Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.6 Exercises - Page 153: 15


(a) $\frac{dV}{dt}=72cm^3 / s$ (b)$\frac{dV}{dt}=1800 cm^3 / s$

Work Step by Step

Step-1: It is given to us that the rate of expansion of edges of the cube is $\frac{ds}{dt}=6 cms^{-1}$. The relation between edge length and the volume is given below: $$V=s^3$$ Step-2: Differentiate the above equation with respect to $t$, $$\frac{dV}{dt}=3s^2\frac{ds}{dt}$$ $$\implies\frac{dV}{dt}=3s^2 \times 6$$ Step-3: When $s=2$, $$\frac{dV}{dt}=3(2)^2\times 6=72 cm^3 s^{-1}$$ Step-4: When $s=10$, $$\frac{dV}{dt}=3(10)^2\times 6=1800 cm^3 s^{-1}$$
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