Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 1 - Section 1.1 - Four Ways to Represent a Function - 1.1 Exercises - Page 20: 70

Answer

$ r(h) = \frac{5\sqrt{\pi h}}{\pi h} $

Work Step by Step

Let $r$ be the radius of the cylinder and $h$ be the height of the cylinder. Volume: $V=\pi r^2 h$ Substituting $V=25$ and isolating $r$: $$ \begin{align*} \displaystyle V& =\pi r^2 h \\\\ 25& =\pi r^2 h\\\\ r^2& =\frac{25}{\pi h} \\\\ r& = \sqrt{\frac{25}{\pi h}}\\\\ r& = \frac{5\sqrt{\pi h}}{\pi h}\\\\ \end{align*} $$
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