\chapter{Polar coordinates and complex numbers}


The rectangular coordinate system is immensely useful, 
but it is not the only way to assign an address to a
point in the plane and sometimes it is not the most useful. 
In applications to physical problems where there is some 
``cylindrical symmetry'', such as a vibrating drum or 
water moving along a pipe, the most natural coordinates are
often polar coordinates rather than rectangular coordinates.
  
In many experimental situations, our location
is fixed and we, using sonar or radar, take 
readings in different directions (Figure \ref{fig:9-1-2}); this
information can be graphed using rectangular coordinates 
(e.g., with the angle on the horizontal axis and the
measurement on the vertical axis). 

\begin{figure}[h!]
\begin{minipage}{\textwidth}
\begin{center}
\vspace{1.0 cm}
\includegraphics[height=4cm,width=13cm]{graphs/fig9-1-1.eps}
\end{center}
\end{minipage}
\caption{Sonar and radar use polar coordinates.}
\label{fig:9-1-1}
\end{figure}

\newpage

\noindent
Sometimes, however, 
it is more useful to plot the information in a way
similar to the way in which it was collected, as 
magnitudes along radial lines (Figure \ref{fig:9-1-2}). This system is
called the Polar Coordinate System.

\begin{figure}[h!]
\begin{minipage}{\textwidth}
\begin{center}
%\vspace{1.0 cm}
\includegraphics[height=4cm,width=5cm]{graphs/fig9-1-2.eps}
\end{center}
\end{minipage}
\caption{Polar coordinates.}
\label{fig:9-1-2}
\end{figure}

\begin{example}
\label{ex:polar1}
SOS! You've just received a
distress signal from a ship located at A
on your radar screen (Figure \ref{fig:9-1-3}). Describe
its location to your captain so your
vessel can speed to the rescue.


\begin{figure}[h!]
\begin{minipage}{\textwidth}
\begin{center}
%\vspace{1.0 cm}
\includegraphics[height=5cm,width=10cm]{graphs/fig9-1-3.eps}
\end{center}
\end{minipage}
\caption{Polar coordinate figure for Example \ref{ex:polar1}.}
\label{fig:9-1-3}
\end{figure}

Solution:    You could convert the relative location 
of the other ship to rectangular coordinates and then 
tell your captain to go due east for 7.5 miles and north for 13
miles, but that certainly is not the quickest way to reach the 
other ship. It is better to tell the captain
to sail for 15 miles in the direction of $60^o$. If the distressed 
ship was at B on the radar screen, your
vessel should sail for 10 miles in the direction $150^o$. 
(Real radar screens have $0^o$ at the top of the
screen, but the convention in mathematics is to put $0^o$ 
in the direction of the positive $x$--axis and to
measure positive angles counterclockwise from there. And 
of course a real sailor speaks of ``bearing'' and ``range''
instead of direction and magnitude.)
\end{example}

\begin{practice}
Describe the locations of the ships at C and D in 
Figure \ref{fig:9-1-3} by giving a distance and a
direction to those ships from your current position at the
center of the radar screen.
\end{practice}      

{\it Points in Polar Coordinates}:
To construct a polar coordinate system we need a starting point (called the
origin or pole) for the magnitude measurements and a starting direction
(called the polar axis) for the angle measurements. 
A {\bf polar coordinate pair} for a point $P$ in the plane 
is an ordered pair $(r,\theta)$, where $r$
is the directed distance along a radial line from $O$ to $P$, 
and $\theta$ is the angle formed by the polar axis and
the segment $OP$. The angle $\theta$ is positive when the angle 
of the radial line $OP$ is measured counterclockwise 
from the polar axis, and $\theta$ is negative when measured clockwise.

{\it Degree or Radian Measure for $\theta$?}:
Either degree or radian measure can be used for the angle in the
polar coordinate system, but when we differentiate and integrate 
trigonometric functions of $\theta$ we will {\it always} want
all of the angles to be given in {\it radian measure}. 
From now on, we will primarily use radian measure. {\it You
should assume that all angles are given in radian measure 
unless the units ``$\, ^o$'' (``degrees'') are shown.}

In the rectangular coordinate system, the derivative $dy/dx$ 
measured both the rate of change of $y$ with respect to $x$ 
and the slope of the tangent line. In the polar coordinate 
system $dr/d\theta$ measures the rate of change of $r$ with 
respect to $\theta$. The sign of $dr/d\theta$ tells us whether
$r$ is increasing or decreasing as $\theta$ increases.


\begin{figure}[h!]
\begin{minipage}{\textwidth}
\begin{center}
%\vspace{1.0 cm}
\includegraphics[height=4cm,width=6cm]{graphs/fig9-2-1.eps}
\end{center}
\end{minipage}
\caption{$r$ changing as a function of $\theta$.}
\label{fig:9-2-1}
\end{figure}


\section{Polar Coordinates}

Rectangular coordinates allow us to describe a point
$(x,y)$ in the plane in a different way, namely
$$
  (x,y) \leftrightarrow (r, \theta),
$$
where $r$ is any real number and $\theta$ is an angle.

