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\title{Year 12 Specialist}
\author{Andrew Lorimer}
\date{2019}

\begin{document}

\pagestyle{fancy}
\fancyhead[LO,LE]{Year 12 Specialist}
\fancyhead[CO,CE]{Andrew Lorimmer}
\maketitle

\section{Complex \& Imaginary Numbers}\label{complex-imaginary-numbers}

\subsection{Imaginary numbers}\label{imaginary-numbers}

\[i^2 = -1 \quad \therefore i = \sqrt {-1}\]

\subsubsection{Simplifying negative
surds}\label{simplifying-negative-surds}

\begin{equation}\begin{split}\sqrt{-2} & = \sqrt{-1 \times 2} \\ & = \sqrt{2}i\end{split}\end{equation}

\subsection{Complex numbers}\label{complex-numbers}

\[\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}\]

General form: \(z=a+bi\)\\
\(\operatorname{Re}(z) = a, \quad \operatorname{Im}(z) = b\)

\subsubsection{Addition}\label{addition}

If \(z_1 = a+bi\) and \(z_2=c+di\), then

\[z_1+z_2 = (a+c)+(b+d)i\]

\subsubsection{Subtraction}\label{subtraction}

If \(z_1=a+bi\) and \(z_2=c+di\), then

\[z_1 - z_2=(a−c)+(b−d)i\]

\subsubsection{Multiplication by a real
constant}\label{multiplication-by-a-real-constant}

If \(z=a+bi\) and \(k \in \mathbb{R}\), then

\[kz=ka+kbi\]

\subsubsection{\texorpdfstring{Powers of
\(i\)}{Powers of i}}\label{powers-of-i}

\begin{itemize}
\tightlist
\item
  \(i^{4n} = 1\)
\item
  \(i^{4n+1} = i\)
\item
  \(i^{4n+2} = -1\)
\item
  \(i^{4n+3} = -i\)
\end{itemize}

For \(i^n\), find remainder \(r\) when \(n \div 4\). Then \(i^n = i^r\).

\subsubsection{Multiplying complex
expressions}\label{multiplying-complex-expressions}

If \(z_1 = a+bi\) and \(z_2=c+di\), then

\[z_1 \times z_2 = (ac-bd)+(ad+bc)i\]

\subsubsection{Conjugates}\label{conjugates}

\[\overline{z} = a \mp bi\]

\subparagraph{Properties}\label{properties}

\begin{itemize}
\tightlist
\item
  \(\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}\)
\item
  \(\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}\)
\item
  \(\overline{kz} = k \overline{z}, \text{ for } k \in \mathbb{R}\)
\item
  \(z \overline{z} = = (a+bi)(a-bi) = a^2+b^2 = |z|^2\)
\item
  \(z + \overline{z} = 2 \operatorname{Re}(z)\)
\end{itemize}

\subsubsection{Modulus}\label{modulus}

Distance from origin.

\[|{z}|=\sqrt{a^2+b^2} \quad  \therefore z \overline{z} = |z|^2\]

Properties

\begin{itemize}
\tightlist
\item
  \(|z_1 z_2| = |z_1| |z_2|\)
\item
  \(|{z_1 \over z_2}| = {|z_1| \over |z_2|}\)
\item
  \(|z_1 + z_2| \le |z_1 + |z_2|\)
\end{itemize}

\subsubsection{Multiplicative inverse}\label{multiplicative-inverse}

\begin{equation}\begin{split}z^{-1} & = {1 \over z} \\ & = {{a-bi} \over {a^2+B^2}} \\ & = {\overline{z} \over {|z|^2}}\end{split}\end{equation}

\subsubsection{Dividing complex numbers}\label{dividing-complex-numbers}

\[{{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}} \quad \text{(multiplicative inverse)}\]

In practice, rationalise denominator:

\[{z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}\]

\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}

\[\cos \theta = {{\boldsymbol{a} \cdot \boldsymbol{b}} \over {|\boldsymbol{a}| |\boldsymbol{b}|}} = {{a_1 b_1 + a_2 b_2} \over {|\boldsymbol{a}| |\boldsymbol{b}|}}\]

\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}

\(b^{m+n}=b^m \cdot b^n\)\\
\((b^m)^n=b^{m \cdot n}\)\\
\((b \cdot c)^n = b^n \cdot c^n\)\\
\({a^m \div a^n} = {a^{m-n}}\)

\subsubsection{\texorpdfstring{\(e\) as a
logarithm}{e as a logarithm}}\label{e-as-a-logarithm}

\[\operatorname{if} y=e^x, \quad \operatorname{then} x=\log_e y\]
\[\ln x = \log_e x\]

\subsubsection{Differentiating
logarithms}\label{differentiating-logarithms}

\[{d(\log_e x)\over dx} = x^{-1} = {1 \over x}\]

\subsection{Derivative rules}\label{derivative-rules}

\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{Reciprocal derivatives}\label{reciprocal-derivatives}

\[{1 \over {dy \over dx}} = {dx \over dy}\]

\subsection{\texorpdfstring{Differentiating
\(x=f(y)\)}{Differentiating x=f(y)}}\label{differentiating-xfy}

