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\def\sol{\medskip\item{}{\bf Solution:}$\ $}
\centerline{Mathematics 241 -- Multivariable Calculus}
\centerline{Solutions for Final Examination -- December 14, 2013}
\bigskip
\noindent
I.  In this problem,
$P = (1,0,1),\ Q = (-2, 3, 2),\ {\rm and}\ R = (1,2,0).$
\medskip
\item{A)}  Find the equation of the plane
containing the points $P,Q,R$ in ${\bf R}^3$.
\sol
The displacement vector from $P$ to $Q$ is ${\bf v} = Q - P = (-3,3,1)$ and
the vector from $P$ to $R$ is ${\bf w} = R - P = (0,2,-1)$.  For the plane
we can take $N = (-3,3,1)\times (0,2,-1) = (-5,-3,-6)$.  Then the equation
of the plane is $0 = N\cdot (x - 1,y - 0, z - 1) = -5x + 5 - 3y - 6z + 6$, or
$5x + 3y + 6z = 11$.
\medskip
\item{B)} At what point does the line containing
$P,Q$ meet the $xy$-plane?
\sol
The line is $(1,0,1) + (-3,3,1)t = (1-3t,3t,1+t)$.  This
meets the $xy$-plane when $z = 1 + t = 0$, so $t = -1$.  The
point of intersection is $(4,-3,0)$.
\medskip
\item{C)}  If ${\bf v}$ is the displacement vector
from $P$ to $Q$ and ${\bf w}$ is the displacement vector
from $P$ to $R$, at what angle do ${\bf v},{\bf w}$
meet?
\sol
The angle $\theta$ satisfies
$\cos\theta = {{\bf v}\cdot {\bf w}\over \|{\bf v}\|\|{\bf w}\|} = {5\over\sqrt{19}\sqrt{5}}$.
So
$$\theta = \cos^{-1}\left(\sqrt{5/19}\right) \doteq 1.032 {\rm\ radians}.$$
\bigskip
\noindent
II.  All parts of this problem refer to the parametric curve
$$\alpha(t) = \left({\cos(t)\over \sin^2(t) + 1},{\cos(t)\sin(t)\over 1 + \sin^2(t)}\right)$$
defined for all $t \in [0,2\pi]$, called a {\it lemniscate}.
\item{A)} Is $\alpha(t)$ a simple closed curve?   (Hint: 
Thinking of $\alpha(t)$ as the position of a moving
object as a function of time, are there different times $t \in [0,2\pi)$ 
at which the object is at the location $(x,y) = (0,0)$?)
\sol
We have $\alpha(0) = \left({1\over 2},0\right) = \alpha(2\pi)$, so this
is a closed curve.  However,
$\cos(t) = 0 = \cos(t)\sin(t)$ for $t = {\pi\over 2}$ and $t = {3\pi\over 2}$
in the given interval.    Since there are two such $t$, the answer to 
the first question is {\it NO}.
\item{B)} What is the tangent vector to the curve at $t = \pi$?
\sol 
By the quotient rule in each component,
$$\eqalign{\alpha'(t) &= {1\over (1 + \sin^2(t))^2} (-(1+\sin^2(t))\sin(t) - 2\cos^2(t)\sin(t),\cr
&\quad (1 + \sin^2(t))(\cos^2(t) - \sin^2(t)) - 2\cos^2(t)\sin^2(t))\cr}$$
When $t - \pi$, we get $\alpha'(\pi) = (0,1)$ 
\item{C)}  A thin wire has the shape of the portion of the 
curve $\alpha$ for $t \in [0,1]$. 
What integral would you use to compute its arc length.  (You do not need
to evaluate!)
\sol
The arc length would be computed by 
$$M = \int_0^1 ds = \int_0^1 \|\alpha'(t)\| dt$$
III.  All parts of this problem refer to $f(x,y) = (x+1)^2 + y^2$.
\medskip
\item{A)} Sketch the level curves of $f(x,y)$ for
the values $c = 1, 4, 9$.
