
\chapter{Combined Parametric Motions}

Sometimes we are interested in the motion of a point $P$ which has a simple
motion relative to some other point $C$, which, in turn, also is moving in
a particular way.  It is not hard to write parametric formulas $x(t)$,
$y(t)$ for the point $P$, if we proceed carefully in a step-by-step manner.



%Example 1
\begin{example} {(The baton)}\index{the baton problem}

Suppose we want to describe the motion of a tip of a baton as it rotates
after being thrown into the air.  More precisely, suppose that at time
$t=0$ a $2$-foot long baton is situated vertically with the tip we are
interested in at height $6$ ft above ground level and the other tip at height
$4$ ft.  At that moment the baton is tossed at $20$ ft/sec at an angle of
$45{}^\circ$ above the horizontal, and it is also set spinning at $3$ rev/sec
clockwise in the plane of its trajectory.  (It could have been set spinning
in a different plane, but that would have made this into a {\it
three}-dimensional problem, and that would be too complicated for our
present purposes.)

\smallskip

To set up the parametric equations for the motion of the tip of the baton,
the basic procedure is to divide the motion into two parts: {\bf the motion
of the center of the baton} and {\bf the rotation of the tip around the
center}.  The first of these is a falling body problem, and the second is a
circular motion problem.  We shall take the $y$-axis to be the vertical
line along which the baton is situated at time $t=0$, and we shall take the
$x$-axis to be the line on the ground under the baton's trajectory.

\smallskip

According to a principle of physics, the center of the baton follows the
same path as a small object (say, a ball) that is thrown with the same
initial velocity.  In other words, the first half of our procedure --- the
motion of the center of the baton --- is like Example 10.6 of the last
chapter.  Let $(x_{\text{cent}}(t),y_{\text{cent}}(t))$ denote the $x$- and
$y$-coordinates of the center of the baton at time $t$.  Repeating the
procedure used in Example 10.6 --- with $g=32$~ft/sec${}^2$,
$v_{0,\text{horiz}}=20\cos 45^\circ=14$~ft/sec (rounded to the nearest
ft/sec), $v_{0,\text{vert}}=20\sin 45^\circ=14$~ft/sec --- we obtain:
\[
   x_{\text{cent}}(t)=14t,\qquad y_{\text{cent}}=-16t^2+14t+5.
\]
(Notice that $y_0=5$ ft, because the initial location of the center is
half-way between 4 and 6.)

\smallskip

Now let $(x(t),y(t))$ denote the $x$- and $y$-coordinates of the tip of the
baton at time $t$.  The second half of our procedure consists in
determining how much $x(t)$ differs from $x_{\text{cent}}(t)$ and how much
$y(t)$ differs from $y_{\text{cent}}(t)$, i.e., how far the tip is from the
center in the $x$- and $y$-directions.  Let $\theta$ be the angle in
radians through which the baton has spun at time $t$.  We have the
following diagram:

%--- figure LA --------------------------------------------------
%DIA Diagram on p. 54.
%

\begin{center}
\begin{psfigure}{1.25in}{1.5in}{figLA.ps}
\puttext{7.5, 15.}{$\theta$}
\puttext{40., 35.}{$(x,y)$}
\puttext{-35., 14}{$(x_{\text{cent}},y_{\text{cent}})$}
\put(-22,7){\vector(3,-1){18}}
\puttext{-10., -17.5}{\Small 1 ft}
\puttext{25., 7.5}{\Small 1 ft}
\end{psfigure}\\
\end{center}
%-----------------------------------------------------

