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\begin{document}
\centerline{Mathematics 241, section 1 -- Multivariable Calculus}
\centerline{Solutions for Final Examination -- December 14, 2010}
\vskip 10pt
\noindent
I.  In this problem,
$$P = (1, -1, 3),\ Q = (0, 3, 1),\ {\rm and}\ R = (-4,1,2).$$
\begin{enumerate}
\item[A)] (10)  Find the equation of the plane
containing the points $P,Q,R$ in ${\bf R}^3$.
\vskip 10pt
\item[] {\it Solution:}  A normal vector is 
\begin{eqnarray*} 
N &=& ((-4,1,2) - (1,-1,3)) \times ((0,3,1) - (1,-1,3))\\
  &=& (-5,2,-1) \times (-1,4,-2) = (0,-9,-18)
\end{eqnarray*}
Since any scalar multiple of $N$ is also normal to the plane, we
can also use $N = (0,1,2)$, and then the equation becomes
$$(0,1,2) \cdot (x - 1, y + 1, z - 3) = 0$$
or 
$$y + 2z - 5 = 0.$$
\item[B)] (10)  At what point does the line containing
$P,Q$ meet the $xy$-plane?
\vskip 10pt
\item[] {\it Solution:}  The line containing $P$ and $Q$ can 
be parametrized as 
$$\alpha(t) = (1,-1,3) + t(-1,4,-2) = (1 - t, -1 + 4t, 3 - 2t).$$
This meets the $xy$ plane when $z = 0$ so $3 - 2t = 0$, or
$t = 3/2$.  The corresponding point is 
$$\alpha(3/2) = (-1/2, 5, 0).$$ 
\item{C)}  (5)  If ${\bf v}$ is the vector
from $P$ to $Q$ and ${\bf w}$ is the vector
from $P$ to $R$, at what angle do ${\bf v},{\bf w}$
meet?
\vskip 10pt
\item[]  {\it Solution:}  The angle is $\theta$ satisfying 
$$\cos\theta = \frac{{\bf v} \cdot {\bf w}}{\|{\bf v}\| \|{\bf w}\|} = 
\frac{15}{\sqrt{30} \sqrt{21}} = \frac{5}{\sqrt{70}}.$$
This says 
$$\theta = \cos^{-1}\left(\frac{5}{\sqrt{70}}\right).$$
\end{enumerate}
\vskip 10pt
\noindent
II.
Suppose you follow a flow
line of the vector field $\nabla f$ for $f(x,y)$ in the $xy$-plane.
\begin{enumerate}
\item[A)] (5)  As you traverse the flow line in the increasing
$t$-direction, is the corresponding
path on the graph $z = f(x,y)$ going uphill or downhill?
Explain.  
\vskip 10pt
\item[] {\it Solution:}  Uphill -- the gradient vector always points
in the direction of maximum rate of increase.
\vskip 10pt
\item[B)]  (5) What does the vector field $\nabla f$ look like
near a local maximum of $f$?  Near a local minimum of $f$?
Make rough sketches by hand to illustrate.
\vskip 10pt
\item[] {\it Solution:}  Near a local maximum, all the arrows
from the vector field $\nabla f$ will be pointing ``in'' -- the
local maximum is a sink of the vector field.  Near a local minimum, 
all the arrows will be pointing ``out'' -- the local minimum is
a source of the vector field.  
\end{enumerate}
\vskip 10pt
\noindent
III.  All parts of this problem refer to $f(x,y) = (x-1)^2 - y^2$.
\begin{enumerate}
\item[A)] (5) Sketch the contours of $f(x,y)$ for
the values $c = -1, 0, 1$.
\vskip 10pt
\item[] {\it Solution:}  The contours for $c = \pm 1$ are hyperbolas
with asymptotes along the lines $y = x - 1$ and $y = -x + 1$.  
The $c = -1$ contour opens up and down, and the $c = 1$
contour opens left and right.  The $c = 0$ contour is the union of the
two asymptotes.  
\vskip 10pt
\item[B)]  (10) At the point $(1,2)$, in which
direction is $f$ increasing the fastest? Express your
answer as a unit direction vector.
\vskip 10pt
\item[] {\it Solution:}  The gradient vector $\nabla f(1,2)$
points in this direction.  $\nabla f(x,y) = (2(x - 1), -2y)$
so we have $\nabla f(1,2) = (0,-4)$.  The unit vector in this
direction is $u = (0,-1)$.  
\end{enumerate}
\vskip 10pt
\noindent
IV. (20) After an ill-fated ``three hour tour'' goes awry, you 
are stranded on a island at the point with coordinates
$(1,1)$.  Fortunately, you have a radio transmitter with you.  
Unfortunately, it has a limited range -- its signal can only be 
received at distances less than or equal to $2/3$ from its position.  
