## Calculus (3rd Edition)

1. Evaluate several points in rectangular coordinates corresponding to $\theta = 0,\frac{\pi }{4},\frac{\pi }{2},...,2\pi$ 2. Then plot the points in rectangular coordinates and sketch the curve by connecting these points.
The rectangular coordinates of a point on the curve corresponding to $\theta$ is given by $\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right)$, $\left( {x,y} \right) = \left( {1 + \cos \theta } \right)\left( {\cos \theta ,\sin \theta } \right)$. 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 a table: $\begin{array}{l} \begin{array}{*{20}{c}} \theta &{\left( {x,y} \right)} \end{array}\\ \begin{array}{*{20}{c}} 0&{\left( {2,0} \right)} \end{array}\\ \begin{array}{*{20}{c}} {\frac{\pi }{4}}&{\left( {\frac{1}{2}\left( {1 + \sqrt 2 } \right),\frac{1}{2}\left( {1 + \sqrt 2 } \right)} \right)} \end{array}\\ \begin{array}{*{20}{c}} {\frac{\pi }{2}}&{\left( {0,1} \right)} \end{array}\\ \begin{array}{*{20}{c}} {\frac{{3\pi }}{4}}&{\left( {\frac{1}{2}\left( {1 - \sqrt 2 } \right),\frac{1}{2}\left( { - 1 + \sqrt 2 } \right)} \right)} \end{array}\\ \begin{array}{*{20}{c}} \pi &{\left( {0,0} \right)} \end{array}\\ \begin{array}{*{20}{c}} {\frac{{5\pi }}{4}}&{\left( {\frac{1}{2}\left( {1 - \sqrt 2 } \right),\frac{1}{2}\left( {1 - \sqrt 2 } \right)} \right)} \end{array}\\ \begin{array}{*{20}{c}} {\frac{{3\pi }}{2}}&{\left( {0, - 1} \right)} \end{array}\\ \begin{array}{*{20}{c}} {\frac{{7\pi }}{4}}&{\left( {\frac{1}{2}\left( {1 + \sqrt 2 } \right),\frac{1}{2}\left( { - 1 - \sqrt 2 } \right)} \right)} \end{array}\\ \begin{array}{*{20}{c}} {2\pi }&{\left( {2,0} \right)} \end{array} \end{array}$ Then we plot the points in rectangular coordinates and sketch the curve by connecting these points.