\documentclass[notes=show,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 Acrobat’s bookmarks
    pdftoolbar=true,        		% show Acrobat’s toolbar?
    pdfmenubar=true,        	% show Acrobat’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

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	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 - Labelling Space}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{August 26, 2019}
\maketitle



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Welcome Page
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\begin{center}
\begin{tcolorbox}[colback=white,colframe=cyan,width=\linewidth,title={\centering \large\textbf{Welcome to Math 213, Fall 2019!}}]


\vskip 0.5cm

\begin{itemize}

\item	Bookmark the course web page

\url{http://www.math.uky.edu/~ma213/213-f19}

\smallskip

\item	Bookmark the instructor webpage

\url{http://www.math.uky.edu/~perry/213-f19-perry}

\smallskip

\item	Familiarize yourself with the 
\href{http://uk.instructure.com}{Canvas Page} for this course

\smallskip

\item	Print out a copy of the \href{http://www.ms.uky.edu/~ma213/213-f19/_assets/calendar.pdf}{Course Calendar} and keep in your notebook  (or store the PDF file on your favorite laptop or mobile device)

\end{itemize}

\vskip 0.5cm

$~$

\end{tcolorbox}
\end{center}
\end{frame}


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

\begin{frame}
\frametitle{Reminders}

\begin{itemize}

\item	Access your WebWork account \emph{only through Canvas}!

\medskip

\item	Homework A1 on Sections 12.1-12.2  is due Friday August 30

\medskip

\item	Applications for an alternate Exam 1 are due \textbf{no later than September 4}
		
		\medskip
		
		Review  your schedule and apply for all alternate exams at once by using the 
		\href{https://docs.google.com/forms/d/e/1FAIpQLSdAMcpIdWP54-vD2UU1F_M6p3DJMphjuXuEjgvGd-y5GSPcPQ/viewform}{Google Form} linked from 
		Canvas or the course home page.
		
	
\end{itemize}

\end{frame}

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

\begin{frame}
\frametitle{Unit I: Geometry and Motion in Space}
\begin{itemize}
\begingroup
\color{red}
\item[{12.1}]		Lecture 1: Three-Dimensional Coordinate Systems
\endgroup
\item[{12.2}] 		Lecture 2: Vectors in the Plane and in Space
\item[{12.3}]		Lecture 3: The Dot Product
\item[{12.4}]		Lecture 4: The Cross Product
\item[{12.5}]		Lecture 5: Equations of Lines and Planes, I
\item[{12.5}]		Lecture 6: Equations of Lines and Planes, II
\item[{12.6}]		Lecture 7: Surfaces in Space
\item[{13.1}]		Lecture 8: Vector Functions and Space Curves
\item[{13.2}]		Lecture 9 Derivatives and Integrals of Vector Functions
\item[{}]				Lecture 10:  Exam I Review
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Learning Goals}
\begin{itemize}
\item	Review basics of Calculus I-II
\item	Preview Calculus III
\item	Introduce 3D coordinate systems
\item	Introduce the \emph{distance formula} in 3D
\item	Find \emph{equations of spheres}
\end{itemize}
\end{frame}

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


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		Make font size on all slides small
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begingroup
\small
\section{Calc I-II}
\begin{frame}
\footnotesize
\frametitle{What Happened in Calculus I-II?}

\bigskip
The \emph{derivative} of a function
$$
f'(x_0) = \lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h}
$$
computes:

\medskip

\begin{itemize}
\item		The slope of the tangent line to the graph of $y=f(x)$	at $x=x_0$
\item		The instantaneous rate of change of a function $f$ at $x=x_0$


\end{itemize}

\bigskip
Using the derivative, you can find: intervals of increase and decrease, local extrema, and global extrema. It will be important to remember the \emph{differential} of $f$, 
$$ df(x) = f'(x) \, dx $$

\end{frame}

\begin{frame}
\frametitle{What Happened in Calculus I-II?}
\footnotesize
The \emph{integral} of a function $f$:
$$\int_a^bf(x) \, dx$$
computes:

