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\begin{document}

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


\title{Math 213 - Linear Approximation}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{September 27, 2019}
\maketitle

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

\begin{frame}
\frametitle{Reminders}

\begin{itemize}
\item  Homework B2 on 14.3 is due tonight!
\end{itemize}

\end{frame}

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

\begin{frame}
\frametitle{Unit II: Functions of Several Variables}
\begin{itemize}

\item[{13.3-4}] 	Lecture 11: Velocity and Acceleration

\medskip

\item[{14.1}] Lecture 12: Functions of Several Variables

\item[{14.3}] Lecture 13: Partial Derivatives
\begingroup
\color{red}
\item[{14.4}] Lecture 14: Linear Approximation
\endgroup
\item[{14.5}] Lecture 15: Chain Rule, Implicit Differentiation
\item[{14.6}] Lecture 16: Directional Derivatives and the Gradient

\medskip

\item[{14.7}] Lecture 17: Maximum and Minimum Values, I
\item[{14.7}] Lecture 18: Maximum and Minimum Values, II
\item[{14.8}] Lecture 19: Lagrange Multipliers

\medskip

\item[{15.1}]	Double Integrals
\item[{15.2}]	Double Integrals over General Regions
\item[{}]			Exam II Review
\end{itemize}
\end{frame}

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

\begin{frame}
\frametitle{Learning Goals}
\begin{itemize}
\item	Understand how the partial derivatives $f_x(a,b)$ and $f_y(a,b)$ define the \emph{tangent plane} to the graph of $z=f(x,y)$ at $(a,b,f(a,b))$

\item	Understand how the partial derivatives $f_x(a,b)$ and $f_y(a,b)$ define the \emph{linear approximation} $L(x,y)$ to $f(x,y)$ near $(x,y)= (a,b)$

\item	Understand the \emph{total differential} $dz$ of a function $z=f(x,y)$ and how it's used to compute percentage change and analyze error

\item	Generalize these ideas to functions of three variables

\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%		New material begins here...
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\begin{frame}
\frametitle{Warm-Up: Linear Functions}
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.4]
\begin{axis}[xlabel={$x$},ylabel={$y$}]
\addplot[domain=-2:2,red]
	{3*x+2};
\end{axis}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
The graph of a line $Ax + By = C$ defines a \emph{linear function}
of one variable
$$ y =f(x) = \frac{C}{B} - \frac{A}{C}x $$
\end{minipage}

\vfill

\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.4]
\begin{axis}[xlabel={$x$},ylabel={$y$}]
\addplot3[surf,domain=-2:2,samples=20]
	{2*x + 3*y};
\end{axis}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
The graph of a plane $ax + by + cz = d $
defines a \emph{linear function} of two variables
$$ z = f(x,y) = \frac{d}{c} - \frac{a}{c} x - \frac{b}{c} y $$
\end{minipage}
\end{frame}



\section{Tangent Plane}

\begin{frame}
\frametitle{Functions of One Variable - Tangent Line}
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.5]
\draw[thick,->,>=stealth]	(-1,0)	--	(4,0);
\draw[thick,->,>=stealth]	(0,-1)	--	(0,6);
\node[above]	at	(0,6)		{\scriptsize{$y$}};
\node[right]	at	(4,0)		{\scriptsize{$x$}};
\draw[red,thick]	plot[domain=0:4] ({\x}, {1+0.5*(\x-1)*(\x-1)});
\node[right] at	(4,5.5)		{\scriptsize{$y=f(x)$}};
\node[below] at (2,0)	{\scriptsize{$a$}};
\draw[black,fill=black]	(2,1.5) circle(0.10cm);
\node[above]	at	(1.2,1.5)	{\scriptsize{$(a,f(a))$}};
\draw[dashed]	(2,0)	--	(2,1.5);
\draw[thick,blue]		plot[domain=0:4]	({\x},{1.5+(\x-2)});
\node[right] at (4,3.5)	{\scriptsize{$y=L(x)$}};
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
\footnotesize
The derivative $f'(a)$ gives the slope of the tangent line to the 
graph of $y=f(x)$ at $(a,f(a))$. 

