[spec] add graphs for Euler's method and bisector theorem

diff --git a/spec/spec-collated.pdf b/spec/spec-collated.pdf
index b59d29231d74cd31e31b0a85572b8e4e2dc0c43f..41d3d5c4543a2fba26306c6b414716be0b84465b 100644 (file)
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 989f7707f20cb52a2481797b2c25d6b66c004b41..13581432233005b8ffb4e0428eba3a934b01a27d 100644 (file)
--- a/spec/spec-collated.tex
+++ b/spec/spec-collated.tex
@@ -26,6 +26,7 @@
 %\usepackage{showframe} % debugging only
 \usepackage{subfiles}
 \usepackage{tabularx}
 %\usepackage{showframe} % debugging only
 \usepackage{subfiles}
 \usepackage{tabularx}

+\usepackage{tabu}

 \usepackage{tcolorbox}
 \usepackage{tikz-3dplot}
 \usepackage{tikz}
 \usepackage{tcolorbox}
 \usepackage{tikz-3dplot}
 \usepackage{tikz}


@@ -357,6 +358,7 @@
       \end{scope}
       \node[black, right] at (2.5,1.5) {\(y\vec{j}\)};
     \end{tikzpicture}\end{center}
       \end{scope}
       \node[black, right] at (2.5,1.5) {\(y\vec{j}\)};
     \end{tikzpicture}\end{center}

+

     \subsection*{Column notation}
 
     \[\begin{bmatrix}x\\ y \end{bmatrix} \iff x\boldsymbol{i} + y\boldsymbol{j}\]
     \subsection*{Column notation}
 
     \[\begin{bmatrix}x\\ y \end{bmatrix} \iff x\boldsymbol{i} + y\boldsymbol{j}\]


@@ -421,7 +423,7 @@
             \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*}
             \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}
 
 
             \subsubsection*{Properties}
 


@@ -584,6 +586,78 @@
                 Area \(=\boldsymbol{c} \cdot \boldsymbol{a}\)
             \end{itemize}
 
                 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}
             \subsubsection*{Useful vector properties}
 
             \begin{itemize}


@@ -830,11 +904,11 @@
                     arc[start angle=220, end angle=320, radius=2cm] 
                     -- cycle;
                     \node {Major Segment};
                     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]
 
                     \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] {} 
                       (320:2cm) node[right] {}
                       -- (0,0) node[above] {}
                       -- (220:2cm) node[left] {} 


@@ -1185,6 +1259,64 @@
 
                   \[f(x) = \frac{P(x)}{Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]
 
 
                   \[f(x) = \frac{P(x)}{Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]
 

+                  \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
                   \subsection*{Fundamental theorem of calculus}
 
                   If \(f\) is continuous on \([a, b]\), then


@@ -1201,29 +1333,65 @@
                   \begin{warning}
                     To verify solutions, find \(\frac{dy}{dx}\) from \(y\) and substitute into original
                   \end{warning}
                   \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}
 
                   \[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\]
 
 
                   \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}
                   \include{calculus-rules}
 
     \section{Kinematics \& Mechanics}