From e0c6872cf40896c7be36b11dcc744620f10adf1d Mon Sep 17 00:00:00 2001
From: Norbert Preining <norbert@preining.info>
Date: Mon, 2 Sep 2019 13:46:59 +0900
Subject: Initial commit

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+%
+% This is the file webeqtst.tex that is distributed with the AeB Bundle
+%
+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{graphicx}
+\usepackage[tight,designi]{web}  % dvipsone, dvips, pdftex, dvipdfm
+\usepackage{exerquiz}
+
+\usepackage[def=lmacs_aeb,js=lmacs_aeb]{lmacs}
+
+\begin{document}
+
+\maketitle
+
+\tableofcontents
+
+
+\section{Introduction}
+
+The \textsf{lmacs} is designed to clean up the preamble of a source file.
+For this file, we have
+\begin{verbatim}
+    \usepackage[def=lmacs_aeb,js=lmacs_aeb]{lmacs}
+\end{verbatim}
+The preamble definitions are in the file \texttt{lmacs\_aeb.def} and a
+document JavaScript is imported with the file \texttt{lmacs\_aeb.js}.
+
+\medskip\noindent We'll test the JavaScript first, press this button:
+\pushButton[\CA{Press Me}\A{\JS{%
+    makeAlert("Hooray for the lmacs package!")
+}}]{alertBtn}{}{11bp}
+
+\medskip\noindent The next section is taken from the file
+\texttt{webeqtst.tex}. The problem environment is defined in the file
+\texttt{lmacs\_aeb.def}, other definitions and customizations can be found
+in that file.
+
+\medskip\noindent Though I am using the \textsf{web} and \textsf{exerquiz} package, lmacs
+does not require them; \textsf{lmacs} is a general purpose package for inputting
+local definitions.
+
+\section{Online Exercises}
+
+A well-designed sequences of exercises can be of aid to the
+student.  The \texttt{exercise} environment makes it easy to
+produce electronic exercises.  By using the \texttt{forpaper}
+option, you can also make a paper version of your exercises.
+
+\begin{exercise}
+Evaluate the integral \(\displaystyle\int x^2 e^{2x}\,dx\).
+\begin{solution}
+We evaluate by \texttt{integration by parts}:\normalsize
+\begin{alignat*}{2}
+ \int x^2 e^{2x}\,dx &
+   = \tfrac12 x^2 e^{2x} - \int x e^{2x}\,dx &&\quad
+           \text{$u=x^2$, $dv=e^{2x}\,dx$}\\&
+   = \tfrac12 x^2 e^{2x} -
+           \Bigl[\tfrac12 x e^{2x}-\int \tfrac12 e^{2x}\,dx\Bigr] &&\quad
+           \text{integration by parts}\\&
+   = \tfrac12 x^2 e^{2x} - \tfrac12 x e^{2x} + \tfrac12\int e^{2x}\,dx &&\quad
+           \text{$u=x^2$, $dv=e^{2x}\,dx$}\\&
+   = \tfrac12 x^2 e^{2x} - \tfrac12 x e^{2x} + \tfrac14 e^{2x} &&\quad
+           \text{integration by parts}\\&
+   = \tfrac14(2x^2-2x+1)e^{2x} &&\quad
+           \text{simplify!}
+\end{alignat*}
+\end{solution}
+\end{exercise}
+
+In the preamble of this document, we defined a \texttt{problem}
+environment with its own counter.  Here is an example of it.
+
+\begin{problem}
+Is $F(t)=\sin(t)$ an antiderivative of $f(x)=\cos(x)$?  Explain
+your reasoning.
+\begin{solution}
+The answer is yes. The definition states that $F$ is an
+antiderivative of $f$ if $F'(x)=f(x)$.  Note that
+$$
+       F(t)=\sin(t) \implies F'(t) = \cos(t)
+$$
+hence, $F(x) = \cos(x) = f(x)$.
+\end{solution}
+\end{problem}
+
+\begin{problem}
+Is $F(t)=\sin(t)$ an antiderivative of $f(x)=\cos(x)$?  Explain
+your reasoning.
+\begin{solution}
+The answer is yes. The definition states that $F$ is an
+antiderivative of $f$ if $F'(x)=f(x)$.  Note that
+$$
+       F(t)=\sin(t) \implies F'(t) = \cos(t)
+$$
+hence, $F(x) = \cos(x) = f(x)$.
+\end{solution}
+\end{problem}
+
+\noindent By modifying the \texttt{exercise} environment, you can
+also create an \texttt{example} environment.  The one defined in
+the preamble of this document has no associated counter.
+
+\begin{example}
+Give an example of a set that is \textit{clopen}.
+\begin{solution}
+The real number line is both closed and open in the usual topology of the
+real line.%
+\end{solution}
+\end{example}
+
+There is a \texttt*-option with the \texttt{exercise} environment,
+using it signals the presence of a multiple part exercise
+question. The following exercise illustrates this option.
+
+\begin{exercise}*\label{ex:parts}
+Suppose a particle is moving along the $s$-axis, and that its position
+at any time $t$ is given by $s=t^2 - 5t + 1$.
+\begin{parts}
+\item[h]\label{item:part} Find the velocity, $v$, of the particle at any time
+$t$.
+\begin{solution}
+Velocity is the rate of change of position with respect to time. In
+symbols:
+$$
+                    v = \frac{ds}{dt}
+$$
+For our problem, we have
+$$
+        v = \frac{ds}{dt} =\frac d{dt}(t^2 - 5t + 1) = 2t-5.
+$$
+The velocity at time $t$ is given by $\boxed{v=2t-5}$.
+\end{solution}
+
+\item Find the acceleration, $a$, of the particle at any time $t$.
+\begin{solution}
+Acceleration is the rate of change of velocity with respect to time.
+Thus,
+$$
+                    a = \frac{dv}{dt}
+$$
+For our problem, we have
+$$
+        a = \frac{dv}{dt} =\frac d{dt}(2t-5)=2.
+$$
+The acceleration at time $t$ is constant: $\boxed{a=2}$.
+\end{solution}
+\end{parts}
+\end{exercise}
+
+References can be made to a particular part of an exercise; for example,
+``see \hyperref[item:part]{Exercise~\ref*{ex:parts}(\ref*{item:part})}.''
+Part (a) is in \textcolor{webblue}{blue}; the solutions for that part is
+``hidden''.  This is a new option for the \texttt{exercise} environment.
+
+There is now an option for listing multipart question in tabular form.
+This problem style does not obey the \texttt{solutions\-after} option.
+
+\begin{exercise}*
+Simplify each of the following expressions in the complex number
+system. \textit{Note}: $\bar z$ is the conjugate of $z$;
+$\operatorname{Re} z$ is the real part of $z$ and
+$\operatorname{Im} z$ is the imaginary part of $z$.
+\begin{parts}[2]
+\item $i^2$
+\begin{solution} $i^2 = -1$ \end{solution}
+&
+\item $i^3$ \begin{solution} $i^3 = i i^2 = -i$\end{solution}
+\\
+\item $z+\bar z$
+\begin{solution} $z+\bar z=\operatorname{Re} z$\end{solution}
+&
+\item[h] $1/z$
+\begin{solution}
+$\displaystyle\frac 1z=\frac 1z\frac{\bar z}{\bar z}=\frac z{z\bar z}=\frac z{|z|^2}$
+\end{solution}
+\end{parts}
+\end{exercise}
+
+
+\end{document}
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