\documentclass{article} \usepackage{amsthm,amsfonts,amsmath,amssymb,latexsym,epsfig} % \title{Math 234} \author{What you should know on day one} \date{\today} % %\setlength{\textheight}{7.5in} \newcommand{\ds}{\displaystyle} \newcommand{\Answer}{\par\medskip\par\noindent{\bf Answer: }} \newcommand{\medskipnoindent}{\par\medskip\par\noindent} % \newcommand{\bfa}{\mathbf{a}} \newcommand{\bfb}{\mathbf{b}} \newcommand{\bfc}{\mathbf{c}} \newcommand{\bfi}{\mathbf{i}} \newcommand{\bfj}{\mathbf{j}} \newcommand{\bfk}{\mathbf{k}} \newcommand{\bfu}{\mathbf{u}} \newcommand{\bfv}{\mathbf{v}} \newcommand{\bfr}{\mathbf{r}} \newcommand{\bfN}{\mathbf{N}} \newcommand{\bfL}{\mathbf{L}} \newcommand{\bfT}{\mathbf{T}} % \newtheorem{PARA}{}%[section] \newcommand{\para}{\begin{PARA}\rm } \newcommand{\arap}{\end{PARA}} \newtheorem{theorem}[PARA]{Theorem} \newtheorem{corollary}[PARA]{Corollary} \newtheorem{lemma}[PARA]{Lemma} \newtheorem{proposition}[PARA]{Proposition} \newtheorem{definition}[PARA]{Definition} \newcommand{\dfn}{\begin{definition}\rm } \newcommand{\nfd}{\end{definition}} \newtheorem{remark}[PARA]{Remark} \newcommand{\rmk}{\begin{remark}\rm } \newcommand{\kmr}{\end{remark}} \newtheorem{example}[PARA]{Example} \newcommand{\xmpl}{\begin{example}\rm } \newcommand{\lpmx}{\end{example}} % % \newcommand{\vect}[1]{\overrightarrow{#1}} % \newcommand{\prob}[1]{\par\bigskip\par\noindent\mathbf{#1} } % \begin{document} \maketitle \para You should be able to use general principles like $$ \mbox{Length} = \int \,ds, \qquad \mbox{Area} = \int \, dA, \qquad \mbox{Volume} = \int\,dV. $$ For example the length of the semi circle $$ x=\cos t, \qquad y=\sin t, \qquad 0\le t\le \pi $$ is $$ L=\int \sqrt{(dx)^2+(dy)^2}= \int_0^\pi\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt= \int_0^\pi1\,dt=\pi $$ \arap \para You should be able to derive a formula like $$ ds = \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta $$ from more general formulas. For example, the last formula can be derived by taking $t=\theta$ in the formula $$ ds = \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt, \qquad x=r\cos\theta,\quad y=r\sin\theta. $$ \arap \para You should be comfortable with the idea of approximating a function by a (Taylor) polynomial. The polynomial of degree $n$ which best approximates the function $f(x)$ near $x=a$ is the Taylor polynomial $$ P_n(x)= \sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k. $$ The linear approximation is the case $n=1$ and is given by $$ L(x)=f(a)+f'(a)(x-a). $$ Thus $y=L(x)$ is the equation for the tangent line to $y=f(x)$ at the point $(x,y)=(a,f(a))$. The quadratic approximation is the case $n=2$ and is given by $$ Q(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2. $$ If $f'(a)=0$ and $f''(a)\ne 0$ then $f(x)$ has a (local) minimum (or maximum) at $x=a$ if and only if $Q$ does. \arap \para You should be comfortable with differentials and simple differential equations. \arap \para You should be comfortable with the distinction between vectors and points. The notation $P(x,y,z)$ means that $P$ is the point with coordinates $x,y,z$. The vector from $P_1(x_1,y_1,z_1)$ to $P_2(x_2,y_2,z_2)$ is $$ \vect{P_1P_2} = (x_2-x_1)\bfi+(y_2-y_1)\bfj+(z_2-z_1)\bfk. $$ A translation which moves $P_1$ to $P_1'$ and $P_2$ to$P_2'$ leaves the vector from $P_1$ to $P_2$ unchanged; i.e. $\vect{P_1P_2} =\vect{P_1'P_2'}$ if and only if $P_1P_2$ and $P_1'P_2'$ are opposite edges of a parallelogram. (In particular, if $\vect{P_1P_2} =\vect{P_1'P_2'}$, then the four points $P_1,P_2,P_1',P_2'$ are coplanar.) \arap \para You should know the fundamental vector operations. Suppose $c$ is a scalar (number) and $$ \bfa = a_1\bfi+a_2\bfj+a_3\bfk, \qquad \bfb = b_1\bfi+b_2\bfj+b_3\bfk, $$ are vectors. There are five vector operations: \begin{center} \renewcommand{\arraystretch}{1.5} \begin{tabular}{ll} {\em\small Vector Addition. } & $\ds\bfa+\bfb= (a_1+b_1)\bfi+(a_2+b_2)\bfj+(a_3+b_3)\bfk$.\\ {\em\small Scalar Product. }& $\ds c\bfa= ca_1\bfi+ca_2\bfj+ca_3\bfk$.\\ {\em\small Dot Product. } & $\ds\bfa\cdot\bfb=a_1b_1+a_2b_2+a_3b_3$.