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: $\boxed{{\displaystyle 13.079}}$