
\chapter{Curves Given Parametrically}

So far we have had two different ways of giving a curve.  The first was
{\bf explicitly}, i.e., in the form $y=f(x)$.  For example, the equation
$y=\sqrt{r^2-x^2}$ gives us the upper semicircle of radius $r$ centered at
the origin.  The second way was {\bf implicitly}, usually in the form
$F(x,y)=$constant.  For example, the equation $x^2+y^2=r^2$ gives us the
circle (both upper and lower semicircles) of radius $r$ centered at the
origin.

\smallskip

There is a third fundamental way a curve can be given: {\bf parametrically}.
This means that we have a third variable, called the  {\bf parameter}, which
is used to determine where we are on the curve in the $xy$-plane.  To give
a curve \index{parametrically} parametrically means to give {\it two} formulas, one for $x$ and one
for $y$, each in terms of the parameter.   This
parameter might have a geometrical meaning --- for example, an angle
denoted $\theta$ --- or it may be the time $t$.    In the case of an angle
parameter, our two formulas take the form $x=g(\theta)$, $y=h(\theta)$; in the
case of a time parameter, the formulas take the form $x=g(t)$, $y=h(t)$.
(Often we don't bother to use separate letters like $g$ and $h$, instead
writing simply $x(t)$ for the formula for $x$ in terms of $t$ and $y(t)$ for
the formula for $y$ in terms of $t$.)
The best way to understand \index{parameterized curves} parameterized curves is through examples.


%Example 1. 
\begin{example} {(Circles)}  
\label{eg:10.1}
Suppose we want to describe the points on the circumference of the circle
of radius $r$ centered at the origin.  A common way to do this is in terms
of the angle $\theta$ which the line from the point to the origin makes
with the positive $x$-axis (measured counterclockwise).  That is, as
$\theta$ goes from $0$ to $2\pi$ radians, we sweep once around the circle:

%--- figure KA --------------------------------------------------
%DIA Diagram on p. 49.
%

\begin{center}
\begin{psfigure}{2.1in}{2.1in}{figKA.ps}
\puttext{9.5,4.8}{$\theta$}
\puttext{47.5,-4.7}{$r$}
\puttext{-47.5,-4.7}{$-r$}
\puttext{28.6,42.9}{$(x(\theta),y(\theta))$}
\end{psfigure}\\
\end{center}
%-----------------------------------------------------

\noindent
What we want to do is express the $x$- and $y$-coordinates of our points on
the circumference in terms of $\theta$.  To do this, we drop the
perpendicular from the point to the $x$-axis, obtaining a triangle with
hypotenuse $r$ and with legs $x$ and $y$.  So we can use the trig functions
$sin$ and $cos$ to express $x$ and $y$ in terms of $\theta$:
\[
  x=x(\theta)=r\cos\theta,\qquad 
  y=y(\theta)=r\sin\theta,\qquad
  0\le\theta\le 2\pi.
\]
This says that, as $\theta$ goes from $0$ to $2\pi$,
if we determine $x$ and $y$ by these two equations, then the corresponding
point $(x,y)$ goes exactly once around the circumference of the circle.

\smallskip

A special case of this is the {\bf unit circle} --- this is simply the case
$r=1$.  Thus, the points on the unit circle can be described parametrically
by the equations 
\[
   x=x(\theta)=\cos\theta,\qquad
   y=y(\theta)=\sin\theta,\qquad 0\le\theta\le 2\pi.
\]
The fact that the
$x$-coordinate is simply the cosine and the $y$-coordinate is simply the
sine is not surprising, since we {\it defined} the cosine and sine of an
angle to be  the $x$- and $y$-coordinates of the point that you get
by going around the unit circle through the given angle.

\smallskip

If we want to shift the circle of radius $r$ so that its center is at the
point $(h,k)$, we can do this by adding $h$ to the $x$-formula and
$k$ to the $y$-formula: 
\[
  x=x(\theta)=h+r\cos\theta,\qquad
  y=y(\theta)=k+r\sin\theta,\qquad 0\le\theta\le 2\pi.
\]
This is the circle which is given implicitly by the equation
$(x-h)^2+(y-k)^2=r^2$.

