Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Section 11.3 - Polar Coordinates - Exercises 11.3 - Page 663: 37

Answer

A line with slope $2$ and intercept is $5$.

Work Step by Step

The conversion of polar coordinates and Cartesian coordinates are described as follows: 1. $r^2=x^2+y^2$ and $r=\sqrt {x^2+y^2}$ 2. $\tan \theta =\dfrac{y}{x} \implies \theta =\tan^{-1} (\dfrac{y}{x})$ 3. $x=r \cos \theta$ and 4. $y=r \sin \theta$ As we know that $x=r \cos \theta$ and $y=r \sin \theta$ Multiply both sides by $\sin \theta -2 \cos \theta $ Now,we have $r\sin \theta -2r \cos \theta=5$ Thus, the Cartesian equation is $y-2x=5$ or, $y=2x+5$ Thus, this shows a line with slope $2$ and intercept is $5$.
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