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%TCIDATA{<META NAME="Title" CONTENT="The Plot Is Implicit">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Fri Feb 14 00:14:10 1997}
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\input{tcilatex}
\begin{document}


\section{The Plot Is Implicit 
\index{Implicit plots}}

\begin{quotation}
Something implied or understood though not directly expressed is \textsl{%
implicit}. An implicit function is determined by an equation that cannot
necessarily be solved in an elementary fashion. The technique of implicit
differentiation allows you to differentiate such functions without
expressing them directly.
\end{quotation}

\hspace{-1in}Make an implicit plot of the equation $2y^{2}+xy=x^{2}+2$ with
viewing rectangle $-5<x<5$ and $-5<y<5$. V\hspace{-0.02in}erify
algebraically that $(1,1)$ is a point on the graph of the equation. Solve
for $y$ in terms of $x$ and plot the curve that goes through the point $%
(1,1) $. Find an equation of the tangent line at $(1,1)$ by differentiating
to find the slope of the tangent line. Differentiate the equation implicitly
to find the slope of the tangent line at $(1,1)$, and compare the results of
the explicit and implicit methods. V\hspace{-0.02in}erify that the equation
of the tangent line is reasonable by graphing the equation together with its
tangent line.

\subsection{Carolyn's Solution}

\FRAME{itbpF}{0.6875in}{0.4013in}{0in}{}{}{solution.wmf}{\special{language
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"XNPEU";}}

The graph of a quadratic equation is a conic section, and an implicit plot
of the equation $2y^{2}+xy=x^{2}+2$ reveals that it is an hyperbola.

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Word";type "MAPLEPLOT";width 2.6039in;height 2.3229in;depth 0pt;display
"PICT";function \TEXUX{$2y^{2}+xy=x^{2}+2$};linecolor "blue";linestyle
1;linethickness 2;pointstyle "point";xmin "-5";xmax "5";ymin "-5";ymax
"5";xviewmin "-5";xviewmax "5";yviewmin "-5";yviewmax
"5";viewset"XY";rangeset"XY";phi 45;theta 45;plottype 12;num-x-gridlines
25;num-y-gridlines 25;axesstyle "normal";xis \TEXUX{x};yis
\TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};}}

\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 equation }$%
2y^{2}+xy=x^{2}+2${\small , choose Plot 2D + Implicit.}}

Solving the equation $2y^{2}+xy=x^{2}+2$ for $y$, we get the two solutions 
\[
\left\{ y=-\frac{1}{4}x+\frac{1}{4}\sqrt{\left( 9x^{2}+16\right) }\right\}
,\allowbreak \left\{ y=-\frac{1}{4}x-\frac{1}{4}\sqrt{\left(
9x^{2}+16\right) }\right\} 
\]
Because 
\[
\left[ -\frac{1}{4}x+\frac{1}{4}\sqrt{\left( 9x^{2}+16\right) }\right]
_{x=1}=\allowbreak -\frac{1}{4}+\frac{1}{4}\sqrt{25}=\allowbreak 1 
\]
it follows that $(1,1)$ is a point on the graph of $y=-\frac{1}{4}x+\frac{1}{%
4}\sqrt{\left( 9x^{2}+16\right) }$.

\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 equation }$%
2y^{2}+xy=x^{2}+2${\small , choose Solve + Exact (Variable(s) to Solve For: }%
$y${\small ).} {\small Select the expression }$-\frac{1}{4}+\frac{9}{4\sqrt{%
\left( 9x^{2}+16\right) }}x${\small \ with the mouse, choose Insert +
Brackets, select Square Brackets, and choose OK. Choose Insert + Subscript
and type }$x=1${\small \ in the input box. With the insertion point in the
expression }$\left[ -\frac{1}{4}+\frac{9}{4\sqrt{\left( 9x^{2}+16\right) }}%
x\right] _{x=1}${\small , choose Evaluate, and then choose Simplify.}}

