Added Slide Rule & Instant Insanity handouts

Intermediate/Slide Rules/main.tex

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+% https://git.betalupi.com/Mark/latex-packages
+% use [nosolutions] flag to hide solutions.
+% use [solutions] flag to show solutions.
+% Last built with version 1.1.0
+\documentclass[
+	solutions
+]{ormc_handout}
+
+\usepackage{pdfpages}
+\usepackage{sliderule}
+
+% Args:
+% x, top scale y, label
+\newcommand{\slideruleind}[3]{
+	\draw[
+		line width=1mm,
+		draw=black,
+		opacity=0.3,
+		text opacity=1
+	]
+		({#1}, {#2 + 1})
+		--
+		({#1}, {#2 - 1.1})
+		node [below] {#3};
+}
+
+\begin{document}
+
+	\maketitle
+		<Intermediate 2>
+		<Summer 2022>
+		{Slide Rules}
+		{
+			Prepared by Mark on \today
+		}
+
+	\vspace{1cm}
+
+	\begin{minipage}{6cm}
+		Dad says that anyone who can't use
+		a slide rule is a cultural illiterate
+		and should not be allowed to vote.
+
+		\vspace{1ex}
+
+		\textit{Have Space Suit -- Will Travel, 1958}
+	\end{minipage}
+	\hfill
+
+	\section{Logarithms}
+
+	\definition{}<logdef>
+	The \textit{logarithm} is the inverse of the exponent. That is, if $b^p = c$, then $\log_b{c} = p$. \\
+	In other words, $\log_b{c}$ asks the question ``what power do I need to raise $b$ to to get $c$?'' \\
+
+	\medskip
+
+	In both $b^p$ and $\log_b{c}$, the number $b$ is called the \textit{base}.
+
+
+	\problem{}
+	Evaluate the following by hand:
+
+	\begin{enumerate}
+		\item $\log_{10}{(1000)}$
+		\vfill
+		\item $\log_2{(64)}$
+		\vfill
+		\item $\log_2{(\frac{1}{4})}$
+		\vfill
+		\item $\log_x{(x)}$ for any $x$
+		\vfill
+		\item $log_x{(1)}$ for any $x$
+		\vfill
+	\end{enumerate}
+
+	\pagebreak
+
+
+	\definition{}
+	There are a few ways to write logarithms:
+	\begin{itemize}
+		\item[] $\log{x} = \log_{10}{x}$
+		\item[] $\lg{x} = \log_{10}{x}$
+		\item[] $\ln{x} = \log_e{x}$
+	\end{itemize}
+
+	\definition{}
+	The \textit{domain} of a function is the set of values it can take as inputs. \\
+	The \textit{range} of a function is the set of values it can produce.
+
+	\medskip
+
+	For example, the domain and range of $f(x) = x$ is $\mathbb{R}$, all real numbers. \\
+	The domain of $f(x) = |x|$ is $\mathbb{R}$, and its range is $\mathbb{R}^+ \cup \{0\}$, all positive real numbers and 0. \\
+
+	\medskip
+
+	Note that the domain and range of a function are not always equal.
+
+	\problem{}<expdomain>
+	What is the domain of $f(x) = 5^x$? \\
+	What is the range of $f(x) = 5^x$?
+	\vfill
+
+	\problem{}<logdomain>
+	What is the domain of $f(x) = \log{x}$? \\
+	What is the range of $f(x) = \log{x}$?
