Calculus with Applications (10th Edition)

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

Chapter 9 - Multiveriable Calculus - Chapter Review - Review Exercises - Page 519: 61

Answer

Thus, $\sqrt 5.1^{2}+12.05^{2} \approx 13 + 0.05= 13.05$ Calculator gives $\sqrt 5.1^{2}+12.05^{2}\approx13.0848$ The error is approximately 0.0348

Work Step by Step

We are given: $\sqrt 5.1^{2}+12.05^{2}$ Let $f(x,y)=\sqrt x^{2}+y^{2}$ with $x=5, dx=0.01, y=12, dy=0.05$ then use dz to approximate $\Delta z$ $dz=f_{x}(x,y).dx+d_{y}(x,y).dy$ $dz=(\frac{1}{2\sqrt x^{2}+y^{2}}.2x)dx+(\frac{1}{2\sqrt x^{2}+y^{2}}.2y)dy$ $dz=(\frac{x}{\sqrt x^{2}+y^{2}})dx+(\frac{y}{\sqrt x^{2}+y^{2}})dy$ $dz=\frac{5}{13}(0.01)+\frac{12}{13}(0.05)$ $dz=0.05$ Thus, $\sqrt 5.1^{2}+12.05^{2} \approx 13 + 0.05= 13.05$ Calculator gives $\sqrt 5.1^{2}+12.05^{2}\approx13.0848$ The error is approximately 0.0348
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