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 - Practice Exercises - Page 687: 11


$a)y= \pm \dfrac{x^{3/2}}{8}-1; b) y=\pm \dfrac{\sqrt {1-x^2}}{x}$

Work Step by Step

(a) Here, we have $x=4t^2$ This gives: $t=\pm \dfrac{\sqrt x}{2}$ Then $ y=t^3-1$ or, $y=(\pm \dfrac{\sqrt x}{2})^3-1=\pm \dfrac{x^{(3/2)}}{8}-1$ (b) Here, we have $\dfrac{1}{x}=\sec t$ or, $\sec^2 t=1+\tan^2 t$ This gives: $ \dfrac{1}{x^2}-1=y^2$ Thus, $y=\pm \dfrac{\sqrt {1-x^2}}{x}$
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