+	\vfill
+
+	\pagebreak
+
+
+	\problem{}<logids>
+	Prove the following identities: \\
+
+	\begin{enumerate}[itemsep=2mm]
+		\item $\log_b{(b^x)} = x$
+		\item $b^{\log_b{x}} = x$
+		\item $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$
+		\item $\log_b{(\frac{x}{y})} = \log_b{(x)} - \log_b{(y)}$
+		\item $\log_b{(x^y)} = y \log_b{(x)}$
+	\end{enumerate}
+
+	\vfill
+
+	\begin{instructornote}
+		A good intro to the following sections is the linear slide rule:
+
+		\begin{center}
+			\begin{tikzpicture}[scale=1]
+				\linearscale{2}{1}{}
+				\linearscale{0}{0}{}
+
+				\slideruleind
+					{5}
+					{1}
+					{2 + 3 = 5}
+		\end{tikzpicture}
+		\end{center}
+
+		Take two linear rulers, offset one, and you add. \\
+		If you do the same with a log scale, you multiply! \\
+		\vspace{1ex}
+		Note that the slide rules above start at 0.
+
+		\linehack{}
+
+		After assembling the paper slide rule, you can make a visor with some transparent tape. Wrap a strip around the slide rule, sticky side out, and stick it to itself to form a ring. Cover the sticky side with another layer of tape, and trim the edges to make them straight. Use the edge of the visor to read your slide rule!
+	\end{instructornote}
+
+	\pagebreak
+
+	\section{Introduction}
+
+	Mathematicians, physicists, and engineers needed to quickly solve complex equations even before computers were invented.
+
+	\medskip
+
+	The \textit{slide rule} is an instrument that uses the logarithm to solve this problem. Before you continue, cut out and assemble your slide rule.
+
+	\medskip
+
+	There are four scales on your slide rule, each labeled with a letter on the left side:
+
+	\def\sliderulewidth{13}
+	\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\tscale{0}{9}{T}
+		\kscale{0}{8}{K}
+		\abscale{0}{7}{A}
+
+		\abscale{0}{5.5}{B}
+		\ciscale{0}{4.5}{CI}
+		\cdscale{0}{3.5}{C}
+
+		\cdscale{0}{2}{D}
+		\lscale{0}{1}{L}
+		\sscale{0}{0}{S}
+	\end{tikzpicture}
+	\end{center}
+
+	Each scale's ``generating function'' is on the right:
+	\begin{itemize}
+		\item T: $\tan$
+		\item K: $x^3$
+		\item A,B: $x^2$
+		\item CI: $\frac{1}{x}$
+		\item C, D: $x$
+		\item L: $\log_{10}(x)$
+		\item S: $\sin$
+	\end{itemize}
+
+	Once you understand the layout of your slide rule, move on to the next page.
+
+	\pagebreak
+
+	\section{Multiplication}
+
+	We'll use the C and D scales of your slide rule to multiply. \\
+
+	Say we want to multiply $2 \times 3$. First, move the \textit{left-hand index} of the C scale over the smaller number, $2$:
+
+	\def\sliderulewidth{10}
+	\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\cdscale{\cdscalefn(2)}{1}{C}
+		\cdscale{0}{0}{D}
+	\end{tikzpicture}
+	\end{center}
+
+	Then we'll find the second number, $3$ on the C scale, and read the D scale under it:
+
+	\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\cdscale{\cdscalefn(2)}{1}{C}
+		\cdscale{0}{0}{D}
+
+		\slideruleind
+			{\cdscalefn(6)}
+			{1}
+			{6}
+
+	\end{tikzpicture}
+	\end{center}
+
+	Of course, our answer is 6.
+
+	\problem{}
+	What is $1.15 \times 2.1$? \\
+	Use your slide rule.
+
+	\begin{solution}
+		\begin{center}
+			\begin{tikzpicture}[scale=1]
+				\cdscale{\cdscalefn(1.15)}{1}{C}
+				\cdscale{0}{0}{D}
+
+				\slideruleind
+					{\cdscalefn(1.15)}
+					{1}
+					{1.15}
+
+				\slideruleind
+					{\cdscalefn(1.15) + \cdscalefn(2.1)}
+					{1}
+					{2.415}
+
+			\end{tikzpicture}
+			\end{center}
+	\end{solution}
+
+	\vfill
+
+	Note that your answer isn't exact. $1.15 \times 2.1 = 2.415$, but an answer accurate within two decimal places is close enough for most practical applications. \\
+
+	\pagebreak
+
+	Look at your C and D scales again. They contain every number between 1 and 10, but no more than that.