\sol
The level curves of $f$ are circles with center at $(-1,0)$ the radii are
$r = 1, 2, 3$ respectively.
\medskip
\item{B)}  At the point $(1,2)$, in which
direction is $f$ increasing the fastest? Express your
answer as a unit direction vector.
\sol
This is in the direction of the gradient vector 
$\nabla f(1,2)$.  The gradient vector
is $\nabla f(x,y) = (2(x+1),2y)$ at a general point.  So $\nabla f(1,2) = (4,4)$.
The unit vector in this direction is ${1\over 4\sqrt{2}}(4,4) = (\sqrt{2}/2,\sqrt{2}/2)$.
\item{C)}  Find the points on the curve
$g(x,y) = {x^2\over 4} + y^2 = 1$
at which $f(x,y)$ takes its largest and smallest
values.  What is true about the vectors
$\nabla f$ and $\nabla g$ at your points?
\sol
Using the Lagrange multiplier method, we must solve
$$\eqalign{ 2(x+1) &= \lambda x/2 \cr
             2y & = 2\lambda y\cr
             {x^2\over 4} + y^2 &= 1\cr}$$
From the second equation, $y = 0$ or $\lambda = 1$.  If $y = 0$, the constraint
equation gives $x = \pm 2$, so we obtain two points  $(\pm 2,0)$.  If
$\lambda = 1$, then from the first equation, $2(x + 1) = x/2$, so $x = -4/3$.
Then from the constraint equation we get $y = \pm \sqrt{5}/3$.
To determine which of these give maximum and minimum values, we substitute into $f(x,y)$:
$$\eqalign{f(2,0) &= 9\ ({\rm maximum})\cr
           f(-2,0) &= 1\cr
           f(-4/3, \pm \sqrt{5}/3) &= 1/9 + 5/9 = 2/3\ ({\rm minimum})\cr}$$
The points we found here are the points where the level curve of $f$ passing through that
point and the constraint curve are {\it tangent}.
\bigskip
\noindent
IV.  Let $f(x,y) = x e^{-2x^2 - y^2}$.
\medskip
\item{A)} Find the equation of the tangent plane
to the graph $z = f(x,y)$ at the point $(1,1,e^{-3})$.
\sol
We must compute the partial derivatives to start:
$$\eqalign{
f_x &= (1 - 4x^2)e^{-2x^2 - y^2}\cr
f_y &= -2xy e^{-2x^2 - y^2}.\cr}$$
At $(x,y) = (1,1)$, $f_x(1,1) = -3e^{-3}$, and 
$f_y(1,1) = -2e^{-3}$, so the tangent plane is
$$z = e^{-3} - 3e^{-3}(x - 1) - 2e^{-3}(y - 1).$$

\item{B)} Find all the critical points of $f(x,y)$.
\sol
The critical points are the solutions of $f_x = 0$
and $f_y = 0$.  Using the formulas for $f_x$, $f_y$
from part A, we see that $f_x = 0$ when $x = \pm 1/2$
and $f_y = 0$ when $x = 0$ or $y = 0$ (Note: the exponential
factor is {\it never zero}.)  Hence the simultaneous solutions
are the two points $(\pm 1/2, 0)$.
\medskip 
\item{C)}  Use the second derivative test (Hessian
criterion) to determine the type of each critical point
you found in part B.
\sol
Now we need the second-order partial derivatives as well:
$$\eqalign{
f_{xx} &= (16x^3 - 12x)e^{-2x^2-y^2}\cr
f_{xy} &= (1-4x^2)(-2y)e^{-2x^2-y^2}\cr
f_{yy} &= -2x(1 - 2y^2)e^{-2x^2-y^2}\cr}$$
So at $(1/2,0)$ the Hessian matrix is
$$ \pmatrix{-4e^{-1/2} & 0\cr
                              0 & -e^{-1/2} \cr}$$
The determinant is $4e^{-1} > 0$ and the upper left
entry is $<0$ so this is a {\it local maximum}.

