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\fancyhead[LO,LE]{Year 12 Specialist}
\fancyhead[CO,CE]{Andrew Lorimer}
-\usepackage{mathtools}
-\usepackage{xcolor} % used only to show the phantomed stuff
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+
+\newtcolorbox{cas}{colframe=cas!75!black, fonttitle=\sffamily\bfseries, title=On CAS, left*=3mm}
+\newtcolorbox{theorembox}[1]{colback=green!10!white, colframe=blue!20!white, coltitle=black, fontupper=\sffamily, fonttitle=\sffamily, #1}
+\newtcolorbox{warning}{colback=white!90!black, leftrule=3mm, colframe=important, coltext=darkgray, fontupper=\sffamily\bfseries}
\begin{document}
+\title{\vspace{-22mm}Year 12 Specialist\vspace{-4mm}}
+\author{Andrew Lorimer}
+\date{}
+\maketitle
+\vspace{-9mm}
\begin{multicols}{2}
\section{Complex numbers}
- \[\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}\]
+ \[\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}\]
+ \begin{align*}
+ \text{Cartesian form: } & a+bi\\
+ \text{Polar form: } & r\operatorname{cis}\theta
+ \end{align*}
+
+ \subsection*{Operations}
+
+ \begin{tabularx}{\columnwidth}{|r|X|X|}
+ \hline
+ \rowcolor{cas}
+ & \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
+ \(z_1 \div z_2\) & \((z_1 \overline{z_2}) \div |z_2|^2\) & \(\left(\frac{r_1}{r_2}\right) \operatorname{cis}(\theta_1 - \theta_2)\) \\
+ \hline
+ \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}
+ \vspace{-7mm} \hfill \colorbox{cas}{\texttt{conjg(a+bi)}}
\begin{align*}
- \text{Cartesian form: } & a+bi\\
- \text{Polar form: } & r\operatorname{cis}\theta
+ \overline{z} &= a \mp bi\\
+ &= r \operatorname{cis}(-\theta)
\end{align*}
- \subsection*{Operations}
-
-\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*}
-
- \noindent \colorbox{cas}{On CAS: \texttt{conjg(a+bi)}}
+ \subsubsection*{Properties}
- \subsubsection*{Properties}
-
- \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*}
+ \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} \> \forall \> k \in \mathbb{R}\\
+ z\overline{z} &= (a+bi)(a-bi)\\
+ &= a^2 + b^2\\
+ &= |z|^2
+ \end{align*}
\subsection*{Modulus}
- \[|z|=|\vec{Oz}|=\sqrt{a^2 + b^2}\]
+ \[|z|=|\vec{Oz}|=\sqrt{a^2 + b^2}\]
- \subsubsection*{Properties}
+ \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*}
+ \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*}
+ \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*}
+ \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}\\
+ & \text{then rationalise denominator}
+ \end{align*}
\subsection*{Polar form}
- \begin{align*}
- z&=r\operatorname{cis}\theta\\
- &=r(\cos \theta + i \sin \theta)
- \end{align*}
+ \[ r \operatorname{cis} \theta = r\left( \cos \theta + i \sin \theta \right) \]
- \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}
+ \begin{itemize}
+ \item{\(r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}\)}
+ \item{\(\theta = \operatorname{arg}(z)\) \hfill \colorbox{cas}{\texttt{arg(a+bi)}}}
+ \item{\(\operatorname{Arg}(z) \in (-\pi,\pi)\) \quad \bf{(principal argument)}}
+ \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}
+
+ \begin{cas}
+ \-\hspace{1em}\verb|compToTrig(a+bi)| \(\iff\) \verb|cExpand{r·cisX}|
+ \end{cas}
\subsection*{de Moivres' theorem}
- \[(r \operatorname{cis} \theta)^n = r^n \operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}\]
+ \begin{theorembox}{}
+ \[(r \operatorname{cis} \theta)^n = r^n \operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}\]
+ \end{theorembox}
\subsection*{Complex polynomials}
-
- Include \(\pm\) for all solutions, incl. imaginary
- \begin{tabularx}{\columnwidth}{ R{0.55} X }
- \hline
- Sum of squares & \(\begin{aligned}
+ 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}
+ &= (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)\\
+ \hline
+ \end{tabularx}
- \subsection*{Roots}
+ \begin{theorembox}{title=Factor theorem}
+ If \(\beta z + \alpha\) is a factor of \(P(z)\), \\
+ \-\hspace{1em}then \(P(-\dfrac{\alpha}{\beta})=0\).
+ \end{theorembox}
- \(n\)th roots of \(z=r\operatorname{cis}\theta\) are:
+ \subsection*{\(n\)th roots}
- \[z = r^{\frac{1}{n}} \operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)\]
+ \(n\)th roots of \(z=r\operatorname{cis}\theta\) are:
- \begin{itemize}
+ \[z = r^{\frac{1}{n}} \operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)\]
- \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}
+ \begin{itemize}
+
+ \item{Same modulus for all solutions}
+ \item{Arguments separated by \(\frac{2\pi}{n} \therefore\) there are \(n\) roots}
+ \item{If one square root is \(a+bi\), the other is \(-a-bi\)}
+ \item{Give one implicit \(n\)th root \(z_1\), function is \(z=z_1^n\)}
+ \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:
+ \noindent For \(0=az^2+bz+c\), use quadratic formula:
- \[z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
+ \[z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\subsection*{Fundamental theorem of algebra}
- A polynomial of degree \(n\) can be factorised into \(n\) linear factors in \(\mathbb{C}\):
+ A polynomial of degree \(n\) can be factorised into \(n\) linear factors in \(\mathbb{C}\):
- \[\implies P(z)=a_n(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)\dots(z-\alpha_n)\]
- \[\text{ where } \alpha_1,\alpha_2,\alpha_3,\dots,\alpha_n \in \mathbb{C}\]
+ \[\implies P(z)=a_n(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)\dots(z-\alpha_n)\]
+ \[\text{ where } \alpha_1,\alpha_2,\alpha_3,\dots,\alpha_n \in \mathbb{C}\]
\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]{\(\begin{aligned}z&=a+bi\\&=r\operatorname{cis}\theta\end{aligned}\)};
+
+ \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}
\node at ($(O)+(20:3mm)$) {$\theta$};
\end{scope}
\filldraw (P) circle (0.5pt);
- \end{tikzpicture}\end{center}
+ \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}
+ \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{\(|z+a|=|z+b| \implies 2(a-b)x=b^2-a^2\)\\Geometric: equidistant from \(a,b\)}
- \end{itemize}
- \subsubsection*{Circles}
+ \subsubsection*{Rays/lines \qquad \(\operatorname{Arg}( z\pm b)=\theta\)}
+
+ \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}
+
+ \begin{itemize}
+ \item \(\operatorname{Arg}(z \pm b) = \theta\) (ray)
+ \item{\(\operatorname{Re}(z)=c\) or \(\operatorname{Im}(z)=c\) (perpendicular bisector)}
+ \item{\(\operatorname{Im}(z)=m\operatorname{Re}(z)\)}
+ \item \(|z - (a+bi)|=|z - (c+di)| \\ \implies \frac{2(c-a)x + a^2 + b^2 - c^2 - d^2}{2(b-d)}\)
+ \item \(\operatorname{Re}(z) \pm \operatorname{Im}(z) = c\)
+ \end{itemize}
+
+ \subsubsection*{Circles}
+
+ \begin{itemize}
+ \item \(|z-z_1|^2=c^2|z_2+2|^2\)
+ \item \(|z-(a+bi)|=c \implies (x-a)^2+_(y-b)^2=c^2\)
+ \item \(z \overline{z} = r^2\)
+ \end{itemize}
+
+ \subsubsection*{Regions \qquad \(\operatorname{Arg}(z) \lessgtr \theta\)}
+
+ \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}
- \begin{itemize}
- \item \(|z-z_1|^2=c^2|z_2+2|^2\)
- \item \(|z-(a+bi)|=c \implies (x-a)^2+_(y-b)^2=c^2\)
- \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}]
+ \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}
+ \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 \(A(x_1,y_1), \> B(x_2,y_2)\)
+ \[\begin{bmatrix}x\\ y \end{bmatrix} \iff x\boldsymbol{i} + y\boldsymbol{j}\]
+ \(\begin{bmatrix}x_2-x_1\\ y_2-y_1 \end{bmatrix}\) \quad between \(A(x_1,y_1), \> B(x_2,y_2)\)
-\subsection*{Scalar multiplication}
+ \subsection*{Scalar multiplication}
-\[k\cdot (x\boldsymbol{i}+y\boldsymbol{j})=kx\boldsymbol{i}+ky\boldsymbol{j}\]
+ \[k\cdot (x\boldsymbol{i}+y\boldsymbol{j})=kx\boldsymbol{i}+ky\boldsymbol{j}\]
-\noindent For \(k \in \mathbb{R}^-\), direction is reversed
+ \noindent For \(k \in \mathbb{R}^-\), direction is reversed
-\subsection*{Vector addition}
-\begin{center}\begin{tikzpicture}[scale=1]
+ \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}
- \item Draw each vector head to tail then join lines
- \item Addition is commutative (parallelogram)
- \item \(\boldsymbol{u}-\boldsymbol{v}=\boldsymbol{u}+(-\boldsymbol{v}) \implies \overrightharp{AB}=\boldsymbol{b}-\boldsymbol{a}\)
-\end{itemize}
-
-\subsection*{Magnitude}
-
-\[|(x\boldsymbol{i} + y\boldsymbol{j})|=\sqrt{x^2+y^2}\]
-
-\subsection*{Parallel vectors}
-
-\[\boldsymbol{u} || \boldsymbol{v} \iff \boldsymbol{u} = k \boldsymbol{v} \text{ where } k \in \mathbb{R} \setminus \{0\}\]
-
-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}
-
-\subsection*{Perpendicular vectors}
-
-\[\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b} = 0\ \quad \text{(since \(\cos 90 = 0\))}\]
-
-\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*{Scalar product \(\boldsymbol{a} \cdot \boldsymbol{b}\)}
-
-
-\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{]})}}
-
-\subsubsection*{Properties}
-
-\begin{enumerate}
-\item
- \(k(\boldsymbol{a\cdot b})=(k\boldsymbol{a})\cdot \boldsymbol{b}=\boldsymbol{a}\cdot (k\boldsymbol{b})\)
-\item
- \(\boldsymbol{a \cdot 0}=0\)
-\item
