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-\title{Year 12 Specialist}
-\author{Andrew Lorimer}
-\date{2019}
-
-\begin{document}
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+\fancypagestyle{plain}{\fancyhead[LO,LE]{} \fancyhead[CO,CE]{}} % rm title & author for first page
\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:
-
-\[{z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}\]
-
-\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}
-
-\[(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}\]
-
-\subsection{Roots of complex numbers}\label{roots-of-complex-numbers}
-
-\(n\)th roots of \(z = r \operatorname{cis} \theta\) are
-
-\[z={r^{1 \over n}} \operatorname{cis}({{\theta + 2 k \pi} \over n})\]
-
-Same modulus for all solutions. Arguments are separated by
-\({2 \pi} \over n\)
-
-The solutions of \(z^n=a \text{ where } a \in \mathbb{C}\) lie on circle
-
-\[x^2 + y^2 = (|a|^{1 \over n})^2\]
-
-\subsection{Sketching complex graphs}\label{sketching-complex-graphs}
-
-\subsubsection{Straight line}\label{straight-line}
-
-\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
- \(|z+a|=|z+b| \longrightarrow 2(a-b)x=b^2-a^2\)
-\end{itemize}
-
-\subsubsection{Circle}\label{circle}
-
-\(|z-z_1|^2 = c^2 |z_2+2|^2\) or \(|z-(a + bi)| = c\)
-
-\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
-
-\[\boldsymbol{u} || \boldsymbol{v} \iff \boldsymbol{u} = k \boldsymbol{v} \text{ where } k \in \mathbb{R} \setminus \{0\}\]
-
-\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
-\(|u|=\sqrt{x^2+y^2}\)
-
-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{\texorpdfstring{Unit vector
-\(|\hat{\boldsymbol{a}}|=1\)}{Unit vector \textbar{}\textbackslash{}hat\{\textbackslash{}boldsymbol\{a\}\}\textbar{}=1}}\label{unit-vector-hatboldsymbola1}
-
-\begin{equation}\begin{split}\hat{\boldsymbol{a}} & = {1 \over {|\boldsymbol{a}|}}\boldsymbol{a} \\ & = \boldsymbol{a} \cdot {|\boldsymbol{a}|}\end{split}\end{equation}
-
-\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}
-
-\[\boldsymbol{a} \cdot \boldsymbol{b} = a_1 b_1 + a_2 b_2\]
-
-\textbf{on CAS:} \texttt{dotP({[}a\ b\ c{]},\ {[}d\ e\ f{]})}
-
-\subsection{Scalar product properties}\label{scalar-product-properties}
-
-\begin{enumerate}
-\def\labelenumi{\arabic{enumi}.}
-\tightlist
-\item
- \(k(\boldsymbol{a\cdot b})=(k\boldsymbol{a})\cdot \boldsymbol{b}=\boldsymbol{a}\cdot (k{b})\)
-\item
- \(\boldsymbol{a \cdot 0}=0\)
-\item
- \(\boldsymbol{a \cdot (b + c)}=\boldsymbol{a \cdot b + a \cdot c}\)
-\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
-\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}
-
-If \(\boldsymbol{a} \cdot \boldsymbol{b} = 0\), then
-\(\boldsymbol{a} \perp \boldsymbol{b}\) (since \(\cos 90 = 0\))
-
-\subsection{Finding angle between
-vectors}\label{finding-angle-between-vectors}
-
-\textbf{positive direction}
+\fancyhead[CO,CE]{Andrew Lorimer}
-\[\cos \theta = {{\boldsymbol{a} \cdot \boldsymbol{b}} \over {|\boldsymbol{a}| |\boldsymbol{b}|}} = {{a_1 b_1 + a_2 b_2} \over {|\boldsymbol{a}| |\boldsymbol{b}|}}\]
-
-\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
-\(x, y, z\) axes:
-\[\cos \alpha = {a_1 \over |\boldsymbol{a}|}, \quad \cos \beta = {a_2 \over |\boldsymbol{a}|}, \quad \cos \gamma = {a_3 \over |\boldsymbol{a}|}\]
-
-\textbf{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}
-
-Vector resolute of \(\boldsymbol{a}\) in direction of \(\boldsymbol{b}\)
-is magnitude of \(\boldsymbol{a}\) in direction of \(\boldsymbol{b}\):
-
-\[\boldsymbol{u}={{\boldsymbol{a}\cdot\boldsymbol{b}}\over |\boldsymbol{b}|^2}\boldsymbol{b}=\left({\boldsymbol{a}\cdot{\boldsymbol{b} \over |\boldsymbol{b}|}}\right)\left({\boldsymbol{b} \over |\boldsymbol{b}|}\right)=(\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}\]
-
-\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}
-
-\[r_s = |\boldsymbol{u}| = \boldsymbol{a} \cdot \hat{\boldsymbol{b}}\]
-
-\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}
-
