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%TCIDATA{<META NAME="Title" CONTENT="What a Difference a Differential Makes">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Fri Feb 14 14:44:52 1997}
%TCIDATA{CSTFile=Math.cst}
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%TCIDATA{<LINK REL="contents" HREF="activity.tex">}
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\input{tcilatex}
\begin{document}


\section{What a Difference a Differential Makes}

\begin{quotation}
The approximation 
\[
f(x)\approx f(a)+f^{\prime }(a)(x-a) 
\]
is called the \textsl{linear approximation} 
\index{Linear approximation}or \textsl{tangent line approximation} 
\index{Tangent line approximation}of $f$ at $a$, and the function 
\[
L(x)=f(a)+f^{\prime }(a)(x-a) 
\]
is called the \textsl{linearization} \label{Linearization} 
\index{Linearization of a function}of $f$ at $a$. The line $y=L(x)$ is the
best approximation of the function $f$ by a line at the point $x=a$, $y=f(a)$%
.

If $y=f(x)$, the differential $dy$ depends on the two variables $x$ and $dx.$
Under most circumstances, \textsl{Scientific Notebook} treats the
expressions $dy$ and $dx$ as the product of $d$ and $y$ or $x$. To use the
differentials as variables, define $%
\limfunc{dx}$ and $\limfunc{dy}$ as \textsl{math names} 
\index{Math name}as follows:
\end{quotation}

\begin{itemize}
\begin{itemize}
\begin{itemize}
\item  
\begin{quotation}
Choose \textsf{Insert + Math Name;} in the \textsf{Name} box, type $%
\limfunc{dx}$; choose \textsf{OK}.

\item  Choose \textsf{Insert + Math Name;} in the \textsf{Name} box, type $%
\limfunc{dy}$; choose \textsf{OK}.
\end{quotation}
\end{itemize}
\end{itemize}
\end{itemize}

\begin{quotation}
(The $\limfunc{dx}$ and $\limfunc{dy}$ should appear in gray on the screen.)
Now \textsf{Define + New Definition} applied to $\limfunc{dy}=f^{\prime }(x)%
\limfunc{dx}$ defines $\limfunc{dy}$ as a differential, and if $f(x)=\sqrt{x}
$, then \textsf{Evaluate} produces 
\[
\left[ f^{\prime }(x)\limfunc{dx}\right] _{x=1.0,\limfunc{dx}%
=0.1}=\allowbreak .0\,5 
\]
\end{quotation}

\hspace{-1in}Make ``math names'' out of $\func{dx}$ and $\func{dy}$ as
described above. Define $f(x)=\sqrt{x}$ and define $\func{dy}=f^{\prime }(x)%
\func{dx}$. Calculate $\func{dy}$ at $x=.2$, $\func{dx}=.1$ by applying 
\textsf{Evaluate} to 
\index{Evaluation} 
\[
\left[ 
\func{dy}\right] _{x=.2,\func{dx}=.1} 
\]
Calculate the difference in the function values $f(.3)$ and $f(.2)$.
Visualize the differentials $\func{dx}$ and $\func{dy}$ by plotting a graph
of $y=\sqrt{x}$ and dragging the expressions 
\[
\left( .2,\sqrt{.2},.3,\sqrt{{.2}},.3,\sqrt{.2}+.1118\right) \text{\quad
and\quad }f(.5)+f^{\prime }(.5)(x-.5) 
\]
to the frame. Explain the connection between the plot and the equation 
\[
\func{dy}=f^{\prime }(x)\func{dx} 
\]

\subsection{Evelyn's Solution}

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Define $f(x)=\sqrt{x}$. Evaluating the differential $\func{dy}=f^{\prime }(x)%
\func{dx}$ at $x=.2$, $\func{dx}=.1$, we get 
\[
\left[ \func{dy}\right] _{x=.2,\func{dx}=.1}=.\,1118 
\]

