From: Andrew Lorimer Date: Tue, 28 May 2019 04:44:26 +0000 (+1000) Subject: [spec] minor editors; add inverse sin&cos graphs X-Git-Tag: yr12~116 X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/f6326f1c24bebb2a4042ee0dc9d9ac489f56bc8e [spec] minor editors; add inverse sin&cos graphs --- diff --git a/spec/spec-collated.pdf b/spec/spec-collated.pdf index 6db681c..6e6dc20 100644 Binary files a/spec/spec-collated.pdf and b/spec/spec-collated.pdf differ diff --git a/spec/spec-collated.tex b/spec/spec-collated.tex index b669b55..e2ccd0e 100644 --- a/spec/spec-collated.tex +++ b/spec/spec-collated.tex @@ -41,164 +41,167 @@ \section{Complex numbers} - \[\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}\] + \[\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}\] - \begin{align*} - \text{Cartesian form: } & a+bi\\ - \text{Polar form: } & r\operatorname{cis}\theta - \end{align*} + \begin{align*} + \text{Cartesian form: } & a+bi\\ + \text{Polar form: } & r\operatorname{cis}\theta + \end{align*} - \subsection*{Operations} + \subsection*{Operations} -\definecolor{shade1}{HTML}{ffffff} -\definecolor{shade2}{HTML}{e6f2ff} + \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 \(k \in \mathbb{R}^-\): - \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\left(\begin{cases}\theta - \pi & |0<\operatorname{Arg}(z)\le \pi \\ \theta + \pi & |-\pi<\operatorname{Arg}(z)\le 0\end{cases}\right)\] + \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*} + \begin{align*} + \overline{z} &= a \mp bi\\ + &= r \operatorname{cis}(-\theta) + \end{align*} - \noindent \colorbox{cas}{On CAS: \texttt{conjg(a+bi)}} + \noindent \colorbox{cas}{On CAS: \texttt{conjg(a+bi)}} - \subsubsection*{Properties} + \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*} + \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} - \[|z|=|\vec{Oz}|=\sqrt{a^2 + b^2}\] + \[|z|=|\vec{Oz}|=\sqrt{a^2 + b^2}\] - \subsubsection*{Properties} + \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*} + \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*} + \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*} + \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{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} + \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} + 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} + &= (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)\\ + \hline + \end{tabularx} - \subsection*{Roots} + \subsection*{\(n\)th roots} - \(n\)th roots of \(z=r\operatorname{cis}\theta\) are: + \(n\)th roots of \(z=r\operatorname{cis}\theta\) are: - \[z = r^{\frac{1}{n}} \operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)\] + \[z = r^{\frac{1}{n}} \operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)\] - \begin{itemize} + \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} + \item{Same modulus for all solutions} + \item{Arguments separated by \(\frac{2\pi}{n} \therefore\) there are \(n\) roots} + \item{If one square root is \(a+bi\), the other is \(-a-bi\)} + \item{Give one implicit \(n\)th root \(z_1\), function is \(z=z_1^n\)} + \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: + \noindent For \(0=az^2+bz+c\), use quadratic formula: - \[z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] + \[z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] \subsection*{Fundamental theorem of algebra} - A polynomial of degree \(n\) can be factorised into \(n\) linear factors in \(\mathbb{C}\): + A polynomial of degree \(n\) can be factorised into \(n\) linear factors in \(\mathbb{C}\): - \[\implies P(z)=a_n(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)\dots(z-\alpha_n)\] - \[\text{ where } \alpha_1,\alpha_2,\alpha_3,\dots,\alpha_n \in \mathbb{C}\] + \[\implies P(z)=a_n(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)\dots(z-\alpha_n)\] + \[\text{ where } \alpha_1,\alpha_2,\alpha_3,\dots,\alpha_n \in \mathbb{C}\] \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]{\(\begin{aligned}z&=a+bi\\&=r\operatorname{cis}\theta\end{aligned}\)}; + + \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} @@ -207,880 +210,901 @@ \node at ($(O)+(20:3mm)$) {$\theta$}; \end{scope} \filldraw (P) circle (0.5pt); - \end{tikzpicture}\end{center} + \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} + \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} + \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}] + \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} + \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 \(A(x_1,y_1), \> B(x_2,y_2)\) + \[\begin{bmatrix}x\\ y \end{bmatrix} \iff x\boldsymbol{i} + y\boldsymbol{j}\] + \(\begin{bmatrix}x_2-x_1\\ y_2-y_1 \end{bmatrix}\) \quad between \(A(x_1,y_1), \> B(x_2,y_2)\) -\subsection*{Scalar multiplication} + \subsection*{Scalar multiplication} -\[k\cdot (x\boldsymbol{i}+y\boldsymbol{j})=kx\boldsymbol{i}+ky\boldsymbol{j}\] + \[k\cdot (x\boldsymbol{i}+y\boldsymbol{j})=kx\boldsymbol{i}+ky\boldsymbol{j}\] -\noindent For \(k \in \mathbb{R}^-\), direction is reversed + \noindent For \(k \in \mathbb{R}^-\), direction is reversed -\subsection*{Vector addition} -\begin{center}\begin{tikzpicture}[scale=1] + \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)} -\[\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u}\] + \end{tikzpicture}\end{center} -\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*} + \[(x\boldsymbol{i}+y\boldsymbol{j}) \pm (a\boldsymbol{i}+b\boldsymbol{j})=(x \pm a)\boldsymbol{i}+(y \pm b)\boldsymbol{j}\] -\subsubsection*{Rectangular (\(\parallel,\perp\)) components} + \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} -\[\boldsymbol{a}=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}+\left(\boldsymbol{a}-\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}\right)\] + \subsection*{Magnitude} + \[|(x\boldsymbol{i} + y\boldsymbol{j})|=\sqrt{x^2+y^2}\] + + \subsection*{Parallel vectors} -\subsection*{Vector proofs} + \[\boldsymbol{u} || \boldsymbol{v} \iff \boldsymbol{u} = k \boldsymbol{v} \text{ where } k \in \mathbb{R} \setminus \{0\}\] + + 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} -\textbf{Concurrent:} intersection of \(\ge\) 3 lines + \[\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b} = 0\ \quad \text{(since \(\cos 90 = 0\))}\] -\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} + \subsection*{Unit vector \(|\hat{\boldsymbol{a}}|=1\)} + \[\begin{split}\hat{\boldsymbol{a}} & = {\frac{1}{|\boldsymbol{a}|}}\boldsymbol{a} \\ & = \boldsymbol{a} \cdot {|\boldsymbol{a}|}\end{split}\] -\subsubsection*{Collinear points} + \subsection*{Scalar product \(\boldsymbol{a} \cdot \boldsymbol{b}\)} -\(\ge\) 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{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{]})}} -\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} + \subsubsection*{Properties} -\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{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 projection/resolute)} + \begin{align*} + s &= |\boldsymbol{u}|\\ + &= \boldsymbol{a} \cdot \hat{\boldsymbol{b}}\\ + &=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}\\ + &= |\boldsymbol{a}| \cos \theta + \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} -\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} + \subsubsection*{Secant} + \begin{center}\includegraphics[width=0.7\columnwidth]{graphics/sec.png}\end{center} -\(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\) 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*} + \[\operatorname{sec} \theta = \frac{1}{\cos \theta} \> \vert \> \cos \theta \ne 0\] -\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. + \begin{itemize} -\subsection*{Three-dimensional vectors} + \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} -Right-hand rule for axes: \(z\) is up or out of page. + \subsubsection*{Cotangent} -\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}] + \begin{center}\includegraphics[width=0.7\columnwidth]{graphics/cot.png}\end{center} -%standard tikz coordinate definition using x, y, z coords -\coordinate (O) at (0,0,0); + \[\operatorname{cot} \theta = {{\cos \theta} \over {\sin \theta}} \> \vert \> \sin \theta \ne 0\] -%tikz-3dplot coordinate definition using x, y, z coords + \begin{itemize} -\pgfmathsetmacro{\ax}{1} -\pgfmathsetmacro{\ay}{1} -\pgfmathsetmacro{\az}{1} + \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} -\coordinate (P) at (\ax,\ay,\az); + \subsubsection*{Symmetry properties} -%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]{$z$}; + \[\begin{split} + \operatorname{sec} (\pi \pm x) & = -\operatorname{sec} x \\ + \operatorname{sec} (-x) & = \operatorname{sec} x \\ + \operatorname{cosec} (\pi \pm x) & = \mp \operatorname{cosec} x \\ + \operatorname{cosec} (-x) & = - \operatorname{cosec} x \\ + \operatorname{cot} (\pi \pm x) & = \pm \operatorname{cot} x \\ + \operatorname{cot} (-x) & = - \operatorname{cot} x + \end{split}\] -%draw a vector from O to P -\draw[vector] (O) -- (P); + \subsubsection*{Complementary properties} -%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} + \[\begin{split} + \operatorname{sec} \left({\pi \over 2} - x\right) & = \operatorname{cosec} x \\ + \operatorname{cosec} \left({\pi \over 2} - x\right) & = \operatorname{sec} x \\ + \operatorname{cot} \left({\pi \over 2} - x\right) & = \tan x \\ + \tan \left({\pi \over 2} - x\right) & = \operatorname{cot} x + \end{split}\] -\subsection*{Parametric vectors} + \subsubsection*{Pythagorean identities} -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: + \[\begin{split} + 1 + \operatorname{cot}^2 x & = \operatorname{cosec}^2 x, \quad \text{where } \sin x \ne 0 \\ + 1 + \tan^2 x & = \operatorname{sec}^2 x, \quad \text{where } \cos x \ne 0 + \end{split}\] -\[\begin{cases}x = x_o + a \cdot t \\ y = y_0 + b \cdot t \\ z = z_0 + c \cdot t\end{cases}\] + \subsection*{Compound angle formulas} -\section{Circular functions} + \[\cos(x \pm y) = \cos x + \cos y \mp \sin x \sin y\] + \[\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\] + \[\tan(x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x \tan y}}\] -\(\sin(bx)\) or \(\cos(bx)\): period \(=\frac{2\pi}{b}\) + \subsection*{Double angle formulas} -\noindent \(\tan(nx)\): period \(=\frac{\pi}{n}\)\\ -\indent\indent\indent asymptotes at \(x=\frac{(2k+1)\pi}{2n} \> \vert \> k \in \mathbb{Z}\) + \[\begin{split} + \cos 2x &= \cos^2 x - \sin^2 x \\ + & = 1 - 2\sin^2 x \\ + & = 2 \cos^2 x -1 + \end{split}\] -\subsection*{Reciprocal functions} + \[\sin 2x = 2 \sin x \cos x\] -\subsubsection*{Cosecant} + \[\tan 2x = {{2 \tan x} \over {1 - \tan^2 x}}\] -\[\operatorname{cosec} \theta = \frac{1}{\sin \theta} \> \vert \> \sin \theta \ne 0\] - -\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} + \subsection*{Inverse circular functions} -\subsubsection*{Secant} + \pgfplotsset{every axis/.append style={ + axis x line=middle, % put the x axis in the middle + axis y line=middle, % put the y axis in the middle + axis line style={<->}, % arrows on the axis + xlabel={$x$}, % default put x on x-axis + ylabel={$y$}, % default put y on y-axis + }} + +% arrows as stealth fighters +\tikzset{>=stealth} +\begin{tikzpicture} + \begin{axis}[domain = -1:1, samples = 500] + \addplot[color = red] {rad(asin(x))} node [pos=0.25, below right] {\(\sin^{-1}x\)}; + \addplot[color = blue] {rad(acos(x))} node [pos=0.25, below left] {\(\cos^{-1}x\)}; + \end{axis} +\end{tikzpicture} -\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} + Inverse functions: \(f(f^{-1}(x)) = x\) (restrict domain) + + \[\sin^{-1}: [-1, 1] \rightarrow \mathbb{R}, \quad \sin^{-1} x = y\] + \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 \(\tan y = x, \> y \in \left(-{\pi \over 2}, {\pi \over 2}\right)\) -\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} + \section{Differential calculus} -\begin{center}\includegraphics[width=0.7\columnwidth]{graphics/cot.png}\end{center} + \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 + \(\sin x\) & \(\cos x\)\\ + \(\sin ax\) & \(a\cos ax\)\\ + \(\cos x\) & \(-\sin x\)\\ + \(\cos ax\) & \(-a \sin ax\)\\ + \(\tan f(x)\) & \(f^2(x) \sec^2f(x)\)\\ + \(e^x\) & \(e^x\)\\ + \(e^{ax}\) & \(ae^{ax}\)\\ + \(ax^{nx}\) & \(an \cdot e^{nx}\)\\ + \(\log_e x\) & \(\dfrac{1}{x}\)\\ + \(\log_e {ax}\) & \(\dfrac{1}{x}\)\\ + \(\log_e f(x)\) & \(\dfrac{f^\prime (x)}{f(x)}\)\\ + \(\sin(f(x))\) & \(f^\prime(x) \cdot \cos(f(x))\)\\ + \(\sin^{-1} x\) & \(\dfrac{1}{\sqrt{1-x^2}}\)\\ + \(\cos^{-1} x\) & \(\dfrac{-1}{sqrt{1-x^2}}\)\\ + \(\tan^{-1} x\) & \(\dfrac{1}{1 + x^2}\)\\ + \(\frac{d}{dy}f(y)\) & \(\dfrac{1}{\frac{dx}{dy}}\) (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} + + \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*} + + \subsubsection*{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*} -\[\operatorname{cot} \theta = {{\cos \theta} \over {\sin \theta}} \> \vert \> \sin \theta \ne 0\] \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} -\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} + \subsection*{Implicit Differentiation} -\[\begin{split} - 1 + \operatorname{cot}^2 x & = \operatorname{cosec}^2 x, \quad \text{where } \sin x \ne 0 \\ - 1 + \tan^2 x & = \operatorname{sec}^2 x, \quad \text{where } \cos x \ne 0 -\end{split}\] + \noindent Used for differentiating circles etc. -\subsection*{Compound angle formulas} + If \(p\) and \(q\) are expressions in \(x\) and \(y\) such that \(p=q\), + for all \(x\) and \(y\), then: -\[\cos(x \pm y) = \cos x + \cos y \mp \sin x \sin y\] -\[\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\] -\[\tan(x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x \tan y}}\] + \[{\frac{dp}{dx}} = {\frac{dq}{dx}} \quad \text{and} \quad {\frac{dp}{dy}} = {\frac{dq}{dy}}\] -\subsection*{Double angle formulas} - -\[\begin{split} - \cos 2x &= \cos^2 x - \sin^2 x \\ - & = 1 - 2\sin^2 x \\ - & = 2 \cos^2 x -1 -\end{split}\] + \noindent \colorbox{cas}{\textbf{On CAS:}}\\ + Action \(\rightarrow\) Calculation \(\rightarrow\) \texttt{impDiff(y\^{}2+ax=5,\ x,\ y)}\\ + Returns \(y^\prime= \dots\). -\[\sin 2x = 2 \sin x \cos x\] + \subsection*{Integration} -\[\tan 2x = {{2 \tan x} \over {1 - \tan^2 x}}\] + \[\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)\] -\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 \(\sin y = x, \> y \in [{-\pi \over 2}, {\pi \over 2}]\) + \subsection*{Integral laws} -\[\cos^{-1}: [-1,1] \rightarrow \mathbb{R}, \quad \cos^{-1} x = y\] -\hfill where \(\cos y = x, \> y \in [0, \pi]\) + \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} -\[\tan^{-1}: \mathbb{R} \rightarrow \mathbb{R}, \quad \tan^{-1} x = y\] -\hfill where \(\tan y = x, \> y \in \left(-{\pi \over 2}, {\pi \over 2}\right)\) + Note \(\sin^{-1} {x \over a} + \cos^{-1} {x \over a}\) is constant \(\forall x \in (-a, a)\) + \subsection*{Definite integrals} -\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} + \[\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)\] -\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} -\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} -\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} -\subsubsection*{Properties} + \[\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx\] -\[\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx\] + \[\int^a_a f(x) \> dx = 0\] -\[\int^a_a f(x) \> dx = 0\] + \[\int^b_a k \cdot f(x) \> dx = k \int^b_a f(x) \> dx\] -\[\int^b_a k \cdot f(x) \> dx = k \int^b_a f(x) \> dx\] + \[\int^b_a f(x) \pm g(x) \> dx = \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx\] -\[\int^b_a f(x) \pm g(x) \> dx = \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx\] + \[\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx\] -\[\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx\] + \subsection*{Integration by substitution} -\subsection*{Integration by substitution} + \[\int f(u) {\frac{du}{dx}} \cdot dx = \int f(u) \cdot du\] -\[\int f(u) {\frac{du}{dx}} \cdot dx = \int f(u) \cdot du\] + \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*} -\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} -\subsubsection*{Definite integrals by substitution} + For \(\int^b_a f(x) {\frac{du}{dx}} \cdot dx\), evaluate new \(a\) and + \(b\) for \(f(u) \cdot du\). -For \(\int^b_a f(x) {\frac{du}{dx}} \cdot dx\), evaluate new \(a\) and -\(b\) for \(f(u) \cdot du\). + \subsubsection*{Trigonometric integration} -\subsubsection*{Trigonometric integration} + \[\sin^m x \cos^n x \cdot dx\] -\[\sin^m x \cos^n x \cdot dx\] + \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{\(m\) 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{\(n\) 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... -\paragraph{\textbf{\(m\) and \(n\) are even:}} -use identities... + \begin{itemize} -\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} -\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} -\subsection*{Partial fractions} + \colorbox{cas}{On CAS:}\\ + \indent Action \(\rightarrow\) Transformation \(\rightarrow\) + \texttt{expand/combine}\\ + \indent Interactive \(\rightarrow\) Transformation \(\rightarrow\) + Expand \(\rightarrow\) Partial -\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} -\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{}..