- \(\boldsymbol{a} \cdot (\boldsymbol{b} + \boldsymbol{c})=\boldsymbol{a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{c}\)
-\item
- \(\boldsymbol{i \cdot i} = \boldsymbol{j \cdot j} = \boldsymbol{k \cdot k}= 1\)
-\item
- \(\boldsymbol{a} \cdot \boldsymbol{b} = 0 \quad \implies \quad \boldsymbol{a} \perp \boldsymbol{b}\)
-\item
- \(\boldsymbol{a \cdot a} = |\boldsymbol{a}|^2 = a^2\)
-\end{enumerate}
-
-\subsection*{Angle between vectors}
-
-\[\cos \theta = \frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}| |\boldsymbol{b}|} = \frac{a_1 b_1 + a_2 b_2}{|\boldsymbol{a}| |\boldsymbol{b}|}\]
-
-\noindent \colorbox{cas}{On CAS:} \texttt{angle([a b c], [a b c])}
-
-(Action \(\rightarrow\) Vector \(\rightarrow\)Angle)
-
-\subsection*{Angle between vector and axis}
-
-\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:
-\[\cos \alpha = \frac{a_1}{|\boldsymbol{a}|}, \quad \cos \beta = \frac{a_2}{|\boldsymbol{a}|}, \quad \cos \gamma = \frac{a_3}{|\boldsymbol{a}|}\]
-
-\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*{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}
+ \end{tikzpicture}\end{center}
+
+ \[(x\boldsymbol{i}+y\boldsymbol{j}) \pm (a\boldsymbol{i}+b\boldsymbol{j})=(x \pm a)\boldsymbol{i}+(y \pm b)\boldsymbol{j}\]
-\subsubsection*{\(\parallel\boldsymbol{b}\) (vector projection/resolute)}
+ \begin{itemize}
+ \item Draw each vector head to tail then join lines
+ \item Addition is commutative (parallelogram)
+ \item \(\boldsymbol{u}-\boldsymbol{v}=\boldsymbol{u}+(-\boldsymbol{v}) \implies \overrightharp{AB}=\boldsymbol{b}-\boldsymbol{a}\)
+ \end{itemize}
-\begin{align*}
- \boldsymbol{u} & = \frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|^2}\boldsymbol{b} \\
- & = \left(\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}\right)\left(\frac{\boldsymbol{b}}{|\boldsymbol{b}|}\right) \\
- & = (\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}
-\end{align*}
+ \subsection*{Magnitude}
-\subsubsection*{\(\perp\boldsymbol{b}\) (perpendicular projection)}
-\[\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u}\]
+ \[|(x\boldsymbol{i} + y\boldsymbol{j})|=\sqrt{x^2+y^2}\]
+
+ \subsection*{Parallel vectors}
-\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*}
+ \[\boldsymbol{u} || \boldsymbol{v} \iff \boldsymbol{u} = k \boldsymbol{v} \text{ where } k \in \mathbb{R} \setminus \{0\}\]
+
+ 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}
+
+ \subsection*{Perpendicular vectors}
-\subsubsection*{Rectangular (\(\parallel,\perp\)) components}
+ \[\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b} = 0\ \quad \text{(since \(\cos 90 = 0\))}\]
-\[\boldsymbol{a}=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}+\left(\boldsymbol{a}-\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}\right)\]
+ \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*{Scalar product \(\boldsymbol{a} \cdot \boldsymbol{b}\)}
-\subsection*{Vector proofs}
-\textbf{Concurrent:} intersection of \(\ge\) 3 lines
+ \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{]})}
-\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);
+ \subsubsection*{Properties}
+
+ \begin{enumerate}
+ \item
+ \(k(\boldsymbol{a\cdot b})=(k\boldsymbol{a})\cdot \boldsymbol{b}=\boldsymbol{a}\cdot (k\boldsymbol{b})\)
+ \item
+ \(\boldsymbol{a \cdot 0}=0\)
+ \item
+ \(\boldsymbol{a} \cdot (\boldsymbol{b} + \boldsymbol{c})=\boldsymbol{a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{c}\)
+ \item
+ \(\boldsymbol{i \cdot i} = \boldsymbol{j \cdot j} = \boldsymbol{k \cdot k}= 1\)
+ \item
+ \(\boldsymbol{a} \cdot \boldsymbol{b} = 0 \quad \implies \quad \boldsymbol{a} \perp \boldsymbol{b}\)
+ \item
+ \(\boldsymbol{a \cdot a} = |\boldsymbol{a}|^2 = a^2\)
+ \end{enumerate}
+
+ \subsection*{Angle between vectors}
+
+ \[\cos \theta = \frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}| |\boldsymbol{b}|} = \frac{a_1 b_1 + a_2 b_2}{|\boldsymbol{a}| |\boldsymbol{b}|}\]
+
+ \noindent \colorbox{cas}{On CAS:} \texttt{angle([a b c], [a b c])}
+
+ (Action \(\rightarrow\) Vector \(\rightarrow\)Angle)
+
+ \subsection*{Angle between vector and axis}
+
+ \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:
+ \[\cos \alpha = \frac{a_1}{|\boldsymbol{a}|}, \quad \cos \beta = \frac{a_2}{|\boldsymbol{a}|}, \quad \cos \gamma = \frac{a_3}{|\boldsymbol{a}|}\]
+
+ \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*{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} & = \frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|^2}\boldsymbol{b} \\
+ & = \left(\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}\right)\left(\frac{\boldsymbol{b}}{|\boldsymbol{b}|}\right) \\
+ & = (\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}
+ \end{align*}
+
+ \subsubsection*{\(\perp\boldsymbol{b}\) (perpendicular projection)}
+ \[\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u}\]
+
+ \subsubsection*{\(|\boldsymbol{u}|\) (scalar projection/resolute)}
+ \begin{align*}
+ s &= |\boldsymbol{u}|\\
+ &= \boldsymbol{a} \cdot \hat{\boldsymbol{b}}\\
+ &=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}\\
+ &= |\boldsymbol{a}| \cos \theta
+ \end{align*}
+
+ \subsubsection*{Rectangular (\(\parallel,\perp\)) components}
+
+ \[\boldsymbol{a}=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}+\left(\boldsymbol{a}-\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}\right)\]
+
+
+ \subsection*{Vector proofs}
+
+ \textbf{Concurrent:} intersection of \(\ge\) 3 lines
+
+ \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*{Collinear points}
+
+ \(\ge\) 3 points lie on the same line
+
+ \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}
+
+ \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*}
+ \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*{Parallelograms}
+
+ \begin{center}\begin{tikzpicture}
+ \coordinate (O) at (0,0) node [below left] {\(O\)};
+ \coordinate (A) at (4,0);
+ \coordinate (B) at (6,2);
+ \coordinate (C) at (2,2);
+ \coordinate (D) at (6,0);
+
+ \draw[postaction={decorate}, decoration={markings, mark=at position 0.6 with {\arrow{>>}}}] (O)--(A) node [below left] {\(A\)};
+ \draw[postaction={decorate}, decoration={markings,mark=at position 0.5 with {\arrow{>}}}] (A)--(B) node [above right] {\(B\)};
+ \draw[postaction={decorate}, decoration={markings, mark=at position 0.6 with {\arrow{>>}}}] (B)--(C) node [above left] {\(C\)};
+ \draw[postaction={decorate}, decoration={markings,mark=at position 0.5 with {\arrow{>}}}] (C)--(O);
+
+ \draw [gray, dashed] (O) -- (B) node [pos=0.75] {\(\diagdown\diagdown\)} node [pos=0.25] {\(\diagdown\diagdown\)};
+ \draw [gray, dashed] (A) -- (C) node [pos=0.75] {\(\diagup\)} node [pos=0.25] {\(\diagup\)};
+ \begin{scope}
+ \path[clip] (C) -- (A) -- (O);
+ \fill[orange, opacity=0.5, draw=black] (0,0) circle (4mm);
+ \node at ($(0,0)+(20:8mm)$) {\(\theta\)};
+ \end{scope}
+ \draw [gray, thick, dotted] (B) -- (D) node [pos=0.5, right, black, font=\footnotesize] {\(|\boldsymbol{c}|\sin\theta\)} (A) -- (D) node [pos=0.5, below, black, font=\footnotesize] {\(|\boldsymbol{c}|\cos\theta\)};
+ \draw pic [draw,thick,red,angle radius=2mm] {right angle=O--D--B};
+ \end{tikzpicture}\end{center}
+
+ \begin{itemize}
+ \item
+ Diagonals \(\overrightharp{OB}, \overrightharp{AC}\) bisect each other
+ \item
+ If diagonals are equal length, it is a rectangle
+ \item
+ \(|\overrightharp{OB}|^2 + |\overrightharp{CA}|^2 = |\overrightharp{OA}|^2 + |\overrightharp{AB}|^2 + |\overrightharp{CB}|^2 + |\overrightharp{OC}|^2\)
+ \item
+ Area \(=\boldsymbol{c} \cdot \boldsymbol{a}\)
+ \end{itemize}
+
+ \subsubsection*{Perpendicular bisectors of a triangle}
+
+\hspace{-1.5cm}\begin{tikzpicture}
+ [
+ scale=3,
+ >=stealth,
+ point/.style = {draw, circle, fill = black, inner sep = 1pt},
+ dot/.style = {draw, circle, fill = black, inner sep = .2pt},
+ thick
+ ]
+
+ \node at (-1,1) [text width=5cm, rounded corners, fill=lblue, inner sep=1ex]
+ {
+ \sffamily The three bisectors meet at the circumcenter \(Z\) where \(|\overrightharp{ZA}| = |\overrightharp{ZB}| = |\overrightharp{ZC}|\).
+ };
+
+ % the circle
+ \def\rad{1}
+ \node (origin) at (0,0) [point, label = {right: {\(Z\)}}]{};
+ \draw [thin] (origin) circle (\rad);
+
+ % triangle nodes: just points on the circle
+ \node (n1) at +(60:\rad) [point, label = above:\(A\)] {};
+ \node (n2) at +(-145:\rad) [point, label = below:\(B\)] {};
+ \node (n3) at +(-45:\rad) [point, label = {below right:\(C\)}] {};
+
+ % triangle edges: connect the vertices, and leave a node at the midpoint
+ \draw[orange] (n3) -- node (a) [label = {above right:\(D\)}] {} (n1);
+ \draw[blue] (n3) -- node (b) [label = {below right:\(F\)}] {} (n2);
+ \draw[red] (n1) -- node (c) [label = {left: \(E\)}] {} (n2);
+
+ % Bisectors
+ % start at the point lying on the line from (origin) to (a), at
+ % twice that distance, and then draw a path going to the point on
+ % the line lying on the line from (a) to the (origin), at 3 times
+ % that distance.