You know that there is a Coast Guard patrol boat that makes a circuit of 
the path $x^2 + y^2 = 4$ every day, and they always carry
a radio receiver and listen for transmissions.  Will the patrol boat 
ever get within $2/3$ of your position and receive your signal?
(Note: Minimizing the distance from a point $(a,b)$ is 
the same as minimizing the function $f(x,y) = \|(x,y) - (a,b)\|^2.$)
\vskip 10pt
\noindent
{\it Solution:}    
One solution of this problem uses the method of Lagrange
multipliers to find the point on the path of the patrol boat
closest to $(1,1)$.  If the distance from the closest
point to $(1,1)$ is less than $2/3$, then the signal will be
heard and you will be rescued.   Using the hint, we want to minimize 
$$f(x,y) = \|(x,y) - (1,1)\|^2 = (x - 1)^2 + (y - 1)^2.$$
The constraint curve is $g(x,y) = x^2 + y^2 - 4 = 0$.  The Lagrange
equations are 
\begin{eqnarray*}
           2(x - 1) &=& 2\lambda x\\
           2(y - 1) &=& 2\lambda y\\
           x^2 + y^2 - 4 &=&0.
\end{eqnarray*}
We can eliminate $\lambda$ between the first two equations by the 
usual method (multiply the first by $y$ and the second by $x$,
then equate the left sides), yielding
$$y(x - 1) = x(y - 1) \Rightarrow y = x.$$  
In the constraint equation, this gives two points $(\sqrt{2},\sqrt{2})$
and $(-\sqrt{2},-\sqrt{2})$.  The values of $f$ at these points are
\begin{eqnarray*}
f(\sqrt{2},\sqrt{2}) &=& 2(\sqrt{2}-1)^2 = 6 - 4\sqrt{2}\\
           f(-\sqrt{2},-\sqrt{2}) &=& 2(-\sqrt{2}-1)^2 = 6 + 4\sqrt{2}.
\end{eqnarray*}
Since $6 - 4\sqrt{2} \doteq .343 < (2/3)^2 \doteq .444$, 
your radio signal {\it will be} received
by the patrol boat and you will be rescued!
\vskip 10pt
\noindent
{\it Notes:} 
\begin{enumerate}
\item  That the closest point to $(1,1)$ on the circle $x^2 + y^2 = 4$
is the point $(\sqrt{2},\sqrt{2})$ can also be seen by elementary geometry.
I gave full credit for solutions that showed this without using Lagrange
multipliers. 
\item The problem has a number of other correct solutions as well.
Another way that many of you thought of 
is to find whether the circle $(x - 1)^2 + (y - 1)^2 = 4/9$
(the boundary of the region in which the signal from your receiver
can be heard) intersects the path of the patrol boat $x^2 + y^2 = 4$.  This 
can be determined by elementary algebra -- the two circles intersect at 
approximately $(1.122,1.655)$ and $(1.655,1.122)$, so there is an arc on
the path of the patrol boat consisting of points from which the signal
can be heard.
\end{enumerate}  
\vskip 10pt
\noindent
V.  Let $f(x,y) = y e^{-x^2-2y^2}$.
\vskip 10pt
\begin{enumerate}
\item[A)] (10) Find the equation of the tangent plane
to the graph $z = f(x,y)$ at the point $(1,1,e^{-3})$.
\vskip 10pt
\item[] {\it Solution:}  We have $f_x = -2xy e^{-x^2-2y^2}$ and 
$f_y = e^{-x^2 - 2y^2}(1 - 4y^2)$.  So the tangent plane is 
\begin{eqnarray*}
z &=& f(1,1) + f_x(1,1) (x - 1) + f_y(1,1)(y - 1)\\
&=& e^{-3} - 2e^{-3}(x-1) - 3e^{-3}(y - 1).
\end{eqnarray*}
\item[B)] (10) Find all the critical points of
$f(x,y)$.
\vskip 10pt
\item[] {\it Solution:}  We have $f_x,f_y$ from part A.  Setting
those equal to zero, we obtain from $f_x = 0$ that $x = 0$ or $y = 0$.
However $f_y = 0$ implies that $y = \pm 1/2$.  So there are only two
critical points $(0, \pm 1/2)$.  
\item[C)]  (20) Use the Second Derivative Test 
to determine the type of each critical point you found in part B.
\vskip 10pt
\item[] {\it Solution:}  We continue to compute the second-order 
partial derivatives.  
\begin{eqnarray*}
f_{xx} &=& y(-2 + 4x^2)e^{-x^2 - 2y^2}\\
f_{xy} &=& x(-2 + 8y^2)e^{-x^2 - 2y^2}\\
f_{yy} &=& (-12y + 16y^3)e^{-x^2 - 2y^2}
\end{eqnarray*}
At $(0,1/2)$, 
\begin{eqnarray*}
A &=& f_{xx}(0,1/2) = -e^{-1/2}\\
B &=& f_{xy}(0,1/2) = 0\\
C &=& f_{yy}(0,1/2) = -4e^{-1/2}.
