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\fancyhead[LO,LE]{Year 12 Specialist}
\fancyhead[CO,CE]{Andrew Lorimer}

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\begin{document}

\begin{multicols}{2}

  \section{Complex numbers}

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

    \begin{align*}
      \text{Cartesian form: } & a+bi\\
      \text{Polar form: } & r\operatorname{cis}\theta
    \end{align*}

    \subsection*{Operations}

\definecolor{shade1}{HTML}{ffffff}
\definecolor{shade2}{HTML}{e6f2ff}
  \definecolor{shade3}{HTML}{cce2ff}
      \begin{tabularx}{\columnwidth}{r|X|X}
        & \textbf{Cartesian} & \textbf{Polar} \\
        \hline
        \(z_1 \pm z_2\) & \((a \pm c)(b \pm d)i\) & convert to \(a+bi\)\\
        \hline
        \(+k \times z\) & \multirow{2}{*}{\(ka \pm kbi\)} & \(kr\operatorname{cis} \theta\)\\
        \cline{1-1}\cline{3-3}
        \(-k \times z\) & & \(kr \operatorname{cis}(\theta\pm \pi)\)\\
        \hline
        \(z_1 \cdot z_2\) & \(ac-bd+(ad+bc)i\) & \(r_1r_2 \operatorname{cis}(\theta_1 + \theta_2)\)\\
        \hline
        \(z_1 \div z_2\) & \((z_1 \overline{z_2}) \div |z_2|^2\) & \(\left(\frac{r_1}{r_2}\right) \operatorname{cis}(\theta_1 - \theta_2)\)
      \end{tabularx}

      \subsubsection*{Scalar multiplication in polar form}
      
        For \(k \in \mathbb{R}^+\):
        \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\theta\]

        \noindent For \(k \in \mathbb{R}^-\):
        \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\left(\begin{cases}\theta - \pi & |0<\operatorname{Arg}(z)\le \pi \\ \theta + \pi & |-\pi<\operatorname{Arg}(z)\le 0\end{cases}\right)\]

    \subsection*{Conjugate}

      \begin{align*}
        \overline{z} &= a \mp bi\\
        &= r \operatorname{cis}(-\theta)
      \end{align*}

      \noindent \colorbox{cas}{On CAS: \texttt{conjg(a+bi)}}

      \subsubsection*{Properties}

        \begin{align*}
          \overline{z_1 \pm z_2} &= \overline{z_1}\pm\overline{z_2}\\
          \overline{z_1 \cdot z_2} &= \overline{z_1}\cdot\overline{z_2}\\
          \overline{kz} &= k\overline{z} \quad | \quad k \in \mathbb{R}\\
          z\overline{z} &= (a+bi)(a-bi)\\
          &= a^2 + b^2\\
          &= |z|^2
        \end{align*}

    \subsection*{Modulus}

      \[|z|=|\vec{Oz}|=\sqrt{a^2 + b^2}\]

      \subsubsection*{Properties}

        \begin{align*}
          |z_1z_2|&=|z_1||z_2|\\
          \left|\frac{z_1}{z_2}\right|&=\frac{|z_1|}{|z_2|}\\
          |z_1+z_2|&\le|z_1|+|z_2|
        \end{align*}

    \subsection*{Multiplicative inverse}

      \begin{align*}
        z^{-1}&=\frac{a-bi}{a^2+b^2}\\
        &=\frac{\overline{z}}{|z|^2}a\\
        &=r \operatorname{cis}(-\theta)
      \end{align*}

    \subsection*{Dividing over \(\mathbb{C}\)}

      \begin{align*}
        \frac{z_1}{z_2}&=z_1z_2^{-1}\\
        &=\frac{z_1\overline{z_2}}{|z_2|^2}\\
        &=\frac{(a+bi)(c-di)}{c^2+d^2}\\
        & \qquad \text{(rationalise denominator)}
      \end{align*}

    \subsection*{Polar form}

      \begin{align*}
        z&=r\operatorname{cis}\theta\\
        &=r(\cos \theta + i \sin \theta)
      \end{align*}

      \begin{itemize}
        \item{\(r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}\)}
        \item{\(\theta = \operatorname{arg}(z)\) \quad \colorbox{cas}{On CAS: \texttt{arg(a+bi)}}}
        \item{\(\operatorname{Arg}(z) \in (-\pi,\pi)\) \quad \bf{(principal argument)}}
        \item{\colorbox{cas}{Convert on CAS:}\\ \verb|compToTrig(a+bi)| \(\iff\) \verb|cExpand{r·cisX}|}
        \item{Multiple representations:\\\(r\operatorname{cis}\theta=r\operatorname{cis}(\theta+2n\pi)\) with \(n \in \mathbb{Z}\) revolutions}
        \item{\(\operatorname{cis}\pi=-1,\qquad \operatorname{cis}0=1\)}
      \end{itemize}

    \subsection*{de Moivres' theorem}

    \[(r \operatorname{cis} \theta)^n = r^n \operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}\]

    \subsection*{Complex polynomials}
    
      Include \(\pm\) for all solutions, incl. imaginary

      \begin{tabularx}{\columnwidth}{ R{0.55} X  }
        \hline
        Sum of squares & \(\begin{aligned} 
        z^2 + a^2 &= z^2-(ai)^2\\
        &= (z+ai)(z-ai) \end{aligned}\) \\
        \hline
        Sum of cubes & \(a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\)\\
        \hline
        Division & \(P(z)=D(z)Q(z)+R(z)\) \\
        \hline
        Remainder theorem & Let \(\alpha \in \mathbb{C}\). Remainder of \(P(z) \div (z-\alpha)\) is \(P(\alpha)\)\\
        \hline
        Factor theorem & \(z-\alpha\) is a factor of \(P(z) \iff P(\alpha)=0\) for \(\alpha \in \mathbb{C}\)\\
        \hline
        Conjugate root theorem & \(P(z)=0 \text{ at } z=a\pm bi\) (\(\implies\) both \(z_1\) and \(\overline{z_1}\) are solutions)
      \end{tabularx}

