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The volume of a cube is decreasing at a constant rate of 1699 cubic meters per second. At the instant when the volume of the cube is 629 cubic meters, what is the rate of change of the surface area of the cube? Round your answer to three decimal places (if necessary).
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$\frac{\mathrm{m}^{2}}{\mathrm{sec}}$
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Step 1 ：The volume \(V\) of a cube is given by \(V = s^3\), where \(s\) is the side length of the cube. The surface area \(A\) of a cube is given by \(A = 6s^2\).

Step 2 ：Given that \(\frac{dV}{dt} = -1699\) m^3/s (the volume is decreasing), we want to find \(\frac{dA}{dt}\) when \(V = 629\) m^3.

Step 3 ：First, we need to find the side length \(s\) when \(V = 629\) m^3. We can do this by taking the cube root of 629: \(s = (629)^{1/3} \approx 8.57\) m.

Step 4 ：Next, we can use the chain rule to find \(\frac{dA}{dt}\). The chain rule states that \(\frac{dA}{dt} = \frac{dA}{ds} \cdot \frac{ds}{dt}\). We know that \(\frac{dA}{ds} = 12s\) (the derivative of \(6s^2\)), so we need to find \(\frac{ds}{dt}\).

Step 5 ：We can find \(\frac{ds}{dt}\) by differentiating the volume equation with respect to time: \(\frac{dV}{dt} = 3s^2 \cdot \frac{ds}{dt}\).

Step 6 ：Solving for \(\frac{ds}{dt}\) gives us: \(\frac{ds}{dt} = \frac{dV}{dt} / (3s^2) = -1699 / (3 \cdot (8.57)^2) \approx -22.02\) m/s.

Step 7 ：Finally, we can substitute these values into the chain rule equation to find \(\frac{dA}{dt}\): \(\frac{dA}{dt} = \frac{dA}{ds} \cdot \frac{ds}{dt} = 12 \cdot 8.57 \cdot -22.02 \approx -2265.84\) m^2/s.

Step 8 ：\(\boxed{-2265.84}\) m^2/s is the rate of change of the surface area of the cube. This means the surface area is decreasing at this rate.