Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.8 - Related Rates - Exercises 3.8 - Page 162: 11


Change in surface area is $\dfrac{dA}{dt}=-180\dfrac{m^2}{min}$ and $\dfrac{dA}{dt}=-135\dfrac{m^3}{min}$

Work Step by Step

As we are given that Side of cube is $x=3m $ and $\dfrac{dx}{dt}=-5\dfrac{m}{min}$ Find surface area. $A=6x^2$ or, $\dfrac{dA}{dt}=12x\dfrac{dx}{dt}= 12\times 3\times{-5}=-180\dfrac{m^2}{min}$ For rate of change in volume, we have : $V=x^3$ Now, differentiating with respect to $t$, we have: Thus, $\dfrac{dV}{dt}=3x^2\dfrac{dx}{dt}=3\times3^2\times(-5)=-135\dfrac{m^3}{min}$ Hence, the Change in surface area is $\dfrac{dA}{dt}=-180\dfrac{m^2}{min}$ and $\dfrac{dA}{dt}=-135\dfrac{m^3}{min}$
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