\noindent
Since the tip is $1$ ft from the center, the hypotenuse of this triangle is
$1$.  Hence $x-x_{\text{cent}}=\sin\theta$ and
$y-y_{\text{cent}}=\cos\theta$, i.e., 
\[
x(t)=14t+\sin\theta,\qquad y(t)=-16t^2+14t+5+\cos\theta.
\]
It remains to write $\theta$ in terms of
$t$.  We do this as in Example 10.3 of the last chapter, obtaining
$\theta=6\pi t$.  We conclude that 
\[
  x(t)=14t+\sin(6\pi t),\qquad
  y(t)=-16t^2+14t+5+\cos(6\pi t).
\]
(Notice that here it is the $x$-coordinate that involves the sine and the
$y$-coordinate that involves the cosine, the reverse of what we got in our
earlier examples of circular motion.  The reason is that in this problem it
was convenient to measure the angle $\theta$ clockwise from the positive
$y$-direction rather than counterclockwise from the positive $x$-direction.
This different way of defining the angle accounts for the reversal of the
sine and cosine.)

\vspace{0.75in}

%--- figure LB --------------------------------------------------
%DIA Diagram at top of p. 55.
%
% psfix -width 5 -height 4 figLB >figLB.ps
%
\begin{center}
\begin{psfigure}{5in}{3.5in}{figLB.ps}

\puttext{-20., 55.}{\begin{minipage}{1.8in}
                 The dotted line is the  path of the center of 
                 the baton:
     
                 $x_{\text{cent}}(t) = 14t$ 

                 $y_{\text{cent}}(t) = -16 t^2 + 14 t + 5$
                 \end{minipage}}

\puttext{45., 45.}{\begin{minipage}{2.25in}
                 The solid line is the path of the top of 
                 the baton:
     
                 $x(t) = 14t + \sin(6\pi t)$ 

                 $y(t) = -16 t^2 + 14 t + 5 + \cos(6\pi t)$
                 \end{minipage}}

\puttext{50., -25.}{\begin{minipage}{1.5in}
                 Numbers indicate how many twentieths
                 of a second have elapsed.
                 \end{minipage}}
\end{psfigure}\\
\end{center}
%-----------------------------------------------------

The graph above shows the \index{trajectory} trajectory of the center of the baton (dotted
line) and the trajectory of the tip of the baton (solid line) between $t=0$
and $t=1$ sec.  The numbered points from 0 to 20 give the location at
intervals of $0.05$ seconds.  If you join the numbers on the dotted line
with the corresponding numbers on the solid line, you will see the spinning
of (the top half of) the baton.


Even though the equations for $x(t)$ and $y(t)$ are not tremendously
complicated, the path of the tip of the baton is a rather elaborate curve.
In this example we cannot find an algebraic or trig formula for $y$ in
terms of $x$, because we cannot eliminate $t$.  So the parametric method is
the only reasonable way to describe this curve.

\end{example}

% Example 2
\begin{example} {(The cycloid)}

Suppose a bicycle wheel of radius $r$ is rolling along the $x$-axis.  A
pebble lying at the origin is picked up in the tread of the tire as it
rolls over the origin.  We would like to describe the path of the pebble as
it moves along the $x$-axis, spinning around the center of the wheel as the
wheel turns.  This path is called a {\bf cycloid}.\index{parameterizing the cycloid}

\smallskip

As in Example 11.1, the procedure is to split the problem into two parts:
describe the motion of the center of the wheel, and then determine the
location of the pebble relative to the center.  The first part is easy,
because the center of the wheel is simply moving along the straight line
$y=r$ (parallel to the $x$-axis at a height of $r$); the second part is a
circular motion problem.

\smallskip

We shall express the coordinates of the center
$(x_{\text{cent}},y_{\text{cent}})$ and the coordinates of the pebble
$(x,y)$ in terms of the following parameter: the angle $\theta$ in radians
through which the wheel has rotated after picking up the pebble at the
origin.  We shall also introduce another letter $s$ to denote the linear
distance the wheel has traveled.  $s$ is the distance along the $x$-axis
from the origin to the point of contact of the wheel, and $s$ is also the
$x$-coordinate of the center of the wheel.