\item{} At $(-1/2,0)$ the Hessian matrix is
$$\pmatrix{4e^{-1/2} & 0\cr
                              0 & e^{-1/2} \cr}$$
The determinant is $4e^{-1} > 0$ and the upper left
entry is $> 0$ so this is a {\it local minimum}.
\bigskip
\noindent
V.  A region $R$ in ${\bf R}^2$ is the set of points
satisfying $x^2 + y^2 \ge 1$, $y \ge x$, $x \ge 0$, and
$y \le 4$.
\medskip
\item{A)} Sketch the region $R$.
\sol
This is the region outside the unit circle with 
center $(0,0)$, to the right of the $y$-axis, 
below the horizontal line $y = 4$, and 
above the line $y = x$.
\medskip
\item{B)}  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 $y$.
\sol
The circle intersects the line $y = x$ at
$(\sqrt{2}/2,\sqrt{2}/2)$.  From there to the 
top of the circle at $y = 1$, the left boundary
of the region is part of the circle.  For $y > 1$,
though, the left boundary is part of the $y$-axis
so we have to split 
the integral at $y = 1$:
$$\int_{\sqrt{2}/2}^1 \int_{\sqrt{1-y^2}}^y f(x,y)\ dx\ dy
+ \int_1^4 \int_0^y f(x,y)\ dx\ dy.$$
\item{C)}  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
$x$.
\sol 
We also need to split the integral this way
since the bottom boundary changes at $x = \sqrt{2}/2$.
The region extends all the way to $x = 4$ on the 
right, where the 
line $y = 4$ intersects $y = x$:
$$\int_0^{\sqrt{2}/2} \int_{\sqrt{1-x^2}}^4 f(x,y)\ dy\ dx 
+\int_{\sqrt{2}/2}^4 \int_x^4 f(x,y)\ dy\ dx.$$
\bigskip
\noindent
VI. (20)  The metal making up a solid half-cone in the shape of
$$H = \{(x,y,z) \in {\bf R}^3 \mid z^2 \ge x^2 + y^2, 0 \le z \le 1,y \ge 0\}$$
has density $\delta(x,y,z) = \sqrt{x^2 + y^2 + z^2}$
at all points.  Determine its total mass.  (Hint: a wise choice of coordinate system will simplify
this one immensely!)
\sol
We will set up the triple integral to compute the 
mass using {\it spherical}
coordinates, since the spherical equation of the boundary 
cone is $\phi = 1$.  The restriction $y \ge 0$ says $0\le \theta \le \pi$.
The plane $z = 1$ is $\rho\cos\phi = 1$, so 
$\rho = \sec\phi$.  In spherical coordinates, the density is just 
$d = \rho$.  So the total mass is 
$$\eqalign{
M &= \int_0^{\pi} \int_0^{\pi/4} \int_0^{\sec\phi}  \rho^3 \sin\phi\ d\rho\ d\phi\ d\theta\cr
&= \int_0^{\pi}\int_0^{\pi/4}\int_0^{\sec\phi}  \rho^3\sin\phi d\rho\ d\phi\ d\theta\cr
&= \int_0^\pi \int_0^{\pi/4} \left. {\rho^4\over 4} \right|_0^{\sec\phi} \sin\phi\ d\phi\ d\theta\cr
&= \int_0^\pi \int_0^{\pi/4} {\sin\phi\over 4\cos^4\phi}\ d\phi\ d\theta \quad (u^{-4} du)\cr
&= \pi \left. {1\over 12 \cos^3(\phi)} \right|_0^{\pi/4}\cr
&= {\pi \over 12}(2\sqrt{2} - 1)\cr}
$$
\bigskip
\noindent
VII.
\medskip
\item{A)} State Green's Theorem.
