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 - Review - Exercises - Page 710: 1

Answer

$x=y^2-8y+12, 1\leq y\leq 6$

Work Step by Step

$x=t^2+4t=(2-y)^2+4(2-y)$ $x=y^2-8y+12$ Therefore, $x=y^2-8y+12, 1\leq y\leq 6$ Also see attached graph.
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