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\textbf{College of the Holy Cross, Fall Semester, 2004} \\
\textbf{Math 131, Lab \#3} (Professor Hwang) \\
Friday, October 29
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\subsection*{Purpose} 

A parametrized curve gives the position of a point as a function of
time. The curve ``swept out'' by the particle depends on the component
functions. On this lab you'll use Maple plot several parametrized
curves that would be difficult to draw by hand. You needn't print out
your plots; just answer the questions. (Not every number has questions
to be handed in.)

\noindent Start by entering the Maple command \texttt{with(plots);}

\medskip

\noindent Once you've done this, commands such as
\begin{verbatim}
  plot([cos(t), sin(t), t=0 .. 2*Pi], scaling=constrained);
\end{verbatim}
will graph the indicated curve. (Note the square brackets around the
component functions and time interval. As in the previous lab, the
``scaling'' option forces Maple to use the same horizontal and
vertical scales.)

Plot the following curves, and use calculus to answer the accompanying
questions. Don't forget to use a ``$*$'' for multiplication, and
``\texttt{Pi}'' (capital~P, small~i) for~$\pi$.

\be

\item $x(t)=2t/(t^2+1)$, $y(t)=(t^2-1)/(t^2+1)$ for $-100\leq
  t\leq 100$. Identify this curve visually, and use algebra to show
  that your guess is correct.

  \sep

\item $x(t)=t-\sin t$, $y(t)=1-\cos t$ for $0\leq t\leq
  6\pi$. Find the values of~$t$ at which the velocity is zero. Where
  on the curve are these points located?

  Explanation: This curve, a \emph{cycloid}, is the path traced by a
  point on the rim of a rolling wheel of radius~1. By changing the
  multiplier on the trig functions, you can simulate motion on other
  wheels. For example, $x(t)=t-1.2\sin t$, $y(t)=1-1.2\cos t$ might
  represent the path of a point on the rim of a train wheel, while
  $x(t)=t-0.9\sin t$, $y(t)=1-0.9\cos t$ might be the path of a
  bicycle reflector. (Plot these curves as well to see the changes.)

  Note that unlike a function graph, a parametrized curve can have
  cusps or corners---even at points where both component functions are
  differentiable---if the velocity is zero. Intuitively, the point can
  come to a stop, then proceed in a different direction, thereby
  tracing a curve with no tangent line. 

  \sep

\item $x(t)=\cos^3t$, $y(t)=\sin^3t$, $0\leq t\leq 2\pi$. Find all
  points on the curve where the velocity is zero.  Where on the curve
  are these points located? Use algebra to show that this curve is
  given implicitly by $x^{2/3}+y^{2/3}=1$.

\vfil\eject

\item $x(t)=e^{t/2\pi}\cos t$, $y(t)=e^{t/2\pi}\sin t$, $-4\leq
  t\leq4$. Try graphing $x(t)=e^{kt}\cos t$, $y(t)=e^{kt}\sin t$ for a
  few values of~$k$ (both positive and negative). What happens as
  $k\to0$? How do the curves with $k>0$ differ from those with $k<0$?
  List four natural forms (from at least two kingdoms) that this
  curve resembles.

  Explanation: Curves in this family are known as \emph{logarithmic
  spirals}, and have the property that they meet every radial line
  through the origin at a constant angle. This property explains their
  appearance in biological growth.

  \sep

\item $x(t)=t\cos t$, $y(t)=t\sin t$, $0\leq t\leq 6\pi$.

  Explanation: This curve, a \emph{spiral of Archimedes}, can be used
  to trisect angles using only a compass and straightedge. It is known
  that a general angle cannot be trisected using \textbf{only} a
  compass and straightedge.

  \sep

\item $x(t)=\cos(3t)\cos t$, $y(t)=\cos(3t)\sin t$, for $0\leq t\leq
  2\pi$. How many ``petals'' do you think $x(t)=\cos(5t)\cos t$,
  $y(t)=\cos(5t)\sin t$, $0\leq t\leq 2\pi$, has? Plot the curve and
  see if your guess is correct.

  What about $x(t)=\cos(2t)\cos t$, $y(t)=\cos(2t)\sin t$, for $0\leq
  t\leq 2\pi$? Again, try it and see. If you can, explain the outcome
  of this experiment. How many petals do you think $x(t)=\cos(4t)\cos
  t$, $y(t)=\cos(4t)\sin t$ has?

  Explanation: These curves are often called ``rosettes'', for the
  obvious reason that they look like stylized flowers.

  \sep

\item Consider a circular ring of inner radius~$96$, inside of which
  is rolled a wheel of radius~$r<96$. (See figure.) A point on the
  wheel 80\%~of the way from the center to the edge traces the
  ``epicycloid'' parametrized by
  \[
  x(t) = (96-r)\cos t + (0.8r/96)\cos(96t/r),\qquad
  y(t) = (96-r)\sin t - (0.8r/96)\sin(96t/r).
  \]
  Plotting over $0\leq t\leq 2\pi$ may not give the entire curve; if
  this happens, use a larger multiple of~$2\pi$ for the plot interval.
  Plot the resulting curves for $r=24$, $30$, $35$, $36$, $40$, $52$,
  $56$, $63$, $64$, $72$, and~$84$.

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