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-\usepackage{harpoon}%
-\pagenumbering{gobble}
+\documentclass[a4paper]{article}
+\usepackage[a4paper,margin=2cm]{geometry}
+\usepackage{multicol}
+\usepackage{multirow}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{harpoon}
+\usepackage{tabularx}
+\usepackage[dvipsnames, table]{xcolor}
+\usepackage{graphicx}
+\usepackage{wrapfig}
+\usepackage{tikz}
+\usepackage{tikz-3dplot}
+\usetikzlibrary{calc}
+\usetikzlibrary{angles}
+\usepgflibrary{arrows.meta}
\usepackage{fancyhdr}
-
-\title{Year 12 Specialist}
-\author{Andrew Lorimer}
-\date{2019}
-
-\begin{document}
-
\pagestyle{fancy}
\fancyhead[LO,LE]{Year 12 Specialist}
-\fancyhead[CO,CE]{Andrew Lorimmer}
-\maketitle
-
-\section{Complex \& Imaginary Numbers}\label{complex-imaginary-numbers}
-
-\subsection{Imaginary numbers}\label{imaginary-numbers}
-
-\[i^2 = -1 \quad \therefore i = \sqrt {-1}\]
-
-\subsubsection{Simplifying negative
-surds}\label{simplifying-negative-surds}
-
-\begin{equation}\begin{split}\sqrt{-2} & = \sqrt{-1 \times 2} \\ & = \sqrt{2}i\end{split}\end{equation}
-
-\subsection{Complex numbers}\label{complex-numbers}
-
-\[\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}\]
-
-General form: \(z=a+bi\)\\
-\(\operatorname{Re}(z) = a, \quad \operatorname{Im}(z) = b\)
-
-\subsubsection{Addition}\label{addition}
-
-If \(z_1 = a+bi\) and \(z_2=c+di\), then
-
-\[z_1+z_2 = (a+c)+(b+d)i\]
-
-\subsubsection{Subtraction}\label{subtraction}
-
-If \(z_1=a+bi\) and \(z_2=c+di\), then
-
-\[z_1 - z_2=(a−c)+(b−d)i\]
-
-\subsubsection{Multiplication by a real
-constant}\label{multiplication-by-a-real-constant}
-
-If \(z=a+bi\) and \(k \in \mathbb{R}\), then
-
-\[kz=ka+kbi\]
-
-\subsubsection{\texorpdfstring{Powers of
-\(i\)}{Powers of i}}\label{powers-of-i}
-
-\begin{itemize}
-\tightlist
-\item
- \(i^{4n} = 1\)
-\item
- \(i^{4n+1} = i\)
-\item
- \(i^{4n+2} = -1\)
-\item
- \(i^{4n+3} = -i\)
-\end{itemize}
-
-For \(i^n\), find remainder \(r\) when \(n \div 4\). Then \(i^n = i^r\).
-
-\subsubsection{Multiplying complex
-expressions}\label{multiplying-complex-expressions}
-
-If \(z_1 = a+bi\) and \(z_2=c+di\), then
-
-\[z_1 \times z_2 = (ac-bd)+(ad+bc)i\]
-
-\subsubsection{Conjugates}\label{conjugates}
-
-\[\overline{z} = a \mp bi\]
-
-\subparagraph{Properties}\label{properties}
-
-\begin{itemize}
-\tightlist
-\item
- \(\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}\)
-\item
- \(\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}\)
-\item
- \(\overline{kz} = k \overline{z}, \text{ for } k \in \mathbb{R}\)
-\item
- \(z \overline{z} = = (a+bi)(a-bi) = a^2+b^2 = |z|^2\)
-\item
- \(z + \overline{z} = 2 \operatorname{Re}(z)\)
-\end{itemize}
-
-\subsubsection{Modulus}\label{modulus}
-
-Distance from origin.
-
-\[|{z}|=\sqrt{a^2+b^2} \quad \therefore z \overline{z} = |z|^2\]
-
-Properties
-
-\begin{itemize}
-\tightlist
-\item
- \(|z_1 z_2| = |z_1| |z_2|\)
-\item
- \(|{z_1 \over z_2}| = {|z_1| \over |z_2|}\)
-\item
- \(|z_1 + z_2| \le |z_1 + |z_2|\)
-\end{itemize}
-
-\subsubsection{Multiplicative inverse}\label{multiplicative-inverse}
-
-\begin{equation}\begin{split}z^{-1} & = {1 \over z} \\ & = {{a-bi} \over {a^2+B^2}} \\ & = {\overline{z} \over {|z|^2}}\end{split}\end{equation}
-
-\subsubsection{Dividing complex numbers}\label{dividing-complex-numbers}
-
-\[{{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}} \quad \text{(multiplicative inverse)}\]
-
-In practice, rationalise denominator:
+\fancyhead[CO,CE]{Andrew Lorimer}
+
+\usepackage{mathtools}
+\usepackage{xcolor} % used only to show the phantomed stuff
+\renewcommand\hphantom[1]{{\color[gray]{.6}#1}} % comment out!
+\setlength\fboxsep{0pt} \setlength\fboxrule{.2pt} % for the \fboxes
+\newcommand*\leftlap[3][\,]{#1\hphantom{#2}\mathllap{#3}}
+\newcommand*\rightlap[2]{\mathrlap{#2}\hphantom{#1}}
+\newcolumntype{L}[1]{>{\hsize=#1\hsize\raggedright\arraybackslash}X}%
+\newcolumntype{R}[1]{>{\hsize=#1\hsize\raggedleft\arraybackslash}X}%
+\definecolor{cas}{HTML}{e6f0fe}
+\linespread{1.5}
+\newcommand{\midarrow}{\tikz \draw[-triangle 90] (0,0) -- +(.1,0);}
-\[{z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}\]
-
-\subsection{Argand planes}\label{argand-planes}
-
-\begin{itemize}
-\tightlist
-\item
- Geometric representation of \(\mathbb{C}\)
-\item
- horizontal \(= \operatorname{Re}(z)\); vertical
- \(= \operatorname{Im}(z)\)
-\item
- Multiplication by \(i\) results in an anticlockwise rotation of
- \(\pi \over 2\)
-\end{itemize}
-
-\vfil \break
-
-\subsection{Complex polynomials}\label{complex-polynomials}
-
-\textbf{Include \(\pm\) for all solutions, including imaginary}
-
-\subsubsection{Sum of two squares
-(quadratics)}\label{sum-of-two-squares-quadratics}
-
-\[z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)\]
-
-Complete the square to get to this point.
-
-\paragraph{Dividing complex
-polynomials}\label{dividing-complex-polynomials}
-
-\(P(z) \div D(z)\) gives quotient \(Q(z)\) and remainder \(R(z)\):
-
-\[P(z) = D(z)Q(z) + R(z)\]
-
-\paragraph{Remainder theorem}\label{remainder-theorem}
-
-Let \(\alpha \in \mathbb{C}\). Remainder of \(P(z) \div (z - \alpha)\)
-is \(P(\alpha)\)
-
-\paragraph{Factor theorem}\label{factor-theorem}
-
-If \(a+bi\) is a solution to \(P(z)=0\), then:
-
-\begin{itemize}
-\tightlist
-\item
- \(P(a+bi)=0\)
-\item
- \(z-(a+bi)\) is a factor of \(P(z)\)
-\end{itemize}
-
-\paragraph{Sum of two cubes}\label{sum-of-two-cubes}
-
-\[a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\]
-
-\subsection{Conjugate root theorem}\label{conjugate-root-theorem}
-
-If \(a+bi\) is a solution to \(P(z)=0\), then the conjugate
-\(\overline{z}=a-bi\) is also a solution.
-
-\subsection{Polar form}\label{polar-form}
-
-\begin{equation}\begin{split}z & =r \operatorname{cis} \theta \\ & = r(\operatorname{cos}\theta+i \operatorname{sin}\theta) \\ & = a + bi \end{split}\end{equation}
-
-\begin{itemize}
-\tightlist
-\item
- \(r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}\)
-\item
- \(\theta=\operatorname{arg}(z)\) (on CAS: \texttt{arg(a+bi)})
-\item
- \textbf{principal argument} is
- \(\operatorname{Arg}(z) \in (-\pi, \pi]\) (note capital
- \(\operatorname{Arg}\))
-\end{itemize}
-
-Each complex number has multiple polar representations:\\
-\(z=r \operatorname{cis} \theta = r \operatorname{cis} (\theta+2 n\pi\))
-with \(n \in \mathbb{Z}\) revolutions
-
-\subsubsection{Conjugate in polar form}\label{conjugate-in-polar-form}
-
-\[(r \operatorname{cis} \theta)^{-1} = r\operatorname{cis} (- \theta)\]
-
-Reflection of \(z\) across horizontal axis.