\medskip

\begin{itemize}
\item	The net area under the graph of $y=f(x)$ between
		$a$ and $b$
\item	The net change in a quantity $F$ with rate of change $f(x) = F'(x)$ between $x=a$ and $x=b$
\end{itemize}

\medskip
The integral is a limit of \emph{Riemann sums}. Any geometric quantity (area, arc length, volume) or physical quantity (displacement given velocity, velocity given acceleration) that can be computed as a limit of Riemann sums can be computed as an integral
\end{frame}

\begin{frame}
\frametitle{The Fundamental Theorem of Calculus}
\footnotesize
\medskip

\textbf{Fundamental Theorem, Part I}  If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$, then
$$ \int_a^b f(x) \, dx = F(b) - F(a) $$

\medskip

\textbf{Fundamental Theorem, Part II} If $f$ is a continuous function on $[a,b]$ then
$$ \frac{d}{dx} \left( \int_a^x f(t) \, dt \right) = f(x)  $$

\bigskip

In other words,
$$ \int df = d \left( \int f \right) = f $$
\end{frame}

\section{Calc III}

\begin{frame}
\frametitle{What will happen in Calculus III?}
\footnotesize
In Calculus III we'll take these concepts of Calculus into \emph{higher dimensions}

\bigskip
\begin{itemize}
\item	We'll consider \emph{vector functions} $\bfv(t)= (x(t),y(t))$ and $\bfw(t) = (x(t),y(t),z(t))$ which describe motion in the plane and in space

\item	We'll consider \emph{functions of several variables}
$f(x,y)$ and $g(x,y,z)$ which describe altitude, temperature distributions, densities, etc.

\item We'll learn about \emph{transformations} $(x(u,v), y(u,v))$ that generalize polar coordinates and describe regions

\item	We'll study \emph{parameterized surfaces} $(x(u,v), y(u,v), z(u,v))$

\item	We'll consider \emph{vector fields} which describe the velocity of a fluid, the force of gravity, the action of electric and magnetic fields, and more!
\end{itemize}
\end{frame}

\section{2D to 3D}

\begin{frame}
\frametitle{What Will Happen Today?}
\footnotesize
We will move into \emph{three-dimensional space}

\bigskip
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[black,thick,->,>=stealth]	(0,0)	--	(4,0);
\draw[black,thick,->,>=stealth]	(0,0)	-- +(225:3.4);
\draw[black,thick,->,>=stealth]	(0,0)	--	(0,4);
%%
\node[right] 		at (4,0)	{$y$};
\node[above]		at	(0,4)	{$z$};
\node at (-2.6,-2.6)			{$x$};
\uncover<2->
{
\draw[black,fill=black]	(1.293,2.293)	circle (0.1cm);
\node[right] at (1.293,2.293)	{$P(1,2,3)$};
}
\uncover<3->
{
\draw[thick,green,->,>=stealth]	
	(0,0)	--	(-0.707,-0.707);
}
\uncover<4->
{
\draw[thick,red, ->, >=stealth]
	(-0.707,-0.707)	--	(1.293,-0.707);
}
\uncover<5->
{
\draw[thick,blue,->,>=stealth]
	(1.293,-0.707)	--	(1.293,2.293);
}
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.45\textwidth}
\footnotesize
\uncover<2->
{
To locate a point $P$ with respect to a chosen origin $O$, 
we specify the $x$, $y$ and $z$ displacements from $O$.
For example, the point $P=(1,2,3)$ is obtained by moving:
}
\begin{itemize}
\uncover<3->{\item $1$ unit in the {\color{green} $x$} direction}
\uncover<4->{\item $2$ units in the {\color{red} $y$} direction}
\uncover<5->{\item $3$ units in the {\color{blue} $z$} direction}
\end{itemize}
\end{minipage}

\medskip

This choice of $x$-, $y$-, $z$-axes forms a \emph{right-handed coordinate system}
\end{frame}

