\documentclass[a4paper]{article}
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+\usepackage{array}
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+\usepackage[a4paper,margin=1.8cm]{geometry}
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+\usepackage{hhline}
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\usepackage{pgfplots}
\usepackage{pst-plot}
+\usepackage{rotating}
+%\usepackage{showframe} % debugging only
\usepackage{subfiles}
\usepackage{tabularx}
+\usepackage{tabu}
\usepackage{tcolorbox}
\usepackage{tikz-3dplot}
\usepackage{tikz}
scopes
}
+\newcommand\given[1][]{\:#1\vert\:}
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+ axis x line=middle,
+ axis y line=middle,
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\pagestyle{fancy}
+\fancypagestyle{plain}{\fancyhead[LO,LE]{} \fancyhead[CO,CE]{}} % rm title & author for first page
\fancyhead[LO,LE]{Year 12 Specialist}
\fancyhead[CO,CE]{Andrew Lorimer}
\newcolumntype{L}[1]{>{\hsize=#1\hsize\raggedright\arraybackslash}X}%
\newcolumntype{R}[1]{>{\hsize=#1\hsize\raggedleft\arraybackslash}X}%
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+\definecolor{cas}{HTML}{cde1fd}
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-\newtcolorbox{warning}{colback=white!90!black, leftrule=3mm, colframe=important, coltext=important, fontupper=\sffamily\bfseries}
-\newtcolorbox{cas}{colframe=cas!75!black, title=On CAS, left*=3mm}
+\newtcolorbox{cas}{colframe=cas!75!black, fonttitle=\sffamily\bfseries, title=On CAS, left*=3mm}
+\newtcolorbox{theorembox}[1]{colback=green!10!white, colframe=blue!20!white, coltitle=black, fontupper=\sffamily, fonttitle=\sffamily, #1}
+\newtcolorbox{warning}{colback=white!90!black, leftrule=3mm, colframe=important, coltext=darkgray, fontupper=\sffamily\bfseries}
\begin{document}
+\title{\vspace{-22mm}Year 12 Specialist\vspace{-4mm}}
+\author{Andrew Lorimer}
+\date{}
+\maketitle
+\vspace{-9mm}
\begin{multicols}{2}
\section{Complex numbers}
\[\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}\]
-
\begin{align*}
\text{Cartesian form: } & a+bi\\
\text{Polar form: } & r\operatorname{cis}\theta
\subsection*{Operations}
- \definecolor{shade1}{HTML}{ffffff}
- \definecolor{shade2}{HTML}{e6f2ff}
- \definecolor{shade3}{HTML}{cce2ff}
- \begin{tabularx}{\columnwidth}{r|X|X}
+ \begin{tabularx}{\columnwidth}{|r|X|X|}
+ \hline
+ \rowcolor{cas}
& \textbf{Cartesian} & \textbf{Polar} \\
\hline
\(z_1 \pm z_2\) & \((a \pm c)(b \pm d)i\) & convert to \(a+bi\)\\
\hline
\(z_1 \cdot z_2\) & \(ac-bd+(ad+bc)i\) & \(r_1r_2 \operatorname{cis}(\theta_1 + \theta_2)\)\\
\hline
- \(z_1 \div z_2\) & \((z_1 \overline{z_2}) \div |z_2|^2\) & \(\left(\frac{r_1}{r_2}\right) \operatorname{cis}(\theta_1 - \theta_2)\)
+ \(z_1 \div z_2\) & \((z_1 \overline{z_2}) \div |z_2|^2\) & \(\left(\frac{r_1}{r_2}\right) \operatorname{cis}(\theta_1 - \theta_2)\) \\
+ \hline
\end{tabularx}
\subsubsection*{Scalar multiplication in polar form}
\[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\left(\begin{cases}\theta - \pi & |0<\operatorname{Arg}(z)\le \pi \\ \theta + \pi & |-\pi<\operatorname{Arg}(z)\le 0\end{cases}\right)\]
\subsection*{Conjugate}
-
+ \vspace{-7mm} \hfill \colorbox{cas}{\texttt{conjg(a+bi)}}
\begin{align*}
\overline{z} &= a \mp bi\\
&= r \operatorname{cis}(-\theta)
\end{align*}
- \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}\\
+ \overline{kz} &= k\overline{z} \> \forall \> k \in \mathbb{R}\\
z\overline{z} &= (a+bi)(a-bi)\\
&= a^2 + b^2\\
&= |z|^2
\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)}
+ & \text{then rationalise denominator}
\end{align*}
\subsection*{Polar form}
- \begin{align*}
- z&=r\operatorname{cis}\theta\\
- &=r(\cos \theta + i \sin \theta)
- \end{align*}
+ \[ r \operatorname{cis} \theta = r\left( \cos \theta + i \sin \theta \right) \]
\begin{itemize}
\item{\(r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}\)}
- \item{\(\theta = \operatorname{arg}(z)\) \quad \colorbox{cas}{On CAS: \texttt{arg(a+bi)}}}
+ \item{\(\theta = \operatorname{arg}(z)\) \hfill \colorbox{cas}{\texttt{arg(a+bi)}}}
\item{\(\operatorname{Arg}(z) \in (-\pi,\pi)\) \quad \bf{(principal argument)}}
- \item{\colorbox{cas}{Convert on CAS:}\\ \verb|compToTrig(a+bi)| \(\iff\) \verb|cExpand{r·cisX}|}
\item{Multiple representations:\\\(r\operatorname{cis}\theta=r\operatorname{cis}(\theta+2n\pi)\) with \(n \in \mathbb{Z}\) revolutions}
\item{\(\operatorname{cis}\pi=-1,\qquad \operatorname{cis}0=1\)}
\end{itemize}
+ \begin{cas}
+ \-\hspace{1em}\verb|compToTrig(a+bi)| \(\iff\) \verb|cExpand{r·cisX}|
+ \end{cas}
+
\subsection*{de Moivres' theorem}
- \[(r \operatorname{cis} \theta)^n = r^n \operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}\]
+ \begin{theorembox}{}
+ \[(r \operatorname{cis} \theta)^n = r^n \operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}\]
+ \end{theorembox}
\subsection*{Complex polynomials}
\hline
\end{tabularx}
+ \begin{theorembox}{title=Factor theorem}
+ If \(\beta z + \alpha\) is a factor of \(P(z)\), \\
+ \-\hspace{1em}then \(P(-\dfrac{\alpha}{\beta})=0\).
