\section{Polar Coordinates}
A coordinate system identifies a point $P$ in space by an ordered list of one or more numbers called coordinates.  We are familiar with the \textbf{Cartesian Coordinate System} \index{Cartesian Coordinate System} in which we identify a point $P$ in two-dimensional space by the ordered pair $(x,y)$ where $x$ is the $x$-coordinate of the point and $y$ is the $y$-coordinate of the point.

In polar coordinates one selects a point to $O$ to be the origin or pole of the coordinate system.  The first coordinate of an arbitrary point $P$ is the distance $r$ from $O$ to $P$.  Next one draws a \textbf{polar axis} \index{polar axis} as a ray emanating from $O$ which, for planar coordinates, is aligned with the positive $x$-axis.  The second coordinate of $P$ is the angle $\theta$ made between this axis and the ray $OP$.   The point $P$ is then represented by the ordered pair $(r,\theta)$ which are the \textbf{polar coordinates} \index{polar coordinates} of $P$.
\begin{center}
\includegraphics{graphics/polar_coordinates1}%
\end{center}

Notes:
\begin{enumerate}
\item Angles $\theta$ are positive if measured in the usual counter-clockwise sense from the polar axis and are negative if measured clockwise.
\item By convention polar coordinates $(r,\theta)$ with a negative $r$ represent the point $P$ at $(|r|,\theta+\pi)$.  In other words, they are at the positive distance $|r|$ directed along the ray opposite the ray subtending the angle $\theta$.
\end{enumerate}

\begin{example}
Plot the point with each of the following polar coordinates.
\begin{multicols}{3}
\begin{enumerate}
\item $\displaystyle \left(3, \frac{\pi}{3}\right)$
\item $\displaystyle \left(2, \frac{9\pi}{4}\right)$
\item $\displaystyle \left(4, \frac{3\pi}{4}\right)$
\end{enumerate}
\end{multicols}
\begin{solution*}
\begin{enumerate}
\item \parbox[t]{.5\textwidth}{Measure an angle of $\pi/3=60^\circ$ counterclockwise from the $x$-axis to determine the terminal ray of the angle followed by a distance of $2$ units along the ray to find the point.} 
\raisebox{-16em}{\includegraphics[width=.4\textwidth]{graphics/polar_coordinates3}}%
\item \parbox[t]{.5\textwidth}{Since $9\pi/4-2\pi=\pi/4$ the point is $3$ units along the ray with terminal axis of $\pi/4=45^\circ$ degrees measured counterclockwise from the polar axis.}
\raisebox{-16em}{\includegraphics[width=.4\textwidth]{graphics/polar_coordinates4}}%
\item\parbox[t]{.5\textwidth}{A negative angle of $-3\pi/4$ is measured clockwise from the polar axis.  We then measure $4$ units along the terminal ray.}
\raisebox{-16em}{\includegraphics[width=.4\textwidth]{graphics/polar_coordinates5}}%
\end{enumerate}
\end{solution*}

\begin{furtherquestions}
Plot the point with each of the following polar coordinates.
\begin{enumerate}
\item $\displaystyle (2,3\pi/4)$
\item $\displaystyle (2,-2\pi/3)$
\item $\displaystyle (-2,3\pi/4)$
\end{enumerate}
\end{furtherquestions}
\end{example}

Unlike in Cartesian Coordinates, the polar coordinates for point $P$ are not unique since the addition of a multiple of $2\pi$ to the angle returns one to the same physical point.
\begin{example}
The following polar coordinates all represent the same point $P$.

\hfil$(2,\pi/4)$,\hfil$(2,9\pi/4)$,\hfil$(2,-7\pi/4)$,\hfil$(-2,5\pi/4)$\hfil
\end{example}

\section{Converting Between Polar and Cartesian Coordinates} \index{polar coordinate system! conversion equations for Cartesian coordinates}

Suppose $P$ has Cartesian coordinates $(x,y)$ and polar coordinates $(r,\theta)$.
\begin{center}
\includegraphics{graphics/polar_coordinates2}%
\end{center}

As suggested by the diagram, one has

\hfil$x=r\cos\theta$\hfil$y=r\sin\theta$~.\hfil

These formulae allow conversion from polar to Cartesian coordinates. 
\begin{example}
Find the Cartesian coordinates of the given point.
\begin{multicols}{2}
\begin{enumerate}
\item $\displaystyle (r, \theta) =\left(2,\frac{\pi}{3}\right)$
\item $\displaystyle (r, \theta)= \left(4, \frac{3\pi}{4}\right)$
\end{enumerate}
\end{multicols}
\begin{solution*}
\begin{enumerate}
 \item For $\displaystyle (r, \theta) =\left(2,\frac{\pi}{3}\right)$\,:
 \begin{align*}
   x&=r \cos \theta  \implies x=2\cos \frac{\pi}{3}  \implies x=2\left(\frac{1}{2}\right) \implies x=1 \\
   y&=r\sin \theta \implies y=2\sin \frac{\pi}{3} \implies y=2\left(\frac{\sqrt{3}}{2}\right) \implies y=\sqrt{3}
 \end{align*}
 Thus the Cartesian coordinates of the point are $(x,y)=\left(1, \sqrt{3}\right)$\,.
 \item For $\displaystyle (r, \theta)= \left(4, \frac{3\pi}{4}\right)$\,:
 \begin{align*}
  x&=4\cos \frac{3\pi}{4} \implies x=4\left(-\frac{\sqrt{2}}{2}\right) \implies x=-2\sqrt{2}   \\
  y&=4\sin \frac{3\pi}{4} \implies y=4 \left(\frac{\sqrt{2}}{2}\right) \implies y= 2\sqrt{2}
 \end{align*}
 Thus the Cartesian coordinates of the points are $(x,y)=\left(-2\sqrt{2},\, 2\sqrt{2}\right)$\,.
\end{enumerate}
\end{solution*}

\begin{furtherquestion}
  Find the Cartesian coordinates of $(r,\theta)=(3,\pi/4)$.
\end{furtherquestion}
\end{example}

 To convert from Cartesian to polar coordinates we solve for $r$ and $\theta$.  Squaring and adding both sides of the two equations gives:
\[x^2+y^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2\]
Dividing both sides of the second equation by the corresponding sides of the first yields:
\[\frac{y}{x}=\frac{r\sin\theta}{r\cos\theta}=\tan\theta\]
In summary we have, to convert from Cartesian to polar coordinates,

\hfil$\displaystyle r=\sqrt{x^2+y^2}$\hfil$\displaystyle\tan\theta=\frac{y}{x}$\hfil