-\[\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u} \> \text{ where } \boldsymbol{u} \text{ is projection } \boldsymbol{a} \text{ on } \boldsymbol{b}\]
-
-\subsection{Vector proofs}\label{vector-proofs}
-
-\subsubsection{Concurrent lines}\label{concurrent-lines}
-
-\(\ge\) 3 lines intersect at a single point
-
-\subsubsection{Collinear points}\label{collinear-points}
-
-\(\ge\) 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
-\(\vec{AC}=m\vec{AB} \text{ where } m \ne 0\)
-
-\subsubsection{Useful vector properties}\label{useful-vector-properties}
-
-\begin{itemize}
-\tightlist
-\item
- If \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are parallel, then
- \(\boldsymbol{b}=k\boldsymbol{a}\) 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
- Two vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are
- perpendicular if \(\boldsymbol{a} \cdot \boldsymbol{b}=0\)
-\item
- \(\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2\)
-\end{itemize}
-
-\subsection{Linear dependence}\label{linear-dependence}
-
-Vectors \(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\) are linearly
-dependent if they are non-parallel and:
-
-\[k\boldsymbol{a}+l\boldsymbol{b}+m\boldsymbol{c} = 0\]
-\[\therefore \boldsymbol{c} = m\boldsymbol{a} + n\boldsymbol{b} \quad \text{(simultaneous)}\]
-
-\(\boldsymbol{a}, \boldsymbol{b},\) 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.
-
-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}
-
-Right-hand rule for axes: \(z\) is up or out of page.
-
-i\includegraphics{graphics/vectors-3d.png}
-
-\subsection{Parametric vectors}\label{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}
-
-Period of \(a\sin(bx)\) is \({2\pi} \over b\)
-
-Period of \(a\tan(nx)\) is \(\pi \over n\)\\
-Asymptotes at \(x={2k+1)\pi \over 2n} \> \vert \> k \in \mathbb{Z}\)
-
-\subsection{Reciprocal functions}\label{reciprocal-functions}
-
-\subsubsection{Cosecant}\label{cosecant}
-
-\begin{figure}
-\centering
-\includegraphics{graphics/csc.png}
-\caption{}
-\end{figure}
-
-\[\operatorname{cosec} \theta = {1 \over \sin \theta} \> \vert \> \sin \theta \ne 0\]
-
-\begin{itemize}
-\tightlist
-\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
- \(\theta = {{(2n + 1)\pi} \over 2} \> \vert \> n \in \mathbb{Z}\)
-\item
- \textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
-\end{itemize}
-
-\subsubsection{Secant}\label{secant}
-
-\begin{figure}
-\centering
-\includegraphics{graphics/sec.png}
-\caption{}
-\end{figure}
-
-\[\operatorname{sec} \theta = {1 \over \cos \theta} \> \vert \> \cos \theta \ne 0\]
-
-\begin{itemize}
-\tightlist
-\item
- \textbf{Domain}
- \(= \mathbb{R} \setminus \{{{(2n + 1) \pi} \over 2 } : n \in \mathbb{Z}\}\)
-\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
- \(\theta = {{(2n + 1) \pi} \over 2} \> \vert \> n \in \mathbb{Z}\)
-\end{itemize}
-
-\subsubsection{Cotangent}\label{cotangent}
-
-\begin{figure}
-\centering
-\includegraphics{graphics/cot.png}
-\caption{}
-\end{figure}
-
-\[\operatorname{cot} \theta = {{\cos \theta} \over {\sin \theta}} \> \vert \> \sin \theta \ne 0\]
-
-\begin{itemize}
-\tightlist
-\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}\label{symmetry-properties}
-
-\begin{equation}\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}\end{equation}
-
-\subsubsection{Complementary properties}\label{complementary-properties}
-
-\begin{equation}\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}\end{equation}
-
-\subsubsection{Pythagorean identities}\label{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}
-
-\[\cos(x \pm y) = \cos x + \cos y \mp \sin x \sin y\]
-\[\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\]
-\[\tan(x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x \tan y}}\]
-
-\subsection{Double angle formulas}\label{double-angle-formulas}
-
-\begin{equation}\begin{split}
- \cos 2x &= \cos^2 x - \sin^2 x \\
- & = 1 - 2\sin^2 x \\
- & = 2 \cos^2 x -1
-\end{split}\end{equation}
-
-\[\sin 2x = 2 \sin x \cos x\]
-
