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Found in: Page 657

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Find the rate of change of $$I$$ when $$R = 400\Omega ,I = 0.08A,\frac{{dV}}{{dt}} = - 0.01\;{\rm{V}}/{\rm{s}}$$ and $$\frac{{dR}}{{dt}} = 0.03\Omega /{\rm{s}}$$.

The current decreases at a rate of$$- 0.000031\;{\rm{A}}/{\rm{s}}$$.

See the step by step solution

## Step 1: Chain rule

"Suppose that $$z = f(x,y)$$ is a differentiable function of $$x$$ and $$y$$, where $$x = g(t)$$ and $$y = h(t)$$ are both differentiable functions of $$t$$. Then, $$z$$ is differentiable function of $$t$$ and$$\frac{{dz}}{{dt}} = \frac{{dz}}{{dx}} \cdot \frac{{dx}}{{dt}} + \frac{{dz}}{{dy}} \cdot \frac{{dy}}{{dt}}$$

## Step 2: Estimate the rate of change of $$I$$

As given, the Ohm's Law is $$V = IR$$ where $$V$$ is voltage, / is current and $$R$$ is resistance.

The values of $$R = 400\Omega ,I = 0.08A,\frac{{dV}}{{dt}} = - 0.01\;{\rm{V}}/{\rm{s}}$$ and$$\frac{{dR}}{{dt}} = 0.03\Omega /{\rm{s}}$$.

The Ohm's Law $$V = IR$$ can be written as,$$I = \frac{V}{R}$$

By using chain rule, obtain the value of $$\frac{{dI}}{{dt}}$$,

\begin{aligned}{l}\frac{{dI}}{{dt}} = \frac{{\partial I}}{{\partial V}} \cdot \frac{{dV}}{{dt}} + \frac{{\partial I}}{{\partial R}} \cdot \frac{{dR}}{{dt}}\\\frac{{dI}}{{dt}} = \frac{1}{R}\frac{{dV}}{{dt}} + \left( { - V{R^{ - 2}}} \right)\frac{{dR}}{{dt}}\\\frac{{dI}}{{dt}} = \frac{1}{R}\frac{{dV}}{{dt}} - \frac{V}{{{R^2}}}\frac{{dR}}{{dt}}\end{aligned}

Thus,

Substitute $$R = 400\Omega ,I = 0.08A,\frac{{dV}}{{dt}} = - 0.01\;{\rm{V}}/{\rm{s}}$$ and $$\frac{{dR}}{{dt}} = 0.03\Omega /{\rm{s}}$$ to the above equation,

\begin{aligned}{l}\frac{{dI}}{{dt}} = \frac{1}{R}\frac{{dV}}{{dt}} - \frac{I}{R}\frac{{dR}}{{dt}}\\\frac{{dI}}{{dt}} = \frac{1}{{400}}( - 0.01) - \frac{{0.08}}{{400}}(0.03)\\\frac{{dI}}{{dt}} = - \frac{1}{{40000}} - \frac{{24}}{{4000000}}\\\frac{{dI}}{{dt}} = \frac{{ - 100 - 24}}{{4000000}}\end{aligned}

Simplify further as follows,

\begin{aligned}{l}\frac{{dI}}{{dt}} = \frac{{ - 124}}{{4000000}}\\\frac{{dI}}{{dt}} = - 0.000031\end{aligned}

Thus, $$\frac{{dI}}{{dt}} = - 0.000031\;{\rm{A}}/{\rm{s}}$$

Therefore, the current is decreasing at a rate of$$- 0.000031\;{\rm{A}}/{\rm{s}}$$.