Calculus (3rd Edition)

Published by W. H. Freeman

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.3 Polar Coordinates - Exercises - Page 618: 29

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.

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