Algebraic Geometry – Physicspupil

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.

February 15, 2024February 15, 2024

202402151329 Solution to 1985-CE-AMATH-II-5

If the equation

$x^2+y^2+kx-(2+k)y=0$

represents a circle with radius $\sqrt{5}$ ,

(a) find the value(s) of $k$ ;
(b) find the equation(s) of the circle(s).

As if sitting an exam in additional mathematics, it certainly not being showing off, we should perhaps involve ourselves with some sort of calculus.

Roughwork.

As always, have in mind some pictures. Hence, we draw:

and also

after the stage got set, we write

$\begin{aligned} 0 & = x^2+y^2+kx-(2+k)y \\ 0 & = 2x\,\mathrm{d}x+2y\,\mathrm{d}y + k\,\mathrm{d}x - (2+k)\,\mathrm{d}y \\ y' & = \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x+k}{k-2y+2} \\ \end{aligned}$

and furthermore,

$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ & \Rightarrow 2x+k=0 \\ & \Rightarrow x = -\frac{k}{2} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \infty \\ & \Rightarrow k-2y+2 = 0 \\ & \Rightarrow y = \frac{k}{2}+1 \\ \end{aligned}$

In the case of $x=-\frac{k}{2}$ :

$\begin{aligned} 0 & = \bigg( -\frac{k}{2}\bigg)^2+y^2+k\bigg( -\frac{k}{2}\bigg) -(2+k)y \\ 0 & = y^2 - (2+k)y - \frac{k^2}{4} \\ \Delta & = \big( -(2+k)\big)^2 - 4(1)\bigg( -\frac{k^2}{4}\bigg) \\ & = 2(k^2+2k+2)\qquad (> 0) \\ y_1,y_3 & = \frac{-\big( -(2+k)\big) \pm \sqrt{\Delta}}{2(1)} \\ 2\sqrt{5} & = y_3-y_1 \\ \sqrt{\Delta} & = 2\sqrt{5} \\ \Delta & = 20 \\ 20 & = 2(k^2+2k+2) \\ 0 & = k^2+2k-8 \\ 0 & = (k-2)(k+4) \\ k & = 2,-4 \\ \end{aligned}$

and the case of $y=\frac{k}{2}+1$ :

$\begin{aligned} 0 & = x^2 + \bigg( \frac{k}{2}+1\bigg)^2 + kx - (2+k)\bigg( \frac{k}{2}+1\bigg) \\ 0 & = x^2+kx - \bigg(\frac{k}{2}+1\bigg)^2\\ \Delta & = (k)^2 - 4(1)\bigg( -\bigg(\frac{k}{2}+1\bigg)^2\bigg) \\ & = 2(k^2+2k+2) \qquad (> 0) \\ x_1,x_3 & = \frac{-(k)\pm\sqrt{\Delta}}{2(1)} \\ 2\sqrt{5} & = x_3 - x_1 \\ \sqrt{\Delta} & = 2\sqrt{5} \\ \Delta & = 20 \\ 20 & = 2(k^2+2k+2) \\ 0 & = k^2+2k-8 \\ 0 & = (k-2)(k+4) \\ k & = 2,-4 \\ \end{aligned}$

Targeting the centre $C(a,b)$ :

$\begin{aligned} a_{i=1,2} & = \frac{x_3 - x_1}{2} = -\frac{k}{2}\bigg|_{k=2,-4} = -1,2 \\ b_{i=1,2} & = \frac{y_3-y_1}{2} = \frac{-\big( -(2+k)\big)}{2}\bigg|_{k=2,-4} = 2,-1 \\ (a,b) & = (-1,2)\cup (2,-1) \\ \end{aligned}$

Of the equation of circle, the standard form is

$\left\{ \begin{aligned} (x+1)^2 + (y-2)^2 & = (\sqrt{5})^2 \\ (x-2)^2 + (y+1)^2 & = (\sqrt{5})^2 \\ \end{aligned}\right\}$

and the general form

$\left\{ \begin{aligned} x^2+y^2+2x-4y & = 0 \\ x^2+y^2-4x+2y & = 0 \\ \end{aligned}\right\}$

This problem is not to be attempted.

October 13, 2023October 16, 2023

202310130920 Exercise 14.1.3

If $\cos A=\frac{4}{5}$ and $\cos B=\frac{12}{13}$ where $A$ and $B$ are both acute angles, find, without using tables or a calculator,

(a) $\sin (A+B)$ ,
(b) $\cos (A-B)$ ,
(c) $\tan (A+B)$ .

