From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Thu, 23 May 2019 08:39:06 +0000 (+1000)
Subject: [spec] start collating notes for SAC
X-Git-Tag: yr12~124
X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/8f6c1394e89e16de8bfb071165a8ec1367d1da4e

[spec] start collating notes for SAC
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+---
+header-includes:
+- \usepackage{harpoon}
+- \usepackage{amsmath}
+- \pagenumbering{gobble}
+---
+
+# Complex & Imaginary Numbers
+
+## Imaginary numbers
+
+$$i^2 = -1 \quad \therefore i = \sqrt {-1}$$
+
+### Simplifying negative surds
+
+\begin{equation}\begin{split}\sqrt{-2} & = \sqrt{-1 \times 2} \\ & = \sqrt{2}i\end{split}\end{equation}
+
+
+## Complex numbers
+
+$$\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}$$
+
+General form: $z=a+bi$  
+$\operatorname{Re}(z) = a, \quad \operatorname{Im}(z) = b$
+
+### Addition
+
+If $z_1 = a+bi$ and $z_2=c+di$, then
+
+$$z_1+z_2 = (a+c)+(b+d)i$$
+
+### Subtraction
+
+If $z_1=a+bi$ and $z_2=c+di$, then
+
+$$z_1−z_2=(a−c)+(b−d)i$$
+
+### Multiplication by a real constant
+
+If $z=a+bi$ and $k \in \mathbb{R}$, then
+
+$$kz=ka+kbi$$
+
+### Powers of $i$
+
+- $i^{4n} = 1$
+- $i^{4n+1} = i$
+- $i^{4n+2} = -1$
+- $i^{4n+3} = -i$
+
+For $i^n$, find remainder $r$ when $n \div 4$. Then $i^n = i^r$.
+
+### Multiplying complex expressions
+
+If $z_1 = a+bi$ and $z_2=c+di$, then
+
+$$z_1 \times z_2 = (ac-bd)+(ad+bc)i$$
+
+### Conjugates
+
+$$\overline{z} = a \mp bi$$
+
+##### Properties
+
+- $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$
+- $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$
+- $\overline{kz} = k \overline{z}, \text{ for } k \in \mathbb{R}$
+- $z \overline{z} = = (a+bi)(a-bi) = a^2+b^2 = |z|^2$
+- $z + \overline{z} = 2 \operatorname{Re}(z)$
+
+### Modulus
+
+Distance from origin.
+
+$$|{z}|=\sqrt{a^2+b^2} \quad  \therefore z \overline{z} = |z|^2$$
+
+###### Properties
+
+- $|z_1 z_2| = |z_1| |z_2|$
+- $|{z_1 \over z_2}| = {|z_1| \over |z_2|}$
+- $|z_1 + z_2| \le |z_1 + |z_2|$
+
+### 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}
+
+### 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:
+
+$${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$$
+
+## Argand planes
+
+- Geometric representation of $\mathbb{C}$
+- horizontal $= \operatorname{Re}(z)$; vertical $= \operatorname{Im}(z)$
+- Multiplication by $i$ results in an anticlockwise rotation of $\pi \over 2$
+
+\vfil \break
+
+## Complex polynomials
+
+**Include $\pm$ for all solutions, including imaginary**
+
+### 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.
+
+#### Dividing complex polynomials
+
+$P(z) \div D(z)$ gives quotient $Q(z)$ and remainder $R(z)$:
+
+$$P(z) = D(z)Q(z) + R(z)$$
+
+#### Remainder theorem
+
+Let $\alpha \in \mathbb{C}$. Remainder of $P(z) \div (z - \alpha)$ is $P(\alpha)$
+
+#### Factor theorem
+
+If $a+bi$ is a solution to $P(z)=0$, then:
+
+- $P(a+bi)=0$
+- $z-(a+bi)$ is a factor of $P(z)$
+
+#### Sum of two cubes
+
+$$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$$
+
+## Conjugate root theorem
+
+If $a+bi$ is a solution to $P(z)=0$, then the conjugate $\overline{z}=a-bi$ is also a solution.
+
+## 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}
+
+- $r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}$
+- $\theta=\operatorname{arg}(z)$ (on CAS: `arg(a+bi)`)
+- **principal argument** is $\operatorname{Arg}(z) \in (-\pi, \pi]$ (note capital $\operatorname{Arg}$)
+
+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
+
+### Conjugate in polar form
+
+$$(r \operatorname{cis} \theta)^{-1} = r\operatorname{cis} (- \theta)$$
+
+Reflection of $z$ across horizontal axis.
