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\begin{document}
%MODULE 1: Epsilon-Delta limits


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%INPUT THE INSTRUCTIONS PAGE 



\title{An intuitive notion of derivatives}


\subtitle{In this lesson you begin to study a topic called the {\hilite \emph{derivative}}.
In Calculus, the derivative is a tool used to describe the \emph{rate
of change} of a function. }

\makebeamertitle

\lyxframeend{}\lyxframe{{A Sample problem}}

An electronic instrument makes a sound by vibrating a speaker's voice
coil according to the function $\ec s(x)=v\sin(x)$, where
\begin{itemize}
\item $\ec v$ is a constant; 
\item $\ec x$ is the time in seconds; and 
\item $\ec s(x)$ is the distance in mm that the coil has moved from equilibrium.
\end{itemize}
\noindent \begin{question}What is the rate of change of the position
of the coil at time $\ec x=1$~second when $\ec v=1$ and when $\ec v=4$?\end{question}


\lyxframeend{}\lyxframe{{SAGE worksheets}}

In this lab you could use a blank SAGE worksheet, although you can
get by with just a calculator, too.


\lyxframeend{}\lyxframe{\noindent {Two kinds of rates of change}}
\begin{overlayarea}
{\textwidth}{6in}

\noindent We said that the derivative is the rate of change. There
are actually two different rates of change that you need to know,
and only one of them is the derivative.


\pause{}
\begin{onlyenv}%{}
<2-3>
\begin{definition}%{}
[(Average Rate of Change)]The \emph{average} rate of change between
\emph{two points} $\ec x=a$ and $\ec x=b$ is the slope of the line
connecting the points $\ec(a,f(a))$ and $\ec(b,f(b))$.
\end{definition}%{}

\pause{}
\begin{onlyenv}%{}
<3>

In the case of the speaker ($\ec s(x)=\sin(x)$), the average rate
of change between $\ec x=0$ and $\ec x=1$ is 
\[
\ec\frac{\sin(1)-\sin(0)}{1-0}\approx0.84.
\]



\vspace{-6ex}
\begin{figure}[h!]
\centering{}\includegraphics[scale=0.2]{figures/avg_chg_sinx_0_1} 
\end{figure}


\end{onlyenv}%{}
\end{onlyenv}%{}

\pause{}
\begin{definition}%{}
The \emph{instantaneous} rate of change at \emph{one point} $\ec x=b$
is the slope of the line \emph{tangent to} the curve at $\ec x=b$.
\end{definition}%{}

\pause{}


In the case of the speaker ($\ec s(x)=\sin(x)$), the instantaneous
rate of change at $\ec x=1$ is $\ec m\approx0.54$.


\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.2]{figures/inst_chg_sinx_1} 
\end{figure}



\noindent (If you have no idea how I computed 0.54, don't worry. You're
not supposed to know that yet.)

\end{overlayarea}

\lyxframeend{}\lyxframe{{Tangent = Limit of Secants}}
\begin{overlayarea}
{\textwidth}{6in}

It might not be obvious at first, but as $\ec a$ approaches $\ec b$,
the secant lines between $\ec x=a$ and $\ec x=b$ usually approach
the tangent line at $\ec x=b$. The next few slides illustrate this.


\pause{}
\begin{onlyenv}%{}
<2>

The secant line between $\ec x=0$ and $\ec x=1$: 
\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.5]{figures/avg_chg_sinx_0_1_tanline} 
\end{figure}


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<3>

The secant line between $\ec x=0.5$ and $\ec x=1$: 
\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.5]{figures/avg_chg_sinx_0_5_1} 
\end{figure}


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<4>

The secant line between $\ec x=0.75$ and $\ec x=1$: 
\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.5]{figures/avg_chg_sinx_0_75_1} 
\end{figure}


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<5>

The secant line between $\ec x=0.9$ and $\ec x=1$: 
\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.5]{figures/avg_chg_sinx_0_9_1} 
\end{figure}


\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<6>

The tangent line at $\ec x=1$: 
\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.5]{figures/inst_chg_sinx_1} 
\end{figure}


\end{onlyenv}%{}
\end{overlayarea}

\lyxframeend{}\lyxframe{{Okay, so which one is the derivative?}}
\begin{definition}%{}
[(The derivative at a point)]The \emph{\hilite derivative} of $\ec f$
at any point $\ec x=b$ is the \emph{instantaneous} rate of change
of $\ec f$.
\end{definition}%{}
As we pointed out earlier, this is the slope of the tangent line.
Thus the derivative is the \emph{slope} of the line tangent to $\ec f$
at $\ec x=b$.