-\[\tan 2x = {{2 \tan x} \over {1 - \tan^2 x}}\]
-
-\subsection{Inverse circular
-functions}\label{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}
-
-\[\sin^{-1}: [-1, 1] \rightarrow \mathbb{R}, \quad \sin^{-1} x = y, \quad \text{where } \sin y = x \text{ and } y \in [{-\pi \over 2}, {\pi \over 2}]\]
-
-\subsubsection{\texorpdfstring{\(\arcos\)}{\textbackslash{}arcos}}\label{arcos}
-
-\[\cos^{-1} \rightarrow \mathbb{R}, \quad \cos^{-1} x = y, \quad \text{where } \cos y = x \text{ and } y \in [0, \pi]\]
-
-\subsubsection{\texorpdfstring{\(\arctan\)}{\textbackslash{}arctan}}\label{arctan}
-
-\[\tan^{-1}: \mathbb{R} \rightarrow \mathbb{R}, \quad \tan^{-1} x = y, \quad \text{where } \tan y = x \text{ and } y \in \left(-{\pi \over 2}, {\pi \over 2}\right)\]
-\# Differential calculus
-
-\subsection{Limits}\label{limits}
-
-\[\lim_{x \rightarrow a}f(x)\]
-
-\(L^-\) - limit from below
-
-\(L^+\) - limit from above
-
-\(\lim_{x \to a} f(x)\) - limit of a point
-
-\begin{itemize}
-\tightlist
-\item
- Limit exists if \(L^-=L^+\)
-\item
- If limit exists, point does not.
-\end{itemize}
-
-Limits can be solved using normal techniques (if div 0, factorise)
-
-\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
- \({\lim_{x \rightarrow a} {f(x) \over g(x)}} = {F \over G}, G \ne 0\)
-\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}
-
-A function is continuous if \(L^-=L^+=f(x)\) for all values of \(x\).
-
-\subsection{Gradients of secants and
-tangents}\label{gradients-of-secants-and-tangents}
-
-Secant (chord) - line joining two points on curve
-
-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
-\({m_{PQ}}={\operatorname{rise} \over \operatorname{run}} = {\delta y \over \delta x}\)
-
-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}
-
-\[{d(ax^n) \over dx}=anx^{n-1}\]
-
-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}
-
-\(\log_b (xy)=\log_b x + \log_b y\)\\
-\(\log_b x^n = n \log_b x\)\\
-\(\log_b y^{x^n} = x^n \log_b y\)
-
-\subsubsection{Index identities}\label{index-identities}
+\renewcommand\hphantom[1]{{\color[gray]{.6}#1}} % comment out!
+\newcommand*\leftlap[3][\,]{#1\hphantom{#2}\mathllap{#3}}
+\newcommand*\rightlap[2]{\mathrlap{#2}\hphantom{#1}}
+\linespread{1.5}
+\setlength{\parindent}{0pt}
+\setlength\fboxsep{0pt} \setlength\fboxrule{.2pt} % for the \fboxes
-\(b^{m+n}=b^m \cdot b^n\)\\
-\((b^m)^n=b^{m \cdot n}\)\\
-\((b \cdot c)^n = b^n \cdot c^n\)\\
-\({a^m \div a^n} = {a^{m-n}}\)
+\newcolumntype{L}[1]{>{\hsize=#1\hsize\raggedright\arraybackslash}X}%
+\newcolumntype{R}[1]{>{\hsize=#1\hsize\raggedleft\arraybackslash}X}%
+\newcolumntype{Y}{>{\centering\arraybackslash}X}
-\subsubsection{\texorpdfstring{\(e\) as a
-logarithm}{e as a logarithm}}\label{e-as-a-logarithm}
+\definecolor{cas}{HTML}{cde1fd}
+\definecolor{important}{HTML}{fc9871}
+\definecolor{dark-gray}{gray}{0.2}
+\definecolor{light-gray}{HTML}{cccccc}
+\definecolor{peach}{HTML}{e6beb2}
+\definecolor{lblue}{HTML}{e5e9f0}
-\[\operatorname{if} y=e^x, \quad \operatorname{then} x=\log_e y\]
-\[\ln x = \log_e x\]
+\newcommand{\tg}{\mathop{\mathrm{tg}}}
+\newcommand{\cotg}{\mathop{\mathrm{cotg}}}
+\newcommand{\arctg}{\mathop{\mathrm{arctg}}}
+\newcommand{\arccotg}{\mathop{\mathrm{arccotg}}}
-\subsubsection{Differentiating
-logarithms}\label{differentiating-logarithms}
+\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}
-\[{d(\log_e x)\over dx} = x^{-1} = {1 \over x}\]
+\begin{document}
-\subsection{Derivative rules}\label{derivative-rules}
+\title{\vspace{-22mm}Year 12 Specialist\vspace{-4mm}}
+\author{Andrew Lorimer}
+\date{}
+\maketitle
+\vspace{-9mm}
+\begin{multicols}{2}
+
+ \section{Complex numbers}
+
+ \[\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}\]
+ \begin{align*}
+ \text{Cartesian form: } & a+bi\\
+ \text{Polar form: } & r\operatorname{cis}\theta
+ \end{align*}
-\begin{longtable}[]{@{}ll@{}}
-\toprule
-\(f(x)\) & \(f^\prime(x)\)\tabularnewline
-\midrule
-\endhead
-\(\sin x\) & \(\cos x\)\tabularnewline