-
-\subsubsection{Multiplication and division in polar
-form}\label{multiplication-and-division-in-polar-form}
-
-\[z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)\]
-
-\[{z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)\]
-
-\subsection{de Moivres' Theorem}\label{de-moivres-theorem}
-
-\[(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}\]
+\begin{document}
-\subsection{Roots of complex numbers}\label{roots-of-complex-numbers}
+\begin{multicols}{2}
+
+ \section{Complex numbers}
+
+ \[\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}\]
-\(n\)th roots of \(z = r \operatorname{cis} \theta\) are
+ \begin{align*}
+ \text{Cartesian form: } & a+bi\\
+ \text{Polar form: } & r\operatorname{cis}\theta
+ \end{align*}
-\[z={r^{1 \over n}} \operatorname{cis}({{\theta + 2 k \pi} \over n})\]
+ \subsection*{Operations}
-Same modulus for all solutions. Arguments are separated by
-\({2 \pi} \over n\)
+\definecolor{shade1}{HTML}{ffffff}
+\definecolor{shade2}{HTML}{e6f2ff}
+ \definecolor{shade3}{HTML}{cce2ff}
+ \begin{tabularx}{\columnwidth}{r|X|X}
+ & \textbf{Cartesian} & \textbf{Polar} \\
+ \hline
+ \(z_1 \pm z_2\) & \((a \pm c)(b \pm d)i\) & convert to \(a+bi\)\\
+ \hline
+ \(+k \times z\) & \multirow{2}{*}{\(ka \pm kbi\)} & \(kr\operatorname{cis} \theta\)\\
+ \cline{1-1}\cline{3-3}
+ \(-k \times z\) & & \(kr \operatorname{cis}(\theta\pm \pi)\)\\
+ \hline
+ \(z_1 \cdot z_2\) & \(ac-bd+(ad+bc)i\) & \(r_1r_2 \operatorname{cis}(\theta_1 + \theta_2)\)\\
+ \hline
+ \(z_1 \div z_2\) & \((z_1 \overline{z_2}) \div |z_2|^2\) & \(\left(\frac{r_1}{r_2}\right) \operatorname{cis}(\theta_1 - \theta_2)\)
+ \end{tabularx}
+
+ \subsubsection*{Scalar multiplication in polar form}
+
+ For \(k \in \mathbb{R}^+\):
+ \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\theta\]
+
+ \noindent For \(k \in \mathbb{R}^-\):
+ \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\left(\begin{cases}\theta - \pi & |0<\operatorname{Arg}(z)\le \pi \\ \theta + \pi & |-\pi<\operatorname{Arg}(z)\le 0\end{cases}\right)\]
+
+ \subsection*{Conjugate}
+
+ \begin{align*}
+ \overline{z} &= a \mp bi\\
+ &= r \operatorname{cis}(-\theta)
+ \end{align*}
-The solutions of \(z^n=a \text{ where } a \in \mathbb{C}\) lie on circle
+ \noindent \colorbox{cas}{On CAS: \texttt{conjg(a+bi)}}
-\[x^2 + y^2 = (|a|^{1 \over n})^2\]
+ \subsubsection*{Properties}
-\subsection{Sketching complex graphs}\label{sketching-complex-graphs}
+ \begin{align*}
+ \overline{z_1 \pm z_2} &= \overline{z_1}\pm\overline{z_2}\\
+ \overline{z_1 \cdot z_2} &= \overline{z_1}\cdot\overline{z_2}\\
+ \overline{kz} &= k\overline{z} \quad | \quad k \in \mathbb{R}\\
+ z\overline{z} &= (a+bi)(a-bi)\\
+ &= a^2 + b^2\\
+ &= |z|^2
+ \end{align*}
-\subsubsection{Straight line}\label{straight-line}
+ \subsection*{Modulus}
+
+ \[|z|=|\vec{Oz}|=\sqrt{a^2 + b^2}\]
+
+ \subsubsection*{Properties}
+
+ \begin{align*}
+ |z_1z_2|&=|z_1||z_2|\\
+ \left|\frac{z_1}{z_2}\right|&=\frac{|z_1|}{|z_2|}\\
+ |z_1+z_2|&\le|z_1|+|z_2|
+ \end{align*}
+
+ \subsection*{Multiplicative inverse}
+
+ \begin{align*}
+ z^{-1}&=\frac{a-bi}{a^2+b^2}\\
+ &=\frac{\overline{z}}{|z|^2}a\\
+ &=r \operatorname{cis}(-\theta)
+ \end{align*}
+
+ \subsection*{Dividing over \(\mathbb{C}\)}
+
+ \begin{align*}
+ \frac{z_1}{z_2}&=z_1z_2^{-1}\\
+ &=\frac{z_1\overline{z_2}}{|z_2|^2}\\
+ &=\frac{(a+bi)(c-di)}{c^2+d^2}\\
+ & \qquad \text{(rationalise denominator)}
+ \end{align*}
+
+ \subsection*{Polar form}
+
+ \begin{align*}
+ z&=r\operatorname{cis}\theta\\
+ &=r(\cos \theta + i \sin \theta)
+ \end{align*}
+
+ \begin{itemize}
+ \item{\(r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}\)}
+ \item{\(\theta = \operatorname{arg}(z)\) \quad \colorbox{cas}{On CAS: \texttt{arg(a+bi)}}}
+ \item{\(\operatorname{Arg}(z) \in (-\pi,\pi)\) \quad \bf{(principal argument)}}
+ \item{\colorbox{cas}{Convert on CAS:}\\ \verb|compToTrig(a+bi)| \(\iff\) \verb|cExpand{r·cisX}|}
+ \item{Multiple representations:\\\(r\operatorname{cis}\theta=r\operatorname{cis}(\theta+2n\pi)\) with \(n \in \mathbb{Z}\) revolutions}
+ \item{\(\operatorname{cis}\pi=-1,\qquad \operatorname{cis}0=1\)}
+ \end{itemize}
+
+ \subsection*{de Moivres' theorem}
+
+ \[(r \operatorname{cis} \theta)^n = r^n \operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}\]
+
+ \subsection*{Complex polynomials}
+
+ Include \(\pm\) for all solutions, incl. imaginary
+
+ \begin{tabularx}{\columnwidth}{ R{0.55} X }
+ \hline
+ Sum of squares & \(\begin{aligned}
+ z^2 + a^2 &= z^2-(ai)^2\\
+ &= (z+ai)(z-ai) \end{aligned}\) \\
+ \hline
+ Sum of cubes & \(a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\)\\
+ \hline
+ Division & \(P(z)=D(z)Q(z)+R(z)\) \\
+ \hline
+ Remainder theorem & Let \(\alpha \in \mathbb{C}\). Remainder of \(P(z) \div (z-\alpha)\) is \(P(\alpha)\)\\
+ \hline
+ Factor theorem & \(z-\alpha\) is a factor of \(P(z) \iff P(\alpha)=0\) for \(\alpha \in \mathbb{C}\)\\
+ \hline
+ Conjugate root theorem & \(P(z)=0 \text{ at } z=a\pm bi\) (\(\implies\) both \(z_1\) and \(\overline{z_1}\) are solutions)
+ \end{tabularx}
+
+ \subsection*{Roots}
+
+ \(n\)th roots of \(z=r\operatorname{cis}\theta\) are:
+
+ \[z = r^{\frac{1}{n}} \operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)\]
+
+ \begin{itemize}
+
+ \item{Same modulus for all solutions}
+ \item{Arguments are separated by \(\frac{2\pi}{n}\)}
+ \item{Solutions of \(z^n=a\) where \(a \in \mathbb{C}\) lie on the circle \(x^2+y^2=\left(|a|^{\frac{1}{n}}\right)^2\) \quad (intervals of \(\frac{2\pi}{n}\))}
+ \end{itemize}
+
+ \noindent For \(0=az^2+bz+c\), use quadratic formula:
+
+ \[z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
+
+ \subsection*{Fundamental theorem of algebra}
+
+ A polynomial of degree \(n\) can be factorised into \(n\) linear factors in \(\mathbb{C}\):
+
+ \[\implies P(z)=a_n(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)\dots(z-\alpha_n)\]
+ \[\text{ where } \alpha_1,\alpha_2,\alpha_3,\dots,\alpha_n \in \mathbb{C}\]
+
+ \subsection*{Argand planes}
+
+ \begin{center}\begin{tikzpicture}[scale=2]
+ \draw [->] (-0.2,0) -- (1.5,0) node [right] {$\operatorname{Re}(z)$};
+ \draw [->] (0,-0.2) -- (0,1.5) node [above] {$\operatorname{Im}(z)$};
+ \coordinate (P) at (1,1);
+ \coordinate (a) at (1,0);
+ \coordinate (b) at (0,1);
+ \coordinate (O) at (0,0);
+ \draw (0,0) -- (P) node[pos=0.5, above left]{\(r\)} node[pos=1, right]{\(\begin{aligned}z&=a+bi\\&=r\operatorname{cis}\theta\end{aligned}\)};
+ \draw [gray, dashed] (1,1) -- (1,0) node[black, pos=1, below]{\(a\)};