+	What should we do if we want to calculate $32 \times 210$? \\
+
+	\problem{}
+	Using your slide rule, calculate $32 \times 210$. \\
+	%\hint{$32 = 3.2 \times 10^1$}
+
+	\begin{solution}
+		\begin{center}
+		\begin{tikzpicture}[scale=1]
+			\cdscale{\cdscalefn(2.1)}{1}{C}
+			\cdscale{0}{0}{D}
+
+			\slideruleind
+				{\cdscalefn(2.1)}
+				{1}
+				{2.1}
+
+			\slideruleind
+				{\cdscalefn(2.1) + \cdscalefn(3.2)}
+				{1}
+				{6.72}
+
+		\end{tikzpicture}
+		\end{center}
+
+			Placing the decimal point correctly is your job. \\
+			$10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$.
+	\end{solution}
+
+	\vfill
+
+	%This method of writing numbers is called \textit{scientific notation}. In the form $a \times 10^b$, $a$ is called the \textit{mantissa}, and $b$, the \textit{exponent}. \\
+
+	%You may also see expressions like $4.3\text{e}2$. This is equivalent to $4.3 \times 10^2$, but is more compact.
+
+
+	\problem{}
+	Compute the following:
+	\begin{enumerate}
+		\item $1.44 \times 52$
+		\item $0.38 \times 1.24$
+		\item $\pi \times 2.35$
+	\end{enumerate}
+
+	\begin{solution}
+		\begin{enumerate}
+			\item $1.44 \times 52 = 74.88$
+			\item $0.38 \times 1.24 = 0.4712$
+			\item $\pi \times 2.35 = 7.382$
+		\end{enumerate}
+	\end{solution}
+
+	\vfill
+	\pagebreak
+
+	\problem{}<provemult>
+	Note that the numbers on your C and D scales are logarithmically spaced.
+
+	\def\sliderulewidth{13}
+	\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\cdscale{0}{1}{C}
+		\cdscale{0}{0}{D}
+	\end{tikzpicture}
+	\end{center}
+
+	Why does our multiplication procedure work? \\
+	%\hint{See \ref{logids}}
+
+	\vfill
+	\pagebreak
+
+	Now we want to compute $7.2 \times 5.5$:
+
+	\def\sliderulewidth{10}
+	\begin{center}
+	\begin{tikzpicture}[scale=0.8]
+		\cdscale{\cdscalefn(5.5)}{1}{C}
+		\cdscale{0}{0}{D}
+
+		\slideruleind
+			{\cdscalefn(5.5)}
+			{1}
+			{5.5}
+
+		\slideruleind
+			{\cdscalefn(5.5) + \cdscalefn(7.2)}
+			{1}
+			{???}
+
+	\end{tikzpicture}
+	\end{center}
+
+	No matter what order we go in, the answer ends up off the scale. There must be another way. \\
+
+	\medskip
+
+	Look at the far right of your C scale. There's an arrow pointing to the $10$ tick, labeled \textit{right-hand index}. Move it over the \textit{larger} number, $7.2$:
+
+	\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
+		\cdscale{0}{0}{D}
+
+		\slideruleind
+			{\cdscalefn(7.2)}
+			{1}
+			{7.2}
+
+	\end{tikzpicture}
+	\end{center}
+
+	Now find the smaller number, $5.5$, on the C scale, and read the D scale under it:
+
+	\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
+		\cdscale{0}{0}{D}
+
+
+		\slideruleind
+			{\cdscalefn(7.2)}
+			{1}
+			{7.2}
+
+		\slideruleind
+			{\cdscalefn(3.96)}
+			{1}
+			{3.96}
+
+	\end{tikzpicture}
+	\end{center}
+
+	Our answer should be about $7 \times 5 = 35$, so let's move the decimal point: $5.5 \times 7.2 = 39.6$. We can do this by hand to verify our answer. \\
+
+	\medskip
+
+	\iftrue
+		\problem{}
+		Why does this work?