-\(\sin ax\) & \(a\cos ax\)\tabularnewline
-\(\cos x\) & \(-\sin x\)\tabularnewline
-\(\cos ax\) & \(-a \sin ax\)\tabularnewline
-\(\tan f(x)\) & \(f^2(x) \sec^2f(x)\)\tabularnewline
-\(e^x\) & \(e^x\)\tabularnewline
-\(e^{ax}\) & \(ae^{ax}\)\tabularnewline
-\(ax^{nx}\) & \(an \cdot e^{nx}\)\tabularnewline
-\(\log_e x\) & \(1 \over x\)\tabularnewline
-\(\log_e {ax}\) & \(1 \over x\)\tabularnewline
-\(\log_e f(x)\) & \(f^\prime (x) \over f(x)\)\tabularnewline
-\(\sin(f(x))\) & \(f^\prime(x) \cdot \cos(f(x))\)\tabularnewline
-\(\sin^{-1} x\) & \(1 \over {\sqrt{1-x^2}}\)\tabularnewline
-\(\cos^{-1} x\) & \(-1 \over {sqrt{1-x^2}}\)\tabularnewline
-\(\tan^{-1} x\) & \(1 \over {1 + x^2}\)\tabularnewline
-\bottomrule
-\end{longtable}
+ \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*}
+ \overline{z} &= a \mp bi\\
+ &= r \operatorname{cis}(-\theta)
+ \end{align*}
+
+ \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} \> \forall \> k \in \mathbb{R}\\
+ z\overline{z} &= (a+bi)(a-bi)\\
+ &= a^2 + b^2\\
+ &= |z|^2
+ \end{align*}
+
+ \subsection*{Modulus}
+
+ \[|z|=|\vec{Oz}|=\sqrt{a^2 + b^2}\]
+
+ \subsubsection*{Properties}
-\subsection{Reciprocal derivatives}\label{reciprocal-derivatives}
+ \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}\\
+ & \text{then rationalise denominator}
+ \end{align*}
+
+ \subsection*{Polar form}
+
+ \[ r \operatorname{cis} \theta = r\left( \cos \theta + i \sin \theta \right) \]
+
+ \begin{itemize}
+ \item{\(r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}\)}
+ \item{\(\theta = \operatorname{arg}(z)\) \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}
+
+ \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 & \(\begin{aligned}
+ z^2 + a^2 &= z^2-(ai)^2\\
+ &= (z+ai)(z-ai) \end{aligned}\) \\
+ \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)\\
+ \hline
+ \end{tabularx}
+
+ \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}
+
+ \subsection*{\(n\)th 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 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:
+
+ \[z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
+
+ \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*{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}
+
+
+ \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}
+ \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}
+
+ \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}
+
+ \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}
+
+ \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}
+
+ \subsubsection*{Cosecant}
+
+ \[\operatorname{cosec} \theta = \frac{1}{\sin \theta} \> \vert \> \sin \theta \ne 0\]
+
+ \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}
+
+ \subsubsection*{Secant}
+
+\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}
+
+ \[\operatorname{sec} \theta = \frac{1}{\cos \theta} \> \vert \> \cos \theta \ne 0\]
+
+ \begin{itemize}
+
+ \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}
+
+ \subsubsection*{Cotangent}
+
+\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}
+
+ \[\operatorname{cot} \theta = {{\cos \theta} \over {\sin \theta}} \> \vert \> \sin \theta \ne 0\]
+
+ \begin{itemize}
+
+ \item
+ \textbf{Domain} \(= \mathbb{R} \setminus \{n \pi: n \in \mathbb{Z}\}\)
+ \item
+ \textbf{Range} \(= \mathbb{R}\)
+ \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}
+
+ \[\begin{split}
+ 1 + \operatorname{cot}^2 x & = \operatorname{cosec}^2 x, \quad \text{where } \sin x \ne 0 \\
+ 1 + \tan^2 x & = \operatorname{sec}^2 x, \quad \text{where } \cos x \ne 0
+ \end{split}\]
+
+ \subsection*{Compound angle formulas}
+
+ \[\cos(x \pm y) = \cos x + \cos y \mp \sin x \sin y\]
+ \[\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\]
+ \[\tan(x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x \tan y}}\]
+
+ \subsection*{Double angle formulas}
+
+ \[\begin{split}
+ \cos 2x &= \cos^2 x - \sin^2 x \\
+ & = 1 - 2\sin^2 x \\
+ & = 2 \cos^2 x -1
+ \end{split}\]
+
+ \[\sin 2x = 2 \sin x \cos x\]
+
