The volume of a cube decreases at a rate of $0.8 \mathrm{ft}^{3} / \mathrm{min}$. What is the rate of change of the side length when the side lengths are $13 \mathrm{ft}$ ?

Step 1 ：We are given that the volume of a cube decreases at a rate of \(-0.8 \, \text{ft}^{3} / \text{min}\). We need to find the rate of change of the side length when the side lengths are \(13 \, \text{ft}\).

Step 2 ：The volume of a cube is given by \(V = s^3\) where \(s\) is the side length.

Step 3 ：We can use the chain rule to differentiate \(V\) with respect to \(t\) to get \(\frac{dV}{dt} = 3s^2 \frac{ds}{dt}\).

Step 4 ：We are given that \(\frac{dV}{dt} = -0.8\) and we need to find \(\frac{ds}{dt}\) when \(s = 13\).

Step 5 ：Substituting the given values into the equation, we get \(3*13^2*\frac{ds}{dt} = -0.8\).

Step 6 ：Solving for \(\frac{ds}{dt}\), we get \(\frac{ds}{dt} = -0.00157790927021696\).

Step 7 ：Rounding to four decimal places, we get \(\frac{ds}{dt} = -0.0016\).

Step 8 ：Final Answer: The rate of change of the side length when the side lengths are \(13 \, \text{ft}\) is approximately \(\boxed{-0.0016 \, \text{ft/min}}\).