Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.1 Parametric Equations - Exercises - Page 603: 10

Answer

$$ y =x^{3/2}+1$$

Work Step by Step

Given $$ x=t^2,\ \ \ \ y= t^3+1$$ Since $ t=\sqrt{x}$, then \begin{align*} y&= t^3+1\\ &= x^{3/2}+1 \end{align*} i.e. $$ y =x^{3/2}+1$$
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