\bigskip
The derivative $f'(a)$ defines a linear function
$$ L(x) = f(a) + f'(a) (x-a) $$
the linear approximation to $f$ near $a$

\bigskip	
The differential of $y=f(x)$ is
$$ dy = f'(x) \, dx $$
\end{minipage}
\end{frame}

\begin{frame}
\frametitle{Functions of One Variable - Differentiability}
\footnotesize
\begin{minipage}{0.45\textwidth}
\begin{tikzpicture}[scale=0.6]
\draw[thick,->,>=stealth]	(-1,0)	--	(4,0);
\draw[thick,->,>=stealth]	(0,-1)	--	(0,6);
\node[above]	at	(0,6)			{\scriptsize{$y$}};
\node[right]	at	(4,0)			{\scriptsize{$x$}};
%%
\draw[red,thick]	plot[domain=0:4] ({\x}, {1+0.5*(\x-1)*(\x-1)});
%%
\node[below] at (2,0)			{\scriptsize{$a$}};
\draw[black,fill=black]			(2,1.5) circle(0.10cm);
%%
\node[above]	at	(1.2,1.5)	{\scriptsize{$(a,f(a))$}};
\draw[dashed]	(2,0)	--	(2,1.5);
%%
\draw[thick,blue]		plot[domain=0:4]	({\x},{1.5+(\x-2)});
%% new stuff
\draw[thick,<->,>=stealth]	(2,-1)		--	(3,-1);
\node[below] at					(2.5,-1)		{\scriptsize{$\Delta x$}};
\draw[black,fill=black]			(3,3) 	circle(0.10cm);
\draw[dashed]					(3,0)	--	(3,3);
%% label increment
\draw[thick,<->,>=stealth]	(5.5,1.5)	--	(5.5,3);
\node[right] at 					(5.5,2.25)	{\scriptsize{$\Delta y$}};
%%
\draw[dashed]	(3,2.5)	--	(4,2.5);
\draw[blue,<->,>=stealth]		(4.5,1.5)	--	(4.5,2.5);
\draw[green,<->,>=stealth]	(4.5,2.5)	--	(4.5,3);
%%
\draw[dashed]	(2,1.5)	--	(4,1.5);
\draw[dashed]	(3,3)		--	(4,3);
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.45\textwidth}
Recall that if $y=f(x)$, the \emph{increment} of $y$ as $x$ changes from $a$ to $a + \Delta x$
is
$$
\Delta y = f(a+\Delta x) - f(a).
$$
If $f$ is differentiable at $a$, then
$$ 
\Delta y = {\color{blue} f'(a) \, \Delta x} + {\color{green} \eps \Delta x}
$$
where
$$  \eps \to 0 \text{ as }\Delta x \to 0
$$

\medskip
That is, the linear approximation is \emph{very} good as $\Delta x \to 0$.
\end{minipage}

\end{frame}


\begin{frame}
\frametitle{Derivatives - Two Variables}
\begin{minipage}{0.4\textwidth}
\begin{tikzpicture}[scale=0.5]
\begin{axis}[title={$z=x^2+y^2$},surf,xlabel={$x$},ylabel={$y$}]
	\addplot3[view={60}{30},domain=-2:2]
		{x^2+y^2};
		\uncover<2->
		{
				\addplot3[view={60}{90},domain=0:2]
						{-2+2*x+2*y};
				\addplot3[mark=*,color=red]  coordinates {(1,1,2)};
		}
\end{axis}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.5\textwidth}
\footnotesize
The derivatives $f_x(a,b)$ and $f_y(a,b)$ define a \emph{tangent plane}
to the graph of $f$ at $(a,b,f(a,b))$

\bigskip
\uncover<2->
{
These derivatives define a linear function
\begin{align*}
 L(x,y) 	&= 	f(a,b) \\
 			&\quad + f_x(a,b) (x-a) \\
 			&\quad + f_y(a,b) (x-b) 
\end{align*}
the linear approximation to $f$ near $(a,b)$
}

