%\usepackage{showframe} % debugging only
\usepackage{subfiles}
\usepackage{tabularx}
+\usepackage{tabu}
\usepackage{tcolorbox}
\usepackage{tikz-3dplot}
\usepackage{tikz}
\end{scope}
\node[black, right] at (2.5,1.5) {\(y\vec{j}\)};
\end{tikzpicture}\end{center}
+
\subsection*{Column notation}
\[\begin{bmatrix}x\\ y \end{bmatrix} \iff x\boldsymbol{i} + y\boldsymbol{j}\]
\end{scope}
\end{tikzpicture}\end{center}
\begin{align*}\boldsymbol{a} \cdot \boldsymbol{b} &= a_1 b_1 + a_2 b_2 \\ &= |\boldsymbol{a}| |\boldsymbol{b}| \cos \theta \\ &\quad (\> 0 \le \theta \le \pi) \text{ - from cosine rule}\end{align*}
- \noindent\colorbox{cas}{On CAS: \texttt{dotP({[}a\ b\ c{]},\ {[}d\ e\ f{]})}}
+ \noindent\colorbox{cas}{On CAS:} \texttt{dotP({[}a\ b\ c{]},\ {[}d\ e\ f{]})}
\subsubsection*{Properties}
Area \(=\boldsymbol{c} \cdot \boldsymbol{a}\)
\end{itemize}
+ \subsubsection*{Perpendicular bisectors of a triangle}
+
+\hspace{-1.5cm}\begin{tikzpicture}
+ [
+ scale=3,
+ >=stealth,
+ point/.style = {draw, circle, fill = black, inner sep = 1pt},
+ dot/.style = {draw, circle, fill = black, inner sep = .2pt},
+ thick
+ ]
+
+ \node at (-1,1) [text width=5cm, rounded corners, fill=lblue, inner sep=1ex]
+ {
+ \sffamily The three bisectors meet at the circumcenter \(Z\) where \(|\overrightharp{ZA}| = |\overrightharp{ZB}| = |\overrightharp{ZC}|\).
+ };
+
+ % the circle
+ \def\rad{1}
+ \node (origin) at (0,0) [point, label = {right: {\(Z\)}}]{};
+ \draw [thin] (origin) circle (\rad);
+
+ % triangle nodes: just points on the circle
+ \node (n1) at +(60:\rad) [point, label = above:\(A\)] {};
+ \node (n2) at +(-145:\rad) [point, label = below:\(B\)] {};
+ \node (n3) at +(-45:\rad) [point, label = {below right:\(C\)}] {};
+
+ % triangle edges: connect the vertices, and leave a node at the midpoint
+ \draw[orange] (n3) -- node (a) [label = {above right:\(D\)}] {} (n1);
+ \draw[blue] (n3) -- node (b) [label = {below right:\(F\)}] {} (n2);
+ \draw[red] (n1) -- node (c) [label = {left: \(E\)}] {} (n2);
+
+ % Bisectors
+ % start at the point lying on the line from (origin) to (a), at
+ % twice that distance, and then draw a path going to the point on
+ % the line lying on the line from (a) to the (origin), at 3 times
+ % that distance.