\smallskip

There is a simple relation between the distance $s$ the wheel has traveled
and the angle $\theta$ in radians through which it has rotated.  Namely,
the distance traveled along the $x$-axis is also equal to the distance
measured around the circumference through which the wheel has turned (here
we are assuming that the wheel rolls without slipping or sliding).  For
example, every time the wheel rotates through 1 revolution (i.e., $2\pi$
radians), the wheel moves a distance equal to the circumference $2\pi r$.
More generally, if the wheel rotates through an angle of $\theta$ radians,
the wheel moves a distance equal to $\theta r$.  This is because the
distance along the circumference of the unit circle of the part of the
circumference corresponding to an angle of $\theta$ radians is simply
$\theta$, by the definition of the measure of an angle in radians.  In a
circle of radius $r$ --- i.e., the unit circle magnified by a factor of $r$
--- the distance along the circumference will be $r$ times as great:

%--- figure LC --------------------------------------------------
%DIA Diagram at bottom of p. 55.
%

\begin{center}
\begin{psfigure}{4in}{1.25in}{figLC.ps}
\puttext{-40., -20.}{$\theta$}
\puttext{77.5, 27.5}{$r \theta$}
\puttext{-42.5, -55.}{$1$}
\puttext{75., -50.}{$r$}
\end{psfigure}\\
\end{center}
%-----------------------------------------------------
\medskip

\noindent
To summarize, we have the relation: 
\[ 
     s=r \, \theta.
\]

\smallskip

The center of the wheel always has the same $y$-coordinate $r$ (a
constant), and its $x$-coordinate is $s=r\, \theta$.  Thus, we have
\[
  x_{\text{cent}}(\theta)=r\theta,\qquad y_{\text{cent}}(\theta)=r.
\]


The second part of our procedure is to find the difference between the
coordinates $(x,y)$ of the pebble and the coordinates
$(x_{\text{cent}},y_{\text{cent}})$ of the center.  Since the wheel is
rolling from left to right, and the radius from the center to the pebble is
pointing down when we start out at the origin, it is convenient to measure
the angle $\theta$ clockwise from the negative $y$-direction.  We have the
following diagram:

\medskip

%--- figure LD --------------------------------------------------
 
%DIA First diagram on p. 56. 
%
\begin{center}
\begin{psfigure}{4.5in}{2.25in}{figLD.ps}
\puttext{-85., -35.}{$O$}
\puttext{-65., -20.}{$\theta$}
\puttext{-35., 0}{$(x_{\text{cent}},y_{\text{cent}})$}
\puttext{-95, -15}{$(x,y)$}
\end{psfigure}\\
\end{center}
%-----------------------------------------------------

\medskip

\noindent
If $\theta$ is an acute angle, as in the diagram, then $\sin\theta$ and
$\cos\theta$ are positive.  Since the pebble is to the {\it left} and {\it
below} the center, we must {\it subtract} the horizontal leg of the
triangle from $x_{\text{cent}}$ to get $x$, and we must {\it subtract} the
vertical leg of the triangle from $y_{\text{cent}}$ to get $y$.  Using the
trig functions of $\theta$ to get the horizontal and vertical legs of the
triangle (whose hypotenuse is equal to $r$), we end up with
\[
 x(\theta)=x_{\text{cent}}(\theta)-r\sin\theta=r\theta-r\sin\theta,\qquad
 y(\theta)=y_{\text{cent}}(\theta)-r\cos\theta=r-r\cos\theta.
\]
This parametric curve is drawn below.  Each time $\theta$ goes through
$2\pi$ radians, the wheel turns around once, and the pebble goes through
one motion from cusp to cusp.

\begin{center}
\begin{psfigure}{5in}{1.5in}{figLE.ps}
\puttext{-67.,-43.3}{$\pi r$}
\puttext{ 29., -43.3}{$2 \pi r$}
\puttext{125., -43.3}{$3\pi r$}
\puttext{-175., -3.}{$r$}
\puttext{-175., 30}{$2 r$}
\end{psfigure}\\
\end{center}
\smallskip

\newpage

The figure below shows the point on the curve for several values of
$\theta$.