Note that when solving $\tan\theta=y/x$ for $\theta$:
\begin{enumerate}
\item There will typically be two solutions in $[0,2\pi)$.  The selection of the correct value is made by looking at the quadrant determined by the original $x$ and $y$ values.
\item If $x=0$ (so $y/x$ is undefined) consider the $y$-value to determine if $\theta=\pi/2$ or $\theta=3\pi/2$.
\end{enumerate}
\begin{example}
Find the polar coordinates of the given point.
\begin{multicols}{2}
\begin{enumerate}
\item $\displaystyle (x,y) =\left(-2\sqrt{3}, \,2\right)$
\item $\displaystyle (x,y)= \left(-3\sqrt{2}, \, -3\sqrt{2}\right)$
\end{enumerate}
\end{multicols}
\begin{solution*}
\begin{enumerate}
 \item For $\displaystyle (x,y) =\left(-2\sqrt{3}, \,2\right)$:
 \begin{align*}
r&=\sqrt{x^2+y^2} \implies r=\sqrt{12+4}\implies r=\sqrt{16} \implies r=4 \\
\tan \theta&=\frac{y}{x} \implies \tan \theta= \frac{2}{-2\sqrt{3}} \implies \tan \theta =\frac{-1}{\sqrt{3}}
 \end{align*}
 To solve the trigonometric equation for $\theta$ we note the (acute) reference angle (the angle of the terminal ray with respect to the $x$-axis) satisfies $\tan\theta'=\left|\frac{-1}{\sqrt{3}}\right|=\frac{1}{\sqrt{3}}$. Thus $\theta'=\frac{\pi}{6}$ using a $30^\circ-60^\circ-90^\circ$ triangle or the arctangent function. Since $\tan\theta<0$ we have two possible angles in $[0,2\pi)$:
 \begin{align*}
   \text{Quadrant II}: \theta&=\pi-\theta'=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\\
   \text{Quadrant IV}: \theta&=2\pi-\theta'=2\pi-\frac{\pi}{6}=\frac{11\pi}{6}
 \end{align*}
 Since $x<0$ and $y>0$ our point lies in Quadrant II and hence the polar coordinates are $\displaystyle (r, \theta)=\left(4, \frac{5\pi}{6}\right)$\,.
\item For $\displaystyle (x,y)= \left(-3\sqrt{2}, \, -3\sqrt{2}\right)$:
\begin{align*}
r&=\sqrt{x^2+y^2}=\sqrt{18+18}= \sqrt{36}=6    \\
\tan \theta&=\frac{y}{x} \implies \tan \theta= \frac{-3\sqrt{2}}{-3\sqrt{2}} 
\implies \tan \theta=1 
\end{align*}
 Reference angle $\theta'$ satisfies $\tan\theta'=|1|=1$. Thus $\theta'=\frac{\pi}{4}$ using a $45^\circ-45^\circ-90^\circ$ triangle or the arctangent function. Since $\tan\theta>0$ we have two possible angles in $[0,2\pi)$:
 \begin{align*}
   \text{Quadrant I}: \theta&=\theta'=\frac{\pi}{4}\\
   \text{Quadrant III}: \theta&=\pi+\theta'=\pi+\frac{\pi}{4}=\frac{5\pi}{4}
 \end{align*}
 Since $x<0$ and $y<0$ our point lies in Quadrant III and hence the polar coordinates are $\displaystyle (r, \theta)=\left(6, \frac{5\pi}{4}\right)$\,.

Note that in both these examples $\theta\neq\tan^{-1}\left(\frac{y}{x}\right)$. Using that formula is only correct if your point happens to lie in Quadrant I or IV.
\end{enumerate}
\end{solution*}

\begin{furtherquestion}
  Find the polar coordinates of $(x,y)=\left(1,-\sqrt{3}\right)$\,.
\end{furtherquestion}
\end{example}

\begin{exercise}
\input{POLAR_COORDINATES/polar-coordinates.tex}
\end{exercise}

\section{Curves in Polar Coordinates} \index{polar coordinates! curve of}

If we consider the coordinate $\theta$ as an independent variable and $r$ a dependent variable determined by function $f(\theta)$ then $r=f(\theta)$ describes a curve in polar coordinates.  More generally the solutions $(r,\theta)$ of the relation $g(r,\theta)=0$ describe a curve.

\begin{example}
For each of the following,
\begin{enumerate}[label=(\alph*)]
\item  Sketch the curve represented by the given polar equation.
\item Find a Cartesian equation for this curve.
\end{enumerate}
\begin{multicols}{2}
\begin{enumerate}
\item $\displaystyle r=\tan\theta$
\item $\displaystyle r^{2}\sin(2\theta)=16$
\end{enumerate}
\end{multicols}  
\begin{solution*}
\begin{enumerate}
\item $\displaystyle r=\tan\theta$
\begin{enumerate}
\item \parbox[t]{.7\textwidth}{Here the curve is explicitly defined as a function $r=f(\theta)=\tan\theta$. In $[0,2\pi)$ the function is defined everywhere except $\pi/2$ and $3\pi/2$ at which values $r=f(\theta)$ approaches $\pm\infty$ from right and left.  To find a few points $(r,\theta)$ in the first quadrant we choose $\theta=0,\pi/6=30^\circ,\pi/4=45^\circ$,and $\pi/3=60^\circ$ at which $r=\tan\theta$ is easily evaluated to be $r=0,1/\sqrt{3},1$, and $\sqrt{3}$ respectively.  Similarly points with reference angles of $30^\circ$, $45^\circ$, $60^\circ$ can be found in the remaining quadrants. One notes that for angles satisfying $\pi/2<\theta<\pi$ and $3\pi/2<\theta<2\pi$ the radius $r$ will be negative.  So for instance we have the points $(r,\theta)=(-1,3\pi/4)$ and $(r,\theta)=(-1,7\pi/4)$.  By convention these points are measured along the opposite direction of the terminal ray as shown in the diagram.  (All hollow points shown have radius $r=f(\theta)<0$.)  Joining these points produces a complete graph of the polar function.}~~~~~~\raisebox{-15em}{
\includegraphics[width=.21\textwidth]{graphics/polar_coordinates6}%
}
\item Since $r=\sqrt{x^2+y^2}$ and $\displaystyle \tan \theta= \frac{y}{x}$, one has
 \begin{align*}
r=\tan \theta
&\implies \sqrt{x^2+y^2}=\frac{y}{x}
\implies x^2+y^2=\frac{y^2}{x^2}
\implies x^4+x^2y^2=y^2 \\
&\implies y^2-x^2y^2=x^4
\implies y^2(1-x^2)=x^4  
\implies y^2=\frac{x^4}{1-x^2}
 \end{align*}
 Thus the Cartesian form of the given curve is $\displaystyle y^{2}=\frac{x^4}{1-x^2}$\,.
\end{enumerate}
\item $\displaystyle r^{2}\sin(2\theta)=16$
  \begin{enumerate}
\item  \parbox[t]{.7\textwidth}{Here $r$ is defined implicitly in terms of $\theta$. One notes that since $\sin(2\theta)=16/r^2>0$ we require $0<2\theta<\pi$ or $2\pi<2\theta<3\pi$ where sine is positive.  This implies $0<\theta<\pi/2$ or $3\pi/2<\theta<\pi$ and hence the graph only has points in Quadrants I and III.  For values of $\theta$ in those quadrants we get two solutions for $r$, namely $r=\pm\frac{4}{\sqrt{\sin(2\theta)}}$, however the negative $r$ points are already generated by positive $r$ points in the opposing quadrant. The closest points to the origin are $(r,\theta)=(4,\pi/4)$ and $(4,5\pi/4)$\,.  Black dots show further points plotted in increments of $\Delta\theta=0.3$ around these points.}~~~~~~\raisebox{-11em}{\includegraphics[width=.3\textwidth]{graphics/polar_coordinates7}}
\item Since $r^2=x^2+y^2$, $\cos\theta=\frac{x}{r}$, and $\sin\theta=\frac{y}{r}$, it follows that
\begin{align*}
r^2\sin(2\theta)=16
&\implies (x^2+y^2)2\sin\theta\cos\theta =16
\implies (x^2+y^2)2\left(\frac{y}{r}\right)\left(\frac{x}{r}\right)=16 \\
&\implies (r^2)2\left(\frac{xy}{r^2}\right)= 16
\implies 2xy=16 \implies xy=8
\end{align*}
Thus the Cartesian form of the given curve is $xy=8$\,. Hence the graph is that of a hyperbola with axis along the diagonal $y=x$.
\end{enumerate}
\end{enumerate}
\end{solution*}

\begin{furtherquestions*}
\begin{enumerate}
\item Sketch the curve represented by the polar equation $r=3\sin\theta$.
\item Find a Cartesian equation for this curve.
\end{enumerate}
\end{furtherquestions*}
\end{example}

