From 0b2f3efe1be4451751dcaeae1fa0c49d5eb0dbde Mon Sep 17 00:00:00 2001 From: Mark Date: Thu, 4 May 2023 11:24:40 -0700 Subject: [PATCH] Cleanup --- Intermediate/Slide Rules/main.tex | 739 +----------------- .../Slide Rules/parts/0 logarithms.tex | 103 +++ Intermediate/Slide Rules/parts/1 intro.tex | 43 + .../Slide Rules/parts/2 multiplication.tex | 284 +++++++ Intermediate/Slide Rules/parts/3 division.tex | 92 +++ Intermediate/Slide Rules/parts/4 squares.tex | 62 ++ Intermediate/Slide Rules/parts/5 inverses.tex | 19 + Intermediate/Slide Rules/parts/6 log.tex | 105 +++ 8 files changed, 720 insertions(+), 727 deletions(-) create mode 100644 Intermediate/Slide Rules/parts/0 logarithms.tex create mode 100644 Intermediate/Slide Rules/parts/1 intro.tex create mode 100644 Intermediate/Slide Rules/parts/2 multiplication.tex create mode 100644 Intermediate/Slide Rules/parts/3 division.tex create mode 100644 Intermediate/Slide Rules/parts/4 squares.tex create mode 100644 Intermediate/Slide Rules/parts/5 inverses.tex create mode 100644 Intermediate/Slide Rules/parts/6 log.tex diff --git a/Intermediate/Slide Rules/main.tex b/Intermediate/Slide Rules/main.tex index 8838305..04f9b73 100755 --- a/Intermediate/Slide Rules/main.tex +++ b/Intermediate/Slide Rules/main.tex @@ -1,7 +1,7 @@ % use [nosolutions] flag to hide solutions. % use [solutions] flag to show solutions. \documentclass[ - solutions + nosolutions ]{../../resources/ormc_handout} \usepackage{pdfpages} @@ -32,8 +32,7 @@ Prepared by Mark on \today } - \vspace{1cm} - + \begin{center} \begin{minipage}{6cm} Dad says that anyone who can't use a slide rule is a cultural illiterate @@ -41,733 +40,24 @@ \vspace{1ex} - \textit{Have Space Suit -- Will Travel, 1958} + \textit{Have Space Suit --- Will Travel, 1958} \end{minipage} + \end{center} \hfill - \section{Logarithms} - - \definition{} - 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{} - What is the domain of $f(x) = 5^x$? \\ - What is the range of $f(x) = 5^x$? - \vfill - - \problem{} - What is the domain of $f(x) = \log{x}$? \\ - What is the range of $f(x) = \log{x}$? - \vfill - - \pagebreak - - - \problem{} - 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{} - 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{} - 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} - - - + \input{parts/0 logarithms.tex} + \input{parts/1 intro.tex} + \input{parts/2 multiplication.tex} + \input{parts/3 division.tex} + \input{parts/4 squares.tex} + \input{parts/5 inverses.tex} + \input{parts/6 log.tex} % 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 + \ifoddpage\else \vspace*{\fill} \begin{center} { @@ -777,13 +67,8 @@ \end{center} \vspace{\fill} \pagebreak - \else - \vfill - \pagebreak \fi - % Slide rule files - \includepdf[ pages=1, fitpaper=true diff --git a/Intermediate/Slide Rules/parts/0 logarithms.tex b/Intermediate/Slide Rules/parts/0 logarithms.tex new file mode 100644 index 0000000..ecf8d18 --- /dev/null +++ b/Intermediate/Slide Rules/parts/0 logarithms.tex @@ -0,0 +1,103 @@ +\section{Logarithms} + +\definition{} +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{} +What is the domain of $f(x) = 5^x$? \\ +What is the range of $f(x) = 5^x$? +\vfill + +\problem{} +What is the domain of $f(x) = \log{x}$? \\ +What is the range of $f(x) = \log{x}$? +\vfill + +\pagebreak + + +\problem{} +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 \ No newline at end of file diff --git a/Intermediate/Slide Rules/parts/1 intro.tex b/Intermediate/Slide Rules/parts/1 intro.tex new file mode 100644 index 0000000..54a8eb7 --- /dev/null +++ b/Intermediate/Slide Rules/parts/1 intro.tex @@ -0,0 +1,43 @@ +\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 diff --git a/Intermediate/Slide Rules/parts/2 multiplication.tex b/Intermediate/Slide Rules/parts/2 multiplication.tex new file mode 100644 index 0000000..da5a7bb --- /dev/null +++ b/Intermediate/Slide Rules/parts/2 multiplication.tex @@ -0,0 +1,284 @@ +\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{} +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 \ No newline at end of file diff --git a/Intermediate/Slide Rules/parts/3 division.tex b/Intermediate/Slide Rules/parts/3 division.tex new file mode 100644 index 0000000..193965b --- /dev/null +++ b/Intermediate/Slide Rules/parts/3 division.tex @@ -0,0 +1,92 @@ +\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 \ No newline at end of file diff --git a/Intermediate/Slide Rules/parts/4 squares.tex b/Intermediate/Slide Rules/parts/4 squares.tex new file mode 100644 index 0000000..6887bf2 --- /dev/null +++ b/Intermediate/Slide Rules/parts/4 squares.tex @@ -0,0 +1,62 @@ +\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 \ No newline at end of file diff --git a/Intermediate/Slide Rules/parts/5 inverses.tex b/Intermediate/Slide Rules/parts/5 inverses.tex new file mode 100644 index 0000000..fdb3554 --- /dev/null +++ b/Intermediate/Slide Rules/parts/5 inverses.tex @@ -0,0 +1,19 @@ +\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 \ No newline at end of file diff --git a/Intermediate/Slide Rules/parts/6 log.tex b/Intermediate/Slide Rules/parts/6 log.tex new file mode 100644 index 0000000..6282882 --- /dev/null +++ b/Intermediate/Slide Rules/parts/6 log.tex @@ -0,0 +1,105 @@ +\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{} +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} + + +\vfill +\pagebreak \ No newline at end of file