479 lines
18 KiB
TeX
479 lines
18 KiB
TeX
\documentclass[10pt,landscape]{article}
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\usepackage[scaled=0.8]{helvet}
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\usepackage{calc}
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\usepackage{multicol}
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\usepackage{ifthen}
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\usepackage[a4paper,margin=3mm,landscape]{geometry}
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\usepackage{amsmath,amsthm,amsfonts,amssymb}
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\usepackage{hyperref}
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\usepackage{newtxtext}
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\usepackage{enumitem}
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\usepackage{amssymb}
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\usepackage[table]{xcolor}
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\usepackage{vwcol}
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\usepackage{tikz}
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\usepackage{wrapfig}
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\usepackage{pgfplots}
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\usepackage{makecell}
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% Testing %
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\usepackage{blindtext}
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\usetikzlibrary{calc}
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\setlist{nosep}
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\graphicspath{ {./images/} }
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\pagestyle{empty}
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\newenvironment{tightcenter}{%
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\setlength\topsep{0pt}
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\setlength\parskip{0pt}
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\begin{center}
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}{%
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\end{center}
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}
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% redefine section commands to use less space
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\makeatletter
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\renewcommand{\section}{\@startsection{section}{1}{0mm}%
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{-1ex plus -.5ex minus -.2ex}%
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{0.5ex plus .2ex}%x
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{\normalfont\large\bfseries}}
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\renewcommand{\subsection}{\@startsection{subsection}{2}{0mm}%
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{-1explus -.5ex minus -.2ex}%
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{0.5ex plus .2ex}%
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{\normalfont\normalsize\bfseries}}
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\renewcommand{\subsubsection}{\@startsection{subsubsection}{3}{0mm}%
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{-1ex plus -.5ex minus -.2ex}%
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{\normalfont\small\bfseries}}%
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\renewcommand{\familydefault}{\sfdefault}
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\renewcommand\rmdefault{\sfdefault}
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% makes nested numbering (e.g. 1.1.1, 1.1.2, etc)
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\renewcommand{\labelenumii}{\theenumii}
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\renewcommand{\theenumii}{\theenumi.\arabic{enumii}.}
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\renewcommand\labelitemii{•}
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% for logical not operator
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\renewcommand{\lnot}{\mathord{\sim}}
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\renewcommand{\bf}[1]{\textbf{#1}}
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\newcommand{\abs}[1]{\vert #1 \vert}
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\newcommand{\Mod}[1]{\ \mathrm{mod}\ #1}
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\newcommand{\vv}[1]{\boldsymbol{#1}}
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\newcommand{\VV}[1]{\overrightarrow{#1}}
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\newcommand{\cvv}[1]{\left(\begin{smallmatrix}#1\end{smallmatrix}\right)}
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\makeatother
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\definecolor{myblue}{cmyk}{1,.72,0,.38}
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% Define BibTeX command
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\everymath\expandafter{\the\everymath \color{myblue}}
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\def\BibTeX{{\rm B\kern-.05em{\sc i\kern-.025em b}\kern-.08em
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T\kern-.1667em\lower.7ex\hbox{E}\kern-.125emX}}
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\let\iff\leftrightarrow
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\let\Iff\Leftrightarrow
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\let\then\rightarrow
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\let\Then\Rightarrow
