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\documentclass[12pt]{article}
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\def\i{{\;\bf i}}
\def\j{{\;\bf j}}
\def\k{{\;\bf k}}

\oddsidemargin 5pt \evensidemargin 5pt \marginparwidth 68pt
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\pagestyle{myheadings}
\def\la{\langle}  \def\ra{\rangle}
\def\header{{\hfill Exam 1 \hfill A. Miller \hfill Fall 2000
                                       \hfill Math 234\hfill}}
\markboth\header\header
\newcount\probno
\newcount\total
\newwrite\examTBLFile

\probno=0
\def\prob#1{\advance\probno by 1  \vfill\newpage
%\def\prob#1{\advance\probno by 1 \par
    \noindent\the\probno . (#1 pts)\advance\total by #1
    \immediate\write\examTBLFile{\the\probno
       \string& #1 \string&
       \string\qquad \string\\ \string\hline}}

\def\probx#1{\advance\probno by 1  \vfill  % \par
    \noindent\the\probno . (#1 pts)\advance\total by #1
    \immediate\write\examTBLFile{\the\probno
       \string& #1 \string&
       \string\qquad \string\\ \string\hline}}

\def\finishtbl{
    \immediate\write\examTBLFile{Total \string& \the\total
    \string& \qquad \string\\ \string\hline}
     }

\newwrite\ans
\immediate\openout\ans=answers
\outer\def\answer{\par\medbreak
  \immediate\write\ans{}
  \immediate\write\ans{\string\bigskip \the\probno . }
  \copytoblankline}
\outer\def\putanswer{\par\medbreak
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\def\copytoblankline{\begingroup\setupcopy\copyans}
\def\setupcopy{\def\do##1{\catcode`##1=12}\dospecials
     \catcode`\|=12\obeylines}
{\obeylines \gdef\copyans#1
  {\def\next{#1}%
  \ifx\next\empty\let\next=\endgroup%
  \else\immediate\write\ans{\next}\let\next=\copyans\fi\next}}
\def\printanswers{\newpage\begin{center} Answers \end{center}
    \immediate\closeout\ans\input answers}

\begin{document}

\bigskip

\begin{quote}
{
No calculator, no cell phones, no pagers, no electronic devices at all.
\par You may bring in a single ``cheat sheet''.  One ordinary 8.5 x 11 sheet
of paper with anything you want written on both sides. } \end{quote}

\par\vskip .25in
\begin{center}
Name\makebox[3in]{\hrulefill} \\
\end{center}

\begin{center}
{\bf Hand in to your TA -- Circle your TA section:} \\
\end{center}
\begin{center}
\begin{tabular}{l|l|l}
DIS 321 & T 8:50    &  Milind Shah            \\
DIS 322 & R 8:50    &  Milind Shah            \\
DIS 323 & T 9:55    &  Milind Shah            \\
DIS 324 & R 9:55    &  Milind Shah            \\
DIS 325 & T 11:00   &  Duygu Unlu            \\
DIS 326 & R 11:00   &  Duygu Unlu            \\
DIS 327 & T 12:05p  &  Duygu Unlu            \\
DIS 328 & R 12:05p  &  Duygu Unlu            \\
DIS 329 & T 1:20p   &  Kiran Manchikanti     \\
DIS 330 & R 1:20p   &  Kiran Manchikanti     \\
DIS 331 & T 2:25p   &  Lipika Deka            \\
DIS 332 & R 2:25p   &  Lipika Deka            \\
DIS 333 & T 3:30p   &  Lipika Deka            \\
DIS 334 & R 3:30p   &  Lipika Deka            \\
\end{tabular}
\end{center}

% end file with \finishtbl
%\bigskip

\begin{center} \Large
 \begin{tabular}{||c|c|c||} \hline\hline
  Problem & Points & Score \\  \hline\hline
  \input{exam.tbl}
  \hline \hline
 \end{tabular}
\end{center}

\immediate\openout\examTBLFile=exam.tbl

\setcounter{page}{0}


\prob{10}  % 11.7-31
Find the domain of the vector function:
$${\bf r}(t)=\ln(9-t^2)\i \;\;+\;\;\sqrt{1+t}\j \;\;+\;\; \sin(t)\k$$
\answer $D=\{t:-1\leq t<3\}$

\prob{10}  % 11.8-37
For what value of $t$ is the normal plane to the curve
$${\bf r}(t)=(t^3,t^2,5t)$$
parallel to the plane $6x + 4y + 10z = 5$?
\answer $t=1$.  ${\bf r}^\prime(1)=(3,2,5)$ which is parallel to
$(6,4,10)$.

