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\begin{document}

\input{213-Lecture-Macros}		%% 	macros, settings, et.c


\title{Math 213 - Fundamental Theorem for Line Integrals}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{November 15, 2019}
\maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%	
%	Due Date Reminders
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Reminders}

\begin{itemize}

\item	Homework C7 is due tonight
\item Thanksgiving is coming!

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	Unit Outline
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Unit IV: Vector Calculus}
\begin{center}
\begin{itemize}
\begingroup
\color{red}
\item[{}]			Fundamental Theorem for Line Integrals
\endgroup
\item[{}]			Green's Theorem
\item[{}]			Curl and Divergence
\item[{}]			Parametric Surfaces and their Areas
\item[{}]			Surface Integrals
\item[{}]			Stokes' Theorem, I
\item[{}]			Stokes' Theorem, II 
\item[{}]			The Divergence Theorem
\item[{}]			
\item[{}]			Review
\item[{}]			Review
\item[{}]			Review
\end{itemize}
\end{center}
\end{frame}


\section{Learning Goals}

\begin{frame}
\frametitle{Goals of the Day}
\begin{itemize}
\item	Learn the Vocabulary for Section 16.3
\item	Learn the Fundamental Theorem for Line Integrals
\item	Learn what it means for a line integral to be \emph{independent of path}
\item	Learn how to tell when a vector field $\bfF$ is \emph{conservative} and how to find the
		function $f$ with $\nabla f = \bfF$
\end{itemize}
\end{frame}

\section{Vocabulary}


\begin{frame}
\frametitle{Vocabulary - Open Regions}

\medskip

\begin{tabular}{ll}
\textbf{open region}	&			A region $D$ of $\mathbb{R}^2$ or $\mathbb{R}^3$ where for every point $P$\\
								&			in the region,  there is a disc or sphere centered at $P$\\
								&			 contained in $D$\\[5pt]
\end{tabular}

\bigskip

Which of the following regions is open?

\begin{center}

\begin{minipage}{0.25\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]	(-3,0)	--	(3,0);
\draw[thick,->,>=stealth]	(0,-3)	--	(0,3);
\path[draw=none,fill=orange!20,fill opacity=0.75]
	(1,-3)		rectangle	(3,3);
\draw[dashed,orange]	(1,-3)	--	(1,3);
\node[below]	at	(0,-3)		{\tiny{$\{(x,y): x > 1 \}$}};
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.25\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]	(-3,0)	--	(3,0);
\draw[thick,->,>=stealth]	(0,-3)	--	(0,3);
\draw[orange,thick,fill=orange!20,opacity=0.75]	
	(0,0)	circle(2);
\node[below] at (0,-3)		{\tiny{$\{(x,y): x^2+y^2 \leq 4\}$}};
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.25\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]	(-3,0)	--	(3,0);
\draw[thick,->,>=stealth]	(0,-3)	--	(0,3);
\begin{scope}
\clip	(-3,-3)	rectangle	(3,3);
\path[draw=none,fill=orange!40,fill opacity=0.5,even odd rule]
	(0,0)	circle(6)
	(0,0) circle(2);
\end{scope}
\draw[orange,dashed]
	(0,0)	circle(2);
\node[below]	at	(0,-3)		{\tiny{$\{ (x,y): x^2+y^2 > 4\}$}};
\end{tikzpicture}
\end{minipage}
\end{center}


\end{frame}


\section{Fundamental Theorem}

\begin{frame}
\frametitle{Chain Rule Puzzler}
\footnotesize
If $f(x,y,z)$ is a function and $\bfr(t)=\langle x(t),y(t),z(t)\rangle$ is a parameterized curve,
what is 
$$ \frac{d}{dt} \left[ f(x(t),y(z),z(t)) \right]$$
in terms of $\nabla f$ and $\bfr'(t)$?