+ \end{theorembox}
+
\subsection*{\(n\)th roots}
\(n\)th roots of \(z=r\operatorname{cis}\theta\) are:
\subsection*{Sketching complex graphs}
- \subsubsection*{Linear}
+ \subsubsection*{Rays/lines \qquad \(\operatorname{Arg}( z\pm b)=\theta\)}
+
+ \begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
+ \draw [->] (-0.75,0) -- (1.5,0) node [right] {$\operatorname{Re}(z)$};
+ \draw [->] (0,-1) -- (0,1) node [above] {$\operatorname{Im}(z)$};
+ \draw [->, thick, brown] (-0.25,0) -- (-0.75,-1);
+ \node [above, font=\footnotesize] at (-0.25,0) {\(\frac{1}{4}\)};
+ \begin{scope}
+ \path[clip] (-0.25,0) -- (-0.75,-1) -- (0,0);
+ \fill[orange, opacity=0.5, draw=black] (-0.25,0) circle (2mm);
+ \end{scope}
+ \node at (-0.08,-0.3) {\(\frac{\pi}{8}\)};
+ \node [font=\footnotesize, left] at (-0.75,-1) {\(\operatorname{Arg}(z+\frac{1}{4})=\frac{\pi}{8}\)};
+ \node [brown, mydot] at (-0.25,0) {};
+ \draw [<->, thick, green] (0,-1) -- (1.5,0.5) node [pos=0.25, black, font=\footnotesize, right] {\(|z-2|=|z-(1+i)|\)};
+ \node [left, font=\footnotesize] at (0,-1) {\(-1\)};
+ \node [below, font=\footnotesize] at (1,0) {\(1\)};
+ \end{tikzpicture}\end{center}
\begin{itemize}
+ \item \(\operatorname{Arg}(z \pm b) = \theta\) (ray)
\item{\(\operatorname{Re}(z)=c\) or \(\operatorname{Im}(z)=c\) (perpendicular bisector)}
\item{\(\operatorname{Im}(z)=m\operatorname{Re}(z)\)}
- \item{\(|z+a|=|z+b| \implies 2(a-b)x=b^2-a^2\)\\Geometric: equidistant from \(a,b\)}
+ \item \(|z - (a+bi)|=|z - (c+di)| \\ \implies \frac{2(c-a)x + a^2 + b^2 - c^2 - d^2}{2(b-d)}\)
+ \item \(\operatorname{Re}(z) \pm \operatorname{Im}(z) = c\)
\end{itemize}
\subsubsection*{Circles}
\begin{itemize}
\item \(|z-z_1|^2=c^2|z_2+2|^2\)
\item \(|z-(a+bi)|=c \implies (x-a)^2+_(y-b)^2=c^2\)
+ \item \(z \overline{z} = r^2\)
\end{itemize}
- \noindent \textbf{Loci} \qquad \(\operatorname{Arg}(z)<\theta\)
+ \subsubsection*{Regions \qquad \(\operatorname{Arg}(z) \lessgtr \theta\)}
\begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
\draw [->] (0,0) -- (1,0) node [right] {$\operatorname{Re}(z)$};
\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}
\end{scope}
\node[black, right] at (2.5,1.5) {\(y\vec{j}\)};
\end{tikzpicture}\end{center}
+
\subsection*{Column notation}
\[\begin{bmatrix}x\\ y \end{bmatrix} \iff x\boldsymbol{i} + y\boldsymbol{j}\]
\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{]})}}
+ \noindent\colorbox{cas}{On CAS:} \texttt{dotP({[}a\ b\ c{]},\ {[}d\ e\ f{]})}
\subsubsection*{Properties}
Area \(=\boldsymbol{c} \cdot \boldsymbol{a}\)
\end{itemize}
+ \subsubsection*{Perpendicular bisectors of a triangle}
+
+\hspace{-1.5cm}\begin{tikzpicture}
+ [
+ scale=3,
+ >=stealth,
+ point/.style = {draw, circle, fill = black, inner sep = 1pt},
+ dot/.style = {draw, circle, fill = black, inner sep = .2pt},
+ thick
+ ]
+
+ \node at (-1,1) [text width=5cm, rounded corners, fill=lblue, inner sep=1ex]
+ {
+ \sffamily The three bisectors meet at the circumcenter \(Z\) where \(|\overrightharp{ZA}| = |\overrightharp{ZB}| = |\overrightharp{ZC}|\).
+ };
+
+ % the circle
+ \def\rad{1}
+ \node (origin) at (0,0) [point, label = {right: {\(Z\)}}]{};
+ \draw [thin] (origin) circle (\rad);
+
+ % triangle nodes: just points on the circle
+ \node (n1) at +(60:\rad) [point, label = above:\(A\)] {};
+ \node (n2) at +(-145:\rad) [point, label = below:\(B\)] {};
+ \node (n3) at +(-45:\rad) [point, label = {below right:\(C\)}] {};
+
+ % triangle edges: connect the vertices, and leave a node at the midpoint
+ \draw[orange] (n3) -- node (a) [label = {above right:\(D\)}] {} (n1);
+ \draw[blue] (n3) -- node (b) [label = {below right:\(F\)}] {} (n2);
+ \draw[red] (n1) -- node (c) [label = {left: \(E\)}] {} (n2);
+
+ % Bisectors
+ % start at the point lying on the line from (origin) to (a), at
+ % twice that distance, and then draw a path going to the point on
+ % the line lying on the line from (a) to the (origin), at 3 times
+ % that distance.
