Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.3 - Polar Coordinates - 10.3 Exercises - Page 666: 3

Answer

(a) We can see the point $(2,\frac{3\pi}{2})$ on the graph. The Cartesian coordinates are $(0,-2)$ (b) We can see the point $(\sqrt{2},\frac{\pi}{4})$ on the graph. The Cartesian coordinates are $(1,1)$ (c) We can see the point $(-1,-\frac{\pi}{6})$ on the graph. The Cartesian coordinates are $(-\frac{\sqrt{3}}{2},\frac{1}{2})$

Work Step by Step

(a) We can see the point $(2,\frac{3\pi}{2})$ on the graph. We can find the Cartesian coordinates: $x = 2~cos~\frac{3\pi}{2} = 0$ $y = 2~sin~\frac{3\pi}{2} = -2$ The Cartesian coordinates are $(0,-2)$ (b) We can see the point $(\sqrt{2},\frac{\pi}{4})$ on the graph. We can find the Cartesian coordinates: $x = \sqrt{2}~cos~\frac{\pi}{4} = 1$ $y = \sqrt{2}~sin~\frac{\pi}{4} = 1$ The Cartesian coordinates are $(1,1)$ (c) We can see the point $(-1,-\frac{\pi}{6})$ on the graph. We can find the Cartesian coordinates: $x = -1~cos~\frac{-\pi}{6} = -\frac{\sqrt{3}}{2}$ $y = -1~sin~\frac{-\pi}{6} = \frac{1}{2}$ The Cartesian coordinates are $(-\frac{\sqrt{3}}{2},\frac{1}{2})$
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