<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[10pt]{beamer}
\usepackage{mathpazo}			%% Beamer stuff
\usepackage{amsmath}			%% AMS math packages
\usepackage{amsfonts}			%%	-	special fonts
\usepackage{amssymb}			%%	-	special symbols
\usepackage{graphicx}			%% graphics
\usepackage{tcolorbox}			%% color boxes

\usepackage{hyperref}			%% for web links
\hypersetup{
%    bookmarks=true,         		% show bookmarks bar?
    unicode=false,          		% non-Latin characters in Acrobatâ€™s bookmarks
    pdftoolbar=true,        		% show Acrobatâ€™s toolbar?
    pdfmenubar=true,        	% show Acrobatâ€™s menu?
    pdffitwindow=false,     		% window fit to page when opened
    pdfstartview={FitH},    		% fits the width of the page to the window
    pdfnewwindow=true,      	% links in new PDF window
    colorlinks=true,       			% false: boxed links; true: colored links
    linkcolor=blue,          		% color of internal links (change box color with linkbordercolor)
    citecolor=PineGreen,    		% color of links to bibliography
    filecolor=magenta,      		% color of file links
    urlcolor=cyan           		% color of external links
}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		TiKZ Ploting Stuff
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\usepackage{tikz}				%% TikZ -graphics

\usetikzlibrary{arrows, positioning, calc, intersections}
\usetikzlibrary{decorations.pathreplacing, decorations.markings}

\usepackage{pgfplots}			%% fancy plotting

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	For 3d plotting, using the tikz-3dplot package so
%	that (x,y,z) coordinates are specified in normal order
%	and one can change the perspective for 3-d plots
%
%	To invoke, use
%
%	\begin{tikzpicture}[tdplot_main_coords,{other options}]
%		...
%	\end{tikzpicture}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\usepackage{tikz-3dplot}		%% Enhanced 3d plotting

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Package forest to draw tree diagrams for chain rule
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\usepackage{forest}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	adjustbox package to force minipage alignment
%
%	Two-column structure is
%
%	\adjustbox{valign=t}
%		{
%			\begin{minipage}{0.45\textwidth}
%			...
%			\end{minipage}
%		}
%	\adjustbox{valign=t}
%		{
%		
%			\begin{minipage}{0.45\textwidth}
%			...
%			\end{minipage}
%		}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\usepackage{adjustbox}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Set Beamer theme
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\usetheme{Dresden}
\usecolortheme{dolphin}

%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Hack to remove subsection bar from our
%		friends at Tex.StackExchange.Com
%
%%%%%%%%%%%%%%%%%%%%%%%%%%
\makeatletter
\beamer@theme@subsectionfalse

\makeatother
%%%%%%%%%%%%%%%%%%%%%%%%%%
%		
%		Hack to make the nice little circles appear
%		from Tex.stackexchange.com
%
%		Reference: 
%
%		https://tex.stackexchange.com/questions/2072/beamer-navigation-circles-without-subsections
%
%%%%%%%%%%%%%%%%%%%%%%%%%%

\usepackage{remreset}
\makeatletter
\@removefromreset{subsection}{section}
\makeatother
\setcounter{subsection}{1}

%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Get rid of triangles, put in circles for 
%		itemize items
%
%%%%%%%%%%%%%%%%%%%%%%%%%%
\setbeamertemplate{items}[circle]

%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Clean up fonts
%
%%%%%%%%%%%%%%%%%%%%%%%%%%

\usefonttheme{professionalfonts}
\usefonttheme{serif}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		tikz plotting options
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\pgfplotsset{my style/.append style={axis x line=middle, axis y line=
           middle, xlabel={}, ylabel={}, axis equal }}
           
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Math Operators
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\DeclareMathOperator{\comp}{comp}
\DeclareMathOperator{\proj}{proj}
\DeclareMathOperator{\Hess}{Hess}

\DeclareMathOperator{\curl}{curl}
\DeclareMathOperator{\diver}{div}
\DeclareMathOperator{\grad}{grad}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	Document begins here...
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
           
\begin{document}

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


\title{Math 213 - Line Integrals, Part II}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{November 11, 2019}
\maketitle

