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: 36

Answer

$dV = 1.96 m^3$

Work Step by Step

$v = \frac{2}{3} \pi r^3$ $\frac{dv}{dr} = 2\pi r^2$ Solve for $dv$: $dv = 2\pi r^2 dr$ $dv = 2\pi (25)^2 (0.0005)$ $dv = 1.96 m^3$
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