## Calculus (3rd Edition) 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.