Functions Modeling Change: A Preparation for Calculus, 5th Edition

Published by Wiley
ISBN 10: 1118583191
ISBN 13: 978-1-11858-319-7

Chapter 8 - Triangle Trigonometry and Polar Coordinates - 8.3 Polar Coordinates - Exercises and Problems for Section 8.3 - Exercises and Problems - Page 347: 17


$$(2,\:-\frac{\pi }{6})$$

Work Step by Step

Converting the Cartesian coordinates to polar coordinates, we find: $$r=\sqrt{\left(-\sqrt{3}\right)^2+1^2} =2 \\ θ=\arctan \left(\frac{1}{-\sqrt{3}}\right) = -\frac{\pi }{6}$$ Thus, it follows: $$(2,\:-\frac{\pi }{6})$$
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