Question The radius of a cylinder is decreasing at a constant rate of 6 meters per second, and the volume is decreasing at a rate of 1348 cubic meters per second. At the instant when the height of the cylinder is 4 meters and the volume is 762 cubic meters, what is the rate of change of the height? The volume of a cylinder can be found with the equation $V=\pi r^{2} h$. Round your answer to three decimal places.

Step 1 ：Given: \( \frac{dr}{dt} = -6 \) meters per second, \( \frac{dV}{dt} = -1348 \) cubic meters per second, \( h = 4 \) meters, \( V = 762 \) cubic meters

Step 2 ：Use the formula for the volume of a cylinder \( V = \pi r^2 h \)

Step 3 ：Differentiate the volume formula with respect to time to find the relationship between the rates of change: \( \frac{dV}{dt} = 2\pi r \frac{dr}{dt} h + \pi r^2 \frac{dh}{dt} \)

Step 4 ：Solve for \( \frac{dh}{dt} \) using the given values: \( \frac{dh}{dt} = \frac{\frac{dV}{dt} - 2\pi r \frac{dr}{dt} h}{\pi r^2} \)

Step 5 ：Substitute the given values into the equation: \( \frac{dh}{dt} = \frac{-1348 - 2\pi (\sqrt{\frac{V}{\pi h}}) (-6) h}{\pi (\sqrt{\frac{V}{\pi h}})^2} \)

Step 6 ：Simplify the equation to find the rate of change of the height: \( \frac{dh}{dt} = \frac{-1348 - 2\pi (\sqrt{\frac{762}{\pi \cdot 4}}) (-6) \cdot 4}{\pi (\sqrt{\frac{762}{\pi \cdot 4}})^2} \)

Step 7 ：Calculate the value of \( \frac{dh}{dt} \) and round to three decimal places: \( \frac{dh}{dt} = 13.079 \)

Step 8 ：Final Answer: \( \boxed{13.079} \)