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: 60


percentage change in radius is 1%

Work Step by Step

We are given that the % error change in volume is 3% x is diameter of sphere radius $r=\frac{x}{2}$ Volume of sphere $V=\frac{4}{3}\pi r^3= (\frac{4}{3})\pi (\frac{x^3}{8})= (\frac{1}{6})\pi {x^3}$ on differentiating the above equation $dV=\frac{\pi}{6}(3x^2dx))$ so %change in volume =change in volume/original volume $\frac{dV}{V}=\frac{\frac{\pi}{6}(3x^2(dx))}{ (\frac{1}{6})\pi x^3} $ $3=\frac{\frac{\pi}{6}(3(100)^2(1))}{ (\frac{1}{6})\pi (100)^3} $ $\frac{dx}{x}=1$ so the percentage change in radius is 1%
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