+ \draw [gray, dashed] (1,1) -- (0,1) node[black, pos=1, left]{\(b\)};
+ \begin{scope}
+ \path[clip] (O) -- (P) -- (a);
+ \fill[red, opacity=0.5, draw=black] (O) circle (2mm);
+ \node at ($(O)+(20:3mm)$) {$\theta$};
+ \end{scope}
+ \filldraw (P) circle (0.5pt);
+ \end{tikzpicture}\end{center}
+
+ \begin{itemize}
+ \item{Multiplication by \(i \implies\) CCW rotation of \(\frac{\pi}{2}\)}
+ \item{Addition: \(z_1 + z_2 \equiv\) \overrightharp{\(Oz_1\)} + \overrightharp{\(Oz_2\)}}
+ \end{itemize}
+
+ \subsection*{Sketching complex graphs}
+
+ \subsubsection*{Linear}
+
+ \begin{itemize}
+ \item{\(\operatorname{Re}(z)=c\) or \(\operatorname{Im}(z)=c\) (perpendicular bisector)}
+ \item{\(\operatorname{Im}(z)=m\operatorname{Re}(z)\)}
+ \item{\(|z+a|=|z+b| \implies 2(a-b)x=b^2-a^2\)}
+ \end{itemize}
+
+ \subsubsection*{Circles}
+
+ \begin{itemize}
+ \item \(|z-z_1|^2=c^2|z_2+2|^2\)
+ \item \(|z-(a+bi)|=c\)
+ \end{itemize}
+
+ \noindent \textbf{Loci} \qquad \(\operatorname{Arg}(z)<\theta\)
+
+ \begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
+ \draw [->] (0,0) -- (1,0) node [right] {$\operatorname{Re}(z)$};
+ \draw [->] (0,-0.5) -- (0,1) node [above] {$\operatorname{Im}(z)$};
+ \draw [<-, dashed, thick, blue] (-1,0) -- (0,0);
+ \draw [->, thick, blue] (0,0) -- (1,1);
+ \fill [gray, opacity=0.2, domain=-1:1, variable=\x] (-1,-0.5) -- (-1,0) -- (0, 0) -- (1,1) -- (1,-0.5) -- cycle;
+ \begin{scope}
+ \path[clip] (0,0) -- (1,1) -- (1,0);
+ \fill[red, opacity=0.5, draw=black] (0,0) circle (2mm);
+ \node at ($(0,0)+(20:3mm)$) {$\frac{\pi}{4}$};
+ \end{scope}
+ \node [font=\footnotesize] at (0.5,-0.25) {\(\operatorname{Arg}(z)\le\frac{\pi}{4}\)};
+ \node [blue, mydot] {};
+ \end{tikzpicture}\end{center}
+
+ \noindent \textbf{Rays} \qquad \(\operatorname{Arg}(z-b)=\theta\)
+
+ \begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
+ \draw [->] (-0.75,0) -- (1.5,0) node [right] {$\operatorname{Re}(z)$};
+ \draw [->] (0,-1) -- (0,1) node [above] {$\operatorname{Im}(z)$};
+ \draw [->, thick, brown] (-0.25,0) -- (-0.75,-1);
+ \node [above, font=\footnotesize] at (-0.25,0) {\(\frac{1}{4}\)};
+ \begin{scope}
+ \path[clip] (-0.25,0) -- (-0.75,-1) -- (0,0);
+ \fill[orange, opacity=0.5, draw=black] (-0.25,0) circle (2mm);
+ \end{scope}
+ \node at (-0.08,-0.3) {\(\frac{\pi}{8}\)};
+ \node [font=\footnotesize, left] at (-0.75,-1) {\(\operatorname{Arg}(z+\frac{1}{4})=\frac{\pi}{8}\)};
+ \node [brown, mydot] at (-0.25,0) {};
+ \draw [<->, thick, green] (0,-1) -- (1.5,0.5) node [pos=0.25, black, font=\footnotesize, right] {\(|z-2|=|z-(1+i)|\)};
+ \node [left, font=\footnotesize] at (0,-1) {\(-1\)};
+ \node [below, font=\footnotesize] at (1,0) {\(1\)};
+ \end{tikzpicture}\end{center}
+
+ \section{Vectors}
+\begin{center}\begin{tikzpicture}
+ \draw [->] (-0.5,0) -- (3,0) node [right] {\(x\)};
+ \draw [->] (0,-0.5) -- (0,3) node [above] {\(y\)};
+ \draw [orange, ->, thick] (0.5,0.5) -- (2.5,2.5) node [pos=0.5, above] {\(\vec{u}\)};
+ \begin{scope}[very thick, every node/.style={sloped,allow upside down}]
+ \draw [gray, dashed, thick] (0.5,0.5) -- (2.5,0.5) node [pos=0.5] {\midarrow} node[black, pos=0.5, below]{\(x\vec{i}\)};
+ \draw [gray, dashed, thick] (2.5,0.5) -- (2.5,2.5) node [pos=0.5] {\midarrow};
+ \end{scope}
+ \node[black, right] at (2.5,1.5) {\(y\vec{j}\)};
+
+\end{tikzpicture}\end{center}
+
+\subsection*{Column notation}
+
+\[\begin{bmatrix}x\\ y \end{bmatrix} \iff x\boldsymbol{i} + y\boldsymbol{j}\]
+\(\begin{bmatrix}x_2-x_1\\ y_2-y_1 \end{bmatrix}\) \quad between \(A(x_1,y_1), \> B(x_2,y_2)\)
+
+\subsection*{Scalar multiplication}
+
+\[k\cdot (x\boldsymbol{i}+y\boldsymbol{j})=kx\boldsymbol{i}+ky\boldsymbol{j}\]
+
+\noindent For \(k \in \mathbb{R}^-\), direction is reversed
+
+\subsection*{Vector addition}
+\begin{center}\begin{tikzpicture}[scale=1]
+ \coordinate (A) at (0,0);
+ \coordinate (B) at (2,2);
+ \draw [->, thick, red] (0,0) -- (2,2) node [pos=0.5, below right] {\(\vec{u}=2\vec{i}+2\vec{j}\)};
+ \draw [->, thick, blue] (2,2) -- (1,4) node [pos=0.5, above right] {\(\vec{v}=-\vec{i}+2\vec{j}\)};
+ \draw [->, thick, orange] (0,0) -- (1,4) node [pos=0.5, left] {\(\vec{u}+\vec{v}=\vec{i}+4\vec{j}\)};
+\end{tikzpicture}\end{center}
+
+\[(x\boldsymbol{i}+y\boldsymbol{j}) \pm (a\boldsymbol{i}+b\boldsymbol{j})=(x \pm a)\boldsymbol{i}+(y \pm b)\boldsymbol{j}\]
\begin{itemize}
-\tightlist
-\item
- \(\operatorname{Re}(z) = c\) or \(\operatorname{Im}(z) = c\)
- (perpendicular bisector)
-\item
- \(\operatorname{Arg}(z) = \theta\)
-\item
- \(|z+a|=|z+bi|\) where \(m={a \over b}\)
-\item
- \(|z+a|=|z+b| \longrightarrow 2(a-b)x=b^2-a^2\)
+ \item Draw each vector head to tail then join lines
+ \item Addition is commutative (parallelogram)
+ \item \(\boldsymbol{u}-\boldsymbol{v}=\boldsymbol{u}+(-\boldsymbol{v})\)
\end{itemize}
-\subsubsection{Circle}\label{circle}
+\subsection*{Magnitude}
-\(|z-z_1|^2 = c^2 |z_2+2|^2\) or \(|z-(a + bi)| = c\)
+\[|(x\boldsymbol{i} + y\boldsymbol{j})|=\sqrt{x^2+y^2}\]
-\subsubsection{Locus}\label{locus}
-
-\(\operatorname{Arg}(z) < \theta\)
-
-\section{Vectors}\label{vectors}
-
-\begin{itemize}
-\tightlist
-\item
- \textbf{vector:} a directed line segment\\
-\item
- arrow indicates direction
-\item
- length indicates magnitude
-\item
- column notation: \(\begin{bmatrix} x \\ y \end{bmatrix}\)
-\item
- vectors with equal magnitude and direction are equivalent
-\end{itemize}
-
-\begin{figure}
-\centering
-\includegraphics[width=0.20000\textwidth]{graphics/vectors-intro.png}
-\caption{}\label{id}
-\end{figure}
-
-\subsection{Vector addition}\label{vector-addition}
-
-\(\boldsymbol{u} + \boldsymbol{v}\) can be represented by drawing each
-vector head to tail then joining the lines.\\
-Addition is commutative (parallelogram)
-
-\subsection{Scalar multiplication}\label{scalar-multiplication}
-
-For \(k \in \mathbb{R}^+\), \(k\boldsymbol{u}\) has the same direction
-as \(\boldsymbol{u}\) but length is multiplied by a factor of \(k\).
-
-When multiplied by \(k < 0\), direction is reversed and length is
-multplied by \(k\).
-
-\subsection{Vector subtraction}\label{vector-subtraction}
-
-To find \(\boldsymbol{u} - \boldsymbol{v}\), add \(\boldsymbol{-v}\) to
-\(\boldsymbol{u}\)
-
-\subsection{Parallel vectors}\label{parallel-vectors}
-
-Same or opposite direction
+\subsection*{Parallel vectors}
\[\boldsymbol{u} || \boldsymbol{v} \iff \boldsymbol{u} = k \boldsymbol{v} \text{ where } k \in \mathbb{R} \setminus \{0\}\]
-\subsection{Position vectors}\label{position-vectors}
-
-Vectors may describe a position relative to \(O\).
