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\title{Math 234}
\author{What you should know on day one}
\date{\today}
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\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 $0<e<1$, then the set of all points
$P$ such that
$$
      d(P,F) = e\,d(P,L)
$$
is the ellipse with eccentricity $e$, directrix $L$, and
corresponding focus at $F$.
\arap

\section{Practice Questions}


\para
The following question  involve the three points
$$
A(-1,0,2), \qquad B(2,1,-1),\qquad C(1,-2,2).
$$

\medskipnoindent{\bf (i)} % 3 page 507
Find the  angle   $ABC$.

\Answer
The angle $\theta$ is given by the formula
$$
 \cos\theta = \left(\frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{
        \left|\overrightarrow{BA}\right|\;\left|\overrightarrow{BC}\right|
}\right)
$$
Now
$$
\begin{array}{lclcl}
\overrightarrow{BA}
        &=&(-1-2)\mathbf{i}+(0-1)\mathbf{j}+(2-2)\mathbf{k}
        &=& -3\mathbf{i}-\mathbf{j} \\
\overrightarrow{BC}
        &=&(1-2)\mathbf{i}+(-2-1)\mathbf{j}+(2-(-1))\mathbf{k}
        &=& -\mathbf{i}-3\mathbf{j}+3\mathbf{k}
\end{array}
$$
so
$$
\overrightarrow{BA}\cdot\overrightarrow{BC}=(-3)(-1)+(-1)(-3)+(0)(3)=6,
$$
$$
\left|\overrightarrow{BA}\right|=\sqrt{9+1+0}=\sqrt{10}, \qquad
\left|\overrightarrow{BC}\right|=\sqrt{1+9+9}=\sqrt{19}
$$
and $\theta=\cos^{-1}(6/\sqrt{190})$.

\medskipnoindent{\bf (ii)} % 3 page 512
Find an equation for the plane containing the three
points $A$, $B$, and $C$.

\Answer The vector
$$
   \mathbf{N} = \overrightarrow{BA}\times \overrightarrow{BC},
$$
is perpendicular to the plane and
$B$ is a point on the plane  so the point
$P(x,y,z)$ lies on the plane if and only if the vector
$\overrightarrow{BP}$ is perpendicular to the vector $\mathbf{N}$ so the equation
of the plane is
$$
 \mathbf{N}\cdot\overrightarrow{BP}=0.
$$
Now
$$
\mathbf{N}=\left|\begin{array}{rrr}
-3 & -1 & 0\\
-1 & -3 & 3\\
\mathbf{i} & \mathbf{j} &\mathbf{k}
\end{array}\right|=
-3\mathbf{i} +9\mathbf{j}+8\mathbf{k}
$$
and $\overrightarrow{BP} =(x-2)\mathbf{i}+(y-1)\mathbf{j}+(z+1)\mathbf{k}$
so the equation is
$$
   -3(x-2)+9(y-1)+8(z+1)=0.
$$

\medskipnoindent{\bf (iii)}
Find the area of the triangle $ABC$.

\Answer
The length of $\mathbf{N}$ is the area of the parallelogram
with edges $\overline{BA}$ and $\overline{BC}$ so the area
of the parallelogram is half this. The area
is $\frac12\sqrt{9+81+64}$.
\arap

\para

Find the distance from the origin to (the nearest point on) the line
$$
 \frac{x-2}{3}=\frac{y-1}{4}=\frac{2-z}{5}.
$$
\Answer There are several possible approaches to the
problem of finding the distance from the point $P_1(x_1,y_1,z_1)$
to the line $L$ whose equation is
$$
 \frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}. \eqno(L)
$$