\end{eqnarray*}
Hence $AC - B^2 > 0$ and $A < 0$, so $f$ has a local maximum at $(0,1/2)$.  

Similarly, at $(0,-1/2)$, 
\begin{eqnarray*}
A &=& f_{xx}(0,-1/2) = +e^{-1/2}\\
B &=& f_{xy}(0,-1/2) = 0\\
C &=& f_{yy}(0,-1/2) = +4e^{-1/2}.
\end{eqnarray*}
Hence $AC - B^2 > 0$ and $A > 0$, so $f$ has a local minimum at $(0,-1/2)$.  
\end{enumerate}
\vskip 10pt
\noindent
VI.  The region $R$ in ${\bf R}^2$ is the set of points
$$R = \{(x,y) \in {\bf R}^2 \mid x^2 + y^2 \le 25, y \ge x, x \ge 0, {\rm\ and\ }
y \le 4\}
$$
and let $f(x,y)$ be some continuous function defined on $R$.  
\vskip 10pt
\begin{enumerate}
\item[A)] (5) Sketch $R$.
\vskip 10pt
\item[] {\it Solution:}  This is the region inside
the circle of radius 5 centered at $(0,0)$, above the line $y = x$, 
to the right of the $y$-axis, but below the line $y = 4$:
\begin{center}
\includegraphics[width=3.0truein,height=3.0truein]{FVI.ps}
\end{center}
The line $y = 4$ intersects the circle at $(3,4)$, and the line
$y = x$ intersects at $(5\sqrt{2}/2,5\sqrt{2}/2)$.  Note that this
has $y = 5\sqrt{2}/2 \doteq 3.54 < 4$.  So the boundary of the 
region we want includes a small arc of the circle $x^2 + y^2 = 25$
between the points $(3,4)$ and $(5\sqrt{2}/2,5\sqrt{2}/2)$.  
\item[B)]  (10) Set up the limits of integration
of iterated integral(s) to
compute $\int\!\int_R f(x,y)\ dA$ integrating with
respect to $x$ first, then $y$.
\item[] {\it Solution:}  This way, we must split the interval of $y$-values
at $y = 5\sqrt{2}/2$, where the line $y = x$ crosses the circle:
\[
\int_0^{5\sqrt{2}/2} \int_0^y f(x,y)\ dx\ dy + \int_{5\sqrt{2}/2}^4 \int_0^{\sqrt{25 - y^2}} f(x,y)\ dx\ dy.
\]
\item[C)]  (10) Now reverse the order of the
variables and set up iterated integral(s) to compute
the same integral, but integrating with respect to $y$ first, then
$x$.
\item[] {\it Solution:}  This one also must be split at $x = 3$ where the 
line $y = 4$ crosses the circle.
\[
\int_0^3 \int_x^4 f(x,y)\ dy\ dx + \int_3^{5\sqrt{2}/2} \int_x^{\sqrt{25-x^2}} f(x,y)\ dy\ dx.
\]
\end{enumerate}
\vskip 10pt
\noindent
VII.  (20)  The metal making up a thin plate
with the shape of the region in
${\bf R}^2$ with $x^2 + y^2 \le 4$ has density $\delta(x,y) = 6 + y$
at all points.  Determine the coordinates of its center of
mass.
\vskip 10pt
\noindent
{\it Solution:}  Since the plate has the shape of a circular disk,
we will set up all the integrals in polar coordinates.  First the 
total mass is 
\begin{eqnarray*}
M = \int\!\int_R 6 + y\ dA &=& 
\int_0^{2\pi} \int_0^2 (6 + r\sin\theta)r\ dr\ d\theta\\
&=& \int_0^{2\pi} 12 + \frac{8}{3}\sin\theta\ d\theta\\
&=& 24\pi.
\end{eqnarray*}
We can see $\overline{x} = 0$ since the plate and the density function
are symmetric under $x \mapsto -x$.  Finally 
\begin{eqnarray*} 
\overline{y} &=& \frac{1}{24\pi} \int_0^{2\pi}\int_0^2 r\sin\theta(6 + r\sin\theta) 
r\ dr\ d\theta\\
&=& \frac{1}{24\pi} \int_0^{2\pi} 16\sin\theta + 4\sin^2\theta\ d\theta\\
&=& \frac{4\pi}{24\pi}\\
 &=& \frac{1}{6}.
\end{eqnarray*}
\vskip 10pt
\noindent
VIII.    Consider the following triple integral:
$$\int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}
\int_{-\sqrt{9-x^2-y^2}}^{\sqrt{9-x^2-y^2}} 1 + x\ dz\ dy\ dx$$
\begin{enumerate}
\item[A)] (5) Describe the solid over which you are 
integrating here.