-
-For a point \(A\), the position vector is \overrightharp{OA}
-
-\vfill\eject
-
-\subsection{Linear combinations of non-parallel
-vectors}\label{linear-combinations-of-non-parallel-vectors}
-
-If two non-zero vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are
-not parallel, then:
-
-\[m\boldsymbol{a} + n\boldsymbol{b} = p \boldsymbol{a} + q \boldsymbol{b}\quad \therefore \quad m = p, \> n = q\]
-
-\includegraphics[width=0.20000\textwidth]{graphics/parallelogram-vectors.jpg}
-\includegraphics[width=0.10000\textwidth]{graphics/vector-subtraction.jpg}
-
-\subsection{Column vector notation}\label{column-vector-notation}
-
-A vector between points \(A(x_1,y_1), \> B(x_2,y_2)\) can be represented
-as \(\begin{bmatrix}x_2-x_1\\ y_2-y_1 \end{bmatrix}\)
-
-\subsection{Component notation}\label{component-notation}
-
-A vector \(\boldsymbol{u} = \begin{bmatrix}x\\ y \end{bmatrix}\) can be
-written as \(\boldsymbol{u} = x\boldsymbol{i} + y\boldsymbol{j}\).\\
-\(\boldsymbol{u}\) is the sum of two components \(x\boldsymbol{i}\) and
-\(y\boldsymbol{j}\)\\
-Magnitude of vector
-\(\boldsymbol{u} = x\boldsymbol{i} + y\boldsymbol{j}\) is denoted by
-\(|u|=\sqrt{x^2+y^2}\)
+For parallel vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\):\\
+\[\boldsymbol{a \cdot b}=\begin{cases}
+|\boldsymbol{a}||\boldsymbol{b}| \hspace{2.8em} \text{if same direction}\\
+-|\boldsymbol{a}||\boldsymbol{b}| \hspace{2em} \text{if opposite directions}
+\end{cases}\]
+%\includegraphics[width=0.2,height=\textheight]{graphics/parallelogram-vectors.jpg}
+%\includegraphics[width=1]{graphics/vector-subtraction.jpg}
-Basic algebra applies:\\
-\((x\boldsymbol{i} + y\boldsymbol{j}) + (m\boldsymbol{i} + n\boldsymbol{j}) = (x + m)\boldsymbol{i} + (y+n)\boldsymbol{j}\)\\
-Two vectors equal if and only if their components are equal.
+\subsection*{Perpendicular vectors}
-\subsection{\texorpdfstring{Unit vector
-\(|\hat{\boldsymbol{a}}|=1\)}{Unit vector \textbar{}\textbackslash{}hat\{\textbackslash{}boldsymbol\{a\}\}\textbar{}=1}}\label{unit-vector-hatboldsymbola1}
+\[\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b} = 0\ \quad \text{(since \(\cos 90 = 0\))}\]
-\begin{equation}\begin{split}\hat{\boldsymbol{a}} & = {1 \over {|\boldsymbol{a}|}}\boldsymbol{a} \\ & = \boldsymbol{a} \cdot {|\boldsymbol{a}|}\end{split}\end{equation}
+\subsection*{Unit vector \(|\hat{\boldsymbol{a}}|=1\)}
+\[\begin{split}\hat{\boldsymbol{a}} & = {\frac{1}{|\boldsymbol{a}|}}\boldsymbol{a} \\ & = \boldsymbol{a} \cdot {|\boldsymbol{a}|}\end{split}\]
-\subsection{\texorpdfstring{Scalar/dot product
-\(\boldsymbol{a} \cdot \boldsymbol{b}\)}{Scalar/dot product \textbackslash{}boldsymbol\{a\} \textbackslash{}cdot \textbackslash{}boldsymbol\{b\}}}\label{scalardot-product-boldsymbola-cdot-boldsymbolb}
+ \subsection*{Scalar product \(\boldsymbol{a} \cdot \boldsymbol{b}\)}
-\[\boldsymbol{a} \cdot \boldsymbol{b} = a_1 b_1 + a_2 b_2\]
-\textbf{on CAS:} \texttt{dotP({[}a\ b\ c{]},\ {[}d\ e\ f{]})}
+\begin{center}\begin{tikzpicture}[scale=2]
+ \draw [->] (0,0) -- (1,0.5) node [pos=0.5, above left] {\(\boldsymbol{b}\)};
+ \draw [->] (0,0) -- (1,0) node [pos=0.5, below] {\(\boldsymbol{a}\)};
+ \begin{scope}
+ \path[clip] (1,0.5) -- (1,0) -- (0,0);
+ \fill[orange, opacity=0.5, draw=black] (0,0) circle (2mm);
+ \node at ($(0,0)+(15:4mm)$) {\(\theta\)};
+ \end{scope}
+\end{tikzpicture}\end{center}
+\begin{align*}\boldsymbol{a} \cdot \boldsymbol{b} &= a_1 b_1 + a_2 b_2 \\ &= |\boldsymbol{a}| |\boldsymbol{b}| \cos \theta \\ &\quad (\> 0 \le \theta \le \pi) \text{ - from cosine rule}\end{align*}
+\noindent\colorbox{cas}{On CAS: \texttt{dotP({[}a\ b\ c{]},\ {[}d\ e\ f{]})}}
-\subsection{Scalar product properties}\label{scalar-product-properties}
+\subsubsection*{Properties}
\begin{enumerate}
-\def\labelenumi{\arabic{enumi}.}
-\tightlist
\item
- \(k(\boldsymbol{a\cdot b})=(k\boldsymbol{a})\cdot \boldsymbol{b}=\boldsymbol{a}\cdot (k{b})\)
+ \(k(\boldsymbol{a\cdot b})=(k\boldsymbol{a})\cdot \boldsymbol{b}=\boldsymbol{a}\cdot (k\boldsymbol{b})\)
\item
\(\boldsymbol{a \cdot 0}=0\)
\item
- \(\boldsymbol{a \cdot (b + c)}=\boldsymbol{a \cdot b + a \cdot c}\)
+ \(\boldsymbol{a} \cdot (\boldsymbol{b} + \boldsymbol{c})=\boldsymbol{a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{c}\)
\item
\(\boldsymbol{i \cdot i} = \boldsymbol{j \cdot j} = \boldsymbol{k \cdot k}= 1\)
\item
- If \(\boldsymbol{a} \cdot \boldsymbol{b} = 0\), \(\boldsymbol{a}\) and
- \(\boldsymbol{b}\) are perpendicular
+ \(\boldsymbol{a} \cdot \boldsymbol{b} = 0 \quad \implies \quad \boldsymbol{a} \perp \boldsymbol{b}\)
\item
\(\boldsymbol{a \cdot a} = |\boldsymbol{a}|^2 = a^2\)
\end{enumerate}
-For parallel vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\):\\
-\[\boldsymbol{a \cdot b}=\begin{cases}
-|\boldsymbol{a}||\boldsymbol{b}| \hspace{2.8em} \text{if same direction}\\
--|\boldsymbol{a}||\boldsymbol{b}| \hspace{2em} \text{if opposite directions}
-\end{cases}\]
-
-\subsection{Geometric scalar products}\label{geometric-scalar-products}
-
-\[\boldsymbol{a} \cdot \boldsymbol{b} = |\boldsymbol{a}| |\boldsymbol{b}| \cos \theta\]
-
-where \(0 \le \theta \le \pi\)
-
-\subsection{Perpendicular vectors}\label{perpendicular-vectors}
+\subsection*{Angle between vectors}
-If \(\boldsymbol{a} \cdot \boldsymbol{b} = 0\), then
-\(\boldsymbol{a} \perp \boldsymbol{b}\) (since \(\cos 90 = 0\))
+\[\cos \theta = {{\boldsymbol{a} \cdot \frac{\boldsymbol{b}}{|\boldsymbol{a}| |\boldsymbol{b}|}} = {\frac{a_1 b_1 + a_2 b_2}{|\boldsymbol{a}| |\boldsymbol{b}|}}\]
-\subsection{Finding angle between
-vectors}\label{finding-angle-between-vectors}
+\noindent \colorbox{cas}{On CAS:} \texttt{angle([a b c], [a b c])}
-\textbf{positive direction}
+(Action \(\rightarrow\) Vector \(\rightarrow\)Angle)
-\[\cos \theta = {{\boldsymbol{a} \cdot \boldsymbol{b}} \over {|\boldsymbol{a}| |\boldsymbol{b}|}} = {{a_1 b_1 + a_2 b_2} \over {|\boldsymbol{a}| |\boldsymbol{b}|}}\]
+\subsection*{Angle between vector and axis}
-\textbf{on CAS:} \texttt{angle({[}a\ b\ c{]},\ {[}a\ b\ c{]})} (Action
--\textgreater{} Vector -\textgreater{} Angle)
-
-\subsection{Angle between vector and
-axis}\label{angle-between-vector-and-axis}
-
-Direction of a vector can be given by the angles it makes with
-\(\boldsymbol{i}, \boldsymbol{j}, \boldsymbol{k}\) directions.