+
+### Multiplication and division in polar form
+
+$$z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)$$
+
+$${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$$
+
+## de Moivres' Theorem
+
+$$(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}$$
+
+## 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
+
+$$x^2 + y^2 = (|a|^{1 \over n})^2$$
+
+## Sketching complex graphs
+
+### Straight line
+
+- $\operatorname{Re}(z) = c$ or $\operatorname{Im}(z) = c$ (perpendicular bisector)
+- $\operatorname{Arg}(z) = \theta$
+- $|z+a|=|z+bi|$ where $m={a \over b}$
+- $|z+a|=|z+b| \longrightarrow 2(a-b)x=b^2-a^2$
+
+### Circle
+
+$|z-z_1|^2 = c^2 |z_2+2|^2$ or $|z-(a + bi)| = c$
+
+### Locus
+
+$\operatorname{Arg}(z) < \theta$
+
+# Vectors
+
+- **vector:** a directed line segment  
+- arrow indicates direction
+- length indicates magnitude
+- notated as $\vec{a}, \widetilde{A}, \overrightharp{a}$
+- column notation: $\begin{bmatrix}
+       x \\ y
+     \end{bmatrix}$
+- vectors with equal magnitude and direction are equivalent
+
+
+![](graphics/vectors-intro.png){#id .class width=20%} 
+
+## Vector addition
+
+$\boldsymbol{u} + \boldsymbol{v}$ can be represented by drawing each vector head to tail then joining the lines.  
+Addition is commutative (parallelogram)
+
+## 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 $k$.
+
+## Vector subtraction
+
+To find $\boldsymbol{u} - \boldsymbol{v}$, add $\boldsymbol{-v}$ to $\boldsymbol{u}$
+
+## Parallel vectors
+
+Same or opposite direction
+
+$$\boldsymbol{u} || \boldsymbol{v} \iff \boldsymbol{u} = k \boldsymbol{v} \text{ where } k \in \mathbb{R} \setminus \{0\}$$ 
+
+## Position vectors
+
+Vectors may describe a position relative to $O$.
+
+For a point $A$, the position vector is $\overrightharp{OA}$
+
+\vfill\eject
+
+## Linear combinations of non-parallel vectors
+
+If two non-zero vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ are not parallel, then:
+
+$$m\boldsymbol{a} + n\boldsymbol{b} = p \boldsymbol{a} + q \boldsymbol{b}\quad \therefore \quad m = p, \> n = q$$
+
+![](graphics/parallelogram-vectors.jpg){#id .class width=20%}
+![](graphics/vector-subtraction.jpg){#id .class width=10%}
+
+## Column vector notation
+
+A vector between points $A(x_1,y_1), \> B(x_2,y_2)$ can be represented as $\begin{bmatrix}x_2-x_1\\ y_2-y_1 \end{bmatrix}$
+
+## 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.
+
+## Unit vector $|\hat{\boldsymbol{a}}|=1$
+
+\begin{equation}\begin{split}\hat{\boldsymbol{a}} & = {1 \over {|\boldsymbol{a}|}}\boldsymbol{a} \\ & = \boldsymbol{a} \cdot {|\boldsymbol{a}|}\end{split}\end{equation}
+
+## Scalar/dot product $\boldsymbol{a} \cdot \boldsymbol{b}$
+
+$$\boldsymbol{a} \cdot \boldsymbol{b} = a_1 b_1 + a_2 b_2$$
+
+**on CAS:** `dotP([a b c], [d e f])`
+
+## Scalar product properties
+
+1. $k(\boldsymbol{a\cdot b})=(k\boldsymbol{a})\cdot \boldsymbol{b}=\boldsymbol{a}\cdot (k{b})$
+2. $\boldsymbol{a \cdot 0}=0$
+3. $\boldsymbol{a \cdot (b + c)}=\boldsymbol{a \cdot b + a \cdot c}$
+4. $\boldsymbol{i \cdot i} = \boldsymbol{j \cdot j} = \boldsymbol{k \cdot k}= 1$
