Answer:
-6π ft³/min ≈ -18.8 ft³/min
Step-by-step explanation:
Volume of a cone is:
V = ⅓ π r² h
Using similar triangles, we can relate the radius and height of the water to the radius and height of the tank.
r / h = R / H
r / h = 2 / 6
r = ⅓ h
Substituting:
V = ⅓ π (⅓ h)² h
V = π h³ / 27
Take the derivative with respect to time:
dV/dt = π h² / 9 dh/dt
Given that dh/dt = -6 ft/min, and h = 3 ft:
dV/dt = π (3)² / 9 (-6)
dV/dt = -6π
dV/dt ≈ -18.8 ft³/min
The water drains from the tank at a rate of 6π ft³/min or approximately 18.8 ft³/min.