+ \[\tan 2x = {{2 \tan x} \over {1 - \tan^2 x}}\]
+
+ \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 \(\cos y = x, \> y \in [0, \pi]\)
+
+ \[\tan^{-1}: \mathbb{R} \rightarrow \mathbb{R}, \quad \tan^{-1} x = y\]
+ \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}
+
+ \subsection*{Mensuration}
+
+ \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*}
-\[{1 \over {dy \over dx}} = {dx \over dy}\]
+ \subsection*{Integration by substitution}
-\subsection{\texorpdfstring{Differentiating
-\(x=f(y)\)}{Differentiating x=f(y)}}\label{differentiating-xfy}
+ \[\int f(u) {\frac{du}{dx}} \cdot dx = \int f(u) \cdot du\]
-Find \(dx \over dy\). Then
-\({dx \over dy} = {1 \over {dy \over dx}} \implies {dy \over dx} = {1 \over {dx \over dy}}\).
+ \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*}
-\[{dy \over dx} = {1 \over {dx \over dy}}\]
-
-\subsection{Second derivative}\label{second-derivative}
-
-\[f(x) \longrightarrow f^\prime (x) \longrightarrow f^{\prime\prime}(x)\]
-
-\[\therefore y \longrightarrow {dy \over dx} \longrightarrow {d({dy \over dx}) \over dx} \longrightarrow {d^2 y \over dx^2}\]
-
-Order of polynomial \(n\)th derivative decrements each time the
-derivative is taken
-
-\subsubsection{Points of Inflection}\label{points-of-inflection}
-
-\emph{Stationary point} - point of zero gradient (i.e.
-\(f^\prime(x)=0\))\\
-\emph{Point of inflection} - point of maximum \(|\)gradient\(|\) (i.e.
-\(f^{\prime\prime} = 0\))
-
-\begin{itemize}
-\tightlist
-\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{figure}
-\centering
-\includegraphics{graphics/second-derivatives.png}
-\caption{}
-\end{figure}
-
-\subsection{Implicit Differentiation}\label{implicit-differentiation}
-
-\textbf{On CAS:} Action \(\rightarrow\) Calculation \(\rightarrow\)
-\texttt{impDiff(y\^{}2+ax=5,\ x,\ y)}. Returns \(y^\prime= \dots\).
-
-Used for differentiating circles etc.
-
-If \(p\) and \(q\) are expressions in \(x\) and \(y\) such that \(p=q\),
-for all \(x\) nd \(y\), then:
-
-\[{dp \over dx} = {dq \over dx} \quad \text{and} \quad {dp \over dy} = {dq \over dy}\]
-
-\subsection{Integration}\label{integration}
-
-\[\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)\]
-
-\[\int x^n \cdot dx = {x^{n+1} \over n+1} + c\]
-
-\begin{itemize}
-\tightlist
-\item
- area enclosed by curves
-\item
- \(+c\) should be shown on each step without \(\int\)
-\end{itemize}
-
-\subsubsection{Integral laws}\label{integral-laws}
-
-\(\int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx\)\\
-\(\int k f(x) dx = k \int f(x) dx\)
-
-\begin{longtable}[]{@{}ll@{}}
-\toprule
-\begin{minipage}[b]{0.42\columnwidth}\raggedright\strut
-\(f(x)\)\strut
-\end{minipage} & \begin{minipage}[b]{0.38\columnwidth}\raggedright\strut
-\(\int f(x) \cdot dx\)\strut
-\end{minipage}\tabularnewline
-\midrule
-\endhead
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\(k\) (constant)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\(kx + c\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\(x^n\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\({x^{n+1} \over {n+1}} + c\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\(a x^{-n}\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\(a \cdot \log_e x + c\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\({1 \over {ax+b}}\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\({1 \over a} \log_e (ax+b) + c\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\((ax+b)^n\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\({1 \over {a(n+1)}}(ax+b)^{n-1} + c\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\(e^{kx}\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\({1 \over k} e^{kx} + c\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\(e^k\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\(e^kx + c\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\(\sin kx\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\(-{1 \over k} \cos (kx) + c\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\(\cos