\bigskip

\uncover<3->
{
The differential of $z=f(x,y)$ is 
$$ dz = \frac{\dee f}{\dee x} \, dx + \frac{\dee f}{\dee y} \, dy $$
}
\end{minipage}
\end{frame}

\begin{frame}
\frametitle{Find the Tangent Plane}
\footnotesize
If $f$ has continuous partial derivatives, the tangent plane to $z=f(x,y)$ at $(a,b,f(a,b))$ is
$$ z-f(a,b) = f_x(a,b) (x-a) + f_y(a,b) (y-b) $$

\medskip

\begin{minipage}{0.4\textwidth}
\only<1>
{
\begin{tikzpicture}[scale=0.5]
\begin{axis}[view={60}{50},title={$z=2x^2+y^2-5y$},xlabel={$x$},ylabel={$y$}]
\addplot3[domain=1:3,surf]
	{-6+4*x - y};
\addplot3[domain=0:3,surf]
	{2*x^2 + y^2 - 5*y};
\end{axis}
\end{tikzpicture}
}
\only<2>
{
\begin{tikzpicture}[scale=0.5]
\begin{axis}[view={60}{50},title={$z=2x^2+y^2-5y$},xlabel={$x$},ylabel={$y$}]
\addplot3[domain=1:3,surf]
	{1+x-y};
\addplot3[domain=0:3,surf]
	{exp(x-y)};
\end{axis}
\end{tikzpicture}
}
\end{minipage}
\qquad
\begin{minipage}{0.5\textwidth}
\footnotesize
\begin{enumerate}
\uncover<1->
{
\item		Find the equation of the tangent plane to the surface 
$$z=2x^2+y^2-5y$$ at $(1,2,-4)$.
}

\bigskip

\uncover<2->
{
\item		Find the equation of the tangent plane to the surface 
			$$z=e^{x-y}$$ at $(2,2,1)$.
}

\end{enumerate}

\end{minipage}
\end{frame}

\begin{frame}
\frametitle{The Tangent Plane Contains Tangent Lines}
\begin{minipage}{0.4\textwidth}
\begin{tikzpicture}[scale=0.5]
\begin{axis}[view={60}{60},xlabel={$x$},ylabel={$y$},zlabel={$z$},colormap/violet]
%% surface
\addplot3[surf,domain=0:3,y domain=0:3,samples=20]
	{2-0.25*(x-0.5)^2 - 0.5*(y-0.5)^2};
%% hold y=1.5 and get curve
\addplot3[domain=0:3,samples=20,samples y=0,red,very thick]
	({x},{1.5},{2-0.25*(x-0.5)^2 - 0.5});
%% hold x=1.5 and get curve
\addplot3[domain=0:3,samples=20,samples y=0,red,very thick]
	({1.5},{x},{2-0.25-0.5*(x-0.5)^2});
%% superimpose tangent plane
\addplot3[surf,domain=0.5:2.5,y domain=0.5:2.5,samples=10]
	{1.25-0.5*(x-1.5)-(y-1.5)};
%% draw tangent lines on tangent plane
\addplot3[domain=0.5:2.5,samples=20,samples y=0,blue,very thick]
	({x},{1.5},{1.25-0.5*(x-1.5)});
\addplot3[domain=0.5:2.5,samples=20,samples y=0,blue,very thick]
	({1.5},{x},{1.25-(x-1.5)});
\end{axis}
\end{tikzpicture}
\end{minipage}
\quad
\begin{minipage}{0.5\textwidth}
\footnotesize
The red curves represent $f(a,y)$ and $f(x,b)$

\bigskip

The blue lines are the tangent lines
$$ r_1(t) = \langle a,b, f(a,b) \rangle + t \langle 1, 0, f_x(a,b) \rangle $$
$$ r_2(t)	=	\langle a,b, f(a,b) \rangle + t \langle 0, 1, f_y(a,b) \rangle $$
\end{minipage}