\sol
If $D$ is a region in ${\bf R}^2$ bounded by a finite
collection of simple closed curves, $\partial D$ is
the positively-oriented boundary of $D$,
and ${\bf F}(x,y) = (F_1(x,y),F_2(x,y))$ is a $C^1$ vector
field on $D$, then
$$\oint_{\partial D} F\cdot T\ ds = \oint_{\partial D} F_1 dx + F_2 dy  
= \int\!\int_D {\partial F_2 \over \partial x} - {\partial F_1\over \partial y}\ dA.$$ 
\item{B)} (10) Let ${\bf F}(x,y) = (x - y^2, x^2 + y)$.
Verify that Green's Theorem holds for the region
$D = \{(x,y) \in {\bf R}^2 : x^2 + y^2 \le 9\}$.
\sol
Using the standard parametrization $(x,y) = (3\cos(t),3\sin(t))$
of the boundary
circle of $D$, 
$$\eqalign{
\oint_{\partial D} F_1 dx + F_2 dy 
&= \int_0^{2\pi} (3\cos(t) - 9\sin^2(t))(-3\sin(t)) \cr
&\qquad + (9\cos^2(t) + 3\sin(t))(3\cos(t))\ dt\cr
&= 27 \int_0^{2\pi} \sin^3(t) + \cos^3(t)\ dt\cr
&= 27\left.\left(-{2\over 3}\cos(t) -{1\over 3}\sin^2(t)\cos(t)
+{2\over 3}\sin(t) + {1\over 3}\cos^2(t)\sin(t)\right)\right|_0^{2\pi}\cr
&=0.\cr}$$ 
The double integral over $D$ is 
$$\int\!\int_D (F_2)_x - (F_1)_y\ dA = \int\!\int_D 2x + 2y\ dA.$$
This can be evaluated in a number of ways.  Switching to 
polar coordinates, for instance,
$$ = \int_0^{2\pi}\int_0^3 2r^2(\cos\theta +\sin\theta)\ dr\ d\theta = 0$$
since both $\int_0^{2\pi} \cos\theta\ d\theta = 0$
and $\int_0^{2\pi} \sin\theta\ d\theta = 0$.
\bigskip
\noindent
VIII.  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 $U$.
\item{A)} 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.
\sol
Every nondegenerate critical point of a 
harmonic function is a {\it saddle point}
because the Hessian matrix is
$$D^2(f) = \pmatrix{f_{xx} & f_{xy} \cr f_{xy} & f_{yy}\cr}
 = \pmatrix{f_{xx} & f_{xy} \cr f_{xy} & -f_{xx}\cr}.$$
 The determinant here is $-(f_{xx})^2 - (f_{xy})^2 < 0$.  
 \medskip
\item{B)} If $f$ is harmonic, what is true about the line
integral of the vector field
$${\bf F}(x,y) =
\left(-f_y, f_x\right)$$
around any simple closed curve in $U$?
\sol
Let $D$ be the region bounded by the simple
closed curve.  By Green's theorem, the integral is equal to
$$\int\!\int_D (f_x)_x - (-f_y)_y\ dA = \int\!\int_D f_{xx} + f_{yy}\ dA = 0.$$ 

\bigskip
\noindent
{\bf Extra Credit}  (10)
\bigskip
Suppose you follow a flow
line of the vector field $-\nabla f$ for $f(x,y)$ in the $xy$-plane.
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.  What does the vector field $-\nabla f$ look like
near a local maxmimum of $f$?  near a local minimum of $f$?
\sol
You are always going {\it downhill} by the most direct route -- 
recall $\nabla f(a,b)$ gives the direction in which 
$f$ is increasing the fastest.  The negative gradient vector
field near a local maximum will have all arrows pointing
away from the critical point (flow lines will diverge from
the maximum).  Near a local minimum, the negative gradient vector
field will be pointing toward from the crirical point
(flow lines will be converging toward the minimum).
\bigskip
\centerline{\it Have a peaceful and joyous holiday season!}
\bye