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% Don't print section numbers
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\setcounter{secnumdepth}{0}
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\setlength{\parindent}{0pt}
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\setlength{\parskip}{0pt plus 0.5ex}
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%% this changes all items (enumerate and itemize)
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\setlength{\leftmargini}{0.5cm}
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\setlength{\leftmarginii}{0.5cm}
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\setlist[itemize,1]{leftmargin=2mm,labelindent=1mm,labelsep=1mm}
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\setlist[itemize,2]{leftmargin=4mm,labelindent=1mm,labelsep=1mm}
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% My Environments
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\newtheorem{example}[section]{Example}
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% -----------------------------------------------------------------------
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\begin{document}
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\raggedright
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\footnotesize
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\begin{multicols*}{4}
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\setlength{\columnseprule}{0.25pt}
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\setlength{\premulticols}{1pt}
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\setlength{\postmulticols}{1pt}
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\setlength{\multicolsep}{1pt}
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\setlength{\columnsep}{2pt}
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\section{Function and Limits}
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\begin{itemize}
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\item $\lim\limits_{x\to \pm \infty}\frac{Ax^\alpha}{Bx^\beta}
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\begin{cases}
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0 & \text{if} \alpha < \beta\\
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\frac{A}{B} & \text{if} \alpha = \beta\\
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\pm \infty & \text{if} \alpha > \beta\\
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\end{cases}$
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\item $\lim\limits_{x\to c}\frac{sin(g(x))}{g(x)} = 1(\lim\limits_{x \to c}g(x) = 0)$
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\item $\lim\limits_{x\to c}\frac{tan(g(x))}{g(x)} = 1$
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\item $\lim\limits_{x\to 0}\frac{sin(x)}{x} = 1$
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\item $\lim\limits_{x\to 0}\frac{tan(x)}{x} = 1$
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\item $\lim\limits_{n \to \infty} \frac{n!}{n^{n}} = 0$
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\end{itemize}
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\section{Differentiation}
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parametric differentiaton: $\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$
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\begin{tabular}{|>{\color{black}}c | >{\color{black}}c|}
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\hline
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$f(x)$ & $f'(x)$
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\\ \hline
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\rule{0pt}{2.3ex} % top spacing
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$\tan x$ & $\sec ^2 x$ \\
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$\csc x$ & $-\csc x \cot x$ \\
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$\sec x$ & $\sec x \tan x$ \\
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$\cot x$ & $- \csc ^2 x$
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\\ \hline
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\rule{0pt}{2.3ex} % top spacing
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$a^{f(x)}$ & $\ln a \cdot f'(x)a^{f(x)}$ \\
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$\log_af(x)$ & $\log_a e \cdot \frac{f'(x)}{f(x)}$
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\\[1ex] \hline
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\rule{0pt}{3ex} % top spacing
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$\sin^{-1} f(x)$ & $\frac{f'(x)}{\sqrt{1-[f(x)]^2}}, \ \ _{\vert f(x) \vert < 1}$ \\[1.5ex]
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$\cos^{-1} f(x)$ & $-\frac{f'(x)}{\sqrt{1-[f(x)]^2}}, \ \ _{\vert f(x) \vert < 1}$ \\[1.5ex]
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$\tan^{-1} f(x)$ & $\frac{f'(x)}{1 + [f(x)]^2}$ \\[1.5ex]
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$\cot^{-1} f(x)$ & $-\frac{f'(x)}{1 + [f(x)]^2}$ \\[1.5ex]
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$\sec^{-1} f(x)$ & $\frac{f'(x)}{\vert f(x) \vert \sqrt{[f(x)]^2-1}}$ \\[1.5ex]
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$\csc^{-1} f(x)$ & $-\frac{f'(x)}{\vert f(x) \vert \sqrt{[f(x)]^2-1}}$ \\[2ex]
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\hline
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\end{tabular}
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\textbf{Second Derivative} Test: $f'(c) = 0, f''(c) < 0$ then local max, $f''(c) > 0$ local min.
