Let's denote the side length of the cube as \( s \) in meters and the volume of the cube as \( V \) in cubic meters.
The volume of a cube as a function of its side length is given by:
\[ V = s^3 \]
Now, we need to find the rate at which the side of the cube changes with respect to time (\( \frac{ds}{dt} \)) when the side length is 3 m, given that the rate of increase of the volume (\( \frac{dV}{dt} \)) is 8 m³/sec.
To relate the rate of change of the volume with the rate of change of the side length, we can use the chain rule from calculus. Taking the derivative of volume \( V \) with respect to time \( t \) gives:
\[ \frac{dV}{dt} = \frac{dV}{ds} \cdot \frac{ds}{dt} \]
We are given that the volume increases at 8 m³/sec, so:
\[ \frac{dV}{dt} = 8 \text{ m³/sec} \]
Next, we differentiate \( V = s^3 \) with respect to \( s \) to find \( \frac{dV}{ds} \):
\[ \frac{dV}{ds} = 3s^2 \]
Now, we can substitute \( \frac{dV}{ds} \) into our chain rule expression and solve for \( \frac{ds}{dt} \):
\[ \frac{dV}{dt} = 3s^2 \cdot \frac{ds}{dt} \]
\[ 8 = 3s^2 \cdot \frac{ds}{dt} \]
The problem asks us to find \( \frac{ds}{dt} \) when the side length is 3 m. Therefore, we substitute \( s = 3 \) m:
\[ 8 = 3 \cdot (3)^2 \cdot \frac{ds}{dt} \]
\[ 8 = 3 \cdot 9 \cdot \frac{ds}{dt} \]
\[ 8 = 27 \cdot \frac{ds}{dt} \]
Now, we can solve for \( \frac{ds}{dt} \) by dividing both sides by 27:
\[ \frac{ds}{dt} = \frac{8}{27} \]
The rate of change of the side length when \( s = 3 \) m is:
\[ \frac{ds}{dt} = \frac{8}{27} \text{ m/sec} \]
Thus, the side of the cube changes at a rate of \( \frac{8}{27} \) meters per second when its length is 3 meters.