\bigskip

\uncover<2->
{
{\color{cyan}  \emph{Answer: }  $ \nabla f (\bfr(t)) \cdot \bfr'(t) $}
}
\end{frame}

\begin{frame}
\frametitle{Remember the Fundamental Theorem of Calculus?}
\footnotesize
What is 
$$ \int_a^b \frac{d}{dt} F(t) \, dt  \ ?$$

\medskip

(Remember the \emph{Net Change Theorem}?)

\bigskip

\uncover<2->
{
{\color{cyan} \emph{Answer}: $F(b) - F(a)$}
}
\end{frame}

\begin{frame}
\frametitle{Line Integral of a Gradient Vector Field}

\footnotesize
Suppose $\bfF = \nabla f$ for a potential function $f(x,y,z)$

\medskip

Suppose $\bfr(t)$, $a \leq t \leq b$ is a parameterized path $C$.

\bigskip

Is there a simple way to compute
$$ \int_C \bfF \cdot d \bfr = \int_a^b \nabla f(\bfr(t)) \cdot \bfr'(t) \, dt = \int_a^b \frac{d}{dt} \left( f(\bfr(t)) \right) \, dt$$
like the one-variable ``net change theorem''?

\bigskip
\uncover<2->
{
{\color{cyan} \emph{Answer}: You bet!}
}
\end{frame}

\begin{frame}
\frametitle{Line Integral of a Gradient Vector Field}
\footnotesize
\begin{center}
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
{\color{cyan} \textbf{Theorem}} \hspace{5pt} Suppose that $\bfF(\bfr) = \nabla f (\bfr)$ is a gradient vector field, and
$C$ is a path parameterized by $\bfr(t)$, $a \leq t \leq b$. Then
$$
\int_C \bfF \cdot d \bfr = f(\bfr(b)) - f(\bfr(a))
$$ 
\end{minipage}
}
\end{center}
\end{frame}

\begin{frame}
\frametitle{How to think about the Fundamental Theorem for Line Integrals}
\footnotesize
\begin{minipage}{0.40\textwidth}
\begin{tikzpicture}
\draw[thick,->,>=stealth]	(-0.5,0)	--	(3,0);
\draw[thick,->,>=stealth]  (0,-0.5)	--	(0,3);
\node[left]		at	(0,3)	{\scriptsize{$y$}};
\node[below]	at (3,0)	{\scriptsize{$x$}};
%% contours as in Stewart problem 1
\begin{scope}
\clip (0,0)	rectangle	(2.9,2.9);
\foreach[
	evaluate=\a as \phi using 	int((\a/0.25)*5)), 
	evaluate=\a as \x	using		\a*cos(55), 
	evaluate=\a as \y	using		(\a/2)*sin(55)+0.15
] 
\a in {1.5, 2,25, 3, 3.5, 4, 4.5} 
{
	\draw[red]	(\a,0)	arc(0:90:{\a} and {\a/2});					%%	lines
	\node[black] 		at ({\x},{\y})		{\tiny{$\phi$}};		%% 	labels
}
\end{scope}
%% a path C
\draw[cyan,thick,->-]	
	plot[domain=1.5:4.5,variable=\a]
		(
			{\a*cos(70 + 6.25- (\a-2)^2)},
			{(\a/2)*sin(70)}
		);
\end{tikzpicture}
%%
\end{minipage}
\quad
\begin{minipage}{0.55\textwidth}
The figure at the left shows a curve $C$ and a contour map of a function $f$ whose gradient is continuous. Find
$\int_C \nabla f  \cdot d\bfr$. 

\medskip

\emph{Hint}: Think of $f$ as a height function, and the contour plot as a contour map. The gradient gives the magnitude and direction of the greatest change in height at any given point.
\end{minipage}

\end{frame}


\section{Path Independence}

\begin{frame}
\frametitle{Vocabulary - Paths and Vector Fields}
\footnotesize

\medskip

\begin{tabular}{ll}
\textbf{path}				&			A piecewise smooth curve\\[5pt]
\textbf{closed path}		&			A curve whose initial and terminal points are the same\\[10pt]
\textbf{conservative}	&			A vector field $\bfF$ which is the gradient of a scalar function $f$, \\
\textbf{vector field}		&			called the \emph{potential}, so that $ \bfF = \nabla f $
\end{tabular}

\bigskip

Which of the following is not a closed path?