\subsection{Symmetry in Polar Curves} \index{polar curves! symmetry in}
Suppose a polar curve is defined either explictly by $r=f(\theta)$ or implicitly with the relation $g(r,\theta)=0$.  Then, just as we saw with Cartesian Coordinates, the curves may have certain symmetries arising from the following transformations:
\begin{description}
\item[$\theta\rightarrow-\theta$~:] If $f(-\theta)=f(\theta)$ or $g(r,-\theta)=g(r,\theta)$, then the curve is symmetric about the polar axis.
\item[$r\rightarrow-r$~:] If $g(-r,\theta)=g(r,\theta)$ then the curve is symmetric about the origin $O$.
\item[$\theta\rightarrow\pi-\theta$~:] If $f(\pi-\theta)=f(\theta)$ or $g(r,\pi-\theta)=g(r,\theta)$, then the curve is symmetric about the vertical $\theta=\pi/2$ line.
\end{description}
\begin{example}
  Sketch the given curve.
\begin{multicols}{2}
\begin{enumerate}
\item $\displaystyle r=5+4\sin \theta$
\item $\displaystyle r^{2}=16\sin(2\theta)$
\end{enumerate}
\end{multicols}
\begin{solution*}
\begin{enumerate}
\item $\displaystyle r=5+4\sin \theta$\\
  Here $r=f(\theta)$ so $r$ is explicitly defined with $f(\theta)=5+4\sin \theta$.
  \begin{align*}
    f(-\theta)&=5+4\sin(-\theta)=5-4\sin(\theta)\neq f(\theta)\\
    f(\pi-\theta)&=5+4\sin(\pi-\theta)=5+4(\sin\pi\cos\theta-\cos\pi\sin\theta)=5+4\sin\theta=f(\theta)
  \end{align*}
  Since $f(\pi-\theta)=f(\theta)$ the graph is symmetric about the vertical line $\theta=\pi/2$ or equivalently the $y$-axis. We need only plot points in quadrants I and IV and reflect across that axis to get the graph of the polar function.
\begin{center}
\includegraphics[width=.35\textwidth]{graphics/polar_coordinates8}%
\end{center}
Note for an explicitly defined polar function $r=f(\theta)$ the transformation $r\to-r$ can never preserve the equality and consequently its graph will never be symmetric about the origin.
\item $\displaystyle r^{2}=16\sin(2\theta)$\\
Here $r$ is implicitly defined and may be written equivalently as $g(r,\theta)=0$ where
  \[g(r,\theta)=16\sin(2\theta)-r^2\,.\]
  Looking for symmetries we evaluate the following:
  \begin{align*}
    g(r,-\theta) & = 16\sin(-2\theta)-r^2=-16\sin(2\theta)-r^2\neq g(r,\theta)\\
    g(-r,\theta) & = 16\sin(2\theta)-(-r)^2=16\sin(2\theta)-r^2=g(r,\theta)\\
    g(r,\pi-\theta) & = 16\sin[2(\pi-\theta)]-r^2=16\sin(2\pi-2\theta)-r^2\\
    &=16\left[\sin(2\pi)\cos(2\theta)-\sin(2\theta)\cos(2\pi)\right]-r^2=-16\sin(2\theta)-r^2\neq g(r,\theta)
  \end{align*}
  Since $g(-r,\theta)=g(r,\theta)$ we conclude that the graph is symmetric about the origin. Since $r^2$ is positive we require that $\sin(2\theta)>0$ which implies $0\le\theta\le\frac{\pi}{2}$ and $\pi\le\theta\le \frac{3\pi}{2}$ so the graph lies only in quadrants I and III. Using the symmetry we need only plot the graph in the first quadrant and reflect the points across the origin to get the following graph:
\begin{center}
\includegraphics[width=.4\textwidth]{graphics/polar_coordinates9}%
\end{center}
\end{enumerate}
\end{solution*}

\begin{furtherquestion}
  Sketch the curve $r=1+\cos\theta$.
\end{furtherquestion}
\end{example}

\subsection{Tangents} \index{polar curves! tangent line to}

The slope of the tangent to a polar curve described by $r=f(\theta)$ may be found by converting to Cartesian Coordinates.  Since $x=r\cos\theta$ and $y=r\sin\theta$ we have