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\textbf{L'Hopital's Rule}: Given $\lim\limits_{x\to c}f(x) $ and $ g(x) = 0 $ or $ \pm \infty$ $ \lim\limits_{x \to c}\frac{f(x)}{g(x)} = \lim\limits_{x \to c}\frac{f'(x)}{g'(x)}$
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\begin{itemize}
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\item Use for $\frac{0}{0}$ or $\frac{\infty}{\infty}$
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\end{itemize}
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\subsection*{Trigo Identities}
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\begin{enumerate}
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\item $\sec^2x - 1 = \tan^2x$
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\item $\csc^2x - 1 = \cot^2x$
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\item $\sin A\cos A = \frac{1}{2}\sin2A$
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\item $\cos^2A = \frac{1}{2}(1+\cos2A)$
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\item $\sin^2A = \frac{1}{2}(1-\cos2A)$
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\item $\sin A\cos B = \frac{1}{2}(\sin(A + B) + \sin(A - B)$
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\item $\cos A\sin B = \frac{1}{2}(\sin(A + B) - \sin(A - B)$
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\item $\cos A\cos B = \frac{1}{2}(\cos(A + B) + \cos(A - B)$
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\item $\sin A\sin B = \frac{1}{2}(\cos(A + B) - \cos(A - B)$
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\end{enumerate}
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\section{Integration}
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\begin{tabular}{|>{\color{black}}c | >{\color{black}}c|}
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\hline
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$f(x)$ & $\int f(x)$\\
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$\tan ax$ & $\frac{1}{a}\ln|\sec(ax)|$\\
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$\cot ax$ & $\frac{1}{a}\ln|\cot(ax)|$\\
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$\sec ax$ & $\frac{1}{a}\ln|\sec(ax) + tan(ax)|$\\
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$\csc ax$ & $\frac{1}{a}\ln|\csc(ax) + cot(ax)|$\\
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\hline
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$\frac{1}{a^2+(x+b)^2}$ & $\frac{1}{a}\tan^{-1}(\frac{x+b}{a})$\\
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$\frac{1}{\sqrt{a^2-(x+b)^2}}$ & $\sin^{-1}(\frac{x+b}{a})$\\
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$\frac{1}{a^2-(x+b)^2}$ & $\frac{1}{2a}\ln|\frac{x+b+a}{x+b-a}|$\\
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$\frac{1}{(x+b)^2-a^2}$ & $\frac{1}{2a}\ln|\frac{x+b-a}{x+b+a}|$\\
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\hline
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\end{tabular}
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\textbf{Substitution} $\int f(g(x)) \cdot g'(x) dx = \int f(u) du, u = g(x)$
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\textbf{By Parts} $\int u v' dx = uv - \int u'v dx$, order: LIATE: Differentiate to integrate
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\subsection{Application of Integration}
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about x axis
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\begin{itemize}
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\item Vol Disk: $V = \pi \int^b_a f(x)^2 - g(x)^2 dx$
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\item Vol Shell: $V = 2\pi\int^b_a x|f(x)-g(x)|dx$ (absolute!!)
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\item Length of curve: $\int^b_a \sqrt{1+f'(x)^2}dx$
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\end{itemize}
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\section{Vectors}
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unit vector: $\hat{p} = \frac{p}{|p|}$, $\VV{AB} = \VV{OB} - \VV{OA}$
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\begin{center}
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\begin{multicols}{2}
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\begin{tikzpicture}[scale=0.8, every node/.style={transform shape}]
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\coordinate[label=below left:O] (O) at (0,0);
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\coordinate[label=A] (A) at (0.3,1.6);
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\coordinate[label=B] (B) at (1.5, 1.4);
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\coordinate[label=P] (P) at (1, 1.5);
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\draw (O)
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-- node[left] {$\vv{a}$} (A)
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-- node[above] {$\lambda$} (P)
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-- node[above] {$\mu$} (B)
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-- node[right] {$\vv{b}$} (O)
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-- node[left] {$\vv{p}$} (P);
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\end{tikzpicture}
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\\ \textbf{ratio theorem}
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\\* $\vv{p} = \frac{\mu\vv{a} + \lambda\vv{b}}{\lambda + \mu}$
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\newline
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\\ \textbf{midpoint theorem}
