From e0c6872cf40896c7be36b11dcc744620f10adf1d Mon Sep 17 00:00:00 2001 From: Norbert Preining Date: Mon, 2 Sep 2019 13:46:59 +0900 Subject: Initial commit --- macros/latex/contrib/lmacs/examples/lmacs_aeb.tex | 181 ++++++++++++++++++++++ 1 file changed, 181 insertions(+) create mode 100644 macros/latex/contrib/lmacs/examples/lmacs_aeb.tex (limited to 'macros/latex/contrib/lmacs/examples/lmacs_aeb.tex') diff --git a/macros/latex/contrib/lmacs/examples/lmacs_aeb.tex b/macros/latex/contrib/lmacs/examples/lmacs_aeb.tex new file mode 100644 index 0000000000..87d6bfec94 --- /dev/null +++ b/macros/latex/contrib/lmacs/examples/lmacs_aeb.tex @@ -0,0 +1,181 @@ +% +% 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} -- cgit v1.2.3