Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.8 - Related Rates - Exercises 3.8 - Page 161: 10



Work Step by Step

$\dfrac{dr}{dt}+2s\dfrac{ds}{dt}+3v^2\dfrac{dv}{dt}=0$ As we are given that ${r+s^2+v^3=12}$,$\dfrac{dr}{dt}=4$ and $\dfrac{ds}{dt}=-3$ Thus, $4+2s(-3)+3v^2\dfrac{dv}{dt}=0$ Plug $r=3$ and $s=1$ Then ${r+s^2+v^3=12}=3+1+v^3$ or, $v=2$ or, $4+2\times1(-3)+12 \dfrac{dv}{dt}=0$ and $\dfrac{dv}{dt}=\dfrac{1}{6}$
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