To find the rate of change of the volume \[V(t)\] of the cube at the instant \[t_0\], we can use the formula for the volume of a cube: \[V(t) = s(t)^3\], where \[s(t)\] represents the length of a side of the cube at time \[t\].
We are given that the side length \[s(t)\] is increasing at a rate of \[2\] kilometers per hour. So, the rate of change of the side length with respect to time is: \[\frac{ds}{dt} = 2 \; km/h\].
Using the chain rule of differentiation, we can find the rate of change of the cube's volume with respect to time:
\[V(t) = s(t)^3\]
Differentiating with respect to time gives us:
\[\frac{dV}{dt} = 3s(t)^2 \frac{ds}{dt}\]
At the instant \[t_0\], the side length \[s(t_0) = 1.5\] kilometers and the rate of change of the side length \[\frac{ds}{dt} = 2\] kilometers per hour, so we can calculate the rate of change of the volume at that instant using these values:
\[\frac{dV}{dt} = 3 \times (1.5^2) \times 2\]
\[\frac{dV}{dt} = 3 \times 2.25 \times 2\]
\[\frac{dV}{dt} = 13.5 \; km^3/h\]
Therefore, at the instant \[t_0\], the rate of change of the volume \[V(t)\] of the cube is \[13.5 \; km^3/h\].