%%%%%

\begin{frame}
\footnotesize
\begin{minipage}{0.50\textwidth}
\begin{tikzpicture}[scale=0.7]
\draw[black,thick,->,>=stealth]	(0,0)	--	(4,0);
\draw[black,thick,->,>=stealth]	(0,0)	-- +(225:3.4);
\draw[black,thick,->,>=stealth]	(0,0)	--	(0,4);
%%
\node[right] 		at (4,0)	{$y$};
\node[above]		at	(0,4)	{$z$};
\node at (-2.6,-2.6)			{$x$};
\uncover<2->
{
\draw[black,fill=black]	(-0.414,-2.414)	circle (0.1cm);
\node[right] at (-0.414,-2.414) {$P(2,1,-1)$};
}
\uncover<3->
{
\draw[thick,green,->,>=stealth]	
	(0,0)	--	(-1.414,-1.414);
}
\uncover<4->
{
\draw[thick,red, ->, >=stealth]
	(-1.414,-1.414)	--	(-0.414,-1.414);
}
\uncover<5->
{
\draw[thick,blue,->,>=stealth]
	(-0.414,-1.414)	--	(-0.414,-2.414);
}
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.40\textwidth}
\footnotesize
\uncover<2->
{
The point $$P=(2,1,-1)$$ is obtained by moving:
}
\begin{itemize}
\uncover<3->{\item $2$ units in the {\color{green} $x$} direction}
\uncover<4->{\item $1$ unit in the {\color{red} $y$} direction}
\uncover<5->{\item $-1$ units in the {\color{blue} $z$} direction}
\end{itemize}
\end{minipage}
\end{frame}

\begin{frame}
\frametitle{The Distance Formula in $\R^2$}
Recall the distance between two points in the $xy$ plane:

\bigskip

\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,black,->,>=stealth]	(-2,-1)	--	(-2,6);
\draw[thick,black,->,>=stealth]	(-2,-1)	--	(6,-1);
\node[above]	at	(-2,6)		{$y$};
\node[right]	at	(6,-1)		{$x$};
%%
\draw[black,fill=black]	(1,1)	circle(0.1cm);
\draw[black,fill=black]	(5,4)	circle(0.1cm);
%%
\draw[thick]				(1,1)	--	(5,4);
%%
\node[below] at (1,1)		{\footnotesize{$P_1(x_1,y_1)$}};
\node[above] at (5,4)		{\footnotesize{$P_2(x_2,y_2)$}};
\uncover<2->
{
\draw[black,fill=black]	(5,1) 	circle(0.1cm);
\node[below] at	(5,1)		{\footnotesize{$Q(x_2,y_1)$}};
\draw[thick,dashed]	(1,1)	--	(5,1);
\draw[thick,dashed]	(5,1)	--	(5,4);
}
%%
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
\uncover<2->
{Add an ``extra'' point $Q$ below $P_2$}

\medskip

\uncover<3->
{
By the Pythagorean Theorem,
$$ |P_1 P_2|^2 = |P_1 Q_1|^2 + |QP_2|^2 $$
so
\begin{align*}
 |P_1 P_2| &= 	\sqrt{|P_1 Q_1|^2 + |QP_2|^2}\\
 				&=	\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}
\end{align*}
}
\end{minipage}
\end{frame}