+5. If $\boldsymbol{a} \cdot \boldsymbol{b} = 0$, $\boldsymbol{a}$ and $\boldsymbol{b}$ are perpendicular
+6. $\boldsymbol{a \cdot a} = |\boldsymbol{a}|^2 = a^2$
+
+For parallel vectors $\boldsymbol{a}$ and $\boldsymbol{b}$:  
+$$\boldsymbol{a \cdot b}=\begin{cases}
+|\boldsymbol{a}||\boldsymbol{b}| \hspace{2.8em} \text{if same direction}\\
+-|\boldsymbol{a}||\boldsymbol{b}| \hspace{2em} \text{if opposite directions}
+\end{cases}$$
+
+## Geometric scalar products
+
+$$\boldsymbol{a} \cdot \boldsymbol{b} = |\boldsymbol{a}| |\boldsymbol{b}| \cos \theta$$
+
+where $0 \le \theta \le \pi$
+
+## Perpendicular vectors
+
+If $\boldsymbol{a} \cdot \boldsymbol{b} = 0$, then $\boldsymbol{a} \perp \boldsymbol{b}$ (since $\cos 90 = 0$)
+
+## Finding angle between vectors
+
+**positive direction**
+
+$$\cos \theta = {{\boldsymbol{a} \cdot \boldsymbol{b}} \over {|\boldsymbol{a}| |\boldsymbol{b}|}} = {{a_1 b_1 + a_2 b_2} \over {|\boldsymbol{a}| |\boldsymbol{b}|}}$$
+
+**on CAS:** `angle([a b c], [a b c])` (Action -> Vector -> Angle)
+
+## 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:
+$$\cos \alpha = {a_1 \over |\boldsymbol{a}|}, \quad \cos \beta = {a_2 \over |\boldsymbol{a}|}, \quad \cos \gamma = {a_3 \over |\boldsymbol{a}|}$$
+
+**on CAS:** `angle([a b c], [1 0 0])` for angle between $a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}$ and $x$-axis
+
+## Vector projections
+
+Vector resolute of $\boldsymbol{a}$ in direction of $\boldsymbol{b}$ is magnitude of $\boldsymbol{a}$ in direction of $\boldsymbol{b}$:
+
+$$\boldsymbol{u}={{\boldsymbol{a}\cdot\boldsymbol{b}}\over |\boldsymbol{b}|^2}\boldsymbol{b}=\left({\boldsymbol{a}\cdot{\boldsymbol{b} \over |\boldsymbol{b}|}}\right)\left({\boldsymbol{b} \over |\boldsymbol{b}|}\right)=(\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}$$
+
+## Scalar resolute of $\boldsymbol{a}$ on $\boldsymbol{b}$
+
+$$r_s = |\boldsymbol{u}| = \boldsymbol{a} \cdot \hat{\boldsymbol{b}}$$
+
+## Vector resolute of $\boldsymbol{a} \perp \boldsymbol{b}$
+
+$$\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u} \> \text{ where } \boldsymbol{u} \text{ is projection } \boldsymbol{a} \text{ on } \boldsymbol{b}$$
+
+## Vector proofs
+
+### Concurrent lines
+
+$\ge$ 3 lines intersect at a single point  
+
+### 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 $\vec{AC}=m\vec{AB} \text{ where } m \ne 0$
+
+### Useful vector properties
+
+- If $\boldsymbol{a}$ and $\boldsymbol{b}$ are parallel, then $\boldsymbol{b}=k\boldsymbol{a}$ for some $k \in \mathbb{R} \setminus \{0\}$
+- If $\boldsymbol{a}$ and $\boldsymbol{b}$ are parallel with at least one point in common, then they lie on the same straight line
+- Two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ are perpendicular if $\boldsymbol{a} \cdot \boldsymbol{b}=0$
+- $\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2$
+
+## 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 $\boldsymbol{v_1}, \boldsymbol{v_2}, \boldsymbol{v_3}$
+
+## Three-dimensional vectors
+
+Right-hand rule for axes: $z$ is up or out of page.