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

\begin{frame}
\frametitle{Reminders}

\begin{itemize}

\item	Exam III is this Wednesday, November 13, 5:00-7:00 PM
\item	The Review Session for Exam III is tonight, 6:00-8:00 PM in room KAS 213

\end{itemize}

\end{frame}

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

\begin{frame}
\frametitle{Unit III: Multiple Integrals, Vector Fields}
\begin{itemize}

\item[{}]			Double Integrals in Polar Coordinates
\item[{}]			Triple Integrals (Part I)
\item[{}]			Triple Integrals (Part II)
\item[{}]			Triple Integrals in Cylindrical Coordinates
\item[{}]			Triple Integrals in Spherical Coordinates



\item[{}]			Change of Variables, Part I

\item[{}]			Change of Variables, Part II

\medskip


\item[{}]			Vector Fields

\item[{}]			Line Integrals (Scalar functions)
\begingroup
\color{red}
\item[{}]			Line Integrals (Vector functions)
\endgroup

\medskip

\item[{}]			Exam III Review
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	Learning Goals
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Learning Goals}

\begin{frame}
\frametitle{Goals of the Day}
\begin{itemize}
\item	Know how to compute line integrals of a vector function in the plane and in space
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		New material begins here...
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Where Are We?}
\footnotesize
We've talked about line integrals of \emph{scalar functions} over plane and space curves. Today we'll talk about line integrals of \emph{vector functions} 
over plane and space curves. If $C$ is a curve in the plane or in space, we've already seen how to compute:

\medskip

\begin{itemize}
\item	$\int_C f(x,y) \, ds$, the integral of a scalar function over a plane curve $C$
\item	$\int_C f(x,y,z) \, ds$, the integral of a scalar function over a space curve $C$
\end{itemize}

\medskip

Today we'll discuss:

\medskip

\begin{itemize}
\item	$\int_C \bfF \cdot d\bfr$, the integal of a vector function $\bfF(x,y)$ over
a plane curve $C$
\item	$\int_C \bfF \cdot d\bfr$, the integral of a vector function $\bfF(x,y,z)$ over
a space curve $C$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Remember Space Curves?}
\footnotesize
A \emph{space curve} is given by 
$$ \bfr(t) = x(t) \ii + y(t) \jj + z(t) \kk $$

\medskip
The \emph{tangent vector} to a space curve is
$$ \bfr'(t) = x'(t) \ii + y'(t) \jj + z'(t) \kk$$
Recall that $\bfr'(t)$ is the velocity, and $| \bfr'(t) |$ is the speed.

\medskip
The \emph{unit tangent vector} is 
$$ \bfT(t) = \frac{\bfr'(t)}{|\bfr(t)|}. $$

\medskip

\hrulefill

\medskip

Find the tangent vector and unit tangent vector to the curve
$$ \bfr(t) = \cos(t) \ii + \sin(t) \jj + t \kk $$
for $t=0$, $t=\pi/2$, and $t=\pi$. 


\end{frame}

\section{Line Integrals of Vector Functions}

\begin{frame}
\begin{minipage}{0.4\textwidth}
\begin{tikzpicture}[scale=0.6]
\draw[thick,-&gt;,&gt;=stealth]	(0,0,0)	--	(3,0,0);
\draw[thick,-&gt;,&gt;=stealth]	(0,0,0)	--	(0,3,0);
\draw[thick,-&gt;,&gt;=stealth]	(0,0,0)	--	(0,0,5);
\node[above] at	(0,3,0)		{\scriptsize{$z$}};
\node[above] at  (3,0,0)		{\scriptsize{$y$}};
\node[below] at 	(0,0,4)		{\scriptsize{$x$}};
%%
\draw[red]	plot[domain=0:360,variable=\t,samples=20,smooth]
		( {cos(\t)}, {sin(\t)}, {3*(\t/360)} );
%%
\draw[black,fill=black]	({cos(120)},{sin(120)},1) circle(0.05cm);
%%
\draw[thick,-&gt;,&gt;=stealth]	({cos(120)},{sin(120)},1) --	++(0,1.5,1.5);
\node[above]	 at ({cos(120)},{sin(120)+1.5},{2.5})	{\scriptsize{\textbf{F}}};
\draw[thick,red,-&gt;,&gt;=stealth]	({cos(120)},{sin(120)},1)	--	++({-sin(120)},{cos(120)},1);
\node[left] at ({cos(120)-sin(120)},{sin(120)+cos(120)},2)	{\scriptsize{\textbf{T}}};
%%
\draw[thick,cyan]	plot[domain=110:130,variable=\t]
		( {cos(\t)}, {sin(\t)}, {3*(\t/360)} );
\node[below] at 	({cos(120)},{sin(120)},1) 		{{\color{cyan}\scriptsize{$\Delta s$}}};
	