\CustomNote[\FRAME{itbpF}{0.4367in}{0.4575in}{0in}{}{}{snbhint.wmf}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "PICT";valid_file "F";width 0.4367in;height 0.4575in;depth
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"1";cropright "1";cropbottom "0";filename 'Snbhint.wmf';file-properties
"XNPEU";}}]{Margin Hint}{{\small With the insertion point in the equation }$%
f(x)=\sqrt{x}${\small , choose Define + New Definition. Choose Insert + Math
Name (type }$dy${\small , and choose OK). Type }$=f^{\prime }(x)$. {\small %
Choose Insert + Math Name (type }$dx${\small , and choose OK). With the
insertion point in the equation }$\func{dy}=f^{\prime }(x)\func{dx}${\small %
, choose Define + New Definition. Choose Insert + Math Name (type }$dy$%
{\small , and choose OK). Select }$\func{dy}${\small \ with the mouse,
choose Insert + Brackets, and select Square Brackets. Choose Insert +
Subscript and type }$x=.2,${\small \ choose Insert + Math Name (type }$dx$%
{\small , and choose OK). }$\ ${\small Type }$=1$. {\small With the
insertion point in the expression }$\left[ \func{dy}\right] _{x=.5,\func{dx}%
=.1}${\small , choose Evaluate.}}

Notice that this is roughly the same as the difference in function values $%
f(.3)-f(.2)=\allowbreak .\,10051$.

In the following figures, the graph of $y=\sqrt{x}$ is drawn with a thick
pen, and the tangent line and both the horizontal and vertical line segments
are drawn with a thin pen. The differential $\func{dx}$ corresponds to the
length of the horizontal leg of the right triangle, and the differential $%
\func{dy}$ corresponds to the length of the vertical leg of the right
triangle.

We know that $f^{\prime }(.2)$ is the slope of the tangent line at $x=.2$,
and indeed the slope of the tangent line is the length $\func{dy}$ of the
opposite side divided by the length $\func{dx}$ of the adjacent side of the
right triangle. Notice that the vertical leg of the ``triangular'' region
bounded above by the graph of $y=\sqrt{x}$ has length $f(.3)-f(.2)=0.0051$,
which is slightly shorter than the length of $\func{dy},$ which is $%
f^{\prime }(.2)\left( .1\right) =0.\,1118$.

\begin{center}
\FRAME{itbpFU}{2.2502in}{1.6181in}{0in}{\Qcb{$f(x)=\protect\sqrt{x}$, $\func{%
dy}=f^{\prime }(x)\func{dx}$ }}{}{}{\special{language "Scientific Word";type
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\TEXUX{$\sqrt{x}$};linecolor "navy";linestyle 1;linethickness 2;pointstyle
"point";function \TEXUX{$\left(
.\,2,.\,44721,.\,3,.\,44721,.\,3,.\,55901\right) $};linecolor
"red";linestyle 1;linethickness 1;pointstyle "circle";function
\TEXUX{$\allowbreak .\,22361+1.\,118x$};linecolor "magenta";linestyle
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"0.1772";xviewmax "0.3229";yviewmin "0.4211";yviewmax
"0.5846";rangeset"X";phi 45;theta 45;plottype 4;constrained TRUE;numpoints
49;axesstyle "normal";xis \TEXUX{x};var1name \TEXUX{$x$};}}
\end{center}

\CustomNote[\FRAME{itbpF}{0.4367in}{0.4575in}{0in}{}{}{snbhint.wmf}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "PICT";valid_file "F";width 0.4367in;height 0.4575in;depth
0in;original-width 29.375pt;original-height 30.875pt;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename 'Snbhint.wmf';file-properties
"XNPEU";}}]{Margin Hint}{{\small With the insertion point in the expression }%
$f(.3)-f(.2)${\small , choose Evaluate. With the insertion point in the
expression }$\sqrt{x}${\small , choose Plot 2D + Rectangular. With the
insertion point in the expression }$\left(
.2,f(.2),.3,f(.2),.3,f(.2)+.1118\right) ${\small , choose Evaluate. Select
and drag to the frame }$\left(
.\,2,.\,44721,.\,3,.\,44721,.\,3,.\,55901\right) ${\small . With the
insertion point in the expression }$f(.2)+f^{\prime }(.2)(x-.2)${\small ,
choose Evaluate. Select and drag to the frame }$.\,22361+1.\,118x${\small .
For the two views shown, change the Domain Interval to }$0<x<.6${\small \
and }$.18<x<.32${\small , respectively, and check Equal Scaling Along Each
Axis.}}

