284 lines
5.5 KiB
TeX
284 lines
5.5 KiB
TeX
\section{Multiplication}
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We'll use the C and D scales of your slide rule to multiply. \\
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Say we want to multiply $2 \times 3$. First, move the \textit{left-hand index} of the C scale over the smaller number, $2$:
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\def\sliderulewidth{10}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(2)}{1}{C}
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\cdscale{0}{0}{D}
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\end{tikzpicture}
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\end{center}
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Then we'll find the second number, $3$ on the C scale, and read the D scale under it:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(2)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(6)}
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{1}
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{6}
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\end{tikzpicture}
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\end{center}
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Of course, our answer is 6.
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\problem{}
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What is $1.15 \times 2.1$? \\
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Use your slide rule.
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\begin{solution}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(1.15)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(1.15)}
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{1}
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{1.15}
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\slideruleind
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{\cdscalefn(1.15) + \cdscalefn(2.1)}
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{1}
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{2.415}
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\end{tikzpicture}
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\end{center}
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\end{solution}
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\vfill
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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. \\
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\pagebreak
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Look at your C and D scales again. They contain every number between 1 and 10, but no more than that.
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What should we do if we want to calculate $32 \times 210$? \\
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\problem{}
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Using your slide rule, calculate $32 \times 210$. \\
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%\hint{$32 = 3.2 \times 10^1$}
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\begin{solution}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(2.1)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(2.1)}
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{1}
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{2.1}
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\slideruleind
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{\cdscalefn(2.1) + \cdscalefn(3.2)}
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{1}
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{6.72}
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\end{tikzpicture}
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\end{center}
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Placing the decimal point correctly is your job. \\
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$10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$.
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\end{solution}
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\vfill
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%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}. \\
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%You may also see expressions like $4.3\text{e}2$. This is equivalent to $4.3 \times 10^2$, but is more compact.
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\problem{}
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Compute the following:
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\begin{enumerate}
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\item $1.44 \times 52$
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\item $0.38 \times 1.24$
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\item $\pi \times 2.35$
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\end{enumerate}
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\begin{solution}
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\begin{enumerate}
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\item $1.44 \times 52 = 74.88$
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\item $0.38 \times 1.24 = 0.4712$
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\item $\pi \times 2.35 = 7.382$
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\end{enumerate}
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\end{solution}
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\vfill
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\pagebreak
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\problem{}<provemult>
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Note that the numbers on your C and D scales are logarithmically spaced.
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\def\sliderulewidth{13}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{0}{1}{C}
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\cdscale{0}{0}{D}
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\end{tikzpicture}
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\end{center}
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Why does our multiplication procedure work? \\
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%\hint{See \ref{logids}}
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\vfill
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\pagebreak
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Now we want to compute $7.2 \times 5.5$:
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\def\sliderulewidth{10}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\cdscale{\cdscalefn(5.5)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(5.5)}
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{1}
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{5.5}
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\slideruleind
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{\cdscalefn(5.5) + \cdscalefn(7.2)}
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{1}
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{???}
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\end{tikzpicture}
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\end{center}
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No matter what order we go in, the answer ends up off the scale. There must be another way. \\
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\medskip
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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, $7.2$:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(7.2)}
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{1}
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{7.2}
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\end{tikzpicture}
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\end{center}
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Now find the smaller number, $5.5$, on the C scale, and read the D scale under it:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(7.2)}
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{1}
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{7.2}
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\slideruleind
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{\cdscalefn(3.96)}
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{1}
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{3.96}
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\end{tikzpicture}
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\end{center}
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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. \\
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\medskip
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\iftrue
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\problem{}
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Why does this work?
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\else
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Why does this work? \\
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\medskip
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Consider the following picture, where I've put two D scales next to each other:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
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\cdscale{0}{0}{}
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\cdscale{-10}{0}{}
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\draw[
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draw=black,
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]
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(0, 0)
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--
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(0, -0.3)
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node [below] {D};
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\draw[
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draw=black,
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]
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(-10, 0)
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--
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(-10, -0.3)
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node [below] {D};
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\slideruleind
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{-10 + \cdscalefn(7.2)}
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{1}
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{7.2}
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\slideruleind
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{\cdscalefn(7.2)}
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{1}
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{7.2}
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\slideruleind
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{\cdscalefn(3.96)}
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{1}
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{3.96}
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\end{tikzpicture}
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\end{center}
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\medskip
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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.
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\medskip
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\medskip
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In other words, the answer we get from reverse multiplication is the following: $\log{a} + \log{b} - \log{10}$. \\
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This reduces to $\log{(\frac{a \times b}{10})}$, which explains the misplaced decimal point in $7.2 \times 5.5$.
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\fi
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\vfill
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\pagebreak
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\problem{}
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Compute the following using your slide rule:
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\begin{enumerate}
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\item $9 \times 8$
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\item $15 \times 35$
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\item $42.1 \times 7.65$
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\item $6.5^2$
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\end{enumerate}
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\begin{solution}
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\begin{enumerate}
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\item $9 \times 8 = 72$
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\item $15 \times 35 = 525$
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\item $42.1 \times 7.65 = 322.065$
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\item $6.5^2 = 42.25$
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\end{enumerate}
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\end{solution}
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\vfill
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\pagebreak |