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\begin{document}
\begin{center}
Math 311 \\
Final Exam Key\\
Dec. 13, 1995
\end{center}
\q{20}The following questions refer to the system of equations given below.
\begin{eqnarray*}
x_1+x_2+x_3&=&5\\
x_1-x_2-2x_3&=&0\\
x_1+x_2+2x_3&=&8\\
x_1-x_2+x_3&=&9
\end{eqnarray*}
\begin{list}{\alph{bean}. }{\usecounter{bean}}
\item Does this system of equations have a solution? If yes, find it,
if not, explain how you came to that conclusion. 
\begin{quote}
The system has a unique solution: $x_1=4$, $x_2=-2$, and $x_3=3$.
\end{quote}
\item Assume the system does not have a solution. Find a least squares
solution by use of a $QR$ factorization of the coefficient matrix
$A$. Be sure to explain what you are doing. 
\begin{quote}
The columns of the matrix $Q$ are an orthonormal basis for the column
space of the coefficient matrix of the system, and the matrix
$R=Q^TA$. Thus,
$$
Q\approx\left[\begin{array}{rrr}0.5&0.5&-0.224\\0.5&-0.5&-0.671\\0.5&0.5&0.224
\\0.5&-0.5&0.671\end{array}\right],\quad
R\approx\left[\begin{array}{rrr}2&0&1\\0&2&2\\0&0&2.236\end{array}\right]\,.
$$
The normal equation, after simplification, becomes
$R\vec{x}=Q^T\vec{b}$. Solving this system, we again get $x_1=4$,
$x_2=-2$, and $x_3=3$.
\end{quote}
\end{list}
\Sk
\q{30}Let $\Omega$ be the  solid sphere centered at the origin of
radius 2. That is, $\Omega$ consists of all points $(x,y,z)$ such that
$x^2+y^2+z^2\leq 4$. Let $\rho(x,y,z) = 3 +x^2+yz$ be the mass density
function of the sphere $\Omega$. Let $F(x,y,z) = (x^2,y^2,z^2)$ be a
force field defined on the sphere $\Omega$. 
\begin{list}{\alph{bean}. }{\usecounter{bean}}
\item Set up the integral which calculates the total mass of the
sphere $\Omega$. Be sure to clearly indicate the limits of
integration. You may use any
coordinate system you please. Your final answer must be in a form so
that it can be calculated by plug and chug methods, i.e., the HP can
do the calculation. 
\begin{quote}
\begin{eqnarray*}
\mbox{Mass}&=&\int_0^2dr\int_0^{2\pi}d\theta\int_0^\pi
[3+(r^2\sin^2\phi\cos^2\phi) +
(r\sin\phi\sin\theta)r\cos\phi]r^2\sin\phi\,d\phi\\ 
&{}&\\
&=&\int_0^2dr\int_0^{2\pi}d\theta\int_0^\pi
\left(3+r^2\sin^2\phi\cos^2\phi +
r^2\sin\phi\cos\phi\sin\theta\right)r^2\sin\phi\,d\phi\\ 
&{}&\\
&\approx&111.254\,.
\end{eqnarray*}
\end{quote}
\item Set up the integral which calculates the flux of the force field
$F$ over the bounding surface of the sphere $\Omega$
($x^2+y^2+z^2=4$). Be sure to clearly indicate the limits of
integration, and to use an {\bf outward} pointing normal. You may use
any coordinate system you please. Your final answer must be in a form
so that it can be calculated by plug and chug methods, i.e., the HP
can do the calculation.
\begin{quote}
The first task is to find a parametrization of the surface. The most
natural way to do this is with spherical coordinates. Thus, set
$T(\theta,\phi)=
(2\sin\phi\cos\theta,\,2\sin\phi\sin\theta,\,2\cos\phi)$. Computing
the cross product of the two tangent vectors to the surface: $\partial
T/\partial\theta$, $\partial T/\partial\phi$, we get:
$$
\frac{\partial T}{\partial\theta}\times\frac{\partial T}{\phi}=
-4\left(\sin^2\phi\cos\theta,\,\sin^2\phi\sin\theta,\,\sin\phi\cos\phi\right)\,.
$$
Note: this particular normal points into the region. Thus, we will
have to take its negative to get an outward pointing normal.
The flux of the force field, $F$, over the surface is given by:
\begin{eqnarray*}
\mbox{Flux}&=&\int_0^{2\pi}d\theta\int_0^\pi
F(T(\theta,\phi))\cdot\left(\frac{\partial T}{\partial\theta}\times
\frac{\partial T}{\phi}\right)d\phi\\
&{}&\\
&=&\int_0^{2\pi}d\theta\int_0^\pi
\left(4\sin^2\phi\cos^2\theta,4\sin^2\phi\sin^2\theta,4\cos^2\phi\right)\cdot\\
&\quad&
4\left(\sin^2\phi\cos\theta,\sin^2\phi\sin\theta,\sin\phi\cos\phi\right)\,d\phi\\
&{}&\\
&=&16\int_0^{2\pi}d\theta\int_0^\pi\left(\sin^4\phi\cos^3\theta+\sin^4\phi\sin^3\theta+\sin\phi\cos^3\phi\right)\,d\phi\\
&=&0\,.