\smallskip

If instead of $0\le\theta\le 2\pi$ we had written, say, $0\le\theta\le\pi$,
then we would have described only half of the circle (the upper
semicircle).  If we had written $0\le\theta\le 4\pi$, then we would have
described going around the circle twice.

\end{example}

% Example 2
\begin{example} {(Ellipses)}
\label{eg:10.2}
Let us suppose that we have put our \index{ellipse} ellipse on the axes in the following
convenient way: its center is at the origin, its longer axis of symmetry
--- of length $2a$ --- is along the $x$-axis, and its shorter axis of
symmetry --- of length $2b$ --- is along the $y$-axis.  Such an ellipse can
be obtained by taking the unit circle and stretching it by a factor of $a$
in the $x$-direction and by a factor of $b$ in the $y$-direction.  (If $a$
or $b$ is less than 1, the ``stretching'' is actually a contraction.)  In
terms of the parametric equations, we can accomplish this stretching simply
by multiplying the $x$-equation for the unit circle by $a$ and multiplying
the $y$-equation for the unit circle by $b$:
\[
   x=x(\theta)=a\cos\theta,\qquad 
   y=y(\theta)=b\sin\theta,\qquad    0\le\theta\le 2\pi.
\]
This is the parametric form for our ellipse --- the
ellipse which can be given implicitly in the form 
$\disp\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

\end{example}

\smallskip

{\bf Note:}
A word of warning, though, about the parameter $\theta$ for an ellipse.  It
no longer has the meaning of the angle formed by the line from the point
$(x,y)$ to the origin.  For example, suppose $a=2$, $b=1$.  Then the point
on the ellipse corresponding to the parameter value $\theta=\pi/4 =
45^{\circ}$ is the point $x=2\cos(\pi/4)=\sqrt{2}$,
$y=\sin(\pi/4)=\sqrt{2}/2$; and it is easy to see that the line joining the
point $(1.414,0.707)$ to the origin does not make a $45^\circ$ angle with
the $x$-axis.

There is a geometric way to interpret the variable $\theta$ as a suitable angle
anyway and this is best illustrated from the following picture of an ellipse
with its auxiliary circle.

%%% To update! Add a picture.

\section{Trajectories}

It is very common to want to describe a curve in terms of the \index{time
parameter} time
parameter $t$.  In that case we sometimes call the curve the ``path'' or
``trajectory'' of our point (which we think of as a small object or
particle).  As $t$ goes from an initial time $a$ to a final time $b$, the
equations $x(t)$, $y(t)$ tell us where the particle is at that particular
time $t$. \index{trajectory}



%Example 3
\begin{example}
\label{eg:10.3}
Suppose a particle is traveling counterclockwise at $3$ rev/sec around the
circle of radius $r$ centered at the origin.  It starts at the point
$(r,0)$ at time $t=0$.  We want to give a parametric description of the
motion in terms of $t$.

\smallskip

To do this, we find an expression in terms of $t$ for the angle $\theta$
through which the particle has rotated, and then we substitute this
expression in the equations in Example~10.1.  Because we shall be measuring
angles in radians, we first translate the \index{angular velocity} angular velocity of $3$ rev/sec
into units of rad/sec.  Since $1$ rev $=~2\pi~$rad, we obtain:
$d\theta/dt=3\cdot 2\pi=6\pi$~rad/sec.  Thus, $\theta$ starts at 0 at time
$t=0$ (because our point starts at $(r,0)$ on the positive $x$-axis), and
it increases at the constant rate of $6\pi$.  Hence, after $t$ seconds it
has value $\theta=6\pi t$.  Finally, substituting this expression for
$\theta$ in the equations in Example~10.1, we obtain:
\[
  x=x(t)=r\cos(6\pi t),\qquad 
  y=y(t)=r\sin(6\pi t),\qquad 0\le t\le 1/3,
\] 
where we chose to take $t$ between 0 and $1/3$ so that we would describe
exactly one rotation of the particle around the circle.

\end{example}


% Example 4
\begin{example} {(Motion at constant velocity between two points)}
\label{eg:10.4}                    

\noindent
Suppose that at time $t=0$ you start at the point $P=(2,5)$ and travel at
constant velocity along the line from $P$ to the point $Q=(6,3)$, arriving
$5$ seconds later at $Q$.  Find parametric formulas for this motion.