\hfil$x=f(\theta)\cos\theta$\hfil$y=f(\theta)\sin\theta$\hfil

The curve is now represented by a parametric equation with parameter $\theta$.  Applying our rules for finding the derivative $dy/dx$ of such equations from Chapter~\ref{ch:parametric_equations} gives $dy/dx=(dy/d\theta)/(dx/d\theta)$ and so
\[\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{f'(\theta)\sin\theta+f(\theta)\cos\theta}{f'(\theta)\cos\theta-f(\theta)\sin\theta}\,.\]
Vertical tangent lines occur where $\frac{dy}{d\theta}=0$ (and $\frac{dx}{d\theta}\neq0$) while for horizontal tangent lines $\frac{dx}{d\theta}=0$ (and $\frac{dy}{d\theta}\neq0$). If both derivatives vanish at some $\theta$, consideration of the limit of $\frac{dy}{dx}$ as one approaches that value must be made to determine the nature of the tangent.
\begin{example}
For each of the following,
\begin{enumerate}[label=(\alph*)]
\item Find the slope of the tangent line to the given curve at the given value.
\item Find the points on the curve where the tangent line is horizontal or vertical. 
\end{enumerate}
\begin{multicols}{2}
\begin{enumerate}
\item $\displaystyle r=5\cos \theta ;\,\, \theta=\frac{\pi}{3}$
\item $\displaystyle r=4\sin(2\theta); \,\, \theta=\frac{\pi}{6}$
\end{enumerate}
\end{multicols} 
\begin{solution*}
\begin{enumerate}
\item $\displaystyle r=5\cos \theta ;\,\, \theta=\frac{\pi}{3}$
\begin{enumerate}
    \item Here $f(\theta)=5\cos \theta, \,\, f'(\theta)=-5\sin\theta$ and so
    \begin{align*}
     \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{f'(\theta)\sin \theta+f(\theta)\cos \theta}{f'(\theta)\cos \theta-f(\theta)\sin \theta}   
     =\frac{-5\sin \theta \sin \theta + 5\cos \theta \cos \theta}{-5\sin \theta \cos \theta-5\cos \theta \sin \theta}
      =\frac{\cos^{2}\theta-\sin^{2}\theta}{-2\sin\theta \cos \theta}\,,
    \end{align*}
    \begin{align*}
    \left.\frac{dy}{dx}\right|_{\theta=\frac{\pi}{3}}=\frac{\cos ^2 \frac{\pi}{3}-\sin^{2}\frac{\pi}{3}}{-2\sin\frac{\pi}{3}\cos\frac{\pi}{3}} 
    =\frac{\frac{1}{4}-\frac{3}{4}}{-2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\,.
    \end{align*}
  \item We note that due to the periodicity of cosine, $\cos(\theta+2\pi)=\cos(\theta)$, the entire curve $r=5\cos\theta$ is generated by $\theta$ in $[0,2\pi)$ so we only need to find solutions in that interval. We solve $dy/d\theta=0$ and $dx/d\theta=0$. From~(a) we have
    \begin{align*}
    \frac{dy}{d\theta}=0 
    &\implies -5\sin\theta\sin\theta+5\cos\theta\cos\theta = 0 \\
    &\implies \cos^{2}\theta -\sin^{2} \theta=0 \implies (\cos\theta+\sin\theta)(\cos\theta-\sin\theta)=0\\
    &\implies \cos \theta= \pm \sin \theta 
      \implies \left\{\begin{array}{l}
                        \cos \theta=\sin \theta \implies \theta= \frac{\pi}{4}, \frac{5\pi}{4}\\[.4em]
                        \cos \theta= -\sin \theta 
                        \implies \theta= \frac{3\pi}{4}, \frac{7\pi}{4}
                      \end{array}\right.\\
      \frac{dx}{d\theta}=0 
      &\implies -5\sin \theta \cos \theta-5\cos \theta \sin \theta=0
        \implies -10\sin \theta \cos \theta=0 \\
      &\implies \sin \theta \cos \theta =0
        \implies \left\{\begin{array}{l}\sin \theta=0 \implies \theta= 0, \pi\\
                          \cos \theta=0 \implies \theta= \frac{\pi}{2}, \frac{3\pi}{2}
                        \end{array}\right.
    \end{align*}
  The points on the curve where the tangent line is horizontal occur when $\frac{dy}{d\theta}=0$ (and $\frac{dx}{d\theta}\neq 0$).  Therefore the following points $(r,\theta)$ have horizontal tangents
  \begin{align*}
    \left(\frac{5}{\sqrt{2}}, \frac{\pi}{4}\right), \left(-\frac{5}{\sqrt{2}}, \frac{3\pi}{4}\right), \left(-\frac{5}{\sqrt{2}}, \frac{5\pi}{4}\right), \left(\frac{5}{\sqrt{2}}, \frac{7\pi}{4}\right)\,,
  \end{align*}
  which simplifies to $\displaystyle \left(\frac{5}{\sqrt{2}},\frac{\pi}{4}\right), \left(\frac{5}{\sqrt{2}}, \frac{7\pi}{4}\right)$ when we remove identical points.
  
    The points on the curve where the tangent line is vertical occur when $\frac{dx}{d\theta}=0$ (and $\frac{dy}{d\theta}\neq 0$).  Therefore the following points $(r,\theta)$ have vertical tangents
  \begin{align*}
  \left(5,0\right), \left(0, \frac{\pi}{2}\right), \left(-5, \pi\right),\left(0, \frac{3\pi}{2}\right)\,,
  \end{align*}
  which simplifies to $\left(5,0\right), \left(0,\frac{\pi}{2}\right)$ once redundant points are removed.
\end{enumerate}
\item $\displaystyle r=4\sin(2\theta); \,\, \theta=\frac{\pi}{6}$
\begin{enumerate}
    \item Here $f(\theta)=4\sin(2\theta), \,\, f'(\theta)=8\cos(2\theta)$ and so
    \begin{align*}
      \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{f'(\theta)\sin \theta+f(\theta)\cos \theta}{f'(\theta)\cos \theta-f(\theta)\sin \theta}=\frac{8\cos(2\theta)\sin \theta + 4\sin(2\theta)\cos \theta}{8\cos(2\theta)\cos\theta-4\sin(2\theta)\sin \theta}\,,
    \end{align*}
    \begin{align*}
     \left.\frac{dy}{dx}\right|_{\theta=\frac{\pi}{6}}&=\frac{8\cos \frac{\pi}{3}\sin \frac{\pi}{6}+4\sin \frac{\pi}{3}\cos \frac{\pi}{6}}{8\cos \frac{\pi}{3}\cos \frac{\pi}{6}-4\sin\frac{\pi}{3}\sin \frac{\pi}{6}} 
     =\frac{8\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+ 4\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)}{8\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-4\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)}\\
     &=\frac{2+3}{2\sqrt{3}-\sqrt{3}}=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}\,.
    \end{align*}
    As an aside, if we desire the actual tangent line at $P(x_0,y_0)$ for $\theta=\frac{\pi}{6}$ we have
    \begin{align*}
     x_0 &=r\cos \theta=4\sin(2\theta)\cos\theta=4\sin\frac{\pi}{3}\cos\frac{\pi}{6} 
         =4\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) \implies x=3 \\
      y_0 &=r\sin\theta=4\sin(2\theta)\sin\theta=4\sin\frac{\pi}{3}\sin\frac{\pi}{6}
          =4\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) \implies y=\sqrt{3}
    \end{align*}
    Thus the tangent line is
    \begin{align*}
     y&=m(x-x_0)+y_0=\frac{5\sqrt{3}}{3}(x-3)+\sqrt{3}=\frac{5\sqrt{3}}{3}x-5\sqrt{3} +\sqrt{3}\\
     \implies y&=\frac{5\sqrt{3}}{3}x-4 \sqrt{3}\,.
    \end{align*}
  \item Here again the periodicity of $\sin(2\theta)$ restricts consideration of solutions of $\theta$ to $[0,2\pi)$. We solve $dy/d\theta=0$ and $dx/d\theta=0$. From~(a) we have
    \begin{align*}
     \frac{dy}{d\theta}=0 &\implies 8\cos(2\theta)\sin \theta+4\sin(2\theta)\cos\theta=0 \\
     &\implies 8(\cos^{2}\!\theta-\sin^{2}\!\theta)\sin \theta +8\sin \theta\cos \theta \cos \theta=0 \\ 
     &\implies\cos^{2}\!\theta\sin \theta-\sin^{3}\!\theta + \sin \theta \cos^{2}\!\theta=0 \\
     &\implies 2\cos^{2}\!\theta\sin \theta-\sin^{3}\!\theta=0 \\
     &\implies \sin \theta(2\cos^{2}\theta-\sin^{2}\theta)=0 \\
     &\implies \sin \theta\left(2\cos^{2}\theta-(1-\cos^{2}\theta)\right)=0 \\
     &\implies \sin \theta(3\cos^{2}\theta-1)=0
    \end{align*}
    Then $\sin\theta=0 \implies \theta=0, \pi$.
    For $3\cos^{2}\theta-1=0\implies\cos\theta=\pm\sqrt{1/3}$ the reference angle is, for both $\pm$ equations, $\theta'=\cos^{-1}(1/\sqrt{3})\approx0.9553$ radians. The $+$ equation gives Quadrant I and IV solutions while the $-$ equation gives Quadrant II and III solutions.  The four solutions of $3\cos^{2}\theta-1=0$ in $[0,2\pi)$ are therefore, in radians,
    \begin{align*}
      \text{I}: \theta&=\theta'=0.9553\\
      \text{II}: \theta&=\pi-\theta'=2.1863\\
      \text{III}: \theta&=\pi+\theta'=4.0969\\
      \text{IV}: \theta&=2\pi-\theta'=5.3279
    \end{align*}
    Thus $\frac{dy}{d\theta}$ vanishes at $\theta$ equal to
    \[0, 0.9553, 2.1863, \pi, 4.0969, 5.3279\,.\]
    Again from (a) we have
    \begin{align*}
      \frac{dx}{d\theta}=0 
      &\implies 8\cos(2\theta)\cos\theta-4\sin(2\theta)\sin\theta=0  \\
      &\implies 8(\cos ^{2}\!\theta-\sin^{2}\theta)\cos \theta-8\sin^{2}\!\theta\cos \theta=0 \\
      &\implies \cos^{3}\theta-\sin^{2}\theta\cos \theta-\sin^{2}\!\theta\cos\theta=0 \\
      &\implies \cos^{3}\theta-2\sin^{2}\theta\cos\theta=0 \\
      &\implies \cos \theta(\cos^{2}\theta-2\sin^{2}\theta)=0 \\
      &\implies \cos \theta(1-\sin^{2}\theta-2\sin^{2}\theta)=0 \\
      &\implies \cos \theta(1-3\sin^{2}\theta)=0
    \end{align*}
    Then $\cos \theta=0 \implies \theta=\frac{\pi}{2}, \frac{3\pi}{2}$.
    For $1-3\sin^{2}\theta=0\implies\sin\theta=\pm\sqrt{1/3}$ the reference angle is, for both $\pm$ equations, $\theta'=\sin^{-1}(1/\sqrt{3})\approx0.6155$ radians. Solving as above the four solutions of $1-3\sin^{2}\theta=0$ in $[0,2\pi)$ are therefore $0.6155, 2.5261, 3.7571,5.6677$ and $\frac{dx}{d\theta}$ vanishes at $\theta$ equal to
    \begin{align*}
      0.6155, \frac{\pi}{2}, 2.5261, 3.7571, \frac{3\pi}{2}, 5.6677\,.
    \end{align*}