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.5\textwidth}
\footnotesize
Recall that the work done by a constant force $\bfF$ moving an object through a displacement $\bfD$ is 
$$ W = \bfF \cdot \bfD$$

\medskip

What if $\bfF$ and the displacement $\bfD$ vary as the force acts through a curve $C$?

\medskip

Write $\bfD = \bfT \Delta s$ where $\bfT$ is the tangent vector and $\Delta s$ is arc length.

\medskip

Then
\begin{align*} 
W 	&amp;	\simeq \sum_{i=1}^n \bfF(x_i^*,y_i^*,z_i^*) \cdot \bfT_i \, \Delta s	\\
		&amp;  \to \int_C \bfF \cdot \bfT \, ds 
\end{align*}
\end{minipage}

\end{frame}

\begin{frame}
\frametitle{How Do You Compute It?}
\footnotesize
The work done by a variable force $\bfF$ moving a particle along a curve $C$ is
$$ W = \int_C \bfF \cdot \bfT \, ds. $$

\medskip

If $C$ is parameterized by $\bfr(t) = x(t) \ii + y(t) \jj + z(t) \kk$ for $ a \leq t \leq b$:
$$ \bfT = \frac{\bfr'(t)}{|\bfr'(t)|} $$
and 
$$ ds = |\bfr'(t)| \, dt $$

\bigskip
So
\begin{align*}
\int_C \bfF \cdot \bfT \, ds	&amp;=	\int_a^b \bfF(x(t),y(t),z(t)) \cdot \frac{\bfr'(t)}{{\color{cyan}|\bfr'(t)|}} 
	{\color{cyan}|\bfr'(t)|} \, dt\\
		&amp;=	\int_C \bfF(x(t),y(t),z(t)) \cdot \bfr'(t)  \, dt
\end{align*}
This line integral is sometimes written
$$ \int_C \bfF \cdot d\bfr $$ 
for short
\end{frame}

\begin{frame}
\frametitle{Now You Try It}
\footnotesize
$$ \int_C \bfF \cdot d \bfr = \int_a^b \bfF(\bfr(t)) \cdot \bfr'(t) \, dt $$

\bigskip

Find $\int_C \bfF \cdot d\bfr$ if:
\begin{enumerate}
\item		$\bfF(x,y,z) = x \ii + y \jj + z \kk$ and $\bfr(t) = t^2 \ii + t^3 \jj - 2t \kk$, $0 \leq t \leq 2$
\item		$\bfF(x,y,z)= yze^x \ii + zxe^y \jj + xye^z \kk$ and $\bfr(t) = \sin t \ii + \cos t \jj + \tan t \kk$, $0 \leq t \leq \pi/4$
\end{enumerate}

\end{frame}

\begin{frame}
\frametitle{Now You Try It}
\begin{minipage}{0.4\textwidth}
\begin{tikzpicture}[scale=0.5]
\begin{axis}
	[
		title={$\bfF(x,y) = x^2 \ii + xy \jj$},
		view={0}{90},
		domain=-2:2,
		axis background/.style={fill=white},
		unit vector ratio*=1 1 1,
	]
\addplot3[blue,quiver={u={x^2},v={x*y},scale arrows=0.2},-stealth,samples=15]
		{0};
\addplot3[red,domain=0:360,smooth]
		({2*cos(x)},{2*sin(x)},{0});
\end{axis}		 
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.5\textwidth}
\footnotesize
Find the work done by the force field $$\bfF(x,y) = x^2 \ii + xy \jj$$ on a particle that moves around the circle $$x^2+y^2=4$$ oriented in the counterclockwise direction
\end{minipage}