+ \draw[orange, dotted]
+ ($ (origin) ! 2 ! (a) $)
+ node [right] {\sffamily Bisector \(AC\)}
+ -- ($(a) ! 3 ! (origin)$ );
+
+ % similarly for origin and b
+ \draw[blue, dotted]
+ ($ (origin) ! 2 ! (b) $)
+ -- ($(b) ! 3 ! (origin)$ )
+ node [right] {\sffamily Bisector \(BC\)};
+
+ \draw[red, dotted]
+ ($ (origin) ! 5 ! (c) $)
+ -- ($(c) ! 7 ! (origin)$ )
+ node [right] {\sffamily Bisector \(AB\)};
+
+ \draw[gray, dashed, thin] (n1) -- (origin) -- (n2);
+ \draw[gray, dashed, thin] (origin) -- (n3);
+
+ % Right angle symbols
+ \def\ralen{.5ex} % length of the short segment
+ \foreach \inter/\first/\last in {a/n3/origin, b/n2/origin, c/n2/origin}
+ {
+ \draw [thin] let \p1 = ($(\inter)!\ralen!(\first)$), % point along first path
+ \p2 = ($(\inter)!\ralen!(\last)$), % point along second path
+ \p3 = ($(\p1)+(\p2)-(\inter)$) % corner point
+ in
+ (\p1) -- (\p3) -- (\p2); % path
+ }
+\end{tikzpicture}
+
+ \begin{theorembox}{title=Perpendicular bisector theorem}
+ If a point \(P\) lies on the perpendicular bisector of line \(\overrightharp{XY}\), then \(P\) is equidistant from the endpoints of the bisected segment
+ \[ \text{i.e. } |\overrightharp{PX}| = |\overrightharp{PY}| \]
+ \end{theorembox}
+
\subsubsection*{Useful vector properties}
\begin{itemize}
arc[start angle=220, end angle=320, radius=2cm]
-- cycle;
\node {Major Segment};
- \node at (-90:2) {Minor Segment};
+ \node at (-90:1.5) {Minor Segment};
\begin{scope}[xshift=4.5cm]
- \draw circle (2cm);
- \filldraw[fill=lblue]
+ \draw [fill=lblue] circle (2cm);
+ \filldraw[fill=white]
(320:2cm) node[right] {}
-- (0,0) node[above] {}
-- (220:2cm) node[left] {}
\[f(x) = \frac{P(x)}{Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]
+ \subsection*{Euler's method}
+
+ \[\dfrac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
+
+ \[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
+
+ \begin{theorembox}{}
+ If \(\dfrac{dy}{dx} = g(x)\) with \(x_0 = a\) and \(y_0 = b\), then:
+ \[\begin{cases}
+ x_{n+1} = x_n + h \\
+ y_{n+1} = y_n + hg(x_n)
+ \end{cases}\]
+ \end{theorembox}
+
+ \[
+ \dfrac{d^2y}{dx^2}
+ \begin{cases}
+ > 0 \implies \text{ underestimate (concave up)} \\
+ < 0 \implies \text{ overestimate (concave down)}
+ \end{cases}
+ \]
+
+ \begin{center}\begin{tikzpicture}
+ \begin{axis}[xmin=0, xmax=1.6, ticks=none, enlargelimits=true, samples=100]
+ \addplot[blue, domain=-0.25:1.5, postaction={decorate,decoration={text along path, text align={align=center, left indent=3cm}, text={|\sffamily|solution curve}}}] {e^(x-3/2)+1/4};
+ \addplot[red] {(x+1/2)*e^(-1)+1/4} (1.7,1.0593) node [above, black] {\(\ell\)};
+ \addplot[mark=*, black] coordinates {(0.5,0.6179)} node[above left]{\((x_0, y_0)\)};
+ \addplot[mark=*, orange] coordinates {(1.4,1.1548)} node[left]{\color{black} \sffamily correct solution};
+ \addplot[mark=*, black] coordinates {(1.4,0.94897)} node[above right] {\((x_1,y_1)\)};