\prob{10}  % 11.9-5
Find the {\bf velocity, acceleration, and speed} of a particle with the
position function:
 $${\bf r}(t)=\cos(t)\i \;\;+\;\; \sin(t)\j \;\;+\;\; t\k$$
Sketch the path of the particle for $t\geq 0$.  Include in your sketch (at
any point of your choice) the tangent ${\bf T}$, normal
${\bf N}$, and binormal ${\bf B}$.
\answer This is a helix see page 705.  The normal vector and
the acceleration vector are the same and point directly at the $z$-axis.
\par ${\bf v}={d{\bf r}\over dt}
=-\sin(t)\i \;\;+\;\; \cos(t)\j \;\;+\;\; \k$
\par ${\bf a}={d{\bf v }\over dt} =-\cos(t)\i \;\;+\;\; -\sin(t)\j$

\prob{10} % 12.1-27
Find and sketch the domain of the function:
$$f(x,y)=\ln(y-x^2)$$
\answer $D=\{(x,y): y>x^2\}$. This is the set of points in the plane
strictly above the parabola.

\prob{10} % 12.2-45
Determine whether the function below is continuous at $(0,0)$
and prove it:
$$f(x,y)=\left\{
\begin{array}{cl}
{xy\over x^2+2y^2}      & \mbox{ if } (x,y)\not=(0,0) \\
         0              & \mbox{ if } (x,y)=(0,0)     \\
\end{array}\right.$$
\answer It is discontinuous at $(0,0)$.  Along the line $x=0$
the limit is $0$ but along the line $y=x$ the limit is $1/3$.

\prob{10}  % 12.3-57
Find all the {\bf first} and {\bf second} order partial derivatives of
the function
$$f(x,y)=Ax^2+Bxy+Cy^2+Dx+Ey+F$$
\answer \par
$f_x=2Ax+By+D$ \par
$f_{xx}=2A$ \par
$f_y=Bx+2Cy+E$ \par
$f_{yy}=2C$ \par
$f_{xy}=f_{yx}=B$


\prob{10}  % 12.4-37
Show that the function $f(x,y)=x^2y$ is (Frechet) differentiable
at every point $(x,y)$ by finding the values $\epsilon_1$ and
$\epsilon_2$ where $\epsilon_1\triangle x+\epsilon_2\triangle y$
is the error term in the definition of differentiable.
\answer $\epsilon_1\triangle x\;\;+\epsilon_2\triangle y=
(\triangle x)^2y \;\;+\;\; 2 x\triangle x\triangle y  \;\;+\;\;
(\triangle x)^2 \triangle y$
\par Many choices are possible, for example,
$\epsilon_1= (\triangle x) y \;\;+\;\; 2 x\triangle y  \;\;+\;\;
\triangle x \triangle y$ and $\epsilon_2=0$.


\prob{10} % 12.5-35
The height $y$ and width $x$ of a rectangle change
with time.  At a certain instant the dimensions are $x=30$
and $y=40$ feet.  At that same time $x$ is increasing at the
rate of $4$ feet per second, and $y$ is increasing
at the rate of $2$ feet per second.  Find the rate at which the
the diagonal of the rectangle is increasing.
\answer 4 feet per second

\prob{10} % 12.6-17
Find the maximum rate of change of $f(x,y)=ye^x+y$ at $(0,1)$.
\answer $\sqrt{5}$

\prob{10} % 12.7-39
Find the point (or points) on the surface $z^2=y-2x+1$ that is
closest to the origin.
\answer The minimum of $g(x,y)=x^2+y^2+(y-2x+1)$ in the region
$$D=\{(x,y): y-2x+1\geq 0\}$$
occurs on the boundary of $D$, i.e. on the line $y-2x+1=0$.
The critical point of $g$ is not in the region $D$. The closest point is
$x=2/5$, $y=-{1/5}$ and $z=0$.

\finishtbl
\printanswers

\end{document}