+
+i![](graphics/vectors-3d.png)
+
+## 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}
+
+# Circular functions
+
+Period of $a\sin(bx)$ is ${2\pi} \over b$
+
+Period of $a\tan(nx)$ is $\pi \over n$  
+Asymptotes at $x={2k+1)\pi \over 2n} \> \vert \> k \in \mathbb{Z}$
+
+## Reciprocal functions
+
+### Cosecant
+
+![](graphics/csc.png)
+
+$$\operatorname{cosec} \theta = {1 \over \sin \theta} \> \vert \> \sin \theta \ne 0$$
+
+- **Domain** $= \mathbb{R} \setminus {n\pi : n \in \mathbb{Z}}$
+- **Range** $= \mathbb{R} \setminus (-1, 1)$
+- **Turning points** at $\theta = {{(2n + 1)\pi} \over 2} \> \vert \> n \in \mathbb{Z}$
+- **Asymptotes** at $\theta = n\pi \> \vert \> n \in \mathbb{Z}$
+
+### Secant
+
+![](graphics/sec.png)
+
+$$\operatorname{sec} \theta = {1 \over \cos \theta} \> \vert \> \cos \theta \ne 0$$
+
+- **Domain** $= \mathbb{R} \setminus \{{{(2n + 1) \pi} \over 2 } : n \in \mathbb{Z}\}$
+- **Range** $= \mathbb{R} \setminus (-1, 1)$
+- **Turning points** at $\theta = n\pi \> \vert \> n \in \mathbb{Z}$
+- **Asymptotes** at $\theta = {{(2n + 1) \pi} \over 2} \> \vert \> n \in \mathbb{Z}$
+
+### Cotangent
+
+![](graphics/cot.png)
+
+$$\operatorname{cot} \theta = {{\cos \theta} \over {\sin \theta}} \> \vert \> \sin \theta \ne 0$$
+
+- **Domain** $= \mathbb{R} \setminus \{n \pi: n \in \mathbb{Z}\}$
+- **Range** $= \mathbb{R}$
+- **Asymptotes** at $\theta = n\pi \> \vert \> n \in \mathbb{Z}$
+
+### 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}
+
+### 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}
+
+### 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}
+
+## Compound angle formulas
+
+$$\cos(x \pm y) = \cos x + \cos y \mp \sin x \sin y$$
+$$\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y$$
+$$\tan(x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x \tan y}}$$
+
+## 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}
+
+$$\sin 2x = 2 \sin x \cos x$$
+
+$$\tan 2x = {{2 \tan x} \over {1 - \tan^2 x}}$$
+
+## 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.
+
+### $\arcsin$
+
+$$\sin^{-1}: [-1, 1] \rightarrow \mathbb{R}, \quad \sin^{-1} x = y, \quad \text{where } \sin y = x \text{ and } y \in [{-\pi \over 2}, {\pi \over 2}]$$
+
+### $\arcos$
+
+$$\cos^{-1} \rightarrow \mathbb{R}, \quad \cos^{-1} x = y, \quad \text{where } \cos y = x \text{ and } y \in [0, \pi]$$
+
+### $\arctan$
+
+$$\tan^{-1}: \mathbb{R} \rightarrow \mathbb{R}, \quad \tan^{-1} x = y, \quad \text{where } \tan y = x \text{ and } y \in \left(-{\pi \over 2}, {\pi \over 2}\right)$$
+# Differential calculus
+
+## 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  
+
+- Limit exists if $L^-=L^+$
+- If limit exists, point does not.
+
+Limits can be solved using normal techniques (if div 0, factorise)
+
+## Limit theorems
+
+1. For constant function $f(x)=k$, $\lim_{x \rightarrow a} f(x) = k$
+2. $\lim_{x \rightarrow a} (f(x) \pm g(x)) = F \pm G$
+3. $\lim_{x \rightarrow a} (f(x) \times g(x)) = F \times G$
+4. ${\lim_{x \rightarrow a} {f(x) \over g(x)}} = {F \over G}, G \ne 0$
+
+Corollary: $\lim_{x \rightarrow a} c \times f(x)=cF$ where $c=$ constant
+
+## Solving limits for $x\rightarrow\infty$
+
+Factorise so that all values of $x$ are in denominators.
+
+e.g.
+
+$$\lim_{x \rightarrow \infty}{{2x+3} \over {x-2}}={{2+{3 \over x}} \over {1-{2 \over x}}}={2 \over 1} = 2$$
+
+
+## Continuous functions
+
+A function is continuous if $L^-=L^+=f(x)$ for all values of $x$.
+
+## 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 $h=\delta x$.