    \subsection*{Roots}

      \(n\)th roots of \(z=r\operatorname{cis}\theta\) are:

      \[z = r^{\frac{1}{n}} \operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)\]

      \begin{itemize}

        \item{Same modulus for all solutions}
        \item{Arguments are separated by \(\frac{2\pi}{n}\)}
        \item{Solutions of \(z^n=a\) where \(a \in \mathbb{C}\) lie on the circle \(x^2+y^2=\left(|a|^{\frac{1}{n}}\right)^2\) \quad (intervals of \(\frac{2\pi}{n}\))}
      \end{itemize}

      \noindent For \(0=az^2+bz+c\), use quadratic formula:

      \[z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    \subsection*{Fundamental theorem of algebra}

      A polynomial of degree \(n\) can be factorised into \(n\) linear factors in \(\mathbb{C}\):

        \[\implies P(z)=a_n(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)\dots(z-\alpha_n)\]
        \[\text{ where } \alpha_1,\alpha_2,\alpha_3,\dots,\alpha_n \in \mathbb{C}\]

    \subsection*{Argand planes}
    
      \begin{center}\begin{tikzpicture}[scale=2]
        \draw [->] (-0.2,0) -- (1.5,0) node [right]  {$\operatorname{Re}(z)$};
        \draw [->] (0,-0.2) -- (0,1.5) node [above] {$\operatorname{Im}(z)$};
        \coordinate (P) at (1,1);
        \coordinate (a) at (1,0);
        \coordinate (b) at (0,1);
        \coordinate (O) at (0,0);
        \draw (0,0) -- (P) node[pos=0.5, above left]{\(r\)} node[pos=1, right]{\(\begin{aligned}z&=a+bi\\&=r\operatorname{cis}\theta\end{aligned}\)};
        \draw [gray, dashed] (1,1) -- (1,0) node[black, pos=1, below]{\(a\)};
        \draw [gray, dashed] (1,1) -- (0,1) node[black, pos=1, left]{\(b\)};
        \begin{scope}
          \path[clip] (O) -- (P) -- (a);
          \fill[red, opacity=0.5, draw=black] (O) circle (2mm);
          \node at ($(O)+(20:3mm)$) {$\theta$};
        \end{scope}
        \filldraw (P) circle (0.5pt);
      \end{tikzpicture}\end{center}

      \begin{itemize}
        \item{Multiplication by \(i \implies\) CCW rotation of \(\frac{\pi}{2}\)}
        \item{Addition: \(z_1 + z_2 \equiv\) \overrightharp{\(Oz_1\)} + \overrightharp{\(Oz_2\)}}
      \end{itemize}

    \subsection*{Sketching complex graphs}
      
      \subsubsection*{Linear}

        \begin{itemize}
          \item{\(\operatorname{Re}(z)=c\) or \(\operatorname{Im}(z)=c\) (perpendicular bisector)}
          \item{\(\operatorname{Im}(z)=m\operatorname{Re}(z)\)}
          \item{\(|z+a|=|z+b| \implies 2(a-b)x=b^2-a^2\)\\Geometric: equidistant from \(a,b\)}
        \end{itemize}

      \subsubsection*{Circles}

        \begin{itemize}
          \item \(|z-z_1|^2=c^2|z_2+2|^2\)
          \item \(|z-(a+bi)|=c \implies (x-a)^2+_(y-b)^2=c^2\)
        \end{itemize}

      \noindent \textbf{Loci} \qquad \(\operatorname{Arg}(z)<\theta\)

        \begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
          \draw [->] (0,0) -- (1,0) node [right]  {$\operatorname{Re}(z)$};
          \draw [->] (0,-0.5) -- (0,1) node [above] {$\operatorname{Im}(z)$};
          \draw [<-, dashed, thick, blue] (-1,0) -- (0,0);
          \draw [->, thick, blue] (0,0) -- (1,1);
          \fill [gray, opacity=0.2, domain=-1:1, variable=\x] (-1,-0.5) -- (-1,0) -- (0, 0) -- (1,1) -- (1,-0.5) -- cycle;
          \begin{scope}
            \path[clip] (0,0) -- (1,1) -- (1,0);
            \fill[red, opacity=0.5, draw=black] (0,0) circle (2mm);
            \node at ($(0,0)+(20:3mm)$) {$\frac{\pi}{4}$};
          \end{scope}
          \node [font=\footnotesize] at (0.5,-0.25) {\(\operatorname{Arg}(z)\le\frac{\pi}{4}\)};
          \node [blue, mydot] {};
        \end{tikzpicture}\end{center}

      \noindent \textbf{Rays} \qquad \(\operatorname{Arg}(z-b)=\theta\)

        \begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
          \draw [->] (-0.75,0) -- (1.5,0) node [right]  {$\operatorname{Re}(z)$};
          \draw [->] (0,-1) -- (0,1) node [above] {$\operatorname{Im}(z)$};
          \draw [->, thick, brown] (-0.25,0) -- (-0.75,-1);
          \node [above, font=\footnotesize] at (-0.25,0) {\(\frac{1}{4}\)};
          \begin{scope}
            \path[clip] (-0.25,0) -- (-0.75,-1) -- (0,0);
            \fill[orange, opacity=0.5, draw=black] (-0.25,0) circle (2mm);
          \end{scope}
          \node at (-0.08,-0.3) {\(\frac{\pi}{8}\)};
          \node [font=\footnotesize, left] at (-0.75,-1) {\(\operatorname{Arg}(z+\frac{1}{4})=\frac{\pi}{8}\)};
          \node [brown, mydot] at (-0.25,0) {};
          \draw [<->, thick, green] (0,-1) -- (1.5,0.5) node [pos=0.25, black, font=\footnotesize, right] {\(|z-2|=|z-(1+i)|\)};
          \node [left, font=\footnotesize] at (0,-1) {\(-1\)};
          \node [below, font=\footnotesize] at (1,0) {\(1\)};
        \end{tikzpicture}\end{center}