+ \draw[orange, dotted]
+ ($ (origin) ! 2 ! (a) $)
+ node [right] {\sffamily Bisector \(AC\)}
+ -- ($(a) ! 3 ! (origin)$ );
+
+ % similarly for origin and b
+ \draw[blue, dotted]
+ ($ (origin) ! 2 ! (b) $)
+ -- ($(b) ! 3 ! (origin)$ )
+ node [right] {\sffamily Bisector \(BC\)};
+
+ \draw[red, dotted]
+ ($ (origin) ! 5 ! (c) $)
+ -- ($(c) ! 7 ! (origin)$ )
+ node [right] {\sffamily Bisector \(AB\)};
+
+ \draw[gray, dashed, thin] (n1) -- (origin) -- (n2);
+ \draw[gray, dashed, thin] (origin) -- (n3);
+
+ % Right angle symbols
+ \def\ralen{.5ex} % length of the short segment
+ \foreach \inter/\first/\last in {a/n3/origin, b/n2/origin, c/n2/origin}
+ {
+ \draw [thin] let \p1 = ($(\inter)!\ralen!(\first)$), % point along first path
+ \p2 = ($(\inter)!\ralen!(\last)$), % point along second path
+ \p3 = ($(\p1)+(\p2)-(\inter)$) % corner point
+ in
+ (\p1) -- (\p3) -- (\p2); % path
+ }
\end{tikzpicture}
-\subsubsection*{Collinear points}
+ \begin{theorembox}{title=Perpendicular bisector theorem}
+ If a point \(P\) lies on the perpendicular bisector of line \(\overrightharp{XY}\), then \(P\) is equidistant from the endpoints of the bisected segment
+ \[ \text{i.e. } |\overrightharp{PX}| = |\overrightharp{PY}| \]
+ \end{theorembox}
-\(\ge\) 3 points lie on the same line
+ \subsubsection*{Useful vector properties}
-\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}
+ \begin{itemize}
+ \item
+ \(\boldsymbol{a} \parallel \boldsymbol{b} \implies \boldsymbol{b}=k\boldsymbol{a}\) for some
+ \(k \in \mathbb{R} \setminus \{0\}\)
+ \item
+ If \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are parallel with at
+ least one point in common, then they lie on the same straight line
+ \item
+ \(\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b}=0\)
+ \item
+ \(\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2\)
+ \end{itemize}
+
+ \subsection*{Linear dependence}
+
+ \(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\) are linearly dependent if they are \(\nparallel\) and:
+ \begin{align*}
+ 0&=k\boldsymbol{a}+l\boldsymbol{b}+m\boldsymbol{c}\\
+ \therefore \boldsymbol{c} &= m\boldsymbol{a} + n\boldsymbol{b} \quad \text{(simultaneous)}
+ \end{align*}
+
+ \noindent \(\boldsymbol{a}, \boldsymbol{b},\) and \(\boldsymbol{c}\) are linearly
+ independent if no vector in the set is expressible as a linear
+ combination of other vectors in set, or if they are parallel.
+
+ \subsection*{Three-dimensional vectors}
+
+ Right-hand rule for axes: \(z\) is up or out of page.
+
+ \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);
-\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*}
-\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*{Parallelograms}
-
-\begin{center}\begin{tikzpicture}
- \coordinate (O) at (0,0) node [below left] {\(O\)};
- \coordinate (A) at (4,0);
- \coordinate (B) at (6,2);
- \coordinate (C) at (2,2);
- \coordinate (D) at (6,0);
-
- \draw[postaction={decorate}, decoration={markings, mark=at position 0.6 with {\arrow{>>}}}] (O)--(A) node [below left] {\(A\)};
- \draw[postaction={decorate}, decoration={markings,mark=at position 0.5 with {\arrow{>}}}] (A)--(B) node [above right] {\(B\)};
- \draw[postaction={decorate}, decoration={markings, mark=at position 0.6 with {\arrow{>>}}}] (B)--(C) node [above left] {\(C\)};
- \draw[postaction={decorate}, decoration={markings,mark=at position 0.5 with {\arrow{>}}}] (C)--(O);
-
- \draw [gray, dashed] (O) -- (B) node [pos=0.75] {\(\diagdown\diagdown\)} node [pos=0.25] {\(\diagdown\diagdown\)};
- \draw [gray, dashed] (A) -- (C) node [pos=0.75] {\(\diagup\)} node [pos=0.25] {\(\diagup\)};
- \begin{scope}
- \path[clip] (C) -- (A) -- (O);
- \fill[orange, opacity=0.5, draw=black] (0,0) circle (4mm);
- \node at ($(0,0)+(20:8mm)$) {\(\theta\)};
- \end{scope}
- \draw [gray, thick, dotted] (B) -- (D) node [pos=0.5, right, black, font=\footnotesize] {\(|\boldsymbol{c}|\sin\theta\)} (A) -- (D) node [pos=0.5, below, black, font=\footnotesize] {\(|\boldsymbol{c}|\cos\theta\)};
- \draw pic [draw,thick,red,angle radius=2mm] {right angle=O--D--B};
-\end{tikzpicture}\end{center}
-
-\begin{itemize}
- \item
- Diagonals \(\overrightharp{OB}, \overrightharp{AC}\) bisect each other
- \item
- If diagonals are equal length, it is a rectangle
- \item
- \(|\overrightharp{OB}|^2 + |\overrightharp{CA}|^2 = |\overrightharp{OA}|^2 + |\overrightharp{AB}|^2 + |\overrightharp{CB}|^2 + |\overrightharp{OC}|^2\)
- \item
- Area \(=\boldsymbol{c} \cdot \boldsymbol{a}\)
-\end{itemize}
-
- \subsubsection*{Useful vector properties}
-
-\begin{itemize}
-\item
- \(\boldsymbol{a} \parallel \boldsymbol{b} \implies \boldsymbol{b}=k\boldsymbol{a}\) for some
- \(k \in \mathbb{R} \setminus \{0\}\)
-\item
- If \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are parallel with at
- least one point in common, then they lie on the same straight line
-\item
- \(\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b}=0\)
-\item
- \(\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2\)
-\end{itemize}
-
-\subsection*{Linear dependence}
-
-\(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\) are linearly dependent if they are \(\nparallel\) and:
-\begin{align*}
- 0&=k\boldsymbol{a}+l\boldsymbol{b}+m\boldsymbol{c}\\
- \therefore \boldsymbol{c} &= m\boldsymbol{a} + n\boldsymbol{b} \quad \text{(simultaneous)}
-\end{align*}
-
-\noindent \(\boldsymbol{a}, \boldsymbol{b},\) and \(\boldsymbol{c}\) are linearly
-independent if no vector in the set is expressible as a linear
-combination of other vectors in set, or if they are parallel.
-
-\subsection*{Three-dimensional vectors}
-
-Right-hand rule for axes: \(z\) is up or out of page.
-
-\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);
-
-%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:
-
-\[\begin{cases}x = x_o + a \cdot t \\ y = y_0 + b \cdot t \\ z = z_0 + c \cdot t\end{cases}\]
-
-\section{Circular functions}
-
-\(\sin(bx)\) or \(\cos(bx)\): period \(=\frac{2\pi}{b}\)
-
-\noindent \(\tan(nx)\): period \(=\frac{\pi}{n}\)\\
-\indent\indent\indent asymptotes at \(x=\frac{(2k+1)\pi}{2n} \> \vert \> k \in \mathbb{Z}\)
-
-\subsection*{Reciprocal functions}
+ %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]{$z$};
-\subsubsection*{Cosecant}
+ %draw a vector from O to P
+ \draw[vector] (O) -- (P);
-\[\operatorname{cosec} \theta = \frac{1}{\sin \theta} \> \vert \> \sin \theta \ne 0\]
+ %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}
-\begin{itemize}
-\item
- \textbf{Domain} \(= \mathbb{R} \setminus {n\pi : n \in \mathbb{Z}}\)
-\item
- \textbf{Range} \(= \mathbb{R} \setminus (-1, 1)\)
-\item
- \textbf{Turning points} at
- \(\theta = \frac{(2n + 1)\pi}{2} \> \vert \> n \in \mathbb{Z}\)
-\item
- \textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
-\end{itemize}
+ \subsection*{Parametric vectors}
-\subsubsection*{Secant}
+ 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:
+ \[\begin{cases}x = x_o + a \cdot t \\ y = y_0 + b \cdot t \\ z = z_0 + c \cdot t\end{cases}\]
-\begin{center}\includegraphics[width=0.7\columnwidth]{graphics/sec.png}\end{center}
+ \section{Circular functions}
-\[\operatorname{sec} \theta = \frac{1}{\cos \theta} \> \vert \> \cos \theta \ne 0\]
+ \(\sin(bx)\) or \(\cos(bx)\): period \(=\frac{2\pi}{b}\)
-\begin{itemize}
+ \noindent \(\tan(nx)\): period \(=\frac{\pi}{n}\)\\
+ \indent\indent\indent asymptotes at \(x=\frac{(2k+1)\pi}{2n} \> \vert \> k \in \mathbb{Z}\)
-\item
- \textbf{Domain}
- \(= \mathbb{R} \setminus \frac{(2n + 1) \pi}{2} : n \in \mathbb{Z}\}\)
-\item
- \textbf{Range} \(= \mathbb{R} \setminus (-1, 1)\)
-\item
- \textbf{Turning points} at
- \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
-\item
- \textbf{Asymptotes} at
- \(\theta = \frac{(2n + 1) \pi}{2} \> \vert \> n \in \mathbb{Z}\)
-\end{itemize}
+ \subsection*{Reciprocal functions}
-\subsubsection*{Cotangent}
+ \subsubsection*{Cosecant}
-\begin{center}\includegraphics[width=0.7\columnwidth]{graphics/cot.png}\end{center}
+ \[\operatorname{cosec} \theta = \frac{1}{\sin \theta} \> \vert \> \sin \theta \ne 0\]
-\[\operatorname{cot} \theta = {{\cos \theta} \over {\sin \theta}} \> \vert \> \sin \theta \ne 0\]
+ \begin{itemize}
+ \item
+ \textbf{Domain} \(= \mathbb{R} \setminus {n\pi : n \in \mathbb{Z}}\)
+ \item
+ \textbf{Range} \(= \mathbb{R} \setminus (-1, 1)\)
+ \item
+ \textbf{Turning points} at
+ \(\theta = \frac{(2n + 1)\pi}{2} \> \vert \> n \in \mathbb{Z}\)
+ \item
+ \textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
+ \end{itemize}
-\begin{itemize}
+ \subsubsection*{Secant}
-\item
- \textbf{Domain} \(= \mathbb{R} \setminus \{n \pi: n \in \mathbb{Z}\}\)
-\item
- \textbf{Range} \(= \mathbb{R}\)
-\item
- \textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
-\end{itemize}
+\begin{tikzpicture}
+ \begin{axis}[ytick={-1,1}, yticklabels={\(-1\), \(1\)}, xmin=-7,xmax=7,ymin=-3,ymax=3,enlargelimits=true, xtick={-6.2830, -3.1415, 3.1415, 6.2830},xticklabels={\(-2\pi\), \(-\pi\), \(\pi\), \(2\pi\)}]
+% \addplot[blue, domain=-6.2830:6.2830,unbounded coords=jump,samples=80] {sec(deg(x))};
+ \addplot[blue, restrict y to domain=-10:10, domain=-7:7,samples=100] {sec(deg(x))} node [pos=0.93, black, right] {\(\operatorname{sec} x\)};
+ \addplot[red, dashed, domain=-7:7,samples=100] {cos(deg(x))};
+ \draw [gray, dotted, thick] ({axis cs:1.5708,0}|-{rel axis cs:0,0}) -- ({axis cs:1.5708,0}|-{rel axis cs:0,1});
+ \draw [gray, dotted, thick] ({axis cs:4.71239,0}|-{rel axis cs:0,0}) -- ({axis cs:4.71239,0}|-{rel axis cs:0,1});
+ \draw [gray, dotted, thick] ({axis cs:-4.71239,0}|-{rel axis cs:0,0}) -- ({axis cs:-4.71239,0}|-{rel axis cs:0,1});
