Thomas' Calculus 13th Edition

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

Chapter 14: Partial Derivatives - Practice Exercises - Page 865: 61

Answer

$$\approx 15.83 \% $$

Work Step by Step

Re-write as: $dI=\dfrac{dV}{R}-\dfrac{V \ dR}{R^2}$ $dI(24,100)=\dfrac{dV}{100}-\dfrac{24 \ dR}{(100)^2}$ Set $v=-1, dR=-20$ $\implies dI=-0.01+480(0.0001)=0.038$ Estimated change in $I$ can be computed as: $$I=\dfrac{dI}{I} \times 100 \\ =\dfrac{0.038}{0.24} (100) \approx 15.83 \% $$
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