Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 8 - Section 8.1 - Polar Coordinates - 8.1 Exercises - Page 593: 30

Answer

The point in rectangular coordinates is $(-3,3\sqrt{3})$

Work Step by Step

$(6,2\pi/3)$ The point has $r=6$ and $\theta=\dfrac{2\pi}{3}$. The formulas for converting polar to rectangular coordinates are $x=r\cos\theta$ and $y=r\sin\theta$ Substitute the known values into the formulas to obtain $x$ and $y$: $x=r\cos\theta$ $x=(6)\cos\Big(\dfrac{2\pi}{3}\Big)=(6)\Big(-\dfrac{1}{2}\Big)=-3$ $y=r\sin\theta$ $y=(6)\sin\Big(\dfrac{2\pi}{3}\Big)=(6)\Big(\dfrac{\sqrt{3}}{2}\Big)=3\sqrt{3}$ The point in rectangular coordinates is $(-3,3\sqrt{3})$
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