## Calculus 10th Edition

(a) $\frac{dV}{dt}=72cm^3 / s$ (b)$\frac{dV}{dt}=1800 cm^3 / s$
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 $s=10$, $$\frac{dV}{dt}=3(10)^2\times 6=1800 cm^3 s^{-1}$$