\subsection{Activities}

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"1";cropright "1";cropbottom "0";filename 'Activity.wmf';file-properties
"XNPEU";}}\hspace{0.36in}

\begin{enumerate}
\item  Define $f(x)=x^{3}$ and define $\limfunc{dy}=f^{\prime }(x)\limfunc{dx%
}$. Calculate $\limfunc{dy}$ at $x=1$, $\limfunc{dx}=.1$ by applying \textsf{%
Evaluate} to 
\[
\left[ \limfunc{dy}\right] _{x=1,\limfunc{dx}=.1}
\]
Visualize the differentials $\limfunc{dx}$ and $\limfunc{dy}$ by plotting a
graph of $y=x^{3}$ and dragging the expressions $\left[
1,1,1.1,1,1.1,1.\,3\right] $ and $f(1)+f^{\prime }(1)(x-1)$ to the frame.
Name the horizontal and vertical legs of the right triangle and explain the
connection between the plot and the equation $\limfunc{dy}=f^{\prime }(x)%
\limfunc{dx}$.

\item  Define $f(x)=\sqrt{x}$. Define $P(x)=A\left( x-1\right) ^{2}+B\left(
x-1\right) +C$ and solve the system 
\begin{eqnarray*}
P(1) &=&f(1) \\
P^{\prime }(1) &=&f^{\prime }(1) \\
P^{\prime \prime }(1) &=&f^{\prime \prime }(1)
\end{eqnarray*}
for $A$, $B$, and $C$. Visualize the quadratic approximation 
\index{Quadratic approximation} to $f$ at $1$ by plotting the graph of $f$
together with the graph of $P$. Zoom in slightly so that the viewing
rectangle is restricted to $-1\leq x\leq 5$. Over what approximate interval
does the quadratic approximation provide a reasonably good fit?

\item  Define $f(x)$ to be a generic function and define 
\[
P(x)=A\left( x-a\right) ^{2}+B\left( x-a\right) +C
\]
If $P(x)$ is the quadratic approximation to $f$ at $a$, find a formula for $P
$ in terms of $f(a)$, $f^{\prime }(a)$, and $f^{\prime \prime }(a)$ by
solving the system 
\begin{eqnarray*}
P(a) &=&f(a) \\
P^{\prime }(a) &=&f^{\prime }(a) \\
P^{\prime \prime }(a) &=&f^{\prime \prime }(a)
\end{eqnarray*}
for $A$, $B$, and $C$.

\item  A section of highway that crosses a gully has the shape of a parabola 
$f(x)=ax^{2}+bx+c$, where $x$ measures horizontal distance from a reference
point and $y$ measures elevation relative to the reference point. The
reference point is chosen as the low point as the highway crosses the gully.
The radius of curvature at the reference point is 2000 feet. This means that
a circle of radius 2000 feet resting on the reference point would have the
same second derivative as the parabola at the reference point. Determine $a$%
, $b$, and $c$, and plot the parabolic highway together with the graph of a
circle of radius 2000 feet that rests on the reference point.%
\FRAME{dtbpF}{3.5362in}{2.3661in}{0pt}{}{}{paracirc.wmf}{\special{language
"Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display
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0pt;original-width 216.8125pt;original-height 144.5625pt;cropleft
"0";croptop "1";cropright "1";cropbottom "0";filename
'Paracirc.wmf';file-properties "XNPEU";}}\emph{Hint}\quad Choose \textsf{%
Equal Scaling Along Each Axis} to make the circle look like a circle.
\end{enumerate}

\end{document}