    \section{Vectors}
\begin{center}\begin{tikzpicture}
  \draw [->] (-0.5,0) -- (3,0) node [right]  {\(x\)};
          \draw [->] (0,-0.5) -- (0,3) node [above] {\(y\)};
          \draw [orange, ->, thick] (0.5,0.5) -- (2.5,2.5) node [pos=0.5, above] {\(\vec{u}\)};
         \begin{scope}[very thick, every node/.style={sloped,allow upside down}]
        \draw [gray, dashed, thick] (0.5,0.5) -- (2.5,0.5) node [pos=0.5] {\midarrow} node[black, pos=0.5, below]{\(x\vec{i}\)};
        \draw [gray, dashed, thick] (2.5,0.5) -- (2.5,2.5) node [pos=0.5] {\midarrow};
         \end{scope}
        \node[black, right] at (2.5,1.5) {\(y\vec{j}\)};
\end{tikzpicture}\end{center}
\subsection*{Column notation}

\[\begin{bmatrix}x\\ y \end{bmatrix} \iff x\boldsymbol{i} + y\boldsymbol{j}\]
\(\begin{bmatrix}x_2-x_1\\ y_2-y_1 \end{bmatrix}\) \quad between \(A(x_1,y_1), \> B(x_2,y_2)\)

\subsection*{Scalar multiplication}

\[k\cdot (x\boldsymbol{i}+y\boldsymbol{j})=kx\boldsymbol{i}+ky\boldsymbol{j}\]

\noindent For \(k \in \mathbb{R}^-\), direction is reversed

\subsection*{Vector addition}
\begin{center}\begin{tikzpicture}[scale=1]
          \coordinate (A) at (0,0);
          \coordinate (B) at (2,2);
          \draw [->, thick, red] (0,0) -- (2,2) node [pos=0.5, below right] {\(\vec{u}=2\vec{i}+2\vec{j}\)};
          \draw [->, thick, blue] (2,2) -- (1,4) node [pos=0.5, above right] {\(\vec{v}=-\vec{i}+2\vec{j}\)};
          \draw [->, thick, orange] (0,0) -- (1,4) node [pos=0.5, left] {\(\vec{u}+\vec{v}=\vec{i}+4\vec{j}\)};
\end{tikzpicture}\end{center}

\[(x\boldsymbol{i}+y\boldsymbol{j}) \pm (a\boldsymbol{i}+b\boldsymbol{j})=(x \pm a)\boldsymbol{i}+(y \pm b)\boldsymbol{j}\]

\begin{itemize}
  \item Draw each vector head to tail then join lines
  \item Addition is commutative (parallelogram)
  \item \(\boldsymbol{u}-\boldsymbol{v}=\boldsymbol{u}+(-\boldsymbol{v}) \implies \overrightharp{AB}=\boldsymbol{b}-\boldsymbol{a}\)
\end{itemize}

\subsection*{Magnitude}

\[|(x\boldsymbol{i} + y\boldsymbol{j})|=\sqrt{x^2+y^2}\]

\subsection*{Parallel vectors}

\[\boldsymbol{u} || \boldsymbol{v} \iff \boldsymbol{u} = k \boldsymbol{v} \text{ where } k \in \mathbb{R} \setminus \{0\}\]

For parallel vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\):\\
\[\boldsymbol{a \cdot b}=\begin{cases}
|\boldsymbol{a}||\boldsymbol{b}| \hspace{2.8em} \text{if same direction}\\
-|\boldsymbol{a}||\boldsymbol{b}| \hspace{2em} \text{if opposite directions}
\end{cases}\]
%\includegraphics[width=0.2,height=\textheight]{graphics/parallelogram-vectors.jpg}
%\includegraphics[width=1]{graphics/vector-subtraction.jpg}

\subsection*{Perpendicular vectors}

\[\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b} = 0\ \quad \text{(since \(\cos 90 = 0\))}\]

\subsection*{Unit vector \(|\hat{\boldsymbol{a}}|=1\)}
\[\begin{split}\hat{\boldsymbol{a}} & = {\frac{1}{|\boldsymbol{a}|}}\boldsymbol{a} \\ & = \boldsymbol{a} \cdot {|\boldsymbol{a}|}\end{split}\]

  \subsection*{Scalar product \(\boldsymbol{a} \cdot \boldsymbol{b}\)}


\begin{center}\begin{tikzpicture}[scale=2]
  \draw [->] (0,0) -- (1,0.5) node [pos=0.5, above left] {\(\boldsymbol{b}\)};
  \draw [->] (0,0) -- (1,0) node [pos=0.5, below] {\(\boldsymbol{a}\)};
          \begin{scope}
            \path[clip] (1,0.5) -- (1,0) -- (0,0);
            \fill[orange, opacity=0.5, draw=black] (0,0) circle (2mm);
            \node at ($(0,0)+(15:4mm)$) {\(\theta\)};
          \end{scope}
\end{tikzpicture}\end{center}
\begin{align*}\boldsymbol{a} \cdot \boldsymbol{b} &= a_1 b_1 + a_2 b_2 \\  &= |\boldsymbol{a}| |\boldsymbol{b}| \cos \theta \\ &\quad (\> 0 \le \theta \le \pi) \text{ - from cosine rule}\end{align*}
\noindent\colorbox{cas}{On CAS: \texttt{dotP({[}a\ b\ c{]},\ {[}d\ e\ f{]})}}

\subsubsection*{Properties}

\begin{enumerate}
\item
  \(k(\boldsymbol{a\cdot b})=(k\boldsymbol{a})\cdot \boldsymbol{b}=\boldsymbol{a}\cdot (k\boldsymbol{b})\)
\item
  \(\boldsymbol{a \cdot 0}=0\)
\item
  \(\boldsymbol{a} \cdot (\boldsymbol{b} + \boldsymbol{c})=\boldsymbol{a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{c}\)
\item
  \(\boldsymbol{i \cdot i} = \boldsymbol{j \cdot j} = \boldsymbol{k \cdot k}= 1\)
\item
  \(\boldsymbol{a} \cdot \boldsymbol{b} = 0 \quad \implies \quad \boldsymbol{a} \perp \boldsymbol{b}\)
\item
  \(\boldsymbol{a \cdot a} = |\boldsymbol{a}|^2 = a^2\)
\end{enumerate}