kx\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\({1 \over k} \sin (kx) + c\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\(\sec^2 kx\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\({1 \over k} \tan(kx) + c\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\(1 \over \sqrt{a^2-x^2}\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\(\sin^{-1} {x \over a} + c \>\vert\> a>0\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\(-1 \over \sqrt{a^2-x^2}\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\(\cos^{-1} {x \over a} + c \>\vert\> a>0\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\(a \over {a^2-x^2}\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\(\tan^{-1} {x \over a} + c\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\({f^\prime (x)} \over {f(x)}\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\(\log_e f(x) + c\)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\(g^\prime(x)\cdot f^\prime(g(x)\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\(f(g(x))\) (chain rule)\strut
-\end{minipage}\tabularnewline
-\begin{minipage}[t]{0.42\columnwidth}\raggedright\strut
-\(f(x) \cdot g(x)\)\strut
-\end{minipage} & \begin{minipage}[t]{0.38\columnwidth}\raggedright\strut
-\(\int [f^\prime(x) \cdot g(x)] dx + \int [g^\prime(x) f(x)] dx\)\strut
-\end{minipage}\tabularnewline
-\bottomrule
-\end{longtable}
-
-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}
-
-\[\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)\]
-
-\begin{itemize}
-\tightlist
-\item
- Signed area enclosed by:
- \(\> y=f(x), \quad y=0, \quad x=a, \quad x=b\).
-\item
- \emph{Integrand} is \(f\).
-\item
- \(F(x)\) may be any integral, i.e. \(c\) is inconsequential
-\end{itemize}
-
-\paragraph{Properties}\label{properties-2}
-
-\[\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx\]
-
-\[\int^a_a f(x) \> dx = 0\]
-
-\[\int^b_a k \cdot f(x) \> dx = k \int^b_a f(x) \> dx\]
-
-\[\int^b_a f(x) \pm g(x) \> dx = \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx\]
-
-\[\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx\]
-
-\subsubsection{Integration by
-substitution}\label{integration-by-substitution}
-
-\[\int f(u) {du \over dx} \cdot dx = \int f(u) \cdot du\]
-
-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 \(u=x+4\)\\
-\(\implies {du \over dx} = 1\)\\
-\(\implies x = u - 4\)\\
-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}
-
-For \(\int^b_a f(x) {du \over dx} \cdot dx\), evaluate new \(a\) and
-\(b\) for \(f(u) \cdot du\).
-
-\subsubsection{Trigonometric
-integration}\label{trigonometric-integration}
-
-\[\sin^m x \cos^n x \cdot dx\]
-
-\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 \(u=\cos x\)
-
-\textbf{\(n\) is odd:}\\
-\(n=2k+1\) where \(k \in \mathbb{Z}\)\\
-\(\implies \cos^{2k+1} x = (\cos^2 x)^k \cos x = (1-\sin^2 x)^k \cos x\)\\
-Subbstitute \(u=\sin x\)
+ \subsubsection*{Definite integrals by substitution}
-\textbf{\(m\) and \(n\) are even:}\\
-Use identities:
+ For \(\int^b_a f(x) {\frac{du}{dx}} \cdot dx\), evaluate new \(a\) and
+ \(b\) for \(f(u) \cdot du\).
-\begin{itemize}
-\tightlist
-\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}
+ \subsubsection*{Trigonometric integration}
-\subsection{Partial fractions}\label{partial-fractions}
+ \[\sin^m x \cos^n x \cdot dx\]
-On CAS: Action \(\rightarrow\) Transformation \(\rightarrow\)
-\texttt{expand/combine}\\
-or Interactive \(\rightarrow\) Transformation \(\rightarrow\)
-\texttt{expand} \(\rightarrow\) Partial
+ \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 \(u=\cos x\)
-\subsection{Graphing integrals on CAS}\label{graphing-integrals-on-cas}
+ \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 \(u=\sin x\)
-In main: Interactive \(\rightarrow\) Calculation \(\rightarrow\)
-\(\int\) (\(\rightarrow\) Definite)\\
-Restrictions: \texttt{Define\ f(x)=...} \(\rightarrow\)
-\texttt{f(x)\textbar{}x\textgreater{}1} (e.g.)
+ \paragraph{\textbf{\(m\) and \(n\) are even:}}
+ use identities...