The graph of this branch of the curve is given by\FRAME{dtbpF}{2.8824in}{%
1.7504in}{0pt}{}{}{}{\special{language "Scientific Word";type
"MAPLEPLOT";width 2.8824in;height 1.7504in;depth 0pt;display "PICT";function
\TEXUX{$-\frac{1}{4}x+\frac{1}{4}\sqrt{\left( 9x^{2}+16\right) }$};linecolor
"magenta";linestyle 1;linethickness 1;pointstyle "point";function
\TEXUX{$(1,1)$};linecolor "black";linestyle 1;linethickness 1;pointstyle
"circle";pointplot TRUE;xmin "-5";xmax "5";xviewmin "-5";xviewmax
"5";yviewmin "0";yviewmax "4";viewset"XY";rangeset"X";phi 45;theta
45;plottype 4;numpoints 49;axesstyle "normal";xis \TEXUX{x};var1name
\TEXUX{$x$};}}We add the point $(1,1)$ to the plot to verify visually that
this point is actually on the curve.

\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 }%
$-\frac{1}{4}x+\frac{1}{4}\sqrt{\left( 9x^{2}+16\right) }${\small , choose
Plot 2D + Rectangular. Select the expression }$(1,1)${\small \ and drag it
to the plot.}}

\noindent The derivative at $x=1$ is given by 
\begin{eqnarray*}
\left[ \frac{d}{dx}\left( -\frac{1}{4}x+\frac{1}{4}\sqrt{\left(
9x^{2}+16\right) }\right) \right] _{x=1} &=&\allowbreak \left[ -\frac{1}{4}+%
\frac{9}{4\sqrt{\left( 9x^{2}+16\right) }}x\right] _{x=1} \\
&=&\frac{1}{5}
\end{eqnarray*}
and hence the slope of the graph at $\left( 1,1\right) $ is $\frac{1}{5}$
and the tangent line is given by $y-1=\frac{1}{5}(x-1)$, which simplifies to 
$y=\frac{4}{5}+\frac{1}{5}x$.\medskip

\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 To show one of the steps in evaluating the
derivative, use Edit + Copy and Paste to make a copy of }$\left[ \frac{d}{dx}%
\left( -\frac{1}{4}x+\frac{1}{4}\sqrt{\left( 9x^{2}+16\right) }\right)
\right] _{x=1}${\small \ and type }$=${\small \ between the two expressions.
For an ``in-place replacement,'' select the copy of }$\frac{d}{dx}\left( -%
\frac{1}{4}x+\frac{1}{4}\sqrt{\left( 9x^{2}+16\right) }\right) ${\small \
with the mouse, and while pressing the }\textsc{ctrl}{\small \ key, choose
Evaluate. With the insertion point in the expression }$\left[ -\frac{1}{4}+%
\frac{9}{4\sqrt{\left( 9x^{2}+16\right) }}x\right] _{x=1}${\small , choose
Evaluate. To simplify the equation for the tangent line, with the insertion
point in the equation }$y-1=\frac{1}{5}(x-1)${\small , choose Solve + Exact
(Variable(s) to Solve For: }$y${\small ).}}

Alternatively, implicit differentiation applied to the equation $%
2y^{2}+xy=x^{2}+2$ yields $4yy^{\prime }+y+xy^{\prime }=2x$, and solving for 
$y^{\prime }$, we get 
\[
y^{\prime }=-\frac{y-2x}{4y+x} 
\]

\strut

\CustomNote[\FRAME{itbpF}{0.4367in}{0.4575in}{0in}{}{}{snbhint.wmf}{\special%
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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 }$%
2y^{2}+xy=x^{2}+2${\small , choose Calculus + Implicit Differentiation
(Differentiation Variable: }$x${\small ). With the insertion point in the
equation }$4yy^{\prime }+y+xy^{\prime }=2x${\small , choose Solve + Exact
(Variable(s) to Solve For: }$y^{\prime }${\small )}$.$}

Because 
\[
\left[ -\frac{y-2x}{4y+x}\right] _{x=1,y=1}=\allowbreak \frac{1}{5} 
\]
it follows that an equation of the tangent line is $y-1=\frac{1}{5}(x-1)$,
which can be rewritten as $x-5y=-4$.

