## Calculus with Applications (10th Edition)

Published by Pearson

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

#### Answer

Thus, $(1.92^{2}+2.1^{2})^{1/3} = 2+ 0.0067= 2.0067$ Calculator gives $(1.92^{2}+2.1^{2})^{1/3}$$\approx2.008$ The error is approximately 0.0013

#### Work Step by Step

We are given $(1.92^{2}+2.1^{2})^{1/3}$ We know that $(2^{2}+2^{2})^{1/3}=2$. Therefore, let $x=2, dx=-0.08,y=2,dy=0.1$ $dz=f_{x}(x,y).dx+f_{y}(x,y).dy$ $dz=(\frac{(x^{2}+y^{2})^{-2/3}.2x}{3})dx+(\frac{(x^{2}+y^{2})^{-2/3}.2y}{3})dy$ $dz=\frac{1}{3}dx+\frac{1}{3}dy$ $dz\approx0.0067$ Thus, $(1.92^{2}+2.1^{2})^{1/3} = 2+ 0.0067= 2.0067$ Calculator gives $(1.92^{2}+2.1^{2})^{1/3} \approx2.008$ The error is approximately 0.0013

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