University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Practice Exercises - Page 593: 26

Answer

$x=-\sqrt 2$

Work Step by Step

Here, we have $r =-\sqrt 2\sec \theta$ This implies that $r=\dfrac{-\sqrt 2}{\cos \theta}$ or, $r \cos \theta =-\sqrt 2$ Thus, we have $x=-\sqrt 2$
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