-
-For
-\(\boldsymbol{a} = a_1 \boldsymbol{i} + a_2 \boldsymbol{j} + a_3 \boldsymbol{k}\)
-which makes angles \(\alpha, \beta, \gamma\) with positive direction of
+\noindent For\(\boldsymbol{a} = a_1 \boldsymbol{i} + a_2 \boldsymbol{j} + a_3 \boldsymbol{k}\)
+which makes angles \(\alpha, \beta, \gamma\) with positive side of
\(x, y, z\) axes:
-\[\cos \alpha = {a_1 \over |\boldsymbol{a}|}, \quad \cos \beta = {a_2 \over |\boldsymbol{a}|}, \quad \cos \gamma = {a_3 \over |\boldsymbol{a}|}\]
+\[\cos \alpha = \frac{a_1}{|\boldsymbol{a}|}, \quad \cos \beta = \frac{a_2}{|\boldsymbol{a}|}, \quad \cos \gamma = \frac{a_3}{|\boldsymbol{a}|}\]
-\textbf{on CAS:} \texttt{angle({[}a\ b\ c{]},\ {[}1\ 0\ 0{]})} for angle
+\noindent \colorbox{cas}{On CAS:} \texttt{angle({[}a\ b\ c{]},\ {[}1\ 0\ 0{]})}\\for angle
between \(a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}\) and
\(x\)-axis
-\subsection{Vector projections}\label{vector-projections}
+\subsection*{Projections \& resolutes}
+
+\begin{tikzpicture}[scale=3]
+ \draw [->, purple] (0,0) -- (1,0.5) node [pos=0.5, above left] {\(\boldsymbol{a}\)};
+ \draw [->, orange] (0,0) -- (1,0) node [pos=0.5, below] {\(\boldsymbol{u}\)};
+ \draw [->, blue] (1,0) -- (2,0) node [pos=0.5, below] {\(\boldsymbol{b}\)};
+ \begin{scope}
+ \path[clip] (1,0.5) -- (1,0) -- (0,0);
+ \fill[orange, opacity=0.5, draw=black] (0,0) circle (2mm);
+ \node at ($(0,0)+(15:4mm)$) {\(\theta\)};
+ \end{scope}
+ \begin{scope}[very thick, every node/.style={sloped,allow upside down}]
+ \draw [gray, dashed, thick] (1,0) -- (1,0.5) node [pos=0.5] {\midarrow} node[black, pos=0.5, right, rotate=-90]{\(\boldsymbol{w}\)};
+ \end{scope}
+\draw (0,0) coordinate (O)
+ (1,0) coordinate (A)
+ (1,0.5) coordinate (B)
+ pic [draw,red,angle radius=2mm] {right angle = O--A--B};
+\end{tikzpicture}
+
+\subsubsection*{\(\parallel\boldsymbol{b}\) (vector projection/resolute)}
+\begin{align*}
+ \boldsymbol{u}&={{\boldsymbol{a}\cdot\boldsymbol{b}}\over |\boldsymbol{b}|^2}\boldsymbol{b}\\
+ &=\left({\boldsymbol{a}\cdot{\boldsymbol{b} \over |\boldsymbol{b}|}}\right)\left({\boldsymbol{b} \over |\boldsymbol{b}|}\right)\\
+ &=(\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}
+\end{align*}
+
+\subsubsection*{\(\perp\boldsymbol{b}\) (perpendicular projection)}
+\[\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u}\]
-Vector resolute of \(\boldsymbol{a}\) in direction of \(\boldsymbol{b}\)
-is magnitude of \(\boldsymbol{a}\) in direction of \(\boldsymbol{b}\):
+\subsubsection*{\(|\boldsymbol{u}|\) (scalar resolute)}
+\begin{align*}
+ r_s &= |\boldsymbol{u}|\\
+ &= \boldsymbol{a} \cdot \hat{\boldsymbol{b}}\\
+ &=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}
+\end{align*}
-\[\boldsymbol{u}={{\boldsymbol{a}\cdot\boldsymbol{b}}\over |\boldsymbol{b}|^2}\boldsymbol{b}=\left({\boldsymbol{a}\cdot{\boldsymbol{b} \over |\boldsymbol{b}|}}\right)\left({\boldsymbol{b} \over |\boldsymbol{b}|}\right)=(\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}\]
+\subsubsection*{Rectangular (\(\parallel,\perp\)) components}
-\subsection{\texorpdfstring{Scalar resolute of \(\boldsymbol{a}\) on
-\(\boldsymbol{b}\)}{Scalar resolute of \textbackslash{}boldsymbol\{a\} on \textbackslash{}boldsymbol\{b\}}}\label{scalar-resolute-of-boldsymbola-on-boldsymbolb}
+\[\boldsymbol{a}=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}+\left(\boldsymbol{a}-\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}\right)\]
-\[r_s = |\boldsymbol{u}| = \boldsymbol{a} \cdot \hat{\boldsymbol{b}}\]
-\subsection{\texorpdfstring{Vector resolute of
-\(\boldsymbol{a} \perp \boldsymbol{b}\)}{Vector resolute of \textbackslash{}boldsymbol\{a\} \textbackslash{}perp \textbackslash{}boldsymbol\{b\}}}\label{vector-resolute-of-boldsymbola-perp-boldsymbolb}
+\subsection*{Vector proofs}
-\[\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u} \> \text{ where } \boldsymbol{u} \text{ is projection } \boldsymbol{a} \text{ on } \boldsymbol{b}\]
+\textbf{Concurrent:} intersection of \(\ge\) 3 lines
-\subsection{Vector proofs}\label{vector-proofs}
+\begin{tikzpicture}
+ \draw [blue] (0,0) -- (1,1);
+ \draw [red] (1,0) -- (0,1);
+ \draw [brown] (0.4,0) -- (0.6,1);
+ \filldraw (0.5,0.5) circle (2pt);
+\end{tikzpicture}
-\subsubsection{Concurrent lines}\label{concurrent-lines}
+\subsubsection*{Collinear points}
-\(\ge\) 3 lines intersect at a single point
+\(\ge\) 3 points lie on the same line
-\subsubsection{Collinear points}\label{collinear-points}
+\begin{tikzpicture}
+ \draw [purple] (0,0) -- (4,1);
+ \filldraw (2,0.5) circle (2pt) node [above] {\(C\)};
+ \filldraw (1,0.25) circle (2pt) node [above] {\(A\)};
+ \filldraw (3,0.75) circle (2pt) node [above] {\(B\)};
+ \coordinate (O) at (2.8,-0.2);
+ \node at (O) [below] {\(O\)};
+ \begin{scope}[->, orange, thick]
+ \draw (O) -- (2,0.5) node [pos=0.5, above, font=\footnotesize, black] {\(\boldsymbol{c}\)};
+ \draw (O) -- (1,0.25) node [pos=0.5, below, font=\footnotesize, black] {\(\boldsymbol{a}\)};
+ \draw (O) -- (3,0.75) node [pos=0.5, right, font=\footnotesize, black] {\(\boldsymbol{b}\)};
+ \end{scope}
+\end{tikzpicture}
-\(\ge\) 3 points lie on the same line\\
-\(\implies \vec{OC} = \lambda \vec{OA} + \mu \vec{OB}\) where
-\(\lambda + \mu = 1\). If \(C\) is between \(\vec{AB}\), then
-\(0 < \mu < 1\)\\
-Points \(A, B, C\) are collinear iff
-\(\vec{AC}=m\vec{AB} \text{ where } m \ne 0\)
+\begin{align*}
+ \text{e.g. Prove that}\\
+ \overrightharp{AC}=m\overrightharp{AB} \iff \boldsymbol{c}&=(1-m)\boldsymbol{a}+m\boldsymbol{b}\\
+ \implies \boldsymbol{c} &= \overrightharp{OA} + \overrightharp{AC}\\
+ &= \overrightharp{OA} + m\overrightharp{AB}\\
+ &=\boldsymbol{a}+m(\boldsymbol{b}-\boldsymbol{a})\\
+ &=\boldsymbol{a}+m\boldsymbol{b}-m\boldsymbol{a}\\
+ &=(1-m)\boldsymbol{a}+m{b}
+\end{align*}
-\subsubsection{Useful vector properties}\label{useful-vector-properties}
+\begin{align*}
+ \text{Also, } \implies \overrightharp{OC} &= \lambda \vec{OA} + \mu \overrightharp{OB} \\
+ \text{where } \lambda + \mu &= 1\\
+ \text{If } C \text{ lies along } \overrightharp{AB}, & \implies 0 < \mu < 1
+\end{align*}
+
+
+ \subsubsection*{Useful vector properties}
\begin{itemize}
-\tightlist
\item
- If \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are parallel, then
- \(\boldsymbol{b}=k\boldsymbol{a}\) for some
+ \(\boldsymbol{a} \parallel \boldsymbol{b} \implies \boldsymbol{b}=k\boldsymbol{a}\) for some
\(k \in \mathbb{R} \setminus \{0\}\)
\item
If \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are parallel with at
\(\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2\)
\end{itemize}
-\subsection{Linear dependence}\label{linear-dependence}
+\subsection*{Linear dependence}
Vectors \(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\) are linearly
dependent if they are non-parallel and:
Vector \(\boldsymbol{w}\) is a linear combination of vectors
\(\boldsymbol{v_1}, \boldsymbol{v_2}, \boldsymbol{v_3}\)
-\subsection{Three-dimensional vectors}\label{three-dimensional-vectors}
+\subsection*{Three-dimensional vectors}
Right-hand rule for axes: \(z\) is up or out of page.
