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: 16


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

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.