\pause{}

So the derivative is the \emph{limit} of the slopes of the secant
lines between $\ec x=a$ and $\ec x=b$ as $\ec x=a$ approaches $\ec x=b$.

\begin{figure}[h!]
 

\centering{}\includegraphics[scale=0.2]{figures/secs_approach_tan} 
\end{figure}



\lyxframeend{}\lyxframe{{Definition of the derivative}}

We write this formally as
\begin{definition}%{}
[(The derivative at a point)]
\[
\ec f'(b)=\lim_{x\rightarrow b}\frac{f(x)-f(b)}{x-b}
\]
where $\ec f'(b)$ is the symbol used for the derivative of $\ec f$
at $\ec x=b$.
\end{definition}%{}
(This assumes that the above limit exists.)


\pause{}
\begin{definition}%{}
 If the derivative of $\ec f$ exists and is finite at $\ec x=b$,
we say that $\ec f$ is \emph{\hilite differentiable at} $\ec x=b$.
If the derivative does not exist, then $\ec f$ is not differentiable
at $\ec x=b$. 
\end{definition}%{}

\lyxframeend{}\lyxframe{\noindent {Differentiability}}

\noindent An important point to remember is that
\begin{theorem}%{}
\noindent  If a function $\ec f$ is differentiable at $\ec x=b$,
then $\ec f$ is continuous at $\ec x=b$. 
\end{theorem}%{}

\pause{}

\noindent Why? The definition of continuity requires that $\ec f(b)$
exists, and $\ec\lim_{x\rightarrow b}f(x)=f(b)$. 
\begin{itemize}
\item If $\ec f(b)$ does not exist, then we cannot compute the derivative
, whose definition requires $\ec f(b)$. (\emph{Go back and check
where!}) 
\item If $\ec\lim_{x\rightarrow b}f(x)\neq f(b)$, then the difference quotient
\[
\ec\frac{f(x)-f(b)}{x-b}
\]
 approaches a non-zero constant in the numerator, and zero in the
denominator. In a previous module, we stated that this was \emph{not
a finite quantity}. Differentiability requires a \emph{finite} limit
for the difference quotient. (\emph{Go back and check where!}) 
\end{itemize}

\lyxframeend{}\lyxframe{{Numerical approach}}
\begin{overlayarea}
{\textwidth}{6in}

We turn our attention to the question of computing derivatives. Because
it can be hard to estimate slopes from looking at a graph, it is often
easy to estimate the derivative using the \emph{numerical approach}
to computing a limit.


\pause{}
\begin{onlyenv}%{}
<2>

Build a table of $a$-$m$ values, where $\ec m$ is the slope of
the secant line of $\ec s(x)=\sin(x)$ between $\ec x=a$ and $\ec x=1$,
where $\ec a=0,0.9,0.99,0.9999,1.0001,1.01,1.1,2$.


{\ec 


\begin{center}
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
$a$  & 0  & 0.9  & 0.99  & 0.9999  & {\hilite 1}  & 1.0001  & 1.01  & 1.1  & 2 \tabularnewline
\hline 
$m$  & ?  & ?  & ?  & ?  & ???  & ?  & ?  & ?  & ? \tabularnewline
\end{tabular}
\par\end{center}


}

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<3>

Build a table of $a$-$m$ values, where $m$ is the slope of the
secant line of $s(x)=\sin(x)$ between $x=a$ and $x=1$, where $a=0,0.9,0.99,0.9999$.


{\ec 


\begin{center}
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
$a$  & 0  & 0.9  & 0.99  & 0.9999  & {\hilite 1}  & 1.0001  & 1.01  & 1.1  & 2 \tabularnewline
\hline 
$m$  & ?  & ?  & ?  & ?  & ???  & ?  & ?  & ?  & ? \tabularnewline
\end{tabular}
\par\end{center}


}


\noindent Remember that you can make SAGE do this computation using
a \texttt{\ttcolor for} loop if you like.

