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%TCIDATA{<META NAME="Title" CONTENT="Multivariable Calculus">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Sat Mar 29 23:23:34 1997}
%TCIDATA{Language=American English}
%TCIDATA{CSTFile=MathHelp.cst}
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%TCIDATA{<LINK REL="end" HREF="DM7-9.tex">}
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\input{tcilatex}
\begin{document}


\section{Multivariable \label{Multivariable calculus}Calculus}

Multivariable calculus 
\index{Multivariable calculus}extends the ideas of calculus to functions of
several variables. The first of these ideas that we examine is optimization.

\subsection{\ Optimization\label{Optimization}\label{Extreme values}}

Optimization of functions of several variables requires special techniques.

\subsubsection{Extreme Value on a Surface}

To find the extreme values of a function such as $f(x,y)=x^{3}-3xy+y^{3}$,
it is sufficient to locate all pairs $(x,y)$ where the partial derivatives
are zero.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$%
\frac{\partial }{\partial x}f(x,y)=\allowbreak 3x^{2}-3y\medskip $

$\frac{\partial }{\partial y}f(x,y)=\allowbreak -3x+3y^{2}\medskip $
\end{quotation}

Enter the equations $3x^{2}-3y=0$ and $-3x+3y^{2}=0$ into a display or a $%
2\times 1$ matrix, then from the \textsf{Solve} submenu, choose \textsf{Exact%
}.

\begin{quote}
$\blacktriangleright $ \textsf{Solve + Exact}
\end{quote}

\begin{quotation}
\begin{eqnarray*}
3x^{2}-3y &=&0 \\
-3x+3y^{2} &=&0
\end{eqnarray*}
Solution is : $\left\{ x=0,y=0\right\} ,\allowbreak \left\{ x=1,y=1\right\}
,\allowbreak \left\{ x=-\rho -1,y=\rho \right\} $ where $\rho $ is a root of 
$Z^{2}+Z+1$
\end{quotation}

Note that $D_{xx}f(x,y)D_{yy}f(x,y)-\left( D_{xy}f(x,y)\right)
^{2}=\allowbreak 36xy-9;$ hence $(0,0)$ represents a saddle point because $%
-9<0$; and $(1,1)$ represents a local minimum because $36-9>0$, and $%
D_{xx}f(x,y)=\allowbreak 6x>0$. The roots of $Z^{2}+Z+1$ are $Z=-\frac{1}{2}+%
\frac{1}{2}i\sqrt{3}$ and $\allowbreak Z=-\frac{1}{2}-\frac{1}{2}i\sqrt{3}$
so there are no other real extreme values.

You can visualize the local minimum at $(1,1)$ by generating a plot of the
surface. To get the following plot, with the insertion point in the
expression $x^{3}-3xy+y^{3}$, click \FRAME{itbpF}{0.3009in}{0.3009in}{%
0.0701in}{}{}{3dplot.wmf}{\special{language "Scientific Word";type
"GRAPHIC";display "PICT";valid-file "F";width 0.3009in;height 0.3009in;depth
0.0701in;original-width 17.8125pt;original-height 16.9375pt;cropleft
"0";croptop "1";cropright "1";cropbottom "0";filename
'3dplot.wmf';file-properties "XNPEU";}}; in the \textsf{Plot Components }%
page of the \textsf{Plot Properties} tabbed dialog set the \textsf{Domain
Intervals} to $-1\leq x\leq 2$ and $-1\leq y\leq 2$; choose \textsf{Patch
and Contour}.

\begin{quote}
$\blacktriangleright $ \textsf{Plot 3D + Rectangular}
\end{quote}

\begin{quotation}
$x^{3}-3xy+y^{3}$
\end{quotation}

\FRAME{dtbpF}{3in}{2.0003in}{0pt}{}{}{}{\special{language "Scientific
Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth 0pt;display
"PICT";function \TEXUX{$x^{3}-3xy+y^{3}$};linecolor "black";linestyle
1;linethickness 1;pointstyle "point";xmin "-1";xmax "2";ymin "-1";ymax
"2";xviewmin "-1.06";xviewmax "2.061";yviewmin "-1.06";yviewmax
"2.061";zviewmin "-5.36";zviewmax "13.37";rangeset"XY";phi 45;theta
45;plottype 5;num-x-gridlines 25;num-y-gridlines 25;plotstyle
"patchcontour";axesstyle "frame";plotshading "Z";xis \TEXUX{x};yis
\TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};}}