+ \draw [gray, dotted, thick] ({axis cs:-1.5708,0}|-{rel axis cs:0,0}) -- ({axis cs:-1.5708,0}|-{rel axis cs:0,1});
+\end{axis}
+ \node [black] at (7,3.5) {\(\cos x\)};
+\end{tikzpicture}
-\subsubsection*{Symmetry properties}
+ \[\operatorname{sec} \theta = \frac{1}{\cos \theta} \> \vert \> \cos \theta \ne 0\]
-\[\begin{split}
- \operatorname{sec} (\pi \pm x) & = -\operatorname{sec} x \\
- \operatorname{sec} (-x) & = \operatorname{sec} x \\
- \operatorname{cosec} (\pi \pm x) & = \mp \operatorname{cosec} x \\
- \operatorname{cosec} (-x) & = - \operatorname{cosec} x \\
- \operatorname{cot} (\pi \pm x) & = \pm \operatorname{cot} x \\
- \operatorname{cot} (-x) & = - \operatorname{cot} x
-\end{split}\]
+ \begin{itemize}
-\subsubsection*{Complementary properties}
+ \item
+ \textbf{Domain}
+ \(= \mathbb{R} \setminus \frac{(2n + 1) \pi}{2} : n \in \mathbb{Z}\}\)
+ \item
+ \textbf{Range} \(= \mathbb{R} \setminus (-1, 1)\)
+ \item
+ \textbf{Turning points} at
+ \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
+ \item
+ \textbf{Asymptotes} at
+ \(\theta = \frac{(2n + 1) \pi}{2} \> \vert \> n \in \mathbb{Z}\)
+ \end{itemize}
-\[\begin{split}
- \operatorname{sec} \left({\pi \over 2} - x\right) & = \operatorname{cosec} x \\
- \operatorname{cosec} \left({\pi \over 2} - x\right) & = \operatorname{sec} x \\
- \operatorname{cot} \left({\pi \over 2} - x\right) & = \tan x \\
- \tan \left({\pi \over 2} - x\right) & = \operatorname{cot} x
-\end{split}\]
+ \subsubsection*{Cotangent}
-\subsubsection*{Pythagorean identities}
+\begin{tikzpicture}
+ \begin{axis}[xmin=-3,xmax=3,ymin=-1.5,ymax=1.5,enlargelimits=true, xtick={-3.1415, -1.5708, 1.5708, 3.1415},xticklabels={\(-\pi\), \(-\frac{\pi}{2}\), \(\frac{\pi}{2}\), \(\pi\)}]
+ \addplot[blue, smooth, domain=-3:-0.1,unbounded coords=jump,samples=105] {cot(deg(x))} node [pos=0.3, left] {\(\operatorname{cot} x\)};
+\addplot[blue, smooth, domain=0.1:3,unbounded coords=jump,samples=105] {cot(deg(x))};
+\addplot[red, smooth, dashed] gnuplot [domain=-1.5:1.5,unbounded coords=jump,samples=105] {tan(x)};
+\addplot[red, smooth, dashed] gnuplot [domain=-3.5:-1.8,unbounded coords=jump,samples=105] {tan(x)} node [pos=0.5, right] {\(\tan x\)};
+\addplot[red, smooth, dashed] gnuplot [domain=1.8:3.5,unbounded coords=jump,samples=105] {tan(x)};
+ \draw [thick, red, dotted] ({axis cs:-1.5708,0}|-{rel axis cs:0,0}) -- ({axis cs:-1.5708,0}|-{rel axis cs:0,1});
+ \draw [thick, blue, dotted] ({axis cs:-3.1415,0}|-{rel axis cs:0,0}) -- ({axis cs:-3.1415,0}|-{rel axis cs:0,1});
+ \draw [thick, blue, dotted] ({axis cs:0,0}|-{rel axis cs:0,0}) -- ({axis cs:0,0}|-{rel axis cs:0,1});
+ \draw [thick, blue, dotted] ({axis cs:3.1415,0}|-{rel axis cs:0,0}) -- ({axis cs:3.1415,0}|-{rel axis cs:0,1});
+ \draw [thick, red, dotted] ({axis cs:1.5708,0}|-{rel axis cs:0,0}) -- ({axis cs:1.5708,0}|-{rel axis cs:0,1});
+\end{axis}
+\end{tikzpicture}
-\[\begin{split}
- 1 + \operatorname{cot}^2 x & = \operatorname{cosec}^2 x, \quad \text{where } \sin x \ne 0 \\
- 1 + \tan^2 x & = \operatorname{sec}^2 x, \quad \text{where } \cos x \ne 0
-\end{split}\]
+ \[\operatorname{cot} \theta = {{\cos \theta} \over {\sin \theta}} \> \vert \> \sin \theta \ne 0\]
-\subsection*{Compound angle formulas}
+ \begin{itemize}
+
+ \item
+ \textbf{Domain} \(= \mathbb{R} \setminus \{n \pi: n \in \mathbb{Z}\}\)
+ \item
+ \textbf{Range} \(= \mathbb{R}\)
+ \item
+ \textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
+ \end{itemize}
+
+ \subsubsection*{Symmetry properties}
+
+ \[\begin{split}
+ \operatorname{sec} (\pi \pm x) & = -\operatorname{sec} x \\
+ \operatorname{sec} (-x) & = \operatorname{sec} x \\
+ \operatorname{cosec} (\pi \pm x) & = \mp \operatorname{cosec} x \\
+ \operatorname{cosec} (-x) & = - \operatorname{cosec} x \\
+ \operatorname{cot} (\pi \pm x) & = \pm \operatorname{cot} x \\
+ \operatorname{cot} (-x) & = - \operatorname{cot} x
+ \end{split}\]
+
+ \subsubsection*{Complementary properties}
+
+ \[\begin{split}
+ \operatorname{sec} \left({\pi \over 2} - x\right) & = \operatorname{cosec} x \\
+ \operatorname{cosec} \left({\pi \over 2} - x\right) & = \operatorname{sec} x \\
+ \operatorname{cot} \left({\pi \over 2} - x\right) & = \tan x \\
+ \tan \left({\pi \over 2} - x\right) & = \operatorname{cot} x
+ \end{split}\]
+
+ \subsubsection*{Pythagorean identities}
-\[\cos(x \pm y) = \cos x + \cos y \mp \sin x \sin y\]
-\[\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\]
-\[\tan(x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x \tan y}}\]
+ \[\begin{split}
+ 1 + \operatorname{cot}^2 x & = \operatorname{cosec}^2 x, \quad \text{where } \sin x \ne 0 \\
+ 1 + \tan^2 x & = \operatorname{sec}^2 x, \quad \text{where } \cos x \ne 0
+ \end{split}\]
+
+ \subsection*{Compound angle formulas}
-\subsection*{Double angle formulas}
+ \[\cos(x \pm y) = \cos x + \cos y \mp \sin x \sin y\]
+ \[\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\]
+ \[\tan(x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x \tan y}}\]
-\[\begin{split}
- \cos 2x &= \cos^2 x - \sin^2 x \\
- & = 1 - 2\sin^2 x \\
- & = 2 \cos^2 x -1
-\end{split}\]
+ \subsection*{Double angle formulas}
+
+ \[\begin{split}
+ \cos 2x &= \cos^2 x - \sin^2 x \\
+ & = 1 - 2\sin^2 x \\
+ & = 2 \cos^2 x -1
+ \end{split}\]
+
+ \[\sin 2x = 2 \sin x \cos x\]
+
+ \[\tan 2x = {{2 \tan x} \over {1 - \tan^2 x}}\]
+
+ \subsection*{Inverse circular functions}
+
+ \begin{tikzpicture}
+ \begin{axis}[ymin=-2, ymax=4, xmin=-1.1, xmax=1.1, ytick={-1.5708, 1.5708, 3.14159},yticklabels={$-\frac{\pi}{2}$, $\frac{\pi}{2}$, $\pi$}]
+ \addplot[color=red, smooth] gnuplot [domain=-2:2,unbounded coords=jump,samples=500] {asin(x)} node [pos=0.25, below right] {\(\sin^{-1}x\)};
+ \addplot[color=blue, smooth] gnuplot [domain=-2:2,unbounded coords=jump,samples=500] {acos(x)} node [pos=0.25, below left] {\(\cos^{-1}x\)};
+ \addplot[mark=*, red] coordinates {(-1,-1.5708)} node[right, font=\footnotesize]{\((-1,-\frac{\pi}{2})\)} ;
+ \addplot[mark=*, red] coordinates {(1,1.5708)} node[left, font=\footnotesize]{\((1,\frac{\pi}{2})\)} ;
+ \addplot[mark=*, blue] coordinates {(1,0)};
+ \addplot[mark=*, blue] coordinates {(-1,3.1415)} node[right, font=\footnotesize]{\((-1,\pi)\)} ;
+ \end{axis}
+ \end{tikzpicture}\\
+
+ Inverse functions: \(f(f^{-1}(x)) = x\) (restrict domain)
+
+ \[\sin^{-1}: [-1, 1] \rightarrow \mathbb{R}, \quad \sin^{-1} x = y\]
+ \hfill where \(\sin y = x, \> y \in [{-\pi \over 2}, {\pi \over 2}]\)
+
+ \[\cos^{-1}: [-1,1] \rightarrow \mathbb{R}, \quad \cos^{-1} x = y\]
+ \hfill where \(\cos y = x, \> y \in [0, \pi]\)
-\[\sin 2x = 2 \sin x \cos x\]
+ \[\tan^{-1}: \mathbb{R} \rightarrow \mathbb{R}, \quad \tan^{-1} x = y\]
+ \hfill where \(\tan y = x, \> y \in \left(-{\pi \over 2}, {\pi \over 2}\right)\)
+
+ \begin{tikzpicture}
+ \begin{axis}[yticklabel style={yshift=1.0pt, anchor=north east},x=0.1cm, y=1cm, ymax=2, ymin=-2, xticklabels={}, ytick={-1.5708,1.5708},yticklabels={\(-\frac{\pi}{2}\),\(\frac{\pi}{2}\)}]
+ \addplot[color=orange, smooth] gnuplot [domain=-35:35, unbounded coords=jump,samples=350] {atan(x)} node [pos=0.5, above left] {\(\tan^{-1}x\)};
+ \addplot[gray, dotted, thick, domain=-35:35] {1.5708} node [black, font=\footnotesize, below right, pos=0] {\(y=\frac{\pi}{2}\)};
+ \addplot[gray, dotted, thick, domain=-35:35] {-1.5708} node [black, font=\footnotesize, above left, pos=1] {\(y=-\frac{\pi}{2}\)};
+ \end{axis}
+ \end{tikzpicture}
-\[\tan 2x = {{2 \tan x} \over {1 - \tan^2 x}}\]
+ \subsection*{Mensuration}
-\subsection*{Inverse circular functions}
+ \begin{tikzpicture}[draw=blue!70,thick]
+ \filldraw[fill=lblue] circle (2cm);
+ \filldraw[fill=white]
+ (320:2cm) node[right] {}
+ -- (220:2cm) node[left] {}
+ arc[start angle=220, end angle=320, radius=2cm]
+ -- cycle;
+ \node {Major Segment};
+ \node at (-90:1.5) {Minor Segment};
+
+ \begin{scope}[xshift=4.5cm]
+ \draw [fill=lblue] circle (2cm);
+ \filldraw[fill=white]
+ (320:2cm) node[right] {}
+ -- (0,0) node[above] {}
+ -- (220:2cm) node[left] {}
+ arc[start angle=220, end angle=320, radius=2cm]
+ -- cycle;
+ \node at (90:1cm) {Major Sector};
+ \node at (-90:1.5) {Minor Sector};
+ \end{scope}
+ \end{tikzpicture}
+
+
+ \begin{align*}
+ \textbf{Sectors: } A &= \pi r^2 \dfrac{\theta}{2\pi} \\
+ &= \dfrac{r^2 \theta}{2}
+ \end{align*}
+
+ \[ \textbf{Segments: } A = \dfrac{r^2}{2} \left(\theta - \sin \theta \right) \]
+
+ \begin{align*}
+ \textbf{Chords: } \operatorname{crd} \theta &= \sqrt{(1 - \cos\theta)^2 + \sin^2 \theta} \\
+ &= \sqrt{2 - 2\cos\theta} \\
+ &= 2 \sin \left(\dfrac{\theta}{2}\right)
+ \end{align*}
+
+ \section{Differential calculus}
+
+ \[f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={\frac{dy}{dx}}\]
+
+ \subsection*{Limits}
+
+ \[\lim_{x \rightarrow a}f(x)\]
+ \(L^-,\quad L^+\) \qquad limit from below/above\\
+ \(\lim_{x \to a} f(x)\) \quad limit of a point\\
+
+ \noindent For solving \(x\rightarrow\infty\), put all \(x\) terms in denominators\\
+ e.g. \[\lim_{x \rightarrow \infty}{{2x+3} \over {x-2}}={{2+{3 \over x}} \over {1-{2 \over x}}}={2 \over 1} = 2\]
+
+ \subsubsection*{Limit theorems}
+
+ \begin{enumerate}
+ \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
+ \(\therefore \lim_{x \rightarrow a} c \times f(x)=cF\) where \(c=\) constant
+ \item
+ \({\lim_{x \rightarrow a} {f(x) \over g(x)}} = {F \over G}, G \ne 0\)
+ \item
+ \(f(x)\) is continuous \(\iff L^-=L^+=f(x) \> \forall x\)
+ \end{enumerate}
+
+ \subsection*{Gradients}
+
+ \textbf{Secant (chord)} - line joining two points on curve\\
+ \textbf{Tangent} - line that intersects curve at one point
+
+ \subsubsection*{Points of Inflection}
+
+ \emph{Stationary point} - i.e.
