Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

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\( - 0.000031\;{\rm{A}}/{\rm{s}}\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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\( - 0.000031\;{\rm{A}}/{\rm{s}}\).

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

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

Step by Step Solution

Step 1: Chain rule

Step 2: Estimate the rate of change of \(I\)

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

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

Step by Step Solution

Step 1: Chain rule

Step 2: Estimate the rate of change of \(I\)

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