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\centerline{Mathematics 241, section 1 -- Multivariable Calculus}
\centerline{Solutions for Exam III -- December 3, 2010}
\vskip 10pt
\noindent
I.  
\begin{enumerate}
\item[A)] (10)  Give a precise statement of the Second Derivative
Test for local extrema of functions of two variables.
\vskip 10pt
\item[] {\it Solution:}  Let $(x_0,y_0)$ be a critical point of 
$f(x,y)$ and assume the partial derivatives $f_x$ and $f_y$ are
continuous in some open set containing $(x_0,y_0)$.  (Note: this
implies that $f(x,y)$ is differentiable at $(x_0,y_0)$.)  If
\begin{enumerate}
\item[1)]  $f_{xx}(x_0,y_0)f_{yy}(x_0,y_0) - (f_{xy}(x_0,y_0))^2 > 0$
and $f_{xx}(x_0,y_0) > 0$, then $f$ has a local minimum at $(x_0,y_0)$.
\item[2)]  $f_{xx}(x_0,y_0)f_{yy}(x_0,y_0) - (f_{xy}(x_0,y_0))^2 > 0$
and $f_{xx}(x_0,y_0) < 0$, then $f$ has a local maximum at $(x_0,y_0)$.   
\item[3)]  $f_{xx}(x_0,y_0)f_{yy}(x_0,y_0) - (f_{xy}(x_0,y_0))^2 < 0$,
then $f$ has a saddle point at $(x_0,y_0)$ (neither a maximum nor a 
minimum).
\item[4)]  If $f_{xx}(x_0,y_0)f_{yy}(x_0,y_0) - (f_{xy}(x_0,y_0))^2 = 0$,
then the test gives no conclusion.
\end{enumerate}
\vskip 10pt
\item[B)] (20) Let $f(x,y) = 3x^4 + y^3 - 4x^3 - 3y$.  Find
all critical points of $f$ and apply the Second
Derivative Test to each.  
\vskip 10pt
\item{} {\it Solution:}  The partial derivatives are
\begin{eqnarray*}
f_x &=& 12x^3 - 12x^2 = 12x^2(x - 1)\\
f_y &=& 3y^2 - 3
\end{eqnarray*}
So the critical points are the solutions of $f_x = f_y = 0$, or 
$$(0,1), (0,-1), (1,1), (1,-1).$$
Next, we compute:
\begin{eqnarray*}
f_{xx} &=& 36x^2 - 24x\\
f_{xy} &=& 0\\
f_{yy} &=& 6y.
\end{eqnarray*}
At $(0,\pm 1)$, 
$$f_{xx}(0,\pm 1)f_{yy}(0,\pm 1) - (f_{xy}(0,\pm 1))^2 = 0\cdot (\pm 6) - 0^2 = 0.$$
Hence the Second Derivative Test gives {\it no conclusion} at those points.  
(By looking at the form of the function, we can see, in fact, that both
of these are saddle points of a sort -- they are neither maxima nor minima.)
At $(1,1)$, $f_{xx}(1,1)f_{yy}(1,1) - (f_{xy}(1,1))^2 = 12 \cdot 6 - 0 = 72 > 0$
and $f_{xx}(1,1) = 12 > 0$.
Therefore, $f$ has a {\it local minimum} at $(1,1)$.  Finally, 
at $(-1,1)$,  
$f_{xx}(1,-1)f_{yy}(1,-1) - (f_{xy}(1,-1))^2 = 12 \cdot -6 - 0 = -72 < 0$.
Therefore $f$ has a {\it saddle point} at $(1,-1)$.  
\end{enumerate}
\vskip 10pt
\noindent
II.  
\begin{enumerate}
\item[A)]  (20) Find the maximum and minimum values of 
$$f(x,y) = xy - y$$
on the closed and bounded set 
$$S = \{(x,y) \in \R^2 \mid x^2 + y^2 \le 4\}.$$
\vskip 10pt
\item[] {\it Solution:}  We have 
\begin{eqnarray*}
f_x &=& y\\
f_y &=& x - 1
\end{eqnarray*}
So $f$ has a critical point at $(1,0) \in S$.  By the Second Derivative
Test, we can see that $f$ has a saddle point here, and $f(1,0) = 0$.  
There must be smaller and larger values on the boundary, so we use 
Lagrange Multipliers next.  With the constraint equation $x^2 + y^2 = 4$,
the Lagrange equations are 
\begin{eqnarray*}
y &=& 2\lambda x\\
x - 1 &=& 2\lambda y\\
x^2 + y^2 - 4 &=&0. 
\end{eqnarray*}
If we multply the first equation by $y$ and the second by $x$, then
we can equate the right sides to obtain an equation in which $\lambda$
has been eliminated:  $y^2 = x(x - 1) = x^2 - x$.  We can then use this
equation to substitute for the $y^2$ and eliminate $y$ in the constraint equation:
$$2x^2 - x - 4 = 0.$$
This equation can be solved by the quadratic formula:
$$x = \frac{1 \pm \sqrt{1 + 32}}{4} \doteq 1.69, -1.19.$$
Then from the constraint equation we get two $y$-values for each of these
$x$-values, so there are four solutions of the Lagrange equations
all together -- approximately
$$(-1.19,\pm 1.61),\ {\rm and}\ (1.69,\pm 1.08).$$
We have 
\begin{eqnarray*}
f(-1.19,-1.61) &\doteq& 3.52\\
f(-1.19,1.61) &\doteq& -3.52\\
f(1.69,-1.08) &\doteq&  -.74\\
f(1.69,1.08) &\doteq& .74
\end{eqnarray*}
So putting everything together, we see that the maximum 
value of $f$ on $S$ is about $3.52$ and the minimum value
of $f$ on $S$ is about $-3.52$. 
