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\begin{document}

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


\title{Math 213 - Calculus for Vector Functions}			%%	title page
\author{Peter A. Perry}
\institute{University of Kentucky}
\date{September 16, 2019}
\maketitle


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

\begin{frame}
\frametitle{Reminders}

\begin{itemize}

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

\medskip

\item	Homework A5 on section 13.1-13.2  is due Wednesday
\item	The Review Session for Exam 1 will take place tonight from 6 PM to 8 PM in KAS 213
\item	On Wednesday September 18 we will have an in-class review for Exam I
\item	Exam 1 takes place next Wednesday, September 18. Section 17 will meet in 
		CB 118, and sections 18 and 19 will meet in CB 122.

\medskip
		
	
\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	Unit Outline
%
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\begin{frame}
\frametitle{Unit I: Geometry and Motion in Space}
\begin{itemize}

\item[{12.1}]		Lecture 1: Three-Dimensional Coordinate Systems
\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
\begingroup
\color{red}
\item[{13.2}]		Lecture 9 Derivatives and Integrals of Vector Functions
\endgroup
\item[{}]				Lecture 10:  Exam I Review
\end{itemize}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%	Learning Goals
%
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\begin{frame}
\frametitle{Learning Goals}
\begin{itemize}
\item		Know how to compute derivatives and integrals of vector functions
\item		Know how to use the derivative to find tangent lines and unit tangents
\item		Know how to compute the arc length of a space curve
\end{itemize}
\end{frame}

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\begingroup
\small

\section{Derivatives and Integrals}

\begin{frame}
\frametitle{Mechanics}
\footnotesize
Derivative: if 
$$ \bfr(t) = \langle f(t), g(t), h(t) \rangle $$
then
$$ \bfr'(t) = \langle f'(t), g'(t), h'(t) \rangle$$

\vfill

\hrulefill

\vfill

Definite Integral: If
$$ \bfr(t) =  \langle f(t), g(t), h(t) \rangle $$
then
$$ \int_a^b \bfr(t) \, dt = \left\langle \int_a^b f(t) \, dt, \int_a^b g(t) \, dt, \int_a^b h(t) \, dt \right\rangle $$

\vfill

\hrulefill

\vfill
Indefinite integral: if
$$ \bfr(t) =  \langle f(t), g(t), h(t) \rangle $$
then
$$ \int \bfr(t) \, dt = \left\langle \int f(t) \, dt, \int g(t) \, dt, \int h(t) \, dt \right\rangle + \mathbf{C}$$
where $\mathbf{C}$ is a constant vector

\end{frame}

\begin{frame}
\frametitle{Meaning}
\footnotesize
\fcolorbox{cyan}{white}
{
\begin{minipage}{0.85\textwidth}
The \textbf{derivative} of a vector-valued function $\bfr(t)$ is given by
$$ \bfr'(t) = \frac{d\bfr}{dt} = \lim_{h \to 0} \frac{\bfr(t+h) - \bfr(t)}{h} $$
\end{minipage}
}
\vfill
\begin{minipage}{0.4\textwidth}
\begin{tikzpicture}[scale=0.5]
\drawxyz{6}
\draw[red]	plot[domain=0:6,smooth]	({\x},{6-0.5*(3-\x)*(3-\x)},{1+0.6*\x});
\uncover<2-3>
{
\draw[cyan,thick,->,>=stealth]	(0,0,0)	--	(3,6,2.8);			%%	r(3)
}
\uncover<3-3>
{
\draw[cyan,thick,->,>=stealth]	(0,0,0)	--	(2,5.5,2.2);			%% 	r(2)
\draw[black,thick,->,>=stealth]	(2,5.5,2.2)	--	(3,6,2.8);		%%	r(3)-r(2)
}
\uncover<4->
{
\draw[cyan,thick,->,>=stealth]	(0,0,0)	--	(2,5.5,2.2);			%% 	r(2)
\draw[red,fill=red]	(2,5.5,2.2)	circle(0.10cm);					%% point at r(2)
\draw[thick,red,->,>=stealth]											%% r'(2)
	(2,5.5,2.2)	--	++(1,1,0.6);
}
\end{tikzpicture}
\end{minipage}
\qquad
\begin{minipage}{0.5\textwidth}
\footnotesize
\uncover<2-3>
{
The vector
$$ \frac{\uncover<2-3>{\bfr(t+h)} \uncover<3-3>{-\bfr(t)}}{\uncover<3-3>{h}} $$
}
\uncover<3-3>
{
\noindent
measures the displacement from $t$ to $t+h$
}