The level curve $x^{3}-3xy+y^{3}=0$ goes through the point $(0,0,0).$ To get
a better view of this level curve, do the following.

\begin{quote}
$\blacktriangleright $ Plot 2D + Implicit
\end{quote}

\begin{quotation}
$x^{3}-3xy+y^{3}=0$
\end{quotation}

\FRAME{dtbpF}{3in}{2.0003in}{0pt}{}{}{}{\special{language "Scientific
Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth 0pt;display
"PICT";function \TEXUX{$x^{3}-3xy+y^{3}=0$};linecolor "red";linestyle
1;linethickness 2;pointstyle "point";function
\TEXUX{$x^{3}-3xy+y^{3}=-.5$};linecolor "black";linestyle 1;linethickness
1;pointstyle "point";xmin "-1";xmax "2";ymin "-1";ymax "2";xviewmin
"-1";xviewmax "2";yviewmin "-1";yviewmax "2";rangeset"XY";phi 45;theta
45;plottype 12;constrained TRUE;num-x-gridlines 25;num-y-gridlines
25;axesstyle "normal";xis \TEXUX{x};yis \TEXUX{y};var1name
\TEXUX{$x$};var2name \TEXUX{$y$};}}

To get a better idea of where the $z$-values are positive and where they are
negative, enter the equation $x^{3}-3xy+y^{3}=-.5$ and drag it to the frame.
Note that the $z$-values on the surface $z=x^{3}-3xy+y^{3}$ are negative
inside the loop in the first quadrant and in the lower left corner of the $%
xy $-plane.\medskip

\begin{example}
\begin{quotation}
A paper cup 
\index{Paper cup problem}\label{Paper cup problem}is to be made from a
circular piece of paper of radius 1 unit. One possible shape is a cone of
slant height 1. Another possibility is to make a circular base out of the
central part of the piece of paper, and then to crumple the remainder of the
paper to form a right circular cylinder. (In this case, the height plus
radius is $1$.) By using an angle less than $90^{\circ }$, you can also form
cups in the shape of a truncated cone. Let 
\[
V=%
\frac{\pi R^{2}H}{3}-\frac{\pi r^{2}h}{3}
\]
denote the volume of a truncated cone, where $R$ and $H$ represent the
radius and height of a cone and $r$ and $h$ represent the radius and height
of the tip of a cone that has been cut off. You obtain a cone by setting $%
r=0.$ Setting $R=r$ gives a right circular cone. Of all such possible
shapes, which will hold the most water? Let $x$ represent the height of the
truncated cone, so that $H=h+x$. Then 
\[
V=\allowbreak \frac{1}{3}\pi R^{2}\left( h+x\right) -\frac{1}{3}\pi r^{2}h
\]
Using similar triangles, $\frac{x}{R-r}=\frac{h}{r}$ and solving for $%
h\allowbreak $ yields $h=\frac{xr}{R-r}$and hence 
\begin{eqnarray*}
V &=&\allowbreak -\frac{1}{3}\pi \left( -x\frac{r}{R-r}R^{2}-xR^{2}+x\frac{%
r^{3}}{R-r}\right)  \\
&=&\allowbreak \frac{1}{3}\left( r^{2}+Rr+R^{2}\right) x\pi 
\end{eqnarray*}
Since the slant height of the truncated cone is $1-r$, it follows from the
Pythagorean theorem that $(1-r)^{2}=x^{2}+(R-r)^{2}$, and hence $x=\sqrt{%
(1-r)^{2}-(R-r)^{2}}$. Rewriting $V$ one more time, we have 
\[
V=\allowbreak \frac{1}{3}\left( r^{2}+Rr+R^{2}\right) \pi \sqrt{%
1-2r-R^{2}+2Rr}
\]
The partial derivatives are given by 
\begin{eqnarray*}
\frac{\partial V}{\partial r} &=&\allowbreak \frac{1}{3}\left( -1+R\right)
\left( Rr-R+5r^{2}-2r\right)  \\
&&\times \frac{\pi }{\sqrt{\left( \left( -1+R\right) \left( -R+2r-1\right)
\right) }} \\
\frac{\partial V}{\partial R} &=&\allowbreak \frac{1}{3}\left(
r^{3}+r-2r^{2}+3rR^{2}+2Rr^{2}-3R^{3}+2R-4Rr\right)  \\
&&\times \frac{\pi }{\sqrt{\left( \left( -1+R\right) \left( -R+2r-1\right)
\right) }}
\end{eqnarray*}