+ \(f^\prime(x)=0\)\\
+ \emph{Point of inflection} - max \(|\)gradient\(|\) (i.e.
+ \(f^{\prime\prime} = 0\))
+
+ \subsubsection*{Strictly increasing/decreasing}
+
+ For \(x_2\) and \(x_1\) where \(x_2 > x_1\):
+
+ \textbf{strictly increasing}\\
+ \-\hspace{1em}where \(f(x_2) > f(x_1)\) or \(f^\prime(x)>0\)
+
+ \textbf{strictly decreasing}\\
+ \hspace{1em}where \(f(x_2) < f(x_1)\) or \(f^\prime(x)<0\)
+ \begin{warning}
+ Endpoints are included, even where \(\boldsymbol{\frac{dy}{dx}=0}\)
+ \end{warning}
+
+
+ \subsection*{Second derivative}
+ \begin{align*}f(x) \longrightarrow &f^\prime (x) \longrightarrow f^{\prime\prime}(x)\\
+ \implies y \longrightarrow &\frac{dy}{dx} \longrightarrow \frac{d^2 y}{dx^2}\end{align*}
+
+ \noindent Order of polynomial \(n\)th derivative decrements each time the derivative is taken
+
+
+ \subsection*{Slope fields}
+
+ \begin{tikzpicture}[declare function={diff(\x,\y) = \x+\y;}]
+ \begin{axis}[axis equal, ymin=-4, ymax=4, xmin=-4, xmax=4, ticks=none, enlargelimits=true, ]
+ \addplot[thick, orange, domain=-4:2] {e^(x)-x-1};
+ \pgfplotsinvokeforeach{-4,...,4}{%
+ \draw[gray] ( {#1 -0.1}, {4 - diff(#1, 4) *0.1}) -- ( {#1 +0.1}, {4 + diff(#1, 4) *0.1});
+ \draw[gray] ( {#1 -0.1}, {3 - diff(#1, 3) *0.1}) -- ( {#1 +0.1}, {3 + diff(#1, 3) *0.1});
+ \draw[gray] ( {#1 -0.1}, {2 - diff(#1, 2) *0.1}) -- ( {#1 +0.1}, {2 + diff(#1, 2) *0.1});
+ \draw[gray] ( {#1 -0.1}, {1 - diff(#1, 1) *0.1}) -- ( {#1 +0.1}, {1 + diff(#1, 1) *0.1});
+ \draw[gray] ( {#1 -0.1}, {0 - diff(#1, 0) *0.1}) -- ( {#1 +0.1}, {0 + diff(#1, 0) *0.1});
+ \draw[gray] ( {#1 -0.1}, {-1 - diff(#1, -1) *0.1}) -- ( {#1 +0.1}, {-1 + diff(#1, -1) *0.1});
+ \draw[gray] ( {#1 -0.1}, {-2 - diff(#1, -2) *0.1}) -- ( {#1 +0.1}, {-2 + diff(#1, -2) *0.1});
+ \draw[gray] ( {#1 -0.1}, {-3 - diff(#1, -3) *0.1}) -- ( {#1 +0.1}, {-3 + diff(#1, -3) *0.1});
+ \draw[gray] ( {#1 -0.1}, {-4 - diff(#1, -4) *0.1}) -- ( {#1 +0.1}, {-4 + diff(#1, -4) *0.1});
+ }
+ \end{axis}
+ \end{tikzpicture}
+
+ \begin{table*}[ht]
+ \centering
+ \begin{tabularx}{\textwidth}{|r|Y|Y|Y|}
+ \hline
+ \rowcolor{lblue}
+ & \adjustbox{margin=0 1ex, valign=m}{\centering\(\dfrac{d^2 y}{dx^2} > 0\)} & \adjustbox{margin=0 1ex, valign=m}{\centering \(\dfrac{d^2y}{dx^2}<0\)} & \adjustbox{margin=0 1ex, valign=m}{\(\dfrac{d^2y}{dx^2}=0\) (inflection)} \\
+ \hline
+ \(\dfrac{dy}{dx}>0\) &
+ \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-3, xmax=0.8, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(e^(x)}; \addplot[red] {x/2.5+0.75}; \end{axis}\end{tikzpicture} \\Rising (concave up)}&
+ \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=0.1, xmax=4, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(ln(x))}; \addplot[red] {x/1.5-0.56}; \end{axis}\end{tikzpicture} \\Rising (concave down)}&
+ \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1.5, xmax=1.5, scale=0.2, samples=100] \addplot[blue] {(sin((deg x)))}; \addplot[red] {x}; \end{axis}\end{tikzpicture} \\Rising inflection point}\\
+ \hline
+ \(\dfrac{dy}{dx}<0\) &
+ \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-.5, xmax=1, ymin=-.5, ymax=.5, scale=0.2, samples=100] \addplot[blue] {1/(x+1)-1}; \addplot[red] {-x}; \end{axis}\end{tikzpicture} \\Falling (concave up)}&
+ \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=0, xmax=1.5, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(2-x*x)^(1/2)}; \addplot[red] {-x+2}; \end{axis}\end{tikzpicture} \\Falling (concave down)}&
+ \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=1.5, xmax=4.5, scale=0.2, samples=100] \addplot[blue] {(sin((deg x)))}; \addplot[red] {-x+3.1415}; \end{axis}\end{tikzpicture} \\Falling inflection point}\\
+ \hline
+ \(\dfrac{dy}{dx}=0\)&
+ \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1, xmax=1, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(x*x)}; \addplot[red, thick] {0}; \end{axis}\end{tikzpicture} \\Local minimum}& \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1, xmax=1, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(-x*x)}; \addplot[red, very thick] {0}; \end{axis}\end{tikzpicture} \\Local maximum}&
+ \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1, xmax=1, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(x*x*x)}; \addplot[red, thick] {0}; \end{axis}\end{tikzpicture} \(\>\) \begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1, xmax=1, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(-x*x*x)}; \addplot[red, thick] {0}; \end{axis}\end{tikzpicture} \\Stationary inflection point}\\
+ \hline
+ \end{tabularx}
+ \end{table*}
+ \begin{itemize}
+ \item
+ \(f^\prime (a) = 0, \> f^{\prime\prime}(a) > 0\) \\
+ \textbf{local min} at \((a, f(a))\) (concave up)
+ \item
+ \(f^\prime (a) = 0, \> f^{\prime\prime} (a) < 0\) \\
+ \textbf{local max} at \((a, f(a))\) (concave down)
+ \item
+ \(f^{\prime\prime}(a) = 0\) \\
+ \textbf{point of inflection} at \((a, f(a))\)
+ \item
+ \(f^{\prime\prime}(a) = 0, \> f^\prime(a)=0\) \\
+ stationary point of inflection at \((a, f(a)\)
+ \end{itemize}
+
+ \subsection*{Implicit Differentiation}
+
+ \noindent Used for differentiating circles etc.
+
+ If \(p\) and \(q\) are expressions in \(x\) and \(y\) such that \(p=q\),
+ for all \(x\) and \(y\), then:
+
+ \[{\frac{dp}{dx}} = {\frac{dq}{dx}} \quad \text{and} \quad {\frac{dp}{dy}} = {\frac{dq}{dy}}\]
+
+ \begin{cas}
+ Action \(\rightarrow\) Calculation \\
+ \-\hspace{1em}\texttt{impDiff(y\^{}2+ax=5,\ x,\ y)}
+ \end{cas}
+
+ \subsection*{Function of the dependent
+ variable}
+
+ 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 \(A\).
+
+ \subsection*{Reciprocal derivatives}
+
+ \[\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}\]
+
+ \subsection*{Differentiating \(x=f(y)\)}
+ Find \(\dfrac{dx}{dy}\), then \(\dfrac{dy}{dx} = \dfrac{1}{\left(\dfrac{dx}{dy}\right)}\)
+
+ \subsection*{Parametric equations}
+
+
+ \begin{align*}
+ \dfrac{dy}{dt} &= \dfrac{dy}{dx} \cdot \dfrac{dx}{dt} \\
+ \therefore \dfrac{dy}{dx} &= \dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)} \text{ provided } \dfrac{dx}{dt} \ne 0 \\
+ \dfrac{d^2y}{dx^2} &= \dfrac{\left(\dfrac{dy^\prime}{dt}\right)}{\left(\dfrac{dx}{dt}\right)} \text{ where } y^\prime = \dfrac{dy}{dx}
+ \end{align*}
+
+ \subsection*{Integration}
+
+ \[\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)\]
+
+ \subsubsection*{Properties}
+
+ \begin{align*}
+ \int^b_a f(x) \> dx &= \int^c_a f(x) \> dx + \int^b_c f(x) \> dx \\
+ \int^a_a f(x) \> dx &= 0 \\
+ \int^b_a k \cdot f(x) \> dx &= k \int^b_a f(x) \> dx \\
+ \int^b_a f(x) \pm g(x) \> dx &= \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx \\
+ \int^b_a f(x) \> dx &= - \int^a_b f(x) \> dx \\
+ \end{align*}
-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.
