% LaTex2e \documentclass[12pt]{article} \usepackage{amssymb,latexsym} \oddsidemargin 5pt \evensidemargin 5pt \marginparwidth 68pt \marginparsep 10pt \topmargin 10pt \headheight 10pt \headsep 15pt \footskip 35pt \textheight = 625pt \textwidth 440pt \columnsep 10pt \columnseprule 5pt \pagestyle{myheadings} \def\la{\langle} \def\ra{\rangle} \def\header{{\hfill Exam 1 \hfill A. Miller \hfill Spring 2001 \hfill Math 217\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 \immediate\write\ans{} \copytoblankline} \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 Show all work. No books, no notes, no calculator, no cell phones, no pagers, no electronic devices at all. Solutions will be posted shortly after the exam: www.math.wisc.edu/$\sim$miller/m217 \par\vskip .25in \begin{center} Name\makebox[3in]{\hrulefill} \\ \end{center} \begin{center} {\bf Hand in to your TA -- Circle your DIScussion section (column one):} \\ \end{center} \begin{center} \begin{tabular}{l|l|l|l} DIS 321 & 7:45 TR & B309 Van Vleck & Forray, Susan Jane \\ DIS 322 & 8:50 TR & 6224 Soc Sci & Bisgard, James William \\ DIS 324 & 9:55 TR & 123 Ingraham & Bisgard, James William \\ DIS 326 & 11:00 TR & 123 Ingraham & Leete, Jennifer A. \\ DIS 327 & 11:00 TR & 113 Ingraham & Lu, Tianshi \\ DIS 328 & 12:05p TR & 4322 Soc Sci & Leete, Jennifer A. \\ DIS 329 & 12:05p TR & 6228 Soc Sci & Lu, Tianshi \\ DIS 330 & 1:20p TR & 6310 Soc Sci & Forray, Susan Jane \\ DIS 332 & 2:25p TR & B135 Van Vleck & Hur, Youngmi \\ DIS 334 & 3:30p TR & B329 Van Vleck & Hur, Youngmi \\ \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} % from M171 example 3 page 143 A square sheet of tin $12$ inches on a side is used to make an open-top box by cutting a small square of tin from each corner and bending up the sides. How large a square should be cut from each corner in order that the box shall have as large a volume as possible? Let $x$ be the lenght of the side of the small square. Give your answer in the form $x=\ldots$. \answer The volume of the box is $V=x(12-2x)^2$ and is a maximum on $[0,12]$ when $x=2$. \prob{10} % 4.2-6 Solve the following differential equation: $${dy\over dx}=3x^2y^2$$ Give your answer in the form $y=\ldots$ \answer $y={-1\over{x^3+C}}\;\;\;$ Remark: there is also the constant solution $y\equiv 0$. \prob{10} % 4.4-18 Evaluate the following indefinite integral: $$\int\sin^3(3x)\cos(3x)dx$$ \answer $\int\sin^3(3x)\cos(3x)dx={\sin^4(3x)\over 12}+C$ \prob{10} % 4.6-6 Establish the formula $$\sum_{k=1}^n(2k-1)=1+3+5+\cdots+(2n-1)= n^2$$ for every positive integer $n$, by showing \bigskip \par (a) the formula is correct for $n=1$, and \bigskip \bigskip \par (b) if true for $n$, the formula is also true for $n+1$. \answer (a) $\sum_{k=1}^1(2k-1)=2\cdot 1 -1=1=1^2$ \par\bigskip (b) $\sum_{k=1}^{n+1}(2k-1)= \sum_{k=1}^{n}(2k-1)\;\;+(2(n+1)-1)= n^2+(2(n+1)-1)=n^2+2n+1=(n+1)^2$ \prob{10} % 4.7-12 Find the area $A$ between the curve $y=9-x^2$ and the $x$-axis. Give your answer in the form $A=\ldots$. \answer $A=36$ \prob{10} % 4.8 Prove the following part of the fundamental theorem of calculus: \bigskip Suppose $f(x)$ is continuous on the interval $[a,b]$, $F(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$ with the $F^\prime(x)=f(x)$ for all $x$ in $(a,b)$. Let $n$ be a large integer. \bigskip (a) Define $\triangle x=$ \bigskip (b) Define $x_k=$ \bigskip (c) Show there exists $c_k$ with $x_{k-1}\leq c_k\leq x_k$ for $k=1,2,\ldots,n$ and such that $$F(b)-F(a)=\sum_{k=1}^n \; f(c_k)\triangle x$$ \answer (a)$\triangle x={b-a\over n}$ \par (b) $x_k=a+k\triangle x$ \par (c) Using the Mean Value Theorem we have that for every $k$ there exists $c_k$ with $x_{k-1}