\bigskip

\begin{center}
\begin{minipage}{0.25\textwidth}
\begin{tikzpicture}[scale=0.7]
\draw[thick,->,>=stealth]	(-0.5,0)	--	(3,0);
\draw[thick,->,>=stealth]  (0,-0.5)	--	(0,3);
\node[left]		at	(0,3)	{\scriptsize{$y$}};
\node[below]	at (3,0)	{\scriptsize{$x$}};
%%
\draw[cyan,thick,->-]		(2.5,1.5)	arc(0:180:1);
\draw[cyan,thick,->-]		(0.5,1.5)	arc(180:360:1);
%%
\node at (1.5,-1)	{\scriptsize{$A$}};
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.25\textwidth}
\begin{tikzpicture}[scale=0.7]
\draw[thick,->,>=stealth]	(-0.5,0)	--	(3,0);
\draw[thick,->,>=stealth]  (0,-0.5)	--	(0,3);
\node[left]		at	(0,3)	{\scriptsize{$y$}};
\node[below]	at (3,0)	{\scriptsize{$x$}};
%%
\begin{scope}
\clip	(0,0)	rectangle (3,3);
\draw[cyan,thick,->-]
	plot[domain=0.5:2.5]	({\x},{\x+0.8*sin(5*\x*(180/3.1415926)});
\end{scope}
%%
\node at (1.5,-1)	{\scriptsize{$B$}};
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.25\textwidth}
\begin{tikzpicture}[scale=0.7]
\draw[thick,->,>=stealth]	(-0.5,0)	--	(3,0);
\draw[thick,->,>=stealth]  (0,-0.5)	--	(0,3);
\node[left]		at	(0,3)	{\scriptsize{$y$}};
\node[below]	at (3,0)	{\scriptsize{$x$}};
%%
\draw[cyan,thick,->-]	
	plot[domain=1.5:4.5,variable=\a]
		(
			{\a*cos(70 + 4.5- 2*(\a-3)^2)},
			{(\a/2)*sin(70)}
		);
%%
\draw[cyan,thick,->-]	
	plot[domain=4.5:1.5,variable=\a]
		(
			{\a*cos(70 - 4.5 +  2*(\a-3)^2)},
			{(\a/2)*sin(70)}
		);
%%
\node at (1.5,-1)	{\scriptsize{$C$}};
\end{tikzpicture}
\end{minipage}
\end{center}
\end{frame}


\begin{frame}
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}
\draw[thick,->,>=stealth]	(-0.5,0)	--	(3,0);
\draw[thick,->,>=stealth]  (0,-0.5)	--	(0,3);
\node[left]		at	(0,3)	{\scriptsize{$y$}};
\node[below]	at (3,0)	{\scriptsize{$x$}};
%% contours as in Stewart problem 1
\begin{scope}
\clip (0,0)	rectangle	(2.9,2.9);
\foreach[
	evaluate=\a as \phi using 	int((\a/0.25)*5)), 
	evaluate=\a as \x	using		\a*cos(55), 
	evaluate=\a as \y	using		(\a/2)*sin(55)+0.15
] 
\a in {1.5, 	2,		2.5, 	3, 	3.5, 	4, 	4.5} 
{
	\draw[red]	(\a,0)	arc(0:90:{\a} and {\a/2});					%%	lines
	\node[black] 		at ({\x},{\y})		{\tiny{$\phi$}};		%% 	labels
}
\end{scope}
%% a path C1
\uncover<2->
{
\node at		(70:1.45)		{\tiny{$C_1$}};
\draw[cyan,thick,->-]	
	plot[domain=1.5:4.5,variable=\a]
		(
			{\a*cos(70 + 4.5- 2*(\a-3)^2)},
			{(\a/2)*sin(70)}
		);
}
\uncover<3->
{
\node at		(46:2.35)		{\tiny{$C_2$}};
\draw[cyan,thick,->-]	
	plot[domain=1.5:4.5,variable=\a]
		(
			{\a*cos(70 - 4.5 +  2*(\a-3)^2)},
			{(\a/2)*sin(70)}
		);
}
\end{tikzpicture}
%%
\end{minipage}
\quad
\begin{minipage}{0.45\textwidth}
\footnotesize
At left is the contour plot for a function $f$ whose gradient is continuous.