-
-\[\sin^{-1}: [-1, 1] \rightarrow \mathbb{R}, \quad \sin^{-1} x = y\]
-\hfill where \(\sin y = x, \> y \in [{-\pi \over 2}, {\pi \over 2}]\)
+ \subsection*{Integration by substitution}
-\[\cos^{-1}: [-1,1] \rightarrow \mathbb{R}, \quad \cos^{-1} x = y\]
-\hfill where \(\cos y = x, \> y \in [0, \pi]\)
+ \[\int f(u) {\frac{du}{dx}} \cdot dx = \int f(u) \cdot du\]
-\[\tan^{-1}: \mathbb{R} \rightarrow \mathbb{R}, \quad \tan^{-1} x = y\]
-\hfill where \(\tan y = x, \> y \in \left(-{\pi \over 2}, {\pi \over 2}\right)\)
+ \begin{warning}
+ \(\boldsymbol{f(u)}\) must be 1:1 \(\boldsymbol{\implies}\) one \(\boldsymbol{x}\) for each \(\boldsymbol{y}\)
+ \end{warning}
+ \begin{align*}\text{e.g. for } y&=\int(2x+1)\sqrt{x+4} \cdot dx\\
+ \text{let } u&=x+4\\
+ \implies& {\frac{du}{dx}} = 1\\
+ \implies& x = u - 4\\
+ \text{then } &y=\int (2(u-4)+1)u^{\frac{1}{2}} \cdot du\\
+ &\text{(solve as normal integral)}
+ \end{align*}
+ \subsubsection*{Definite integrals by substitution}
-\section{Differential calculus}
+ For \(\int^b_a f(x) {\frac{du}{dx}} \cdot dx\), evaluate new \(a\) and
+ \(b\) for \(f(u) \cdot du\).
-\subsection*{Limits}
+ \subsubsection*{Trigonometric integration}
-\[\lim_{x \rightarrow a}f(x)\]
-\(L^-,\quad L^+\) \qquad limit from below/above\\
-\(\lim_{x \to a} f(x)\) \quad limit of a point\\
-
-\noindent For solving \(x\rightarrow\infty\), put all \(x\) terms in denominators\\
- e.g. \[\lim_{x \rightarrow \infty}{{2x+3} \over {x-2}}={{2+{3 \over x}} \over {1-{2 \over x}}}={2 \over 1} = 2\]
-
-\subsubsection*{Limit theorems}
-
-\begin{enumerate}
-\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
-\(\therefore \lim_{x \rightarrow a} c \times f(x)=cF\) where \(c=\) constant
-\item
- \({\lim_{x \rightarrow a} {f(x) \over g(x)}} = {F \over G}, G \ne 0\)
-\item
- \(f(x)\) is continuous \(\iff L^-=L^+=f(x) \> \forall x\)
-\end{enumerate}
-
-\subsection*{Gradients of secants and tangents}
-
-\textbf{Secant (chord)} - line joining two points on curve\\
-\textbf{Tangent} - line that intersects curve at one point
-
-\subsection*{First principles derivative}
-
-\[f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={\frac{dy}{dx}}\]
-
-\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}
-
-\(b^{m+n}=b^m \cdot b^n\)\\
-\((b^m)^n=b^{m \cdot n}\)\\
-\((b \cdot c)^n = b^n \cdot c^n\)\\
-\({a^m \div a^n} = {a^{m-n}}\)
+ \[\sin^m x \cos^n x \cdot dx\]
-\subsection*{Derivative rules}
-
-\renewcommand{\arraystretch}{1.4}
-\begin{tabularx}{\columnwidth}{rX}
- \hline
-\(f(x)\) & \(f^\prime(x)\)\\
-\hline
-\(\sin x\) & \(\cos x\)\\
-\(\sin ax\) & \(a\cos ax\)\\
-\(\cos x\) & \(-\sin x\)\\
-\(\cos ax\) & \(-a \sin ax\)\\
-\(\tan f(x)\) & \(f^2(x) \sec^2f(x)\)\\
-\(e^x\) & \(e^x\)\\
-\(e^{ax}\) & \(ae^{ax}\)\\
-\(ax^{nx}\) & \(an \cdot e^{nx}\)\\
- \(\log_e x\) & \(\dfrac{1}{x}\)\\
- \(\log_e {ax}\) & \(\dfrac{1}{x}\)\\
- \(\log_e f(x)\) & \(\dfrac{f^\prime (x)}{f(x)}\)\\
-\(\sin(f(x))\) & \(f^\prime(x) \cdot \cos(f(x))\)\\
- \(\sin^{-1} x\) & \(\dfrac{1}{\sqrt{1-x^2}}\)\\
- \(\cos^{-1} x\) & \(\dfrac{-1}{sqrt{1-x^2}}\)\\
- \(\tan^{-1} x\) & \(\dfrac{1}{1 + x^2}\)\\
- \hline
-\end{tabularx}
-
-\subsection*{Reciprocal derivatives}
-
-\[\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}\]
-
-\subsection*{Differentiating \(x=f(y)\)}
-\begin{align*}
- \text{Find }& \frac{dx}{dy}\\
- \text{Then, }\frac{dx}{dy} &= \frac{1}{\frac{dy}{dx}} \\
- \implies {\frac{dy}{dx}} &= \frac{1}{\frac{dx}{dy}}\\
- \therefore {\frac{dy}{dx}} &= \frac{1}{\frac{dx}{dy}}
-\end{align*}
-
-\subsection*{Second derivative}
-\begin{align*}f(x) \longrightarrow &f^\prime (x) \longrightarrow f^{\prime\prime}(x)\\
-\implies y \longrightarrow &\frac{dy}{dx} \longrightarrow \frac{d^2 y}{dx^2}\end{align*}
-
-\noindent Order of polynomial \(n\)th derivative decrements each time the derivative is taken
-
-\subsubsection*{Points of Inflection}
-
-\emph{Stationary point} - i.e.
-\(f^\prime(x)=0\)\\
-\emph{Point of inflection} - max \(|\)gradient\(|\) (i.e.
-\(f^{\prime\prime} = 0\))
-%\begin{table*}[ht]
-%\centering
-% \begin{tabularx}{\textwidth}{XXXX}
-%\hline
-% \rowcolor{shade2}
-% & \(\dfrac{d^2 y}{dx^2} > 0\) & \(\dfrac{d^2y}{dx^2}<0\) & \(\dfrac{d^2y}{dx^2}=0\) (inflection) \\
-%\hline
-% \(\frac{dy}{dx}>0\) & \begin{tikzpicture} \draw[domain=1:2,smooth,variable=\x,blue] plot ({\x},{(1/10)*\x*\x*\x}) plot ({\x},{0.675*\x-0.677}); \end{tikzpicture} & cell 3\\
-%cell 1 & cell 2 & cell 3\\
-%\hline
-%\end{tabularx}
-%\end{table*}
-\begin{itemize}
-
-\item
- if \(f^\prime (a) = 0\) and \(f^{\prime\prime}(a) > 0\), then point
- \((a, f(a))\) is a local min (curve is concave up)
-\item
- if \(f^\prime (a) = 0\) and \(f^{\prime\prime} (a) < 0\), then point
- \((a, f(a))\) is local max (curve is concave down)
-\item
- if \(f^{\prime\prime}(a) = 0\), then point \((a, f(a))\) is a point of
- inflection
-\item
- if also \(f^\prime(a)=0\), then it is a stationary point of inflection
-\end{itemize}
-
-\begin{table*}[ht]
- \centering
- \includegraphics[width=0.7\textwidth]{graphics/second-derivatives.png}
-\end{table*}
-
-\subsection*{Implicit Differentiation}
-
-\noindent Used for differentiating circles etc.
-
-If \(p\) and \(q\) are expressions in \(x\) and \(y\) such that \(p=q\),
-for all \(x\) and \(y\), then:
-
-\[{\frac{dp}{dx}} = {\frac{dq}{dx}} \quad \text{and} \quad {\frac{dp}{dy}} = {\frac{dq}{dy}}\]
-
-\noindent \colorbox{cas}{\textbf{On CAS:}}\\
-Action \(\rightarrow\) Calculation \(\rightarrow\) \texttt{impDiff(y\^{}2+ax=5,\ x,\ y)}\\
-Returns \(y^\prime= \dots\).
-
-\subsection*{Integration}
-
-\[\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)\]
-
-\subsection*{Integral laws}
-
-\renewcommand{\arraystretch}{1.4}
-\begin{tabularx}{\columnwidth}{rX}
-\hline
- \(f(x)\) & \(\int f(x) \cdot dx\) \\
- \hline
- \(k\) (constant) & \(kx + c\)\\
- \(x^n\) & \(\dfrac{1}{n+1} x^{n+1}\) \\
- \(a x^{-n}\) &\(a \cdot \log_e x + c\)\\
- \(\dfrac{1}{ax+b}\) &\(\dfrac{1}{a} \log_e (ax+b) + c\)\\
- \((ax+b)^n\) & \(\dfrac{1}{a(n+1)}(ax+b)^{n-1} + c\)\\
- \(e^{kx}\) & \(\dfrac{1}{k} e^{kx} + c\)\\
- \(e^k\) & \(e^kx + c\)\\
- \(\sin kx\) & \(\dfrac{-1}{k} \cos (kx) + c\)\\
- \(\cos kx\) & \(\frac{1}{k} \sin (kx) + c\)\\
- \(\sec^2 kx\) & \(\frac{1}{k} \tan(kx) + c\)\\
- \(\dfrac{1}{\sqrt{a^2-x^2}}\) & \(\sin^{-1} \dfrac{x}{a} + c \>\vert\> a>0\)\\
- \(\dfrac{-1}{\sqrt{a^2-x^2}}\) & \(\cos^{-1} \dfrac{x}{a} + c \>\vert\> a>0\)\\
- \(\frac{a}{a^2-x^2}\) & \(\tan^{-1} \frac{x}{a} + c\)\\
- \(\frac{f^\prime (x)}{f(x)}\) & \(\log_e f(x) + c\)\\
- \(g^\prime(x)\cdot f^\prime(g(x)\) & \(f(g(x))\) (chain rule)\\
- \(f(x) \cdot g(x)\) & \(\int [f^\prime(x) \cdot g(x)] dx + \int [g^\prime(x) f(x)] dx\)\\
- \hline
-\end{tabularx}
-
-Note \(\sin^{-1} {x \over a} + \cos^{-1} {x \over a}\) is constant \(\forall x \in (-a, a)\)
-
-\subsection*{Definite integrals}
-
-\[\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)\]
-
-\begin{itemize}
-
-\item
- Signed area enclosed by\\
- \(\> y=f(x), \quad y=0, \quad x=a, \quad x=b\).