\item[] {\it Solution:}  It is the solid sphere of radius 3
centered at $(0,0,0)$ in ${\bf R}^3$.  
\vskip 10pt
\item[B)] (20) Using any convenient coordinate system,
evaluate this integral.
\item[] {\it Solution:}  Spherical coordinates are probably the 
best choice:
\begin{eqnarray*}
&&\int_0^{2\pi} \int_0^\pi \int_0^3 (1 + \rho\sin\phi\cos\theta)\rho^2\sin\phi\ d\rho\ d\phi\ d\theta\\
&=& \int_0^{2\pi} \int_0^\pi 9\sin\phi + \frac{81}{4}\sin^2\phi\cos\theta\ d\phi\ d\theta\\
&=& \int_0^{2\pi} 18 + \frac{81\pi}{8}\cos\theta\ d\theta\\
&=& 36\pi.
\end{eqnarray*}
This can also be done relatively easily in cylindrical coordinates, of course:
$$\int_0^{2\pi} \int_0^3 \int_{-\sqrt{9 - r^2}}^{\sqrt{9-r^2}} (1 + r\cos\theta)r\ dz\ dr\ d\theta.$$
\end{enumerate}
\vskip 10pt
\noindent
IX.
\vskip 10pt
\begin{enumerate}
\item[A)] (10)  State Green's Theorem.
\item[] {\it Solution:}  If $\alpha$ is a differentiable 
simple, closed, positively oriented
curve enclosing a region $R$ and $F(x,y) = (u(x,y),v(x,y))$
is a vector field whose component functions $u,v$ have continuous
first order partial derivatives on the region $R$, then
$$\int_\alpha F\cdot T\ ds = \int\!\int_R v_x - u_y\ dA.$$
\item[B)] (10) Let ${\bf F}(x,y) = (x - y^2, x^2 + y)$.
Verify that Green's Theorem holds for the region
$D = \{(x,y) \in {\bf R}^2 : x^2 + y^2 \le 9\}$ 
by computing both sides of the equation in the 
theorem and showing that they are equal. 
\item[] {\it Solution:}  The line integral can be computed
using $\alpha(t) = (3\cos(t), 3\sin(t))$. This gives:
\begin{eqnarray*}
\int_\alpha F\cdot T\ ds &=& \int_0^{2\pi} (3\cos(t) - 9\sin^2(t),9\cos^2(t) + 3\sin(t))\cdot (-3\sin(t),3\cos(t))\ dt\\
&=& \int_0^{2\pi} 27\sin^3 t + 27\cos^3 t\ dt\\
&=& 0.
\end{eqnarray*} 
The double integral can be computed using polar coordinates:
\begin{eqnarray*}
\int\!\int_R 2x + 2y\ dA &=& \int_0^{2\pi} \int_0^3 2r^2(\cos\theta + \sin\theta)\ dr\ d\theta\\
&=& \int_0^{2\pi} 18(\cos\theta + \sin\theta)\ d\theta\\
&=& 0.
\end{eqnarray*}
\end{enumerate}
\vskip 10pt
\noindent
{\it Extra Credit}  
\vskip 10pt
\noindent
A function $f(x,y)$ is said to be {\it harmonic}
on an open set $U$ in ${\bf R}^2$ if it satisfies the equation
$$f_{xx} + f_{yy} = 0$$
at all points in $U$.  (Here $f_x, f_y, f_{xx}, f_{yy}$
are the partial derivatives of the harmonic function $f$ with
respect to the indicated variables.)
\begin{enumerate}
\item[A)] (5)  How does a nondegenerate
critical point of a harmonic function
fit into our classification?  Is it a local maximum,
local minimum, or a saddle point?  Explain how you can
tell from the Second Derivative Test.
\item[] {\it Solution:}  Since $f_{yy} = -f_{xx}$ at all 
critical points, 
$$f_{xx} f_{yy} - (f_{xy})^2 = - (f_{xx})^2 - (f_{xy})^2 < 0$$
at every nondegenerate critical point.  They are all saddle points(!)
\item[B)] (5)  If $f$ is harmonic, what is true about the 
{\it total flux} of the vector field
$${\bf F} (x,y) = \nabla f (x,y) $$
across any simple closed curve in $U$?  Explain.
\item[] {\it Solution:}  The divergence of $\nabla f = (f_x,f_y)$ is 
$$(f_x)_x + (f_y)_y = f_{xx} + f_{yy} = 0.$$
Hence 
$$\int_\alpha F\cdot N\ ds = \int\!\int_R  f_{xx} + f_{yy}\ dA = 0$$
by Green's Theorem.
\end{enumerate}
\end{document} 