+
+	\else
+		Why does this work? \\
+
+		\medskip
+
+		Consider the following picture, where I've put two D scales next to each other:
+
+		\begin{center}
+		\begin{tikzpicture}[scale=0.7]
+			\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
+			\cdscale{0}{0}{}
+			\cdscale{-10}{0}{}
+
+			\draw[
+				draw=black,
+			]
+				(0, 0)
+				--
+				(0, -0.3)
+				node [below] {D};
+
+			\draw[
+				draw=black,
+			]
+				(-10, 0)
+				--
+				(-10, -0.3)
+				node [below] {D};
+
+			\slideruleind
+				{-10 + \cdscalefn(7.2)}
+				{1}
+				{7.2}
+
+			\slideruleind
+				{\cdscalefn(7.2)}
+				{1}
+				{7.2}
+
+			\slideruleind
+				{\cdscalefn(3.96)}
+				{1}
+				{3.96}
+
+		\end{tikzpicture}
+		\end{center}
+
+		\medskip
+
+		The second D scale has been moved to the right by $(\log{10})$, so every value on it is $(\log{10})$ smaller than it should be.
+
+		\medskip
+
+		\medskip
+		In other words, the answer we get from reverse multiplication is the following: $\log{a} + \log{b} - \log{10}$. \\
+		This reduces to $\log{(\frac{a \times b}{10})}$, which explains the misplaced decimal point in $7.2 \times 5.5$.
+	\fi
+
+	\vfill
+	\pagebreak
+
+	\problem{}
+	Compute the following using your slide rule:
+	\begin{enumerate}
+		\item $9 \times 8$
+		\item $15 \times 35$
+		\item $42.1 \times 7.65$
+		\item $6.5^2$
+	\end{enumerate}
+
+	\begin{solution}
+		\begin{enumerate}
+			\item $9 \times 8 = 72$
+			\item $15 \times 35 = 525$
+			\item $42.1 \times 7.65 = 322.065$
+			\item $6.5^2 = 42.25$
+		\end{enumerate}
+	\end{solution}
+
+	\vfill
+	\pagebreak
+
+	\section{Division}
+
+	Now that you can multiply, division should be easy. All you need to do is work backwards. \\
+	Let's look at our first example again: $3 \times 2 = 6$.
+
+	\medskip
+
+	We can easily see that $6 \div 3 = 2$
+
+	\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\cdscale{\cdscalefn(2)}{1}{C}
+		\cdscale{0}{0}{D}
+
+		\slideruleind
+			{\cdscalefn(6)}
+			{1}
+			{Align here}
+
+		\slideruleind
+			{\cdscalefn(2)}
+			{1}
+			{2}
+	\end{tikzpicture}
+	\end{center}
+
+	and that $6 \div 2 = 3$:
+	\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\cdscale{\cdscalefn(3)}{-3}{C}
+		\cdscale{0}{-4}{D}
+
+
+		\slideruleind
+			{\cdscalefn(6)}
+			{-3}
+			{Align here}
+
+		\slideruleind
+			{\cdscalefn(3)}
+			{-3}
+			{3}
+
+	\end{tikzpicture}
+	\end{center}
+
+	If your left-hand index is off the scale, read the right-hand one. \\
+	Consider $42.25 \div 6.5 = 6.5$:
+
+	\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\cdscale{\cdscalefn(6.5) - \cdscalefn(10)}{1}{C}
+		\cdscale{0}{0}{D}
+
+
+		\slideruleind
+			{\cdscalefn(4.225)}
+			{1}
+			{Align here}
+
+		\slideruleind
+			{\cdscalefn(6.5)}
+			{1}
+			{6.5}
+
+	\end{tikzpicture}
+	\end{center}
+
+	Place your decimal points carefully.