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\\* $\vv{p} = \frac{\vv{a} + \vv{b}}{2}$
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\end{multicols}
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\end{center}
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\subsection{Dot Product}
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\begin{itemize}
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\item $\VV{a} \cdot \VV{b} = a_1b_1 + a_2b_2 + a_3b_3 = |a||b|\cos\theta$
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\item $a \perp b \Then a \cdot b = 0$
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\item $a \parallel b \Then a \cdot b = |a||b|$
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\end{itemize}
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\subsection{Cross Product}
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$
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\vv{a} \times \vv{b} =
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\begin{vmatrix} \vv{i} & \vv{j} & \vv{k} \\ a_i & a_2 & a_3 \\ b_i & b_2 & b_3 \end{vmatrix} =
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\begin{pmatrix} (a_2b_3 - a_3b_2) \\ -(a_1b_3 - a_3b_1) \\ (a_1b_2 - a_2b_1) \end{pmatrix}
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$
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\begin{center}
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\begin{multicols}{2}
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$|\vv{a} \times \vv{b}| = |\vv{a}||\vv{b}|\sin\theta$
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$a \perp b \Then a \times b = |a||b|$
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$a \parallel b \Then a \times b = 0$
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Area Parallelogram = $|\vv{a} \times \vv{b}|$
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\end{multicols}
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\end{center}
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\subsection{Projection}
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\begin{multicols}{2}
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\begin{tikzpicture}[scale=0.7, every node/.style={transform shape}]
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\coordinate[label=below left:O] (O) at (0,0);
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\coordinate[label=right:A] (A) at (2, 1);
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\coordinate[label=below:B] (B) at (3, 0);
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\coordinate[label=below:N] (N) at (2, 0);
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\draw (A)
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-- node[above] {$\vv{a}$} (O)
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-- node[below] {$\vv{b}$} (B);
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\draw[shorten >=0pt, dashed] (A) -- (N);
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\end{tikzpicture}
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$\triangle ANO = \frac{1}{2} \abs{\VV{OA} \times \VV{ON}}$
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\columnbreak
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$\text{comp}_{\vv{b}}\vv{a} = |\vv{b}|\cos\theta = \frac{\vv{a}\cdot \vv{b}}{|\vv{a}|}$
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$\text{proj}_{\vv{b}}\vv{a} = \text{comp}_{\vv{b}}\vv{a} \cdot \frac{a}{|a|} = \VV{ON} = \frac{\vv{a}\cdot \vv{b}}{\vv{a}\cdot \vv{a}}\vv{a} = \frac{\vv{a} \cdot \vv{b}}{|\vv{a}|^2}\vv{b}$
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\end{multicols}
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\subsection{Lines}
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\begin{multicols}{2}
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$\vv{r} = \vv{r}_0 + t\vv{v} = \langle x,y,z\rangle$
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$\langle x_0,y_0,z_0\rangle + t\langle a,b,c\rangle$
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$\begin{pmatrix}
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x_0 + at \\
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y_0 + bt \\
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z_0 + ct \\
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\end{pmatrix}$
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\end{multicols}
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\subsection{Planes}
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$\vv{n} = \langle a, b, c \rangle, \vv{r} = \langle x, y, z \rangle,\vv{r}_0\langle x_0, y_0, c_0 \rangle$\\
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Scalar: $\vv{n} \cdot \vv{r} = \vv{n} \cdot \vv{r}_0$\\
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Cartesian: $ax + by + cz = d$
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\subsection{Distance from Point to Plane}
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$\frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$
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\section{Partial Derivatives}
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\subsection{Chain Rule}
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For $z(t) = f(x(t), y(t))$,
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\\* $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$
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For $z(s, t) = f(x(s,t), y(s,t))$,
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\\* $\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$
