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the side \[s(t)\] of a cube is increasing at a rate of \[2\] kilometers per hour. at a certain instant \[t 0\], the side is \[1.5\] kilometers. what is the rate of change of the volume \[v(t)\] of the cube at that instant?

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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 \[13.5 \; km^3/h\].

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