-\subsection{Applications of
-antidifferentiation}\label{applications-of-antidifferentiation}
+ \begin{itemize}
-\begin{itemize}
-\tightlist
-\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}
+ \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 find stationary points of a function, substitute \(x\) value of given
-point into derivative. Solve for \({dy \over dx}=0\). Integrate to find
-original function.
+ \subsection*{Separation of variables}
-\subsection{Solids of revolution}\label{solids-of-revolution}
+ If \({\frac{dy}{dx}}=f(x)g(y)\), then:
-Approximate as sum of infinitesimally-thick cylinders
+ \[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\]
-\subsubsection{\texorpdfstring{Rotation about
-\(x\)-axis}{Rotation about x-axis}}\label{rotation-about-x-axis}
+ \subsection*{Partial fractions}
-\begin{align*}
- V &= \int^{x=b}_{x-a} \pi y^2 \> dx \\
- &= \pi \int^b_a (f(x))^2 \> dx
-\end{align*}
+ 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*}
-\subsubsection{\texorpdfstring{Rotation about
-\(y\)-axis}{Rotation about y-axis}}\label{rotation-about-y-axis}
+ \subsubsection*{Repeated linear factors}
-\begin{align*}
- V &= \int^{y=b}_{y=a} \pi x^2 \> dy \\
- &= \pi \int^b_a (f(y))^2 \> dy
-\end{align*}
+ \[ \dfrac{p(x)}{(x-a)^n} = \dfrac{A_1}{(x-a)} + \dfrac{A_2}{(x-a)^2} + \dots + \dfrac{A_n}{(x-a)^n} \]
-\subsubsection{\texorpdfstring{Regions not bound by
-\(y=0\)}{Regions not bound by y=0}}\label{regions-not-bound-by-y0}
+ \subsubsection*{Irreducible quadratic factors}
-\[V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx\]\\
-where \(f(x) > g(x)\)
+ \[ \text{e.g. } \dfrac{3x-4}{(2x-3)(x^2+5)} = \dfrac{A}{2x-3} + \dfrac{Bx+C}{x^2+5} \]
-\subsection{Length of a curve}\label{length-of-a-curve}
+ \begin{cas}
+ Action \(\rightarrow\) Transformation:\\
+ \-\hspace{1em} \texttt{expand(..., x)}
-\[L = \int^b_a \sqrt{1 + ({dy \over dx})^2} \> dx \quad \text{(Cartesian)}\]
+ To reverse, use \texttt{combine(...)}
+ \end{cas}
-\[L = \int^b_a \sqrt{{dx \over dt} + ({dy \over dt})^2} \> dt \quad \text{(parametric)}\]
+ \subsection*{Integrating \(\boldsymbol{\dfrac{dy}{dx} = g(y)}\)}
-Evaluate on CAS. Or use Interactive \(\rightarrow\) Calculation
-\(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen}.
+ \[ \text{if } \dfrac{dy}{dx} = g(y), \text{ then } x = \int{\dfrac{1}{g(y)}} \> dy \]
-\subsection{Rates}\label{rates}
+ \subsection*{Graphing integrals on CAS}
-\subsubsection{Related rates}\label{related-rates}
+ \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}
-\[{da \over db} \quad \text{(change in } a \text{ with respect to } b)\]
+ \subsection*{Solids of revolution}
-\subsubsection{Gradient at a point on parametric
-curve}\label{gradient-at-a-point-on-parametric-curve}
+ Approximate as sum of infinitesimally-thick cylinders
-\[{dy \over dx} = {{dy \over dt} \div {dx \over dt}} \> \vert \> {dx \over dt} \ne 0\]
+ \subsubsection*{Rotation about \(\boldsymbol{x}\)-axis}
-\[{d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \div {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}\]
+ \[ V = \pi\int^{x=b}_{x=a} f(x)^2 \> dx \]
-\subsection{Rational functions}\label{rational-functions}
+ \subsubsection*{Rotation about \(\boldsymbol{y}\)-axis}
-\[f(x) = {P(x) \over Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]
+ \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{Addition of ordinates}\label{addition-of-ordinates}
+ \subsubsection*{Rotating the area between two graphs}
-\begin{itemize}
-\tightlist
-\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}
+ \[V = \pi \int^b_a \left( f(x)^2 - g(x)^2 \right) \> dx\]
+ \hfill where \(f(x) > g(x)\)
-\subsection{Fundamental theorem of
-calculus}\label{fundamental-theorem-of-calculus}
+ \subsection*{Length of a curve}
-If \(f\) is continuous on \([a, b]\), then
+ 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*}
-\[\int^b_a f(x) \> dx = F(b) - F(a)\]
+ \begin{cas}
+ \begin{enumerate}[label=\alph*), leftmargin=5mm]
+ \item Evaluate formula
+ \item Interactive \(\rightarrow\) Calculation \(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen}
+ \end{enumerate}
+ \end{cas}
-where \(F\) is any antiderivative of \(f\)
+ \subsection*{Applications of antidifferentiation}
-\subsection{Differential equations}\label{differential-equations}
+ \begin{itemize}
-One or more derivatives
+ \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}
-\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
+ 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.