The tangent line is horizontal if $\frac{dy}{d\theta}=0$ (and $\frac{dx}{d\theta}\neq 0$).  Therefore the following points in polar form $(r,\theta)$ have a horizontal tangent line
    \begin{align*}
    (0,0), (3.7712, 0.9553), (-3.7712, 2.1863), (0, \pi), (3.7712, 4.0969),(-3.7712, 5.3279)\,.
    \end{align*}
    Removing the identical point at the origin we are left with
    \begin{align*}
    (0,0), (3.7712, 0.9553), (-3.7712, 2.1863), (3.7712, 4.0969),(-3.7712, 5.3279)\,.
    \end{align*}
    The tangent line is vertical if $\frac{dx}{d\theta}=0$ (and $\frac{dy}{d\theta}\neq 0$).  Therefore the following points $(r,\theta)$ have a vertical tangent line
\begin{align*}
 (3.7712, 0.6155), \left(0, \frac{\pi}{2}\right),(-3.7712, 2.5261),
    (3.7712, 3.7571), \left(0, \frac{3\pi}{2}\right),(-3.7712, 5.6677)\,.
\end{align*}
 Removing the identical point at the origin we are left with
 \begin{align*}
   (3.7712, 0.6155), \left(0, \frac{\pi}{2}\right),(-3.7712, 2.5261),
   (3.7712, 3.7571),(-3.7712, 5.6677)\,.
 \end{align*}
\end{enumerate}
\end{enumerate}
\end{solution*}

\begin{furtherquestions}
Consider the curve $r=1+\cos\theta$.
\begin{enumerate}
\item Find the slope of the tangent line to the curve where $\theta=\pi/2$.
\item Find the points on the curve where the tangent line is horizontal or vertical.
\end{enumerate}
\end{furtherquestions}
\end{example}

\subsection{Areas}
Recall that a sector of a circle of radius $r$ is a pie-shaped wedge between two radii subtended by an angle $\theta$ as shown:
\begin{center}
\includegraphics[width=.3\textwidth]{graphics/polar_coordinates16}%
\end{center}
Since there are $2\pi$ radians in a circle the fraction of the area contained in the sector of angle $\theta$ must equal $\frac{\theta}{2\pi}$. As the area of the circle is $\pi r^2$ it follows that $\frac{A}{\pi r^2}=\frac{\theta}{2\pi}$ and hence the area of the sector is
\begin{align*}
  A=\frac{1}{2}r^2\theta\,.
\end{align*}
Note $\theta$ must be in radians.

In general we may be interested in the area of a region $S$ generated by a continuous polar function $r=f(\theta)\ge 0$ with $a<\theta<b$ and $b-a<2\pi$ as shown below on the left.
\begin{center}
\includegraphics[height=.3\textwidth]{graphics/polar_coordinates17}\hspace{.13\textwidth}\includegraphics[height=.3\textwidth]{graphics/polar_coordinates18}%
\end{center}
We proceed as we did when we originally derived the area under a curve but here using the area of sectors rather than rectangles to approximate the area.  First partition the $\theta$ interval $[a,b]$ into $n$ intervals of equal width $\Delta\theta=\frac{b-a}{n}$,
\begin{align*}
  [\theta_0,\theta_1],   [\theta_1,\theta_2],   \ldots, [\theta_{i-1},\theta_i],\ldots   [\theta_{n-1},\theta_n], 
\end{align*}
where $\theta_0=a$, $\theta_n=b$, and, in general, $\theta_i=\theta_0+i\Delta\theta$. Within the $i^{\text{th}}$ interval $[\theta_{i-1},\theta_i]$ select a value $\theta_i^*$ at which to evaluate the radius of the sector, $r_i=f(\theta_i^*)$.  (In the diagram the midpoint of the interval has been chosen.)  The $i^{\text{th}}$ sector is subtended by angle $\Delta\theta$ and its area is therefore
\begin{align*}
A_i=\frac{1}{2}r_i^2\Delta\theta\,.
\end{align*}
The area $A$ of $S$ will be approximated by the sum of the sector areas,
\begin{align*}
  A\approx\sum_{i=1}^nA_i=\sum_{i=1}^n\frac{1}{2}[f(\theta_i)]^2\Delta\theta\,.
\end{align*}
Recognizing this as a Riemann sum of the function $\frac{1}{2}[f(\theta)]^2$, let the number of intervals $n$ approach infinity to get the exact result
\begin{align*}
  A=\lim_{n\to\infty}\sum_{n=1}^\infty\frac{1}{2}[f(\theta_i)]^2\Delta\theta\,.
\end{align*}
The area of region $S$ is therefore the definite integral
\begin{align*}
  A=\int_a^b\frac{1}{2}[f(\theta)]^2d\theta\,.
\end{align*}
Since by assumption $f$ is continuous on $[a,b]$, so is $\frac{1}{2}[f(\theta)]^2$, and the latter is therefore integrable and the area is finite.  The formula is easy to remember since $A=\int\!dA$ where the differential area $dA$ at angle $\theta$ is, analogous to the area of the sector,
\begin{align*}
dA=\frac{1}{2}r^2d\theta=\frac{1}{2}[f(\theta)]^2d\theta\,.
\end{align*}

