\magnification=\magstep1
\overfullrule=0pt
\def\sol{\medskip\item{}{\bf Solution:}$\ $}
\centerline{Mathematics 241 -- Multivariable Calculus}
\centerline{Final Examination -- December 11, 2007}
\bigskip
\noindent
I.  In this problem,
$P = (1,0,1),\ Q = (-2, 3, 2),\ {\rm and}\ R = (1,2,0).$
\medskip
\item{A)} (10)  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)} (10)  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)}  (5)  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
$$\theta = \cos^{-1}\left(\sqrt{5/19}\right) \doteq 1.032 {\rm\ radians}.$$
\noindent
II.
\medskip
\item{A)} (7.5) Let $F: {\bf R}^2_{u,v} \to {\bf R}^2_{x,y}$
be $F(u,v) = (x(u,v),y(u,v)) = (u^2 - v^2, 2uv)$
and let $g: {\bf R}^2_{x,y} \to {\bf R}$ be
$g(x,y) = \sin(x)\cos(y).$  Find the derivative matrix
$D(g\circ F)$ by direct substitution and differentiation.
\sol
The function $(g\circ F)(u,v) = \sin(u^2 - v^2)\cos(2uv)$, so
computing partial derivatives by the product rule
$$\eqalign{{\partial (g\circ F)\over \partial u} &= 2u\cos(u^2 - v^2)\cos(2uv) - 2v\sin(u^2-v^2)\sin(2uv)\cr
{\partial (g\circ F)\over \partial u} &= -2v\cos(u^2 - v^2)\cos(2uv) - 2u\sin(u^2-v^2)\sin(2uv)\cr}$$
and the derivative matrix is the $1\times 2$ matrix with these entries.
\item{B)} (7.5) Compute $D(g\circ F)$ for
the functions in part A using the Chain Rule
and show you get the same result as in part A.
\sol
The Chain Rule says $D(g\circ F) = D(g)(F(u,v))D(F)(u,v)$ (matrix product).  So we compute
$D(g) = \pmatrix{\cos(x)\cos(y) & -\sin(x)\sin(y)\cr}$ and
$$D(F) = \pmatrix{2u & -2v\cr 2v & 2u\cr}.$$
Hence the product matrix to be computed is
$$\pmatrix{\cos(u^2 - v^2)\cos(2uv) & -\sin(u^2-v^2)\sin(2uv)} \pmatrix{2u & -2v\cr 2v & 2u\cr}$$
which gives a $1\times 2$ matrix with first entry
$$2u\cos(u^2 - v^2)\cos(2uv) - 2v\sin(u^2-v^2)\sin(2uv)$$
and second entry
$$-2v\cos(u^2 - v^2)\cos(2uv) - 2u\sin(u^2-v^2)\sin(2uv)$$
as in the matrix $D(g\circ F)$.
\medskip
\item{C)} (10)  Now let $z = g(x,y)$ for a {\it general
function $g$ -- not specifically $g(x,y)$ as above in part A},
and $x = u^2 - v^2$,
$y = 2uv$ as in the definition of $F$ above. Show
that
$${1\over 4(u^2+v^2)}\left((z_u)^2 + (z_v)^2\right)
= (z_x)^2 + (z_y)^2.$$
\sol
Now we use the expanded form of the Chain Rule:  We have
$z_u = z_x x_u + z_y y_u$ and $z_v = z_x x_v + z_y y_v$ so
$$\eqalign{(z_u)^2 + (z_v)^2 &= (z_x x_u + z_y y_u)^2 + (z_x x_v + z_y y_v)^2\cr
&= (2u z_x + 2v z_y)^2 + (-2v z_x + 2u z_y)^2\cr
&= 4u^2 (z_x)^2  + 8uv z_x z_y + 4v^2 (z_y)^2 + 4v^2 (z_x)^2 - 8uv z_x z_y + 4u^2 (z_y)^2\cr
&= 4(u^2 + v^2)((z_x)^2 + (z_y)^2)\cr}$$
Dividing by $4(u^2 + v^2)$ yields the desired equation.
\bigskip
\noindent
III.  All parts of this problem refer to $f(x,y) = (x+1)^2 + y^2$.
\medskip
\item{A)} (5) 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)}  (10) 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)}  (15)  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)} (10) 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)} (10) 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)}  (20) 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
$$D^2(f)(1/2,0) = \pmatrix{-4e^{-1/2} & 0\cr
                              0 & -e^{-1/2} \cr}$$
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
$$D^2(f)(-1/2,0) = \pmatrix{4e^{-1/2} & 0\cr
                              0 & e^{-1/2} \cr}$$
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)} (5) 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)}  (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 $y$.
\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)}  (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
$x$.
\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 slug
having the shape of the region in
${\bf R}^3$ with $x^2 + y^2 \le 4$ and
$-1\le z \le 1$ has density $\delta(x,y,z) = 5 + x$
at all points.  Determine the coordinates of its center of
mass.