\subsection*{Angle between vectors}

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

\noindent \colorbox{cas}{On CAS:} \texttt{angle([a b c], [a b c])}

(Action \(\rightarrow\) Vector \(\rightarrow\)Angle)

\subsection*{Angle between vector and axis}

\noindent For\(\boldsymbol{a} = a_1 \boldsymbol{i} + a_2 \boldsymbol{j} + a_3 \boldsymbol{k}\)
which makes angles \(\alpha, \beta, \gamma\) with positive side of
\(x, y, z\) axes:
\[\cos \alpha = \frac{a_1}{|\boldsymbol{a}|}, \quad \cos \beta = \frac{a_2}{|\boldsymbol{a}|}, \quad \cos \gamma = \frac{a_3}{|\boldsymbol{a}|}\]

\noindent \colorbox{cas}{On CAS:} \texttt{angle({[}a\ b\ c{]},\ {[}1\ 0\ 0{]})}\\for angle
between \(a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}\) and
\(x\)-axis

\subsection*{Projections \& resolutes}

\begin{tikzpicture}[scale=3]
  \draw [->, purple] (0,0) -- (1,0.5) node [pos=0.5, above left] {\(\boldsymbol{a}\)};
  \draw [->, orange] (0,0) -- (1,0) node [pos=0.5, below] {\(\boldsymbol{u}\)};
  \draw [->, blue] (1,0) -- (2,0) node [pos=0.5, below] {\(\boldsymbol{b}\)};
          \begin{scope}
            \path[clip] (1,0.5) -- (1,0) -- (0,0);
            \fill[orange, opacity=0.5, draw=black] (0,0) circle (2mm);
            \node at ($(0,0)+(15:4mm)$) {\(\theta\)};
          \end{scope}
         \begin{scope}[very thick, every node/.style={sloped,allow upside down}]
        \draw [gray, dashed, thick] (1,0) -- (1,0.5) node [pos=0.5] {\midarrow} node[black, pos=0.5, right, rotate=-90]{\(\boldsymbol{w}\)};
          \end{scope}
\draw (0,0) coordinate (O)
  (1,0) coordinate (A)
  (1,0.5) coordinate (B)
  pic [draw,red,angle radius=2mm] {right angle = O--A--B};
\end{tikzpicture}

\subsubsection*{\(\parallel\boldsymbol{b}\) (vector projection/resolute)}

\begin{align*}
  \boldsymbol{u} & = \frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|^2}\boldsymbol{b} \\
  & = \left(\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}\right)\left(\frac{\boldsymbol{b}}{|\boldsymbol{b}|}\right) \\
  & = (\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}
\end{align*}

\subsubsection*{\(\perp\boldsymbol{b}\) (perpendicular projection)}
\[\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u}\]

\subsubsection*{\(|\boldsymbol{u}|\) (scalar resolute)}
\begin{align*}
  r_s &= |\boldsymbol{u}|\\
  &= \boldsymbol{a} \cdot \hat{\boldsymbol{b}}\\
  &=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}
\end{align*}

\subsubsection*{Rectangular (\(\parallel,\perp\)) components}

\[\boldsymbol{a}=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}+\left(\boldsymbol{a}-\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}\right)\]


\subsection*{Vector proofs}

\textbf{Concurrent:} intersection of \(\ge\) 3 lines

\begin{tikzpicture}
  \draw [blue] (0,0) -- (1,1);
  \draw [red] (1,0) -- (0,1);
  \draw [brown] (0.4,0) -- (0.6,1);
        \filldraw (0.5,0.5) circle (2pt);
\end{tikzpicture}

\subsubsection*{Collinear points}

\(\ge\) 3 points lie on the same line

\begin{tikzpicture}
  \draw [purple] (0,0) -- (4,1);
  \filldraw (2,0.5) circle (2pt) node [above] {\(C\)};
  \filldraw (1,0.25) circle (2pt) node [above] {\(A\)};
  \filldraw (3,0.75) circle (2pt) node [above] {\(B\)};
  \coordinate (O) at (2.8,-0.2);
  \node at (O) [below] {\(O\)}; 
         \begin{scope}[->, orange, thick] 
           \draw (O) -- (2,0.5) node [pos=0.5, above, font=\footnotesize, black] {\(\boldsymbol{c}\)};
           \draw (O) -- (1,0.25) node [pos=0.5, below, font=\footnotesize, black] {\(\boldsymbol{a}\)};
           \draw (O) -- (3,0.75) node [pos=0.5, right, font=\footnotesize, black] {\(\boldsymbol{b}\)};
         \end{scope}
\end{tikzpicture}

\begin{align*}
  \text{e.g. Prove that}\\
  \overrightharp{AC}=m\overrightharp{AB} \iff \boldsymbol{c}&=(1-m)\boldsymbol{a}+m\boldsymbol{b}\\
  \implies \boldsymbol{c} &= \overrightharp{OA} + \overrightharp{AC}\\
  &= \overrightharp{OA} + m\overrightharp{AB}\\
  &=\boldsymbol{a}+m(\boldsymbol{b}-\boldsymbol{a})\\
  &=\boldsymbol{a}+m\boldsymbol{b}-m\boldsymbol{a}\\
  &=(1-m)\boldsymbol{a}+m{b}
\end{align*}
\begin{align*}
  \text{Also, } \implies \overrightharp{OC} &= \lambda \vec{OA} + \mu \overrightharp{OB} \\
  \text{where } \lambda + \mu &= 1\\
  \text{If } C \text{ lies along } \overrightharp{AB}, & \implies 0 < \mu < 1
\end{align*}