Find \(dx \over dy\). Then
\({dx \over dy} = {1 \over {dy \over dx}} \implies {dy \over dx} = {1 \over {dx \over dy}}\).

\[{dy \over dx} = {1 \over {dx \over dy}}\]

\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 \(u=\sin x\)

\textbf{\(m\) and \(n\) are even:}\\
Use identities:

\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}

\subsection{Partial fractions}\label{partial-fractions}

On CAS: Action \(\rightarrow\) Transformation \(\rightarrow\)
\texttt{expand/combine}\\
or Interactive \(\rightarrow\) Transformation \(\rightarrow\)
\texttt{expand} \(\rightarrow\) Partial

\subsection{Graphing integrals on CAS}\label{graphing-integrals-on-cas}

In main: Interactive \(\rightarrow\) Calculation \(\rightarrow\)
\(\int\) (\(\rightarrow\) Definite)\\
Restrictions: \texttt{Define\ f(x)=...} \(\rightarrow\)
\texttt{f(x)\textbar{}x\textgreater{}1} (e.g.)

\subsection{Applications of
antidifferentiation}\label{applications-of-antidifferentiation}

\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}

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{Solids of revolution}\label{solids-of-revolution}

Approximate as sum of infinitesimally-thick cylinders

\subsubsection{\texorpdfstring{Rotation about
\(x\)-axis}{Rotation about x-axis}}\label{rotation-about-x-axis}

\begin{align*}
  V &= \int^{x=b}_{x-a} \pi y^2 \> dx \\
    &= \pi \int^b_a (f(x))^2 \> dx
\end{align*}

\subsubsection{\texorpdfstring{Rotation about
\(y\)-axis}{Rotation about y-axis}}\label{rotation-about-y-axis}

\begin{align*}
  V &= \int^{y=b}_{y=a} \pi x^2 \> dy \\
    &= \pi \int^b_a (f(y))^2 \> dy
\end{align*}

\subsubsection{\texorpdfstring{Regions not bound by
\(y=0\)}{Regions not bound by y=0}}\label{regions-not-bound-by-y0}

\[V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx\]\\
where \(f(x) > g(x)\)

\subsection{Length of a curve}\label{length-of-a-curve}

\[L = \int^b_a \sqrt{1 + ({dy \over dx})^2} \> dx \quad \text{(Cartesian)}\]

\[L = \int^b_a \sqrt{{dx \over dt} + ({dy \over dt})^2} \> dt \quad \text{(parametric)}\]

Evaluate on CAS. Or use Interactive \(\rightarrow\) Calculation
\(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen}.

\subsection{Rates}\label{rates}

\subsubsection{Related rates}\label{related-rates}

\[{da \over db} \quad \text{(change in } a \text{ with respect to } b)\]

\subsubsection{Gradient at a point on parametric
curve}\label{gradient-at-a-point-on-parametric-curve}

\[{dy \over dx} = {{dy \over dt} \div {dx \over dt}} \> \vert \> {dx \over dt} \ne 0\]

\[{d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \div {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}\]

\subsection{Rational functions}\label{rational-functions}

\[f(x) = {P(x) \over Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]

\subsubsection{Addition of ordinates}\label{addition-of-ordinates}

\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}

\subsection{Fundamental theorem of
calculus}\label{fundamental-theorem-of-calculus}

If \(f\) is continuous on \([a, b]\), then

\[\int^b_a f(x) \> dx = F(b) - F(a)\]

where \(F\) is any antiderivative of \(f\)

\subsection{Differential equations}\label{differential-equations}

One or more derivatives

\textbf{Order} - highest power inside derivative\\
\textbf{Degree} - highest power of highest derivative\\
e.g. \({\left(dy^2 \over d^2 x\right)}^3\): order 2, degree 3

\subsubsection{Verifying solutions}\label{verifying-solutions}

Start with \(y=\dots\), and differentiate. Substitute into original
equation.

\subsubsection{Function of the dependent
variable}\label{function-of-the-dependent-variable}

If \({dy \over dx}=g(y)\), then
\({dx \over dy} = 1 \div {dy \over dx} = {1 \over g(y)}\). Integrate
both sides to solve equation. Only add \(c\) on one side. Express
\(e^c\) as \(A\).

\subsubsection{Mixing problems}\label{mixing-problems}

\[\left({dm \over dt}\right)_\Sigma = \left({dm \over dt}\right)_{\text{in}} - \left({dm \over dt}\right)_{\text{out}}\]

\subsubsection{Separation of variables}\label{separation-of-variables}

If \({dy \over dx}=f(x)g(y)\), then:

\[\int f(x) \> dx = \int {1 \over g(y)} \> dy\]

\subsubsection{Using definite integrals to solve
DEs}\label{using-definite-integrals-to-solve-des}

Used for situations where solutions to \({dy \over dx} = f(x)\) is not
required.

In some cases, it may not be possible to obtain an exact solution.

Approximate solutions can be found by numerically evaluating a definite
integral.

\subsubsection{Using Euler's method to solve a differential
equation}\label{using-eulers-method-to-solve-a-differential-equation}

\[{{f(x+h) - f(x)} \over h } \approx f^\prime (x) \quad \text{for small } h\]

\[\implies f(x+h) \approx f(x) + hf^\prime(x)\]

\end{document}