\smallskip

To do this, we use a \index{general principle for setting up  parametric equations} general principle for setting up  parametric equations:
{\bf work with the $x$- and $y$-directions separately.} In the
$x$-direction in this example we are starting at $x=2$ and traveling to
$x=6$ in $5$ seconds, moving at constant speed.  This constant speed ---
called the ``horizontal component of velocity'' and denoted
$\dot x =\frac{dx}{ dt}$ --- is the distance traveled divided by time:
$(6-2)/5=0.8$.  Thus, the formula for $x(t)$ is $2+0.8t$ (i.e., starting
$x$-value plus speed times time).  Similarly, in the $y$-direction we
compute the vertical component of velocity $\dot
y=\frac{dy}{ dt}$ to be $(3-5)/5=-0.4$, so that $y(t)$ is given by the
formula $5-0.4t$.  This is all going on between $t=0$ and $t=5$.  Thus, our
parametric description of the motion is: 
\[
  x=x(t)=2+0.8t,\qquad
  y=y(t)=5-0.4t,\qquad 0\le t\le 5.
\]
We can check these formulas by noting
that when $t=0$ they give us $x(0)=2$, $y(0)=5$ (i.e., we're at the point
$P$), and when $t=5$ they give us $x(5)=6$, $y(5)=3$ (i.e., we're at the
point $Q$).

\end{example}

% Example 5
\begin{example} {(Falling body in two dimensions)}
\label{eg:10.5}

Again we are interested in an object falling under the \index{influence of gravity} influence of gravity
(we neglect air resistance).  In the past (see Example~\ref{eg:6.9}) our object was
moving only in one dimension --- up and down.  For example, if we drop a
ball from a window $19.6$ meters high, its height $y$ at time $t$ is given
by the formula $y=-\frac{1}{ 2}gt^2+y_0 =-4.9t^2+19.6$ (here $g=9.8$
m/sec${}^2$; the acceleration is negative, i.e., it is $-g$, because we are
choosing up to be the positive direction).

\smallskip

Now suppose that, instead of dropping the ball from the window, we throw it
horizontally out from the window at 10 m/sec.  The ball then moves in the
$xy$-plane, where we have chosen the side of the building as the $y$-axis
and the line along the ground under the ball's path as the $x$-axis.  Its
path is a parabola, as we shall soon see.
Our task in this example is to write \index{parametric equations} parametric equations for this motion.

\smallskip

\begin{rightindent}{2.5in}
To do this, we again consider the $x$- and $y$-directions separately.  In
the $x$-direction, there is no force affecting the velocity, and so the 10
m/sec horizontal component of velocity of the ball remains unchanged.  That
is, as in Example~10.4, in the $x$-direction we are simply traveling at
constant speed.
Since we start at initial location $x=0$, the formula for $x$ at time $t$
is: $x(t)=10t$.
\end{rightindent}
\hfill
%--- figure KB --------------------------------------------------
%DIA Diagram at top of p. 51.
%
\begin{psfigure}{2.0in}{2.2in}{figKB.ps}
\puttext{45,-27}{$x$}
\puttext{-30,40.}{$y$}
\end{psfigure}
%-----------------------------------------------------

\smallskip

In the vertical direction we have a falling body problem, and so can use
the formula $y=-\frac{1}{ 2}gt^2+v_0t+y_0$.  Here it is important to
understand what $v_0$ means.  $v_0$ is the initial velocity
\underline{in} \underline{the} $y$-\underline{direction}.  
In this example $v_0$ is $0$, since the ball's \index{initial velocity} initial velocity is entirely
horizontal.  In other words, in the $y$-direction the problem is exactly as
if we had simply dropped the ball from a height of 19.6.  Thus, we have
$y(t)=-4.9t^2+19.6$.