\subsubsection*{Parallelograms}

\begin{center}\begin{tikzpicture}
  \coordinate (O) at (0,0) node [below left] {\(O\)};
  \coordinate (A) at (4,0);
  \coordinate (B) at (6,2);
  \coordinate (C) at (2,2);
  \coordinate (D) at (6,0);

    \draw[postaction={decorate}, decoration={markings, mark=at position 0.6 with {\arrow{>>}}}] (O)--(A) node [below left] {\(A\)};
    \draw[postaction={decorate}, decoration={markings,mark=at position 0.5 with {\arrow{>}}}] (A)--(B) node [above right] {\(B\)};
    \draw[postaction={decorate}, decoration={markings, mark=at position 0.6 with {\arrow{>>}}}] (B)--(C) node [above left] {\(C\)};
    \draw[postaction={decorate}, decoration={markings,mark=at position 0.5 with {\arrow{>}}}] (C)--(O);

    \draw [gray, dashed] (O) -- (B) node [pos=0.75] {\(\diagdown\diagdown\)} node [pos=0.25] {\(\diagdown\diagdown\)};
    \draw [gray, dashed] (A) -- (C) node [pos=0.75] {\(\diagup\)} node [pos=0.25] {\(\diagup\)};
          \begin{scope}
            \path[clip] (C) -- (A) -- (O);
            \fill[orange, opacity=0.5, draw=black] (0,0) circle (4mm);
            \node at ($(0,0)+(20:8mm)$) {\(\theta\)};
          \end{scope}
          \draw [gray, thick, dotted] (B) -- (D) node [pos=0.5, right, black, font=\footnotesize] {\(|\boldsymbol{c}|\sin\theta\)} (A) -- (D) node [pos=0.5, below, black, font=\footnotesize] {\(|\boldsymbol{c}|\cos\theta\)};
  \draw pic [draw,thick,red,angle radius=2mm] {right angle=O--D--B};
\end{tikzpicture}\end{center}

\begin{itemize}
  \item
    Diagonals \(\overrightharp{OB}, \overrightharp{AC}\) bisect each other
  \item
    If diagonals are equal length, it is a rectangle
  \item
    \(|\overrightharp{OB}|^2 + |\overrightharp{CA}|^2 = |\overrightharp{OA}|^2 + |\overrightharp{AB}|^2 + |\overrightharp{CB}|^2 + |\overrightharp{OC}|^2\)
  \item
    Area \(=\boldsymbol{c} \cdot \boldsymbol{a}\)
\end{itemize}

  \subsubsection*{Useful vector properties}

\begin{itemize}
\item
  \(\boldsymbol{a} \parallel \boldsymbol{b} \implies \boldsymbol{b}=k\boldsymbol{a}\) for some
  \(k \in \mathbb{R} \setminus \{0\}\)
\item
  If \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are parallel with at
  least one point in common, then they lie on the same straight line
\item
  \(\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b}=0\)
\item
  \(\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2\)
\end{itemize}

\subsection*{Linear dependence}

\(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\) are linearly dependent if they are \(\nparallel\) and:
\begin{align*}
  0&=k\boldsymbol{a}+l\boldsymbol{b}+m\boldsymbol{c}\\
  \therefore \boldsymbol{c} &= m\boldsymbol{a} + n\boldsymbol{b} \quad \text{(simultaneous)}
\end{align*}

\noindent \(\boldsymbol{a}, \boldsymbol{b},\) and \(\boldsymbol{c}\) are linearly
independent if no vector in the set is expressible as a linear
combination of other vectors in set, or if they are parallel.

\subsection*{Three-dimensional vectors}

Right-hand rule for axes: \(z\) is up or out of page.

\tdplotsetmaincoords{60}{120} 
\begin{center}\begin{tikzpicture} [scale=3, tdplot_main_coords, axis/.style={->,thick}, 
vector/.style={-stealth,red,very thick}, 
vector guide/.style={dashed,gray,thick}]

%standard tikz coordinate definition using x, y, z coords
\coordinate (O) at (0,0,0);

%tikz-3dplot coordinate definition using x, y, z coords

\pgfmathsetmacro{\ax}{1}
\pgfmathsetmacro{\ay}{1}
\pgfmathsetmacro{\az}{1}

\coordinate (P) at (\ax,\ay,\az);

%draw axes
\draw[axis] (0,0,0) -- (1,0,0) node[anchor=north east]{$x$};
\draw[axis] (0,0,0) -- (0,1,0) node[anchor=north west]{$y$};
\draw[axis] (0,0,0) -- (0,0,1) node[anchor=south]{$z$};

%draw a vector from O to P
\draw[vector] (O) -- (P);

%draw guide lines to components
\draw[vector guide]         (O) -- (\ax,\ay,0);
\draw[vector guide] (\ax,\ay,0) -- (P);
\draw[vector guide]         (P) -- (0,0,\az);
\draw[vector guide] (\ax,\ay,0) -- (0,\ay,0);
\draw[vector guide] (\ax,\ay,0) -- (0,\ay,0);
\draw[vector guide] (\ax,\ay,0) -- (\ax,0,0);
\node[tdplot_main_coords,above right]
at (\ax,\ay,\az){(\ax, \ay, \az)};
\end{tikzpicture}\end{center}

\subsection*{Parametric vectors}

Parametric equation of line through point \((x_0, y_0, z_0)\) and
parallel to \(a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}\) is:

\[\begin{cases}x = x_o + a \cdot t \\ y = y_0 + b \cdot t \\ z = z_0 + c \cdot t\end{cases}\]

\section{Circular functions}

\(\sin(bx)\) or \(\cos(bx)\): period \(=\frac{2\pi}{b}\)

\noindent \(\tan(nx)\): period \(=\frac{\pi}{n}\)\\
\indent\indent\indent asymptotes at \(x=\frac{(2k+1)\pi}{2n} \> \vert \> k \in \mathbb{Z}\)