\end{frame}

\begin{frame}
\frametitle{Real Science}
\footnotesize
\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.7]
%% green background
\draw[green,fill=green!10]
	(-2,0,2)	--	(2,0,2)	--	(2,0,-2)	--	(-2,0,-2)		--	cycle;
%% circles
\draw[black,thick]	plot[domain=0:360,samples=30,smooth]
		({1.5*cos(\x)},0,{1.5*sin(\x)});
\draw[black,thick]	plot[domain=0:360,samples=30,smooth]
		({0.75*cos(\x)},0,{0.75*sin(\x)});
\draw[blue,left color=blue!10,right color=blue!70]
		plot[domain=-90:90,smooth]
			({0.25*sin(\x)},0,{0.25*cos(\x)})	
		--	++(0,2,0)	
		--	plot[domain=90:-90]
			({0.25*sin(\x)},2,{0.25*cos(\x)})
		--	++(0,-2,0);
%% vectors
\foreach \theta in {0,90,180,270}
	{	
		\draw[red,thick,-&gt;,&gt;=stealth]	({1.5*cos(\theta)},0,{1.5*sin(\theta)})		--	++({sin(\theta)},0,{-cos(\theta)});
	}
\foreach \theta in {30,120,210,300}
	{	
		\draw[red,thick,-&gt;,&gt;=stealth]	({0.75*cos(\theta)},0,{0.75*sin(\theta)})		--	++({0.5*sin(\theta)},0,{-0.5*cos(\theta)});
	}
%% wire
\draw[blue,left color=blue!10,right color=blue!70]
		plot[domain=-90:90,smooth]
			({0.25*sin(\x)},0,{0.25*cos(\x)})	
		--	++(0,2,0)	
		--	plot[domain=90:-90]
			({0.25*sin(\x)},2,{0.25*cos(\x)})
		--	++(0,-2,0);
%%
\draw[blue,densely dashed]
			plot[domain=90:270]
			({0.25*sin(\x)},2,{0.25*cos(\x)});
%% current		
\draw[thick,-&gt;,&gt;=stealth]	(0,2,0)	--	++(0,0.5,0);
\node[above] at (0,2.5,0)		{\scriptsize{$I$}};
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.55\textwidth}
Steady current in a wire generates a magnetic field $\bfB$ tangent to any circle that
lies in the plane perpendicular to the wire centered on the wire. According to 
\emph{Ampere's law},
$$ \int_C \bfB \cdot d \bfr = \mu_0 I $$
where

\smallskip

\begin{itemize}
\item	$I$ is the net current flowing through the wire
\item $\mu_0$ is a physical constant
\end{itemize}

\medskip

What is the magnitude of the magnetic field at a distance $r$
from the wire?
\end{minipage}
\end{frame}



\section{Summary of Line Integrals}

\begin{frame}
\frametitle{Summary}
\scriptsize
\begin{align*}
\intertext{Arc length differential}
ds							&amp;=	\sqrt{x'(t)^2+y'(t)^2+z'(t)^2} \, dt\\
\intertext{Line integral with respect to arc length}
\int_C f(x,y,z)  \,ds	&amp;=	\int_a^b f(x(t),y(t),z(t)) ds\\
\intertext{Line integral with respect to $x$, $y$, $z$}
\int_C f(x,y,z) \, dx	&amp;=	\int_a^b f(x(t),y(t),z(t))  x'(t) \, dt\\
\int_C f(x,y,z) \, dy	&amp;=	\int_a^b f(x(t),y(t),z(t))  y'(t) \, dt\\
\int_C f(x,y,z) \, dy	&amp;=	\int_a^b f(x(t),y(t),z(t))  z'(t) \, dt\\
\intertext{Line integral of a vector field}
\int_C \bfF \cdot d\bfr	&amp;=	\int_a^b \bfF(x(t),y(t),z(t)) \cdot \bfr'(t) \, dt 
\end{align*}

\end{frame}

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

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

\begin{frame}
\frametitle{Line Integral of a Gradient Vector Field}
Suppose $\bfF = \nabla \phi$ for a potential function $\phi(x,y,z)$

\medskip

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

\bigskip

Is there a simple way to compute
$$ \int_C \bfF \cdot d \bfr ?$$
\end{frame}



\end{document}
</pre></body></html>