The above results can be generalized.  With regard to $r=f(\theta)$, it is sufficient for the function $f$ to be piecewise continous on $[a,b]$.  If the function is discontinuous at an endpoint of the interval or has, say, an infinite discontinuity at a point $c$ within the interval, one can generate an improper integral of the second kind to represent the area. The latter may or may not converge.
\begin{example}
  Find the area of the region bounded by one loop of the graph of the given functions.
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle r=2\cos(3\theta)$
    \item $\displaystyle r^2=16\cos(4\theta)$
    \end{enumerate}
  \end{multicols}
\begin{solution*}
\begin{enumerate}
\item $\displaystyle r=2\cos(3\theta)$\\
  The graph will start and end a loop when it goes through the origin ($r=0$).  Solving for $\theta$ gives
  \begin{align*}
    r=0 &\implies 2\cos(3\theta)=0 \\
        &\implies \cos 3\theta=0 \implies 3\theta=\left\{\begin{array}{c}
                                                           \frac{\pi}{2}\pm 2n\pi \\[.5em]
                                                           \frac{3\pi}{2}\pm2n\pi \end{array}\right.
    \implies \theta=\left\{\begin{array}{c}
                             \frac{\pi}{6}\pm \frac{2n\pi}{3} \\[.5em]
                             \frac{\pi}{2}\pm\frac{2n\pi}{3}\end{array}\right.\ (n\ \text{an integer})
  \end{align*}
  Since $\cos(3\theta)$ is periodic solutions of interest are those in  $[0,2\pi)$:
  \begin{align*}
    \frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}\,.
  \end{align*}
  A graph of the polar function is
\begin{center}
\includegraphics[width=.4\textwidth]{graphics/polar_coordinates10}%
\end{center}
The area of the shaded loop is double the area from $\theta=0$ to $\theta=\frac{\pi}{6}$ and therefore
\begin{align*}
  A&=2\int_{0}^{\frac{\pi}{6}}  \frac{1}{2}\left[2\cos(3\theta)\right]^2\, d\theta
  =\int_{0}^{\frac{\pi}{6}} 4\cos^{2}(3\theta)\, d\theta
    =4\int_{0}^{\frac{\pi}{6}}\frac{1}{2} \left[1+ \cos(6\theta)\right]\,d\theta \\
  &=2\left.\left[\theta + \frac{1}{6}\sin(6\theta) \right]\right|_0^{\frac{\pi}{6}}
    =2\left\{\left[\frac{\pi}{6}+0\right]-\left[0+0\right]\right\}=\frac{\pi}{3}\,\,\text{square units}.
\end{align*}
    \item $\displaystyle r^2=16\cos(4\theta)$
    \begin{align*}
       r=0 &\implies 16\cos(4\theta)=0
       \implies 4\theta=\left\{\begin{array}{c}
                                                           \frac{\pi}{2}\pm 2n\pi \\[.5em]
                                                           \frac{3\pi}{2}\pm2n\pi \end{array}\right.\implies\theta=\left\{\begin{array}{c}
                                                           \frac{\pi}{8}\pm \frac{n\pi}{2} \\[.5em]
                                                                                                                            \frac{3\pi}{8}\pm \frac{n\pi}{2} \end{array}\right.\ (n\ \text{an integer})
    \end{align*}
    Solutions for $\theta$ in $[0,2\pi)$ are
    \begin{align*}
      \frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8},\frac{13\pi}{8},\frac{15\pi}{8}\,.
    \end{align*}
    A graph of the polar function is
\begin{center}
\includegraphics[width=.4\textwidth]{graphics/polar_coordinates11}%
\end{center}
\begin{align*}
  A&=2\int_{0}^{\frac{\pi}{8}}\left(\frac{1}{2}\right)16\cos (4 \theta)\,d\theta 
  =16 \int_{0}^{\frac{\pi}{8}} \cos(4\theta)\, d\theta
  =\left.4\sin(4\theta)\right|^{\frac{\pi}{8}}_{0}
  =4\left[\sin\left(\frac{\pi}{2}\right)-\sin(0)\right]\\
  &= 4\,\,\text{square units}.
\end{align*}
\end{enumerate}
\end{solution*}

\begin{furtherquestion}
Find the area of the curve $r=1+e^{-\theta}$ between $\theta=0$ and $\theta=2$\,.
\end{furtherquestion}
\end{example}

