Finding the rate of change of the surface area of a cube given volume

Question

the volume V of a cube increases at a rate of

4 cm³sec^-1. Find the rate of change of the surface area of the cube at the instant when the volume of the cube is

8 cm³.

Answer 1

If the variable x is the dimension of a side of the cube, the the volume V = x³ cm³

The surface area of the cube is A = 6x² cm²

Then, dV/dt = 3x² dx/dt = 4 cm³/sec From this equation, we know that dx/dt = 4/3x²

and dA/dt = 12x dx/dt So, dA/dt = 12x(4/3x²)

When the volume 8 cm³, this tells you that x = 2 at this instant.

Substituting into dA/dt we find

dA/dt = 12(2)(4/3(2)²) Therefore, dA/dt = 8 cm²/sec