\vfill
\uncover<4->
{
The vector $\bfr'(t)$ gives the instantaneous change in displacement


\medskip

The magnitude $|r'(t)|$ gives instantaneous \emph{speed}

}

\end{minipage}
\end{frame}

\begin{frame}
\frametitle{Tangent Vectors}
\footnotesize
Sketch the plane curve $\bfr(t) = \langle t-2,t^2+1 \rangle$ and sketch the tangent vector at $t=-1$

\begin{tikzpicture}[scale=0.4]
\draw[help lines]		(-4,-1)		grid		(1,5);
\draw[thick,->,>=stealth]	(-4,0)		--	(2,0);
\draw[thick,->,>=stealth]	(0,-1)	--	(0,6);
\begin{scope}
\clip	(-4,0) rectangle (2,5);
\draw[red]	plot[domain=-3:2]	({\x-2},{\x*\x+1});
\end{scope}
\uncover<2->
{
	\draw[red,fill=red]	(-3,2) circle(0.10cm);
}
\uncover<3->
{
	\draw[thick,cyan,->,>=stealth]	(-3,2)	--	++(1,-2);	
}
\end{tikzpicture}
\end{frame}

\begin{frame}
\frametitle{Lots of Rules}
\begin{center}
\fcolorbox{cyan}{white}
{
	\begin{minipage}{0.8\textwidth}
	\footnotesize
	\begin{align*}
	\frac{d}{dt} \left(\bfu(t) + \bfv(t)\right)	&=	\bfu'(t) + \bfv'(t)\\
	\frac{d}{dt}	\left( c\bfu(t) \right)			&=	c \bfu'(t)\\
	\frac{d}{dt} \left( f(t) \bfu(t) \right)		&=	f'(t) \bfu(t) + f(t) \bfu'(t)\\
	\frac{d}{dt} \left( \bfu(t) \cdot \bfv(t) \right)	&=	\bfu'(t) \cdot \bfv(t) + \bfu(t) \cdot \bfv'(t)\\
	\dfrac{d}{dt} \left( \bfu(t) \times \bfv(t) \right) &= \bfu'(t) \times \bfv(t) + \bfu(t) \times \bfv'(t)\\
	\dfrac{d}{dt} \left( \bfu(f(t)) \right)	&=	f'(t) \bfu'(f(t))
	\end{align*}
	\end{minipage}
}
\footnotesize

\vfill
There are three different versions of the ``product rule''!
\end{center}
\end{frame}



\section{Tangent lines, Unit Tangent}

\begin{frame}
\frametitle{Tangent Lines}
\footnotesize
Find parametric equations for the tangent line to the curve
$$ x = t, \quad y = e^{-t}, \quad z = 2t - t^2 $$
at $(0,1,0)$.

\vfill

\begin{itemize}
\item		What value of $t$ corresponds to $(0,1,0)$?

\bigskip
\item		What is $\bfr'(t)$ for this value of $t$?

\bigskip
\item		What are the point on the line and the vector along the line used to derive the parametric equations?