+
+	\vfill
+	\pagebreak
+
+	\problem{}
+	Compute the following using your slide rule. \\
+
+	\begin{enumerate}
+		\item $135 \div 15$
+		\item $68.2 \div 0.575$
+		\item $(118 \times 0.51) \div 6.6$
+	\end{enumerate}
+
+	\begin{solution}
+		\begin{enumerate}
+			\item $135 \div 15 = 9$
+			\item $68.2 \div 0.575 = 118.609$
+			\item $(118 \times 0.51) \div 6.6 = 9.118$
+		\end{enumerate}
+	\end{solution}
+
+	\vfill
+	\pagebreak
+
+
+
+	\section{Squares, Cubes, and Roots}
+
+	Now, take a look at scales A and B, and note the label on the right: $x^2$. If C, D are $x$, A and B are $x^2$, and K is $x^3$.
+
+	\medskip
+
+	Finding squares of numbers up to ten is straightforward: just read the scale. \\
+	Square roots are also easy: find your number on B and read its pair on C. \\
+
+
+	\def\sliderulewidth{13}
+	\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\abscale{0}{1}{B}
+		\cdscale{0}{0}{C}
+	\end{tikzpicture}
+	\end{center}
+
+	\problem{}
+	Compute the following.
+	\begin{enumerate}
+		\item $1.5^2$
+		\item $3.1^2$
+		\item $7^3$
+		\item $\sqrt{14}$
+		\item $\sqrt[3]{150}$
+	\end{enumerate}
+
+	\begin{solution}
+		\begin{enumerate}
+			\item $1.5^2 = 2.25$
+			\item $3.1^2 = 9.61$
+			\item $7^3 = 343$
+			\item $\sqrt{14} = 3.74$
+			\item $\sqrt[3]{150} = 5.313$
+		\end{enumerate}
+	\end{solution}
+
+	\vfill
+	\problem{}
+	Compute the following.
+	\begin{enumerate}
+		\item $42^2$
+		\item $\sqrt{200}$
+		\item $\sqrt{2000}$
+		\item $\sqrt{0.9}$
+		\item $\sqrt[3]{0.12}$
+	\end{enumerate}
+
+	\begin{solution}
+		\begin{enumerate}
+			\item $42^2 = 1,764$
+			\item $\sqrt{200} = 14.14$
+			\item $\sqrt{2000} = 44.72$
+			\item $\sqrt{0.9} = 0.948$
+			\item $\sqrt[3]{0.12} = 0.493$
+		\end{enumerate}
+	\end{solution}
+
+
+	\vfill
+	\pagebreak
+
+	\section{Inverses}
+
+	Try finding $1 \div 32$ using your slide rule. \\
+	The procedure we learned before doesn't work!
+
+	\medskip
+
+	This is why we have the CI scale, or the ``C Inverse'' scale.
+
+	\problem{}
+	Figure out how the CI scale works and compute the following:
+	\begin{enumerate}[itemsep=1mm]
+		\item $\frac{1}{7}$
+		\item $\frac{1}{120}$
+		\item $\frac{1}{\pi}$
+	\end{enumerate}
+
+	\vfill
+	\pagebreak
+
+	\section{Logarithms Base 10}
+
+	When we take a logarithm, the resulting number has two parts: the \textit{characteristic} and the \textit{mantissa}. \\
+	The characteristic is the integral (whole-numbered) part of the answer, and the mantissa is the fractional part (what comes after the decimal). \\
+
+	\medskip
+
+	For example, $\log_{10}{18} = 1.255$, so in this case the characteristic is $1$ and the mantissa is $0.255$.
+
+	\problem{}
+	Approximate the following logs without a slide rule. Find the exact characteristic, and approximate the mantissa.
