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: 48


$dS=\pi {\frac{2r^2+h^2}{ (\sqrt{(r^2+h^2)})}}dr$

Work Step by Step

We are given: $S= \pi r \sqrt{(r^2+h^2)}$ on differentiating both sides and applying the product rule: $dS= \pi r (\sqrt{(r^2+h^2)})'+\pi (\sqrt{(r^2+h^2)}){(r)'}$ $dS=\pi r{\frac{1}{2}\frac{1}{ (\sqrt{(r^2+h^2)})}(r^2+h^2)'}+\pi (\sqrt{(r^2+h^2)})dr$ $dS=\pi r{\frac{1}{2}\frac{1}{ (\sqrt{(r^2+h^2)})}(2rdr+2hdh)}+\pi (\sqrt{(r^2+h^2)})dr$ $dh=0$ $dS=\pi r{\frac{1}{2}\frac{1}{ (\sqrt{(r^2+h^2)})}2rdr}+\pi (\sqrt{(r^2+h^2)})dr$ $dS=\pi r^2{\frac{1}{ (\sqrt{(r^2+h^2)})}dr}+\pi (\sqrt{(r^2+h^2)})dr$ $dS=\pi {\frac{2r^2+h^2}{ (\sqrt{(r^2+h^2)})}}dr$
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