Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.1 Polar Cordinates - 8.1 Assess Your Understanding - Page 592: 67


$r=\dfrac{\sqrt 6}{2}\quad \quad \text{or}\quad \quad r^2=\dfrac{3}{2}$

Work Step by Step

Recall that $x=r\cos\theta$ and $y=r\sin\theta$. Substitute these in the given equation $2x^{2}+2y^{2}=3$ to obtain: $2(r\cos\theta)^{2}+2(r\sin\theta)^{2}=3$ $2r^2\cos^2\theta+2r^2\sin^2\theta=3$ Factor out $2r^2$ to obtain: $2r^{2}(\cos^{2}\theta+\sin^{2}\theta)=3$ Since $\cos^{2}\theta+\sin^{2}\theta=1$, then the equation above becomes: $2r^{2}=3$ $r^{2}=\frac{3}{2} \implies r=\frac{\sqrt 3}{\sqrt 2}=\frac{\sqrt 3}{\sqrt2}\cdot \frac{\sqrt2}{\sqrt2}=\frac{\sqrt 6}{2}$ Thus, the polar form of the given equation is: $$r=\dfrac{\sqrt 6}{2}\quad \quad \text{or}\quad \quad r^2=\dfrac{3}{2}$$
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