\begin{frame}
\frametitle{The Distance Formula in $\R^3$}

\begin{minipage}{0.4\textwidth}
\begin{tikzpicture}[scale=0.6]
\draw[black,thick,->,>=stealth]	(0,0)	--	(5,0);
\draw[black,thick,->,>=stealth]	(0,0)	-- +(225:3);
\draw[black,thick,->,>=stealth]	(0,0)	--	(0,5);
%%
\node[right] 		at (5,0)	{$y$};
\node[above]		at	(0,5)	{$z$};
\node at (225:3.2)			{$x$};
%%
\draw[black,fill=black]	(2,2)		circle(0.1cm);
\draw[black,fill=black]	(4,3.5)		circle(0.1cm);
\draw[thick]				(2,2)	--	(4,3.5);
%%
\node[below] at (1.5,2)	{\scriptsize{$P_1(x_1,y_1,z_1)$}};
\node[above] at (4.3,3.5)	{\scriptsize{$P_2(x_2,y_2,z_2)$}};
%%
\uncover<2->
{
\draw[black,fill=white]	(4,1.5)	circle(0.1cm);
\node[below]	at	(4.3,1.5)	{\scriptsize{$Q(x_2,y_2,z_1)$}};
\draw[dashed]	(2,2)	--	(4,1.5);
\draw[dashed]	(4,1.5) -- (4,3.5);
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
\uncover<2->
{
Add an ``extra'' point $Q$
}
\medskip
\begin{itemize}
\uncover<3->
{
\item  By the Pythagorean Theorem,
$$
|P_1 P_2|^2	=	|P_1 Q|^2 + |Q P_2|^2 	
$$
}
\uncover<4->
{\item	By the two-dimensional distance formula
$$
|P_1 Q|^2 = (x_1-x_2)^2 + (y_1-y_2)^2
$$
while
$$ |QP_2|^2 = (z_2 - z_1)^2 $$
}
\end{itemize}
\end{minipage}

\uncover<5->
{ 	  
\footnotesize
So
$$ |P_1 P_2| = \sqrt{ (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} $$
}
\end{frame}

%%%%%%%%%%%%%%%%%%%%
%
%		Line versus plane
%
%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Two and Three Dimensions }

%%%%%%%%%%%%%%%%%%%%
%
%		Two-dimensional example (line)
%
%%%%%%%%%%%%%%%%%%%%

\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]	(0,-2)	--	(0,2);
\draw[thick,->,>=stealth]	(0,0)	--	(4,0);
\node[right]	at	(4,0)		{\footnotesize{$x$}};
\node[above] at (0,2)		{\footnotesize{$y$}};
\uncover<2->
{
\draw[thick,blue!50]		(2,-2)		--	(2,2);
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.55\textwidth}
\footnotesize
Find the set of all points $(x,y)$ that satisfy the equation
$x=2$

\bigskip

\uncover<2->
{\emph{Answer:} A vertical line through $x=2$}
\end{minipage}

\bigskip

%%%%%%%%%%%%%%%%%%%%
%
%		Three-dimensional example (plane)
%
%%%%%%%%%%%%%%%%%%%%

\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.5]
\uncover<3->
{
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(4,0,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(0,4,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(0,0,4);
\node[below] 	at		(0,0,4)	{\footnotesize{$x$}};
\node[right]	 at		(4,0,0)	{\footnotesize{$y$}};
\node[above]	 at		(0,4,0)	{\footnotesize{$z$}};
}
\uncover<4->
{
\draw[blue!80,fill=blue!20,opacity=0.5]
	(2,0,0)	--	++(0,4,0)	--	++(0,0,4)	--	++(0,-4,0)	
				--	++(0,0,-4)	--	cycle;
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.55\textwidth}
\footnotesize
\uncover<3->
{Find the set of all points $(x,y,z)$ that obey the equation $y=2$}

\bigskip

\uncover<4->
{\emph{Answer}: A vertical plane through $y=2$}
\end{minipage}
\end{frame}

%%%%%%%%%%%%%%%%%%%%
%
%	Circle versus Cylinder
%
%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Two and Three Dimensions}

%%%%%%%%%%%%%%%%%%%%
%
%		Two-dimensional example
%
%%%%%%%%%%%%%%%%%%%%

\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.55]
%%
\draw[thick,->,>=stealth]	(0,-2)	--	(0,2);
\draw[thick,->,>=stealth]	(-2,0)	--	(2,0);
\node[right]	at	(2,0)		{\footnotesize{$x$}};
\node[above] at (0,2)		{\footnotesize{$y$}};
\uncover<2->
{
\draw[blue!40]	(0,0)	circle(1cm);
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.55\textwidth}
\footnotesize
Find the set of all points $(x,y)$ that satisfy the equation 
$$x^2+y^2=1$$