-i\includegraphics{graphics/vectors-3d.png}
+\tdplotsetmaincoords{60}{120}
+\begin{center}\begin{tikzpicture} [scale=3, tdplot_main_coords, axis/.style={->,thick},
+vector/.style={-stealth,red,very thick},
+vector guide/.style={dashed,gray,thick}]
+
+%standard tikz coordinate definition using x, y, z coords
+\coordinate (O) at (0,0,0);
+
+%tikz-3dplot coordinate definition using x, y, z coords
+
+\pgfmathsetmacro{\ax}{1}
+\pgfmathsetmacro{\ay}{1}
+\pgfmathsetmacro{\az}{1}
+
+\coordinate (P) at (\ax,\ay,\az);
-\subsection{Parametric vectors}\label{parametric-vectors}
+%draw axes
+\draw[axis] (0,0,0) -- (1,0,0) node[anchor=north east]{$x$};
+\draw[axis] (0,0,0) -- (0,1,0) node[anchor=north west]{$y$};
+\draw[axis] (0,0,0) -- (0,0,1) node[anchor=south]{$z$};
+
+%draw a vector from O to P
+\draw[vector] (O) -- (P);
+
+%draw guide lines to components
+\draw[vector guide] (O) -- (\ax,\ay,0);
+\draw[vector guide] (\ax,\ay,0) -- (P);
+\draw[vector guide] (P) -- (0,0,\az);
+\draw[vector guide] (\ax,\ay,0) -- (0,\ay,0);
+\draw[vector guide] (\ax,\ay,0) -- (0,\ay,0);
+\draw[vector guide] (\ax,\ay,0) -- (\ax,0,0);
+\node[tdplot_main_coords,above right]
+at (\ax,\ay,\az){(\ax, \ay, \az)};
+\end{tikzpicture}\end{center}
+
+\subsection*{Parametric vectors}
Parametric equation of line through point \((x_0, y_0, z_0)\) and
parallel to \(a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}\) is:
\begin{equation}\begin{cases}x = x_o + a \cdot t \\ y = y_0 + b \cdot t \\ z = z_0 + c \cdot t\end{cases}\end{equation}
-\section{Circular functions}\label{circular-functions}
+\section{Circular functions}
-Period of \(a\sin(bx)\) is \({2\pi} \over b\)
+Period of \(a\sin(bx)\) is \(\frac{{2\pi}{b}\)
-Period of \(a\tan(nx)\) is \(\pi \over n\)\\
-Asymptotes at \(x={2k+1)\pi \over 2n} \> \vert \> k \in \mathbb{Z}\)
+Period of \(a\tan(nx)\) is \(\frac{\pi}{n}\)\\
+Asymptotes at \(x=\frac{2k+1)\pi}{2n} \> \vert \> k \in \mathbb{Z}\)
-\subsection{Reciprocal functions}\label{reciprocal-functions}
+\subsection*{Reciprocal functions}
-\subsubsection{Cosecant}\label{cosecant}
+\subsubsection*{Cosecant}
\begin{figure}
\centering
\caption{}
\end{figure}
-\[\operatorname{cosec} \theta = {1 \over \sin \theta} \> \vert \> \sin \theta \ne 0\]
+\[\operatorname{cosec} \theta = \frac{1}{\sin \theta} \> \vert \> \sin \theta \ne 0\]
\begin{itemize}
\tightlist
\textbf{Range} \(= \mathbb{R} \setminus (-1, 1)\)
\item
\textbf{Turning points} at
- \(\theta = {{(2n + 1)\pi} \over 2} \> \vert \> n \in \mathbb{Z}\)
+ \(\theta = {\frac{(2n + 1)\pi}{2} \> \vert \> n \in \mathbb{Z}\)
\item
\textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
\end{itemize}
-\subsubsection{Secant}\label{secant}
+\subsubsection*{Secant}
\begin{figure}
\centering
\caption{}
\end{figure}
-\[\operatorname{sec} \theta = {1 \over \cos \theta} \> \vert \> \cos \theta \ne 0\]
+\[\operatorname{sec} \theta = \frac{1}{\cos \theta} \> \vert \> \cos \theta \ne 0\]
\begin{itemize}
\tightlist
\(\theta = {{(2n + 1) \pi} \over 2} \> \vert \> n \in \mathbb{Z}\)
\end{itemize}
-\subsubsection{Cotangent}\label{cotangent}
+\subsubsection*{Cotangent}
\begin{figure}
\centering
\textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
\end{itemize}
-\subsubsection{Symmetry properties}\label{symmetry-properties}
+\subsubsection*{Symmetry properties}
\begin{equation}\begin{split}
\operatorname{sec} (\pi \pm x) & = -\operatorname{sec} x \\
\operatorname{cot} (-x) & = - \operatorname{cot} x
\end{split}\end{equation}
-\subsubsection{Complementary properties}\label{complementary-properties}
+\subsubsection*{Complementary properties}
\begin{equation}\begin{split}
\operatorname{sec} \left({\pi \over 2} - x\right) & = \operatorname{cosec} x \\
\tan \left({\pi \over 2} - x\right) & = \operatorname{cot} x
\end{split}\end{equation}
-\subsubsection{Pythagorean identities}\label{pythagorean-identities}
+\subsubsection*{Pythagorean identities}
\begin{equation}\begin{split}
1 + \operatorname{cot}^2 x & = \operatorname{cosec}^2 x, \quad \text{where } \sin x \ne 0 \\
1 + \tan^2 x & = \operatorname{sec}^2 x, \quad \text{where } \cos x \ne 0
\end{split}\end{equation}
-\subsection{Compound angle formulas}\label{compound-angle-formulas}
+\subsection*{Compound angle formulas}
\[\cos(x \pm y) = \cos x + \cos y \mp \sin x \sin y\]
\[\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\]
\[\tan(x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x \tan y}}\]
-\subsection{Double angle formulas}\label{double-angle-formulas}
+\subsection*{Double angle formulas}
\begin{equation}\begin{split}
\cos 2x &= \cos^2 x - \sin^2 x \\
\[\tan 2x = {{2 \tan x} \over {1 - \tan^2 x}}\]
-\subsection{Inverse circular
-functions}\label{inverse-circular-functions}
+\subsection*{Inverse circular functions}
Inverse functions: \(f(f^{-1}(x)) = x, \quad f(f^{-1}(x)) = x\)\\
Must be 1:1 to find inverse (reflection in \(y=x\)
Domain is restricted to make functions 1:1.
-\subsubsection{\texorpdfstring{\(\arcsin\)}{\textbackslash{}arcsin}}\label{arcsin}
+\subsubsection*{\(\arcsin\)}
\[\sin^{-1}: [-1, 1] \rightarrow \mathbb{R}, \quad \sin^{-1} x = y, \quad \text{where } \sin y = x \text{ and } y \in [{-\pi \over 2}, {\pi \over 2}]\]
-\subsubsection{\texorpdfstring{\(\arcos\)}{\textbackslash{}arcos}}\label{arcos}
+\subsubsection*{\(\arccos\)}
\[\cos^{-1} \rightarrow \mathbb{R}, \quad \cos^{-1} x = y, \quad \text{where } \cos y = x \text{ and } y \in [0, \pi]\]
-\subsubsection{\texorpdfstring{\(\arctan\)}{\textbackslash{}arctan}}\label{arctan}
+\subsubsection*{\(\arctan\)}
\[\tan^{-1}: \mathbb{R} \rightarrow \mathbb{R}, \quad \tan^{-1} x = y, \quad \text{where } \tan y = x \text{ and } y \in \left(-{\pi \over 2}, {\pi \over 2}\right)\]
-\# Differential calculus
-\subsection{Limits}\label{limits}
+
+\section{Differential calculus}
+
+\subsection*{Limits}
\[\lim_{x \rightarrow a}f(x)\]
\(\lim_{x \to a} f(x)\) - limit of a point
\begin{itemize}
-\tightlist
\item
Limit exists if \(L^-=L^+\)
\item
If limit exists, point does not.
-\end{itemize}
-
+\item
+ For solving \(x\rightarrow\infty\), factorise so that all \(x\) terms are in denominators\\
+ e.g. \[\lim_{x \rightarrow \infty}{{2x+3} \over {x-2}}={{2+{3 \over x}} \over {1-{2 \over x}}}={2 \over 1} = 2\]
+ \item
Limits can be solved using normal techniques (if div 0, factorise)
+\end{itemize}
-\subsection{Limit theorems}\label{limit-theorems}
\begin{enumerate}
-\def\labelenumi{\arabic{enumi}.}
-\tightlist
\item
For constant function \(f(x)=k\), \(\lim_{x \rightarrow a} f(x) = k\)
\item
\(\lim_{x \rightarrow a} (f(x) \pm g(x)) = F \pm G\)
\item
\(\lim_{x \rightarrow a} (f(x) \times g(x)) = F \times G\)
-\item
+ \item
+\(\therefore \lim_{x \rightarrow a} c \times f(x)=cF\) where \(c=\) constant
+\ite
\({\lim_{x \rightarrow a} {f(x) \over g(x)}} = {F \over G}, G \ne 0\)
-\end{enumerate}
-
-Corollary: \(\lim_{x \rightarrow a} c \times f(x)=cF\) where \(c=\)
-constant
-
-\subsection{\texorpdfstring{Solving limits for
-\(x\rightarrow\infty\)}{Solving limits for x\textbackslash{}rightarrow\textbackslash{}infty}}\label{solving-limits-for-xrightarrowinfty}
-
-Factorise so that all values of \(x\) are in denominators.
-
-e.g.
-
-\[\lim_{x \rightarrow \infty}{{2x+3} \over {x-2}}={{2+{3 \over x}} \over {1-{2 \over x}}}={2 \over 1} = 2\]
-
-\subsection{Continuous functions}\label{continuous-functions}
-
+\item
A function is continuous if \(L^-=L^+=f(x)\) for all values of \(x\).