_{Extracted from J. F. Talbert & H. H. Heng. (1995). Additional Mathematics Pure and Applied (6e).}

Roughwork.

(a)

Not yet to approach a solution, I wish to get a feel of the problem from three perspectives— i. algebraic, ii. geometric, and iii. differential:

(i. algebraic)

$\begin{aligned} \cos A & =\frac{4}{5} \\ \cos B & = \frac{12}{13} \\ & \\ \cos^2A & = \frac{16}{25} \\ \cos^2B & = \frac{144}{169} \\ & \\ \sin^2A & = 1-\cos^2A\\ & = \frac{9}{25} \\ \sin^2B & = 1-\cos^2B\\ & = \frac{25}{169} \\ & \\ \sin A & = \sqrt{\frac{9}{25}} \\ & = \frac{3}{5} \\ \sin B & = \sqrt{\frac{25}{169}} \\ & = \frac{5}{13} \\ \end{aligned}$

(ii. geometric)

(iii. differential)

Let $f(x)=\sin x$ . Then, recall that

$\displaystyle{f(x)\approx x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots }$

because of Taylor series/expansion of a function $f(x)$ at some point $a$ :

$\displaystyle{f(x)\big|_{x=a} = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\bigg|_{x=a}}$

such that in our situation, you know, we have $a=0$ (the Maclaurin series):

$\displaystyle{f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n}$

Now that we have enough, here begins the

Solution.

(i. algebraic)

$\begin{aligned} \sin (A+B) & = \sin A\cos B + \cos A\sin B \\ & = \bigg(\frac{3}{5}{\bigg)\bigg(\frac{12}{13}\bigg) + \bigg(\frac{4}{5}\bigg)\bigg(\frac{5}{13}\bigg) \\ & = \frac{56}{65} \\ \end{aligned}$

(ii. geometric)

$\begin{aligned} \frac{h}{14} & = \sin (90^\circ -B) \\ h & = 14\cos B \\ & \\ \sin (A+B) & = \frac{h}{15} \\ & = \frac{14\cos B}{15} \\ & = \frac{14}{15}\bigg( \frac{12}{13} \bigg) \\ & = \frac{56}{65} \\ \end{aligned}$

(iii. differential)

That angles $A$ and $B$ are acute angles, i.e.,

$0<B<A<90^\circ$

does not imply $A+B$ is also an acute angle, but that

$0<A+B<180^\circ$ .

such that

$\begin{aligned} 0 & <\sin (A+B)<1 \\ -1 & <\cos (A+B)<1 \\ 0 & <\sin B<\sin A<\cos A<\cos B < 1\\ \end{aligned}$

Letting $f(x)=\sin x$ ,

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [0,A+B] \\ & = \int_{0}^{A+B}f(x)\,\mathrm{d}x \\ & = \int_{0}^{A+B}\sin (x)\,\mathrm{d}x \\ & = \big[-\cos x\big]_{0}^{A+B} \\ & = 1-\cos (A+B) \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [0,A] \\ & = \int_{0}^{A}f(x)\,\mathrm{d}x \\ & = \int_{0}^{A}\sin (x)\,\mathrm{d}x \\ & = \big[-\cos x\big]_{0}^{A} \\ & = 1-\cos A \\ & = 1-\frac{4}{5} \\ & = \frac{1}{5} \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [0,B] \\ & = \int_{0}^{B}f(x)\,\mathrm{d}x \\ & = \int_{0}^{B}\sin (x)\,\mathrm{d}x \\ & = \big[-\cos x\big]_{0}^{B} \\ & = 1-\cos B \\ & = 1-\frac{12}{13} \\ & = \frac{1}{13} \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [B,A] \\ & = \frac{1}{5}-\frac{1}{13} \\ & = \frac{8}{65} \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [A,A+B] \\ & = 1-\cos (A+B) - \frac{1}{5} \\ & = \frac{4}{5}-\cos (A+B) \\ \end{aligned}$

then letting $g(x)=\cos x$ , and noting in a right-angled triangle the cosine of an angle admits of no negative values, so can it further be said

$\begin{aligned} & \quad\enspace 0<g(x)<1 \\ &\Rightarrow 0<g(A+B)<1 \\ &\Rightarrow 0<A+B<90^\circ \\ \end{aligned}$

I lost my way…

Let $w(u(t),v(t))=u(t)-v(t)$ be the difference in degrees between time-varying angles $u$ and $v$ where assumed is $u(t)>v(t)$ without loss of generality.