Setting the partials equal to zero and solving, we get

\begin{eqnarray*}
\frac{\partial V}{\partial r} &=&0 \\
\frac{\partial V}{\partial R} &=&0
\end{eqnarray*}
The solution is 
\[
\left\{ R=0,r=0\right\} ,\allowbreak \left\{ R=7\rho ^{2}-\frac{5}{3}\rho -%
\frac{1}{3},r=\rho \right\} 
\]
where $\rho $ is a root of $21Z^{3}-11Z^{2}-2Z+1$, which has three roots 
\begin{eqnarray*}
Z &=&-.3041369802 \\
Z &=&.2923064976 \\
Z &=&.5356400064
\end{eqnarray*}
The following table includes all of the possible extreme values. 
\[
\begin{tabular}{ccc}
$V$ & $r$ & $R$ \\ \hline
\multicolumn{1}{l}{$.3569050888$} & \multicolumn{1}{l}{$-.3041369802$} & 
\multicolumn{1}{l}{$.8210567527$} \\ 
\multicolumn{1}{l}{$3.555092622\times 10^{-2}$} & \multicolumn{1}{l}{$%
.2923064976$} & \multicolumn{1}{l}{$-.2224092096$} \\ 
\multicolumn{1}{l}{$.5429909915$} & \multicolumn{1}{l}{$.5356400064$} & 
\multicolumn{1}{l}{$.7823048379$}
\end{tabular}
\]
Since $r$ and $R$ must both be positive, the only possible extreme value is $%
.5429909915.\smallskip $

As an aside, the formula 
\[
V=\frac{1}{3}\left( r^{2}+rR+R^{2}\right) \pi x 
\]
for the volume of a truncated cone states that its volume is the height $x$
multiplied by the average of the area $\pi r^{2}$ of one face, the area $\pi
R^{2}$ of the other face, and the area $\pi rR$ of an ellipse with major
semi-axis $R$ and minor semi-axis $r$.
\end{quotation}
\end{example}

\subsubsection{Lagrange Multipliers\label{Lagrange multipliers}\label%
{Gradient operator}}

You can use Lagrange multipliers 
\index{Lagrange multipliers} to find constrained optima. To find extreme
values of $f(x,y)$ subject to a constraint $g(x,y)=k$, it is sufficient to
find all values of $x,$ $y,$ and $\lambda $ such that 
\[
\nabla f(x,y)=\lambda \nabla g(x,y) 
\]
and $g(x,y)=k$ where $\nabla $ is the gradient operator 
\[
\nabla f(x,y)=\left( 
\frac{\partial f}{\partial x}(x,y),\frac{\partial f}{\partial y}(x,y)\right) 
\]
The variable $\lambda $ is called the \emph{Lagrange multiplier}.

\strut

\begin{example}
Let $f(x,y)=xy$ and $g(x,y)=x+y$. Then 
\[
\nabla f(x,y)=\allowbreak \left( y,x\right) 
\]
and 
\[
\nabla g(x,y)=\allowbreak \left( 1,1\right) 
\]
so the equation $\nabla f(x,y)=\lambda g(x,y)$ is equivalent to the two
equations $y=\lambda $ and $x=\lambda $. If $k=5$, then the third condition
is $x+y=5$. With the three equations $x+y=5$, $y=\lambda $, and $x=\lambda $
in a $3\times 1$ matrix, from the \textsf{Solve} submenu, choose \textsf{%
Exact} to get the solution $\left\{ \lambda =\frac{5}{2},x=\frac{5}{2},y=%
\frac{5}{2}\right\} $. Thus, two real numbers $x$ and $y$ whose sum is $5$
and whose product is as large as possible are given by $x=5/2$ and $y=5/2$.
\end{example}