\end{eqnarray*}
\end{quote}  
\item Find the maximum and minimum values of the mass density function
on the sphere $\Omega$. 
\begin{quote}
The locations of the extreme values of the density function must occur
at a place where its gradient is zero or on the boundary of the
region $\Omega$. The gradient of the density function $\rho$ is equal
to $\del\rho=(2x,z,y)$. This is zero only at the origin. The Hessian
of the density function evaluated at the origin has three real
eigenvalues not all of the same sign. Thus, the origin is a saddle
point and the extreme values of $\rho$ must lie on the surface
$g(x,y,z)=x^2+y^2+z^2=4$. Using Lagrange multipliers we have the
equation:
$$
\del\rho +\lambda\del g= (2x,z,y)+\lambda(2x,2y,2z)=\vec{0}\,.
$$
This leads to the scalar equations: $2x(1+\lambda)=0$, $z+2\lambda
y=0$, and $y+2\lambda z=0$. We will analyze these equations by
considering the two cases $x=0$ and $x\not=0$.
\item Case $x=0$. This implies that both $y$ and $z$ must be
nonzero. Solving the last two equations for $\lambda$ and then setting
the resulting two equations equal to each other, we have $z/2y = y/2z$. This implies
that $y^2=z^2$, and the constraint equation $g=4$ forces
$y=\pm\sqrt{2}$ and $z=\pm\sqrt{2}$. Thus, we need to evaluate the
function $\rho$ at the points $(0,\pm\sqrt2,\pm\sqrt2)$. Since the
function $\rho$ has some symmetries, we will only need to evaluate the
function at the points $(0,\sqrt2,\sqrt2)$ and $(0,-\sqrt2,\sqrt2)$.
\item Case $x\not=0$ This forces $\lambda = -1$, which then forces
$y=z=0$. This then implies that $x=\pm2$. 
\item We now have three points to evaluate $\rho$ at, and we know that
the extreme values of 
$\rho$ must occur at these points. $\rho(2,0,0)=7$,
$\rho(0,\sqrt2,\sqrt2)=5$, and $\rho(0,-\sqrt2,\sqrt2)=1$. Thus, the
maximum value of the density function on the set $\Omega$ is 7, and
its minimum value is 1.
\end{quote}
\end{list}
\Sk
\q{25}Let $V=\{\,(x_1,x_2,x_3,x_4):\, x_1-x_2+x_3-x_4=0,\
x_1+2x_3+3x_4=0\,\}$. Let $L$ be the linear transformation which
reflects $R^4$ through the subspace $V$.
\begin{list}{\alph{bean}. }{\usecounter{bean}}
\item Show that $V$ is a subspace of $R^4$.
\begin{quote}
$V$ is the null space of the matrix $A=\left[\begin{array}{rrrr}
1&-1&1&-1\\1&0&2&3\end{array}\right]$. Thus, $V$ is a subspace.
\end{quote}
\item Find the matrix representation of $L$ with respect to the
standard basis.
\begin{quote}
We first need to find a ``nice'' basis for the reflection
operator. Let $v_1=(-2,-1,1,0)$ and $v_2=(-3,-4,0,1)$. These two
vectors form a basis for the vector space $V$. We next want to extend
these vectors to a basis of $R^4$. The two rows of the matrix $A$ are
linearly independent and orthogonal to the null space of the
matrix. That is, they are orthogonal to the subspace $V$. Denote these
two rows by $v_3$ and $v_4$ respectively. Let $B$ be the basis of
$R^4$ which consists of the four vectors $v_i$. The matrix
representation of the transformation with respect to this basis is 
$$
{\cal A}=\left[\begin{array}{rrrr}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1\end{array}\right]\,.
$$
If $P$ denotes the change of basis matrix such that
$[\vec{x}]_S=P[\vec{x}]_B$, then the matrix representation of the
linear transformation with respect to the standard basis is:
$$
A=P{\cal
A}P^{-1}\approx\left[\begin{array}{rrrr}0.357&0.5&-0.786&0.071\\
0.500&0.5&0.500&-0.500\\-0.786&0.5&-0.071&-0.357\\0.071&-0.5&-0.357&-0.786\end{array}
\right]\,.
$$
\end{quote}
\item $L([1,1,2,0])=$?
\begin{quote}
$L(1,1,2,0)=A[1,1,2,0]^T\approx (-0.714,2.0,-0.429,-1.143)$.