} -\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} -\subsection*{Applications of antidifferentiation} + \begin{itemize} -\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} -\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. -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} -\subsection*{Solids of revolution} + Approximate as sum of infinitesimally-thick cylinders -Approximate as sum of infinitesimally-thick cylinders + \subsubsection*{Rotation about \(x\)-axis} -\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*} -\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} -\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*} -\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 \(y=0\)} -\subsubsection*{Regions not bound by \(y=0\)} + \[V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx\] + \hfill where \(f(x) > g(x)\) -\[V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx\] -\hfill where \(f(x) > g(x)\) + \subsection*{Length of a curve} -\subsection*{Length of a curve} + \[L = \int^b_a \sqrt{1 + ({\frac{dy}{dx}})^2} \> dx \quad \text{(Cartesian)}\] -\[L = \int^b_a \sqrt{1 + ({\frac{dy}{dx}})^2} \> dx \quad \text{(Cartesian)}\] + \[L = \int^b_a \sqrt{{\frac{dx}{dt}} + ({\frac{dy}{dt}})^2} \> dt \quad \text{(parametric)}\] -\[L = \int^b_a \sqrt{{\frac{dx}{dt}} + ({\frac{dy}{dt}})^2} \> dt \quad \text{(parametric)}\] + \noindent \colorbox{cas}{On CAS:}\\ + \indent Evaluate formula,\\ + \indent or Interactive \(\rightarrow\) Calculation + \(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen} -\noindent \colorbox{cas}{On CAS:}\\ -\indent Evaluate formula,\\ -\indent or Interactive \(\rightarrow\) Calculation -\(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen} + \subsection*{Rates} -\subsection*{Rates} + \subsubsection*{Gradient at a point on parametric curve} -\subsubsection*{Gradient at a point on parametric curve} + \[{\frac{dy}{dx}} = {{\frac{dy}{dt}} \div {\frac{dx}{dt}}} \> \vert \> {\frac{dx}{dt}} \ne 0\] -\[{\frac{dy}{dx}} = {{\frac{dy}{dt}} \div {\frac{dx}{dt}}} \> \vert \> {\frac{dx}{dt}} \ne 0\] + \[\frac{d^2}{dx^2} = \frac{d(y^\prime)}{dx} = {\frac{dy^\prime}{dt} \div {\frac{dx}{dt}}} \> \vert \> y^\prime = {\frac{dy}{dx}}\] -\[\frac{d^2}{dx^2} = \frac{d(y^\prime)}{dx} = {\frac{dy^\prime}{dt} \div {\frac{dx}{dt}}} \> \vert \> y^\prime = {\frac{dy}{dx}}\] + \subsection*{Rational functions} -\subsection*{Rational functions} + \[f(x) = \frac{P(x)}{Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\] -\[f(x) = \frac{P(x)}{Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\] + \subsubsection*{Addition of ordinates} -\subsubsection*{Addition of ordinates} + \begin{itemize} -\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} -\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} -\subsection*{Fundamental theorem of calculus} + If \(f\) is continuous on \([a, b]\), then -If \(f\) is continuous on \([a, b]\), then + \[\int^b_a f(x) \> dx = F(b) - F(a)\] + \hfill where \(F = \int f \> dx\) -\[\int^b_a f(x) \> dx = F(b) - F(a)\] -\hfill where \(F = \int f \> dx\) + \subsection*{Differential equations} -\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 -\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} -\subsubsection*{Verifying solutions} + Start with \(y=\dots\), and differentiate. Substitute into original + equation. -Start with \(y=\dots\), and differentiate. Substitute into original -equation. + \subsubsection*{Function of the dependent + variable} -\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 \(A\). -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 \(A\). + \begin{table*}[ht] + \centering + \includegraphics[width=0.7\textwidth]{graphics/second-derivatives.png} + \end{table*} -\subsubsection*{Mixing problems} + \subsubsection*{Mixing problems} -\[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\] + \[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\] -\subsubsection*{Separation of variables} + \subsubsection*{Separation of variables} -If \({\frac{dy}{dx}}=f(x)g(y)\), then: + If \({\frac{dy}{dx}}=f(x)g(y)\), then: -\[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\] + \[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\] -\subsubsection*{Euler's method for solving DEs} + \subsubsection*{Euler's method for solving DEs} -\[\frac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\] + \[\frac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\] -\[\implies f(x+h) \approx f(x) + hf^\prime(x)\] + \[\implies f(x+h) \approx f(x) + hf^\prime(x)\] - \end{multicols} -\end{document} + \end{multicols} + \end{document}