\medskip


Compute the following:
\begin{itemize}
\uncover<2->
{
\item	$\int_{C_1} \nabla f \cdot d \bfr$
}
\uncover<3->
{
\item	$\int_{C_2} \nabla f \cdot d \bfr$
}
\uncover<4->
{
\item		Does it matter what path connects the endpoints?
}
\end{itemize}
\end{minipage}

\bigskip
\footnotesize

\uncover<5->
{
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
{\color{cyan} \textbf{Definition}} \hspace{5pt} A line integral $\int_C \bfF \cdot d \bfr$ is \emph{independent of path} in a domain $D$ f $$\int_{C_1} \bfF \cdot d \bfr = \int_{C_2} \bfF \cdot d \bfr $$ for any two paths $C_1$ and $C_2$ that have the same initial and terminal points. 
\end{minipage}
}
}
\end{frame}

\begin{frame}
\frametitle{Path Independence and Closed Paths}
\footnotesize
\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}
\draw[thick,->,>=stealth]	(-0.5,0)	--	(3,0);
\draw[thick,->,>=stealth]  (0,-0.5)	--	(0,3);
\node[left]		at	(0,3)	{\scriptsize{$y$}};
\node[below]	at (3,0)	{\scriptsize{$x$}};
%% paths
\node at		(70:1.45)		{\tiny{$C_1$}};
\draw[cyan,thick,->-]	
	plot[domain=1.5:4.5,variable=\a]
		(
			{\a*cos(70 + 4.5- 2*(\a-3)^2)},
			{(\a/2)*sin(70)}
		);
\only<1>
{
\node at		(46:2.35)		{\tiny{$C_2$}};
\draw[cyan,thick,->-]	
	plot[domain=1.5:4.5,variable=\a]
		(
			{\a*cos(70 - 4.5 +  2*(\a-3)^2)},
			{(\a/2)*sin(70)}
		);
}
\uncover<2->
{
\node at		(40:2.15)		{\tiny{$-C_2$}};
\draw[cyan,thick,->-]	
	plot[domain=4.5:1.5,variable=\a]
		(
			{\a*cos(70 - 4.5 +  2*(\a-3)^2)},
			{(\a/2)*sin(70)}
		);
}
\end{tikzpicture}

\end{minipage}
\quad
\begin{minipage}{0.55\textwidth}
If 
$$ \int_{C_1} \bfF \cdot d \bfr = \int_{C_2} \bfF \cdot d \bfr $$
and we reverse the direction of $C_2$ $\ldots$

\medskip

\uncover<2->
{
\noindent
Then
$$ \int_C \bfF \cdot d \bfr = 0 $$
where $C$ is the closed loop path that starts with $C_1$ and ends with $-C_2$.
}
\end{minipage}

\medskip

\uncover<3->
{
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
{\color{cyan} \textbf{Theorem}} The integral $\int_C \bfF \cdot d \bfr$ is independent of path for all paths in a domain $D$ if and only if 
$\int_C \bfF \cdot d \bfr = 0$ for every closed path in $D$.
\end{minipage}
}
}
\end{frame}

\section{Conservative Fields}

\begin{frame}
\frametitle{Vocabulary -  Connected Regions}
\footnotesize

\medskip

\begin{tabular}{ll}
\textbf{connected region}	&		A region  $D$ of $\mathbb{R}^2$ or $\mathbb{R}^3$  where any points $P$ and $Q$ \\
								&			can be connected by a path contained in $D$\\[5pt]
\textbf{domain}			&			An open, connected region of $\mathbb{R}^2$ or $\mathbb{R}^3$\\[5pt]
\end{tabular}

\bigskip

Which of these regions is \emph{not} connected?