\smallskip

Finally, we want to know the time interval where the formulas for $x$ and
$y$ apply.  The motion starts at time $t=0$, and it ends when the ball hits
the ground.  The ball hits the ground when $y=0$, and so to find when this
happens we set $y=-4.9t^2+19.6$ equal to 0 and solve for $t$.  We obtain
$t^2=4$, $t=\pm 2$, and so $t=2$ (since $-2$ does not make practical
sense).  To summarize, we have the following parametric description of the
motion: 
\[
      x=x(t)=10t,\qquad y=y(t)=-4.9t^2+19.6,\qquad 0\le t\le 2.
\]

\smallskip

The actual trajectory in Example~10.4 (see the picture above) is a graph in
the $xy$-plane, that is, a function of the form $y=f(x)$.  In this example,
we can easily find the formula for this function by {\it eliminating} $t$
in the formulas for $x(t)$ and $y(t)$.  That is, since $x=10t$, we can
write $t=0.1x$, and then substitute this in place of $t$ in the formula for
$y$: $y=-4.9(0.1x)^2+19.6= -0.049x^2+19.6$.  Thus, the path $f(x)$ is a
parabola (more precisely, part of a downward-opening parabola).

\end{example}



\section{The Velocity Vector}

The velocity of an object moving in the plane is an example of a \index{vector} vector. A
{\bf vector} is something which has \index{magnitude and direction} magnitude and direction.  It can be
visualized as an arrow of a certain length (or {\bf magnitude}) pointing in
a certain direction. The magnitude of the \index{velocity vector} velocity vector is called the
{\bf speed} of the object.


% Example 6
\begin{example}
\label{eg:10.6}

The same as Example~10.5, except that at time $t=0$ the ball is thrown at an
angle of $60^\circ$ above the horizontal.

\smallskip

Here the initial velocity vector makes an angle of $60^{\circ}$ above the
horizontal and has a magnitude of 10 m/sec.  That is, it is partly
horizontal and partly vertical.  The first step in solving this problem is
to {\bf resolve the initial velocity vector into its horizontal and
vertical components}.  To resolve the velocity vector, we draw the {\bf
velocity triangle} having the velocity vector as its hypotenuse (see

\noindent
\begin{rightindent}{3.5in}
\noindent
figure). This triangle has hypotenuse $10$ m/sec, making an angle of
$60{}^\circ$ above the horizontal.  Then the horizontal leg of the triangle
--- the initial $x$-velocity $v_{0,\text{horiz}}=\dot{x}(0)$ --- can be
found using the cosine, and the vertical leg of the triangle --- the
initial $y$-velocity $v_{0,\text{vert}}=\dot{y}(0)$ --- can be found using
the sine.  We thus have:
\end{rightindent}
\hfill
%--- figure KC --------------------------------------------------
%DIA Diagram at bottom of p. 51.
\begin{psfigure}{1.75in}{1.0in}{figKC.ps}
%\putcrosshair{0,0}
\puttext{0,-30}{\Small$60^{\circ}$}
\puttext{-28,5}{\Small$10$ m/sec}
\puttext{77,-5}{\Small$10 \sin 60^{\circ}$ m/sec}
\puttext{0,-57}{\Small$10 \cos 60^{\circ}$ m/sec}
\end{psfigure}
%-----------------------------------------------------
\begin{align*}
  v_{0,\text{horiz}}&=\dot{x}(0)
     =10\cos 60^\circ=10\cdot 0.5=5\,\text{m/sec}, \\
  v_{0,\text{vert}}&=\dot{y}(0)=10\sin 60^\circ
     =10\cdot 0.866=8.66\,\text{m/sec}.
\end{align*}

\bigskip

We are now ready to write the equations for $x(t)$ and $y(t)$, as we did in
Example~\ref{eg:10.5}.  The \index{horizontal component} horizontal component of
velocity (which remains constant) is not $10$ m/sec, but rather only the
horizontal part, namely, $5$ m/sec.  So $x=5t$.  In the 

\begin{rightindent}{2.5in}
\noindent
falling body formula $y=-\frac{1}{ 2}gt^2+v_0t+y_0$, we now have a nonzero initial
vertical component of velocity $v_0=v_{0,\text{vert}}=8.66$ m/sec.
Thus, we obtain: 
\begin{align*}
     x&=x(t)=5t,\\
     y&=y(t)=-4.9t^2+8.66t+19.6.
\end{align*}
The trajectory of the ball is shown in the figure at the right.