\subsection*{Reciprocal functions}

\subsubsection*{Cosecant}

\[\operatorname{cosec} \theta = \frac{1}{\sin \theta} \> \vert \> \sin \theta \ne 0\]

\begin{itemize}
\item
  \textbf{Domain} \(= \mathbb{R} \setminus {n\pi : n \in \mathbb{Z}}\)
\item
  \textbf{Range} \(= \mathbb{R} \setminus (-1, 1)\)
\item
  \textbf{Turning points} at
    \(\theta = \frac{(2n + 1)\pi}{2} \> \vert \> n \in \mathbb{Z}\)
\item
  \textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
\end{itemize}

\subsubsection*{Secant}


\begin{center}\includegraphics[width=0.7\columnwidth]{graphics/sec.png}\end{center}

\[\operatorname{sec} \theta = \frac{1}{\cos \theta} \> \vert \> \cos \theta \ne 0\]

\begin{itemize}

\item
  \textbf{Domain}
    \(= \mathbb{R} \setminus \frac{(2n + 1) \pi}{2} : n \in \mathbb{Z}\}\)
\item
  \textbf{Range} \(= \mathbb{R} \setminus (-1, 1)\)
\item
  \textbf{Turning points} at
  \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
\item
  \textbf{Asymptotes} at
    \(\theta = \frac{(2n + 1) \pi}{2} \> \vert \> n \in \mathbb{Z}\)
\end{itemize}

\subsubsection*{Cotangent}

\begin{center}\includegraphics[width=0.7\columnwidth]{graphics/cot.png}\end{center}

\[\operatorname{cot} \theta = {{\cos \theta} \over {\sin \theta}} \> \vert \> \sin \theta \ne 0\]

\begin{itemize}

\item
  \textbf{Domain} \(= \mathbb{R} \setminus \{n \pi: n \in \mathbb{Z}\}\)
\item
  \textbf{Range} \(= \mathbb{R}\)
\item
  \textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
\end{itemize}

\subsubsection*{Symmetry properties}

\[\begin{split}
  \operatorname{sec} (\pi \pm x) & = -\operatorname{sec} x \\
  \operatorname{sec} (-x) & = \operatorname{sec} x \\
  \operatorname{cosec} (\pi \pm x) & = \mp \operatorname{cosec} x \\
  \operatorname{cosec} (-x) & = - \operatorname{cosec} x \\
  \operatorname{cot} (\pi \pm x) & = \pm \operatorname{cot} x \\
  \operatorname{cot} (-x) & = - \operatorname{cot} x
\end{split}\]

\subsubsection*{Complementary properties}

\[\begin{split}
  \operatorname{sec} \left({\pi \over 2} - x\right) & = \operatorname{cosec} x \\
  \operatorname{cosec} \left({\pi \over 2} - x\right) & = \operatorname{sec} x \\
  \operatorname{cot} \left({\pi \over 2} - x\right) & = \tan x \\
  \tan \left({\pi \over 2} - x\right) & = \operatorname{cot} x
\end{split}\]

\subsubsection*{Pythagorean identities}

\[\begin{split}
  1 + \operatorname{cot}^2 x & = \operatorname{cosec}^2 x, \quad \text{where } \sin x \ne 0 \\
  1 + \tan^2 x & = \operatorname{sec}^2 x, \quad \text{where } \cos x \ne 0
\end{split}\]

\subsection*{Compound angle formulas}

\[\cos(x \pm y) = \cos x + \cos y \mp \sin x \sin y\]
\[\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\]
\[\tan(x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x \tan y}}\]

\subsection*{Double angle formulas}

\[\begin{split}
  \cos 2x &= \cos^2 x - \sin^2 x \\
  & = 1 - 2\sin^2 x \\
  & = 2 \cos^2 x -1
\end{split}\]

\[\sin 2x = 2 \sin x \cos x\]

\[\tan 2x = {{2 \tan x} \over {1 - \tan^2 x}}\]

\subsection*{Inverse circular functions}

Inverse functions: \(f(f^{-1}(x)) = x, \quad f(f^{-1}(x)) = x\)\\
Must be 1:1 to find inverse (reflection in \(y=x\)).\\
Domain is restricted to make functions 1:1.

\[\sin^{-1}: [-1, 1] \rightarrow \mathbb{R}, \quad \sin^{-1} x = y\]
\hfill where \(\sin y = x, \> y \in [{-\pi \over 2}, {\pi \over 2}]\)

\[\cos^{-1}: [-1,1] \rightarrow \mathbb{R}, \quad \cos^{-1} x = y\]
\hfill where \(\cos y = x, \> y \in [0, \pi]\)

\[\tan^{-1}: \mathbb{R} \rightarrow \mathbb{R}, \quad \tan^{-1} x = y\]
\hfill where \(\tan y = x, \> y \in \left(-{\pi \over 2}, {\pi \over 2}\right)\)


\section{Differential calculus}

\subsection*{Limits}

\[\lim_{x \rightarrow a}f(x)\]
\(L^-,\quad L^+\) \qquad limit from below/above\\
\(\lim_{x \to a} f(x)\) \quad limit of a point\\

\noindent For solving \(x\rightarrow\infty\), put all \(x\) terms in denominators\\
    e.g. \[\lim_{x \rightarrow \infty}{{2x+3} \over {x-2}}={{2+{3 \over x}} \over {1-{2 \over x}}}={2 \over 1} = 2\]

\subsubsection*{Limit theorems}

\begin{enumerate}
\item
  For constant function \(f(x)=k\), \(\lim_{x \rightarrow a} f(x) = k\)
\item
  \(\lim_{x \rightarrow a} (f(x) \pm g(x)) = F \pm G\)
\item
  \(\lim_{x \rightarrow a} (f(x) \times g(x)) = F \times G\)
    \item
\(\therefore \lim_{x \rightarrow a} c \times f(x)=cF\) where \(c=\) constant
\item
  \({\lim_{x \rightarrow a} {f(x) \over g(x)}} = {F \over G}, G \ne 0\)
\item
  \(f(x)\) is continuous \(\iff L^-=L^+=f(x) \> \forall x\)
\end{enumerate}