\strut

\begin{example}
Define $f(x,y)=x+2y$ and $g(x,y)=ye^{x}+xe^{y}$. Then 
\[
\nabla f(x,y)=\allowbreak \left[ 
\begin{array}{c}
1 \\ 
2
\end{array}
\right] \quad \text{and\quad }\nabla g(x,y)=\allowbreak \left[ 
\begin{array}{c}
ye^{x}+e^{y} \\ 
e^{x}+xe^{y}
\end{array}
\right] 
\]
so the condition $\nabla f(x,y)=\lambda g(x,y)$ is equivalent to the two
equations 
\begin{eqnarray*}
1 &=&\lambda \left( ye^{x}+e^{y}\right) \\
2 &=&\lambda \left( e^{x}+xe^{y}\right)
\end{eqnarray*}
If $k=g(x,y)=5$, then this gives a third equation $ye^{x}+xe^{y}=5$. You can
find a solution for the three equations, with $0\leq x\leq 10$ and $0\leq
y\leq 10,$ by applying \textsf{Solve + Numeric} to the system 
\begin{eqnarray*}
ye^{x}+xe^{y} &=&5 \\
\lambda \left( ye^{x}+e^{y}\right) &=&1 \\
\lambda \left( e^{x}+xe^{y}\right) &=&2 \\
x &\in &\left( 0,10\right) \\
y &\in &\left( 0,10\right)
\end{eqnarray*}
The solution is 
\[
x=1.666459547,y=.4505619968,\lambda =.252895924 
\]
These numbers give 
\[
g(1.666459547,.4505619968)=\allowbreak 4.999999999 
\]
and 
\[
f(1.666459547,.4505619968)=\allowbreak 2.567583541 
\]
This gives a possible extreme value for $f(x,y)$ satisfying the constraint $%
g(x,y)=5$.\medskip
\end{example}

\subsection{\ Iterated Integrals\label{Iterated integrals}}

You can enter and evaluate iterated integrals 
\index{Integrals!iterated}. If $a\leq b$, $f(x)\leq g(x)$ for all $x\in
\left[ a,b\right] $, and $k(x,y)\geq 0$ for all $x\in \left[ a,b\right] $
and all $y\in \left[ f(x),g(x)\right] $ then the iterated integral 
\[
\int_{a}^{b}\int_{f(x)}^{g(x)}k(x,y)\,dy\,dx 
\]
can be interpreted as the volume of the solid bounded by 
\begin{eqnarray*}
&&a\leq x\leq b \\
&&f(x)\leq y\leq g(x) \\
&&0\leq z\leq k(x,y)
\end{eqnarray*}

\begin{example}
Find the volume of the solid under the surface $z=1+xy$ and above the
triangle with vertices $(1,1)$, $(4,1)$, and $(3,2)\smallskip $

\begin{enumerate}
\item  Plot the triangle with the given vertices.%
\FRAME{dtbpF}{3in}{2.0003in}{0pt}{}{}{}{\special{language "Scientific
Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth 0pt;display
"PICT";function \TEXUX{$\left( 1,1,4,1,3,2,1,1\right) $};linecolor
"black";linestyle 1;linethickness 1;pointstyle "point";xmin "-5";xmax
"5";xviewmin "0.0";xviewmax "4.061";yviewmin "0.0";yviewmax
"2.02";viewset"XY";rangeset"X";phi 45;theta 45;plottype 4;constrained
TRUE;numpoints 49;axesstyle "normal";xis \TEXUX{x};var1name \TEXUX{$x$};}}

\item  Find the equations of the bounding lines 
\begin{eqnarray*}
y &=&\frac{1}{2}x+\frac{1}{2} \\
y &=&5-x \\
y &=&1
\end{eqnarray*}

\item  Solve for $x$ in terms of $y$%
\begin{eqnarray*}
x &=&2y-1 \\
x &=&5-y
\end{eqnarray*}