\begin{description}
\item[(i)] The line has parametric equations
$$
   x = at+x_0,\qquad y=bt+y_0, \qquad z=ct+z_0,
$$
and the distance from $P_1$ to $P(x,y,z)$ is
$$
   r(t)=\sqrt{(at+x_0-x_1)^2+(bt+y_0-y_1)^2+(ct+z_0-z_1)^2}.
$$
The minimum value of the function $r$ is the distance from $P_1$ to $L$.
To find this solve the equation $r'(t)=0$ for $t$ and plug the answer
into $r(t)$.
\item[(ii)] The vector
$$
\mathbf{v}=a\mathbf{i}+b\mathbf{j}+c\mathbf{k}
$$
is parallel to the line.
Let $P(x,y,z)$ be the closest point of the line~$L$ to the point $P_1$.
Then the vectors $\overrightarrow{P_1P}$ and $\mathbf{v}$ are
orthogonal so
$$
  0 =
\overrightarrow{P_1P}\cdot\mathbf{v}=a(x-x_1)+b(y-y_1)+c(z-z_1).\eqno(*)
$$
The two equations~$(L)$ and the equation~$(*)$ give three
equations in three unknowns $x,y,z$. Solve to find $P$.
The distance from $P_1$ to $L$ is $|\overrightarrow{P_1P}|$.
\item[(iii)] Pick a point $P_0$ on the line; for example,
the point $P_0(x_0,y_0,z_0)$ and let
$P$ be the closest point of the line~$L$ to the point $P_1$
as in~(ii). Then $P_1PP_0$ is a right triangle so
$$
     |\overrightarrow{P_1P_0}|^2=|\overrightarrow{P_1P}|^2+|\overrightarrow{PP_0}|^2.
$$
Then $|\overrightarrow{P_1P_0}|^2=(x_1-x_0)^2+(y_1-y_0)^2+(z_1-z_0)^2$,
$$
|\overrightarrow{PP_0}| =\frac{\overrightarrow{PP_0}\cdot\mathbf{v}}{|\mathbf{v}|},
$$
and the distance from $P_1$ to $L$ is  $|\overrightarrow{P_1P}|$.
\item[(iv)] The vector $\mathbf{M}=\mathbf{v}\times\overrightarrow{P_1P_0}$
is orthogonal to the plane   containing the line and the point
$P_1$ so the vector $\mathbf{N}=\mathbf{v}\times\mathbf{M}$ lies in this
plane and is orthogonal to $\mathbf{v}$ and hence the line. The distance from $P_1$
to the line is the length of the projection of $\overrightarrow{P_1P_0}$ on
$\mathbf{N}$; i.e. the distance from $P_1$ to $L$ is
$$
   |\overrightarrow{P_1P}|=  \frac{\mathbf{N}\cdot\overrightarrow{P_1P_0}}{|\mathbf{N}|}.
$$
\item[(v)] Let $P_0$ and $Q$ be two points on the line, e.g.
$P_0(x_0,y_0,z_0)$ and $Q(x_0+a,y_0,+b,z_0+c)$.
The area of the parallelogram with vertices $P_0$, $P_1$, and $Q$
is
$|\overrightarrow{P_0P_1}\times\overrightarrow{P_0Q}|$.
But this area is also the length of the base $|\overrightarrow{P_0Q}|$
times the height and the height is the distance from $P_1$ to the line.
Hence the  distance from $P_1$ to $L$  is
$$
 \frac{|\overrightarrow{P_0P_1}\times\overrightarrow{P_0Q}|}{|\overrightarrow{P_0Q}|}.
$$
\end{description}
\arap

\para
The following questions concern the four points
$$
   P_1(1,2,3),\quad P_2(4,1,1), \qquad
   Q_1(2,3,4), \quad Q_2(1,5,1).
$$

\begin{description}
\item[(i)] Find parametric equations
for the line $P_1P_2$ through $P_1$ and $P_2$,
and
for the line $Q_1Q_2$ through $Q_1$ and $Q_2$.

\item[(ii)] Find a unit vector $\mathbf{u}$ perpendicular to both
the lines $P_1P_2$ and $Q_1Q_2$.


\item[(iii)] Find the distance $d(P_1P_2,Q_1Q_2)$
between the lines $P_1P_2$ and $Q_1Q_2$.

\item[(iv)] Find an equation for the plane $A$ which contains
the line $P_1P_2$ and is perpendicular to $\mathbf{u}$. Find an
equation for the plane $B$ which contains the line $Q_1Q_2$ and is
perpendicular to $\mathbf{u}$.

\item[(v)] The planes $A$ and $B$ are parallel. Find the distance
between them.

\item[(vi)] Find an equation for the plane $M$ which contains
the line $P_1P_2$ and is parallel to $\mathbf{u}$. Find an
equation for the plane $N$ which contains the line $Q_1Q_2$ and is
parallel to $\mathbf{u}$. (The planes $M$ and $N$ intersect.)


\item[(vii)] Find the point $P$ on the line $P_1P_2$ which is closest
to the line $Q_1Q_2$.  Find the point $Q$ on the line $Q_1Q_2$ which is closest
to the line $P_1P_2$.

\item[(viii)] Find the distance $d(P,Q)$ from the point $P$ to the point
$Q$.