+\end{enumerate}
-\subsection{Gradients of secants and
-tangents}\label{gradients-of-secants-and-tangents}
-
-Secant (chord) - line joining two points on curve
+\subsection{Gradients of secants and tangents}
-Tangent - line that intersects curve at one point
+\textbf{Secant (chord)} - line joining two points on curve\\
+\textbf{Tangent} - line that intersects curve at one point
-given \(P(x,y) \quad Q(x+\delta x, y + \delta y)\): gradient of chord
-joining \(P\) and \(Q\) is
-\({m_{PQ}}={\operatorname{rise} \over \operatorname{run}} = {\delta y \over \delta x}\)
+\(m\left(\overrightharp{PQ}\right){m_{PQ}}={\operatorname{rise} \over \operatorname{run}} = {\delta y \over \delta x} \text{ for } P(x,y),\quad Q(x+\delta x, y+ \delta y)\)
As \(Q \rightarrow P, \delta x \rightarrow 0\). Chord becomes tangent
(two infinitesimal points are equal).
-Can also be used with functions, where \(h=\delta x\).
-
-\subsection{First principles
-derivative}\label{first-principles-derivative}
-
-\[f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={dy \over dx}\]
-
-\[m_{\tan}=\lim_{h \rightarrow 0}f^\prime(x)\]
-
-\[m_{\vec{PQ}}=f^\prime(x)\]
-
-first principles derivative:
-\[{m_{\text{tangent at }P} =\lim_{h \rightarrow 0}}{{f(x+h)-f(x)}\over h}\]
-
-\subsection{Gradient at a point}\label{gradient-at-a-point}
-
-Given point \(P(a, b)\) and function \(f(x)\), the gradient is
-\(f^\prime(a)\)
-
-\subsection{\texorpdfstring{Derivatives of
-\(x^n\)}{Derivatives of x\^{}n}}\label{derivatives-of-xn}
+\subsection{First principles derivative}
-\[{d(ax^n) \over dx}=anx^{n-1}\]
+\[f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={\frac{dy}{dx}}\]
-If \(x=\) constant, derivative is \(0\)
-
-If \(y=ax^n\), derivative is \(a\times nx^{n-1}\)
-
-If
-\(f(x)={1 \over x}=x^{-1}, \quad f^\prime(x)=-1x^{-2}={-1 \over x^2}\)
-
-If
-\(f(x)=^5\sqrt{x}=x^{1 \over 5}, \quad f^\prime(x)={1 \over 5}x^{-4/5}={1 \over 5 \times ^5\sqrt{x^4}}\)
-
-If \(f(x)=(x-b)^2, \quad f^\prime(x)=2(x-b)\)
-
-\[f^\prime(x)=\lim_{h \rightarrow 0}{{f(x+h)-f(x)} \over h}\]
-
-\subsection{\texorpdfstring{Derivatives of
-\(u \pm v\)}{Derivatives of u \textbackslash{}pm v}}\label{derivatives-of-u-pm-v}
-
-\[{dy \over dx}={du \over dx} \pm {dv \over dx}\] where \(u\) and \(v\)
-are functions of \(x\)
-
-\subsection{Euler's number as a limit}\label{eulers-number-as-a-limit}
-
-\[\lim_{h \rightarrow 0} {{e^h-1} \over h}=1\]
-
-\subsection{\texorpdfstring{Chain rule for
-\((f\circ g)\)}{Chain rule for (f\textbackslash{}circ g)}}\label{chain-rule-for-fcirc-g}
-
-If \(f(x) = h(g(x)) = (h \circ g)(x)\):
-
-\[f^\prime(x) = h^\prime(g(x)) \cdot g^\prime(x)\]
-
-If \(y=h(u)\) and \(u=g(x)\):
-
-\[{dy \over dx} = {dy \over du} \cdot {du \over dx}\]
-\[{d((ax+b)^n) \over dx} = {d(ax+b) \over dx} \cdot n \cdot (ax+b)^{n-1}\]
-
-Used with only one expression.
-
-e.g. \(y=(x^2+5)^7\) - Cannot reasonably expand\\
-Let \(u-x^2+5\) (inner expression)\\
-\({du \over dx} = 2x\)\\
-\(y=u^7\)\\
-\({dy \over du} = 7u^6\)
-
-\subsection{\texorpdfstring{Product rule for
-\(y=uv\)}{Product rule for y=uv}}\label{product-rule-for-yuv}
-
-\[{dy \over dx} = u{dv \over dx} + v{du \over dx}\]
-
-\subsection{\texorpdfstring{Quotient rule for
-\(y={u \over v}\)}{Quotient rule for y=\{u \textbackslash{}over v\}}}\label{quotient-rule-for-yu-over-v}
-
-\[{dy \over dx} = {{v{du \over dx} - u{dv \over dx}} \over v^2}\]
-
-\[f^\prime(x)={{v(x)u^\prime(x)-u(x)v^\prime(x)} \over [v(x)]^2}\]
-
-\subsection{Logarithms}\label{logarithms}
-
-\[\log_b (x) = n \quad \operatorname{where} \hspace{0.5em} b^n=x\]
-
-Wikipedia:
-
-\begin{quote}
-the logarithm of a given number \(x\) is the exponent to which another
-fixed number, the base \(b\), must be raised, to produce that number
-\(x\)
-\end{quote}
-
-\subsubsection{Logarithmic identities}\label{logarithmic-identities}
+\subsubsection*{Logarithmic identities}
\(\log_b (xy)=\log_b x + \log_b y\)\\
\(\log_b x^n = n \log_b x\)\\
\(\log_b y^{x^n} = x^n \log_b y\)
-\subsubsection{Index identities}\label{index-identities}
+\subsubsection*{Index identities}}
\(b^{m+n}=b^m \cdot b^n\)\\
\((b^m)^n=b^{m \cdot n}\)\\
\[\operatorname{if} y=e^x, \quad \operatorname{then} x=\log_e y\]
\[\ln x = \log_e x\]
-\subsubsection{Differentiating
-logarithms}\label{differentiating-logarithms}
-
-\[{d(\log_e x)\over dx} = x^{-1} = {1 \over x}\]
-
-\subsection{Derivative rules}\label{derivative-rules}
+\subsection*{Derivative rules}
\begin{longtable}[]{@{}ll@{}}
\toprule
\bottomrule
\end{longtable}
-\subsection{Reciprocal derivatives}\label{reciprocal-derivatives}
+\subsection*{Reciprocal derivatives}
+
+\[\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}\]
-\[{1 \over {dy \over dx}} = {dx \over dy}\]
+\subsection*{Differentiating \(x=f(y)\)}
-\subsection{\texorpdfstring{Differentiating
-\(x=f(y)\)}{Differentiating x=f(y)}}\label{differentiating-xfy}
+Find \(\frac{dx}{dy}\). Then:
-Find \(dx \over dy\). Then
-\({dx \over dy} = {1 \over {dy \over dx}} \implies {dy \over dx} = {1 \over {dx \over dy}}\).
+\begin{align*}
+ {\frac{dx}{dy}} =& {1 \over {\frac{dy}{dx}}} \\
+ \implies {\frac{dy}{dx}} &= {1 \over {\frac{dx}{dy}}}\).
-\[{dy \over dx} = {1 \over {dx \over dy}}\]
+\[{\frac{dy}{dx}} = {1 \over {\frac{dx}{dy}}}\]
-\subsection{Second derivative}\label{second-derivative}
+\subsection*{Second derivative}}
\[f(x) \longrightarrow f^\prime (x) \longrightarrow f^{\prime\prime}(x)\]
-\[\therefore y \longrightarrow {dy \over dx} \longrightarrow {d({dy \over dx}) \over dx} \longrightarrow {d^2 y \over dx^2}\]
+\[\therefore y \longrightarrow {\frac{dy}{dx}} \longrightarrow {d({\frac{dy}{dx}}) \over dx} \longrightarrow {d^2 y \over dx^2}\]
Order of polynomial \(n\)th derivative decrements each time the
derivative is taken
-\subsubsection{Points of Inflection}\label{points-of-inflection}
+\subsubsection*{Points of Inflection}
\emph{Stationary point} - point of zero gradient (i.e.
\(f^\prime(x)=0\))\\
\caption{}
\end{figure}
-\subsection{Implicit Differentiation}\label{implicit-differentiation}
+\subsection*{Implicit Differentiation}
\textbf{On CAS:} Action \(\rightarrow\) Calculation \(\rightarrow\)
\texttt{impDiff(y\^{}2+ax=5,\ x,\ y)}. Returns \(y^\prime= \dots\).
If \(p\) and \(q\) are expressions in \(x\) and \(y\) such that \(p=q\),
for all \(x\) nd \(y\), then:
-\[{dp \over dx} = {dq \over dx} \quad \text{and} \quad {dp \over dy} = {dq \over dy}\]
+\[{\frac{dp}{dx}} = {\frac{dq}{dx}} \quad \text{and} \quad {\frac{dp}{dy}} = {\frac{dq}{dy}}\]
-\subsection{Integration}\label{integration}
+\subsection*{Integration}
\[\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)\]
\(+c\) should be shown on each step without \(\int\)
\end{itemize}
-\subsubsection{Integral laws}\label{integral-laws}
+\subsubsection*{Integral laws}
\(\int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx\)\\
\(\int k f(x) dx = k \int f(x) dx\)
Note \(\sin^{-1} {x \over a} + \cos^{-1} {x \over a}\) is constant for
all \(x \in (-a, a)\).
