Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 9 - Multiveriable Calculus - 9.5 Total Differentials and Approximations - 9.5 Exercises - Page 503: 10

Answer

Thus, $(2.93^{2}-0.94^{2})^{1/3} = 2+(-0.025)= 1.975$ Calculator $(2.93^{2}-0.94^{2})^{1/3} \approx1.9748$ The error is approximately 0.0002

Work Step by Step

We are given $(2.93^{2}-0.94^{2})^{1/3}$ We know that $(3^{2}-1^{2})^{1/3}=2$. Therefore, let $x=3, dx=-0.07,y=1,dy=-0.06$ $dz=f_{x}(x,y).dx+f_{y}(x,y).dy$ $dz=(\frac{(x^{2}-y^{2})^{-2/3}.2x}{3})dx+(\frac{(x^{2}-y^{2})^{-2/3}.(-2y)}{3})dy$ $dz=\frac{1}{2}dx-\frac{1}{6}dy$ $dz\approx -0.025$ Thus, $(2.93^{2}-0.94^{2})^{1/3} = 2+(-0.025)= 1.975$ Calculator $(2.93^{2}-0.94^{2})^{1/3} \approx1.9748$ The error is approximately 0.0002
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