+
+## 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:
+$${m_{\text{tangent at }P} =\lim_{h \rightarrow 0}}{{f(x+h)-f(x)}\over h}$$
+
+## Gradient at a point
+
+Given point $P(a, b)$ and function $f(x)$, the gradient is $f^\prime(a)$
+
+
+## Derivatives of $x^n$
+
+$${d(ax^n) \over dx}=anx^{n-1}$$
+
+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 $f(x)=(x-b)^2, \quad f^\prime(x)=2(x-b)$
+
+$$f^\prime(x)=\lim_{h \rightarrow 0}{{f(x+h)-f(x)} \over h}$$
+
+## Derivatives of $u \pm v$
+
+$${dy \over dx}={du \over dx} \pm {dv \over dx}$$
+where $u$ and $v$ are functions of $x$
+
+## Euler's number as a limit
+
+$$\lim_{h \rightarrow 0} {{e^h-1} \over h}=1$$
+
+## Chain rule for $(f\circ g)$
+
+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)  
+${du \over dx} = 2x$  
+$y=u^7$  
+${dy \over du} = 7u^6$  
+
+## Product rule for $y=uv$
+
+$${dy \over dx} = u{dv \over dx} + v{du \over dx}$$
+
+## Quotient rule for $y={u \over v}$
+
+$${dy \over dx} = {{v{du \over dx} - u{dv \over dx}} \over v^2}$$
+
+$$f^\prime(x)={{v(x)u^\prime(x)-u(x)v^\prime(x)} \over [v(x)]^2}$$
+
+## Logarithms
+
+$$\log_b (x) = n \quad \operatorname{where} \hspace{0.5em} b^n=x$$
+
+Wikipedia:
+
+> 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 $x$
+
+### Logarithmic identities
+
+$\log_b (xy)=\log_b x + \log_b y$  
+$\log_b x^n = n \log_b x$  
+$\log_b y^{x^n} = x^n \log_b y$
+
+### Index identities
+
+$b^{m+n}=b^m \cdot b^n$  
+$(b^m)^n=b^{m \cdot n}$  
+$(b \cdot c)^n = b^n \cdot c^n$  
+${a^m \div a^n} = {a^{m-n}}$
+
+### $e$ as a logarithm
+
+$$\operatorname{if} y=e^x, \quad \operatorname{then} x=\log_e y$$
+$$\ln x = \log_e x$$
+
+### Differentiating logarithms
+$${d(\log_e x)\over dx} = x^{-1} = {1 \over x}$$
+
+## Derivative rules
+
+| $f(x)$ | $f^\prime(x)$ |
+| ------ | ------------- |
+| $\sin x$ | $\cos x$ |
+| $\sin ax$ | $a\cos ax$ |
+| $\cos x$ | $-\sin x$ |
+| $\cos ax$ | $-a \sin ax$ |
+| $\tan f(x)$ | $f^2(x) \sec^2f(x)$ |
+| $e^x$ | $e^x$ |
+| $e^{ax}$ | $ae^{ax}$ |
+| $ax^{nx}$ | $an \cdot e^{nx}$ |
+| $\log_e x$ | $1 \over x$ |
+| $\log_e {ax}$ | $1 \over x$ |
+| $\log_e f(x)$ | $f^\prime (x) \over f(x)$ |
+| $\sin(f(x))$ | $f^\prime(x) \cdot \cos(f(x))$ |
+| $\sin^{-1} x$ | $1 \over {\sqrt{1-x^2}}$ |
+| $\cos^{-1} x$ | $-1 \over {sqrt{1-x^2}}$ |
+| $\tan^{-1} x$ | $1 \over {1 + x^2}$ |
+
+## Reciprocal derivatives
+
+$${1 \over {dy \over dx}} = {dx \over dy}$$
+
+## Differentiating $x=f(y)$
+
+Find $dx \over dy$. Then ${dx \over dy} = {1 \over {dy \over dx}} \implies {dy \over dx} = {1 \over {dx \over dy}}$.
+
+$${dy \over dx} = {1 \over {dx \over dy}}$$
+
+## 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
+
+### Points of Inflection
+
+*Stationary point* - point of zero gradient (i.e. $f^\prime(x)=0$)  
+*Point of inflection* - point of maximum $|$gradient$|$ (i.e.  $f^{\prime\prime} = 0$)
+
+* if $f^\prime (a) = 0$ and $f^{\prime\prime}(a) > 0$, then point $(a, f(a))$ is a local min (curve is concave up)
+* if $f^\prime (a) = 0$ and $f^{\prime\prime} (a) < 0$, then point $(a, f(a))$ is local max (curve is concave down)
+* if $f^{\prime\prime}(a) = 0$, then point $(a, f(a))$ is a point of inflection
+  + if also $f^\prime(a)=0$, then it is a stationary point of inflection
+
+![](graphics/second-derivatives.png)
+
+## Implicit Differentiation
+
+**On CAS:** Action $\rightarrow$ Calculation $\rightarrow$ `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:
+
+$${dp \over dx} = {dq \over dx} \quad \text{and} \quad {dp \over dy} = {dq \over dy}$$
+
+## Integration
+
+$$\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)$$
+