To find the time interval until the ball hits the ground, we again set
$y=0$, and use the \index{quadratic formula} quadratic formula to find the time $t$ when we're on the
ground.  There will be $2$ roots, of which we want the positive one:
\end{rightindent}
\hfill
%--- figure KDa --------------------------------------------------
\begin{psfigure}{2.5in}{2.5in}{figKDa.ps}
\puttext{45,-40}{$x$}
\puttext{-30,48}{$y$}
\end{psfigure}
%-----------------------------------------------------
\[
  (-8.66-\sqrt{8.66^2-4\cdot(-4.9)\cdot 19.6})/(2\cdot (-4.9))=3.07 \text{
sec}.
\]
So the time interval is $0\le t\le 3.07$.


\end{example}


%Example 7
\begin{example}
\label{eg:10.7}

The same as Example~\ref{eg:10.6}, except that the ball is thrown at an angle of
$60{}^\circ$ {\it below} the horizontal.

\smallskip


Here the only change is that the tip of the velocity triangle points
downward, i.e., its vertical leg $\dot{y}(0)$ points in the {\it negative}
direction.  Thus, in the equations of motion the only thing that changes is
the sign of $v_{0,\text{vert}}$ in the \index{falling body formula} falling body formula:
\[
      x=x(t)=5t,\qquad y=y(t)=-4.9t^2-8.66t+19.6.
\]
Of course, the ball hits the ground much sooner (see figure on the
following page).

\end{example}
In all three cases in Examples~\ref{eg:10.5}--\ref{eg:10.7}, the trajectory is part of a parabola,
as we can see by eliminating $t$ and writing $y$ in terms of $x$ rather
than in terms of $t$.

\bigskip

Suppose that we have a \index{trajectory} trajectory $x(t)$, $y(t)$, as in Examples~\ref{eg:10.3}--\ref{eg:10.7}.
There are three types of derivatives we might be interested in.  One is the
{\bf horizontal component of velocity} $dx/dt=\dot{x}(t)$.  The second is
the {\bf \index{vertical component of
velocity} vertical component of
velocity} $dy/dt=\dot{y}(t)$.  The third is not a velocity at all,
but rather the slope of the tangent to the 
\begin{rightindent}{3.5in}
\noindent
curve in the $xy$-plane: $dy/dx$.  If we find
$dy/dx$ at some instant, we can determine the direction in which the object
is headed at that instant.


How do we find $dy/dx$ if we do not have $y$ written in terms of $x$, i.e.,
if we do not have a formula for the function $y=f(x)$?  Well, according to
the chain rule, at any instant our three derivatives are connected by the
relation 
\[
\frac{dy}{ dt}=\frac{dy}{ dx}\cdot\frac{dx}{ dt}.
\]
In the past we
have used the chain rule to find the left hand side.  Now we want to find
$dy/dx$.  So the form of the chain rule we need is: 

\smallskip
\[
\frac{dy}{ dx}=\frac{dy/dt}{ dx/dt}=\frac{\dot{y}}{\dot{x}}.
\]

\end{rightindent}
\hfill
%--- figure KDb --------------------------------------------------
\begin{psfigure}{1.5in}{2.5in}{figKDb.ps}
\puttext{20., -35.}{\Small$x$}
\puttext{-15., 45.}{\Small$y$}
\end{psfigure}
%-----------------------------------------------------

\smallskip

\noindent
In words, this says that {\it the slope of the tangent to the path is 
equal to the vertical
component of velocity divided by the horizontal component of
velocity}.  This makes sense: if the horizontal component of velocity is
small and the vertical component of velocity is large, we are traveling at a
steep angle; while if the horizontal component of velocity is large and the
vertical component of velocity is small, we are traveling at a shallow angle.

\smallskip

In finding $dy/dx$ it is always important to remember that you take the
ratio of {\it velocities} $\dot{y}$ over $\dot{x}$, {\bf NOT} simply
the ratio of $y$ to $x$.  In other words, you must take the derivatives
(with respect to $t$) of the $y(t)$ and $x(t)$ formulas before taking
their ratio. 


