#### Answer

1. First, convert the polar coordinates to rectangular coordinates.
2. Then plot the points in rectangular coordinates and sketch the curve by connecting these points.

#### Work Step by Step

The rectangular coordinates of a point on the curve corresponding to $\theta$ is given by
$\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right)$,
$\left( {x,y} \right) = \frac{\theta }{2}\left( {\cos \theta ,\sin \theta } \right)$.
For the interval $0 \le \theta \le 2\pi $, we evaluate several points in rectangular coordinates corresponding to $\theta = 0,\frac{\pi }{4},\frac{\pi }{2},...,2\pi $ and list them on the following table:
$\begin{array}{*{20}{c}}
\theta \\
0\\
{\frac{\pi }{4}}\\
{\frac{\pi }{2}}\\
{\frac{{3\pi }}{4}}\\
\pi \\
{\frac{{5\pi }}{4}}\\
{\frac{{3\pi }}{2}}\\
{\frac{{7\pi }}{4}}\\
{2\pi }
\end{array}\begin{array}{*{20}{c}}
{\left( {x,y} \right)}\\
{\left( {0,0} \right)}\\
{\left( {\frac{\pi }{{16}}\sqrt 2 ,\frac{\pi }{{16}}\sqrt 2 } \right)}\\
{\left( {0,\frac{\pi }{4}} \right)}\\
{\left( { - \frac{{3\pi }}{{16}}\sqrt 2 ,\frac{{3\pi }}{{16}}\sqrt 2 } \right)}\\
{\left( { - \frac{\pi }{2},0} \right)}\\
{\left( { - \frac{{5\pi }}{{16}}\sqrt 2 , - \frac{{5\pi }}{{16}}\sqrt 2 } \right)}\\
{\left( {0, - \frac{{3\pi }}{4}} \right)}\\
{\left( {\frac{{7\pi }}{{16}}\sqrt 2 , - \frac{{7\pi }}{{16}}\sqrt 2 } \right)}\\
{\left( {\pi ,0} \right)}
\end{array}$
Then we plot the points in rectangular coordinates and sketch the curve by connecting these points.