\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 Select the expression }$-\frac{y-2x}{4y+x}$%
{\small \ with the mouse and choose Insert + Brackets (select Square
Brackets). Choose Insert + Subscript and type }$x=1,y=1${\small \ in the
input box. With the insertion point in the expression }$\left[ -\frac{y-2x}{%
4y+x}\right] _{x=1,y=1}${\small , choose Evaluate.\medskip }}

The implicit and explicit techniques lead to equivalent answers, but the two
forms for the derivative have quite a different appearance: 
\begin{eqnarray*}
y^{\prime } &=&-\frac{1}{4}+\frac{9}{4\sqrt{\left( 9x^{2}+16\right) }}x \\
y^{\prime } &=&-\frac{y-2x}{4y+x}
\end{eqnarray*}
The implicit form appears simpler, but one must remember that $x$ and $y$
remain related by the original equation.

The following is a graph of the equation $2y^{2}+xy=x^{2}+2$ together with
the line $x-5y=-4$.\FRAME{dtbpFU}{3.3909in}{2.0003in}{0pt}{\Qcb{$%
2y^{2}+xy=x^{2}+2$, $x-5y=-4$}}{}{}{\special{language "Scientific Word";type
"MAPLEPLOT";width 3.3909in;height 2.0003in;depth 0pt;display "PICT";function
\TEXUX{$2y^{2}+xy=x^{2}+2$};linecolor "blue";linestyle 1;linethickness
2;pointstyle "point";function \TEXUX{$x-5y=-4$};linecolor "red";linestyle
1;linethickness 1;pointstyle "point";xmin "-5";xmax "5";ymin
"-2.021977";ymax "2.978023";xviewmin "-5.2";xviewmax "5.204";yviewmin
"-2.021977";yviewmax "2.978023";viewset"XY";rangeset"X";phi 45;theta
45;plottype 12;num-x-gridlines 25;num-y-gridlines 25;axesstyle "normal";xis
\TEXUX{x};yis \TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};}}

{\small 
\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 equation $%
2y^{2}+xy=x^{2}+2$, choose Plot 2D + Implicit. Select the equation $x-5y=4$\
and drag it to the frame.}}}

From the appearance of this graph, the equation we used for the tangent line
is reasonable.

\subsection{Activities}

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"XNPEU";}}

\begin{enumerate}
\item  Consider the equation $x^{3}+y^{3}=6xy$. V\hspace{-0.02in}erify that $%
(3,3)$ is a point on the graph of the equation. Find an equation of the
tangent line at $(3,3)$ by solving for $y$ in terms of $x$ and
differentiating. Differentiate the equation implicitly to find the slope of
the tangent line at $(1,1).$ V\hspace{-0.02in}erify that the equation of the
tangent line is reasonable by plotting the equation with its tangent line.
Discuss the pros and cons of using implicit differentiation versus solving
for $y$ in terms of $x$ and differentiating explicitly.

\item  Apply \textsf{Plot 2D + Implicit} to get a graph of $%
x^{2}y^{2}=(y+1)^{2}(4-y^{2})$. V\hspace{-0.02in}erify that $(0,-1)$ is a
point on the graph$.$ Find two lines that are ``tangent'' to the graph at $%
(0,-1)$. Plot the curve together with these two tangent lines.

\begin{quotation}
\QTR{NoteLeadin}{Hint}\quad Solve for $x$ in terms of $y$ and assume that $%
\dfrac{dy}{dx}=\dfrac{1}{dx/dy}$.
\end{quotation}

\item  Apply \textsf{Plot 2D + Implicit} to get a graph of $xy=1$. Select
and drag $x^{2}-y^{2}=4$ to the frame. On the \textsf{Axes} page of the 
\textsf{Plot Properties} tabbed dialog, check \textsf{Equal Scaling Along
Each Axis}. Find the point in the first quadrant where the two curves
intersect, and attach tangent lines to each curve at that point. Do the
tangent lines appear to be orthogonal? Explain why they are perpendicular or
show that they are not.
\end{enumerate}

\label{Higher derivative}

\end{document}