+ \draw[orange, dotted]
+ ($ (origin) ! 2 ! (a) $)
+ node [right] {\sffamily Bisector \(AC\)}
+ -- ($(a) ! 3 ! (origin)$ );
+
+ % similarly for origin and b
+ \draw[blue, dotted]
+ ($ (origin) ! 2 ! (b) $)
+ -- ($(b) ! 3 ! (origin)$ )
+ node [right] {\sffamily Bisector \(BC\)};
+
+ \draw[red, dotted]
+ ($ (origin) ! 5 ! (c) $)
+ -- ($(c) ! 7 ! (origin)$ )
+ node [right] {\sffamily Bisector \(AB\)};
+
+ \draw[gray, dashed, thin] (n1) -- (origin) -- (n2);
+ \draw[gray, dashed, thin] (origin) -- (n3);
+
+ % Right angle symbols
+ \def\ralen{.5ex} % length of the short segment
+ \foreach \inter/\first/\last in {a/n3/origin, b/n2/origin, c/n2/origin}
+ {
+ \draw [thin] let \p1 = ($(\inter)!\ralen!(\first)$), % point along first path
+ \p2 = ($(\inter)!\ralen!(\last)$), % point along second path
+ \p3 = ($(\p1)+(\p2)-(\inter)$) % corner point
+ in
+ (\p1) -- (\p3) -- (\p2); % path
+ }
+\end{tikzpicture}
+
+ \begin{theorembox}{title=Perpendicular bisector theorem}
+ If a point \(P\) lies on the perpendicular bisector of line \(\overrightharp{XY}\), then \(P\) is equidistant from the endpoints of the bisected segment
+ \[ \text{i.e. } |\overrightharp{PX}| = |\overrightharp{PY}| \]
+ \end{theorembox}
+
\subsubsection*{Useful vector properties}
\begin{itemize}
\addplot[gray, dotted, thick, domain=-35:35] {-1.5708} node [black, font=\footnotesize, above left, pos=1] {\(y=-\frac{\pi}{2}\)};
\end{axis}
\end{tikzpicture}
-\columnbreak
+
+ \subsection*{Mensuration}
+
+ \begin{tikzpicture}[draw=blue!70,thick]
+ \filldraw[fill=lblue] circle (2cm);
+ \filldraw[fill=white]
+ (320:2cm) node[right] {}
+ -- (220:2cm) node[left] {}
+ arc[start angle=220, end angle=320, radius=2cm]
+ -- cycle;
+ \node {Major Segment};
+ \node at (-90:1.5) {Minor Segment};
+
+ \begin{scope}[xshift=4.5cm]
+ \draw [fill=lblue] circle (2cm);
+ \filldraw[fill=white]
+ (320:2cm) node[right] {}
+ -- (0,0) node[above] {}
+ -- (220:2cm) node[left] {}
+ arc[start angle=220, end angle=320, radius=2cm]
+ -- cycle;
+ \node at (90:1cm) {Major Sector};
+ \node at (-90:1.5) {Minor Sector};
+ \end{scope}
+ \end{tikzpicture}
+
+
+ \begin{align*}
+ \textbf{Sectors: } A &= \pi r^2 \dfrac{\theta}{2\pi} \\
+ &= \dfrac{r^2 \theta}{2}
+ \end{align*}
+
+ \[ \textbf{Segments: } A = \dfrac{r^2}{2} \left(\theta - \sin \theta \right) \]
+
+ \begin{align*}
+ \textbf{Chords: } \operatorname{crd} \theta &= \sqrt{(1 - \cos\theta)^2 + \sin^2 \theta} \\
+ &= \sqrt{2 - 2\cos\theta} \\
+ &= 2 \sin \left(\dfrac{\theta}{2}\right)
+ \end{align*}
+
\section{Differential calculus}
+ \[f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={\frac{dy}{dx}}\]
+
\subsection*{Limits}
\[\lim_{x \rightarrow a}f(x)\]
\(f(x)\) is continuous \(\iff L^-=L^+=f(x) \> \forall x\)
\end{enumerate}
- \subsection*{Gradients of secants and tangents}
+ \subsection*{Gradients}
\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}
+ \subsubsection*{Points of Inflection}
- \(\log_b (xy)=\log_b x + \log_b y\)\\
- \(\log_b x^n = n \log_b x\)\\
- \(\log_b y^{x^n} = x^n \log_b y\)
+ \emph{Stationary point} - i.e.
+ \(f^\prime(x)=0\)\\
+ \emph{Point of inflection} - max \(|\)gradient\(|\) (i.e.
