be attempted.

May 10, 2023May 11, 2023

202305101606 Problem 3.16.1

Let $G=\{ \mathbf{e},a,a^2,a^3\}$ be a cyclic group of order $4$ and $G'$ be the multiplicative group $\{1,-1,\mathbf{i},-\mathbf{i}\}$ of $4$ complex numbers.

i. Tabulate two isomorphisms from $G$ to $G'$ .
ii. Explain why there are no other isomorphisms from $G$ to $G'$ .

_{Extracted from A. P. Hillman. (1999). Abstract Algebra A First Undergraduate Course.}

Background.

The group $G$ is cyclic if and only if every element of $G$ can be expressed as the power of one element of $G$ :

$\exists\, g\in G,\,\forall\, h\in G:\enspace h=g^n\textrm{ for some }n\in\mathbb{Z}$ .

if and only if it is generated by one element $g\in G$ called a generator of $G$ :

$G=\langle g\rangle$ .

_{ProofWiki on Cyclic Group}

Let $(F,+,\times )$ be a field and let $F^*:=F\backslash\{ 0\}$ be the set $F$ less its zero. The group $(F^*,\times )$ is known as the multiplicative group of $F$ .

_{ProofWiki on Multiplicative Group}

A bijective homomorphism is called an isomorphism; that is, a group isomorphism $\theta$ from $G$ to $G'$ is a bijection $\theta$ such that $\theta (ab)=\theta (a)\theta (b)$ for all $a$ and $b$ in $G$ .

_{Text on 3.2 Group Isomorphism, pg.141}

Let $\theta$ be a group isomorphism from $G$ to $G'$ . Then:

(a) $\mathbf{e}\stackrel{\theta}{\mapsto}\mathbf{e}'$ , where $\mathbf{e}$ and $\mathbf{e}'$ are the identities of $G$ and $G'$ .
(b) If $a\stackrel{\theta}{\mapsto}a'$ , then $a^n\stackrel{\theta}{\mapsto}(a')^n$ for all integers $n$ . In particular,

$a^{-1}\stackrel{\theta}{\mapsto}(a')^{-1}$ .

(c) If $a\stackrel{\theta}{\mapsto}a'$ , then $a$ and $a'$ have equal orders.

_{Text on 3.2 Group Isomorphism, pg.143}

Roughwork.

$\begin{array}{c|cccc} (G,*) & \mathbf{e} & a & a^2 & a^3 \\\hline \mathbf{e} & \mathbf{e} & a & a^2 & a^3 \\ a & a & a^2 & a^3 & \mathbf{e} \\ a^2 & a^2 & a^3 & \mathbf{e} & a \\ a^3 & a^3 & \mathbf{e} & a & a^2 \\ \end{array}$

$\begin{array}{c|cccc} (G',\times ) & 1 & -1 & \mathbf{i} & -\mathbf{i} \\\hline 1 & 1 & -1 & \mathbf{i} & -\mathbf{i} \\ -1 & -1 & 1 & -\mathbf{i} & \mathbf{i} \\ \mathbf{i} & \mathbf{i} & -\mathbf{i} & -1 & 1 \\ -\mathbf{i} &-\mathbf{i} & \mathbf{i}} & 1 & -1 \\ \end{array}$

Observe that

$G=\langle a\rangle$ ; $G'=\langle \mathbf{i}\rangle=\langle -\mathbf{i}\rangle$ .

This problem is not to be attempted.

April 13, 2023April 22, 2023

202304131118 Problem 2.1.1

For each of the following parts, tell whether it specifies $\theta$ as a permutation on $\mathbf{X_5}=\{ 1,2,3,4,5\}$ . If not, explain how Definition 1 is not satisfied.

(a) $\theta:1\mapsto 1,2\mapsto 3,3\mapsto 4,4\mapsto 5,5\mapsto 1$ .
(b) $\theta:1\mapsto 5,2\mapsto 1,3\mapsto 2,4\mapsto 3,5\mapsto 4$ .
(c) $\theta:1\mapsto 1,2\mapsto 2,3\mapsto 3,4\mapsto 4,5\mapsto 5$ .

_{Extracted from A. P. Hillman. (1999). Abstract Algebra A First Undergraduate Course.}

Roughwork.

(a)

The arrow form

$\theta:1\mapsto 1,2\mapsto 3,3\mapsto 4,4\mapsto 5,5\mapsto 1$

may also be expressed as in function form

$\begin{aligned} \theta (1) & = 1\\ \theta (2) & = 3\\ \theta (3) & = 4\\ \theta (4) & = 5\\ \theta (5) & = 1\\ \end{aligned}$

For any finite non-empty set $S$ , $A(S)$ the set of all $\textrm{1--1}$ transformations (mappings) of $S$ onto $S$ forms a group called permutation group and any element of $A(S)$ , i.e., a mapping from $S$ onto itself, is called permutation.

_{From Wikibooks on Permutation groups}

As is seen, function $\theta$ here is $\textrm{\scriptsize{NOT}}$ injective (i.e., $\textrm{\scriptsize{NOT}}$ $\textrm{1--1}$ ), for $\theta$ sends both $1$ and $5$ to $1$ . Hence it cannot specify a permutation.

(b)

The arrow form

$\theta:1\mapsto 5,2\mapsto 1,3\mapsto 2,4\mapsto 3,5\mapsto 4$

or two-line form

$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 5 & 1 & 2 & 3 & 4 \end{pmatrix}$

or cyclic form

$(15432)$

also in diagram

is obviously a one-to-one function (/an injection). Thus function $\theta$ here specifies a permutation.