$\begin{aligned} f(u,v,w) & \stackrel{\textrm{def}}{=} \sin (u+v) \\ & = \sin (2v+w ) \\ & = \sin (2u-w ) \\ & \\ g(u,v,w) &\stackrel{\textrm{def}}{=} \cos (u+v) \\ & = \cos (2v+w ) \\ & = \cos (2u-w ) \\ \end{aligned}$

Taking partial derivatives wrt $u$ , $v$ , $w$ , and $t$ :

$\begin{aligned} \partial_uf(u,v,w) & = 2\cos (2u-w) \\ \partial_vf(u,v,w) & = 2\cos (2v+w) \\ \partial_wf(u,v,w) & = 0 \\ \partial_tf(u,v,w) & = (u'+v')\cos (u+v)\\ \end{aligned}$

$\begin{aligned} \mathrm{d}f(u,v,w) & = \partial_uf\,\mathrm{d}u + \partial_vf\,\mathrm{d}v + \partial_wf\,\mathrm{d}w \\ & = 2\cos (2u-w )\,\mathrm{d}u + 2\cos (2v+w )\,\mathrm{d}v + 0 \\ \dots\,\mathrm{d}u & = \mathrm{d}v+\mathrm{d}w\,\dots \\ & = 4\cos (u+v)\,\mathrm{d}v + 2\cos (u+v)\,\mathrm{d}w \\ & = 2\big(\cos (u+v)\big)(2\,\mathrm{d}v+\mathrm{d}w) \\ \end{aligned}$

I lost my faith…

$f(x,y)\stackrel{\textrm{(1)}}{=}\sin (x+y)$

then

$f(x'=x+y,0)\stackrel{\textrm{(2)}}{=} f(x,y)\stackrel{\textrm{(3)}}{=} f(0,x+y=y')$ .

By equality $\textrm{(1)}$ it states that the output value $f(x,y)$ of $f(\texttt{var1},\texttt{var2})$ is determined by input values of two arguments $(\texttt{var1},\texttt{var2})=(x,y)$ ;

by equality $\textrm{(2)}$ that the output value $f(x,y)$ of $f(\texttt{var1},\texttt{var2})$ is determined by input values $(\texttt{var1},\texttt{var2})=(x+y,0)$ , i.e., by one significant argument $\texttt{var1}=x+y$ , and another trivial argument $\texttt{var2}=0$ ; and

by equality $\textrm{(3)}$ that the output value $f(x,y)$ of $f(\texttt{var1},\texttt{var2})$ is determined by input values $(\texttt{var1},\texttt{var2})=(0,x+y)$ , i.e., by one trivial argument $\texttt{var1}=0$ , and another significant argument $\texttt{var2}=x+y$ .

Hence, assuming a partial variation

$\begin{aligned} &\quad\enspace \sin (\texttt{var1}+\texttt{var2}) \\ & = k_1(\texttt{var1},\texttt{var2})\sin (\texttt{var1}) + k_2(\texttt{var1},\texttt{var2})\sin (\texttt{var2}) \\ \end{aligned}$

we WTS the following

$\begin{aligned} k_1(\texttt{var1},\texttt{var2}) & = \cos (\texttt{var2}) \\ k_2(\texttt{var1},\texttt{var2}) & = \cos (\texttt{var1}) \\ \end{aligned}$

LHS:

$\begin{aligned} &\quad\enspace \frac{\mathrm{d}}{\mathrm{d}t}\sin (\texttt{var1}+\texttt{var2}) \\ & = \big[\cos (\texttt{var1}+\texttt{var2})\big] (\texttt{var1}'+\texttt{var2}') \\ \end{aligned}$

RHS:

$\begin{aligned} &\quad\enspace \frac{\mathrm{d}}{\mathrm{d}t}(k_1\sin (\texttt{var1})+k_2\sin (\texttt{var2})) \\ & = \big(k_1'\cos (\texttt{var1})\big) \texttt{var1}' + \big(k_2'\cos (\texttt{var2})\big) \texttt{var2}' \\ \end{aligned}$

and LHS=RHS implies:

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}k_1 & = \frac{\cos (\texttt{var1}+\texttt{var2})}{\cos (\texttt{var1})} \\ \frac{\mathrm{d}}{\mathrm{d}t}k_2 & = \frac{\cos (\texttt{var1}+\texttt{var2})}{\cos (\texttt{var2})} \\ \dots & \dots \dots \dots \\ k_1' & = \bigg( \frac{\cos (\texttt{var2})}{\cos (\texttt{var1})}\bigg) k_2' \\ \end{aligned}$

(to be continued)

October 12, 2021October 24, 2022

202110121137 Exercise 1.1.1

(a) Show that $x^2\in\langle x-y^2,xy\rangle$ in $k[x,y]$ ( $k$ any field).