\\ {\em\small Cross Product. } & $\ds\bfa\times\bfb=(a_2b_3-a_3b_2)\bfi+(a_3b_1-a_1b_2)\bfj+(a_1b_2-a_2b_1)\bfk$.\\ {\em\small Length. } & $\ds|\bfa| = \sqrt{\bfa\cdot\bfa}=\sqrt{a_1^2+a_2^2+a_3^2}$. \end{tabular} \end{center} The formula for the cross product can best be remembered with the determinant formula $$ \bfa\times\bfb=\left|\begin{array}{ccc} a_1&a_2&a_3\\ b_1&b_2&b_3\\ \bfi&\bfj&\bfk \end{array}\right|. $$ \arap \para You should be familiar with the geometric interpretations of the vector operations. \begin{itemize} \item To add $\bfa$ and $\bfb$ translate $\bfb$ so its tail is the tip of $\bfa$ and draw the vector for the tail of $\bfa$ to the tip of $\bfb$. In other words, $\vect{P_1P_2}+\vect{P_2P_3}=\vect{P_1P_3}$. \item The vector $c\bfa$ points in the same direction as $\bfa$ if $c>0$ and in the opposite direction if $c<0$. Its length is $|c|$ times the length of $\bfa$, i.e. $|c\bfa|=|c|\,|\bfa|$. \item If $\theta$ is the angle between $\bfa$ and $\bfb$ then $$ \bfa\cdot\bfb = |\bfa|\,|\bfb|\,\cos\theta $$ and $$ |\bfa\times\bfb| = |\bfa|\,|\bfb|\,\sin\theta. $$ \item $\bfa\times\bfb\perp\bfa$, $\bfa\times\bfb\perp\bfb$, and $\bfa\times\bfb=-\bfb\times\bfa$. \item $|\bfa|$ is the distance from the tail of $\bfa$ to the tip of $\bfa$; i.e. $d(P_1,P_2)=|\vect{P_1P_2}|$ is the distance from the point $P_1$ to the point $P_2$. \end{itemize} \arap \para You should know how to use the geometric interpretation ro guide calculations. For example, \begin{itemize} \item The point $P$ lies on the line parallel to the vector $\bfL$ through the point $P_0$ if and only if $\vect{P_0P}=t\bfL$ for some number $t$. In particular the vector equation $\vect{P_0P}=t\vect{P_0P_1}$ gives parametric equations for the line through the points $P_0$ and $P_1$. \item Thus the line through the points $P_0(x_0,y_0,z_0)$ and $P_1(x_1,y_1,z_1)$ has the parametric equations $$ x=x_0+t(x_1-x_0),\qquad y=y_0+t(y_1-y_0), \qquad z=z_0+t(z_1-z_0), $$ i.e. a point $P(x,y,z)$ lies on this line if and only if its coordinates satisfy these parametric equations for some value of $t$. Eliminating $t$ gives the ``symmetric equations'' for the line: $$ \frac{x-x_0}{x_1-x_0}=\frac{y-y_0}{y_1-y_0}=\frac{z-z_0}{z_1-z_0}. $$ \item The point $P$ lies on the plane perpendicular to the vector $\bfN$ through the point $P_0$ if and only if $\vect{P_0P}\perp\bfN$, i.e. $\bfN\cdot\vect{P_0P}=0$. Thus the point $P(x,y,x)$ lies on plane perpendicular to $\bfN=a\bfi+b\bfj+c\bfk$ and through $P_0(x_0,y_0,z_0)$ if and only if its coordinates satisfy the equation $$ a(x-x_0)+b(y-y_0)+c(z-z_0)=0. $$ \item If the angle between $\bfa$ and $\bfb$ is $\theta$, the length of the projection of $\bfb$ along $\bfa$ is $|\bfb|\cos\theta$ and the area of the parallelogram with edges $\bfa$ and $\bfb$ is $|\bfa|\,|\bfb|\,\sin\theta$, The area of a triangle is one half the area of the corresponding parallelogram, and the volume of a parallelopiped is the area of the base time the altitude. (The altitude is the length of the projection of the third edge along the perpendicular to the first two.) \end{itemize} \arap \para You should be familiar with parametric representation of a curve. When the point $P$ is a function of time (or some other parameter) $t$ the equations $$ x=x(t),\qquad y=y(t),\qquad z=z(t), $$ are parametric equations for the curve. The position vector for the point $P(x,y,z)$ is $$ \bfr = \vect{OP} = x\bfi+y\bfj+z\bfk $$ where $O(0,0,0)$ is the origin. The {\em velocity vector} of the parametric curve is $$ \bfv=\frac{d\bfr}{dt}=\frac{dx}{dt}\bfi+\frac{dy}{dt}\bfj+\frac{dz}{dt}\bfk. $$ The {\em acceleration vector} is $$ \bfa=\frac{d\bfv}{dt}=\frac{d^2x}{dt^2}\bfi+\frac{d^2y}{dt^2}\bfj+\frac{d^2z}{dt^2}\bfk. $$ \arap \para You should have seen conic sections. (We will meet conic sections in~234: a planet moves in an ellipse with the sun at a focus.) If $F_1$ and $F_2$ are given points (in a plane) and $2a>d(F_1,F_2)$, then the set of all points $P$ such that $$ d(P,F_1)+d(P,F_2)=2a $$ is the ellipse with foci $F_1$ and $F_2$ and major axis $2a$. If $F$ is a point, $L$ is a line, and $0