\medskip

\uncover<2->
{\emph{Answer}: A circle of radius $1$ centered at $(0,0)$}
\end{minipage}

\bigskip


%%%%%%%%%%%%%%%%%%%%
%
%		Three-dimensional example
%
%%%%%%%%%%%%%%%%%%%%

\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.4]
\uncover<3->
{
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(4,0,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(0,5,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(0,0,6);
\node[below] 	at		(0,0,6)	{\footnotesize{$x$}};
\node[right]	 at		(4,0,0)	{\footnotesize{$y$}};
\node[above]	 at		(0,5,0)	{\footnotesize{$z$}};
}
\uncover<4->
{
\draw[blue]
	(-1,3.5) arc (180:360:1cm and 0.3cm)		--
	++(0,-6)	--	++(0,0) arc(360:180:1cm and 0.3cm)		--
	++(0,6);
\shade[left color=blue,right color=blue!10,opacity=0.4]
	(-1,3.5) arc (180:360:1cm and 0.3cm)		--
	++(0,-6)	--	++(0,0) arc(360:180:1cm and 0.3cm)		--
	++(0,6);
\draw[blue,dashed] (-1,-2.5) 	arc(180:0:1 cm and 0.3cm);
\draw[blue,dashed] (-1,0) 		arc(180:-180:1 cm and 0.3cm);
\draw[blue,dashed] (-1,3.5)  	arc(180:0:1 cm and 0.3cm);
\draw[dashed]		(0,0)	--	(0,-4);
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.55\textwidth}
\footnotesize
\uncover<3->
{Find the set of all points $(x,y,z)$ that satisfy the equation 
$$x^2+y^2=1$$}

\medskip

\uncover<4->
{\emph{Answer}: A cylinder of radius $1$ centered at $(0,0,0)$ whose axis of symmetry is the $z$-axis}
\end{minipage}
\end{frame}

%%%%%%%%%%%%%%%%%%%%
%
%		Circle versus sphere
%
%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Two  and Three Dimensions}


%%%%%%%%%%%%%%%%%%%%
%
%		Two-dimensional example
%
%%%%%%%%%%%%%%%%%%%%

\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.55]
%%
\draw[thick,->,>=stealth]	(0,-2)	--	(0,2);
\draw[thick,->,>=stealth]	(-2,0)	--	(2,0);
\node[right]	at	(2,0)		{\footnotesize{$x$}};
\node[above] at (0,2)		{\footnotesize{$y$}};
\uncover<2->
{
\draw[blue!40]	(0,0)	circle(1cm);
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.55\textwidth}
\footnotesize
Find the set of all points $(x,y)$ that satisfy the equation 
$$x^2+y^2=1$$

\medskip

\uncover<2->
{\emph{Answer}: A circle of radius $1$ centered at $(0,0)$}
\end{minipage}

%%%%%%%%%%%%%%%%%%%%
%
%		Three-dimensional example
%
%%%%%%%%%%%%%%%%%%%%

\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.5]
\uncover<3->
{
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(4,0,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(0,4,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(0,0,6);
\node[below] 	at		(0,0,6)	{\footnotesize{$x$}};
\node[right]	 at		(4,0,0)	{\footnotesize{$y$}};
\node[above]	 at		(0,4,0)	{\footnotesize{$z$}};
}
\uncover<4->
{
\shade[ball color=blue!40, opacity=0.4] (0,0) circle(2cm);
\draw[blue!40]	(0,0)	circle(2cm);
\draw[blue!80]				(-2,0)			arc(180:360:2 and 0.6);
\draw[dashed,blue!80]		(2,0)			arc(0:180:2 and 0.6);
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.55\textwidth}
\footnotesize
\uncover<3->
{
Find the set of all points $(x,y,z)$ that satisfy the equation
$$ x^2+y^2+z^2 = 1 $$
}

