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: 52


change in diameter = $\frac{2}{\pi}$ change in cross section = $10\ in^2$

Work Step by Step

The given change in circumference of a tree: (dC)=2 We need to find the change in diameter (dD) circumference(c)=$\pi d$ change in circumference $(dC)=\pi dD$ so $2=\pi dD$ and $dD=\frac{2}{\pi}$ change in cross-sectional area, diameter(D)=10in. area(A)=$\pi r^2=\pi {\frac{D^2}{4}}$ change in cross section (dA) = $\frac{2\pi}{4}DdD=\frac{2\pi\times2\times10}{4\pi}=10in^2$ Thus, the final answer is: change in diameter = $\frac{2}{\pi}$ change in cross section = $10\ in^2$
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