Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 13 - Vector Geometry - 13.7 Cylindrical and Spherical Coordinates - Exercises - Page 700: 77

Answer

$z = \pm r\sqrt {\cos 2\theta } $

Work Step by Step

We have ${z^2} = {x^2} - {y^2}$. The relations between rectangular and cylindrical coordinates are given by $x = r\cos \theta $, ${\ \ }$ $y = r\sin \theta $, ${\ \ }$ $z=z$. Substituting $x$ and $y$ in ${z^2} = {x^2} - {y^2}$ gives ${z^2} = {r^2}{\cos ^2}\theta - {r^2}{\sin ^2}\theta $ ${z^2} = {r^2}\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)$ But $\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta $. So, ${z^2} = {r^2}\cos 2\theta $ $z = \pm r\sqrt {\cos 2\theta } $.
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