\subsection*{Gradients of secants and tangents}

\textbf{Secant (chord)} - line joining two points on curve\\
\textbf{Tangent} - line that intersects curve at one point

\subsection*{First principles derivative}

\[f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={\frac{dy}{dx}}\]

\subsubsection*{Logarithmic identities}

\(\log_b (xy)=\log_b x + \log_b y\)\\
\(\log_b x^n = n \log_b x\)\\
\(\log_b y^{x^n} = x^n \log_b y\)

\subsubsection*{Index identities}

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

\subsection*{Derivative rules}

\renewcommand{\arraystretch}{1.4}
\begin{tabularx}{\columnwidth}{rX}
  \hline
\(f(x)\) & \(f^\prime(x)\)\\
\hline
\(\sin x\) & \(\cos x\)\\
\(\sin ax\) & \(a\cos ax\)\\
\(\cos x\) & \(-\sin x\)\\
\(\cos ax\) & \(-a \sin ax\)\\
\(\tan f(x)\) & \(f^2(x) \sec^2f(x)\)\\
\(e^x\) & \(e^x\)\\
\(e^{ax}\) & \(ae^{ax}\)\\
\(ax^{nx}\) & \(an \cdot e^{nx}\)\\
  \(\log_e x\) & \(\dfrac{1}{x}\)\\
  \(\log_e {ax}\) & \(\dfrac{1}{x}\)\\
  \(\log_e f(x)\) & \(\dfrac{f^\prime (x)}{f(x)}\)\\
\(\sin(f(x))\) & \(f^\prime(x) \cdot \cos(f(x))\)\\
  \(\sin^{-1} x\) & \(\dfrac{1}{\sqrt{1-x^2}}\)\\
  \(\cos^{-1} x\) & \(\dfrac{-1}{sqrt{1-x^2}}\)\\
  \(\tan^{-1} x\) & \(\dfrac{1}{1 + x^2}\)\\
  \hline
\end{tabularx}

\subsection*{Reciprocal derivatives}

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

\subsection*{Differentiating \(x=f(y)\)}
\begin{align*}
  \text{Find }& \frac{dx}{dy}\\
  \text{Then, }\frac{dx}{dy} &= \frac{1}{\frac{dy}{dx}} \\
  \implies {\frac{dy}{dx}} &= \frac{1}{\frac{dx}{dy}}\\
  \therefore {\frac{dy}{dx}} &= \frac{1}{\frac{dx}{dy}}
\end{align*}

\subsection*{Second derivative}
\begin{align*}f(x) \longrightarrow &f^\prime (x) \longrightarrow f^{\prime\prime}(x)\\
\implies y \longrightarrow &\frac{dy}{dx} \longrightarrow \frac{d^2 y}{dx^2}\end{align*}

\noindent Order of polynomial \(n\)th derivative decrements each time the derivative is taken

\subsubsection*{Points of Inflection}

\emph{Stationary point} - i.e.
\(f^\prime(x)=0\)\\
\emph{Point of inflection} - max \(|\)gradient\(|\) (i.e.
\(f^{\prime\prime} = 0\))
%\begin{table*}[ht]
%\centering
%  \begin{tabularx}{\textwidth}{XXXX}
%\hline
%    \rowcolor{shade2}
%    & \(\dfrac{d^2 y}{dx^2} > 0\)  & \(\dfrac{d^2y}{dx^2}<0\) & \(\dfrac{d^2y}{dx^2}=0\) (inflection) \\
%\hline
%    \(\frac{dy}{dx}>0\) & \begin{tikzpicture} \draw[domain=1:2,smooth,variable=\x,blue] plot ({\x},{(1/10)*\x*\x*\x}) plot ({\x},{0.675*\x-0.677}); \end{tikzpicture} & cell 3\\
%cell 1 & cell 2 & cell 3\\
%\hline
%\end{tabularx}
%\end{table*}
\begin{itemize}

\item
  if \(f^\prime (a) = 0\) and \(f^{\prime\prime}(a) > 0\), then point
  \((a, f(a))\) is a local min (curve is concave up)
\item
  if \(f^\prime (a) = 0\) and \(f^{\prime\prime} (a) < 0\), then point
  \((a, f(a))\) is local max (curve is concave down)
\item
  if \(f^{\prime\prime}(a) = 0\), then point \((a, f(a))\) is a point of
  inflection
\item
  if also \(f^\prime(a)=0\), then it is a stationary point of inflection
\end{itemize}

\begin{table*}[ht]
  \centering
  \includegraphics[width=0.7\textwidth]{graphics/second-derivatives.png}
\end{table*}

\subsection*{Implicit Differentiation}

\noindent Used for differentiating circles etc.

If \(p\) and \(q\) are expressions in \(x\) and \(y\) such that \(p=q\),
for all \(x\) and \(y\), then:

\[{\frac{dp}{dx}} = {\frac{dq}{dx}} \quad \text{and} \quad {\frac{dp}{dy}} = {\frac{dq}{dy}}\]

\noindent \colorbox{cas}{\textbf{On CAS:}}\\
Action \(\rightarrow\) Calculation \(\rightarrow\) \texttt{impDiff(y\^{}2+ax=5,\ x,\ y)}\\
Returns \(y^\prime= \dots\).