\bigskip

\begin{center}

\begin{minipage}{0.3\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]	(-3,0)	--	(3,0);
\draw[thick,->,>=stealth]	(0,-3)	--	(0,3);
\path[draw=none,fill=orange!20,fill opacity=0.5,even odd rule]
	(0,0)	circle(2)
	(0,0) circle(1);
\draw[orange,thick]
	(0,0) circle(2);
\draw[orange,thick]
	(0,0)	circle(1);
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.3\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]	(-3,0)	--	(3,0);
\draw[thick,->,>=stealth]	(0,-3)	--	(0,3);
\draw[orange,thick,fill=orange!20,opacity=0.75]
	(-1.5,0)		circle(0.75);
\draw[orange,thick,fill=orange!20,opacity=0.75]
	(1.5,0)		circle(0.75);
\uncover<2->
{
\draw[black,fill=black]	(-1.5,0)	circle(0.05cm);
\node[below]	at	(-1.5,0)	{\tiny{$P$}};
\draw[black,fill=black]	(1.5,0)	circle(0.05cm);
\node[below]	at	(1.5,0)	{\tiny{$Q$}};
}
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.3\textwidth}
\begin{tikzpicture}[scale=0.6]
\draw[thick,->,>=stealth]	(-3,0)	--	(3,0);
\draw[thick,->,>=stealth]	(0,-3)	--	(0,3);
\draw[orange,thick,fill=orange!20,opacity=0.5]
	(0,0)	circle(2);
\end{tikzpicture}
\end{minipage}

\end{center}
\end{frame}

\begin{frame}
\frametitle{Vocabulary - Simply Connected Regions}
\medskip

\footnotesize

\begin{tabular}{ll}
\textbf{simple curve}		&	A curve that doesn't intersect itself\\[5pt]
\textbf{simply connected}	&	A connected region so that every simple closed curve in $D$ \\
								&			surrounds only points of $D$
\end{tabular}

\bigskip

Which of these regions is \emph{not} simply connected?

\bigskip

\begin{center}

\begin{minipage}{0.3\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]	(-3,0)	--	(3,0);
\draw[thick,->,>=stealth]	(0,-3)	--	(0,3);
\path[draw=none,fill=orange!20,fill opacity=0.5,even odd rule]
	(0,0)	circle(2)
	(0,0) circle(1);
\draw[orange,thick]
	(0,0) circle(2);
\draw[orange,thick]
	(0,0)	circle(1);
\node[below] at (0,-3)	
	{\tiny{$\{ (x,y): 1 \leq x^2+y^2 \leq 2 \}$}};
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.3\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]	(-3,0)	--	(3,0);
\draw[thick,->,>=stealth]	(0,-3)	--	(0,3);
\begin{scope}
\clip	(-3,-3) rectangle (3,3);
\path[draw=none,fill=orange!20,fill opacity=0.5,even odd rule]
	(0,0)	circle(6)
	(0,0) circle(0.1);
\draw[white,fill=white]
	(0,0)	circle(0.1);
\end{scope}
\node[below] at	(0,-3)
	{\tiny{$\{ (x,y): (x,y) \neq (0,0)\}$}};
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.3\textwidth}
\begin{tikzpicture}[scale=0.6]
\draw[thick,->,>=stealth]	(-3,0)	--	(3,0);
\draw[thick,->,>=stealth]	(0,-3)	--	(0,3);
\draw[orange,thick,fill=orange!20,opacity=0.5]
	(0,0)	circle(2);
\node[below] at (0,-3)
	{\tiny{$\{(x,y): x^2+y^2 \leq 4\}$}};
\end{tikzpicture}
\end{minipage}