\end{onlyenv}%{}

\pause{}
\begin{onlyenv}%{}
<4>

Build a table of $a$-$m$ values, where $m$ is the slope of the
secant line of $s(x)=\sin(x)$ between $x=a$ and $x=1$, where $a=0,0.9,0.99,0.9999$.


{\ec 


\begin{center}
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
$a$  & 0  & 0.9  & 0.99  & 0.9999  & {\hilite 1}  & 1.0001  & 1.01  & 1.1  & 2 \tabularnewline
\hline 
$m$  & 0.84  & 0.58  & 0.54  & 0.54  & ???  & 0.54  & 0.54  & 0.5  & 0.07 \tabularnewline
\end{tabular}
\par\end{center}


}


\noindent Notice that the slopes do indeed approach 0.54, which is
what we had predicted before.

\end{onlyenv}%{}
\end{overlayarea}

\lyxframeend{}\lyxframe{{Example Problem}}

Now that we've touched on the concepts, let's look at the example
problem again. It was:

\noindent An electronic instrument makes a sound by vibrating a speaker's
voice coil according to the function $\ec s(x)=v\sin(x)$, where
\begin{itemize}
\item $\ec v$ is a constant; 
\item $\ec x$ is the time in seconds; and 
\item $\ec s(x)$ is the distance in mm that the coil moves.
\end{itemize}
\begin{question}What is the rate of change of the position of the
coil at time $\ec x=1$~second when $\ec v=1$ and when $\ec v=4$?\end{question}


\pause{}

If you look back, you will see that we have in fact solved the problem
for $\ec v=1$. We found that the derivative was $\ec s'(1)\approx0.54$.


\lyxframeend{}\lyxframe{{Example Problem: Units}}

\begin{question}What are the \emph{units} of the derivative? mm?
sec? something else?\end{question}


\pause{}

\noindent The units are mm/sec. \emph{Why?}


\pause{}

Remember that the derivative is the limit of the slopes of the secant
lines. These slopes represent the ratio \emph{change in $\ec s$ to
change in $\ec x$}. The problem told us that 
\begin{itemize}
\item $\ec s$ represents distance in mm, and 
\item $\ec x$ represents time in secondes. 
\end{itemize}
Thus the units should be mm/sec.


\lyxframeend{}\lyxframe{{Example Problem: $v=4$}}

Now that we've touched on the concepts, let's look at the example
problem again. It was:

An electronic instrument makes a sound by vibrating a speaker's voice
coil according to the function $s(x)=v\sin(x)$, where
\begin{itemize}
\item $v$ is a constant; 
\item $x$ is the time in seconds; and 
\item $s(x)$ is the distance in mm that the coil moves.
\end{itemize}
\begin{question}What is the rate of change of the position of the
coil at time $x=1$~second when $v=1$ and when $v=4$?\end{question}

\noindent Since we solved the problem for $\ec v=1$, take a few moments
on your own to solve the problem for $\ec v=4$.


\pause{}

You should find that $\ec s'(4)\approx2.16$.


\lyxframeend{}\lyxframe{{Conclusion}}
\begin{itemize}
\item The \emph{\hilite derivative} of $\ec f(x)$ at $\ec x=b$ is the
\emph{slope of the line tangent to $\ec f(x)$} at $\ec x=b$. This
is the limit of the slopes of the secant lines. 
\item The derivative of $\ec f$ at $\ec x=b$ is formally defined to be\setlength{\shadeboxwidth}{4in}

\begin{definition}%{}
[(The derivative at a point)]
\[
\ec f'(b)=\lim_{x\rightarrow b}\frac{f(x)-f(b)}{x-b}
\]
where $\ec f'(b)$ is the symbol used for the derivative of $\ec f$
at $\ec x=b$.
\end{definition}%{}
\item If the derivative of a function exists at a point, we say that the
function is {\hilite differentiable} at that point. 
\item We can estimate this value graphically or numerically. 
\end{itemize}

\lyxframeend{}
\end{document}