\end{frame}

\section{Linear Approximation}

\begin{frame}
\frametitle{The Tangent Plane Defines a Linear Approximation}


\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.4]
\draw[thick,->,>=stealth]	(-1,0)	--	(4,0);
\draw[thick,->,>=stealth]	(0,-1)	--	(0,4);
\draw[help lines]				(-1,-1) grid	(4,4);
\node[above] 	at	(0,4)		{\scriptsize{$y$}};
\node[right] at (4,0)			{\scriptsize{$x$}};
\draw[thick,red]	plot[domain=-1:3]	({\x},{1+0.5*(\x-1)*(\x-1)});
%% tgt line at x=2
\draw[thick,blue]	plot[domain=0:3]	({\x},{1.5+(\x-2)});
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.55\textwidth}
\footnotesize
The tangent line is the graph of a linear function
$$ L(x) = f(a) + f'(a)(x-a) $$
that approximates $f(x)$ near $x=a$
\end{minipage}

\vfill

\begin{minipage}{0.35\textwidth}
\begin{tikzpicture}[scale=0.4]
\begin{axis}[view={60}{60},xlabel={$x$},ylabel={$y$},zlabel={$z$},colormap/violet]
%% surface
\addplot3[surf,domain=0:3,y domain=0:3,samples=20]
	{2-0.25*(x-0.5)^2 - 0.5*(y-0.5)^2};
%% hold y=1.5 and get curve
\addplot3[domain=0:3,samples=20,samples y=0,red,very thick]
	({x},{1.5},{2-0.25*(x-0.5)^2 - 0.5});
%% hold x=1.5 and get curve
\addplot3[domain=0:3,samples=20,samples y=0,red,very thick]
	({1.5},{x},{2-0.25-0.5*(x-0.5)^2});
%% superimpose tangent plane
\addplot3[surf,domain=0.5:2.5,y domain=0.5:2.5,samples=10]
	{1.25-0.5*(x-1.5)-(y-1.5)};
%% draw tangent lines on tangent plane
\addplot3[domain=0.5:2.5,samples=20,samples y=0,blue,very thick]
	({x},{1.5},{1.25-0.5*(x-1.5)});
\addplot3[domain=0.5:2.5,samples=20,samples y=0,blue,very thick]
	({1.5},{x},{1.25-(x-1.5)});
\end{axis}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.55\textwidth}
\footnotesize
The tangent plane is the graph of a linear function
\begin{multline*}
L(x,y) = f(a,b) +\\[5pt]
 f_x(a,b)(x-a) + f_y(a,b)(y-b)
\end{multline*}
that approximates $f(x,y)$ near $(x,y) = (a,b)$
\end{minipage}
\end{frame}

\begin{frame}
\frametitle{The Linear Approximation}
\footnotesize
The linear approximation to $f(x,y)$ at $(a,b)$ is
$$ L(x,y) = 	f(a,b)	+ f_x(a,b) (x-a) + f_y(a,b) (y-b) $$

\medskip

\begin{enumerate}
\item		Show that the linear approximation to $f(x,y) = e^x \cos(xy)$ at $(0,0)$ is
			$L(x,y) = x+1$
\item		Suppose that  $f(2,5)=6$, $f_x(2,5)=1$, and $f_y(2,5)=-1$.
			Use a linear approximation to estimate $f(2.2,4.9)$
\end{enumerate}
\end{frame}

\begin{frame}
\frametitle{Differentiability}
\footnotesize
If $z=f(x,y)$, the \emph{increment} of $z$ as $x$ changes from $a$ to $a+\Delta x$ and 
$y$ changes from $b$ to $b+\Delta y$ is:
$$ \Delta z = f(a+\Delta x, b+ \Delta y) - f(a,b) $$

\bigskip

$f$ is \emph{differentiable} at $(a,b)$ if 
$$ \Delta z  = f_x(a,b) \Delta x + f_y(a,b) \Delta y + \eps_1 \Delta x + \eps_2 \Delta y $$
where $\eps_1$ and $\eps_2$ approach $0$ as $(\Delta x, \Delta y) \to (0,0)$. 