Polar coordinates are extremely useful, especially
when thinking about complex numbers.   Note, however,
that the $(r,\theta)$ representation of a point is
very non-unique. 

First, $\theta$ is not determined by the point. You could add $2\pi$
to it and get the same point:
$$\left(2,\frac{\pi}{4}\right)= \left(2,\frac{9\pi}{4}\right) 
= \left(2,\frac{\pi}{4}+389\cdot 2\pi\right).
= \left(2,\frac{-7\pi}{4}\right)
$$

Also that $r$ can be negative introduces further non-uniqueness:
$$ 
    \left(1, \frac{\pi}{2}\right) = \left(-1, \frac{3\pi}{2}\right).
$$
Think about this as follows: facing in the direction $3\pi/2$ and
backing up 1 meter gets you to the same point as looking in the
direction $\pi/2$ and walking forward 1 meter.

We can convert back and forth between cartesian and polar
coordinates using that

\begin{align}\label{eqn:polar1}
  x&=r\cos(\theta)\\
  y&=r\sin(\theta),
\end{align}
and in the other direction
\begin{align}\label{eqn:polar2}
  r^2&=x^2 + y^2\\
  \tan(\theta)&=\frac{y}{x}
\end{align}
(Thus $r = \pm \sqrt{x^2+y^2}$ and $\theta = \tan^{-1}(y/x).$)


\begin{figure}[h!]
\begin{minipage}{\textwidth}
\begin{center}
\vspace{1.0 cm}
\includegraphics[height=4cm,width=13cm]{graphs/fig9-1-19.eps}
\end{center}
\end{minipage}
\caption{Rectangular to polar coordinate conversion.}
\label{fig:9-1-19}
\end{figure}

\begin{example}
Sketch $r=\sin(\theta)$, which 
is a  circle sitting on top the $x$ axis.

\begin{figure}[h!]
\begin{minipage}{\textwidth}
\begin{center}
%\vspace{-1.0 cm}
\begin{tabular}{cc}
\includegraphics[height=5cm,width=6cm]{graphs/sage-polar1.eps} &
\includegraphics[height=5cm,width=6cm]{graphs/sage-polar2.eps} \\
\end{tabular}
\end{center}
\end{minipage}
\caption{Plot of (a) $r=1$, $0<\theta<\pi/6$, and (b) $r=\sin(\theta)$, $0<\theta<2\pi$.}
\label{fig:sage-polar1}
\end{figure}

We plug in points for one period of the function we are
graphing---in this case $[0,2\pi]$:
\begin{center}
\begin{tabular}{|l|l|}\hline
  0 & $\sin(0) = 0$\\
  $\pi/6$ & $\sin(\pi/6) = 1/2$\\
  $\pi/4$& $\sin(\pi/4) =\frac{\sqrt{2}}{2}$\\
  $\pi/2$ & $\sin(\pi/2) = 1$\\
  $3\pi/4$  & $\sin(3\pi/4) = \frac{\sqrt{2}}{2}$\\
  $\pi$  & $\sin(\pi) = 0$\\
  $\pi + \pi/6$  & $\sin(\pi+\pi/6) = -1/2$\\\hline
\end{tabular}
\end{center}
Notice it is nice to allow $r$ to be negative, so we don't have to
restrict the input.  BUT it is really painful to draw this
graph by hand.  

To more accurately draw the graph, let's try converting the equation to
one involving polar coordinates.  This is easier if we multiply both 
sides by $r$:
$$
  r^2 = r\sin(\theta).
$$
Note that the new equation has the extra solution 
$(r=0,\theta=\text{anything})$,  so
we have to be careful not to include this point.
Now convert to cartesian coordinates using (\ref{eqn:polar1})
to obtain (\ref{eqn:polar2}):
\begin{equation}
\label{eqn:polar3}
  x^2 + y^2 = y.
\end{equation}
The graph of (\ref{eqn:polar3})
is the same as that of $r=\sin(\theta)$.  To confirm
this we complete the square:
\begin{align*}
  x^2 + y^2 &= y\\
  x^2 + y^2 - y &= 0\\
  x^2 + (y-1/2)^2 &= 1/4\\
\end{align*}
Thus the graph of (\ref{eqn:polar3})
is a circle of radius $1/2$ centered at $(0,1/2)$.
\end{example}


Actually {\em any} polar graph\index{polar graph} 
of the form $r=a\sin(\theta) +b\cos(\theta)$ is a circle
(exercise for the interested reader).




\section{Areas in Polar Coordinates}

The previous section introduced the polar coordinate system and 
discussed how to plot points, how to create
graphs of functions (from data, a rectangular graph, or a formula), 
and how to convert back and forth
between the polar and rectangular coordinate systems. This 
section examines calculus in polar coordinates:
rates of changes, slopes of tangent lines, areas, and lengths of 
curves. The results we obtain may look
different, but they all follow from the approaches used in the 
rectangular coordinate system.

