%% LyX 1.1 created this file.  For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[11pt,oneside]{amsart}
\usepackage[T1]{fontenc}
\usepackage{geometry}
\geometry{verbose,letterpaper,tmargin=1in,bmargin=1in,lmargin=1in,rmargin=1in}
\setlength\parskip{\medskipamount}
\setlength\parindent{0pt}

\makeatletter


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% LyX specific LaTeX commands.
\providecommand{\LyX}{L\kern-.1667em\lower.25em\hbox{Y}\kern-.125emX\@}
%% Special footnote code from the package 'stblftnt.sty'
%% Author: Robin Fairbairns -- Last revised Dec 13 1996
\let\SF@@footnote\footnote
\def\footnote{\ifx\protect\@typeset@protect
    \expandafter\SF@@footnote
  \else
    \expandafter\SF@gobble@opt
  \fi
}
\expandafter\def\csname SF@gobble@opt \endcsname{\@ifnextchar[%]
  \SF@gobble@twobracket
  \@gobble
}
\edef\SF@gobble@opt{\noexpand\protect
  \expandafter\noexpand\csname SF@gobble@opt \endcsname}
\def\SF@gobble@twobracket[#1]#2{}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Textclass specific LaTeX commands.
 \theoremstyle{plain}    
 \newtheorem{thm}{Theorem}[section]
 \numberwithin{equation}{section} %% Comment out for sequentially-numbered
 \numberwithin{figure}{section} %% Comment out for sequentially-numbered
 \theoremstyle{definition}
  \newtheorem{example}[thm]{Example}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
\setcounter{section}{6}
\setcounter{subsection}{3}
\makeatother

\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Arclength\footnote{%
This section starts out all wrong in the books. They always talk about \emph{defining}
the arclength of a curve by some formula. We are not \textit{really} defining
the idea of the \textit{length} of a curve. It seems rather condescending to
say that we don't already know what the length of a curve is. That's just a
mathematician's way of talking. What they mean to say is that we need to come
up with a way of \textit{measuring} the length of a curve.  
}}

The idea that would occur to me, if I had to measure the length of a curve,
would be to drive a bunch of nails in the blackboard along the curve, and loop
a string through those nails (keeping it taut and on the ``right'' side of the
nails). Then I'd unwrap it and hold it alongside a ruler. If I wanted to be
more accurate, I'd go get more nails. 

Let's see how that works without really driving nails into the blackboard. If
the curve is \( y=f(x) \), \( x\in [a,b] \), then we can, as usual, break
the interval from \( a \) to \( b \) up into a number (\( n \)) of subintervals,
\( a=x_{0}<x_{1}<\dots <x_{n-1}<x_{n}=b \). The nails would then be placed
at the points \( \{(x_{0},f(x_{0})),\dots ,(x_{n},f(x_{n}))\} \). The length
of a line segment from \( (x_{i-1},f(x_{i-1})) \) to \( (x_{i},f(x_{i})) \)
is
\[
\Delta s=\sqrt{(x_{i}-x_{i-1})^{2}+(f(x_{i})-f(x_{i-1}))^{2}}=\sqrt{\Delta x^{2}+\Delta y^{2}},\]
and so the length of the string will be:
\[
\sum _{i=1}^{n}\sqrt{(x_{i}-x_{i-1})^{2}+(f(x_{i})-f(x_{i-1}))^{2}}=\sum _{i=1}^{n}\sqrt{\Delta x^{2}+\Delta y^{2}}.\]
This is the first of these applications of integration formulas that doesn't
seem to be approaching an integral. There is no \( dx \). But we can make it
appear by:
\[
length:\cong \sum _{i=1}^{n}\sqrt{\Delta x^{2}+\Delta y^{2}}=\sum _{i=1}^{n}\sqrt{dx^{2}+dy^{2}}=\sum _{i=1}^{n}\sqrt{1+\left( \frac{dy}{dx}\right) ^{2}}\, dx\]
Now we should see the standard idea of an application of integration coming
out of this: take the number of terms to \( \infty  \), or the distance between
the subintervals to 0, and you'd get:
\[
length:=\lim _{n\rightarrow \infty }\sum _{i=1}^{n}\sqrt{1+\left( \frac{dy}{dx}\right) ^{2}}\, dx=\int _{a}^{b}\sqrt{1+\left( \frac{dy}{dx}\right) ^{2}}\, dx.\]
The problems have to be very carefully chosen or else the integrals will be
impossible. That is a hint, since the problems in the text are carefully chosen
so that the integrals will not be impossible. If your solution is impossible
to integrate, perhaps you set the problem up incorrectly. If the set-up is right,
you might be missing something in doing the integral. 

\begin{example}
Find the arc-length of the curve \( y=x^{3/2} \), \( x\in [0,4] \). 
\end{example}

\subsubsection*{Solution}


\begin{eqnarray*}
length & = & \int _{0}^{4}\sqrt{1+(\frac{3}{2}x^{1/2})^{2}}dx\\
 & = & \int _{0}^{4}\sqrt{1+\frac{9x}{4}}dx\\
 & = & \left. \frac{2}{3}\cdot \frac{4}{9}\left( 1+\frac{9x}{4}\right) ^{3/2}\right| _{0}^{4}\\
 & = & \frac{8}{27}\left( 10^{3/2}-1\right) .
\end{eqnarray*}


\begin{example}
Find the arc-length of the curve \( y=\frac{x^{2}}{2}-\frac{\ln (x)}{4} \),
\( x\in (2,4] \). 
\end{example}

\subsubsection*{Solution}

\begin{eqnarray*}
length & = & \int _{2}^{4}\sqrt{1+\left( x-\frac{1}{4x}\right) ^{2}}dx\\
 & = & \int _{2}^{4}\sqrt{1+\left( x^{2}-\frac{1}{2}+\frac{1}{16x^{2}}\right) }dx\\
 & = & \int _{2}^{4}\sqrt{x^{2}+\frac{1}{2}+\frac{1}{16x^{2}}}dx\\
 & = & \int _{2}^{4}\sqrt{\left( x+\frac{1}{4x}\right) ^{2}}dx\\
 & = & \int _{2}^{4}\left( x+\frac{1}{4x}\right) dx\\
 & = & \left. \frac{x^{2}}{2}+\frac{1}{4}\ln x\right| _{2}^{4}\\
 & = & \left( 8+\frac{1}{4}\ln 4\right) -\left( 2+\frac{1}{4}\ln 2\right) .
\end{eqnarray*}
This is a standard trick, where the \( +1 \) was exactly what was needed to
change the one perfect square to another, by changing the cross-term from \( -1/2 \)
to \( 1/2 \).

\begin{example}
Find the arc-length of the parabola \( y=x^{2} \), \( x\in [-2,2] \)

\end{example}
\end{document}