\subsection*{Integration}

\[\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)\]

\subsection*{Integral laws}

\renewcommand{\arraystretch}{1.4}
\begin{tabularx}{\columnwidth}{rX}
\hline
  \(f(x)\) & \(\int f(x) \cdot dx\) \\
  \hline
  \(k\) (constant) & \(kx + c\)\\
  \(x^n\) & \(\dfrac{1}{n+1} x^{n+1}\) \\
  \(a x^{-n}\) &\(a \cdot \log_e x + c\)\\
  \(\dfrac{1}{ax+b}\) &\(\dfrac{1}{a} \log_e (ax+b) + c\)\\
  \((ax+b)^n\) & \(\dfrac{1}{a(n+1)}(ax+b)^{n-1} + c\)\\
  \(e^{kx}\) & \(\dfrac{1}{k} e^{kx} + c\)\\
  \(e^k\) & \(e^kx + c\)\\
  \(\sin kx\) & \(\dfrac{-1}{k} \cos (kx) + c\)\\
  \(\cos kx\) & \(\frac{1}{k} \sin (kx) + c\)\\
  \(\sec^2 kx\) & \(\frac{1}{k} \tan(kx) + c\)\\
  \(\dfrac{1}{\sqrt{a^2-x^2}}\) & \(\sin^{-1} \dfrac{x}{a} + c \>\vert\> a>0\)\\
  \(\dfrac{-1}{\sqrt{a^2-x^2}}\) & \(\cos^{-1} \dfrac{x}{a} + c \>\vert\> a>0\)\\
  \(\frac{a}{a^2-x^2}\) & \(\tan^{-1} \frac{x}{a} + c\)\\
  \(\frac{f^\prime (x)}{f(x)}\) & \(\log_e f(x) + c\)\\
  \(g^\prime(x)\cdot f^\prime(g(x)\) & \(f(g(x))\) (chain rule)\\
  \(f(x) \cdot g(x)\) & \(\int [f^\prime(x) \cdot g(x)] dx + \int [g^\prime(x) f(x)] dx\)\\
  \hline
\end{tabularx}

Note \(\sin^{-1} {x \over a} + \cos^{-1} {x \over a}\) is constant \(\forall x \in (-a, a)\)

\subsection*{Definite integrals}

\[\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)\]

\begin{itemize}

\item
  Signed area enclosed by\\
  \(\> y=f(x), \quad y=0, \quad x=a, \quad x=b\).
\item
  \emph{Integrand} is \(f\).
\end{itemize}

\subsubsection*{Properties}

\[\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx\]

\[\int^a_a f(x) \> dx = 0\]

\[\int^b_a k \cdot f(x) \> dx = k \int^b_a f(x) \> dx\]

\[\int^b_a f(x) \pm g(x) \> dx = \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx\]

\[\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx\]

\subsection*{Integration by substitution}

\[\int f(u) {\frac{du}{dx}} \cdot dx = \int f(u) \cdot du\]

\noindent Note \(f(u)\) must be 1:1 \(\implies\) one \(x\) for each \(y\)
\begin{align*}\text{e.g. for } y&=\int(2x+1)\sqrt{x+4} \cdot dx\\
  \text{let } u&=x+4\\
  \implies& {\frac{du}{dx}} = 1\\
  \implies& x = u - 4\\
  \text{then } &y=\int (2(u-4)+1)u^{\frac{1}{2}} \cdot du\\
  &\text{(solve as  normal integral)}
\end{align*}

\subsubsection*{Definite integrals by substitution}

For \(\int^b_a f(x) {\frac{du}{dx}} \cdot dx\), evaluate new \(a\) and
\(b\) for \(f(u) \cdot du\).

\subsubsection*{Trigonometric integration}

\[\sin^m x \cos^n x \cdot dx\]

\paragraph{\textbf{\(m\) is odd:}}
\(m=2k+1\) where \(k \in \mathbb{Z}\)\\
\(\implies \sin^{2k+1} x = (\sin^2 z)^k \sin x = (1 - \cos^2 x)^k \sin x\)\\
Substitute \(u=\cos x\)

\paragraph{\textbf{\(n\) is odd:}}
\(n=2k+1\) where \(k \in \mathbb{Z}\)\\
\(\implies \cos^{2k+1} x = (\cos^2 x)^k \cos x = (1-\sin^2 x)^k \cos x\)\\
Substitute \(u=\sin x\)

\paragraph{\textbf{\(m\) and \(n\) are even:}}
use identities...

\begin{itemize}

\item
  \(\sin^2x={1 \over 2}(1-\cos 2x)\)
\item
  \(\cos^2x={1 \over 2}(1+\cos 2x)\)
\item
  \(\sin 2x = 2 \sin x \cos x\)
\end{itemize}

\subsection*{Partial fractions}

\colorbox{cas}{On CAS:}\\
\indent Action \(\rightarrow\) Transformation \(\rightarrow\)
\texttt{expand/combine}\\
\indent Interactive \(\rightarrow\) Transformation \(\rightarrow\)
Expand \(\rightarrow\) Partial

\subsection*{Graphing integrals on CAS}

\colorbox{cas}{In main:} Interactive \(\rightarrow\) Calculation \(\rightarrow\)
\(\int\) (\(\rightarrow\) Definite)\\
Restrictions: \texttt{Define\ f(x)=..} then \texttt{f(x)\textbar{}x\textgreater{}..}

\subsection*{Applications of antidifferentiation}

\begin{itemize}

\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 \({\frac{dy}{dx}}=0\). Integrate to find
original function.

\subsection*{Solids of revolution}

Approximate as sum of infinitesimally-thick cylinders

\subsubsection*{Rotation about \(x\)-axis}

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

\subsubsection*{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*{Regions not bound by \(y=0\)}

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

\subsection*{Length of a curve}

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

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

\noindent \colorbox{cas}{On CAS:}\\
\indent Evaluate formula,\\
\indent or Interactive \(\rightarrow\) Calculation
\(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen}

\subsection*{Rates}

\subsubsection*{Gradient at a point on parametric curve}

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

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

\subsection*{Rational functions}

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

\subsubsection*{Addition of ordinates}

\begin{itemize}

\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}

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

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

\subsection*{Differential equations}

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

\subsubsection*{Verifying solutions}

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

\subsubsection*{Function of the dependent
variable}

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

\subsubsection*{Mixing problems}

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

\subsubsection*{Separation of variables}

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

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

\subsubsection*{Euler's method for solving DEs}

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

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

  \end{multicols}
\end{document}