Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.9 - Related Rates - 3.9 Exercises: 7

Answer

$S'(8) = 128\pi \frac{cm^{2}}{min}$

Work Step by Step

$S(r) = 4\pi r^{2}$ Use the chain rule: $\frac{dS}{dt} = \frac{d(4\pi r^{2})}{dr}$ $\frac{dS}{dt} = 8\pi r \times \frac{dr}{dt}$ $ \frac{dr}{dt} = 2 \frac{cm}{min}$ $\frac{dS}{dt} = 8\pi (8) \times 2$ $\frac{dS}{dt} = 64\pi \times 2$ $\frac{dS}{dt} = 128\pi \frac{cm^{2}}{min}$
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