Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.5 - The Chain Rule. - 14.5 Exercise - Page 945: 40


$-0.000031 Amp/s$ or, $-3.1 \times 10^{-5}$

Work Step by Step

Given: $V=IR$ Chain rule : $\dfrac{dV}{dt}=V_I(\dfrac{dI}{ dt})+V_r(\dfrac{dR}{dt})$ This implies that $\dfrac{dV}{dt}=R(\dfrac{dI}{ dt})+I(\dfrac{dR}{dt})$ Now, we have $-(0.1)=(400) \cdot (\dfrac{dI}{ dt})+(0.08) \cdot (0.03)$ or, $-(0.1)=(0.00240)+(400) \cdot (\dfrac{dI}{ dt})$ This gives: $\dfrac{dI}{dt}=-0.000031 Amp/s$ or, $\dfrac{dI}{dt}=-3.1 \times 10^{-5}$
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