Identifying the given you'll have:
the change of volume per time,
dV/dt = 6cm/min
the length of a side
s = 12cm
Volume of a cube, V =[tex] s^{2} [/tex]
differentiating it with time,
[tex] \frac{dV}{dt} [/tex] = 3s² [tex] \frac{ds}{dt}[/tex]
substituting the values given:
6 = 3(12)²[tex] \frac{ds}{dt} [/tex]
[tex] \frac{ds}{dt} [/tex] = 0.0139 cm/min