(c)

In arrow form

$\theta:1\mapsto 1,2\mapsto 2,3\mapsto 3,4\mapsto 4,5\mapsto 5$

or matrix form

$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \end{pmatrix}$

or cyclic form

$()$

is the identity permutation.

This problem is not to be attempted.

April 13, 2023April 13, 2023

202304131021 Problem 1.2.1

Which of the following integers are in the set $6\mathbb{Z}$ of integral multiples of $6$ ?

(a) $10$ ;
(b) $-10$ ;
(c) $12$ ;
(d) $-12$ ;
(e) $4002$ ;
(f) $-4002$ ;
(g) $4003$ ;
(h) $-4003$ .

_{Extracted from A. P. Hillman. (1999). Abstract Algebra A First Undergraduate Course.}

Roughwork.

(a), (b)

$\begin{aligned} \pm 10\div 6 & = \frac{\pm 10}{6} \\ & = \pm 1\frac{2}{3}\enspace\notin \mathbb{Z} \\ \end{aligned}$

$\begin{aligned} \because \quad & 6\nmid \pm 10 \\ \therefore \quad & \mathrm{\pm 10}\notin 6\mathbb{Z} \\ \end{aligned}$

Because $6$ is not a divisor (/factor) of $\pm 10$ in $\mathbb{Z}$ , integers $10$ and $-10$ are $\textrm{\scriptsize{NOT}}$ in the set $6\mathbb{Z}$ of integral multiples of $6$ .

(c), (d)

$\begin{aligned} \pm 12\div 6 & = \frac{\pm 12}{6} \\ & = \pm 2\enspace\in \mathbb{Z} \end{aligned}$

$\begin{aligned} \because\quad & 6\mid \pm12 \\ \therefore\quad & \mathrm{\pm 12}\in 6\mathbb{Z} \\ \end{aligned}$

Since $6$ is an integral divisor of $\pm 12$ , integers $12$ and $-12$ are in the set $6\mathbb{Z}$ of integral multiples of $6$ .

(e), (f)

$\begin{aligned} \pm 4002\div 6 & = \frac{\pm 4002}{6} \\ & = \pm 667\enspace \in\mathbb{Z} \\ \end{aligned}$

$\begin{aligned} \because\quad & 6\mid \pm 4002 \\ \therefore\quad & \mathrm{\pm 4002}\in 6\mathbb{Z} \\ \end{aligned}$

As $\pm 4002$ is a multiple of $6$ in $\mathbb{Z}$ , integers $4002$ and $-4002$ are in the set $6\mathbb{Z}$ of integral multiples of $6$ .

(g), (h)

$\begin{aligned} \pm 4003\div 6 & = \frac{\pm 4003}{6} \\ & = \pm 667\frac{1}{6}\enspace \notin\mathbb{Z} \\ \end{aligned}$

$\begin{aligned} \because\quad & 6\nmid \pm 4003 \\ \therefore\quad & \mathrm{\pm 4003} \notin 6\mathbb{Z} \\ \end{aligned}$

Now that $\pm 4003$ is not an integral multiple of $6$ , integers $4003$ and $-4003$ are $\textrm{\scriptsize{NOT}}$ in the set $6\mathbb{Z}$ of integral multiples of $6$ .

This problem is not to be attempted.

March 27, 2023March 27, 2023

202303270928 Problem 1.1.1

Let $F_1$ , $F_2$ , $F_3$ , … be the Fibonacci sequence. Prove that

$F_{n+1}F_{n+2}-F_{n}F_{n+3}=(-1)^n$

for all positive integers $n$ .

_{Extracted from A. P. Hillman. (1999). Abstract Algebra A First Undergraduate Course.}

Roughwork.

Let $P(n)$ be the statement to prove. When $n=1$ ,

$P(1):\quad F_{2}F_{3}-F_{1}F_{4}\stackrel{\textrm{?}}{=}-1$

is to check.

$\begin{aligned} F_1 & = 1\\ F_2 & = 1\\ F_3 & = 2\\ F_4 & = 3\\ \dots & \dots \\ F_{n} & = F_{n-1} + F_{n-2}\quad\textrm{for integers }n\geqslant 3 \\ \end{aligned}$

First, $P(1)$ is true.

Next, be $P(n)$ true. WTS $P(n+1)$ true whenever $P(n)$ true:

$\begin{aligned} P(n+1): & \quad F_{(n+1)+1}F_{(n+1)+2}-F_{n+1}F_{(n+1)+3}\stackrel{\textrm{?}}{=}(-1)^{n+1} \\ \textrm{LHS} & = F_{n+2}F_{n+3}-F_{n+1}F_{n+4} \\ & = (F_{n+1}+F_n)F_{n+3}-F_{n+1}(F_{n+3}+F_{n+2})\\ & = F_{n}F_{n+3}-F_{n+1}F_{n+2} \\ & = -(F_{n+1}F_{n+2}-F_{n}F_{n+3}) \\ & \stackrel{.}{=} -((-1)^n) \\ & = (-1)^{n+1} \\ & = \textrm{RHS} \\ \end{aligned}$

Lastly, as $P(1)$ true and $P(n)\Rightarrow P(n+1)$ , by principle of mathematical induction, the statement is proven for all $n\in\mathbb{Z}^+$ .

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.