-\subsubsection{Definite integrals}\label{definite-integrals}
+\subsubsection*{Definite integrals}}
\[\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)\]
\(F(x)\) may be any integral, i.e. \(c\) is inconsequential
\end{itemize}
-\paragraph{Properties}\label{properties-2}
+\paragraph{Properties}\label{properties}
\[\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx\]
\[\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx\]
-\subsubsection{Integration by
-substitution}\label{integration-by-substitution}
+\subsubsection{Integration by substitution}
-\[\int f(u) {du \over dx} \cdot dx = \int f(u) \cdot du\]
+\[\int f(u) {\frac{du}{dx}} \cdot dx = \int f(u) \cdot du\]
Note \(f(u)\) must be one-to-one \(\implies\) one \(x\) value for each
\(y\) value
e.g.~for \(y=\int(2x+1)\sqrt{x+4} \cdot dx\):\\
let \(u=x+4\)\\
-\(\implies {du \over dx} = 1\)\\
+\(\implies {\frac{du}{dx}} = 1\)\\
\(\implies x = u - 4\)\\
then \(y=\int (2(u-4)+1)u^{1 \over 2} \cdot du\)\\
Solve as a normal integral
-\paragraph{Definite integrals by
-substitution}\label{definite-integrals-by-substitution}
+\subsubsection*{Definite integrals by substitution}
-For \(\int^b_a f(x) {du \over dx} \cdot dx\), evaluate new \(a\) and
+For \(\int^b_a f(x) {\frac{du}{dx}} \cdot dx\), evaluate new \(a\) and
\(b\) for \(f(u) \cdot du\).
-\subsubsection{Trigonometric
-integration}\label{trigonometric-integration}
+\subsubsection{Trigonometric integration}
\[\sin^m x \cos^n x \cdot dx\]
\(\sin 2x = 2 \sin x \cos x\)
\end{itemize}
-\subsection{Partial fractions}\label{partial-fractions}
+\subsection*{Partial fractions}
On CAS: Action \(\rightarrow\) Transformation \(\rightarrow\)
\texttt{expand/combine}\\
or Interactive \(\rightarrow\) Transformation \(\rightarrow\)
\texttt{expand} \(\rightarrow\) Partial
-\subsection{Graphing integrals on CAS}\label{graphing-integrals-on-cas}
+\subsection*{Graphing integrals on CAS}
In main: Interactive \(\rightarrow\) Calculation \(\rightarrow\)
\(\int\) (\(\rightarrow\) Definite)\\
Restrictions: \texttt{Define\ f(x)=...} \(\rightarrow\)
\texttt{f(x)\textbar{}x\textgreater{}1} (e.g.)
-\subsection{Applications of
-antidifferentiation}\label{applications-of-antidifferentiation}
+\subsection{Applications of antidifferentiation}
\begin{itemize}
\tightlist
\end{itemize}
To find stationary points of a function, substitute \(x\) value of given
-point into derivative. Solve for \({dy \over dx}=0\). Integrate to find
+point into derivative. Solve for \({\frac{dy}{dx}}=0\). Integrate to find
original function.
-\subsection{Solids of revolution}\label{solids-of-revolution}
+\subsection*{Solids of revolution}}
Approximate as sum of infinitesimally-thick cylinders
-\subsubsection{\texorpdfstring{Rotation about
-\(x\)-axis}{Rotation about x-axis}}\label{rotation-about-x-axis}
+\subsubsection{Rotation about \(x\)-axis}
\begin{align*}
V &= \int^{x=b}_{x-a} \pi y^2 \> dx \\
&= \pi \int^b_a (f(x))^2 \> dx
\end{align*}
-\subsubsection{\texorpdfstring{Rotation about
-\(y\)-axis}{Rotation about y-axis}}\label{rotation-about-y-axis}
+\subsubsection{Rotation about \(y\)-axis}
\begin{align*}
V &= \int^{y=b}_{y=a} \pi x^2 \> dy \\
&= \pi \int^b_a (f(y))^2 \> dy
\end{align*}
-\subsubsection{\texorpdfstring{Regions not bound by
-\(y=0\)}{Regions not bound by y=0}}\label{regions-not-bound-by-y0}
+\subsubsection{Regions not bound by\(y=0\)}
\[V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx\]\\
where \(f(x) > g(x)\)
-\subsection{Length of a curve}\label{length-of-a-curve}
+\subsection*{Length of a curve}
-\[L = \int^b_a \sqrt{1 + ({dy \over dx})^2} \> dx \quad \text{(Cartesian)}\]
+\[L = \int^b_a \sqrt{1 + ({\frac{dy}{dx}})^2} \> dx \quad \text{(Cartesian)}\]
-\[L = \int^b_a \sqrt{{dx \over dt} + ({dy \over dt})^2} \> dt \quad \text{(parametric)}\]
+\[L = \int^b_a \sqrt{{\frac{dx}{dt}} + ({\frac{dy}{dt}})^2} \> dt \quad \text{(parametric)}\]
Evaluate on CAS. Or use Interactive \(\rightarrow\) Calculation
\(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen}.
-\subsection{Rates}\label{rates}
+\subsection*{Rates}
-\subsubsection{Related rates}\label{related-rates}
+\subsubsection*{Related rates}
-\[{da \over db} \quad \text{(change in } a \text{ with respect to } b)\]
+\[{\frac{da}{db}} \quad \text{(change in } a \text{ with respect to } b)\]
-\subsubsection{Gradient at a point on parametric
-curve}\label{gradient-at-a-point-on-parametric-curve}
+\subsubsection{Gradient at a point on parametric curve}
-\[{dy \over dx} = {{dy \over dt} \div {dx \over dt}} \> \vert \> {dx \over dt} \ne 0\]
+\[{\frac{dy}{dx}} = {{\frac{dy}{dt}} \div {\frac{dx}{dt}}} \> \vert \> {\frac{dx}{dt}} \ne 0\]
-\[{d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \div {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}\]
+\[\frac{d^2}{dx^2} = \frac{d(y^\prime)}{dx} = {\frac{dy^\prime}{dt} \div {\frac{dx}{dt}}} \> \vert \> y^\prime = {\frac{dy}{dx}}\]
-\subsection{Rational functions}\label{rational-functions}
+\subsection*{Rational functions}
-\[f(x) = {P(x) \over Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]
+\[f(x) = \frac{P(x)}{Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]
-\subsubsection{Addition of ordinates}\label{addition-of-ordinates}
+\subsubsection*{Addition of ordinates}
\begin{itemize}
\tightlist
other ordinate
\end{itemize}
-\subsection{Fundamental theorem of
-calculus}\label{fundamental-theorem-of-calculus}
+\subsection{Fundamental theorem of calculus}
If \(f\) is continuous on \([a, b]\), then
where \(F\) is any antiderivative of \(f\)
-\subsection{Differential equations}\label{differential-equations}
+\subsection*{Differential equations}}
One or more derivatives
\textbf{Order} - highest power inside derivative\\
\textbf{Degree} - highest power of highest derivative\\
-e.g. \({\left(dy^2 \over d^2 x\right)}^3\): order 2, degree 3
+e.g. \({\left(\frac{dy^2}{d^2} x\right)}^3\): order 2, degree 3
-\subsubsection{Verifying solutions}\label{verifying-solutions}
+\subsubsection*{Verifying solutions}
Start with \(y=\dots\), and differentiate. Substitute into original
equation.
\subsubsection{Function of the dependent
variable}\label{function-of-the-dependent-variable}
-If \({dy \over dx}=g(y)\), then
-\({dx \over dy} = 1 \div {dy \over dx} = {1 \over g(y)}\). Integrate
+If \({\frac{dy}{dx}}=g(y)\), then
+\(\frac{{dx}{dy} = 1 \div {\frac{dy}{dx}} = \frac{1}{g(y)}\). Integrate
both sides to solve equation. Only add \(c\) on one side. Express
\(e^c\) as \(A\).
-\subsubsection{Mixing problems}\label{mixing-problems}
+\subsubsection*{Mixing problems}
-\[\left({dm \over dt}\right)_\Sigma = \left({dm \over dt}\right)_{\text{in}} - \left({dm \over dt}\right)_{\text{out}}\]
+\[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left({\frac{dm}{dt}\)_{\text{out}}\]
-\subsubsection{Separation of variables}\label{separation-of-variables}
+\subsubsection*{Separation of variables}
-If \({dy \over dx}=f(x)g(y)\), then:
+If \({\frac{dy}{dx}}=f(x)g(y)\), then:
-\[\int f(x) \> dx = \int {1 \over g(y)} \> dy\]
+\[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\]
-\subsubsection{Using definite integrals to solve
-DEs}\label{using-definite-integrals-to-solve-des}
+\subsubsection{Using definite integrals to solve DEs}
-Used for situations where solutions to \({dy \over dx} = f(x)\) is not
+Used for situations where solutions to \({\frac{dy}{dx}} = f(x)\) is not
required.
In some cases, it may not be possible to obtain an exact solution.
Approximate solutions can be found by numerically evaluating a definite
integral.
-\subsubsection{Using Euler's method to solve a differential
-equation}\label{using-eulers-method-to-solve-a-differential-equation}
+\subsubsection{Using Euler's method to solve a differential equation}
-\[{{f(x+h) - f(x)} \over h } \approx f^\prime (x) \quad \text{for small } h\]
+\[\frac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
\[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
+ \end{multicols}
\end{document}