%% 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}
\usepackage{graphics}

\makeatletter


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% LyX specific LaTeX commands.
\providecommand{\LyX}{L\kern-.1667em\lower.25em\hbox{Y}\kern-.125emX\@}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 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}{2}
\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{Applications of the derivative.}

\begin{example}
A boater has his boat tied to the pier by a  long rope. He feeds the pier-end
of the rope through a winch and pulls his boat in towards the pier. The tide
is out, so the bow of the boat is 6 feet below the level of the pier. If the
winch is pulling in the line at a constant rate of 2 feet per second, how fast
is the boat moving towards the pier when there is 10 feet of rope between the
boat and the winch?  
\end{example}
\vspace{0.3cm}
{\par\centering \includegraphics{boat.ps} \par}
\vspace{0.3cm}


\subsubsection*{Solution}

The first thing to do is to identify what information you \textbf{know}, what
information you \textbf{want} (well, OK, what you are asked to find), and \textbf{when}.
The {}``when{}'' will usually be expressed in terms of the time when one or
more of the quantities has a set value, but sometimes it is a certain time during
the course of the problem. 

To do this, label with variables those quantities that vary. Only label as constants
things that stay constant throughout. Also, as pretty as the animated graphics
may be, it helps to reduce the drawing to the relevant essentials. Since the
boat moves horizontally, the speed of the boat is the rate of change of the
horizontal distance between the boat and the dock. Call that distance \( x \).
Of course, you should think of it as a function \( x(t) \) of time, but just
label it as \( x \). Similarly, call the length of the rope (between the winch
and the bow) \( s \). You need these variables because you need to know something
about how \( x \) changes, and you know something about the rate of change
of \( s \). Of course, the distance down from the winch to the boat is a constant
6 feet. 

Then, in this example, we:


\subsubsection*{Know }

that \( ds/dt=-2 \). It's negative because the length of the rope is decreasing.
Note that we keep units consistent; here we make the natural choices that time
is measured in seconds and distance in feet, since that is the way the problem
is given,


\subsubsection*{Want }

\( dx/dt \) 


\subsubsection*{When}

\( s=10. \) Be careful. Don't label the length of the rope as {}``10{}''.
It is not constant, and only after you differentiate should you replace \( s \)
by 10. 

We display all this by this diagram

\vspace{0.3cm}
{\par\centering \includegraphics{boat-measurements.ps} \par}
\vspace{0.3cm}

The horizontal path of the boat, the line described by the rope, and the vertical
drop from the winch to the boat's path form a right triangle. That triangle
gives the relationship between the variables as well, according to the Pythagorean
theorem (which gets used a lot in these problems) \[
x^{2}+6^{2}=s^{2}.\]
To find the relationship between the rates of change of these quantities, differentiate
both sides of the equation:\[
2x\frac{dx}{dt}+0=2s\frac{ds}{dt}.\]
At this point you should be able to simply plug in the values of what you know:
\( ds/dt=-2 \). But you need more to solve for what you want (\( dx/dt \)).
You of course plug in the when, but here the {}``when{}'' only directly gives
\( s=10 \), and you still can't solve for what you want, since there is that
remaining \( x \). To find \( x \) you re-use the relationship between the
variables, this time plugging in for the when:\begin{eqnarray*}
x^{2}+6^{2} & = & s^{2}\\
x^{2}+36 & = & 100,\: \textrm{so}\\
x^{2} & = & 100-36=64,\: \textrm{or}\\
x & = & 8.
\end{eqnarray*}
Actually, if you remember your geometry there is an easier way to do this. This
is a {}``3-4-5{}'' triangle -- well, a {}``6-8-10{}'', but it's the same
idea.


\subsubsection*{Aside}

Now, plug in for \( x \) and \( s \), as well as \( ds/dt, \) and solve for
\( dx/dt \):\begin{eqnarray*}
x^{2}+6^{2} & = & s^{2}\\
2x\frac{dx}{dt}+0 & = & 2s\frac{ds}{dt}\\
2\cdot 8\frac{dx}{dt} & = & 2\cdot 10\cdot (-2),\: \textrm{or}\\
\frac{dx}{dt} & = & \frac{-5}{2}.
\end{eqnarray*}

\end{document}
