Draining conical filter
An inverted conical filter has height 30 cm and top radius 10 cm. Water drains through a valve at a constant 4 cm³/s. A depth sensor measures the water depth h cm from the vertex. At the instant h = 12 cm, an operator records that the surface is falling, and maintenance suggests doubling the drain rate to shorten cleaning time. The cone is not a cylinder, so the trainee's prediction using V = π(10²)h is suspect. There is no inflow. The supervisor wants the correct volume-depth relation, the instantaneous rate at h = 12 cm, and a check on the total emptying time from a full filter if the valve stays at 4 cm³/s. The filter is treated as empty when h = 0, and the cone is full when h = 30 cm.
1. Derive V in terms of h and find dh/dt at h = 12 cm under the 4 cm³/s drain rate.
Read the worked answer
Similar triangles give r/h = 10/30 = 1/3, so r = h/3. Thus V = (1/3)πr²h = (1/3)π(h/3)²h = πh³/27. Differentiate: dV/dt = (πh²/9) dh/dt. Draining means dV/dt = -4 cm³/s. At h=12, -4 = (π(144)/9) dh/dt = 16π dh/dt, so dh/dt = -1/(4π) ≈ -0.0796 cm/s. The negative sign means the depth is decreasing.
2. Find the time to empty the full filter and compare it with the trainee's cylinder model. Also find the remaining time at h = 12 cm.
Read the worked answer
Full filter: V(30) = π(30³)/27 = 1000π cm³. At the constant drain rate, emptying time = 1000π/4 = 250π s ≈ 785 s.
The trainee's cylinder model uses V = π(10²)h = 3000π cm³ at h=30 and gives 750π s ≈ 2356 s, three times too long. The error is caused by ignoring the conical taper: the cone holds one third of the cylinder of the same height and radius. At h=12, remaining volume is π(12³)/27 = 64π cm³, so about 64π/4 = 16π s ≈ 50.3 s remain.