\end{center}
\end{frame}

\begin{frame}
\frametitle{First Theorem of the Day}
\footnotesize
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
{\color{cyan} \textbf{Theorem}} \hspace{5pt} Suppose $\bfF$ is a vector field that is continuous on an open, {\color{red}simply connected}  region $D$. If $\int_C \bfF \cdot d \bfr$ is independent of path in $D$, then $\bfF$ is a conservative vector field on $D$; that is, there is a function $f$ so that $\nabla f = \bfF$
\end{minipage}
}

\bigskip
How do you find the function $f$ (two dimensions)?

\bigskip

\begin{itemize}
\item	Pick a point $(a,b)$ in the domain $D$
\item	Compute 
		$$ f(x,y) = \int_{(a,b)}^{(x,y)}  \bfF \cdot d \bfr$$
\item	In fact, you can show that this function $f$ satisfies
		$$ \bfF(x,y) = \frac{\dee f}{\dee x}(x,y) \ii + \frac{\dee f}{\dee y}(x,y) \jj $$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{How You (Almost) Tell when $\bfF$ is Conservative}
\footnotesize
\textbf{Key Observation} If $F = \nabla f$ then
$$ \bfF(x,y) = P(x,y) \ii + Q(x,y) \jj = \frac{\dee f}{\dee x}(x,y) \ii + \frac{\dee f}{\dee y} \jj $$

\medskip
Compute $\dee P/\dee y$ and $\dee Q /\dee x$ as a second derivative of $f$:

\medskip
\begin{center}
\uncover<2->
{
$\dfrac{\dee P}{\dee y} = \dfrac{\dee}{\dee y} \dfrac{\dee f}{\dee x} = \dfrac{\dee^2 f}{\dee y \dee x}$
}
\qquad
\uncover<3->
{
$\dfrac{\dee Q}{\dee x} = \dfrac{\dee}{\dee x} \dfrac{\dee f}{\dee y} = \dfrac{\dee^2 f}{\dee x \dee y}$
}
\end{center}

\medskip
\uncover<4->
{
So, by Clairaut's Theorem, for a conservative vector field:
\begin{center}
\fcolorbox{cyan}{white}
{
	\begin{minipage}{0.4\textwidth}
	$$ \dfrac{\dee P}{\dee y} = \dfrac{\dee Q}{\dee x}$$
	\end{minipage}
}
\end{center}
}
\end{frame}

\begin{frame}
\frametitle{Find the  Conservative Vector Field}
\footnotesize
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
{\color{cyan} \textbf{Theorem}} \hspace{5pt}	If $\bfF(x,y) = P(x,y) \ii + Q(x,y) \jj$ is a conservative vector field, 
and $P,Q$ have continuous first-order partials on a domain $D$, then throughout $D$
$$ \dfrac{\dee P}{\dee y} = \dfrac{\dee Q}{\dee x}$$
\end{minipage}
}

\bigskip
Which of the following vector fields are definitely \emph{not} conservative?
\begin{enumerate}
\item	$\bfF(x,y) = -y \ii + x \jj$
\item	$\bfF(x,y) = x^3 \ii + y^2 \jj$
\item	$\bfF(x,y) = ye^x \ii + (e^x + e^y) \jj$
\item	$\bfF(x,y) = \dfrac{-y}{x^2+y^2}\ii + \dfrac{x}{x^2+y^2} \jj$, \quad $(x,y) \neq (0,0)$
\end{enumerate}
\end{frame}

\begin{frame}
\frametitle{There's One in Every Crowd}
\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.5]
\begin{axis}
	[
		title={$\bfF(x,y) = \dfrac{-y}{x^2+y^2} \ii + \dfrac{x}{x^2+y^2} \jj$},
		view={0}{90},
		domain=-2:2,
		axis background/.style={fill=white}
	]
\addplot3[blue,quiver={u={(-y)/(x^2+y^2+0.1)},v={(x)/(x^2+y^2+0.1)},scale arrows=0.2},-stealth,samples=15]
		{0};
\end{axis}		 
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.55\textwidth}
\footnotesize

\begin{enumerate}
\item	Does $\bfF$ satisfy the ``conservative vector field'' condition?

