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\title{ESP Workshop, Worksheet \#}
\date{\today}
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\begin{center}
{\bf \Large ESP Workshop, Worksheet \#20}

{\bf \large Tuesday \today}

{\bf \large AI: Eric Katerman}
\end{center}
\vspace{.15in}
\begin{enumerate}
\item Recall the second derivative test to find local minima/maxima of
  multivariable functions: Suppose that $f_x (a,b) = 0$ and $f_y(a,b)
  = 0$ (that is, $(a,b)$ is a critical point of $f$). Let
\begin{displaymath}
D = D(a,b) = f_{xx} (a,b) f_{yy}(a,b) -[f_{xy} (a,b) ]^2
\end{displaymath}
\begin{enumerate}
\item If $D > 0$ and $f_{xx} (a,b) > 0$, then $f(a,b)$ is a local minimum.
\item If $D > 0$ and $f_{xx} (a,b) < 0$, then $f(a,b)$ is a local maximum.
\item If $D < 0$, then $(a,b,f(a,b)) \in \mathbb{R}^3$ is a saddle
  point of the graph of $f$.
\end{enumerate}
Let's use this to find the local minimum and maximum values and saddle
points of the following functions:
\begin{enumerate}
\item $f(x,y) = 9-2x +4y -x^2 -4y^2$
\item $f(x,y) = x^4 +y^4 -4xy +2$
\item $f(x,y) = 1+2xy -x^2 -y^2$
\end{enumerate}

\item (Optional; requires technology) If you have a graphing
  calculator that graphs three-dimensional things, plot the functions
  from question 1 and identify local minima, maxima, and saddle
  points.

\item Evaluate the following double integrals; note that for some of
  them, it will help to first identify the integral as the volume of a
  solid.  Draw a picture of each solid in $xyz$-space.
\begin{enumerate}
\item $\iint _R 3\ dA,\ R=\{ (x,y) | -2 \leq x \leq 2, 1\leq y \leq 6 \}$
\item $\iint _R (5-x)\ dA,\ R=\{(x,y) | 0\leq x \leq 5,\ 0\leq y \leq 3 \}$
\item $\iint _R \sqrt{9-y^2}\ dA,\ R= [0,4] \times [0,2]$
\end{enumerate}

\item Recall that the average value of a function $f$ of one variable
  is
\begin{displaymath}
\frac{1}{b-a} \int _a ^b f(x) dx
\end{displaymath}
That is, we divide the integral by the length of the interval we're
integrating along. Similarly, the average value of a function $f$ of
two variables defined on a region $R$ is
\begin{displaymath}
\frac{1}{A(R)} \iint _R f(x,y) dA
\end{displaymath}
where $A(R)$ is the area of $R$.

Suppose that the temperature in Texas yesterday at noon was
\begin{displaymath}
T(x,y) = 60 + y^2 -xy
\end{displaymath}
where Austin is the origin and 1 unit in the positive $x$ and $y$
directions represent 100 miles east and north, respectively.
Approximate the average temperature in Texas by finding the average
value for $T$ in the region $R = [-6,3] \times [-3,3]$. How well does
$R$ approximate Texas? Can you define a better region $R'$ for Texas?
How do the averages compare for the different regions?

\item First try to evaluate the integral in the way it is presented,
  and then reverse the order of integration and try it that way. Don't
  forget to change the limits of integration (drawing a picture will
  help)! Which integral was easier?
\begin{enumerate}
\item $\int _0 ^1 \int _{3y} ^3 e^{x^2}\ dx\ dy$
\item $\int _0 ^9 \int _0 ^{\sqrt{x}} y \cos (x^2)\ dy\ dx$
\item $\int _0 ^1 \int _{\arcsin y} ^{\pi /2} \cos x \sqrt{1+\cos^2 x}\ 
  dx\ dy$
\end{enumerate}

\item (Challenge!) The Riemann zeta function is one of the most famous
  functions in all of mathematics, and it is the subject of one of the
  unsolved million dollar prize problems. It is defined as an infinite
  series:
\begin{displaymath}
\zeta (x) = \sum _{n=1} ^\infty \frac{1}{n^x}
\end{displaymath}
Now, we can finally prove that $\zeta (2) = \pi ^2 /6$, as promised!
Here we go!
\begin{enumerate}
\item The double integral $\int _0 ^1 \int _0 ^1 \frac{1}{1-xy} dx\
  dy$ is an improper integral and could be defined as the limit of
  double integrals over the rectangle $[0,t] \times [0,t]$ as
  $t\rightarrow 1^-$. But if we expand the integrand as a geometric
  series, we can express the integral as the sum of an infinite
  series. Show that
\begin{displaymath}
\int _0 ^1 \int _0 ^1 \frac{1}{1-xy} dx\ dy = \zeta (2)
\end{displaymath}

\item Evaluate that double integral by first making the substitution
\begin{displaymath}
x= \frac{u-v}{\sqrt{2}}\quad y= \frac{u+v}{\sqrt{2}}
\end{displaymath}
(Note: this substitution gives a rotation about the origin through the
angle $\pi /4$.) Sketch the corresponding region in the $uv$-plane.
Hint: if, in evaluating the integral, you encounter either of the
expressions
\begin{displaymath}
\frac{1-\sin \theta}{\cos \theta}\quad \textrm{or} \quad \frac{\cos \theta}{1+\sin \theta}
\end{displaymath}
you might like to use the identity $\cos \theta = \sin ((\pi/2)
-\theta)$ and the corresponding identity for $\sin \theta$.
\end{enumerate}

\end{enumerate}

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