\bigskip

\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
{\textbf{\color{cyan} Theorem}} If the partial derivatives $f_x$ and $f_y$ of $f$ exist near $(a,b)$, and are continuous at $(a,b)$, then $f$ is differentiable at $(a,b)$. 
\end{minipage}
}

\bigskip

\begin{enumerate}
\item		Explain why the function $f(x,y) = \sqrt{xy}$ is differentiable at $(1,4)$ and 
			find its linearization
\end{enumerate}
\end{frame}

\begin{frame}
\frametitle{What Happens if $f$ is not differentiable?}
\footnotesize
\begin{minipage}{0.40\textwidth}
\begin{tikzpicture}[scale=0.5]
\begin{axis}[view={60}{60},xlabel={$x$},ylabel={$y$},zlabel={$z$},colormap/cool]
\addplot3[surf,domain=-2:2,samples=51]
	{x*y/(x^2+y^2)};
\end{axis}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.50\textwidth}
\footnotesize
Let
$$ f(x,y) = 
\begin{cases}
\dfrac{xy}{x^2+y^2},	&	(x,y) \neq (0,0)	\\[5pt]
0							&	(x,y)  = (0,0)
\end{cases}
$$
\end{minipage}

\vfill

Use the definition to check that $f_x(0,0) = f_y(0,0) =0$ 

\medskip

Show that $f_x(x,y)$ and $f_y(x,y)$ are not continuous at $(0,0)$


\end{frame}


\section{The Differential}

\begin{frame}
\frametitle{Differentials}
\footnotesize
For a function $f$ of one variable, the differential of $y=f(x)$ is
given by
$$ dy = f'(x) \ dx $$

\medskip
For a function $f$ of two variables, the differential of $z=f(x,y)$ is
$$ dz = f_x(x,y) \, dx + f_y(x,y) \, dy = \frac{\dee z}{\dee x} \, dx + \frac{\dee z}{\dee y} \, dy $$

\bigskip

\hrulefill

\bigskip

\begin{enumerate}
\item[1.]		The radius of a circle is measured as $10$cm with an error of at most $0.2$cm. What is the maximum calculated area of the circle?

\medskip

\item[2.] 	The length and width of a rectangle are measured
as $30$cm and $24$cm, with an error of at most $0.1$cm each. What is the maximum error in the calculated area of the rectangle?
\end{enumerate}

\end{frame}

\section{Three Variables}

\begin{frame}
\frametitle{Three Variables}
\footnotesize
If $w = f(x,y,z)$:
\begin{itemize}
\item	The \emph{linear approximation} of $f$ at $(a,b,c)$ is
\begin{multline*}
L(x,y,z) 	= f(a,b,c) \\
	+ f_x(a,b,c) (x-a)	
	+ f_y(a,b,c) (y-a) 
	+ f_z(a,b,c) (z-c)
\end{multline*}
\item	The \emph{increment} of $w$ is
$$ \Delta w = f(x+\Delta x, y+ \Delta y, z+ \Delta z) - f(x,y,z) $$
\item	The \emph{differential} $dw$ is
$$ dw	= 	\frac{\dee w}{\dee x} \, dx + 
				\frac{\dee w}{\dee y} \, dy + 
				\frac{\dee w}{\dee z} \, dz
$$
\end{itemize}

\end{frame}

\begin{frame}
\frametitle{Three Variables}
\footnotesize
The \emph{linear approximation} of $f$ at $(a,b,c)$ is
\begin{multline*}
L(x,y,z) 	= f(a,b,c)	\\ 
		+ f_x(a,b,c) (x-a)	
		+ f_y(a,b,c) (y-a) 
		+ f_z(a,b,c) (z-c)
\end{multline*}

\bigskip
Find the linear approximation to
$$ f(x,y,z) = \sqrt{x^2+y^2+z^2} $$
at $(x,y,z)=(3,2,6)$ and estimate
$$ \sqrt{(3.02)^2 + (1.97)^2 + (5.99)^2} $$

\end{frame}


\begin{frame}
\frametitle{Summary}
\footnotesize
\begin{itemize}
\item	We showed how the partial derivatives $f_x(a,b)$ and $f_y(a,b)$ determine the tangent plane to the graph of $f$ at $(a,b,f(a,b))$
\item	We saw how the tangent plane defines a linear approximation $L(x,y)$ to $f(x,y)$ near $(x,y) = (a,b)$
\item	We saw how these ideas generalize to functions of three variables
\end{itemize}
\end{frame}


\end{document}