Answer is: option3
1.5 ft per hrSolution:
We are given the volume of water in a spherical tank as:
\[ V = \frac{\pi h^2}{3} (18 - h) \]where \( V \) is the volume of water and \( h \) is the depth of the water. The rate of change of volume is given as:
\[ \frac{dV}{dt} = 30\pi \quad \text{cubic feet per hour} \]We need to find \( \frac{dh}{dt} \) when \( h = 2 \) feet.
Expanding the given equation:
\[ V = \frac{\pi}{3} (18h^2 - h^3) \]Differentiate both sides with respect to \( t \):
\[ \frac{dV}{dt} = \frac{\pi}{3} (36h \frac{dh}{dt} - 3h^2 \frac{dh}{dt}) \]Factor out \( \frac{dh}{dt} \):
\[ \frac{dV}{dt} = \frac{\pi}{3} (36h - 3h^2) \frac{dh}{dt} \] \[ \frac{dV}{dt} = \frac{\pi}{3} (36h - 3h^2) \frac{dh}{dt} \]Solve for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{3}{\pi(36h - 3h^2)} \frac{dV}{dt} \]Substituting \( \frac{dV}{dt} = 30\pi \):
\[ \frac{dh}{dt} = \frac{3 \times 30\pi}{\pi (36h - 3h^2)} \]Cancel \( \pi \):
\[ \frac{dh}{dt} = \frac{90}{36h - 3h^2} \]Evaluate at \( h = 2 \):
\[ \frac{dh}{dt} = \frac{90}{36(2) - 3(2^2)} \] \[ = \frac{90}{72 - 12} \] \[ = \frac{90}{60} \] \[ = 1.5 \text{ ft per hour} \]Final Answer: 1.5 ft per hr.