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\\* $\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$
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Arc Length of $r(t)$: $\int^b_a |\vv{r}'(t)|dt$
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\subsection{Implicit Differentiation}
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$\frac{\partial z}{\partial x} =- \frac{F_x}{F_z}$
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$\frac{\partial z}{\partial y} =- \frac{F_y}{F_z}$
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\subsection{Directional Derivative}
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Gradient vector at $f(x,y): \triangledown f = f_x\vv{i} + f_y\vv{j}$
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$D_uf(x, y) = \langle f_x, f_y \rangle \cdot \langle a, b \rangle = \langle f_x, f_y\rangle \cdot \hat{\vv{u}} = \ \triangledown f \cdot \hat{\vv{u}}$ (Unit Vector)
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Tangent Plane: $\langle f_{x}, f_{y} -1 \rangle \cdot \langle x-x_0, y-y_0,z-z_0\rangle = 0$
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\subsection{Critical Points}
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$f_x = 0$ and $f_y = 0$, OR ($f_x$ or $f_y$ does not exist)
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$D = f_{xx}(a,b)f_{yy}(a,b) - (f_{xy}(a,b))^2$
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\def\arraystretch{1.2}
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\begin{tabular}{| c | c | c |}
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\hline $D$ & $f_{xx}(a,b)$ & \textbf{local}
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\\\hline + & + & \text{min}
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\\\hline + & - & \text{max}
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\\\hline - & \text{any} & \text{saddle point}
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\\\hline 0 & \text{any} & \text{no conclusion}
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\\\hline
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\end{tabular}
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\section{Double Integrals}
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\subsection{Type I}
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\begin{multicols}{2}
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\includegraphics[width=\linewidth]{Type I}
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\columnbreak
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$\int^b_a\int^{g_2(x)}_{g_1(x)}f(x,y)dydx$
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\newline
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\newline
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$D = \{(x,y): a \leq x \leq b,$ $g_1(x) \leq y \leq g_2(x)\}$
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\end{multicols}
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\subsection{Type II}
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\begin{multicols}{2}
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\includegraphics[width=\linewidth]{Type II}
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\columnbreak
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$\int^d_c\int^{h_2(y)}_{h_1(y)}f(x,y)dxdy$
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\newline
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\newline
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$D = \{(x,y): c \leq y \leq d,$ $ h_1(y) \leq x \leq h_2(y)\}$
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\end{multicols}
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\subsection{Polar Coordinates}
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\begin{multicols}{2}
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\includegraphics[width=\linewidth]{polar}
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\columnbreak
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$x = r\cos\theta$\\
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$y = r\sin\theta$\\
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$R = \{(r, \theta): 0 \leq a \leq r \leq b,$ $\alpha \leq \theta \leq \beta\}$
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\newline
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\newline
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$\int^\beta_\alpha\int^b_af(r\cos\theta, r\sin\theta)rdrd\theta$
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\end{multicols}
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\subsection{Surface Area}
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$S = \iint_R\sqrt{f_x^2 + f_y^2 + 1} dA$, get in the form of $z = f(x,y)$ first
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\section{ODE}
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\begin{tabular}{|>{\color{black}}c | >{\color{black}}c|}
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\hline
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form & change of variable \\
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\hline
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$\frac{dy}{dx} = f(x)g(y)$ & $\int \frac{1}{g(y)}dy = \int f(x)dx + C$\\
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\hline
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$y'=g(\frac{y}{x})$ & \makecell{Set $v = \frac{y}{x}$ \\ $\Then y' = v + xv' $}\\
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\hline