-\item
- \emph{Integrand} is \(f\).
-\end{itemize}
-
-\subsubsection*{Properties}
+ \paragraph{\textbf{\(m\) is odd:}}
+ \(m=2k+1\) where \(k \in \mathbb{Z}\)\\
+ \(\implies \sin^{2k+1} x = (\sin^2 z)^k \sin x = (1 - \cos^2 x)^k \sin x\)\\
+ Substitute \(u=\cos x\)
-\[\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx\]
+ \paragraph{\textbf{\(n\) is odd:}}
+ \(n=2k+1\) where \(k \in \mathbb{Z}\)\\
+ \(\implies \cos^{2k+1} x = (\cos^2 x)^k \cos x = (1-\sin^2 x)^k \cos x\)\\
+ Substitute \(u=\sin x\)
-\[\int^a_a f(x) \> dx = 0\]
-
-\[\int^b_a k \cdot f(x) \> dx = k \int^b_a f(x) \> dx\]
-
-\[\int^b_a f(x) \pm g(x) \> dx = \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx\]
+ \paragraph{\textbf{\(m\) and \(n\) are even:}}
+ use identities...
-\[\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx\]
+ \begin{itemize}
-\subsection*{Integration by substitution}
+ \item
+ \(\sin^2x={1 \over 2}(1-\cos 2x)\)
+ \item
+ \(\cos^2x={1 \over 2}(1+\cos 2x)\)
+ \item
+ \(\sin 2x = 2 \sin x \cos x\)
+ \end{itemize}
-\[\int f(u) {\frac{du}{dx}} \cdot dx = \int f(u) \cdot du\]
+ \subsection*{Separation of variables}
-\noindent Note \(f(u)\) must be 1:1 \(\implies\) one \(x\) for each \(y\)
-\begin{align*}\text{e.g. for } y&=\int(2x+1)\sqrt{x+4} \cdot dx\\
- \text{let } u&=x+4\\
- \implies& {\frac{du}{dx}} = 1\\
- \implies& x = u - 4\\
- \text{then } &y=\int (2(u-4)+1)u^{\frac{1}{2}} \cdot du\\
- &\text{(solve as normal integral)}
-\end{align*}
+ If \({\frac{dy}{dx}}=f(x)g(y)\), then:
-\subsubsection*{Definite integrals by substitution}
+ \[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\]
-For \(\int^b_a f(x) {\frac{du}{dx}} \cdot dx\), evaluate new \(a\) and
-\(b\) for \(f(u) \cdot du\).
+ \subsection*{Partial fractions}
-\subsubsection*{Trigonometric integration}
+ To factorise \(f(x) = \frac{\delta}{\alpha \cdot \beta}\):
+ \begin{align*}
+ \dfrac{\delta}{\alpha \cdot \beta \cdot \gamma} &= \dfrac{A}{\alpha} + \dfrac{B}{\beta} + \dfrac{C}{\gamma} \tag{1} \\
+ \text{Multiply by } & (\alpha \cdot \beta \cdot \gamma) \text{:} \\
+ \delta &= \beta\gamma A + \alpha\gamma B +\alpha\beta C \tag{2} \\
+ \text{Substitute } x &= \{\alpha, \beta, \gamma\} \text{ into (2) to find denominators}
+ \end{align*}
-\[\sin^m x \cos^n x \cdot dx\]
+ \subsubsection*{Repeated linear factors}
-\paragraph{\textbf{\(m\) is odd:}}
-\(m=2k+1\) where \(k \in \mathbb{Z}\)\\
-\(\implies \sin^{2k+1} x = (\sin^2 z)^k \sin x = (1 - \cos^2 x)^k \sin x\)\\
-Substitute \(u=\cos x\)
+ \[ \dfrac{p(x)}{(x-a)^n} = \dfrac{A_1}{(x-a)} + \dfrac{A_2}{(x-a)^2} + \dots + \dfrac{A_n}{(x-a)^n} \]
-\paragraph{\textbf{\(n\) is odd:}}
-\(n=2k+1\) where \(k \in \mathbb{Z}\)\\
-\(\implies \cos^{2k+1} x = (\cos^2 x)^k \cos x = (1-\sin^2 x)^k \cos x\)\\
-Substitute \(u=\sin x\)
+ \subsubsection*{Irreducible quadratic factors}
-\paragraph{\textbf{\(m\) and \(n\) are even:}}
-use identities...
+ \[ \text{e.g. } \dfrac{3x-4}{(2x-3)(x^2+5)} = \dfrac{A}{2x-3} + \dfrac{Bx+C}{x^2+5} \]
-\begin{itemize}
+ \begin{cas}
+ Action \(\rightarrow\) Transformation:\\
+ \-\hspace{1em} \texttt{expand(..., x)}
-\item
- \(\sin^2x={1 \over 2}(1-\cos 2x)\)
-\item
- \(\cos^2x={1 \over 2}(1+\cos 2x)\)
-\item
- \(\sin 2x = 2 \sin x \cos x\)
-\end{itemize}
+ To reverse, use \texttt{combine(...)}
+ \end{cas}
-\subsection*{Partial fractions}
+ \subsection*{Integrating \(\boldsymbol{\dfrac{dy}{dx} = g(y)}\)}
-\colorbox{cas}{On CAS:}\\
-\indent Action \(\rightarrow\) Transformation \(\rightarrow\)
-\texttt{expand/combine}\\
-\indent Interactive \(\rightarrow\) Transformation \(\rightarrow\)
-Expand \(\rightarrow\) Partial
+ \[ \text{if } \dfrac{dy}{dx} = g(y), \text{ then } x = \int{\dfrac{1}{g(y)}} \> dy \]
-\subsection*{Graphing integrals on CAS}
+ \subsection*{Graphing integrals on CAS}
-\colorbox{cas}{In main:} Interactive \(\rightarrow\) Calculation \(\rightarrow\)
-\(\int\) (\(\rightarrow\) Definite)\\
-Restrictions: \texttt{Define\ f(x)=..} then \texttt{f(x)\textbar{}x\textgreater{}..}
+ \begin{cas}
+ \textbf{In main:} Interactive \(\rightarrow\) Calculation \(\rightarrow\) \(\int\)\\
+ For restrictions, \texttt{Define\ f(x)=...} then \texttt{f(x)\textbar{}x\textgreater{}...}
+ \end{cas}
-\subsection*{Applications of antidifferentiation}
+ \subsection*{Solids of revolution}
-\begin{itemize}
+ Approximate as sum of infinitesimally-thick cylinders
-\item
- \(x\)-intercepts of \(y=f(x)\) identify \(x\)-coordinates of
- stationary points on \(y=F(x)\)
-\item
- nature of stationary points is determined by sign of \(y=f(x)\) on
- either side of its \(x\)-intercepts
-\item
- if \(f(x)\) is a polynomial of degree \(n\), then \(F(x)\) has degree
- \(n+1\)
-\end{itemize}
+ \subsubsection*{Rotation about \(\boldsymbol{x}\)-axis}
-To find stationary points of a function, substitute \(x\) value of given
-point into derivative. Solve for \({\frac{dy}{dx}}=0\). Integrate to find
-original function.
+ \[ V = \pi\int^{x=b}_{x=a} f(x)^2 \> dx \]
-\subsection*{Solids of revolution}
+ \subsubsection*{Rotation about \(\boldsymbol{y}\)-axis}
-Approximate as sum of infinitesimally-thick cylinders
+ \begin{align*}
+ V &= \pi \int^{y=b}_{y=a} x^2 \> dy \\
+ &= 2\pi \int^{x=b}_{x=a} x|f(x)| \> dx
+ \end{align*}
-\subsubsection*{Rotation about \(x\)-axis}
+ \subsubsection*{Rotating the area between two graphs}
-\begin{align*}
- V &= \int^{x=b}_{x-a} \pi y^2 \> dx \\
- &= \pi \int^b_a (f(x))^2 \> dx
-\end{align*}
+ \[V = \pi \int^b_a \left( f(x)^2 - g(x)^2 \right) \> dx\]
+ \hfill where \(f(x) > g(x)\)
-\subsubsection*{Rotation about \(y\)-axis}
+ \subsection*{Length of a curve}
-\begin{align*}
- V &= \int^{y=b}_{y=a} \pi x^2 \> dy \\
- &= \pi \int^b_a (f(y))^2 \> dy
-\end{align*}
+ For length of \(f(x)\) from \(x=a \rightarrow x=b\):
+ \begin{align*}
+ &\text{Cartesian} \> & L &= \int^b_a \sqrt{1 + \left(\dfrac{dy}{dx}\right)^2} \> dx \\
+ &\text{Parametric} \> & L & = \int^b_a \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2} \> dt
+ \end{align*}
-\subsubsection*{Regions not bound by \(y=0\)}
+ \begin{cas}
+ \begin{enumerate}[label=\alph*), leftmargin=5mm]
+ \item Evaluate formula
+ \item Interactive \(\rightarrow\) Calculation \(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen}
+ \end{enumerate}
+ \end{cas}
-\[V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx\]
-\hfill where \(f(x) > g(x)\)
+ \subsection*{Applications of antidifferentiation}
-\subsection*{Length of a curve}
+ \begin{itemize}
-\[L = \int^b_a \sqrt{1 + ({\frac{dy}{dx}})^2} \> dx \quad \text{(Cartesian)}\]
+ \item
+ \(x\)-intercepts of \(y=f(x)\) identify \(x\)-coordinates of
+ stationary points on \(y=F(x)\)
+ \item
+ nature of stationary points is determined by sign of \(y=f(x)\) on
+ either side of its \(x\)-intercepts
+ \item
+ if \(f(x)\) is a polynomial of degree \(n\), then \(F(x)\) has degree
+ \(n+1\)
+ \end{itemize}
-\[L = \int^b_a \sqrt{{\frac{dx}{dt}} + ({\frac{dy}{dt}})^2} \> dt \quad \text{(parametric)}\]
+ To find stationary points of a function, substitute \(x\) value of given
+ point into derivative. Solve for \({\frac{dy}{dx}}=0\). Integrate to find
+ original function.
