Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - Review - Exercises - Page 730: 7


(a) Ponit in Cartesian coordinates is $(-2, 2\sqrt 3)$ The point $(4, \dfrac{2\pi}{3})$ can be shown on polar graph as depicted. (b) $$r=3\sqrt 2$$ and $$ \theta =3\pi/4$$

Work Step by Step

(a) from the graph it can be seen that the point $(4, 2\pi/3)$ is lie on a polar graph. Cartesian coordinates are shown as: $$x=rcos\theta=4 cos (2\pi/3)=-2$$ and $$y=rsin\theta=4 sin (2\pi/3)=2\sqrt 3$$ Points in Cartesian coordinates is: $(-2, 2\sqrt 3)$ (b) As we have $r=\sqrt {x^2+y^2}=3\sqrt 2$ $$x=rcos\theta=4 cos (2\pi/3)=-2$$ and $$y=rsin\theta=4 sin (2\pi/3)=2\sqrt 3$$ Thus, $-2=3\sqrt 2cos\theta$ $cos\theta=-\frac{1}{\sqrt 2}$ and $sin\theta=\frac{1}{\sqrt 2}$ As sine is positive and cosine is negative , and the angle $\theta$ lies in second quadrant. $$ \implies \theta =3\pi/4$$ Hence, the result is: $r=3\sqrt 2$ and $ \theta =3\pi/4$
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