+ \(f^{\prime\prime} = 0\))
- \subsubsection*{Index identities}
+ \subsubsection*{Strictly increasing/decreasing}
- \(b^{m+n}=b^m \cdot b^n\)\\
- \((b^m)^n=b^{m \cdot n}\)\\
- \((b \cdot c)^n = b^n \cdot c^n\)\\
- \({a^m \div a^n} = {a^{m-n}}\)
+ For \(x_2\) and \(x_1\) where \(x_2 > x_1\):
- \subsection*{Derivative rules}
+ \textbf{strictly increasing}\\
+ \-\hspace{1em}where \(f(x_2) > f(x_1)\) or \(f^\prime(x)>0\)
- \renewcommand{\arraystretch}{1.4}
- \begin{tabularx}{\columnwidth}{rX}
- \hline
- \(f(x)\) & \(f^\prime(x)\)\\
- \hline
- \(\sin x\) & \(\cos x\)\\
- \(\sin ax\) & \(a\cos ax\)\\
- \(\cos x\) & \(-\sin x\)\\
- \(\cos ax\) & \(-a \sin ax\)\\
- \(\tan f(x)\) & \(f^2(x) \sec^2f(x)\)\\
- \(e^x\) & \(e^x\)\\
- \(e^{ax}\) & \(ae^{ax}\)\\
- \(ax^{nx}\) & \(an \cdot e^{nx}\)\\
- \(\log_e x\) & \(\dfrac{1}{x}\)\\
- \(\log_e {ax}\) & \(\dfrac{1}{x}\)\\
- \(\log_e f(x)\) & \(\dfrac{f^\prime (x)}{f(x)}\)\\
- \(\sin(f(x))\) & \(f^\prime(x) \cdot \cos(f(x))\)\\
- \(\sin^{-1} x\) & \(\dfrac{1}{\sqrt{1-x^2}}\)\\
- \(\cos^{-1} x\) & \(\dfrac{-1}{sqrt{1-x^2}}\)\\
- \(\tan^{-1} x\) & \(\dfrac{1}{1 + x^2}\)\\
- \(\frac{d}{dy}f(y)\) & \(\dfrac{1}{\frac{dx}{dy}}\) (reciprocal)\\
- \(uv\) & \(u \frac{dv}{dx}+v\frac{du}{dx} (product rule)\)\\
- \(\dfrac{u}{v}\) & \(\dfrac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\) (quotient rule)\\
- \(f(g(x))\) & \(f^\prime(g(x))\cdot g^\prime(x)\)\\
- \hline
- \end{tabularx}
+ \textbf{strictly decreasing}\\
+ \hspace{1em}where \(f(x_2) < f(x_1)\) or \(f^\prime(x)<0\)
+ \begin{warning}
+ Endpoints are included, even where \(\boldsymbol{\frac{dy}{dx}=0}\)
+ \end{warning}
- \subsection*{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)\\
\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.
- \(f^{\prime\prime} = 0\))
+ \subsection*{Slope fields}
+
+ \begin{tikzpicture}[declare function={diff(\x,\y) = \x+\y;}]
+ \begin{axis}[axis equal, ymin=-4, ymax=4, xmin=-4, xmax=4, ticks=none, enlargelimits=true, ]
+ \addplot[thick, orange, domain=-4:2] {e^(x)-x-1};
+ \pgfplotsinvokeforeach{-4,...,4}{%
+ \draw[gray] ( {#1 -0.1}, {4 - diff(#1, 4) *0.1}) -- ( {#1 +0.1}, {4 + diff(#1, 4) *0.1});
+ \draw[gray] ( {#1 -0.1}, {3 - diff(#1, 3) *0.1}) -- ( {#1 +0.1}, {3 + diff(#1, 3) *0.1});
+ \draw[gray] ( {#1 -0.1}, {2 - diff(#1, 2) *0.1}) -- ( {#1 +0.1}, {2 + diff(#1, 2) *0.1});
+ \draw[gray] ( {#1 -0.1}, {1 - diff(#1, 1) *0.1}) -- ( {#1 +0.1}, {1 + diff(#1, 1) *0.1});
+ \draw[gray] ( {#1 -0.1}, {0 - diff(#1, 0) *0.1}) -- ( {#1 +0.1}, {0 + diff(#1, 0) *0.1});
+ \draw[gray] ( {#1 -0.1}, {-1 - diff(#1, -1) *0.1}) -- ( {#1 +0.1}, {-1 + diff(#1, -1) *0.1});
+ \draw[gray] ( {#1 -0.1}, {-2 - diff(#1, -2) *0.1}) -- ( {#1 +0.1}, {-2 + diff(#1, -2) *0.1});
+ \draw[gray] ( {#1 -0.1}, {-3 - diff(#1, -3) *0.1}) -- ( {#1 +0.1}, {-3 + diff(#1, -3) *0.1});
+ \draw[gray] ( {#1 -0.1}, {-4 - diff(#1, -4) *0.1}) -- ( {#1 +0.1}, {-4 + diff(#1, -4) *0.1});
+ }
+ \end{axis}
+ \end{tikzpicture}
\begin{table*}[ht]
\centering
- \begin{tabularx}{\textwidth}{rXXX}
+ \begin{tabularx}{\textwidth}{|r|Y|Y|Y|}
\hline
- \rowcolor{shade2}
- & \centering\(\dfrac{d^2 y}{dx^2} > 0\) & \centering \(\dfrac{d^2y}{dx^2}<0\) & \(\dfrac{d^2y}{dx^2}=0\) (inflection) \\
+ \rowcolor{lblue}
+ & \adjustbox{margin=0 1ex, valign=m}{\centering\(\dfrac{d^2 y}{dx^2} > 0\)} & \adjustbox{margin=0 1ex, valign=m}{\centering \(\dfrac{d^2y}{dx^2}<0\)} & \adjustbox{margin=0 1ex, valign=m}{\(\dfrac{d^2y}{dx^2}=0\) (inflection)} \\
\hline
\(\dfrac{dy}{dx}>0\) &
- \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-3, xmax=0.8, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(e^(x))}; \addplot[red] {x/2.5+0.75}; \end{axis}\end{tikzpicture} \\Rising (concave up)}&
+ \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-3, xmax=0.8, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(e^(x)}; \addplot[red] {x/2.5+0.75}; \end{axis}\end{tikzpicture} \\Rising (concave up)}&
\makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=0.1, xmax=4, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(ln(x))}; \addplot[red] {x/1.5-0.56}; \end{axis}\end{tikzpicture} \\Rising (concave down)}&
\makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1.5, xmax=1.5, scale=0.2, samples=100] \addplot[blue] {(sin((deg x)))}; \addplot[red] {x}; \end{axis}\end{tikzpicture} \\Rising inflection point}\\
\hline
\(\dfrac{dy}{dx}<0\) &
- \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-.5, xmax=1, ymin=-.5, ymax=.5, scale=0.2, samples=100] \addplot[blue] {(1/(x+1)-1}; \addplot[red] {-x}; \end{axis}\end{tikzpicture} \\Falling (concave up)}&
+ \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-.5, xmax=1, ymin=-.5, ymax=.5, scale=0.2, samples=100] \addplot[blue] {1/(x+1)-1}; \addplot[red] {-x}; \end{axis}\end{tikzpicture} \\Falling (concave up)}&
\makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=0, xmax=1.5, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(2-x*x)^(1/2)}; \addplot[red] {-x+2}; \end{axis}\end{tikzpicture} \\Falling (concave down)}&
\makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=1.5, xmax=4.5, scale=0.2, samples=100] \addplot[blue] {(sin((deg x)))}; \addplot[red] {-x+3.1415}; \end{axis}\end{tikzpicture} \\Falling inflection point}\\
\hline
\(\dfrac{dy}{dx}=0\)&
- \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1, xmax=1, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(x*x))}; \addplot[red, thick] {0}; \end{axis}\end{tikzpicture} \\Local minimum}& \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1, xmax=1, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(-x*x))}; \addplot[red, very thick] {0}; \end{axis}\end{tikzpicture} \\Local maximum}&
- \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1, xmax=1, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(x*x*x))}; \addplot[red, thick] {0}; \end{axis}\end{tikzpicture} \(\>\) \begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1, xmax=1, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(-x*x*x))}; \addplot[red, thick] {0}; \end{axis}\end{tikzpicture} \\Stationary inflection point}\\
+ \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1, xmax=1, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(x*x)}; \addplot[red, thick] {0}; \end{axis}\end{tikzpicture} \\Local minimum}& \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1, xmax=1, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(-x*x)}; \addplot[red, very thick] {0}; \end{axis}\end{tikzpicture} \\Local maximum}&
+ \makecell{\\\begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1, xmax=1, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(x*x*x)}; \addplot[red, thick] {0}; \end{axis}\end{tikzpicture} \(\>\) \begin{tikzpicture}\begin{axis}[axis x line=none, axis y line=none, xmin=-1, xmax=1, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(-x*x*x)}; \addplot[red, thick] {0}; \end{axis}\end{tikzpicture} \\Stationary inflection point}\\
\hline
\end{tabularx}
\end{table*}
\begin{itemize}
\item
- if \(f^\prime (a) = 0\) and \(f^{\prime\prime}(a) > 0\), then point
- \((a, f(a))\) is a local min (curve is concave up)
+ \(f^\prime (a) = 0, \> f^{\prime\prime}(a) > 0\) \\
+ \textbf{local min} at \((a, f(a))\) (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)
+ \(f^\prime (a) = 0, \> f^{\prime\prime} (a) < 0\) \\
+ \textbf{local max} at \((a, f(a))\) (concave down)
\item
- if \(f^{\prime\prime}(a) = 0\), then point \((a, f(a))\) is a point of
- inflection
+ \(f^{\prime\prime}(a) = 0\) \\
+ \textbf{point of inflection} at \((a, f(a))\)
\item
- if also \(f^\prime(a)=0\), then it is a stationary point of inflection
+ \(f^{\prime\prime}(a) = 0, \> f^\prime(a)=0\) \\
+ stationary point of inflection at \((a, f(a)\)
\end{itemize}
\subsection*{Implicit Differentiation}
\[{\frac{dp}{dx}} = {\frac{dq}{dx}} \quad \text{and} \quad {\frac{dp}{dy}} = {\frac{dq}{dy}}\]
- \noindent \colorbox{cas}{\textbf{On CAS:}}\\
- Action \(\rightarrow\) Calculation \(\rightarrow\) \texttt{impDiff(y\^{}2+ax=5,\ x,\ y)}\\
- Returns \(y^\prime= \dots\).
-
- \subsection*{Integration}
+ \begin{cas}
+ Action \(\rightarrow\) Calculation \\
+ \-\hspace{1em}\texttt{impDiff(y\^{}2+ax=5,\ x,\ y)}
+ \end{cas}
- \[\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)\]
-
- \subsection*{Integral laws}
-
- \renewcommand{\arraystretch}{1.4}
- \begin{tabularx}{\columnwidth}{rX}
- \hline
- \(f(x)\) & \(\int f(x) \cdot dx\) \\
- \hline
- \(k\) (constant) & \(kx + c\)\\
- \(x^n\) & \(\dfrac{1}{n+1} x^{n+1}\) \\
- \(a x^{-n}\) &\(a \cdot \log_e |x| + c\)\\
- \(\dfrac{1}{ax+b}\) &\(\dfrac{1}{a} \log_e (ax+b) + c\)\\
- \((ax+b)^n\) & \(\dfrac{1}{a(n+1)}(ax+b)^{n-1} + c\>|\>n\ne 1\)\\
- \((ax+b)^{-1}\) & \(\dfrac{1}{a}\log_e |ax+b|+c\)\\
- \(e^{kx}\) & \(\dfrac{1}{k} e^{kx} + c\)\\
- \(e^k\) & \(e^kx + c\)\\
- \(\sin kx\) & \(\dfrac{-1}{k} \cos (kx) + c\)\\
- \(\cos kx\) & \(\dfrac{1}{k} \sin (kx) + c\)\\
- \(\sec^2 kx\) & \(\dfrac{1}{k} \tan(kx) + c\)\\
- \(\dfrac{1}{\sqrt{a^2-x^2}}\) & \(\sin^{-1} \dfrac{x}{a} + c \>\vert\> a>0\)\\
- \(\dfrac{-1}{\sqrt{a^2-x^2}}\) & \(\cos^{-1} \dfrac{x}{a} + c \>\vert\> a>0\)\\
- \(\frac{a}{a^2-x^2}\) & \(\tan^{-1} \frac{x}{a} + c\)\\
- \(\frac{f^\prime (x)}{f(x)}\) & \(\log_e f(x) + c\)\\
- \(\int f(u) \cdot \frac{du}{dx} \cdot dx\) & \(\int f(u) \cdot du\) (substitution)\\
- \(f(x) \cdot g(x)\) & \(\int [f^\prime(x) \cdot g(x)] dx + \int [g^\prime(x) f(x)] dx\)\\
- \hline
- \end{tabularx}
+ \subsection*{Function of the dependent
+ variable}
- Note \(\sin^{-1} {x \over a} + \cos^{-1} {x \over a}\) is constant \(\forall x \in (-a, a)\)
+ If \({\frac{dy}{dx}}=g(y)\), then
+ \(\frac{dx}{dy} = 1 \div \frac{dy}{dx} = \frac{1}{g(y)}\). Integrate both sides to solve equation. Only add \(c\) on one side. Express
+ \(e^c\) as \(A\).
