University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 199: 56

Answer

% error in area is 1% % error in volume is 1.5%

Work Step by Step

The given % error change in side length: $\frac{dx}{x}$=0.5% % error change=$\frac{change\ in\ dimension}{original\ dimension} $ $A=6x^2$ $dA=12x dx$ $V=x^3$ $dV=3x^2 dx$ % error in Area=$\frac{dA}{A}=\frac{12x dx}{6x^2}=2\times0.005=0.01$ % error in volume=$\frac{dV}{V}=\frac{3x^2dx}{x^3}=0.015$
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