+	\begin{enumerate}
+		\item $\log_{10}{20}$
+		\item $\log_{2}{18}$
+	\end{enumerate}
+
+	\begin{solution}
+		\begin{enumerate}
+			\item $\log_{10}{20} = 1.30$
+			\item $\log_{2}{18} = 4.17$
+		\end{enumerate}
+	\end{solution}
+
+	\vfill
+
+	Now, find the L scale on your slide rule. As you can see on the right, its generating function is $\log_{10}{x}$.
+
+	\problem{}
+	Compute the following logarithms using your slide rule. \\
+	You'll have to find the characteristic yourself, but your L scale will give you the mantissa. \\
+	Don't forget your log identities!
+
+	\begin{enumerate}
+		\item $\log_{10}{20}$
+		\item $\log_{10}{15}$
+		\item $\log_{10}{150}$
+		\item $\log_{10}{0.024}$
+	\end{enumerate}
+
+	\begin{solution}
+		Careful with number 4.
+
+		\begin{enumerate}
+			\item $\log_{10}{20} = 1.30$
+			\item $\log_{10}{15} = 1.176$
+			\item $\log_{10}{150} = 2.176$
+			\item $\log_{10}{0.024} = -1.6197$
+		\end{enumerate}
+	\end{solution}
+
+	\vfill
+	\pagebreak
+
+	%\problem{}
+	%Find the following.
+	%\begin{enumerate}[itemsep=2mm]
+	%	\item $\frac{118 \times 0.51}{6.6}$
+	%	\item $\sqrt{33.8} \times \sqrt[3]{226}$
+	%	\item $\frac{\sqrt{152}}{\sqrt[3]{4.95}}$
+	%	\item $\frac{\sqrt{96 \times 250}}{\sqrt{7 \times 0.88}}$
+	%	\item The area of a circle with radius $1.47$
+	%	\item The circumference of a circle with radius $31.4$
+	%	\item The radius of a circle with area $6\pi$
+	%	\item $\log_{10}{17.38}$
+	%\end{enumerate}
+	%\vfill
+	%\pagebreak
+
+	\section{Logarithms in Any Base}
+
+	Our slide rule easily computes logarithms in base 10, but we can also use it to find logarithms in \textit{any} base.
+
+	\proposition{}<logcob>
+	This is usually called the \textit{change-of-base} formula:
+
+	\[
+		\log_{b}{a} = \frac{\log_c{a}}{\log_c{b}}
+	\]
+
+	\problem{}
+	Using log identities, prove \ref{logcob}.
+
+	\vfill
+
+	\problem{}
+	Approximate the following:
+	\begin{enumerate}
+		\item $\log_{2}{56}$
+		\item $\log_{5.2}{26}$
+		\item $\log_{12}{500}$
+		\item $\log_{43}{134}$
+	\end{enumerate}
+
+	\begin{solution}
+		\begin{enumerate}
+			\item $\log_{2}{56} = 5.81$
+			\item $\log_{5.2}{26} = 1.97$
+			\item $\log_{12}{500} = 2.50$
+			\item $\log_{43}{134} = 1.30$
+		\end{enumerate}
+	\end{solution}
+
+
+
+
+	% Make sure the slide rule is on an odd page,
+	% so that double-sided printing won't require
+	% students to tear off problems.
+	\checkoddpage
+	\ifoddpage
+		\vfill
+		\pagebreak
+		\vspace*{\fill}
+		\begin{center}
+		{
+			\Large
+			\textbf{This page unintentionally left blank.}
+		}
+		\end{center}
+		\vspace{\fill}
+		\pagebreak
+	\else
+		\vfill
+		\pagebreak
+	\fi
+
+	% Slide rule files
+
+	\includepdf[
+		pages=1,
+		fitpaper=true
+	]{resources/rule.pdf}
+
+\end{document}