+$$\int x^n \cdot dx = {x^{n+1} \over n+1} + c$$
+
+- area enclosed by curves
+- $+c$ should be shown on each step without $\int$
+
+### Integral laws
+
+$\int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx$  
+$\int k f(x) dx = k \int f(x) dx$  
+
+| $f(x)$                          | $\int f(x) \cdot dx$         |
+| ------------------------------- | ---------------------------- |
+| $k$ (constant) | $kx + c$ |
+| $x^n$ | ${x^{n+1} \over {n+1}} + c$ |
+| $a x^{-n}$ | $a \cdot \log_e x + c$ |
+| ${1 \over {ax+b}}$ | ${1 \over a} \log_e (ax+b) + c$ |
+| $(ax+b)^n$ | ${1 \over {a(n+1)}}(ax+b)^{n-1} + c$ |
+| $e^{kx}$ | ${1 \over k} e^{kx} + c$ |
+| $e^k$ | $e^kx + c$ |
+| $\sin kx$ | $-{1 \over k} \cos (kx) + c$ |
+| $\cos kx$ | ${1 \over k} \sin (kx) + c$ |
+| $\sec^2 kx$ | ${1 \over k} \tan(kx) + c$ |
+| $1 \over \sqrt{a^2-x^2}$ | $\sin^{-1} {x \over a} + c \>\vert\> a>0$ |
+| $-1 \over \sqrt{a^2-x^2}$ | $\cos^{-1} {x \over a} + c \>\vert\> a>0$ |
+| $a \over {a^2-x^2}$ | $\tan^{-1} {x \over a} + c$ |
+| ${f^\prime (x)} \over {f(x)}$ | $\log_e f(x) + c$ |
+| $g^\prime(x)\cdot f^\prime(g(x)$ | $f(g(x))$ (chain rule)|
+| $f(x) \cdot g(x)$ | $\int [f^\prime(x) \cdot g(x)] dx + \int [g^\prime(x) f(x)] dx$ |
+
+Note $\sin^{-1} {x \over a} + \cos^{-1} {x \over a}$ is constant for all $x \in (-a, a)$.
+
+### Definite integrals
+
+$$\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)$$
+
+- Signed area enclosed by: $\> y=f(x), \quad y=0, \quad x=a, \quad x=b$.
+- *Integrand* is $f$.
+- $F(x)$ may be any integral, i.e. $c$ is inconsequential
+
+#### Properties
+
+$$\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx$$
+
+$$\int^a_a f(x) \> dx = 0$$
+
+$$\int^b_a k \cdot f(x) \> dx = k \int^b_a f(x) \> dx$$
+
+$$\int^b_a f(x) \pm g(x) \> dx = \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx$$
+
+$$\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx$$
+
+### 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
+
+#### Definite integrals by substitution
+
+For $\int^b_a f(x) {du \over dx} \cdot dx$, evaluate new $a$ and $b$ for $f(u) \cdot du$.
+
+### Trigonometric integration
+
+$$\sin^m x \cos^n x \cdot dx$$
+
+**$m$ 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 $u=\cos x$
+
+**$n$ 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 $u=\sin x$
+
+**$m$ and $n$ are even:**  
+Use identities:
+
+- $\sin^2x={1 \over 2}(1-\cos 2x)$
+- $\cos^2x={1 \over 2}(1+\cos 2x)$
+- $\sin 2x = 2 \sin x \cos x$
+
+## Partial fractions
+
+On CAS: Action $\rightarrow$ Transformation $\rightarrow$ `expand/combine`  
+or Interactive $\rightarrow$ Transformation $\rightarrow$ `expand` $\rightarrow$ Partial
+
+## Graphing integrals on CAS
+
+In main: Interactive $\rightarrow$ Calculation $\rightarrow$ $\int$ ($\rightarrow$ Definite)  
+Restrictions: `Define f(x)=...` $\rightarrow$ `f(x)|x>1` (e.g.)
+
+## Applications of antidifferentiation
+
+- $x$-intercepts of $y=f(x)$ identify $x$-coordinates of stationary points on $y=F(x)$
+- nature of stationary points is determined by sign of $y=f(x)$ on either side of its $x$-intercepts
+- if $f(x)$ is a polynomial of degree $n$, then $F(x)$ has degree $n+1$
+
+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.
+
+## Solids of revolution
+
+Approximate as sum of infinitesimally-thick cylinders
+
+### 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*}
+
+### 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*}
+
+### Regions not bound by $y=0$
+
+$$V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx$$  
+where $f(x) > g(x)$
+
+## 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 $\rightarrow$ `arcLen`.
