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\title[Linear functions]
{Linear functions}

\institute[College of the Holy Cross]
{
   MATH 133 -- Calculus with Fundamentals, section 1, Prof. Little\\
}

%\date[September 7, 2015]
%{September 7, 2015}

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

\begin{frame}
{\bf Simple but extremely useful}

\begin{itemize}
\item<1-> A \emph{linear} function $f$ is one given by an equation of the form
$f(x) = m x + b$, where $m,b$ are constants.
\item<2-> If $m = 0$, then $f(x) = b$ is a \emph{constant function}.  So these
are special cases.
\item<3-> The graph $y = mx + b$ is a (non-vertical)
\emph{straight line}, and this is the \emph{slope-intercept form}
of the equation of a line.  
\item<4->  The constant $b = f(0)$, so the line contains $y$-axis \emph{intercept} point $(0,b)$.
\item<5-> The constant $m$ is called the \emph{slope} (``rise over run'').
\item<5->  Meaning of $m$: Let $(x_1,y_1) = (x_1,m x_1 + b)$ and 
$(x_2,y_2) = (x_2, m x_2 + b)$ be \emph{any two distinct
points} on the graph $y = m x + b$ (so $x_1 \ne x_2$).
\end{itemize}
\end{frame}

\begin{frame}
{\bf The meaning of the slope}

\begin{itemize}
\item<2-> Continuing from the last slide, the ``rise over run'' is
\begin{eqnarray*}
\frac{y_2 - y_1}{x_2 - x_1} &=& \frac{(mx_1 + b) - (mx_2 + b)}{x_2 - x_1}\\
&=& \frac{mx_1 - mx_2}{x_2 - x_1}\\
&=& \frac{m(x_2 - x_1)}{x_2 - x_1}\\
&=& m.
\end{eqnarray*}
\item<3-> So if we write $\Delta y = y_2 - y_1$ and $\Delta x = x_2 - x_1$
for the \emph{changes} in $y$ and $x$ from the first point to the second, then
\[
m = \frac{\Delta y}{\Delta x}.
\]
\item<4-> This gives the \emph{key property of linear functions}:  If we change
$x$ to $x + \Delta x$, $\Delta y$ is always \emph{the same multiple} $\Delta y = m \Delta x$ 
(it doesn't depend on what $x$ is).
\end{itemize}
\end{frame}

\begin{frame}
{\bf Geometric meaning}

This is really just another way of saying that a straight line in the plane is ``straight'' --
this diagram from p. 12 of our text ``says it all'' -- no matter which points $(x_1,y_1)$ and $(x_2,y_2)$
on the line we take, the value of $m$ is the same:

\begin{figure}
\includegraphics[trim = 200 100 0 0, scale=.3]{Line.pdf}
\end{figure}

\end{frame}

\begin{frame}
{\bf Example -- Telling whether a function is linear}

In a previous screencast, we looked at table of values for a function giving the measured period 
$T$ of a pendulum as a function of the length $L$:

\[
\begin{array}{ccccccccc}
      L (cm) &\vert& 20 &\vert& 30 &\vert& 40 &\vert& 50 \\ \hline
     T (sec) &\vert& 0.9 &\vert& 1.1 &\vert& 1.27 &\vert& 1.42 
\end{array}
\]

{\bf Question:} 
Is $T$ \emph{linear}, or perhaps \emph{approximately linear} as a function of $L$? 
\end{frame}

\begin{frame}
{\bf Example, continued}

Here $L$ values correspond to the $x$'s in the equation for a line (the $L$
is the ``independent variable'') and the $T$ values are the $y$'s ($T$ is a function of 
$L$).  Say the four entries from the table are $(L_1,T_1), \ldots, (L_4,T_4)$. 
With successive pairs:
\begin{eqnarray*}
\frac{T_2 - T_1}{L_2 - L_1} &=& \frac{1.1 - 0.9}{30 - 20} = .02\\
\frac{T_3 - T_2}{L_3 - L_2} &=& \frac{1.27 - 1.1}{40 - 30} = .017\\
\frac{T_4 - T_3}{L_4 - L_3} &=& \frac{1.42 - 1.27}{50 - 40} = .015.
\end{eqnarray*}
{\bf First Observation:}
Since the $\frac{\Delta T}{\Delta L}$ are not constant, the four data
points $(L_i,T_i)$ \emph{do not lie on any one line} -- the function is \emph{not (exactly) linear}. 
\end{frame}

\begin{frame}
{\bf Comments}

\begin{itemize}
\item<2-> But in fact the slope values are all \emph{small} and the differences are 
\emph{very small} -- indeed, the $\frac{\Delta T}{\Delta L}$ values all 
round to one decimal place as $m \doteq .2$ (approximately $.2$).   
So the data certainly \emph{is approximately linear}:
\item<3->  Here is a point plot of the $(L_i,T_i)$.  It's actually quite hard to tell visually that
these are \emph{not} on one line:
\begin{figure}
\includegraphics[scale=.3]{PointPlot.png}
\end{figure}
\end{itemize}
\end{frame}

\begin{frame}
{\bf Comments, continued}
\begin{itemize}
\item<1-> Nevertheless, $\frac{\Delta T}{\Delta L}$ seems to be steadily decreasing
as $L$ increases, so this is a good hint that $T$ is \emph{not a linear function of $L$}.  
(If the data was essentially linear, but contained experimental error or other
``noise'' that made the points noncollinear, the $\frac{\Delta T}{\Delta L}$ values 
would vary randomly on both sides of the actual slope.)
\item<2-> \emph{We should not draw any conclusions} without more data 
points or a theoretical analysis of the physics involved.
\item<3-> In a physics course, you see that $T$ proportional to 
$\sqrt{L}$:
$$T = 2\pi \sqrt{\frac{L}{g}},$$
where $g$ is the constant acceleration of gravity on the surface of the earth (i.e. 9.8 $m/sec^2$ or 
about 16 $ft/sec^2$).   So this function \emph{is actually not linear}.
\end{itemize}
\end{frame}

\end{document}