\medskip

\item	Suppose $C$ is the circle $x^2+y^2=1$. What is 
$\int_C \bfF \cdot d \bfr$ for the vector field shown?

\medskip
\item	Is the domain 
$$ \{ (x,y) : x^2 + y^2 \neq 0 \}$$
simply connected?
\end{enumerate}


\end{minipage}
\end{frame}

\begin{frame}
\frametitle{Second Theorem of the Day}
\footnotesize
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
{\color{cyan} \textbf{Theorem}} \hspace{0.5cm}  Let $\bfF = P \ii + Q \jj$ be a vector field defined on an open, simply connected region $D$. Suppose that $P$ and $Q$ have continuous partial derivatives and
$$ \frac{\dee P}{\dee y}  = \frac{\dee Q}{\dee x} $$
throughout $D$. Then $\bfF$ is conservative.
\end{minipage}
}
\bigskip

Which of the following vector fields are conservative?
\begin{enumerate}
\item	$\bfF(x,y) = -y \ii + x \jj$
\item	$\bfF(x,y) = x^3 \ii + y^2 \jj$
\item	$\bfF(x,y) = ye^x \ii + (e^x + e^y) \jj$
\item	$\bfF(x,y) = \dfrac{-y}{x^2+y^2}\ii + \dfrac{x}{x^2+y^2} \jj$, \quad $(x,y) \neq (0,0)$
\end{enumerate}
\end{frame}

\begin{frame}
\frametitle{How to Find the Potential $f$}
\footnotesize
Recall that if $\bfF =P \ii + Q \jj= \nabla f$, then
$$ P = \frac{\dee f}{\dee x}, \quad Q = \frac{\dee f}{\dee y} $$

\medskip

\textbf{Example} Find $f$ if $\bfF(x,y) = (y^2-2x)\ii + 2xy \jj $

\medskip
\begin{enumerate}
\uncover<2->
{
\item		$\dfrac{\dee f}{\dee x} = y^2 - 2x$ so taking antiderivatives in $x$
			$$ f(x,y) = y^2 x - x^2 + {\color{red}C(y)} $$
			where $C(y)$ is a constant \emph{that may depend on $y$}
}
\smallskip
\uncover<3->
{
\item		From the answer we found in step 1, 
			$\dfrac{\dee f}{\dee y} = 2xy +C'(y) = 2xy$ so $C'(y) = 0$
}
\smallskip
\uncover<4->
{
\item		Finally, $f(x,y) = xy^2 - x^2 + C$
}
\end{enumerate}

\end{frame}

\begin{frame}
\frametitle{Line Integrals of Conservative Vector Fields}
\footnotesize
Recall that if $\bfF =P \ii + Q \jj= \nabla f$, then
$$ P = \frac{\dee f}{\dee x}, \quad Q = \frac{\dee f}{\dee y} $$

\medskip

{\textbf{Example}}:  \hspace{5pt}	Find $\int_C \bfF \cdot d\bfr$ by finding $f$ so that $\nabla f = \bfF$
if:
\begin{align*}
	\bfF(x,y) 	&	= (1+xy) e^{xy} \ii + x^2 e^{xy} \jj \\
	 C: \, \bfr(t) &	= \cos t \ii + 2 \sin t \jj, \quad 0 \leq t \leq \pi/2
\end{align*}

\end{frame}


\end{document}

