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Math Exam

Mathematics examination template with point values and solution spaces

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Math

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Free to use (MIT)

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math-exam/main.tex

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\documentclass[11pt,letterpaper]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{geometry}
\usepackage{amsmath,amssymb}
\usepackage{enumitem}
\usepackage{fancyhdr}
\usepackage{lastpage}
\usepackage{array}
\usepackage{tabularx}
\usepackage{xcolor}
\usepackage{tikz}
\usepackage{pgffor}

\geometry{margin=1in}

\pagestyle{fancy}
\fancyhf{}
\fancyhead[L]{\textbf{MATH 241 --- Calculus III}}
\fancyhead[R]{\textbf{Midterm Examination}}
\fancyfoot[L]{\small Page \thepage\ of \pageref{LastPage}}
\fancyfoot[R]{\small \textit{Continue on next page} $\rightarrow$}
\renewcommand{\headrulewidth}{0.4pt}
\renewcommand{\footrulewidth}{0.4pt}

\newcommand{\answerspace}[1]{\vspace{#1}}
\newcommand{\answerline}{\vspace{8pt}\noindent\rule{\linewidth}{0.4pt}\vspace{8pt}}
\newcommand{\pointbox}[1]{\hfill\framebox[1.2cm]{\hfill\textbf{/#1}}}

\newcounter{problemnum}
\newcommand{\problem}[1]{%
  \stepcounter{problemnum}%
  \vspace{10pt}%
  \noindent\textbf{Problem \theproblemnum.} \hfill \textbf{[#1 points]}\\[4pt]%
}

\begin{document}

\thispagestyle{fancy}

\begin{center}
  \rule{\textwidth}{1.5pt}\\[10pt]
  {\LARGE\bfseries MATH 241 --- Calculus III}\\[6pt]
  {\Large Midterm Examination}\\[4pt]
  {\large Fall 2025}\\[10pt]
  \rule{\textwidth}{1.5pt}
\end{center}

\vspace{8pt}

\noindent
\begin{tabularx}{\textwidth}{|X|X|}
\hline
\textbf{Name:} \rule{0pt}{20pt} & \textbf{Student ID:} \\[10pt]
\hline
\textbf{Section:} \rule{0pt}{20pt} & \textbf{Instructor:} \\[10pt]
\hline
\end{tabularx}

\vspace{12pt}

\noindent\textbf{Instructions:}
\begin{itemize}[nosep,leftmargin=*]
  \item This exam consists of \textbf{three sections} with a total of \textbf{100 points}.
  \item You have \textbf{75 minutes} to complete the exam.
  \item Show all work for full credit. Partial credit will be awarded where appropriate.
  \item No calculators, notes, or electronic devices are permitted.
  \item Write clearly and box your final answers.
\end{itemize}

\vspace{8pt}

\begin{center}
\renewcommand{\arraystretch}{1.4}
\begin{tabular}{|c|c|c|}
\hline
\textbf{Section} & \textbf{Points Possible} & \textbf{Score} \\
\hline
I --- Multiple Choice & 24 & \hspace{1.5cm} \\
\hline
II --- Short Answer & 36 & \hspace{1.5cm} \\
\hline
III --- Long Problems & 40 & \hspace{1.5cm} \\
\hline
\textbf{Total} & \textbf{100} & \hspace{1.5cm} \\
\hline
\end{tabular}
\end{center}

\vspace{12pt}
\hrule
\vspace{12pt}

% ============================================================
\section*{Section I: Multiple Choice \hfill (24 points)}
{\small Circle the single best answer. Each question is worth \textbf{4 points}. No partial credit.}

\vspace{8pt}

\setcounter{problemnum}{0}

\begin{enumerate}[label=\textbf{\arabic*.},leftmargin=*,itemsep=14pt]

\item The gradient of $f(x,y) = x^2 y - 3xy^2$ at the point $(1,2)$ is:

\begin{enumerate}[label=(\alph*),itemsep=2pt]
  \item $(-8, -11)$
  \item $(-8, -7)$
  \item $(0, -11)$
  \item $(-4, -11)$
\end{enumerate}

\item Which of the following vector fields is conservative?

\begin{enumerate}[label=(\alph*),itemsep=2pt]
  \item $\mathbf{F} = \langle y, -x \rangle$
  \item $\mathbf{F} = \langle 2xy, x^2 \rangle$
  \item $\mathbf{F} = \langle y^2, x \rangle$
  \item $\mathbf{F} = \langle x+y, x-y \rangle$
\end{enumerate}

\item The directional derivative of $f(x,y,z) = xyz$ at $(1,1,1)$ in the direction of $\mathbf{v} = \langle 1,1,1 \rangle$ is:

\begin{enumerate}[label=(\alph*),itemsep=2pt]
  \item $\sqrt{3}$
  \item $3$
  \item $3/\sqrt{3}$
  \item $1$
\end{enumerate}

\item The equation $x^2 + y^2 - z^2 = 1$ describes a:

\begin{enumerate}[label=(\alph*),itemsep=2pt]
  \item Ellipsoid
  \item Hyperboloid of one sheet
  \item Hyperboloid of two sheets
  \item Cone
\end{enumerate}

\item The curvature of $\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle$ at any point is:

\begin{enumerate}[label=(\alph*),itemsep=2pt]
  \item $1$
  \item $1/2$
  \item $\sqrt{2}/2$
  \item $1/\sqrt{2}$
\end{enumerate}

\item $\displaystyle\int_0^1\!\int_0^{\sqrt{x}} dy\, dx$ equals:

\begin{enumerate}[label=(\alph*),itemsep=2pt]
  \item $1/2$
  \item $2/3$
  \item $1/3$
  \item $1$
\end{enumerate}

\end{enumerate}

\newpage

% ============================================================
\section*{Section II: Short Answer \hfill (36 points)}
{\small Show your work. Write your final answer in the box provided. Each question is worth \textbf{6 points}.}

\vspace{8pt}

\setcounter{problemnum}{0}

\begin{enumerate}[label=\textbf{\arabic*.},leftmargin=*,itemsep=6pt]

\item Find the equation of the tangent plane to $z = x^2 + y^2$ at the point $(1, 2, 5)$.

