Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.9 Related Rates - Exercises - Page 159: 7


$9600\pi\,cm^{2}/min $

Work Step by Step

We know that the surface area of a sphere is given by: $S=4\pi r^2$ We can find the rate of change of the surface area with respect to time as: $\frac{dS}{dt}=\frac{dS}{dr}\times\frac{dr}{dt}=\frac{d}{dr}(4\pi r^{2})\times30\,cm/min $ $=8\pi r\times 30\,cm/min $ Thus, when $ r=40\,cm $, $\frac{dS}{dt}=8\pi(40\,cm)\times30\,cm/min $ $=9600\pi\,cm^{2}/min $
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