\item  Set up and evaluate an iterated integral 
\[
\int_{1}^{2}\int_{2y-1}^{5-y}\left( 1+xy\right) \,dx\,dy=\frac{55}{8} 
\]
Here is a plot that shows the top of the surface $z=1+xy$ with the bounding
triangle.\FRAME{dtbpF}{3in}{2.0003in}{0pt}{}{}{Plot }{\special{language
"Scientific Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth
0pt;display "PICT";function \TEXUX{$\left[ x,y,1+xy\right] $};linecolor
"black";linestyle 1;linethickness 1;pointstyle "point";function
\TEXUX{$\left[ x,1,8y\right] $};linecolor "black";linestyle 1;linethickness
1;pointstyle "point";function \TEXUX{$\left[
x,\frac{1}{2}x+\frac{1}{2},8y\right] $};linecolor "black";linestyle
1;linethickness 1;pointstyle "point";function \TEXUX{$\left[ x,5-x,8y\right]
$};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";xmin
"0";xmax "4";ymin "0";ymax "2";xviewmin "0";xviewmax "4";yviewmin
"0";yviewmax "2";zviewmin "0";zviewmax "8";viewset"XYZ";rangeset"XYZ";phi
19;theta -80;plottype 5;num-x-gridlines 25;num-y-gridlines 25;plotstyle
"hidden";axesstyle "normal";plotshading "Z";xis \TEXUX{x};yis
\TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};}}To make this plot,
start with $\left[ x,y,1+xy\right] $ and choose \textsf{Plot 3D Rectangular}%
. Add the items $\left[ x,1,8y\right] $, $\left[ x,\frac{1}{2}x+\frac{1}{2}%
,8y\right] $, and $\left[ x,5-x,8y\right] $. In the \textsf{Plot Properties}
dialog, change \textsf{Plot Style} to \textsf{Hidden Line}, set \textsf{Turn}
to -80 and \textsf{Tilt} to 19, and set the \textsf{Domain} and \textsf{View
Intervals} to $0<x<4$, $0<y<2$, and $0<z<8$.

Here is another way to show the surface that uses the Maple Heaviside
function. \FRAME{dtbpF}{3.0441in}{2.0384in}{0pt}{}{}{volume1.wmf}{\special%
{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file
"F";width 3.0441in;height 2.0384in;depth 0pt;original-width
216.8125pt;original-height 144.5625pt;cropleft "0";croptop "1";cropright
"1";cropbottom "0";filename 'Volume1.wmf';file-properties "XNPEU";}}To make
this plot, choose \textsf{Define + Define Maple Name}, enter \textsc{%
Heaviside(x)} as the Maple name and $H(x)$ as the Scientific Notebook name.
Type the expression $(1+xy)H\left( \frac{1}{2}x+\frac{1}{2}-y\right) H\left(
5-x-y\right) H\left( y-1\right) $ and choose \textsf{Plot 3D + Rectangular}.
In the \textsf{Plot Properties} dialog, change \textsf{Plot Style} to 
\textsf{Hidden Line}, set \textsf{Turn} to -80 and \textsf{Tilt} to 19, and
set the \textsf{Domain Intervals} to $0<x<4$ and $0<y<2$. Change the \textsf{%
Axes Type} to \textsf{Boxed}.
\end{enumerate}
\end{example}

\strut

\strut

Here are a few examples of iterated integrals.

\[
\begin{tabular}{lll}
\textbf{Integral} & \textsf{Evaluate} & \textsf{Evaluate Numerically} \\ 
\hline
$\vspace{6pt}\dint_{0}^{1}\dint_{0}^{x}x^{2}\cos y\,dy\,dx$ & $=\allowbreak
\cos 1+2\sin 1-2$ & $=\allowbreak .2232442755$ \\ 
$\dint_{0}^{1}\dint_{3y}^{3}e^{x^{2}}\,dx\,dy\vspace{6pt}$ & $=\allowbreak 
\frac{1}{6}e^{9}-\frac{1}{6}$ & $=\allowbreak 1350.347321$ \\ 
$\dint_{0}^{1}\dint_{0}^{x}\sin x^{2}\,dy\,dx\vspace{6pt}$ & $=\allowbreak -%
\frac{1}{2}\cos 1+\frac{1}{2}$ & $=\allowbreak .2298488471$ \\ 
$\dint_{0}^{1}\dint_{0}^{1-x}\dfrac{2y}{x+1}\,dy\,dx$ & $=\allowbreak -\frac{%
5}{2}+4\ln 2$ & $=\allowbreak .2725887222$%
\end{tabular}
\]