\answerspace{3cm}
\noindent\fbox{\parbox{\dimexpr\linewidth-2\fboxsep-2\fboxrule}{%
  \textbf{Answer:}\vspace{1.5cm}}}

\vspace{6pt}

\item Evaluate $\displaystyle\int_0^2\!\int_0^x (x^2 + y)\, dy\, dx$.

\answerspace{3cm}
\noindent\fbox{\parbox{\dimexpr\linewidth-2\fboxsep-2\fboxrule}{%
  \textbf{Answer:}\vspace{1.5cm}}}

\vspace{6pt}

\item Find all critical points of $f(x,y) = x^3 - 3x + y^2 - 2y$ and classify each as a local maximum, local minimum, or saddle point.

\answerspace{3.5cm}
\noindent\fbox{\parbox{\dimexpr\linewidth-2\fboxsep-2\fboxrule}{%
  \textbf{Answer:}\vspace{1.5cm}}}

\vspace{6pt}

\item Convert the integral $\displaystyle\int_{-1}^{1}\!\int_{0}^{\sqrt{1-x^2}} (x^2+y^2)\, dy\, dx$ to polar coordinates and evaluate.

\answerspace{3cm}
\noindent\fbox{\parbox{\dimexpr\linewidth-2\fboxsep-2\fboxrule}{%
  \textbf{Answer:}\vspace{1.5cm}}}

\vspace{6pt}

\item Compute $\nabla \times \mathbf{F}$ where $\mathbf{F} = \langle yz,\, xz,\, xy \rangle$.

\answerspace{3cm}
\noindent\fbox{\parbox{\dimexpr\linewidth-2\fboxsep-2\fboxrule}{%
  \textbf{Answer:}\vspace{1.5cm}}}

\vspace{6pt}

\item Find the arc length of $\mathbf{r}(t) = \langle 2t,\, t^2,\, \tfrac{1}{3}t^3 \rangle$ for $0 \le t \le 1$.

\answerspace{3cm}
\noindent\fbox{\parbox{\dimexpr\linewidth-2\fboxsep-2\fboxrule}{%
  \textbf{Answer:}\vspace{1.5cm}}}

\end{enumerate}

\newpage

% ============================================================
\section*{Section III: Long Problems \hfill (40 points)}
{\small Show all work clearly. Justify your answers. Partial credit is available.}

\vspace{10pt}

\setcounter{problemnum}{0}

\noindent\textbf{Problem 1.} \hfill \textbf{[15 points]}

\medskip
\noindent Use Lagrange multipliers to find the maximum and minimum values of $f(x,y) = 2x + 3y$ subject to the constraint $x^2 + y^2 = 13$.

\begin{enumerate}[label=(\alph*),itemsep=4pt]
  \item Set up the Lagrange equations. \hfill [5 pts]
  \answerspace{4cm}
  \item Solve the system for all critical points. \hfill [6 pts]
  \answerspace{5cm}
  \item Identify which points give the maximum and minimum values. \hfill [4 pts]
  \answerspace{3cm}
\end{enumerate}

\vspace{12pt}
\hrule
\vspace{12pt}

\noindent\textbf{Problem 2.} \hfill \textbf{[12 points]}

\medskip
\noindent Let $D$ be the region bounded by $y = x$ and $y = x^2$ for $0 \le x \le 1$.

\begin{enumerate}[label=(\alph*),itemsep=4pt]
  \item Sketch the region $D$ and determine the limits of integration. \hfill [3 pts]
  \answerspace{4cm}
  \item Evaluate $\displaystyle\iint_D (x + 2y)\, dA$. \hfill [9 pts]
  \answerspace{6cm}
\end{enumerate}

\vspace{12pt}
\hrule
\vspace{12pt}

\noindent\textbf{Problem 3.} \hfill \textbf{[13 points]}

\medskip
\noindent Let $\mathbf{F} = \langle P, Q \rangle = \langle x^2 - y,\, x + y^2 \rangle$ and let $C$ be the boundary of the triangle with vertices $(0,0)$, $(1,0)$, and $(0,1)$, oriented counterclockwise.

\begin{enumerate}[label=(\alph*),itemsep=4pt]
  \item State Green's Theorem. \hfill [3 pts]
  \answerspace{2.5cm}
  \item Verify that $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ is computed correctly. \hfill [3 pts]
  \answerspace{2.5cm}
  \item Use Green's Theorem to evaluate $\displaystyle\oint_C \mathbf{F} \cdot d\mathbf{r}$. \hfill [7 pts]
  \answerspace{5cm}
\end{enumerate}

\vfill
\begin{center}
  \rule{0.5\textwidth}{1pt}\\[4pt]
  \textbf{END OF EXAM}\\
  {\small Total Points: 100}
\end{center}

\end{document}
Bibby Mascot

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