We know how to compute the area of a sector, i.e., piece of a circle
with angle $\theta$. [[draw picture]]. This is the basic polar region.
The area is
$$
  A = \text{(fraction of the circle)}\cdot \text{(area of circle)} 
    = \left(\frac{\theta}{2\pi}\right) \cdot \pi r^2 = \frac{1}{2} r^2 \theta.
$$

We now imitate what we did before with Riemann sums.  We chop
up, approximate, and take a limit.
Break the interval of angles from $a$ to $b$ into $n$ subintervals.
Choose $\theta_i^*$ in each interval.
The area of each slice is approximately
$(1/2) f(\theta_i^*)^2 \theta_i^2$.
Thus 
$$
A = \text{Area of the shaded region} 
  \sim \sum_{i=1}^n \frac{1}{2} f(\theta_i^*)^2 \Delta(\theta).
$$
Taking the limit, we see that
$$
A = \lim_{n\to\infty} \sum_{i=1}^n \frac{1}{2} f(\theta_i^*)^2 \Delta(\theta)
    = \frac{1}{2} \cdot \int_{a}^{b} f(\theta)^2 d\theta.
$$
Amazing!  By understanding the definition of Riemann sum, we've 
derived a formula for areas swept out by a polar graph.  But does it work in
practice?

\begin{example}
Find the area enclosed by one leaf of the
four-leaved rose $r=\cos(2\theta)$.
%\figtwo{Graph of $y=\cos(2x)$ and $r=\cos(2\theta)$}%
%{example_4rosecos}{example_4rose}
%sage: gnuplot.plot('set polar; plot [0:2*pi] cos(2*t)', 'example_4rose')
%sage: gnuplot.plot('plot [0:2*pi] cos(2*x)', 'example_4rosecos')


\begin{figure}[h!]
\begin{minipage}{\textwidth}
\begin{center}
%\vspace{1.0 cm}
\includegraphics[height=4cm,width=12cm]{graphs/sage-polar4.eps}
\end{center}
\end{minipage}
\caption{Graph of $y=\cos(2x)$ and $r=\cos(2\theta)$.}
\label{fig:polar4}
\end{figure}


This was plotted in \sage using these commands:

\vskip .1in

\begin{Verbatim}[fontsize=\scriptsize,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: P1 = polar_plot(lambda x:cos(2*x), 0, 2*pi, rgbcolor=(0,0,1))
sage: P2 = plot(lambda x:cos(2*x), 0, 2*pi, rgbcolor=(1,0,0),linestyle=":")
sage: show(P1+P2)

\end{Verbatim}

\vskip .1in
\noindent
To find the area using the methods we know so far, we would 
need to find a function $y=f(x)$ that gives the ``height'' of the leaf. 

Multiplying both sides of the equation $r=\cos(2\theta)$ by $r$ yields
$$
  r^2 = r\cos(2\theta) = r(\cos^2 \theta - \sin^2 \theta) = 
\frac{1}{r}((r\cos\theta)^2 - (r\sin\theta)^2).
$$
Because $r^2 = x^2 + y^2$ and $x=r\cos(\theta)$ and $y=r\sin(\theta)$,
we have
$$ 
  x^2 + y^2 = \frac{1}{\sqrt{x^2+y^2}} (x^2 - y^2).
$$
Solving for $y$ is a crazy mess, and then integrating?  It seems
impossible!

But it isn't... if we remember the basic idea of calculus: subdivide
and take a limit.

We need the boundaries of integration.  Start at
$\theta=-\pi/4$ and go to $\theta=\pi/4$.  As a check, note that
$\cos((-\pi/4) \cdot 2) = 0 = \cos((\pi/4) \cdot 2).$ We evaluate
\begin{align*}
\frac{1}{2} \cdot \int_{-\pi/4}^{\pi/4} \cos(2\theta)^2 d\theta
 &= \int_{0}^{\pi/4} \cos(2\theta)^2 d\theta \qquad \text{(even function)}\\
 &= \frac{1}{2} \int_{0}^{\pi/4} (1+\cos(4\theta))  d\theta \\
 &= \frac{1}{2} \left[ \theta + \frac{1}{4}\cdot \sin(4\theta)\right]_{0}^{\pi/4} \\
 &= \frac{\pi}{8}.
\end{align*}
We used that 
\begin{equation}
\label{eqn:costwosintwo}
\cos^2(x) = (1+\cos(2x))/2
\qquad\text{and}\qquad
\sin^2(x) = (1-\sin(2x))/2,
\end{equation}
which follow from 
$$\cos(2x) = \cos^2(x) - \sin^2(x)
  = 2\cos^2(x) - 1 = 1-2\sin^2(x).$$

\end{example}