\label{Order of integration}Other multiple integrals look innocent enough
but require a bit more effort to evaluate. Attempting to evaluate the double
integral 
\[
\int_{0}^{1}\int_{\sqrt{y}}^{1}\sqrt{x^{3}+1}\,dx\,dy 
\]
exactly leads to frustration. However, you can reverse the order of
integration by looking carefully at the region of integration in the plane. 
\FRAME{dtbpF}{3in}{2.0003in}{0pt}{}{}{}{\special{language "Scientific
Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth 0pt;display
"PICT";function \TEXUX{$x^{2}$};linecolor "green";linestyle 1;linethickness
1;pointstyle "point";function \TEXUX{$(1,0,1,1)$};linecolor
"green";linestyle 1;linethickness 1;pointstyle "point";xmin "0";xmax
"1";xviewmin "-0.02";xviewmax "1.02";yviewmin "-0.02";yviewmax
"1.02";rangeset"X";phi 45;theta 45;plottype 4;numpoints 49;axesstyle
"normal";xis \TEXUX{x};var1name \TEXUX{$x$};}}This region is bounded above
by $y=x^{2}$ and bounded below by $y=0$. The new integral is 
\[
\int_{0}^{1}\int_{0}^{x^{2}}\sqrt{x^{3}+1}\,dy\,dx 
\]
This double integral can be solved by iterated integration. The inner
integral is just 
\[
\int_{0}^{x^{2}}\sqrt{x^{3}+1}\,dy=\allowbreak \sqrt{\left( x^{3}+1\right) }%
x^{2} 
\]
You can integrate the resulting outer integral $\int_{0}^{1}\sqrt{\left(
x^{3}+1\right) }x^{2}\,dx$ by applying \textsf{Calculus + Change Variable},
say with $u^{2}=x^{3}+1$. Then choose \textsf{Evaluate and Evaluate
Numerically}, to get

\begin{eqnarray*}
\int_{0}^{1}\sqrt{\left( x^{3}+1\right) }x^{2}\,dx &=&\allowbreak \int_{1}^{%
\sqrt{2}}\frac{2}{3}u^{2}\,du \\
&=&\allowbreak \frac{4}{9}\sqrt{2}-\frac{2}{9} \\
&=&\allowbreak .4063171388
\end{eqnarray*}

\subsection{\ Related topics}

\begin{itemize}
\item  \hyperref{Limits}{}{}{DM7-2.tex#Limit}

\item  \hyperref{Differentiation}{}{}{DM7-3.tex#Derivative}

\item  \hyperref{Indefinite Integration}{}{}{DM7-4.tex#Indefinite integral}

\item  \hyperref{Definite Integrals}{}{}{DM7-6.tex#Definite integral}

\item  \hyperref{Sequences and Series}{}{}{DM7-7.tex#Sequences}

\item  \hyperref{Plotting Curves and Surfaces}{}{}{DM6.tex}

\item  \hyperref{Matrix Algebra}{}{}{DM8.tex}

\item  \hyperref{Vector Calculus}{}{}{DM9.tex}

\item  \hyperref{Differential Equations}{}{}{DM10.tex}

\item  \hyperref{Exercises and Solutions}{}{}{DM7-9.tex#Exercises}
\end{itemize}

\rule{0.67in}{0.01in}

\hyperref{\hyperref{\FRAME{itbpF}{0.5275in}{0.3399in}{0.0519in}{}{}{index.wmf%
}{\special{language "Scientific Word";type "GRAPHIC";display
"PICT";valid-file "F";width 0.5275in;height 0.3399in;depth
0.0519in;original-width 36.125pt;original-height 22.5625pt;cropleft
"0";croptop "1";cropright "1";cropbottom "0";filename
'index.wmf';file-properties "XNPEU";}}}{}{}{DMIndex.tex#Math Index}}{}{}{%
DMIndex.tex#Math Index}\quad {\small Index entries: \hyperref{Integral}{}{}{%
DMIndex.tex#Integral}, \hyperref{Multivariable calculus}{}{}{%
DMIndex.tex#Multivariable calculus}, \hyperref{Optimization}{}{}{%
DMIndex.tex#Optimization}}

\end{document}
