Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 10 - Parametric Equations and Polar Coordinates - 10.3 Exercises - Page 686: 4

Answer

$a.\quad (1,1)$ $b.\quad (0, 1)$ $c.\quad (-\sqrt{3},1)$

Work Step by Step

Polar coordinates $(r,\theta)$ Cartesian coordinates:$ \quad(r\cos\theta,r\sin\theta)$ $a.$ Plot point A: r is negative, terminal side is $\displaystyle \frac{5\pi}{4}$+$\displaystyle \pi=\frac{9\pi}{4}$ ($ \pi/$4 $above$ the +x axis, at distance $\sqrt{2}$) Cartesian coordinates:$\quad $ $(-\displaystyle \sqrt{2}\cos\frac{5\pi}{4},-\sqrt{2}\sin\frac{5\pi}{4})= (-\displaystyle \sqrt{2}(-\frac{\sqrt{2}}{2}),-\sqrt{2}(-\frac{\sqrt{2}}{2}))=(1, 1)$. $b.$ Plot point B: r is positive, terminal side: $5\pi/2$ ($ \pi/2$ above the +x axis, at distance 1) Cartesian coordinates:$\quad $ $(1\displaystyle \cos\frac{5\pi}{2},1\sin\frac{5\pi}{2})=(0, 1).$ $c.$ Plot point C: r is positive, terminal side: $-7\pi/6$ ($ \pi/6$ above the -x axis, at distance 2) Cartesian coordinates:$\quad $ $(2\displaystyle \cos(-\frac{7\pi}{6}),2\sin(-\frac{7\pi}{6}))= (2(-\displaystyle \frac{\sqrt{3}}{2}),2(\frac{1}{2}))=(-\sqrt{3},1)$
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