\end{quote}
\newpage
\item What does the kernel of $L$ equal?
\begin{quote}
The kernel of $L$ consists of just the zero vector.
\end{quote}
\item What does the range of $L$ equal?
\begin{quote}
The range of $L$ is all of $R^4$.
\end{quote}
\end{list}
\Sk
\q{15}Suppose that $A$ is a $3\times3$ matrix with eigenvalues 1, 1/2,
and 1/3, and has corresponding eigenvectors: $(1,2,3)$, $(0,1,2)$, and
$(0,2,1)$. 
\begin{list}{\alph{bean}. }{\usecounter{bean}}
\item $A^2\left[\begin{array}{c}0\\1\\2\end{array}\right]=$?
\begin{quote}
$A^2\left[\begin{array}{c}0\\1\\2\end{array}\right]=\frac14
\left[\begin{array}{c}0\\1\\2\end{array}\right]=
\left[\begin{array}{c}0\\1/4\\1/2\end{array}\right]$.
\end{quote}
\item $A\left[\begin{array}{c}2\\12\\7\end{array}\right]=$?
\begin{quote}
One way to compute this vector is to write $[2,12,7]$ as a
linear combination of the eigenvectors of $A$. The easiest way
to do this is to use a change of basis matrix. Thus, we have
\begin{eqnarray*}
A\left[\begin{array}{c}2\\12\\7\end{array}\right]&=&A\left(
2\left[\begin{array}{c}1\\2\\3\end{array}\right]
-2\left[\begin{array}{c}0\\1\\2\end{array}\right]
+5\left[\begin{array}{c}0\\2\\1\end{array}\right]\right)\\
&=&
\left(\left[\begin{array}{c}2\\4\\6\end{array}\right]
-\left[\begin{array}{c}0\\1\\2\end{array}\right]
+\frac53\left[\begin{array}{c}0\\2\\1\end{array}\right]\right)\\
&=&\left[\begin{array}{c}2\\19/3\\17/3\end{array}\right]\,.
\end{eqnarray*}
\end{quote}
\newpage
\item $\ds
\lim_{n\to\infty}A^n\left[\begin{array}{c}2\\12\\7\end{array}\right]=$?
\begin{quote}
\begin{eqnarray*}
\lim_{n\to\infty}A^n\left[\begin{array}{c}2\\12\\7\end{array}\right]&=&
\lim_{n\to\infty}A^n
\left(2\left[\begin{array}{c}1\\2\\3\end{array}\right]
-2\left[\begin{array}{c}0\\1\\2\end{array}\right]
+5\left[\begin{array}{c}0\\2\\1\end{array}\right]\right)\\
&=&\lim_{n\to\infty}
\left(2\left[\begin{array}{c}1\\2\\3\end{array}\right]
-2\left(\frac12\right)^n\left[\begin{array}{c}0\\1\\2\end{array}\right]
+5\left(\frac13\right)^n\left[\begin{array}{c}0\\2\\1\end{array}\right]\right)\\
&=&\left[\begin{array}{c}2\\4\\6\end{array}\right]\,.
\end{eqnarray*}
\end{quote}
\end{list}
\Sk
\q{10}Let $\Gamma(t)=(t+\cos t,\sin t)$. Let $F(x,y)=(x^2,y^2)$.
\begin{list}{\alph{bean}. }{\usecounter{bean}}
\item Sketch the curve $\Gamma(t)$ for $0\leq t \leq \pi$. Note the
limits on $t$.
\begin{center}
\epsfig{file=fig1-final-F95.eps,height=2in,angle=270} \\
\end{center}
\item Compute the work done by the force field $F$ on a particle that
moves along the curve $\Gamma(t)$, for $0\leq t \leq 2\pi$. Note the
limits on $t$.
\begin{quote}
\begin{eqnarray*}
\mbox{Work}&=&\int F\cdot d\Gamma = \int_0^{2\pi}F(\Gamma(t))\cdot
\Gamma^\prime(t)\,dt\\ 
&=&\int_0^{2\pi}\left((t+\cos t)^2,\sin^2t\right)
\cdot\left(1-\sin t,\cos t\right)\,dt \\
&=&\int_0^{2\pi}(t+\cos t)^2(1-\sin t)+\sin^2t(\cos t)\,dt\\
&\approx&128.445\,.
\end{eqnarray*}
One can also evaluate the integral by observing that the force field
is the gradient of the function $\phi(x,y)=(x^3+y^3)/3$. The value of
the path integral equals 
\hfill\break
$\phi(\Gamma(2\pi))-\phi(\Gamma(0))=\phi(2\pi+1,0)-\phi(1,0)$.
\end{quote}
\end{list}
\end{document}