University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Section 10.3 - Polar Coordinates - Exercises - Page 578: 51

Answer

The graph is a line with slope $\dfrac{-1}{\sqrt 3}$ and y-intercept $\dfrac{4}{\sqrt 3}$.

Work Step by Step

Conversion of polar coordinates and Cartesian coordinates are as follows: a) $r^2=x^2+y^2 \implies r=\sqrt {x^2+y^2}$ b) $\tan \theta =\dfrac{y}{x} \implies \theta =\tan^{-1} (\dfrac{y}{x})$ c) $x=r \cos \theta$ d) $y=r \sin \theta$ Here, we have $r^2=\dfrac{\sqrt 3}{2}r \sin \theta+\dfrac{1}{2}r \cos \theta $ Therefore, our Cartesian equation is $y=\dfrac{-1}{\sqrt 3}x+\dfrac{4}{\sqrt 3}$ This shows that the graph is a line with slope $\dfrac{-1}{\sqrt 3}$ and y-intercept $\dfrac{4}{\sqrt 3}$.
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