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\begin{document}
	
	\mbox{}
	\title{Vector Calculus (MATH 53)}
	
	\maketitle
	
	
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%\section{Introduction} \label{sec:intro}
	%\noindent {\footnotesize Note: Most of the problems were taken from the textbook} \cite{jS15}.
	
\section{Parametric Curves}

TODO: add to surfaces problem: Is the surface $x^2+4xy+3yz+z^2=0 x 2 + 4 x y + 3 y z + z 2 = 0$  some kind of cone (a union of lines through the origin)? 

\begin{problem} [Cal Final, Fall 09]
	Consider the parameterized curve
	\[
		x = te^{-t}, \quad \quad y = e^{3t}.
	\]
	\begin{enumerate}
		\item What is the equation of the tangent line of the curve at the point $(0,1)$?
		\item Find all points $(x,y)$ on the curve at which the curve is vertical.
	\end{enumerate}
\end{problem}

\noindent \underline{Solution:}
	(1) First let us compute the slope:
	\[
		\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3e^{3t}}{e^{-t}(1-t)} = \frac{3e^{4t}}{1-t}.
	\]
	At $x = 0$, we have that $t = 0$. Therefore $(dy/dx)(0) = 3$, which implies that the equation of the line is $y = 3x + 1$.
	
	(2) The curve is vertical when $dx/dt = 0$ and $dy/dt \neq 0$. Then $e^{-t}(1-t) = 0$, which implies $t=1$. As $(dy/dt)(1) \neq 0$, it follows that the only point where the curve is vertical is $(e^{-1}, e^3)$ when $t=1$.
\hfill \qed

\begin{problem}
	Find the length of the curve
	\[
		x = e^t + e^{-t}, \quad \quad y = 2018 + 2t.
	\]
\end{problem}

\noindent \underline{Solution:}
	Since $dx/dt = e^t - e^{-t}$ and $dy/dt = 2$, the length of the curve is given by
	\[
		\int_0^1 \sqrt{(e^t - e^{-t})^2 + 4} \, dt = \int_0^1 \sqrt{e^{2t} + 2 + e^{-2t}} \, dt = \int_0^1 (e^t + e^{-t}) \, dt = e - e^{-1}.
	\]
\hfill \qed

\section{Polar Coordinates}

\begin{problem}
	Suppose that the polar curve $r = f(\theta)$, where $0 \le \theta \le 2\pi$ has length $L$ and encloses a region of area $A$.
	
	\begin{enumerate}
		\item What is the area of the region enclosed by the polar curve $r = 4 f(\theta)$, where $0 \le \theta \le 2\pi$?
		\item What is the length of the polar curve $r = 5 f(\theta)$, where $0 \le \theta \le 2\pi$?
	\end{enumerate}
\end{problem}

\noindent \underline{Solution:}
	Let $A'$ be the area of the region enclosed by the polar curve $r = 4 f(\theta)$, where $0 \le \theta \le 2\pi$, and let $L'$ be the length of the polar curve $r = 5 f(\theta)$, where $0 \le \theta \le 2\pi$. Then
	\[
		(1) \quad \quad \quad \quad A' = \int_0^{2\pi} \frac 12 [4f(\theta)]^2 \, d\theta = 16A 
	\]
	and
	\[
		(2) \quad L' = \int_0^{2\pi} \sqrt{[5 f(\theta)]^2 + \big[\frac{d(5f(\theta))}{d\theta}\big]^2} \, d\theta=  5\int_0^{2\pi} \sqrt{[f(\theta)]^2 + \big[\frac{d(f(\theta))}{d\theta}\big]^2} \, d\theta = 5L.
	\]
\hfill \qed

\begin{problem}
	\begin{enumerate}
		\item Sketch the region that lies inside both of the polar curves $r = 1 + \cos \theta$ and $r = 1 - \cos \theta$.
		\item Compute the area of the region.
	\end{enumerate}
	
\end{problem}

\noindent \underline{Solution:}
	\begin{enumerate}
		\item The red part of the figure shows the desired enclosed region:
		\begin{center}
			\begin{figure}[h]
				\includegraphics[width=10cm]{images/AreaBetweenCosines.JPG} \\
				\caption{Colored red, the region between $r = 1 + \cos \theta$ and $r = 1 - \cos \theta$.}
			\end{figure}
		\end{center}
		\vspace{5pt}
		\item By symmetry, we that the area enclosed by the the two curves is
		\[
			4 \int_0^{\pi/2} \frac 12 \big(1 - \cos \theta \big)^2 \, d\theta = \frac{3 \pi}{2} - 4.
		\]
	\end{enumerate}
	
\hfill \qed
	
\vspace{20pt}

\begin{thebibliography}{20}
	
	\bibitem{jS15} J.~Stewart: \emph{Calculus} 8th Edition, Cengage Learning, Boston 2016.
	
\end{thebibliography}
	
\end{document}