\vfill
\end{itemize}
\end{frame}



\begin{frame}
\frametitle{Tangent Lines, Unit Tangent Vector}
\footnotesize
The \textbf{unit tangent} to $\bfr(t)$ is the vector
$$ \bfT(t) = \frac{\bfr'(t)}{|\bfr'(t)|}. $$

\medskip

\hrulefill

\medskip

\begin{enumerate}

\item		If $\bfr(t) = \langle t, t^2, t^3 \rangle$, find $\bfr'(t)$, $\bfT(1)$, $\bfr''(t)$, and $\bfr'(t) \times \bfr''(t)$

\medskip

\item		If $\bfr(t) = \langle e^{2t}, e^{-2t}, te^{2t} \rangle$, find $\bfT(0)$, $\bfr''(0)$, and
			$\bfr'(t) \cdot \bfr''(t)$
			
\vfill
			
\item		Find the intersection of the curves $\bfr_1(t) = \langle t, t^2, t^3 \rangle$ and 
			$\bfr_2(t) = \langle \sin t, \sin 2t, t \rangle$ and compute their angle of intersection.

\end{enumerate}

\end{frame}

\section{Arc Length}

\begin{frame}
\frametitle{Arc Length - Two Dimensions}
\footnotesize
The arc length of a plane curve $x=f(t)$, $y=g(t)$, $a \leq t \leq b$ is
$$ L	=	\int_a^b \sqrt{[f'(t)]^2 + [g'(t)]^2 } \, dt $$

\medskip
Notice that:
\begin{itemize}
\uncover<2->
{
\item		If $\bfr(t) = \langle f(t), g(t) \rangle$, then $\bfr'(t) = \langle f'(t), g'(t) \rangle$

\medskip
}
\
\uncover<2->
{
\item		So $$|\bfr'(t)| = \sqrt{[f'(t)]^2 + [g'(t)]^2} $$

\medskip

}
\uncover<3->
{
\item		So we can write the arc length formula s
			$$ L = \int_a^b |\bfr'(t) | \, dt $$
}
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Arc Length - Three Dimensions}
\footnotesize
The arc length of the space curve $$x=f(t), \quad y=g(t), \quad z=h(t)$$ for $a \leq t \leq b$ is
\begin{align*}
L	&=	\int_a^b	\sqrt{ [f'(t)]^2 + [g'(t)]^2 + [h'(t)]^2 } \, dt \
\end{align*}
which is easiest to remember as
$$
L = \int_a^b \left| \bfr'(t) \right| \, dt
$$
After all, distance travelled should be the integral of speed!

\end{frame}

\begin{frame}
\frametitle{Arc Length - Three Dimensions}
\footnotesize
$$
L = \int_a^b \left| \bfr'(t) \right| \, dt
$$

\medskip

\hrulefill

\medskip

\begin{enumerate}

\item		Find the arc length of the curve
$$ \bfr(t) = \langle t, 3 \cos t, 3 \sin t \rangle $$
for $-5 \leq t \leq 5$.

\medskip

\item		Find the arc length of the curve
$$ \ii + t^2 \jj + t^3 \kk $$
for $0 \leq t \leq 1$. 

\end{enumerate}
\end{frame}

\begin{frame}
\frametitle{The Arc Length Function}
\footnotesize
If $C$ is a space curve given by a vector function
$$ \bfr(t) = f(t) \ii + g(t) \jj + h(t) \kk $$
for $a \leq t \leq b$,
the \textbf{arc length function} for $C$ is given by
$$ s(t) = \int_a^t \left| \bfr'(u) \right| \, dt $$
By the Fundamental Theorem of Calculus,
$$ \frac{ds}{dt} = \left| \bfr'(t) \right| . $$

\medskip

\hrulefill

\medskip
Find the arc length function for the curve
$$ \bfr(t) = \cos t \ii + \sin t \jj + t \kk, $$
$0 \leq t \leq 4\pi$ and re-parameterize this curve by arc length. 
\end{frame}


\section{Summary}

\begin{frame}
\frametitle{Summary}
\begin{itemize}
\item	We learned how to compute derivatives and integrals of vector functions
\item	We learned how to use the derivative to find the tangent line and unit tangent to a space curve at a given point
\item	We learned how to compute the arc length of a space curve (the distance travelled along the curve)
\end{itemize}

\end{frame}

\begin{frame}
\frametitle{Homework}
\begin{itemize}
\item	Continue reviewing for Exam I
\item	Re-read section 13.2
\item	Continue on Webwork A5 which is due no later than Wednesday
\end{itemize}
\end{frame}

\endgroup

\end{document}