% Example 8
\begin{example}
\label{eg:10.8}

Suppose we want to know the instant when the ball in Example~\ref{eg:10.6} reaches the
peak of its trajectory.  At that instant the path has a horizontal tangent
line, i.e., $dy/dx=0$.  Since $dy/dx =\dot{y}/\dot{x}$, this slope is zero
if and only if the numerator $\dot{y}$ is $0$.  Thus, to find the instant
when the ball reaches its highest point we set the vertical velocity
$\dot{y}$ equal to 0 and solve for $t$.  This is the same principle that we
saw in our earlier one-dimensional falling body problems: the ball is at
the peak when its vertical velocity is zero.  (Warning: Be sure not to
confuse this procedure, where we set $\dot{y}$ equal to zero, with setting
$y$ itself equal to 0, which we do to find the time when the ball hits the
ground.)  Since $\dot{y}=\frac{d}{ dt}(-4.9t^2+8.66t+19.6)=-9.8t+8.66$, we
have $-9.8t+8.66=0$, i.e., $t=8.66/9.8=0.88$~sec.  If we also want to know
the maximum height reached by the ball, we take this value of $t$ and
substitute it in the formula for the height $y$, i.e., 
\[
y_{\text{max}}=y(0.88)=-4.9\cdot(0.88)^2+8.66\cdot 0.88+19.6=23.4~\text{m}.
\]
\end{example}


% Example 9
\begin{example}
\label{eg:10.9}

Suppose that we want to know the direction in which the ball in Example~\ref{eg:10.6}
is headed after 0.5 sec, and also at the instant when it hits the ground.
We first take the time derivatives of $y$ and $x$:
\[
\frac{dy}{dt}=\frac{d}{ dt}(-4.9t^2+8.66t+19.6)=-9.8t+8.66
\text{ and} \frac{dx}{dt}=\frac{d}{ dt}(5t)=5\,.
\]
Then we have: 
\[
   \frac{dy}{dx}= \frac{\dot{y}(t)}{\dot{x}(t)}= (-9.8t+8.66)/5 \,.
\]
Thus, when $t=0.5$ we have $dy/dx=(-9.8\cdot 0.5+8.66)/5= 0.752$.  At that
instant the ball's location is $(x(0.5),y(0.5))=(2.5,22.7)$, so that the
tangent line to the path at the point $(2.5,22.7)$ has slope $0.752$.  We can
also determine the angle at which the ball is traveling at time $t=0.5$.
By this we mean the angle above the horizontal of the tangent line to the
path.  To do that we observe that the slope of a line $\Delta y/\Delta x$
is opposite over adjacent, i.e., the tan of the angle that the line makes
with the horizontal.  So to determine the angle we use the inverse-tan on
our calculator: $\tan^{-1}0.752=37^\circ$.  At the same time $t=0.5$ we
could also construct the velocity triangle, having horizontal leg
$\dot{x}(0.5)=5$ and vertical leg $\dot{y}(0.5)=3.76$, and compute the
hypotenuse (which is the speed):
\[
\text{speed at time 0.5} = \sqrt{5^2+3.76^2}=6.26 \text{ m/sec.}
\]

%--- figure KE --------------------------------------------------
%DIA Two diagrams side by side on p. 53.

\begin{center}
\begin{psfigure}{5in}{3.0in}{figKE.ps}
\puttext{-66.7, 36.7}{$37^{\circ}$}
\puttext{-50., 16.7}{$t = 0.5 \text{ sec}$}
\puttext{-60., 26.6667}{$5 \text{ m/sec}$}
\puttext{-35, 36.7}{$3.76 \text{ m/sec}$}
\puttext{43.3, 30.}{$-77^{\circ}$}
\puttext{33.3, 16.7}{$t = 3.07 \text{ sec}$}
\puttext{60., 40.}{$5 \text{ m/sec}$}
\puttext{83.3, 16.7}{$-21.4 \text{ m/sec}$}
\end{psfigure}\\
\end{center}
%-----------------------------------------------------

Next, we make the same computations at the instant when the ball hits the
ground, i.e., at $t=3.07$ sec.  We have $\dot{x}(3.07)=5$,
$\dot{y}(3.07)=-9.8\cdot 3.07+8.66=-21.4$, and so
speed$=\sqrt{5^2+(-21.4)^2}=22.0$.  The angle at which the ball hits the
ground is $\tan^{-1}(-21.4/5)=-77^\circ$.  Thus, the ball hits the ground
traveling at $22$ m/sec at an angle of $77{}^\circ$ below the horizontal.

\end{example}

 


