%% 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}
 \theoremstyle{definition}
  \newtheorem{xca}[section]{Exercise}%%Delete [section] for sequential numbering

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
\setcounter{section}{2}
\setcounter{subsection}{2}

\makeatother

\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


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

\maketitle

\subsection{Higher Derivatives}

The \textit{second derivative} of a function \( f \), written \( f'' \), is
defined to be \[
f''(x):=\left( f'(x)\right) '.\]
 It's just the derivative of the derivative of \( f \). 

It is also written in a number of different ways, if \( f(x)=y \): \[
f''(x)=\frac{d^{2}f}{dx^{2}}=\frac{d^{2}y}{dx^{2}}=D^{2}(f)=D^{2}(y)=f^{(2)}=f_{xx}.\]


\begin{example}
Find the following derivative:\[
\displaystyle \left( \sin (x)\right) ''.\]

\end{example}

\subsubsection*{Solution}

\begin{xca}
Find the second derivative\[
\displaystyle \left( \sqrt{x^{2}+2x-1}\right) ''\]

\end{xca}
%

\begin{xca}
Find the indicated third derivative:\[
\left( x^{2}+4x+1\right) '''\]
 The \textit{third} derivative of \( f(x) \), \( f'''(x) \), is defined by
\[
f'''(x)\, :=\left( f''(x)\right) '.\]

\end{xca}
I'll leave to you to guess how to denote third derivatives. As to what a third
derivative \textit{means}, it's actually not that clear. But we haven't yet
discussed what the \emph{second} derivative means.


\subsubsection{Yeah, so what does the second derivative \emph{mean}? }

You can easily see what the first derivative of a function \( f \) means, in
terms of the slope of the curve, or the speed of motion. The second derivative
\( f'' \)  is a bit more subtle. It measures the slope \textit{of} \( f' \),
and since the slope of \( f' \) determines how \( f' \) is changing (increasing,
decreasing, etc.), then \( f'' \) tells you how the slope is \textit{changing}.
You can see that in the graph; in a region where \( f''(x)>0 \), the curve
will be ``curving'' up (or \textit{concave up}). 

\vspace{0.3cm}
{\par\centering \includegraphics{concave-up.ps} \par}
\vspace{0.3cm}

On the other hand, in a region where \( f''(x)<0 \), the curve will be curving
downward (or \textit{concave down}). 

\vspace{0.3cm}
{\par\centering \includegraphics{concave-down.ps} \par}
\vspace{0.3cm}

I want to emphasize that it is possible for \( f(x) \) to be decreasing \( f'(x)<0 \)
and concave up \( f''(x)>0 \), as well as any other combination of the two
phenomena.  

\vspace{0.3cm}
{\par\centering \includegraphics{concavity.ps} \par}
\vspace{0.3cm}

A common way that we use \( f''(x) \) casually is given by the line ``The rate
of increase of unemployment is slowing''. This does not say that more people
are employed, but that the number of unemployed is growing more slowly than
before. 

But the real meaning of the second derivative \( f''(x) \) relates more to
motion. If the position of a particle at time \( t \) is given by \( s(t) \),
then we saw that its velocity was \( v=s'(t) \). The second derivative of the
position, the rate of change of the velocity, is the \textit{acceleration} \[
a(t)=v'(t)=s''(t).\]
 The standard examples of a ball being thrown up in the air, with a position
of, say, \( s(t)=-16t^{2}+40t+5 \), reflect the standard elementary physics
idea that, without air, all falling objects (near the surface of the Earth)
will have a constant acceleration of \( 32\, \, ft/(sec)^{2} \), straight down.
For this example:\begin{eqnarray*}
s(t) & = & -16t^{2}+40t+5\\
v(t)=s'(t) & = & -32t+40,\, \mathrm{and}\\
a(t)=v'(t)=s''(t) & = & -32,
\end{eqnarray*}
in feet per second (the negative value indicates downward acceleration).

\begin{example}
Egbert tosses a baseball straight up, and measures its position at each time
\( t \). He finds that the position \( s(t) \) (in feet above his hand) at
time \( t \) seconds \( t \) is \[
s(t)=-16t^{2}+24t+7.\]
Find the position and velocity at time \( t=0 \), when he let go of the ball. 
\end{example}

\subsubsection*{Solution}

{[}Idea for an interactive thing. The user chooses a number \( a \) which gives
the initial velocity, and the animation then runs through the position at time
\( t \) according to \( s=-16t^{2}+v_{0}t \) with appropriate scale.{]} Text:
Choose a value for the initial velocity \( v_{0} \). Pressing the {}``toss{}''
button will show what happens when you toss a ball upward from its rest position
with that initial velocity (velocity at \( t=0 \)). \( v_{0} \) should be
between 1 and 40, so that you don't break anything. 
\end{document}