-\subsubsection{Verifying solutions}\label{verifying-solutions}
+ \subsection*{Rates}
-Start with \(y=\dots\), and differentiate. Substitute into original
-equation.
+ \subsubsection*{Gradient at a point on parametric curve}
-\subsubsection{Function of the dependent
-variable}\label{function-of-the-dependent-variable}
+ \[{\frac{dy}{dx}} = {{\frac{dy}{dt}} \div {\frac{dx}{dt}}} \> \vert \> {\frac{dx}{dt}} \ne 0 \text{ (chain rule)}\]
-If \({dy \over dx}=g(y)\), then
-\({dx \over dy} = 1 \div {dy \over dx} = {1 \over g(y)}\). Integrate
-both sides to solve equation. Only add \(c\) on one side. Express
-\(e^c\) as \(A\).
+ \[\frac{d^2}{dx^2} = \frac{d(y^\prime)}{dx} = {\frac{dy^\prime}{dt} \div {\frac{dx}{dt}}} \> \vert \> y^\prime = {\frac{dy}{dx}}\]
-\subsubsection{Mixing problems}\label{mixing-problems}
+ \subsection*{Rational functions}
-\[\left({dm \over dt}\right)_\Sigma = \left({dm \over dt}\right)_{\text{in}} - \left({dm \over dt}\right)_{\text{out}}\]
+ \[f(x) = \frac{P(x)}{Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]
-\subsubsection{Separation of variables}\label{separation-of-variables}
+ \subsection*{Euler's method}
-If \({dy \over dx}=f(x)g(y)\), then:
+ \[\dfrac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
-\[\int f(x) \> dx = \int {1 \over g(y)} \> dy\]
+ \[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
-\subsubsection{Using definite integrals to solve
-DEs}\label{using-definite-integrals-to-solve-des}
+ \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}
-Used for situations where solutions to \({dy \over dx} = f(x)\) is not
-required.
+ \[
+ \dfrac{d^2y}{dx^2}
+ \begin{cases}
+ > 0 \implies \text{ underestimate (concave up)} \\
+ < 0 \implies \text{ overestimate (concave down)}
+ \end{cases}
+ \]
+
+ \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}
+
+ \[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\]
+
+ \include{calculus-rules}
+
+ \section{Kinematics \& Mechanics}
+
+ \subsection*{Constant acceleration}
+
+ \begin{itemize}
+ \item \textbf{Position} - relative to origin
+ \item \textbf{Displacement} - relative to starting point
+ \end{itemize}
+
+ \subsubsection*{Velocity-time graphs}
-In some cases, it may not be possible to obtain an exact solution.
+ \begin{description}[nosep, labelindent=0.5cm, leftmargin=0.5\columnwidth]
+ \item[Displacement:] \textit{signed} area
+ \item[Distance travelled:] \textit{total} area
+ \end{description}
-Approximate solutions can be found by numerically evaluating a definite
-integral.
+ \[ \text{acceleration} = \frac{d^2x}{dt^2} = \frac{dv}{dt} = v\frac{dv}{dx} = \frac{d}{dx}\left(\frac{1}{2}v^2\right) \]
-\subsubsection{Using Euler's method to solve a differential
-equation}\label{using-eulers-method-to-solve-a-differential-equation}
+ \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}
+
+ \[ v_{\text{avg}} = \frac{\Delta\text{position}}{\Delta t} \]
+ \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*}
+
+ \noindent \textbf{Shortest distance between \(\boldsymbol{r}(t_0)\) and \(\boldsymbol{r}(t_1)\):}
+ \[ = |\boldsymbol{r}(t_1) - \boldsymbol{r}(t_2)| \]
+
+ \subsection*{Vector functions}
+
+ \[ \boldsymbol{r}(t) = x \boldsymbol{i} + y \boldsymbol{j} + z \boldsymbol{k} \]
+
+ \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}
-\[{{f(x+h) - f(x)} \over h } \approx f^\prime (x) \quad \text{for small } h\]
+ \subsection*{Vector calculus}
-\[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
+ \subsubsection*{Derivative}
+ 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}