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\makecell{ $y'=f(ax + by + c)$\\ $\Then y' = \frac{ax+by+c}{\alpha x + \beta y + \gamma}$} & Set $v = ax+by$ \\
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\hline
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$y' + P(x)y = Q(x)$ & \makecell{$R = e^{\int P(x)dx}$ \\ $\Then y \cdot R = \int Q \cdot R dx $}\\
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$y' + P(x)y = Q(x)y^n$ & \makecell{$z = y^{1-n}$ \\ $\Then$ sub in Z \\ solve linear}\\
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\end{tabular}
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\section{Population Models}
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\begin{center}
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$N_{\infty} = \frac{B}{s}$, $\hat{N} = $ Population Now
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\begin{multicols}{2}
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\textbf{Malthus}\\
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$N(t) = \hat{N}e^{kt}$\\
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$k = B - D$
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\columnbreak
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\textbf{Logistic}\\
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$\frac{1}{N} = \frac{1}{N_{\infty}} + (\frac{1}{\hat{N}} - \frac{1}{N_{\infty}})e^{-Bt}$\\
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$N = \frac{N_{\infty}}{1+(\frac{N_{\infty}}{N} - 1)e^{-Bt}}$
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\end{multicols}
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\end{center}
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\subsection{Uranium Decay into Thorium}
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$U(t) = U_{0}e^{-k_ut}$, $k = \frac{\ln2}{\text{halflife}}, \frac{dU}{dt} = -k_{u}U$\\
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Thorium: $T(t) = \frac{K_{u}U_{0}}{K_{t}-K_{u}}(e^{-k_{u}t} - e^{-k_{t}t}), \frac{dT}{dt} = k_{u}U - k_{T}T$
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\end{multicols*}
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\newpage
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\begin{multicols*}{4}
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\setlength{\columnseprule}{0.25pt}
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\setlength{\premulticols}{1pt}
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\setlength{\postmulticols}{1pt}
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\setlength{\multicolsep}{1pt}
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\setlength{\columnsep}{2pt}
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\section{Series}
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\subsection{Geometric Series}
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$\sum_{n=1}^{\infty}ar^{n-1}, a \ne 0$ converges to $\frac{a}{1-r}$ when $|r| < 1$, diverges otherwise
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If series $\sum_{n=1}^{\infty} a_{n}$ is convergent, then $\lim\limits_{n \to \infty} a_{n} = 0$
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\subsection{Tests}
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Decreasing function -> differentiate and see the range where $x < 0$
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\begin{center}
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\begin{tabular}{|>{\color{black}}p{0.2\linewidth} | >{\color{black}}p{0.7\linewidth}|}
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\hline
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Test & Method \\\hline
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$n^{th}$ term & $\lim\limits_{n \to \infty} a_{n} \ne 0$ or does not exist, then divergent \\\hline
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Integral & $f(n)=a_{n}$ is continuous, positive, decreasing function $\forall x\geq 1$ and $\int_{1}^{\infty}f(x)dx$ converges else divergent \\\hline
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p-series & $\sum_{n=1}^{\infty} \frac{1}{n^{p}}$convergent $\leftrightarrow p > 1$ \\\hline
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Harmonic Series & $\sum_{n=1}^{\infty} \frac{1}{n}$ divergent \\\hline
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Ratio \tiny{If Factorial} & $0 \geq \lim\limits_{n \to \infty} |\frac{a_{n+1}}{a_{n}}|=L < 1$ abs. convergent, $> 1$ divergent, $= 1$ inconclusive \\\hline
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Root \tiny{If nth power} & $0 \geq \lim\limits_{n \to \infty} \sqrt[n]{a_{n}}=L < 1$ abs. convergent, $> 1$ divergent, $= 1$ inconclusive \\\hline
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Alternating series & $b_{n}$ decreasing, $\lim\limits_{n \to \infty}b_{n} = 0$, then $\sum_{n=1}^{\infty}(-1)^{n-1}b_{n} = b_{1}-b_{2}+b_{3}... $ is convergent \\\hline
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Power Series & $b_{n}$ decreasing, $\lim\limits_{n \to \infty}b_{n} = 0$, then $\sum_{n=1}^{\infty}(-1)^{n-1}b_{n} = b_{1}-b_{2}+b_{3}... $ is convergent \\\hline
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Comparison Test & $\sum a_{n} $ and $ \sum b_{n} $ s.t. $a_{n} \leq b_{n}$ Then if $\sum b_{n}$ convergent, $\sum a_{n}$ convergent. If $\sum a_{n}$ divergent, $\sum b_{n}$ divergent\\\hline
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\end{tabular}
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\end{center}
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\subsection{Power Series}