(b) Show that $\langle x-y^2,xy,y^2\rangle = \langle x,y^2\rangle$ .

(c) Is $\langle x-y^2,xy\rangle = \langle x^2,xy\rangle$ ? Why or why not?

Definition.

Let $f_1,\dots ,f_s\in k[x_1,\dots ,x_n]$ . We let $\langle f_1,\dots ,f_s\rangle$ denote the collection $\langle f_1,\dots ,f_s\rangle = \{ p_1f_1+\cdots +p_sf_s:p_i\in k[x_1,\dots ,x_n]\}$ for $i= 1,\dots ,s$ .

(a)

$\begin{aligned} x^2 & = (x)(x-y^2) + (y)(xy) \\ & \in \langle x-y^2,xy\rangle \\ \end{aligned}$

(b)

First, we want to show

$\langle x-y^2,xy,y^2\rangle \in \langle x,y^2\rangle$ .

For any $p_1,p_2,p_3\in k[x,y]$ ,

$\begin{aligned} & \quad \langle x-y^2,xy,y^2\rangle \\ & = (p_1)(x-y^2)+(p_2)(xy)+(p_3)(y^2) \\ & = p_1x-p_1y^2+p_2xy+p_3y^2 \\ & = (p_1+p_2y)(x)+(p_3-p_1)(y^2) \\ & \in \langle x,y^2\rangle \\ \end{aligned}$

Next, we want to show

$\langle x,y^2\rangle\in\langle x-y^2,xy,y^2\rangle$ .

For any $p_1,p_2\in k[x,y]$ ,

$\begin{aligned} &\quad \langle x,y^2\rangle \\ & = (p_1)(x)+(p_2)(y^2) \\ & = (p_1)(x-y^2)+(0)(xy)+(p_1+p_2)(y^2) \\ & \in \langle x-y^2,xy,y^2\rangle \\ \end{aligned}$

All in all,

$\langle x-y^2,xy,y^2\rangle = \langle x,y^2\rangle$ .

(c) I guess $\textrm{\scriptsize{NOT}}$ .

September 28, 2021November 29, 2021

202109281636 SQL Table 001

CREATE TABLE SQL_Table_001
    (ID int,
    Subject varchar(255),
    Object varchar(255),
    Category varchar(255)
    );
INSERT INTO SQL_Table_001
    (Subject, Object, Category)
    VALUES
    ('algebraic geometry', 'regular (polynomial) functions', 'algebraic varieties')
    ('topology', 'continuous functions', 'topological spaces')
    ('differential topology', 'differentiable functions', 'differentiable manifolds')
    ('complex analysis', 'analytic (power series) functions', 'complex manifolds');

CREATE TABLE SQL_Table_001

(ID int,

Subject varchar(255),

Object varchar(255),

Category varchar(255)

);

INSERT INTO SQL_Table_001

(Subject, Object, Category)

VALUES

('algebraic geometry', 'regular (polynomial) functions', 'algebraic varieties')

('topology', 'continuous functions', 'topological spaces')

('differential topology', 'differentiable functions', 'differentiable manifolds')

('complex analysis', 'analytic (power series) functions', 'complex manifolds');

Output.
+------+-------------------------+-------------------------------------+----------------------------+
|  ID  |  Subject                |  Object                             |  Category                  |
+------+-------------------------+-------------------------------------+----------------------------+
|  1   |  algebraic geometry     |  regular (polynomial) functions     |  algebraic varieties       |
|  2   |  topology               |  continuous functions               |  topological spaces        |
|  3   |  differential topology  |  differentiable functions           |  differentiable manifolds  |
|  4   |  complex analysis       |  analytic (power series) functions  |  complex manifolds         |
+------+-------------------------+-------------------------------------+----------------------------+
(J.S. Milne on Algebraic Geometry (2017), pg. 7)

Output.

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

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

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

(J.S. Milne on Algebraic Geometry (2017), pg. 7)