Similar to our study of the area between curves $y=f(x)$ and $y=g(x)$, one may be interested in the area between an outer polar curve $r_o=f(\theta)$ and an inner one $r_i=g(\theta)$ over angular interval $[a,b]$.  Then $f(\theta)\ge g(\theta)\ge0$ on $[a,b]$ and the area between the polar curves is just the difference of the areas,
\begin{align*}
  A=\int_a^b\frac{1}{2}[f(\theta)]^2d\theta-\int_a^b\frac{1}{2}[g(\theta)]^2d\theta=\int_a^b\frac{1}{2}\left([f(\theta)]^2-[g(\theta)]^2\right)d\theta\,.
\end{align*}
One may consider this formula as $A=\int\!dA$ where $dA$ is the differential area at $\theta$ between the two sectors given by
\begin{align*}
  dA=\frac{1}{2}(r_o^2-r_i^2)d\theta=\frac{1}{2}\left([f(\theta)]^2-[g(\theta)]^2\right)d\theta\,.
\end{align*}
As with the Cartesian equivalent of this problem, the endpoints $a$ and $b$ may need to be determined by finding the values of $\theta$ where the curves intersect by solving $r_0=f(\theta)=g(\theta)=r_i$.
\begin{example}
\begin{enumerate}
    \item Find the area of the region that lies inside the cardioid $\displaystyle r=2-2\cos\theta$ and outside the circle $r=3$.
    \item  Find the area of the region that is outside of $r=2$ and inside the graph of $r^2=8\sin(2\theta)$.
\end{enumerate}
\begin{solution*}
\begin{enumerate}
    \item The polar curves $\displaystyle r=2-2\cos\theta$ and $r=3$ intersect when they have the same radius $r$ at the same angle $\theta$, we therefore equate the functions and solve for $\theta$.
    \begin{align*}
      2-2\cos\theta=3 &\implies -2\cos\theta=1 \implies
                        \cos \theta=-\frac{1}{2}\implies\ \text{ref angle}\ \theta'=\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\\
 \text{cosine negative}                     & \implies \theta=\left\{\begin{array}{ll}
                                        \pi-\frac{\pi}{3}+2n\pi\\[.5em]
                                        \pi+\frac{\pi}{3}+2n\pi\end{array}\right.\implies \theta=\left\{\begin{array}{ll}
                                        \frac{2\pi}{3}+2n\pi\\[.5em]
                                        \frac{4\pi}{3}+2n\pi\end{array}\right.\ (n\ \text{an integer})
    \end{align*}
    Both functions of $\theta$ have period $2\pi$ and the curves are generated entirely by $\theta$ in $[0,2\pi)$.  Intersection points in this interval are therefore $(r,\theta)=\left(3,\frac{2\pi}{3}\right),\left(3,\frac{4\pi}{3}\right)$. Here we evaluated $r=3$ for both angles using each equation.  At $\theta=\pi$ between these values we find $r=4$ for the first curve and $r=3$ for the second.  This confirms that the outer curve $r_o=f(\theta)=2-2\cos\theta$ and the inner curve is $r_i=g(\theta)=3$\,.  A graph of the polar functions follows.
\begin{center}
\includegraphics[width=.6\textwidth]{graphics/polar_coordinates12}%
\end{center}
 The required area is given by 
 \begin{align*}
    A&=\frac{1}{2}\int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} \left[(2-2\cos\theta)^2-3^2\right]\,d\theta 
    =\frac{1}{2} \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} \left[4-8\cos\theta + 4\cos^{2}\!\theta-9\right]\, d\theta \\
    &=\frac{1}{2}\int^{\frac{4\pi}{3}}_{\frac{2\pi}{3}} \left[-5-8\cos\theta+2(1+\cos(2\theta))\right]\,d\theta 
    =\frac{1}{2}\int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} \left[-3-8\cos\theta + 2\cos(2\theta)\right]\,d\theta \\
    &=\frac{1}{2}\left[\rule{0em}{1.2em}-3\theta-8\sin\theta+\sin(2\theta)\right|^{\frac{4\pi}{3}}_{\frac{2\pi}{3}} \\
     &=\frac{1}{2}\left(\left[-4\pi -8\sin\frac{4\pi}{3}+ \sin\frac{8\pi}{3}\right]-\left[-2\pi-8\sin\frac{2\pi}{3}+\sin\frac{4\pi}{3}\right]
       \right) \\
    &=\frac{1}{2}\left[-2\pi -9\sin\frac{4\pi}{3}+  \sin\frac{8\pi}{3} + 8\sin \frac{2\pi}{3}\right] \\
    &=\frac{1}{2}\left[-2\pi-9\left(-\frac{\sqrt{3}}{2}\right)+\frac{\sqrt{3}}{2}+8\frac{\sqrt{3}}{2}\right] \\
    &=\frac{1}{2} \left[-2\pi +\frac{9\sqrt{3}}{2}+ \frac{9\sqrt{3}}{2}\right]
    =\frac{9\sqrt{3}}{2}-\pi\,\,\text{square units}.
 \end{align*}
    \item The two curves $r=2$ and $r^2=8\sin(2\theta)$ intersect when
      \begin{align*}
        8\sin 2\theta =2^2&\implies \sin(2\theta)=\frac{1}{2}
                            \implies 2\theta=\left\{\begin{array}{ll}
                                                      0+\frac{\pi}{6}+2n\pi\\[.5em]
                                                      \pi-\frac{\pi}{6}+2n\pi\end{array}\right.
        \implies \theta=\left\{\begin{array}{ll}
                                 \frac{\pi}{12}+n\pi\\[.5em]
                                 \frac{5\pi}{12}+n\pi\end{array}\right.\ (n\ \text{integer})
      \end{align*}
      Periodicity of the curves ensures they are generated entirely by $\theta$ in $[0,2\pi)$.  Intersection points in this interval are $(r,\theta)=\left(2,\frac{\pi}{12}\right),\left(2,\frac{5\pi}{12}\right),\left(2,\frac{13\pi}{12}\right)$, and $\left(2,\frac{17\pi}{12}\right)$.  At $\theta=\pi/4$ which lies within the first interval  $\left[\frac{\pi}{12},\frac{5\pi}{12}\right]$ the first curve evaluates to $r=2$ and the second to $r=2\sqrt{2}$.  Thus the inner curve is $r_i=2$ and the outer curve is $r_o^2=8\sin(2\theta)$\,.  The same holds over the interval $\left[\frac{13\pi}{12},\frac{17\pi}{12}\right]$.  A graph of the polar functions confirms this.
\begin{center}
\includegraphics[width=.4\textwidth]{graphics/polar_coordinates13}%
\end{center}
    The required area (shaded) is twice that in the first quadrant.
    \begin{align*}
     A&=2\int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{1}{2}(r_o^2-r_i^2)\,d\theta=\int_{\frac{\pi}{12}}^{\frac{5\pi}{12}}\left[8\sin(2\theta)-4\right]\,d\theta 
     =\left[\rule{0em}{1.2em}-4\cos(2\theta)-4\theta \right|^{\frac{5\pi}{12}}_{\frac{\pi}{12}} \\
     &=\left[-4\cos\frac{5\pi}{6}-4\left(\frac{5\pi}{12}\right)\right]-\left[-4\cos\frac{\pi}{6} - 4\left(\frac{\pi}{12}\right)\right]\\
      &
        =-4\left(-\frac{\sqrt{3}}{2}\right)-\frac{5\pi}{3} + 4\left(\frac{\sqrt{3}}{2}\right) + \frac{\pi}{3}
     =4\sqrt{3}-\frac{4\pi}{3}\,\,\text{square units}.
    \end{align*}
\end{enumerate}
\end{solution*}

\begin{furtherquestion}
  Find the area in the first quadrant between the polar axis and the curves $r=\cos\theta$ and $r=\sin\theta$.
\end{furtherquestion}
\end{example}

