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 604: 49


$$ \frac{d y}{d x}= \frac{2}{3t} $$ and at $t=-4$, we have $$ \frac{d y}{d x}= -\frac{1}{6}. $$

Work Step by Step

Since $x=t^3$ and $y=t^2-1$ then we have $$ \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{y^{\prime}(t)}{x^{\prime}(t)}=\frac{2t}{3t^2}=\frac{2}{3t} $$ and at $t=-4$, we have $$ \frac{d y}{d x}= \frac{2}{3(-4)}=-\frac{2}{12}=-\frac{1}{6}. $$
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