\item[B)] (5)  Give a sketch of the contours of $f$ for the 
maximum and minimum values you found in part A.  
\item[] {\it Solution:}  The function is $f(x,y) = (x-1)y$.   Hence
the contours of $f$ for values not equal to zero are {\it rectangular hyperbolas}
with asymptotes equal to the lines $x = 1$ and $y = 0$.  
\begin{center}
\includegraphics[width=3.0truein,height=3.0truein]{II.ps}
\end{center}
The sketch
should show (branches of) two hyperbolas {\it tangent to} the boundary circle of $S$
passing through the points where the maximum and minimum values are 
attained.  The relevant portions of the branches of the hyperbolas are in 
the second and third quadrants of the plane.  (Recall the geometry of Lagrange Multipliers we saw in Lab 4.)
\end{enumerate}
\vskip 10pt
\noindent
III.  All parts of this problem refer to the region
$$R = \{(x,y) \in \R^2 \mid x^2 + y^2 \le 9, y \ge 0, y \le 2x\}.$$
and the solid $S \subset \R^3$ with ``shadow'' $R$ in the $xy$-plane
and $0 \le z \le x^2$.  
\vskip 10pt
\begin{enumerate}
\item[A)] (5) Sketch the region $R$ in the plane.
\item[] {\it Solution:} This is the region inside a circle of radius 3, above the 
$x$-axis, and below the line $y = 2x$:
\begin{center}
\includegraphics[width=3.0truein,height=3.0truein]{III.ps}
\end{center}
\item[B)] (10)  Set up iterated integral(s) of the form $\int\int f(x,y)\ dy\ dx$
to compute the volume of $S$.  Do not compute the value.
\vskip 10pt
\item[] {\it Solution:}  Since the top boundary changes when $y = 2x$
crosses the circle, we
need the sum of two interated integrals.
The line $y = 2x$ crosses $x^2 + y^2 = 9$ when $5x^2 = 9$ so $x = 3/\sqrt{5}$.
$$\int_0^{3/\sqrt{5}} \int_0^{2x} x^2\ dy\ dx + 
\int_{3/\sqrt{5}}^3 \int_0^{\sqrt{9-x^2}} x^2\ dy\ dx.$$
\item[C)]  (10)  Now set up iterated integral(s) of the form $\int\int f(x,y)\ dx\ dy$
to compute the volume of $S$.  Do not compute the value.
\vskip 10pt
\item[] {\it Solution:}  This one is slightly tricky because you 
have to notice that, even though the integral does not 
need to be split, the region $R$ does not extend up to 
$y = 3$.  The range of $y$-values is determined like this:  
The line $y = -2x$ crosses $x^2 + y^2 = 9$ when $5x^2 = 9$ so $x = 3/\sqrt{5}$
and $y = 6/\sqrt{5}$.  The integral is:
$$\int_0^{6/\sqrt{5}} \int_{y/2}^{\sqrt{9-y^2}} x^2\ dx\ dy.$$
\item[D)]  (10)  Finally, set up a polar coordinate integral that 
will compute the volume of $S$.  Do not compute the value.
\vskip 10pt
\item[] {\it Solution:} With polar coordinates, you do not need to 
split into two integrals, and the limits of integration are all constant.
The line $y = 2x$ is the ray $\theta = \arctan(2)$.   
With the ``extra $r$'' from the $dA$ in polar coordinates: 
$$\int_0^{\arctan(2)} \int_0^3 r^3\cos^2\theta\ dr\ d\theta.$$
\item[E)]  (10)  Finally, compute the volume of $S$ using any one
of your answers from parts B,C,D (your choice). 
\vskip 10pt
\item[] {\it Solution:}  Don't know about you, but the first two
look pretty bad to me!  Let's evaluate the last one:
\begin{eqnarray*}
\int_0^{\arctan(2)} \int_0^3 r^3\cos^2\theta\ dr\ d\theta 
&=& \int_0^3 r^3\ dr \cdot \int_0^{\arctan(2)} \cos^2\theta\ d\theta\\
&=& \left. \frac{r^4}{4}\right|_0^3 \cdot \left(\left. \frac{\theta}{2} + \frac{\sin(2\theta)}{4}\right)\right|_0^{\arctan(2)}\\
&=& \frac{81}{4} \cdot \left(\frac{1}{2}\arctan(2) + \frac{1}{5}\right).
\end{eqnarray*}

\end{enumerate}
\vskip 10pt     
\noindent
{\it Extra Credit} (10)  
Evaluate the improper double integral
$$\int\int_{{\R}^2} e^{-x^2-y^2}\ dA = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-x^2 -y^2}\ dy\ dx.$$
\vskip 10pt
\noindent
We can evaluate this by switching to polar coordinates!  The integral over
all of $\R^2$ is 
$$\int_0^{2\pi} \int_0^\infty e^{-r^2} r\ dr\ d\theta$$
This equals 
$$
2\pi \cdot \lim_{B\to\infty} \left. -\frac{1}{2} e^{-r^2}\right|_0^B
= 2\pi \cdot \lim_{B\to\infty} \frac{1}{2} - \frac{1}{2} e^{-B^2}
= \pi.
$$
Wow! 

\end{document} 
