For the large conical water tank at your warehouse, the volume \( V \) of water in the tank measured in cubic feet is related to the height \( h \) of water in the tank in feet by the formula \( V=\frac{1}{9} \pi h^{3} \). The tank is being filled with water at a constant rate of 1.6 cubic fe per second. Find the rate at which the height is increasing when the height is 7 feet. (Round your answer to two decimal places.)

To find the rate at which the height \( h \) of water in the conical tank is increasing, we start with the volume formula \( V = \frac{1}{9} \pi h^3 \). We need to differentiate this with respect to time \( t \). 1. Differentiate both sides with respect to \( t \): \[ \frac{dV}{dt} = \frac{d}{dt} \left(\frac{1}{9} \pi h^3\right) \] Using the chain rule, this becomes: \[ \frac{dV}{dt} = \frac{1}{9} \pi \cdot 3h^2 \frac{dh}{dt} = \frac{1}{3} \pi h^2 \frac{dh}{dt} \] 2. We know the tank is being filled at a constant rate, which gives us \( \frac{dV}{dt} = 1.6 \) cubic feet per second. 3. Set the two expressions equal: \[ 1.6 = \frac{1}{3} \pi h^2 \frac{dh}{dt} \] 4. Now, we can substitute \( h = 7 \) feet into the equation to find \( \frac{dh}{dt} \): \[ 1.6 = \frac{1}{3} \pi (7^2) \frac{dh}{dt} \] Simplifying this gives: \[ 1.6 = \frac{49\pi}{3} \frac{dh}{dt} \] 5. Now solve for \( \frac{dh}{dt} \): \[ \frac{dh}{dt} = \frac{1.6 \cdot 3}{49\pi} = \frac{4.8}{49\pi} \] 6. Calculating \( \frac{dh}{dt} \) using the value \( \pi \approx 3.14 \): \[ \frac{dh}{dt} \approx \frac{4.8}{49 \cdot 3.14} \approx \frac{4.8}{153.86} \approx 0.0311 \] 7. Rounding to two decimal places, we find: \[ \frac{dh}{dt} \approx 0.03 \text{ feet per second} \] Thus, the rate at which the height is increasing when the height is 7 feet is approximately \( 0.03 \) feet per second.