University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.10 - Related Rates - Exercises - Page 187: 5

Answer

$\frac{dy}{dt} = -6$

Work Step by Step

Given that $\frac{dx}{dt} = 3$ and $y=x^2$ on differentiating y with respect to time $\frac{d(y)}{dt}$ = $\frac{d x^2}{dt}$ $\frac{dy}{dt} = \frac{2xdx}{dt}$ so $\frac{dy}{dt} = 2x\times 3$ $\frac{dy}{dt} = 6x$ when x=-1 then $ \frac{dy}{dt} = 6\times (-1)$ Thus, the answer is $\frac{dy}{dt} = -6$
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