+
+## Rates
+
+### Related rates
+
+$${da \over db} \quad \text{(change in } a \text{ with respect to } b)$$
+
+### Gradient at a point on parametric curve
+
+$${dy \over dx} = {{dy \over dt} \div {dx \over dt}} \> \vert \> {dx \over dt} \ne 0$$
+
+$${d^2 \over dx^2} = {d(y^\prime) \over dx} = {{dy^\prime \over dt} \div {dx \over dt}} \> \vert \> y^\prime = {dy \over dx}$$
+
+## Rational functions
+
+$$f(x) = {P(x) \over Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}$$
+
+### Addition of ordinates
+
+- when two graphs have the same ordinate, $y$-coordinate is double the ordinate
+- when two graphs have opposite ordinates, $y$-coordinate is 0 i.e. ($x$-intercept)
+- when one of the ordinates is 0, the resulting ordinate is equal to the other ordinate
+
+## 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 $f$
+
+## Differential equations
+
+One or more derivatives
+
+**Order** - highest power inside derivative  
+**Degree** - highest power of highest derivative  
+e.g. ${\left(dy^2 \over d^2 x\right)}^3$: order 2, degree 3
+
+### Verifying solutions
+
+Start with $y=\dots$, and differentiate. Substitute into original equation.
+
+### 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 $A$.
+
+### Mixing problems
+
+$$\left({dm \over dt}\right)_\Sigma = \left({dm \over dt}\right)_{\text{in}} - \left({dm \over dt}\)_{\text{out}}$$
+
+### Separation of variables
+
+If ${dy \over dx}=f(x)g(y)$, then:
+
+$$\int f(x) \> dx = \int {1 \over g(y)} \> dy$$
+
+### Using definite integrals to solve DEs
+
+Used for situations where solutions to ${dy \over dx} = f(x)$ is not required.
+
+In some cases, it may not be possible to obtain an exact solution.
+
+Approximate solutions can be found by numerically evaluating a definite integral.
+
+### Using Euler's method to solve a differential equation
+
+$${{f(x+h) - f(x)} \over h } \approx f^\prime (x) \quad \text{for small } h$$
+
+$$\implies f(x+h) \approx f(x) + hf^\prime(x)$$
+
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+\documentclass[]{article}
+\usepackage{lmodern}
+\usepackage{amssymb,amsmath}
+\usepackage{ifxetex,ifluatex}
+\usepackage{fixltx2e} % provides \textsubscript
+\ifnum 0\ifxetex 1\fi\ifluatex 1\fi=0 % if pdftex
+  \usepackage[T1]{fontenc}
+  \usepackage[utf8]{inputenc}
+\else % if luatex or xelatex
+  \ifxetex
+    \usepackage{mathspec}
+  \else
+    \usepackage{fontspec}
+  \fi
+  \defaultfontfeatures{Ligatures=TeX,Scale=MatchLowercase}
+\fi
+% use upquote if available, for straight quotes in verbatim environments
+\IfFileExists{upquote.sty}{\usepackage{upquote}}{}
+% use microtype if available
+\IfFileExists{microtype.sty}{%
+\usepackage[]{microtype}
+\UseMicrotypeSet[protrusion]{basicmath} % disable protrusion for tt fonts
+}{}
+\PassOptionsToPackage{hyphens}{url} % url is loaded by hyperref
+\usepackage[unicode=true]{hyperref}
+\hypersetup{
+            pdfborder={0 0 0},
+            breaklinks=true}
+\urlstyle{same}  % don't use monospace font for urls
+\usepackage{longtable,booktabs}
+% Fix footnotes in tables (requires footnote package)
+\IfFileExists{footnote.sty}{\usepackage{footnote}\makesavenoteenv{long table}}{}
+\usepackage{graphicx,grffile}
+\makeatletter
+\def\maxwidth{\ifdim\Gin@nat@width>\linewidth\linewidth\else\Gin@nat@width\fi}
+\def\maxheight{\ifdim\Gin@nat@height>\textheight\textheight\else\Gin@nat@height\fi}
+\makeatother
+% Scale images if necessary, so that they will not overflow the page
+% margins by default, and it is still possible to overwrite the defaults
+% using explicit options in \includegraphics[width, height, ...]{}
+\setkeys{Gin}{width=\maxwidth,height=\maxheight,keepaspectratio}
+\IfFileExists{parskip.sty}{%
+\usepackage{parskip}
+}{% else
+\setlength{\parindent}{0pt}