\medskip

\uncover<4->
{
\emph{Answer:} A sphere of radius $1$ centered at $(0,0,0)$.
}
\end{minipage}

\end{frame}

%%%%%%%%%%%%%%%%%%%%
%
%		Annulus and spherical shell
%
%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Two and Three Dimensions}


%%%%%%%%%%%%%%%%%%%%
%
%		Two-dimensional example
%
%%%%%%%%%%%%%%%%%%%%

\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.55]
%%
\draw[thick,->,>=stealth]	(0,-2.5)	--	(0,2.5);
\draw[thick,->,>=stealth]	(-2.5,0)	--	(2.5,0);
\node[right]	at	(2.5,0)		{\footnotesize{$x$}};
\node[above] at (0,2.5)		{\footnotesize{$y$}};
\uncover<2->
{
%% annulus
\draw[white,fill=blue!40, opacity=0.4]	circle(2cm);
\draw[white,fill=white]		circle(1cm);
%% redraw axes
\draw[thick,->,>=stealth]	(0,-2.5)	--	(0,2.5);
\draw[thick,->,>=stealth]	(-2.5,0)	--	(2.5,0);
%% put dashed circles
\draw[blue!40,dashed	]	(0,0)	circle(1cm);
\draw[blue!40,dashed]		(0,0)	circle(2cm);
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.55\textwidth}
\footnotesize
Find the set of points $(x,y)$ that satisfy the \emph{inequality}
$$ 1 < x^2+y^2 < 4$$

\medskip

\uncover<2->
{
\emph{Answer}: The annulus centered at $(0,0)$ and bounded 
by circles of radii $1$ and $2$
}
\end{minipage}

\bigskip

%%%%%%%%%%%%%%%%%%%%
%
%		Three-dimensional example
%
%%%%%%%%%%%%%%%%%%%%

\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.5]
\uncover<3->
{
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(4,0,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(0,4,0);
\draw[thick,black,->,>=stealth]	(0,0,0)	--	++(0,0,6);
\node[below] 	at		(0,0,6)	{\footnotesize{$x$}};
\node[right]	 at		(4,0,0)	{\footnotesize{$y$}};
\node[above]	 at		(0,4,0)	{\footnotesize{$z$}};
}
\uncover<4->
{
\begin{scope}
	\shade[ball color=blue!40, opacity=0.4] (0,0) circle(2cm);
\end{scope}
%% equator
\path[draw=none,fill=white,shade,opacity=0.7] 	
						plot[domain=0:1.57] ({cos(\x r)},{sin(\x r)}) --
						plot[domain=0:1.57] ({-0.4*sin(\x r)},{cos(\x r)- 0.4*sin(\x r)}) --
						plot[domain=0:1.57] ({sin(\x r)-0.4*cos(\x r)},{-0.4*cos(\x r)});
\path[draw=blue!80]	plot[domain=0:1.57] ({2*cos(\x r)},{2*sin(\x r)}) --
						plot[domain=0:1.57] ({ -0.8*sin(\x r) } ,{2*cos(\x r)-0.8*sin(\x r)} )--
						plot[domain=0:1.57] ({2*sin(\x r)-0.8*cos(\x r)},{-0.8*cos(\x r)});
\path[draw=none,fill=blue!80,shade,even odd rule,opacity=0.5]
						plot[domain=0:1.57] ({cos(\x r)},{sin(\x r)}) --
						plot[domain=0:1.57] ({-0.4*sin(\x r)},{cos(\x r)- 0.4*sin(\x r)}) --
						plot[domain=0:1.57] ({sin(\x r)-0.4*cos(\x r)},{-0.4*cos(\x r)}) 
						plot[domain=0:1.57] ({2*cos(\x r)},{2*sin(\x r)}) --
						plot[domain=0:1.57] ({ -0.8*sin(\x r) } ,{2*cos(\x r)-0.8*sin(\x r)} )--
						plot[domain=0:1.57] ({2*sin(\x r)-0.8*cos(\x r)},{-0.8*cos(\x r)});
%% cut away some of outer sphere

