Answer
(a) When the side length is $3~mm$, the instantaneous rate of change of the volume with respect to the side length is $27~mm^2$
(b) The rate of change of the volume with respect to the side length is half the surface area.
Work Step by Step
(a) $V(x) = x^3$
$\frac{dV}{dx} = 3x^2$
When $x = 3~mm$:
$\frac{dV}{dx} = 3(3)^2 = 27$
When the side length is $3~mm$, the instantaneous rate of change of the volume with respect to the side length is $27~mm^2$
(b) The area of each face of a cube is $x^2$
The surface area of a cube is $6x^2$
Since $\frac{dV}{dx} = 3x^2$, the rate of change of the volume with respect to the side length is half the surface area.
Suppose the original volume of a cube is $x^2$
After we increase the sides by $\Delta x$, the new volume is $(x+\Delta x)^3$
We can find the change in volume:
$\Delta V = (x+\Delta x)^3-x^3$
$\Delta V = x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3-x^3$
$\Delta V = 3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3$
$\Delta V \approx 3x^2\Delta x$
Note that we can say this because $3x(\Delta x)^2+(\Delta x)^3 \approx 0$
Then:
$\Delta V \approx 3x^2\Delta x$
$\frac{\Delta V}{\Delta x} \approx 3x^2$