\sol
We will set up the triple integrals to compute the 
coordinates of the center of mass using {\it cylindrical}
coordinates, since the slug is just a cylinder with
axis along the $z$-axis.  The total mass is 
$$\eqalign{
M &= \int_0^{2\pi}\int_0^2\int_{-1}^1 (5 + r\cos\theta) r dz\ dr\ d\theta\cr
&= 2\int_0^{2\pi} \int_0^2 5r + r^2\cos\theta\ dr\ d\theta\cr
&= 2 \int_0^{2\pi} \left. {5\over 2}r^2 + {1\over 3}r^3\cos\theta\right|_0^2\ d\theta\cr
&=2\int_0^{2\pi} 10 + {8\over 3}\cos\theta\ d\theta\cr
&= \left. 20\theta + {8\over 2}\sin\theta\right|_0^{2\pi}\cr 
&=40\pi.}$$
Since the density does not depend on $y$ or $z$, by the 
symmetry of the cylinder, it can be seen that 
$\overline{y} = \overline{z} = 0$.  To compute the $x$-coordinate,
$$\eqalign{
\overline{x} &= {1\over 40\pi} \int_0^{2\pi} \int_0^2 \int_{-1}^1
r\cos\theta(5 + r\cos\theta) r\ dz\ dr\ d\theta\cr
&={1\over 40\pi} \cdot 2\int_0^{2\pi}\int_0^2 
5r^2\cos\theta + r^3\cos^2\theta\ dr\ d\theta\cr
&= {1\over 20\pi} \int_0^{2\pi} \left.
{5\over 3} r^3\cos\theta + {1\over 4}r^4 \cos^2\theta\right|_0^2\ d\theta\cr
&={1\over 20\pi}\int_0^{2\pi} {40\over 3}\cos\theta + 4\cos^2\theta\ d\theta\cr
&={1\over 20\pi}\left.\left({40\over 3}\sin\theta + 
2\theta + \cos(2\theta)\right)\right|_0^{2\pi}\cr
&={1\over 20\pi} \left(0 + 4\pi + 1 - 0 - 0 - 1\right)\cr
&={1\over 5}.\cr}$$
The values of $\overline{y}$ and $\overline{z}$ can 
also be computed directly as above, replacing the 
$x = r\cos\theta$ by $y = r\sin\theta$ and $z$ respectively.
\bigskip
\noindent
VII.
\medskip
\item{A)} (10)  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 ${\bf F}(x,y) = (M(x,y),N(x,y))$ is a $C^1$ vector
field on $D$, then
$$\oint_{\partial D} M dx + Ndy  = \int\!\int_D N_x - M_y\ dA.$$ 
\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 $D$, 
$$\eqalign{
\oint_{\partial D} M dx + N dy 
&= \int_0^{2\pi} (3\cos(t) - 9\sin^2(t))(-3\sin(t)) \cr
&\qquad + (9\cos^2(t) + 3\sin(t))(3\cos(t))\ dt\cr
&= 27 \int_0^{2\pi} \sin^3(t) + \cos^3(t)\ dt\cr
&= 27\left.\left(-{2\over 3}\cos(t) -{1\over 3}\sin^2(t)\cos(t)
+{2\over 3}\sin(t) + {1\over 3}\cos^2(t)\sin(t)\right)\right|_0^{2\pi}\cr
&=0.\cr}$$ 
The double integral over $D$ is 
$$\int\!\int_D N_x - M_y\ dA = \int\!\int_D 2x + 2y\ dA.$$
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 $\int_0^{2\pi} \sin\theta\ d\theta = 0$.
\medskip
\item{C)} (5) We can think of a vector field on ${\bf R}^2$
as a vector field on ${\bf R}^3$ by
making the last component equal to {\it zero} at all points.
Compute ${\rm curl}({\bf F})$ for ${\bf F} = (M(x,y),N(x,y),0)$
and relate your result to
the statement of Green's Theorem.
\sol We have
$${\rm curl}({\bf F}) = \nabla \times (M(x,y),N(x,y),0)$$
Since $M,N$ do not depend on $z$ and because of the 
zero in the last component,
$${\rm curl}({\bf F}) = (0,0,N_x(x,y) - M_y(x,y))$$
The integrand in the double integral in Green's
theorem is just the last component here.  
\medskip
\item{}
{\it Comment:} In fact, Green's theorem is a special
case of a more general result called {\it Stokes'} theorem
in which the plane region is replaced by a surface $S$ in ${\bf R}^3$.
Stokes' theorem then asserts the equality of the 
line integral of ${\bf F}$ over the boundary
of $S$ and the surface integral of ${\rm curl}({\bf F})$
over $S$.  The surface
integral involves integrating ${\rm curl}({\bf F})\cdot {\bf N}$. 
where ${\bf N}$ is a normal vector to the surface.
The normal vector for $D$ in ${\bf R}^2$
is just ${\bf k} = (0,0,1)$ at each point, and ${\rm curl}({\bf F})\cdot {\bf k}$ is exactly the integrand of the double integral in 
Green's theorem.
\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)} (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 Hessian criterion.
\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)} (5)  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}  (20)
\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 uphill} by the most direct route -- 
recall $\nabla f(a,b)$ gives the direction in which 
$f$ is increasing the fastest.  The gradient vector
field near a local maximum will have all arrows pointing
in toward the critical point (flow lines will converge toward
the maximum).  Near a local minimum, the gradient vector
field will be pointing away from the crirical point
(flow lines will be diverging away from the minimum).
\bigskip
\centerline{\it Have a peaceful and joyous holiday season!}
\bye