\item[(ix)] How to the parts of this question relate to one another?
\end{description}
\arap

\para

The motion of a particle in the plane is governed by the equations
$$
  \frac{dx}{dt}= x, \qquad \frac{dy}{dt}=-x^2.
$$

\medskipnoindent{\bf (i)}
Find parametric equations for its coordinates $(x,y)$ at time  $t$ if
$(x,y)=(1,-4)$ when $t=0$.
(Hint: Find $x$ first.)

\Answer
We solve the equation $dx/dt=x$ by separation of variables:
$$
   \frac{dx}{x}=dt\implies \ln(x)=t+C\implies x=e^{t+C}.
$$
When $t=0$ we have $x=1$ so $1=e^{0+C}$ so $C=0$ so
$$
   x=e^t.
$$
Thus $\displaystyle\frac{dy}{dt}=-x^2=-e^{2t}$
so $y=-e^{2t}/2+ C_1$.
When $t=0$ we have $y=-4$ so $-4=-e^0/2+C_1$ so $C_1=-7/2$
$$
      y =-\frac{e^{2t}+7}{2}.
$$

\medskipnoindent{\bf (ii)}
Find the velocity vector $\mathbf{v}$
and the acceleration vector $\mathbf{a}$.

\Answer
$$
 \mathbf{v}=
\frac{dx}{dt}\mathbf{i}+\frac{dy}{dt}\mathbf{j}
=x\mathbf{i}-x^2\mathbf{j}
=  e^t\mathbf{i}-e^{2t}\mathbf{j}
$$
and
$$
 \mathbf{a}=
\frac{d^2x}{dt^2}\mathbf{i}+\frac{d^2y}{dt^2}\mathbf{j}
=  e^t\mathbf{i}-2e^{2t}\mathbf{j}.
$$
\arap

\para
Find the coefficients $a_k$ in the Maclaurin series
$\ds
       f(x)=\sum_{k=0}^\infty a_kx^k
$
for the function
$
     f(x)=(1+x)^{2/3}.
$

\Answer The general formula is Taylor's series
$$
    f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(a)(x-a)^k}{k!}.
$$
Maclaurin series arises by taking $a=0$
$$
   f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(0)x^k}{k!}.
$$
In case $f(x)=(1+x)^p$ this becomes the binomial formula

$$
   (1+x)^p=\sum_{k=0}^\infty \binom{p}{k} x^k.
$$
where
$$
\binom{p}{k} =\frac{p(p-1)(p-2)\cdots(p-k+1)}{k!}
$$
so
$$
a_k=\binom{\frac23}{k} =\frac{\frac23(\frac23-1)(\frac23-2)\cdots(\frac23-k+1)}{k!}.
$$
\arap

\end{document}

Here are the important quantities associated with a parametric curve.
\begin{itemize}
\item The velocity vector $\ds\bfv=\frac{d\bfr}{dt}$.
\item The acceleration vector $\ds\bfa=\frac{d\bfv}{dt}$.
\item The speed $\ds \frac{ds}{dt}=|\bfv|$.
\item The unit tangent vector $\ds\bfT = \frac{d\bfr}{|d\bfr|}=\frac{\bfv}{|\bfv|}$.
\item The curvature vector $\ds\frac{d\bfT}{ds}$.
\item The curvature $\ds\kappa=\left|\frac{d\bfT}{ds}\right|$.
\item The unit normal vector $\ds\bfN = \frac{d\bfT}{|d\bfT|}=\frac{d\bfT/ds}{|d\bfT/ds|}$.
\end{itemize}
\arap

\para Since all the products (scalar, dot, and cross)
are distributive over addition, the derivative of a product obeys the product
rule from first semester calculus. If $\bfu$ is a unit vector,
then $\bfu\cdot\bfu=|\bfu|^2=1$ so
$$
        \frac{d\bfu}{dt}\cdot\bfu = 0.
$$
Using this principle and the definitions you should be able to prove
equations like
$$
  \bfv =  \frac{ds}{dt}\bfT, \qquad
  \frac{d\bfT}{ds} = \kappa\bfN,
$$
$$
    \bfa = a_T\bfT+a_N\bfN,\qquad
    a_T=\frac{d^2s}{dt^2},\quad
    a_N =\left(\frac{ds}{dt}\right)^2\kappa,
$$
$$
     \kappa = \frac{|\bfv\times\bfa|}{(ds/dt)^3}.
$$
\arap