- \subsection*{Definite integrals}
+ \subsection*{Reciprocal derivatives}
- \[\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)\]
+ \[\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}\]
- \begin{itemize}
+ \subsection*{Differentiating \(x=f(y)\)}
+ Find \(\dfrac{dx}{dy}\), then \(\dfrac{dy}{dx} = \dfrac{1}{\left(\dfrac{dx}{dy}\right)}\)
- \item
- Signed area enclosed by\\
- \(\> y=f(x), \quad y=0, \quad x=a, \quad x=b\).
- \item
- \emph{Integrand} is \(f\).
- \end{itemize}
+ \subsection*{Parametric equations}
- \subsubsection*{Properties}
- \[\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx\]
+ \begin{align*}
+ \dfrac{dy}{dt} &= \dfrac{dy}{dx} \cdot \dfrac{dx}{dt} \\
+ \therefore \dfrac{dy}{dx} &= \dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)} \text{ provided } \dfrac{dx}{dt} \ne 0 \\
+ \dfrac{d^2y}{dx^2} &= \dfrac{\left(\dfrac{dy^\prime}{dt}\right)}{\left(\dfrac{dx}{dt}\right)} \text{ where } y^\prime = \dfrac{dy}{dx}
+ \end{align*}
- \[\int^a_a f(x) \> dx = 0\]
+ \subsection*{Integration}
- \[\int^b_a k \cdot f(x) \> dx = k \int^b_a f(x) \> dx\]
+ \[\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)\]
- \[\int^b_a f(x) \pm g(x) \> dx = \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx\]
+ \subsubsection*{Properties}
- \[\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx\]
+ \begin{align*}
+ \int^b_a f(x) \> dx &= \int^c_a f(x) \> dx + \int^b_c f(x) \> dx \\
+ \int^a_a f(x) \> dx &= 0 \\
+ \int^b_a k \cdot f(x) \> dx &= k \int^b_a f(x) \> dx \\
+ \int^b_a f(x) \pm g(x) \> dx &= \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx \\
+ \int^b_a f(x) \> dx &= - \int^a_b f(x) \> dx \\
+ \end{align*}
\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{warning}
+ \(\boldsymbol{f(u)}\) must be 1:1 \(\boldsymbol{\implies}\) one \(\boldsymbol{x}\) for each \(\boldsymbol{y}\)
+ \end{warning}
\begin{align*}\text{e.g. for } y&=\int(2x+1)\sqrt{x+4} \cdot dx\\
\text{let } u&=x+4\\
\implies& {\frac{du}{dx}} = 1\\
\(\sin 2x = 2 \sin x \cos x\)
\end{itemize}
+ \subsection*{Separation of variables}
+
+ If \({\frac{dy}{dx}}=f(x)g(y)\), then:
+
+ \[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\]
+
\subsection*{Partial fractions}
- \colorbox{cas}{On CAS:}\\
- \indent Action \(\rightarrow\) Transformation \(\rightarrow\)
- \texttt{expand/combine}\\
- \indent Interactive \(\rightarrow\) Transformation \(\rightarrow\)
- Expand \(\rightarrow\) Partial
+ To factorise \(f(x) = \frac{\delta}{\alpha \cdot \beta}\):
+ \begin{align*}
+ \dfrac{\delta}{\alpha \cdot \beta \cdot \gamma} &= \dfrac{A}{\alpha} + \dfrac{B}{\beta} + \dfrac{C}{\gamma} \tag{1} \\
+ \text{Multiply by } & (\alpha \cdot \beta \cdot \gamma) \text{:} \\
+ \delta &= \beta\gamma A + \alpha\gamma B +\alpha\beta C \tag{2} \\
+ \text{Substitute } x &= \{\alpha, \beta, \gamma\} \text{ into (2) to find denominators}
+ \end{align*}
- \subsection*{Graphing integrals on CAS}
+ \subsubsection*{Repeated linear factors}
- \colorbox{cas}{In main:} Interactive \(\rightarrow\) Calculation \(\rightarrow\)
- \(\int\) (\(\rightarrow\) Definite)\\
- Restrictions: \texttt{Define\ f(x)=..} then \texttt{f(x)\textbar{}x\textgreater{}..}
+ \[ \dfrac{p(x)}{(x-a)^n} = \dfrac{A_1}{(x-a)} + \dfrac{A_2}{(x-a)^2} + \dots + \dfrac{A_n}{(x-a)^n} \]
- \subsection*{Applications of antidifferentiation}
+ \subsubsection*{Irreducible quadratic factors}
- \begin{itemize}
+ \[ \text{e.g. } \dfrac{3x-4}{(2x-3)(x^2+5)} = \dfrac{A}{2x-3} + \dfrac{Bx+C}{x^2+5} \]
- \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}
+ \begin{cas}
+ Action \(\rightarrow\) Transformation:\\
+ \-\hspace{1em} \texttt{expand(..., x)}
- 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.
