Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.8 Related Rates - 2.8 Exercises - Page 185: 7

Answer

$$128\pi*cm^2/min$$

Work Step by Step

We can write the surface area as, $$A = 4*\pi*r^2$$ Taking the derivative of both sides with respect to t, $$dA/dt = 8\pi r*dr/dt$$ We know $dr/dr$ is 2 cm/min, and $r=8 cm$. Plugging those in, $$dA/dt = 8*\pi*8*2 cm^2/min = 128\pi*cm^2/min$$
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