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$\sum_{n=0}^{\infty} c_{n}(x-a)^{n}$ converges at \textbf{ONE OF}
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\begin{itemize}
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\item $x=a$
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\item For all $x$
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\item converges if $|x-a| < R$ and diverges if $|x-a| > R$ (R is radius of convergence)
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\end{itemize}
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If $\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_{n}} = L \right|$ or $\lim_{n \to \infty} \sqrt[n]{|c_{n}|}=L$, $L \in \mathbb{R}$ or $\infty$, then $R = \frac{1}{L}$
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If power series $\sum_{n=0}^{\infty} c_{n}(x-a)^{n}$ has radius of convergence $R>0$, then function $f$ is differentiable on interval $|x-a| < R$ and
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\begin{itemize}
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\item $f'(x) = \sum_{n=1}^{\infty} nc_{n}(x-a)^{n-1}$, for $|x-a| < R$
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\item $\int f(x) = \sum_{n=0}^{\infty} c_{n}\frac{(x-a)^{n+1}}{n+1}+C$ for $|x-a| < R$
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\end{itemize}
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\subsection{Taylor and Maclaurin Series}
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If f has power series repr @ $f = a$, $f(x) = \sum_{n=0}^{\infty} c_{n}(x-a)^{n}, |x-a| < R, R > 0$, then $c_{n} = \frac{f^{(n)}(a)}{n!}$. \\
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Maclaurin Series: $f(x) = \sum_{n=0}^{\infty} \frac{f^{n}(0)}{n!}x^{n}$
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For $-\infty < x < \infty$
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\\ \; % spacing
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\setlength\tabcolsep{1.5pt} % default value: 6pt
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\begin{tabular}{rl}
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$\sin x$ & $= \sum\limits^\infty_{n = 0} \frac{(-1)^nx^{2n + 1}}{(2n+1)!} $
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\\ $\cos x$ & $= \sum\limits^\infty_{n = 0} \frac{(-1)^nx^{2n}}{(2n)!}$
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\\ $e^x$ & $= \sum\limits^\infty_{n = 0} \frac{x^n}{n!}$
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\end{tabular}
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For $-1 < x < 1$
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\\ \; % spacing
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\setlength\tabcolsep{1.5pt} % default value: 6pt
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\begin{tabular}{rl}
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$\frac{1}{1 - x}$ & $= \sum\limits^\infty_{n = 0} x^n $
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\\ $\frac{1}{1 + x}$ & $= \sum\limits^\infty_{n = 0} (-1)^nx^n $
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\\ $\frac{1}{1 + x^2}$ & $= \sum\limits^\infty_{n = 0} (-1)^nx^{2n} $
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\\ $\ln(1 + x)$ & $= \sum\limits^\infty_{n = 1} \frac{(-1)^{n - 1}x^n}{n} $
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\\ $\tan^{-1}x$ & $= \sum\limits^\infty_{n = 0} \frac{(-1)^n}{2n + 1} x^{2n+1}$
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\\ $\frac{1}{(1+x)^2}$ & $= \sum\limits^\infty_{n = 1} (-1)^{n-1}nx^{n-1}$
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\\ $\frac{1}{(1-x)^2}$ & $= \sum\limits^\infty_{n = 1} nx^{n-1}$
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\\ $\frac{1}{(1-x)^3}$ & $= \frac{1}{2} \sum\limits^\infty_{n = 2} n(n - 1)x^{n-2}$
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\\ $(1 + x)^k$ & $= \sum\limits^\infty_{n = 0} \binom{k}{n}x^n$
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\\ & $= 1 + kx + \frac{k(k-1)}{2!}x^2 + \dots$
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\end{tabular}
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\subsection{Useful Math}
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\begin{itemize}
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\item Line: $y-y_{1} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$
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\item $\int \sqrt{a^{2}-x^{2}}dx = \frac{a^{2}}{2}\sin^{-1}(\frac{x}{a}) + \frac{x}{2}\sqrt{a^{2}-x^{2}}, x = a\sin\theta, dx = a\cos\theta d\theta, $A
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\item $\sqrt{a^{2}+x^{2}}dx = \frac{1}{2}\left(x\sqrt{a^{2}+x^{2}} + a^{2}\ln\left|\frac{x+\sqrt{a^{2}+x^{2}}}{a}\right|\right), x = a\tan\theta, \frac{-\pi}{2} < \theta < \frac{\pi}{2}$
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\item $\int\cos^{2}x = \frac{1}{4} \sin2x + \frac{x}{2} = \frac{1}{2}\cos x \sin x + \frac{1}{2}x$
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\item $\int\sin^{2}x = -\frac{1}{4} \sin2x + \frac{x}{2}$
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\item $(x-y)^{3} = x^{3} - 3x^{2}y + 3xy^{2}-y^{3}$
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\item $(x+y)^{3} = x^{3} + 3x^{2}y + 3xy^{2}+y^{3}$
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\end{itemize}
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\end{multicols*}
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\end{document}
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