%--- figure LF --------------------------------------------------
%DIA Second diagram on p. 56.
%
\begin{center}
\begin{psfigure}{3in}{6in}{figLF.ps}
\puttext{-30,42}{$\theta = 0$}
\puttext{-30,35}{$\theta = 60^{\circ}$}
\puttext{-30,28}{$\theta = 120^{\circ}$}
\puttext{-30,21}{$\theta = 180^{\circ}$}
\puttext{-30,14}{$\theta = 240^{\circ}$}
\puttext{-30, 6}{$\theta = 300^{\circ}$}
\puttext{-30, -1}{$\theta = 360^{\circ}$}
\puttext{-30,-8}{$\theta = 420^{\circ}$}
\puttext{-30,-15}{$\theta = 480^{\circ}$}
\puttext{-30,-22}{$\theta = 540^{\circ}$}
\puttext{-30,-30}{$\theta = 600^{\circ}$}
\puttext{-30,-37}{$\theta = 660^{\circ}$}
\puttext{-30,-44}{$\theta = 720^{\circ}$}
\end{psfigure}\\
\end{center}
%-----------------------------------------------------

Now suppose that we know how the bicycle is moving with time.  For example,
suppose that it is traveling at constant speed $v$, and it passes the
origin at time $t=0$.  That is, $ds/dt=v$.  To express the position of the
pebble in terms of $t$, all we have to do is express $\theta$ in terms of
$t$, and then substitute in the above equations for $(x,y)$.  Since
$\theta$ has a simple relation to $s$ --- namely, $\theta=\frac{1}{ r}\,s$
--- it follows that $d\theta/dt$ has a simple relation to $ds/dt$, namely
(taking $d/dt$ of both sides):
\[
\frac{d\theta}{ dt}=\frac{1}{ r}\,\frac{ds}{ dt}=\frac{1}{ r}\>v.
\]
Because $\theta$ is $0$ at time $t=0$, and it increases as the constant rate
$v/r$, it follows that we have: $\theta=(v/r)t.$ Thus, when the bicycle is
traveling at constant velocity $v$, we obtain the following equations for
the motion of the pebble:
\[
  x(t)=r(v/r)t-r\sin((v/r)t)=vt-r\sin(vt/r),\qquad y(t)=r-r\cos(vt/r).
\]

\end{example}


% Example 3
\begin{example} {(The trochoid)} 

The same as in Example 11.2, except that when the wheel rolls over the origin,
instead of getting stuck in the tread on the outside of the tire, the
pebble flies up and gets stuck in a spoke at a distance $a$ directly below
the center (where $a<r$).  Thus, the pebble again rotates around the
center, but at a distance $a$ rather than $r$ from the center.  The path of
this pebble is called a {\bf trochoid}. \index{parameterizing the trochoid}

\smallskip

In this problem everything is just like in Example 11.2, except that in the
circular motion part the hypotenuse of the triangle (which is the line from
the center to the pebble after the wheel has rotated through an angle of
$\theta$ radians) is now $a$ rather than $r$.  This leads to the formulas
\[
  x(\theta)=x_{\text{cent}}(\theta)-a\sin\theta=r\theta-a\sin\theta,\qquad
  y(\theta)=y_{\text{cent}}(\theta)-a\cos\theta=r-a\cos\theta.
\] 
Here is a picture of the trochoid path in the case when $a=r/2$:


%--- figure LG --------------------------------------------------
%DIA Diagram on p. 57.
%
 
\begin{center}
\begin{psfigure}{5in}{1.5in}{figLG.ps}
\puttext{-53.3, -43.3}{$\pi r$}
\puttext{53.3, -43.3}{$2 \pi r$}
\puttext{160., -43.3}{$3\pi r$}
\puttext{-167., 0.}{$r$}
\puttext{-167., 33.3}{$2 r$}
\end{psfigure}\\
\end{center}
%-----------------------------------------------------

\end{example}