\subsection{Arc Length}
A curve represented in polar coordinates coordinates by the function $r=f(\theta)$ can be written in Cartesian coordinates, since $(x,y)=(r\cos\theta,r\sin\theta)$, as the \emph{parametric curve}
\begin{align*}
  x(\theta) & = f(\theta)\cos\theta \\
  y(\theta) & = f(\theta)\sin\theta
\end{align*}
where here $\theta$ is the parameter of the curve.  If we are interested in the arc length of the curve over the interval $[a,b]$ of $\theta$ we can immediately use our result from Section~\ref{subsec:parametricarclength}, with $t\to\theta$ to get
\begin{align*}
  s&=\int_a^b\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}\,d\theta\,.
\end{align*}
Evaluating the terms in the integrand gives
\begin{align*}
\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2
&=\left(f'(\theta)\cos\theta-f(\theta)\sin\theta\right)^2+\left(f'(\theta)\sin\theta+f(\theta)\cos\theta\right)^2\\
&=[f'(\theta)]^2\cos^2\!\theta+[f(\theta)]^2\sin^2\!\theta-2f'(\theta)f(\theta)\cos\theta\sin\theta\\
&~~~+[f'(\theta)]^2\sin^2\!\theta+[f(\theta)]^2\cos^2\!\theta+2f'(\theta)f(\theta)\sin\theta\cos\theta\\
&=[f(\theta)]^2(\sin^2\!\theta+\cos^2\!\theta)+[f'(\theta)]^2(\cos^2\!\theta+\sin^2\!\theta)\\
&=[f(\theta)]^2+[f'(\theta)]^2\,.
\end{align*}
The final result for the arc length of the polar curve $r=f(\theta)$ is therefore
\begin{align*}
  s=\int_a^b\sqrt{[f(\theta)]^2+[f'(\theta)]^2}\,d\theta=\int_a^b\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta\,.
\end{align*}
One way to remember the formula is to recall $s=\int ds$, equate the integrands, and formally rearrange the differentials to get
\begin{align*}
ds=\sqrt{(rd\theta)^2+(dr)^2}\,.
\end{align*}
The interpretation is that as one proceeds along the curve a change of coordinates by an infinitessimal amount $d\theta$ in angle and $dr$ in radius induces a change of $rd\theta$ in a circular direction and a (perpendicular) change of position by $dr$ radially.  The overall change in distance $ds$ is then found by the Pythagorean formula.
\begin{example}
  Find the length of the given polar curves.
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle r=\theta^{2}; \,\,  0\leq \theta \leq \pi$
    \item  $\displaystyle r=\cos^{2}\left(\frac{\theta}{2}\right); \,\, 0\leq \theta \leq \pi$
    \end{enumerate}
  \end{multicols}
\begin{solution*}
\begin{enumerate}
\item 
A graph of the polar function $\displaystyle r=f(\theta)=\theta^{2}$ on $0\leq \theta \leq \pi$ follows.
\begin{center}
\includegraphics[width=.6\textwidth]{graphics/polar_coordinates14}%
\end{center}
The derivative $\displaystyle\frac{dr}{d\theta}=f'(\theta)=2\theta$ and so the arc length is
\begin{align*}
  s&=\int_{0}^{\pi} \sqrt{\theta^4+4\theta^2}\,d\theta
     =\int_{0}^{\pi} \sqrt{\theta^2}\sqrt{\theta^2+4}\,d\theta
     =\int_{0}^{\pi} |\theta|\sqrt{\theta^2+4}\,d\theta
     =\int_{0}^{\pi} \theta \sqrt{\theta^2+4}\,d\theta\,,
\end{align*}
where here we used that $\theta>0$ on $[0,\pi]$ to evaluate the absolute value.
Let $u=\theta^2+4$, so $du=2\theta \, d\theta \implies \frac{1}{2}du= \theta\,d\theta$.  Under the change of variable the limits become
\begin{align*}
    \theta= \pi &\implies u= \pi^{2}+4\\
  \theta=0 &\implies u=4\,,
\end{align*}
and we have
\begin{align*}
    s &=\int_{4}^{\pi^2+4}\sqrt{u}\left(\frac{1}{2}\,du\right)
    =\left.\left(\frac{1}{2}\right)\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right|^{\pi^2+4}_{4} 
    =\left.\frac{1}{3}u^{\frac{3}{2}}\right |^{\pi^2+4}_{4}
    =\frac{1}{3}\left[(\pi^2+4)^{\frac{3}{2}}-4^{\frac{3}{2}}\right]\\
    &=\frac{1}{3}\left[(\pi^2+4)^{\frac{3}{2}}-8\right]\,\text{units}.
\end{align*}
\item  A graph of the polar function $\displaystyle r=f(\theta)=\cos^{2}\left(\frac{\theta}{2}\right)$ on $0\leq \theta \leq \pi$ is as follows.
\begin{center}
\includegraphics[width=.6\textwidth]{graphics/polar_coordinates15}%
\end{center}
Differentiating gives $\displaystyle\frac{dr}{d\theta}=f'(\theta)=2\cos\left(\frac{\theta}{2}\right)\left[-\sin\left(\frac{\theta}{2}\right)\right]\left(\frac{1}{2}\right)=-\cos\left(\frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)$.

The arc length is then
\begin{align*}
s&=\int_{0}^{\pi}\sqrt{\cos^{4}\!\left(\frac{\theta}{2}\right)+\cos^{2}\!\left(\frac{\theta}{2}\right)\sin^2\!\left(\frac{\theta}{2}\right)}\,d\theta
   =\int_{0}^{\pi}\sqrt{\cos^2\left(\frac{\theta}{2}\right)}\sqrt{\cos^{2}\!\left(\frac{\theta}{2}\right)+\sin^{2}\!\left(\frac{\theta}{2}\right)}\,d\theta \\
  &=\int_{0}^{\pi}\left|\cos\left(\frac{\theta}{2}\right)\right|\sqrt{1}\,d\theta
= \int_{0}^{\pi}\! \cos\!\left(\frac{\theta}{2}\right) \,d\theta \ \ (\Leftarrow\text{since}\ \cos\left(\frac{\theta}{2}\right)>0\ \text{on}\ [0,\pi]\,)\\
&=2\sin\!\left.\left(\frac{\theta}{2}\right) \right|^{\pi}_{0}
=2\sin\frac{\pi}{2}-2\sin0=2-0=2\,\,\text{units}.
\end{align*}
\end{enumerate}
\end{solution*}

\begin{furtherquestion}
  Find the arc length of the polar curve given by $\displaystyle r=\tan\theta$\,; $\displaystyle 0\le\theta\le\frac{\pi}{4}$\,.
\end{furtherquestion}
\end{example}

\begin{exercise}
\input{POLAR_COORDINATES/polar-coordinate-curves.tex}
\end{exercise}

\section{Other Applications of Polar Coordinates}
A point in two dimensions can be represented by Cartesian coordinates $(x,y)$ or polar coordinates $(r,\theta)$. Physically, for example, we can locate the point $P$ on a drum membrane with two such coordinates.  Often in physical applications we are interested in the value of some quantity at such a point.   At a point on a drum we could be interested in the vertical displacement $z$ of the membrane at that point.  A function $z=f(x,y)$ or $z=f(r,\theta)$ could be introduced to represent the shape of the drum surface at a particular instant in time. The motion of the drum in time would require that $z$ further depend on time in addition to space, so $z=f(x,y,t)$ or $z=f(r,\theta,t)$. It will turn out that choosing a spatial coordinate system which reflects the symmetry of the physical system makes calculating such motion easier.  In the case of a drum, most drums have a circular boundary and hence polar coordinates are preferred.  If we made square drums, then Cartesian coordinates would work better. We can introduce more coordinate systems to reflect other physical situations.  A drum with a boundary that was an ellipse might be analyzed easiest using an elliptic coordinate system.

In general one is able to introduce many coordinate systems that allow one to identify a location with a tuple of numbers.  In three dimensions a point can be located with Cartesian coordinates $(x,y,z)$.  Polar coordinates may be generalized in three dimensions to spherical-polar coordinates $(r,\theta,\phi)$ where a further angle is needed to identify the point.  The temperature at every point inside the earth at a particular instant of time could be represented as a function of Cartesian coordinates as $T=f(x,y,z)$.  However since our planet is, in rough approximation, a sphere, a function of spherical coordinates, $T=f(r,\theta,\phi)$ would likely be preferred.  One might expect, for instance, that the temperature inside the earth would depend essentially only on its radial distance $r$ from the centre.  A geologist studying the evolution of the earth's temperature over geological time scales would require a time coordinate as well, $T=f(r,\theta,\phi,t)$.

The above applications involving functions at points in space and time are the domain of multivariate calculus.  Spatial coordinate systems (Cartesian, polar, etc.) are fundamental to those problems.  A coordinate system chosen to reflect the symmetry of the physical problem in question makes solving it much easier. 

\begin{reviewexercise}
\input{POLAR_COORDINATES/polar-coordinates-review.tex}
\end{reviewexercise}

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