Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Review - Exercises - Page 690: 7

Answer

(a) The point in Cartesian coordinates is $(-2, 2\sqrt 3)$ The point $(4, 2\pi/3)$ is shown on a polar graph. (b) $r=3\sqrt 2$ and $ \theta =3\pi/4$ or: $r=-3\sqrt 2$ and $ \theta =7\pi/4$
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Work Step by Step

(a) The point $(4, 2\pi/3)$ is shown on a polar graph. The Cartesian coordinates are given by: $x=rcos\theta=4 cos (2\pi/3)=-2$ $y=rsin\theta=4 sin (2\pi/3)=2\sqrt 3$ The point in Cartesian coordinates is $(-2, 2\sqrt 3)$ (b) $r=\sqrt {x^2+y^2}=3\sqrt 2$ $x=rcos\theta=4 cos (2\pi/3)=-2$ $y=rsin\theta=4 sin (2\pi/3)=2\sqrt 3$ Therefore, $-2=3\sqrt 2cos\theta$ $cos\theta=-\frac{1}{\sqrt 2}$ $sin\theta=\frac{1}{\sqrt 2}$ Since sine is positive and cosine is negative, $\theta$ is in the 2nd quadrant. and $\theta =3\pi/4$ Hence, $r=3\sqrt 2$ and $ \theta =3\pi/4$
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