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Intermediate/Slide Rules/parts/2 multiplication.tex Normal file
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\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