+ To reverse, use \texttt{combine(...)}
+ \end{cas}
+
+ \subsection*{Integrating \(\boldsymbol{\dfrac{dy}{dx} = g(y)}\)}
+
+ \[ \text{if } \dfrac{dy}{dx} = g(y), \text{ then } x = \int{\dfrac{1}{g(y)}} \> dy \]
+
+ \subsection*{Graphing integrals on CAS}
+
+ \begin{cas}
+ \textbf{In main:} Interactive \(\rightarrow\) Calculation \(\rightarrow\) \(\int\)\\
+ For restrictions, \texttt{Define\ f(x)=...} then \texttt{f(x)\textbar{}x\textgreater{}...}
+ \end{cas}
\subsection*{Solids of revolution}
Approximate as sum of infinitesimally-thick cylinders
- \subsubsection*{Rotation about \(x\)-axis}
+ \subsubsection*{Rotation about \(\boldsymbol{x}\)-axis}
- \begin{align*}
- V &= \int^{x=b}_{x-a} \pi y^2 \> dx \\
- &= \pi \int^b_a (f(x))^2 \> dx
- \end{align*}
+ \[ V = \pi\int^{x=b}_{x=a} f(x)^2 \> dx \]
- \subsubsection*{Rotation about \(y\)-axis}
+ \subsubsection*{Rotation about \(\boldsymbol{y}\)-axis}
\begin{align*}
- V &= \int^{y=b}_{y=a} \pi x^2 \> dy \\
- &= \pi \int^b_a (f(y))^2 \> dy
+ V &= \pi \int^{y=b}_{y=a} x^2 \> dy \\
+ &= 2\pi \int^{x=b}_{x=a} x|f(x)| \> dx
\end{align*}
- \subsubsection*{Regions not bound by \(y=0\)}
+ \subsubsection*{Rotating the area between two graphs}
- \[V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx\]
+ \[V = \pi \int^b_a \left( f(x)^2 - g(x)^2 \right) \> dx\]
\hfill where \(f(x) > g(x)\)
\subsection*{Length of a curve}
- \[L = \int^b_a \sqrt{1 + ({\frac{dy}{dx}})^2} \> dx \quad \text{(Cartesian)}\]
+ For length of \(f(x)\) from \(x=a \rightarrow x=b\):
+ \begin{align*}
+ &\text{Cartesian} \> & L &= \int^b_a \sqrt{1 + \left(\dfrac{dy}{dx}\right)^2} \> dx \\
+ &\text{Parametric} \> & L & = \int^b_a \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2} \> dt
+ \end{align*}
+
+ \begin{cas}
+ \begin{enumerate}[label=\alph*), leftmargin=5mm]
+ \item Evaluate formula
+ \item Interactive \(\rightarrow\) Calculation \(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen}
+ \end{enumerate}
+ \end{cas}
- \[L = \int^b_a \sqrt{{\frac{dx}{dt}} + ({\frac{dy}{dt}})^2} \> dt \quad \text{(parametric)}\]
+ \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}
- \noindent \colorbox{cas}{On CAS:}\\
- \indent Evaluate formula,\\
- \indent or Interactive \(\rightarrow\) Calculation
- \(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen}
+ 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*{Rates}
\[f(x) = \frac{P(x)}{Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]
- \subsubsection*{Addition of ordinates}
+ \subsection*{Euler's method}
- \begin{itemize}
+ \[\dfrac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
- \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}
+ \[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
+
+ \begin{theorembox}{}
+ If \(\dfrac{dy}{dx} = g(x)\) with \(x_0 = a\) and \(y_0 = b\), then:
+ \[\begin{cases}
+ x_{n+1} = x_n + h \\
+ y_{n+1} = y_n + hg(x_n)
+ \end{cases}\]
+ \end{theorembox}
+
+ \[
+ \dfrac{d^2y}{dx^2}
+ \begin{cases}
+ > 0 \implies \text{ underestimate (concave up)} \\
+ < 0 \implies \text{ overestimate (concave down)}
+ \end{cases}
+ \]
+
+ \begin{center}\begin{tikzpicture}
+ \begin{axis}[xmin=0, xmax=1.6, ticks=none, enlargelimits=true, samples=100]
+ \addplot[blue, domain=-0.25:1.5, postaction={decorate,decoration={text along path, text align={align=center, left indent=3cm}, text={|\sffamily|solution curve}}}] {e^(x-3/2)+1/4};
+ \addplot[red] {(x+1/2)*e^(-1)+1/4} (1.7,1.0593) node [above, black] {\(\ell\)};
+ \addplot[mark=*, black] coordinates {(0.5,0.6179)} node[above left]{\((x_0, y_0)\)};
+ \addplot[mark=*, orange] coordinates {(1.4,1.1548)} node[left]{\color{black} \sffamily correct solution};
+ \addplot[mark=*, black] coordinates {(1.4,0.94897)} node[above right] {\((x_1,y_1)\)};
+ \draw [gray, dashed] (0.5,0) -- (0.5,0.6179) -- (1.6,0.6179);
+ \draw [gray, dashed] (1.4,0) -- (1.4, 1.1548);
+ \draw [<->] (0.5,0.48) -- (1.4,0.48) node[midway, fill=white] {\(h\)};
+ \draw [gray, dashed] (1.4,0.94897) -- (1.6,0.94897);
+ \draw [<->] (1.5,0.94897) -- (1.5,0.6179) node[midway, rotate=90, below] {\(hg(x_0)\)};
+ \end{axis}
+ \end{tikzpicture}\end{center}
+
+ \begin{cas}
+ Menu \(\rightarrow\) Sequence \(\rightarrow\) Recursive
+
+ \textbf{To generate \(\boldsymbol{x}\)-values:}
+ \begin{itemize}
+ \item Enter \(a_{n+1}=a_n + h\) where \(h\) is the step size \\
+ (input \(a_n\) from menu bar)
+ \item In \(a_0\), set the initial value \(x_0\) as a constant
+ \end{itemize}
+
+ \textbf{To generate \(\boldsymbol{y}\)-values:}
+ \begin{itemize}
+ \item In \(b_{n+1}\), enter \(\dfrac{dy}{dx}\), replacing \(x\) with \(a_n\)
+ \item Set \(b_0 = y(x_0)\) as a constant
+ \end{itemize}
+
+ To view table of values, tap table icon (top left) \\
+ To compare approximations with actual values, enter in \(c_{n+1} = a_{n+1} - f(a_{n+1})\) where \(f(x) = \int \dfrac{dy}{dx} \> dx\)
+
+ \end{cas}
\subsection*{Fundamental theorem of calculus}
\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}
+ \begin{warning}
+ To verify solutions, find \(\frac{dy}{dx}\) from \(y\) and substitute into original
+ \end{warning}
+
+ \vspace*{1cm}
+ \hspace*{-1cm}
- Start with \(y=\dots\), and differentiate. Substitute into original
- equation.
