Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.9 Exercises - Page 257: 33

Answer

(a) The maximum possible error is $270~cm^3$ The relative error is $0.01$ The percentage error is $1\%$ (b) The maximum possible error is $36~cm^2$ The relative error is $0.0067$ The percentage error is $0.67\%$

Work Step by Step

(a) Let $x$ be the length of the edge. $x = 30~cm$ $dx = 0.1~cm$ We can find the maximum possible error: $V = x^3$ $dV = 3x^2~dx$ $dV = 3(30~cm)^2~(0.1~cm)$ $dV = 270~cm^3$ We can find the relative error: $\frac{dV}{V} = \frac{3x^2~dx}{x^3} = 3~(\frac{dx}{x})$ $\frac{dV}{V} = 3~(\frac{0.1~cm}{30~cm})$ $\frac{dV}{V} = 0.01$ The percentage error is $1\%$ (b) Let $x$ be the length of the edge. $x = 30~cm$ $dx = 0.1~cm$ We can find the maximum possible error: $A = 6x^2$ $dA = 12x~dx$ $dA = 12(30~cm)~(0.1~cm)$ $dA = 36~cm^2$ We can find the relative error: $\frac{dA}{A} = \frac{12x~dx}{6x^2} = 2~(\frac{dx}{x})$ $\frac{dA}{A} = 2~(\frac{0.1~cm}{30~cm})$ $\frac{dA}{A} = 0.0067$ The percentage error is $0.67\%$
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