## Calculus: Early Transcendentals 8th Edition

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