}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.55\textwidth}
\footnotesize
\uncover<3->
{
Find the set of all points $(x,y,z)$ that satisfy the \emph{inequality}
$$ 1 < x^2 + y^2 + z^2 < 4 $$
}

\medskip

\uncover<4->
{\emph{Answer}: The spherical shell centered at $(0,0)$ with inner radius $1$ and outer radius $2$}
\end{minipage}

\end{frame}

\begin{frame}
\frametitle{The Two Most Important Formulas in this Lecture}
\footnotesize
\begin{center}
\fbox{
\begin{minipage}{0.8\textwidth}
\textbf{Distance Formula in Three Dimensions}  The distance $|P_1P_2|$ between $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ is
$$
|P_1 P_2| = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
$$
\end{minipage}
}

\vfill

\fbox{
\begin{minipage}{0.8\textwidth}
\textbf{Equation of a Sphere}  The equation of a sphere with center $(h,k,\ell)$ and radius $r$ is
$$
(x-h)^2 + (y-k)^2 + (z-\ell)^2 = r^2
$$
\end{minipage}
}
\end{center}
\end{frame}

\section{Spheres}

\begin{frame}
\frametitle{Some Examples}
\footnotesize
Find the equation of a sphere with center at $(-9,4,8)$ and radius $3$.

\bigskip

\uncover<2->
{
\emph{Answer:} Using the distance formula on $P_1(-9,4,8)$ and $P_2(x,y,z)$ we see that
$$ (x+9)^2 + (y-4)^2 + (z-8)^2 = 3^2 $$
}
\end{frame}

\begin{frame}
\frametitle{Some Examples, Part II}
\footnotesize

Find the equation of a sphere if one of its diameters has endpoints $P_1(9,1,-8)$ and $P_2(11,5,-2)$.

\medskip

Here we'll need to use the given information to find the radius and the center.

\bigskip

\uncover<2->
{
Find the radius by finding the distance between the endpoints of the diameter:
}

\uncover<3->
{
$$
|P_1 P_2| 	= \sqrt{ (11-9)^2 + (5-1)^2 + (-2 - (-8))^2 }
				= \sqrt{56}
$$
so $r^2 = d^2/4 = 14$ (why?) 
}

\bigskip

\uncover<4->
{
Find the center $P(h,k,\ell)$ by finding the midpoint between $P_1$ and $P_2$:
}

\uncover<5->
{
$$
(h,k,\ell) =	 \left(\frac{9+11}{2}, \frac{1+5}{2}, \frac{-8-2}{2} \right)  =	(10,3,-5)
$$

You should now be able to find the equation of the sphere.
}
\end{frame}

\section{Review}

\begin{frame}
\frametitle{Lecture Review}
%\footnotesize
\begin{center}
\begin{minipage}{0.9\textwidth}
\begin{itemize}
\item	We introduced `right-handed' coordinate systems in three-dimensional ($xyz$) space
\smallskip
\item	We derived the \emph{distance formula} for the distance between two points in 
		three-dimensional space
		$$|P_1P_2| =  \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$
\smallskip
\item	We worked out and graphed equations for planes, cylinders, and spheres in three-dimensional space
\medskip
\item	I reminded you to access WebWork \emph{only through Canvas}!
\end{itemize}
\end{minipage}
\end{center}

\end{frame}



\begin{frame}
\frametitle{Homework}

\begin{center}
\fbox{
\begin{minipage}{0.8\textwidth}
Be sure to prepare for recitation tomorrow:

\begin{itemize}
\item Study section 12.1, pp.\ 792--796
\item	Download recitation worksheet and review Worksheet 1 on section 12.1
\item	Create your Webwork account by \emph{logging in through Canvas}
\item Begin Webwork Assignment A1 -- Remember to access WebWork \emph{only through Canvas!}
\end{itemize}

\medskip

For Wednesday, read and study section 12.2, pp.\ 798--804.
\end{minipage}
}
\end{center}
\end{frame}

\endgroup

\end{document}
