Water is drained out of an inverted cone, having the dimensions depicted in the figure below. If the water level drops at 6 ft divided by min, at what rate is water (in ft cubed divided by min) draining from the tank when the water depth is 3 ft?the figure is a cone with a radius of 2 and a height of 6
Accepted Solution
A:
Answer:-6π ft³/min ≈ -18.8 ft³/minStep-by-step explanation:Volume of a cone is:V = ⅓ π r² hUsing similar triangles, we can relate the radius and height of the water to the radius and height of the tank.r / h = R / Hr / h = 2 / 6r = ⅓ hSubstituting:V = ⅓ π (⅓ h)² hV = π h³ / 27Take the derivative with respect to time:dV/dt = π h² / 9 dh/dtGiven that dh/dt = -6 ft/min, and h = 3 ft:dV/dt = π (3)² / 9 (-6)dV/dt = -6πdV/dt ≈ -18.8 ft³/minThe water drains from the tank at a rate of 6π ft³/min or approximately 18.8 ft³/min.