+ \draw [gray, dashed] (0.5,0) -- (0.5,0.6179) -- (1.6,0.6179);
+ \draw [gray, dashed] (1.4,0) -- (1.4, 1.1548);
+ \draw [<->] (0.5,0.48) -- (1.4,0.48) node[midway, fill=white] {\(h\)};
+ \draw [gray, dashed] (1.4,0.94897) -- (1.6,0.94897);
+ \draw [<->] (1.5,0.94897) -- (1.5,0.6179) node[midway, rotate=90, below] {\(hg(x_0)\)};
+ \end{axis}
+ \end{tikzpicture}\end{center}
+
+ \begin{cas}
+ Menu \(\rightarrow\) Sequence \(\rightarrow\) Recursive
+
+ \textbf{To generate \(\boldsymbol{x}\)-values:}
+ \begin{itemize}
+ \item Enter \(a_{n+1}=a_n + h\) where \(h\) is the step size \\
+ (input \(a_n\) from menu bar)
+ \item In \(a_0\), set the initial value \(x_0\) as a constant
+ \end{itemize}
+
+ \textbf{To generate \(\boldsymbol{y}\)-values:}
+ \begin{itemize}
+ \item In \(b_{n+1}\), enter \(\dfrac{dy}{dx}\), replacing \(x\) with \(a_n\)
+ \item Set \(b_0 = y(x_0)\) as a constant
+ \end{itemize}
+
+ To view table of values, tap table icon (top left) \\
+ To compare approximations with actual values, enter in \(c_{n+1} = a_{n+1} - f(a_{n+1})\) where \(f(x) = \int \dfrac{dy}{dx} \> dx\)
+
+ \end{cas}
+
\subsection*{Fundamental theorem of calculus}
If \(f\) is continuous on \([a, b]\), then
\begin{warning}
To verify solutions, find \(\frac{dy}{dx}\) from \(y\) and substitute into original
\end{warning}
-
-
+
+ \vspace*{1cm}
+ \hspace*{-1cm}
+
+ { \tabulinesep=1.2mm
+ \begin{tabu}{|c|c|}
+
+ \hline
+ \taburowcolors 2{gray..white}
+ \textbf{DE} & \textbf{Method} \\
+ \hline
+
+ \tabureset
+ \(\dfrac{dy}{dx} = f(x)\)
+ &
+ {\(\begin{aligned}
+ y &= \int f(x) \> dx \\
+ &= F(x) + c \quad \text{where } F^\prime(x) = f(x)
+ \end{aligned}\)} \\
+
+ \hline
+
+ \(\dfrac{d^2y}{dx^2} = f(x)\)
+ &
+ {\(\begin{aligned}
+ \dfrac{dy}{dx} &= \int f(x) \> dx \\
+ &= F(x) + c \quad \text{where } F^\prime(x) = f(x) \\
+ \therefore y &= \iint f(x) \> dx = \int \left( F(x) + c \right) \> dx \\
+ &= G(x) + cx + d \\
+ & \text{where } G^\prime(x) = F(x)
+ \end{aligned}\)} \\
+
+ \hline
+
+ \(\dfrac{dy}{dx} = g(y)\)
+ &
+ {\(\begin{aligned}
+ \dfrac{dx}{dy} &= \dfrac{1}{g(y)} \\
+ \therefore x &= \int \dfrac{1}{g(y)} \> dy \\
+ &= F(y) + c \\
+ & \text{where } F^\prime(y) = \dfrac{1}{g(y)}
+ \end{aligned}\)} \\
+
+ \hline
+
+ \(\dfrac{dy}{dx} = f(x) g(y)\)
+ &
+ {\(\begin{aligned}
+ f(x) &= \dfrac{1}{g(y)} \cdot \dfrac{dy}{dx} \\
+ \int f(x) \> dx &= \int \dfrac{1}{g(y)} \> dy
+ \end{aligned}\)} \\
+
+ \hline
+ \end{tabu}}
\subsubsection*{Mixing problems}
\[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\]
- \subsection*{Euler's method}
-
- \[\dfrac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
-
- \[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
-
- \begin{theorembox}{}
- If \(\dfrac{dy}{dx} = g(x)\) with \(x_0 = a\) and \(y_0 = b\), then:
- \begin{align*}
- x_{n+1} &= x_n + h \\
- y_{n+1} &= y_n + hg(x_n)
- \end{align*}
- \end{theorembox}
-
-
-
\include{calculus-rules}
\section{Kinematics \& Mechanics}