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: 74

Answer

The longitude and latitude for $\left( {\theta ,\varphi } \right) = \left( {\pi /8,7\pi /12} \right)$: (15$^\circ $ S, 22.5$^\circ $ E). The longitude and latitude for $\left( {\theta ,\varphi } \right) = \left( {4,2} \right)$: (24.59$^\circ $ S, 130.82$^\circ $ W).

Work Step by Step

1. $\left( {\theta ,\varphi } \right) = \left( {\pi /8,7\pi /12} \right)$ We have $\left( {\theta ,\varphi } \right) = \left( {\pi /8,7\pi /12} \right) = \left( {22.5^\circ ,105^\circ } \right)$ The latitude of the point is $105^\circ - 90^\circ = 15^\circ $ south of the equator. The longitude of the point is 22.5$^\circ $ to the east of Greenwich. Thus, the longitude and latitude for this point is (15$^\circ $ S, 22.5$^\circ $ E). 2. $\left( {\theta ,\varphi } \right) = \left( {4,2} \right)$ We have $\left( {\theta ,\varphi } \right) = \left( {4,2} \right) = \left( {229.18^\circ ,114.59^\circ } \right)$ The latitude of the point is $114.59^\circ - 90^\circ = 24.59^\circ $ south of the equator. The longitude of the point is $360^\circ - 229.18^\circ = 130.82^\circ $ to the west of Greenwich. Thus, the longitude and latitude for this point is (24.59$^\circ $ S, 130.82$^\circ $ W).
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