Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 8 - Section 8.1 - Polar Coordinates - 8.1 Exercises - Page 593: 25

Answer

$(3\displaystyle \sqrt{2}, \frac{3\pi}{4}),\ (-3\displaystyle \sqrt{2}, \frac{7\pi}{4})$

Work Step by Step

The point P, with rectangular coordinates ($-3,3)$ lies in the 3rd quadrant. Change from rectangular to polar coordinates with $r^{2}=x^{2}+y^{2}$ and $\displaystyle \tan\theta=\frac{y}{x}$. $r^{2}=3^{2}+3^{2}=18$ $r=\pm 3\sqrt{2}$ $\displaystyle \tan\theta=\frac{3}{-3}=-1,$ and with $\theta$ in quadrant II, $\displaystyle \theta=\frac{3\pi}{4}$, r is positive: so the appropriate polar coordinates are $(3\displaystyle \sqrt{2}, \frac{3\pi}{4}).$ With $\theta$ in quadrant IV, $\displaystyle \theta=\frac{7\pi}{4}$, r is negative: so the appropriate polar coordinates are $(-3\displaystyle \sqrt{2}, \frac{7\pi}{4}).$
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.