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%  Polar Coordinate Curves                         %
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\problemgrouptitle{Polar Coordinate Curves}

\begin{problemblock}[2]{Find a Cartesian equation for the given curve and identify it.}

\item $r=10$
\answer{
circle: $x^2+y^2=100$
}

\item $r=5 \cos \theta$
\answer{
circle: $\displaystyle \left(x-\frac{5}{2}\right)^2+y^2=\frac{25}{4}$
}

\item $r^2\left( 3\sin^2\!\theta-7\cos^2\!\theta\right)=42$
\answer{
hyperbola: $\displaystyle \frac {y^2}{14}-\frac {x^2}{6}=1$
}

\item $r^2\cos 2\theta =10$
\answer{
hyperbola: $x^2-y^2=10$
}

\end{problemblock}

\begin{problemblock}[2]{Find a polar equation that has the same graph as the given Cartesian equation.}

\item $x^2+y^2=64$
\answer{
$r=8$
}

\item $xy=5$
\answer{
$r^2\sin(2\theta)=10$
}

\item $9x^2+16y^2=36$
\answer{
$\displaystyle r^2\left(9+7\sin^2\!\theta \right)=36$
}

\end{problemblock}

\begin{problemblock}[2]{Find the slope of the tangent line to the given curve at the indicated value of $\theta$.}

\item $r=25 \sin \theta;\,\,\,\, \theta=\frac{\pi}{3}$
\answer{
$-\sqrt{3}$
}

\item $r=3 \sin \theta+5;\,\,\,\, \theta=\frac{\pi}{4}$
\answer{
$\displaystyle -\frac {6+5\sqrt 2}{5\sqrt 2}$
}

\item $r=1 -\sin\theta+2\cos\theta;\,\,\,\, \theta=0$
\answer{
$-3$
}

\end{problemblock}

\begin{problemblock}[2]{Find the points on the curve at which the tangent line is horizontal or vertical.}

\item $r=100 \cos \theta$
\answer{
Horizontal if $\displaystyle\theta=\frac{\pi}{4}+n\pi,\,\,\, \theta=\frac{3\pi}4+n\pi$ \\  Vertical if $\displaystyle\theta=n\pi,\,\,\, \theta=\frac{\pi}{2}+2n\pi,\,\,\, \theta=\frac{3\pi}{2}+2n\pi$
}

\item $r=5\sin \theta-5$
\answer{
Horizontal if $\displaystyle\theta=\frac{3\pi}{2}+2n\pi, \,\,\, \theta=\frac{\pi}{6}+2n\pi,\,\,\, \theta=\frac{5\pi}{6}+2n\pi$ \\ Vertical if $\displaystyle\theta=\frac{\pi}{2}+2n\pi,\,\,\, \theta=\frac{7\pi}{6}+2n\pi,\,\,\, \theta=\frac{11\pi}{6}+2n\pi$
}

\end{problemblock}

\begin{problemblock}[2]{Sketch the graph of the equation and find the area of the region bounded by the graph.}

\item $r=5 \cos \theta$
\answer{
$\displaystyle A=\frac {25\pi}{4}$
}

\item $r=4+4\sin \theta$
\answer{
$ A=24\pi$
}

\end{problemblock}

\begin{problemblock}[2]{Find the area of the region bounded by one loop of the graph of the give equation.}

\item $r=3 \sin 2\theta$  
\answer{
$\displaystyle A=\frac{9\pi}{8}$
}

\item $r^2=16 \sin(4\theta)$
\answer{
$A=4$
}

\end{problemblock}

\begin{problemblock}[2]{Find the area of the region that lies inside both curves.}

\item $r=4+4\sin\theta,\,\,\, r=4-4\sin \theta$
\answer{
$\displaystyle A=24\pi-64$
}

\item $r=4+3\sin\theta,\,\,\, r=4+3\cos \theta$
\answer{
$\displaystyle A=\frac{41\pi}2-24\sqrt2$
}

\end{problemblock}

\begin{problemblock}[2]{Find the length of the given polar curve.}

\item $r=e^{2\theta}$; $0\le\theta\le3$
\answer{
$\displaystyle s=\frac{\sqrt 5}{2}\left(e^6-1\right)$
}

\item $r=3+3\cos\theta$
\answer{
$s=24$
}

\end{problemblock}

\item We derived a formula for the arc length of a polar curve described by the function $r=f(\theta)$. Here $\theta$ is the independent variable and and $r$ is the dependent variable.
  \begin{enumerate}
  \item Derive a formula for the arc length of a polar curve described by the function $\theta=g(r)$ for $c\le r\le d$. Now $r$ is the independent variable and the angle $\theta$ is the dependent variable.
  \item Use your formula from (a) to find the length of the curve $\theta=\ln(r)$; $1\le r\le 4$\,.
  \end{enumerate}
  \answer{\begin{enumerate}
      \item $\displaystyle s=\int_c^d\!\sqrt{1+r^2[g'(r)]^2}\,dr=\int_c^d\sqrt{1+r^2\left(\frac{d\theta}{dr}\right)^2}\,dr$\\
        Note that formally the integrand is what is expected if we ``pulled out'' the $dr$ from $ds=\sqrt{(rd\theta)^2+(dr)^2}$.
      \item $s=3\sqrt{2}$
      \end{enumerate}
    }

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