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calc1n-derivative.tex
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calc1n-derivative.tex
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%%%%%%%%%%%%%%%%%%%%
% SEMESTER START.
% Send to repro asap;
% print in color!
%
% DAY OF.
% Do 2 together.
% Write f(x+h) =
% Then add f(x+h) - f(x) =
% Then add (f(x+h) - f(x))(1/h) =
%%%%%%%%%%%%%%%%%%%%
\RequirePackage{luatex85}
\documentclass{article}
\usepackage{tikz, amsmath, mathtools, multicol, setspace}
\usepackage[margin=0.5in, column sep=1in]{geometry}
\setlength\parindent{0mm}
\thispagestyle{empty}
\pagestyle{empty}
%\everymath{\displaystyle}
%% define macro \Repeat{integer}{content}
%\usepackage{expl3}
%\ExplSyntaxOn
%\cs_new_eq:NN \Repeat \prg_replicate:nn
%\ExplSyntaxOff
\newcommand\blob[2][15mm]{\(\begin{matrix}\begin{tikzpicture}
\node[text width=#1, align=flush center]{\begin{spacing}{0.7}#2\end{spacing}};
\end{tikzpicture}\end{matrix}\)}
\newcommand\myoverset[2]{\smash{\large$\overset
{#1}{\vphantom{\bigg(}\hspace{#2mm}}
$}}
\def\yscale{0.8}
\newcommand\diagram[3]{
\begin{tikzpicture}[very thick, scale=2.1, yscale=\yscale]
#3
% axes
\draw[<->] (0,2.75) node[below left] {distance [mi]}
to (0,0)
to (5.75,0) node[right] {time [hr]};
\draw[domain=0.5:5.65]
plot (\x, {(\x-1)^4*5/8/100+15/100}) #1;
% secant
\draw[red, domain=2.5:6]
plot (\x, {0.75*(\x-3)+0.25})
node [right] {\,\,\large secant};
% tangent
\draw[red, domain=1.25:6] plot (\x, {0.2*(\x-3)+0.25})
node [right] {\,\,\large tangent};
#2
\end{tikzpicture}}
\newcommand\pageone{
% margin=0.5in, row sep=1in, so textheight=4.5in
\begin{minipage}[t][4.5in]{\textwidth}
N1. \textbf{Definition.} The Limit Definition of Derivative
%\vspace{-1ex}
Suppose the distance a train has traveled is a function of time $y=f(x)$.
\vspace{-2ex}
\hfil
\diagram{node [right] {$y=f(x)\phantom{{}=15+\frac58x^4}$}}{
% tick marks and coordinates for closed circles
\foreach \x/\n in {3/x, 5/x+h}{
\draw (\x, {(\x-1)^4*5/8/100+15/100}) coordinate (\x);
\draw (\x, 3pt) -- ++(down:6pt) node[below] {\large$\n\vphantom h$};
}
% closed circles (yscale counteracts global yscale)
\foreach \x in {3, 5}{
\draw[black, fill=black, yscale=1/\yscale] (\x) circle (1.5pt);
}
% fillin (x1, y1) = (x, f(x))
\draw (3) node [above left] {
\bfseries\huge(\myoverset{x_1}{14},\myoverset{y_1}{14})\!
};
% fillin (x2, y2) = (x+h, f(x+h))
\draw (5) node [above left] {
\bfseries\huge(\myoverset{x_2}{18},\myoverset{y_2}{18})\!
};
}{
}
\vfill
\blob{avera\smash ge velocity over $[x, x{+}h]$}
= \blob{avera\smash ge rate of change over $[x, x{+}h]$}
= \blob{slope of secant between $x, x{+}h$}
=\hfill$\frac{\Delta y}{\Delta x}$\hfill%
=\hfill$\frac{y_2 - y_1}{x_2 - x_1}$\hfill%
=\hfill{\LARGE\boxed{\phantom{\frac{f(x+h)-f(x)}h}}}\hfill%
= \blob[22mm]{definition\vphantom q of difference quotient}
\vfill
\blob{\,\,\clap{\smash{\small(``instantaneous")}}\,\, velocity at $x$}
= \blob{rate of change at $x$}
= \blob{slope of tangent at $x$}
=\hfill{\LARGE\boxed{\lim_{h\to\quad}\phantom{\frac{f(x+h)-f(x)}h}}}\hfill%
=\hfill\!$\underset{
\text{written } y' \text{ or } \frac{dy}{dx} \text{ or } f'(x) \text{ or } \frac d{dx}f(x)
}{\text{definition of derivative}}$
\vfill
\end{minipage}
}
\newcommand\pagetwo{
\begin{minipage}[t][4.5in]{\textwidth}
\begin{tabbing}
N2. \textbf{Exercise.}\,\, \= Estimate\={} from the d\=iagram: the slope of the \textbf{secant} through 2 and 4. \\[3.5ex]
N3. \textbf{Exercise.} \> Estimate from the diagram: the slope of the \textbf{tangent} at 2. \\[0.5ex]
\diagram{node [right, yshift=-2mm] {$y=f(x)$}}{
% tick marks and coordinates for closed circles
\foreach \x in {0, ..., 4}{
\draw (\x, {(\x-1)^4*5/8/100+15/100}) coordinate (\x);
\draw (\x+1, 3pt) -- ++(down:6pt) node[below] {\large$\x$};
}
% closed circles (yscale counteracts global yscale)
\foreach \x in {3, 5}{
\draw[black, fill=black, yscale=1/\yscale] (\x) circle (1.5pt);
}
% y-tick marks
\foreach \y in {0, 100, 200}{
\draw (3pt, \y/100) -- ++(left:6pt) node[left] {\large$\y$};
}
}{
% grid
\draw[gray, semithick, dotted, step=0.25] (0,0) grid (5.75,2.75);
\draw[gray, thick] (0,0) grid (5.75,2.75);
}\\[-0.5ex]
N4. \scshape{Example.} \> Suppose \>$f(x)=15+\frac58x^4$.
Compute the slope of the \textbf{secant} through $x$ and $x+h$. \\[0.5ex]
\>\>\color{gray} $\scriptstyle f(x+h)$
\>\color{gray} = $15+\frac58(x+h)^4$
\quad\,\, = $15 + \frac58x^4+\frac{20}8x^3h+\frac{30}8x^2h^2+\frac{20}8xh^3+\frac58h^4$ \\[0.5ex]
\>\>\color{gray} $\frac{f(x+h) - f(x)}h$
\>\color{gray} =
\\[1ex]
N5. \textbf{Exercise.} \> Suppose \>$f(x)=15+\frac58x^4$.
Compute the slope of the \textbf{tangent} line at $x$. \\[4.5ex]
N6. \textbf{Exercise.} \> Use your answer to N5 to compute $f'(2)$.
\end{tabbing}
\end{minipage}
\vspace{1in}
}
\begin{document}
\large
\pageone\par\vspace{1in}
\pagetwo\par\vspace{1in}
\newpage
\rotatebox{180}{\pagetwo}\par\vspace{1in}
\rotatebox{180}{\pageone}\par\vspace{1in}
\end{document}