+\setlength{\parskip}{6pt plus 2pt minus 1pt}
+}
+\setlength{\emergencystretch}{3em}  % prevent overfull lines
+\providecommand{\tightlist}{%
+  \setlength{\itemsep}{0pt}\setlength{\parskip}{0pt}}
+\setcounter{secnumdepth}{0}
+% Redefines (sub)paragraphs to behave more like sections
+\ifx\paragraph\undefined\else
+\let\oldparagraph\paragraph
+\renewcommand{\paragraph}[1]{\oldparagraph{#1}\mbox{}}
+\fi
+\ifx\subparagraph\undefined\else
+\let\oldsubparagraph\subparagraph
+\renewcommand{\subparagraph}[1]{\oldsubparagraph{#1}\mbox{}}
+\fi
+
+% set default figure placement to htbp
+\makeatletter
+\def\fps@figure{htbp}
+\makeatother
+
+\usepackage{harpoon}%
+\pagenumbering{gobble}
+\usepackage{fancyhdr}
+
+\title{Year 12 Specialist}
+\author{Andrew Lorimer}
+\date{2019}
+
+\begin{document}
+
+\pagestyle{fancy}
+\fancyhead[LO,LE]{Year 12 Specialist}
+\fancyhead[CO,CE]{Andrew Lorimmer}
+\maketitle
+
+\section{Complex \& Imaginary Numbers}\label{complex-imaginary-numbers}
+
+\subsection{Imaginary numbers}\label{imaginary-numbers}
+
+\[i^2 = -1 \quad \therefore i = \sqrt {-1}\]
+
+\subsubsection{Simplifying negative
+surds}\label{simplifying-negative-surds}
+
+\begin{equation}\begin{split}\sqrt{-2} & = \sqrt{-1 \times 2} \\ & = \sqrt{2}i\end{split}\end{equation}
+
+\subsection{Complex numbers}\label{complex-numbers}
+
+\[\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}\]
+
+General form: \(z=a+bi\)\\
+\(\operatorname{Re}(z) = a, \quad \operatorname{Im}(z) = b\)
+
+\subsubsection{Addition}\label{addition}
+
+If \(z_1 = a+bi\) and \(z_2=c+di\), then
+
+\[z_1+z_2 = (a+c)+(b+d)i\]
+
+\subsubsection{Subtraction}\label{subtraction}
+
+If \(z_1=a+bi\) and \(z_2=c+di\), then
+
+\[z_1 - z_2=(a−c)+(b−d)i\]
+
+\subsubsection{Multiplication by a real
+constant}\label{multiplication-by-a-real-constant}
+
+If \(z=a+bi\) and \(k \in \mathbb{R}\), then
+
+\[kz=ka+kbi\]
+
+\subsubsection{\texorpdfstring{Powers of
+\(i\)}{Powers of i}}\label{powers-of-i}
+
+\begin{itemize}
+\tightlist
+\item
+  \(i^{4n} = 1\)
+\item
+  \(i^{4n+1} = i\)
+\item
+  \(i^{4n+2} = -1\)
+\item
+  \(i^{4n+3} = -i\)
+\end{itemize}
+
+For \(i^n\), find remainder \(r\) when \(n \div 4\). Then \(i^n = i^r\).
+
+\subsubsection{Multiplying complex
+expressions}\label{multiplying-complex-expressions}
+
+If \(z_1 = a+bi\) and \(z_2=c+di\), then
+
+\[z_1 \times z_2 = (ac-bd)+(ad+bc)i\]
+
+\subsubsection{Conjugates}\label{conjugates}
+
+\[\overline{z} = a \mp bi\]
+
+\subparagraph{Properties}\label{properties}
+
+\begin{itemize}
+\tightlist
+\item
+  \(\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}\)
+\item
+  \(\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}\)
+\item
+  \(\overline{kz} = k \overline{z}, \text{ for } k \in \mathbb{R}\)
+\item
+  \(z \overline{z} = = (a+bi)(a-bi) = a^2+b^2 = |z|^2\)
+\item
+  \(z + \overline{z} = 2 \operatorname{Re}(z)\)
+\end{itemize}
+
+\subsubsection{Modulus}\label{modulus}
+
+Distance from origin.
+
+\[|{z}|=\sqrt{a^2+b^2} \quad  \therefore z \overline{z} = |z|^2\]
+
+Properties
+
+\begin{itemize}
+\tightlist
+\item
+  \(|z_1 z_2| = |z_1| |z_2|\)
+\item
+  \(|{z_1 \over z_2}| = {|z_1| \over |z_2|}\)
+\item
+  \(|z_1 + z_2| \le |z_1 + |z_2|\)
+\end{itemize}
+
+\subsubsection{Multiplicative inverse}\label{multiplicative-inverse}
+
+\begin{equation}\begin{split}z^{-1} & = {1 \over z} \\ & = {{a-bi} \over {a^2+B^2}} \\ & = {\overline{z} \over {|z|^2}}\end{split}\end{equation}
+
+\subsubsection{Dividing complex numbers}\label{dividing-complex-numbers}
+
+\[{{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}} \quad \text{(multiplicative inverse)}\]
+
+In practice, rationalise denominator:
+
+\[{z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}\]
+
+\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}