+ { \tabulinesep=1.2mm
+ \begin{tabu}{|c|c|}
- \subsubsection*{Function of the dependent
- variable}
+ \hline
+ \taburowcolors 2{gray..white}
+ \textbf{DE} & \textbf{Method} \\
+ \hline
- 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 \(A\).
+ \tabureset
+ \(\dfrac{dy}{dx} = f(x)\)
+ &
+ {\(\begin{aligned}
+ y &= \int f(x) \> dx \\
+ &= F(x) + c \quad \text{where } F^\prime(x) = f(x)
+ \end{aligned}\)} \\
+ \hline
+ \(\dfrac{d^2y}{dx^2} = f(x)\)
+ &
+ {\(\begin{aligned}
+ \dfrac{dy}{dx} &= \int f(x) \> dx \\
+ &= F(x) + c \quad \text{where } F^\prime(x) = f(x) \\
+ \therefore y &= \iint f(x) \> dx = \int \left( F(x) + c \right) \> dx \\
+ &= G(x) + cx + d \\
+ & \text{where } G^\prime(x) = F(x)
+ \end{aligned}\)} \\
- \subsubsection*{Mixing problems}
+ \hline
- \[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\]
+ \(\dfrac{dy}{dx} = g(y)\)
+ &
+ {\(\begin{aligned}
+ \dfrac{dx}{dy} &= \dfrac{1}{g(y)} \\
+ \therefore x &= \int \dfrac{1}{g(y)} \> dy \\
+ &= F(y) + c \\
+ & \text{where } F^\prime(y) = \dfrac{1}{g(y)}
+ \end{aligned}\)} \\
- \subsubsection*{Separation of variables}
+ \hline
- If \({\frac{dy}{dx}}=f(x)g(y)\), then:
+ \(\dfrac{dy}{dx} = f(x) g(y)\)
+ &
+ {\(\begin{aligned}
+ f(x) &= \dfrac{1}{g(y)} \cdot \dfrac{dy}{dx} \\
+ \int f(x) \> dx &= \int \dfrac{1}{g(y)} \> dy
+ \end{aligned}\)} \\
- \[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\]
+ \hline
+ \end{tabu}}
- \subsubsection*{Euler's method for solving DEs}
+ \subsubsection*{Mixing problems}
- \[\frac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
+ \[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\]
- \[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
+ \include{calculus-rules}
-
\section{Kinematics \& Mechanics}
\subsection*{Constant acceleration}
\subsubsection*{Velocity-time graphs}
- \begin{itemize}
- \item Displacement: \textit{signed} area between graph and \(t\) axis
- \item Distance travelled: \textit{total} area between graph and \(t\) axis
- \end{itemize}
+ \begin{description}[nosep, labelindent=0.5cm, leftmargin=0.5\columnwidth]
+ \item[Displacement:] \textit{signed} area
+ \item[Distance travelled:] \textit{total} area
+ \end{description}
\[ \text{acceleration} = \frac{d^2x}{dt^2} = \frac{dv}{dt} = v\frac{dv}{dx} = \frac{d}{dx}\left(\frac{1}{2}v^2\right) \]
\begin{center}
\renewcommand{\arraystretch}{1}
\begin{tabular}{ l r }
- \hline & no \\ \hline
- \(v=u+at\) & \(x\) \\
- \(v^2 = u^2+2as\) & \(t\) \\
- \(s = \frac{1}{2} (v+u)t\) & \(a\) \\
- \(s = ut + \frac{1}{2} at^2\) & \(v\) \\
- \(s = vt- \frac{1}{2} at^2\) & \(u\) \\ \hline
- \end{tabular}
+ \hline & no \\ \hline
+ \(v=u+at\) & \(x\) \\
+ \(v^2 = u^2+2as\) & \(t\) \\
+ \(s = \frac{1}{2} (v+u)t\) & \(a\) \\
+ \(s = ut + \frac{1}{2} at^2\) & \(v\) \\
+ \(s = vt- \frac{1}{2} at^2\) & \(u\) \\ \hline
+ \end{tabular}
\end{center}
\[ v_{\text{avg}} = \frac{\Delta\text{position}}{\Delta t} \]
\end{align*}
\noindent \textbf{Distance travelled between \(t=a \rightarrow t=b\):}
- \[= \int^b_a \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \cdot dt \]
+ \begin{align*}
+ &= \int^{b}_{a}{\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}} \> dt \tag{2D} \\
+ &= \int^{t=b}_{t=a}{\dfrac{dx}{dt}} \> dt \tag{linear}
+ \end{align*}
\noindent \textbf{Shortest distance between \(\boldsymbol{r}(t_0)\) and \(\boldsymbol{r}(t_1)\):}
\[ = |\boldsymbol{r}(t_1) - \boldsymbol{r}(t_2)| \]