-\noindent \colorbox{cas}{On CAS:}\\
-\indent Evaluate formula,\\
-\indent or Interactive \(\rightarrow\) Calculation
-\(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen}
+ \subsection*{Rates}
-\subsection*{Rates}
+ \subsubsection*{Gradient at a point on parametric curve}
-\subsubsection*{Gradient at a point on parametric curve}
+ \[{\frac{dy}{dx}} = {{\frac{dy}{dt}} \div {\frac{dx}{dt}}} \> \vert \> {\frac{dx}{dt}} \ne 0 \text{ (chain rule)}\]
-\[{\frac{dy}{dx}} = {{\frac{dy}{dt}} \div {\frac{dx}{dt}}} \> \vert \> {\frac{dx}{dt}} \ne 0\]
+ \[\frac{d^2}{dx^2} = \frac{d(y^\prime)}{dx} = {\frac{dy^\prime}{dt} \div {\frac{dx}{dt}}} \> \vert \> y^\prime = {\frac{dy}{dx}}\]
-\[\frac{d^2}{dx^2} = \frac{d(y^\prime)}{dx} = {\frac{dy^\prime}{dt} \div {\frac{dx}{dt}}} \> \vert \> y^\prime = {\frac{dy}{dx}}\]
+ \subsection*{Rational functions}
-\subsection*{Rational functions}
+ \[f(x) = \frac{P(x)}{Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]
-\[f(x) = \frac{P(x)}{Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]
+ \subsection*{Euler's method}
-\subsubsection*{Addition of ordinates}
+ \[\dfrac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
-\begin{itemize}
+ \[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
-\item
- when two graphs have the same ordinate, \(y\)-coordinate is double the
- ordinate
-\item
- when two graphs have opposite ordinates, \(y\)-coordinate is 0 i.e.
- (\(x\)-intercept)
-\item
- when one of the ordinates is 0, the resulting ordinate is equal to the
- other ordinate
-\end{itemize}
+ \begin{theorembox}{}
+ If \(\dfrac{dy}{dx} = g(x)\) with \(x_0 = a\) and \(y_0 = b\), then:
+ \[\begin{cases}
+ x_{n+1} = x_n + h \\
+ y_{n+1} = y_n + hg(x_n)
+ \end{cases}\]
+ \end{theorembox}
-\subsection*{Fundamental theorem of calculus}
+ \[
+ \dfrac{d^2y}{dx^2}
+ \begin{cases}
+ > 0 \implies \text{ underestimate (concave up)} \\
+ < 0 \implies \text{ overestimate (concave down)}
+ \end{cases}
+ \]
+
+ \begin{center}\begin{tikzpicture}
+ \begin{axis}[xmin=0, xmax=1.6, ticks=none, enlargelimits=true, samples=100]
+ \addplot[blue, domain=-0.25:1.5, postaction={decorate,decoration={text along path, text align={align=center, left indent=3cm}, text={|\sffamily|solution curve}}}] {e^(x-3/2)+1/4};
+ \addplot[red] {(x+1/2)*e^(-1)+1/4} (1.7,1.0593) node [above, black] {\(\ell\)};
+ \addplot[mark=*, black] coordinates {(0.5,0.6179)} node[above left]{\((x_0, y_0)\)};
+ \addplot[mark=*, orange] coordinates {(1.4,1.1548)} node[left]{\color{black} \sffamily correct solution};
+ \addplot[mark=*, black] coordinates {(1.4,0.94897)} node[above right] {\((x_1,y_1)\)};
+ \draw [gray, dashed] (0.5,0) -- (0.5,0.6179) -- (1.6,0.6179);
+ \draw [gray, dashed] (1.4,0) -- (1.4, 1.1548);
+ \draw [<->] (0.5,0.48) -- (1.4,0.48) node[midway, fill=white] {\(h\)};
+ \draw [gray, dashed] (1.4,0.94897) -- (1.6,0.94897);
+ \draw [<->] (1.5,0.94897) -- (1.5,0.6179) node[midway, rotate=90, below] {\(hg(x_0)\)};
+ \end{axis}
+ \end{tikzpicture}\end{center}
+
+ \begin{cas}
+ Menu \(\rightarrow\) Sequence \(\rightarrow\) Recursive
+
+ \textbf{To generate \(\boldsymbol{x}\)-values:}
+ \begin{itemize}
+ \item Enter \(a_{n+1}=a_n + h\) where \(h\) is the step size \\
+ (input \(a_n\) from menu bar)
+ \item In \(a_0\), set the initial value \(x_0\) as a constant
+ \end{itemize}
+
+ \textbf{To generate \(\boldsymbol{y}\)-values:}
+ \begin{itemize}
+ \item In \(b_{n+1}\), enter \(\dfrac{dy}{dx}\), replacing \(x\) with \(a_n\)
+ \item Set \(b_0 = y(x_0)\) as a constant
+ \end{itemize}
+
+ To view table of values, tap table icon (top left) \\
+ To compare approximations with actual values, enter in \(c_{n+1} = a_{n+1} - f(a_{n+1})\) where \(f(x) = \int \dfrac{dy}{dx} \> dx\)
+
+ \end{cas}
+
+ \subsection*{Fundamental theorem of calculus}
+
+ If \(f\) is continuous on \([a, b]\), then
+
+ \[\int^b_a f(x) \> dx = F(b) - F(a)\]
+ \hfill where \(F = \int f \> dx\)
+
+ \subsection*{Differential equations}
+
+ \noindent\textbf{Order} - highest power inside derivative\\
+ \textbf{Degree} - highest power of highest derivative\\
+ e.g. \({\left(\dfrac{dy^2}{d^2} x\right)}^3\) \qquad order 2, degree 3
+
+ \begin{warning}
+ To verify solutions, find \(\frac{dy}{dx}\) from \(y\) and substitute into original
+ \end{warning}
+
+ \vspace*{1cm}
+ \hspace*{-1cm}
+
+ { \tabulinesep=1.2mm
+ \begin{tabu}{|c|c|}
+
+ \hline
+ \taburowcolors 2{gray..white}
+ \textbf{DE} & \textbf{Method} \\
+ \hline
+
+ \tabureset
+ \(\dfrac{dy}{dx} = f(x)\)
+ &
+ {\(\begin{aligned}
+ y &= \int f(x) \> dx \\
+ &= F(x) + c \quad \text{where } F^\prime(x) = f(x)
+ \end{aligned}\)} \\
+
+ \hline
+
+ \(\dfrac{d^2y}{dx^2} = f(x)\)
+ &
+ {\(\begin{aligned}
+ \dfrac{dy}{dx} &= \int f(x) \> dx \\
+ &= F(x) + c \quad \text{where } F^\prime(x) = f(x) \\
+ \therefore y &= \iint f(x) \> dx = \int \left( F(x) + c \right) \> dx \\
+ &= G(x) + cx + d \\
+ & \text{where } G^\prime(x) = F(x)
+ \end{aligned}\)} \\
+
+ \hline
+
+ \(\dfrac{dy}{dx} = g(y)\)
+ &
+ {\(\begin{aligned}
+ \dfrac{dx}{dy} &= \dfrac{1}{g(y)} \\
+ \therefore x &= \int \dfrac{1}{g(y)} \> dy \\
+ &= F(y) + c \\
+ & \text{where } F^\prime(y) = \dfrac{1}{g(y)}
+ \end{aligned}\)} \\
+
+ \hline
+
+ \(\dfrac{dy}{dx} = f(x) g(y)\)
+ &
+ {\(\begin{aligned}
+ f(x) &= \dfrac{1}{g(y)} \cdot \dfrac{dy}{dx} \\
+ \int f(x) \> dx &= \int \dfrac{1}{g(y)} \> dy
+ \end{aligned}\)} \\
+
+ \hline
+ \end{tabu}}
+
+ \subsubsection*{Mixing problems}
+
+ \[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\]
-If \(f\) is continuous on \([a, b]\), then
+ \include{calculus-rules}
-\[\int^b_a f(x) \> dx = F(b) - F(a)\]
-\hfill where \(F = \int f \> dx\)
+ \section{Kinematics \& Mechanics}
-\subsection*{Differential equations}
+ \subsection*{Constant acceleration}
-\noindent\textbf{Order} - highest power inside derivative\\
-\textbf{Degree} - highest power of highest derivative\\
-e.g. \({\left(\dfrac{dy^2}{d^2} x\right)}^3\) \qquad order 2, degree 3
+ \begin{itemize}
+ \item \textbf{Position} - relative to origin
+ \item \textbf{Displacement} - relative to starting point
+ \end{itemize}
-\subsubsection*{Verifying solutions}
+ \subsubsection*{Velocity-time graphs}
-Start with \(y=\dots\), and differentiate. Substitute into original
-equation.
+ \begin{description}[nosep, labelindent=0.5cm, leftmargin=0.5\columnwidth]
+ \item[Displacement:] \textit{signed} area
+ \item[Distance travelled:] \textit{total} area
+ \end{description}
-\subsubsection*{Function of the dependent
-variable}
+ \[ \text{acceleration} = \frac{d^2x}{dt^2} = \frac{dv}{dt} = v\frac{dv}{dx} = \frac{d}{dx}\left(\frac{1}{2}v^2\right) \]
-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 \(A\).
+ \begin{center}
+ \renewcommand{\arraystretch}{1}
+ \begin{tabular}{ l r }
+ \hline & no \\ \hline
+ \(v=u+at\) & \(x\) \\
+ \(v^2 = u^2+2as\) & \(t\) \\
+ \(s = \frac{1}{2} (v+u)t\) & \(a\) \\
+ \(s = ut + \frac{1}{2} at^2\) & \(v\) \\
+ \(s = vt- \frac{1}{2} at^2\) & \(u\) \\ \hline
+ \end{tabular}
+ \end{center}
-\subsubsection*{Mixing problems}
+ \[ v_{\text{avg}} = \frac{\Delta\text{position}}{\Delta t} \]
+ \begin{align*}
+ \text{speed} &= |{\text{velocity}}| \\
+ &= \sqrt{v_x^2 + v_y^2 + v_z^2}
+ \end{align*}
+
+ \noindent \textbf{Distance travelled between \(t=a \rightarrow t=b\):}
+ \begin{align*}
+ &= \int^{b}_{a}{\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}} \> dt \tag{2D} \\
+ &= \int^{t=b}_{t=a}{\dfrac{dx}{dt}} \> dt \tag{linear}
+ \end{align*}
-\[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\]
+ \noindent \textbf{Shortest distance between \(\boldsymbol{r}(t_0)\) and \(\boldsymbol{r}(t_1)\):}
+ \[ = |\boldsymbol{r}(t_1) - \boldsymbol{r}(t_2)| \]
-\subsubsection*{Separation of variables}
+ \subsection*{Vector functions}
-If \({\frac{dy}{dx}}=f(x)g(y)\), then:
+ \[ \boldsymbol{r}(t) = x \boldsymbol{i} + y \boldsymbol{j} + z \boldsymbol{k} \]
-\[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\]
+ \begin{itemize}
+ \item If \(\boldsymbol{r}(t) \equiv\) position with time, then the graph of endpoints of \(\boldsymbol{r}(t) \equiv\) Cartesian path
+ \item Domain of \(\boldsymbol{r}(t)\) is the range of \(x(t)\)
+ \item Range of \(\boldsymbol{r}(t)\) is the range of \(y(t)\)
+ \end{itemize}
-\subsubsection*{Euler's method for solving DEs}
+ \subsection*{Vector calculus}
-\[\frac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
+ \subsubsection*{Derivative}
-\[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
+ Let \(\boldsymbol{r}(t)=x(t)\boldsymbol{i} + y(t)\boldsymbol(j)\). If both \(x(t)\) and \(y(t)\) are differentiable, then:
+ \[ \boldsymbol{r}(t)=x(t)\boldsymbol{i}+y(t)\boldsymbol{j} \]
+ \subfile{dynamics}
+ \subfile{statistics}
\end{multicols}
\end{document}