Filling a Conical Tank: A Rate That Will Not Stay Constant
An inverted conical tank has height 6 m and top radius 3 m. Water is pumped in at a constant 0.5 m³ per minute and the tank starts empty. A technician is asked how fast the water level rises and divides the flow rate by the area of the base, obtaining a constant answer. That is wrong: the water surface widens as the depth increases, so the level rises quickly at first and more slowly later. Let h be the depth of water at time t and use similar triangles to relate the surface radius to h. The technician also wants to know at what depth the level is rising at exactly 0.05 m/min, and whether the tank takes longer to fill over its last metre than over its first. Assume the tank is a right circular cone with its vertex at the bottom and ignore any pipe or wall effects.
1. Express the volume of water in terms of the depth h, and find dh/dt when h = 2 m.
Read the worked answer
Similar triangles. At depth h the water surface radius r satisfies r/h = 3/6, so r = h/2.
Volume. V = (1/3)πr²h = (1/3)π(h²/4)h = πh³/12.
Related rates. Differentiating with respect to time using the chain rule:
dV/dt = (dV/dh)(dh/dt) = (πh²/4)(dh/dt).
At h = 2 m, dV/dh = π(4)/4 = π m². With dV/dt = 0.5 m³/min:
0.5 = π(dh/dt) → dh/dt = 0.159 m/min.
The key judgement is that dh/dt is the flow rate divided by the area of the water surface at that instant (π m² here), not by any fixed base area. The technician's constant answer uses a fixed area and so cannot be right for a cone.
2. Find the depth at which the level rises at exactly 0.05 m/min, and comment on the technician's constant-rate assumption.
Read the worked answer
General expression. Rearranging the related-rates result:
dh/dt = 0.5/(πh²/4) = 2/(πh²).
Target rate. Set 2/(πh²) = 0.05:
πh² = 40 → h² = 12.73 → h = 3.57 m.
Comment. dh/dt falls as 1/h², so the level rises rapidly when the tank is nearly empty and very slowly near the top. A constant inflow gives a constant rate of rise only in a tank of constant cross-section. Dividing 0.5 by the base area is doubly flawed here, because the cone has zero radius at its base - the method is meaningless at the vertex and simply wrong elsewhere. The rate of rise must always be evaluated from the instantaneous surface area.
3. Find the total time to fill the tank and compare the time taken for the first metre with the time taken for the last metre.
Read the worked answer
Total time. Since V = πh³/12 and volume is supplied at 0.5 m³/min:
t = V/0.5 = πh³/6.
Full tank (h = 6 m): t = π(216)/6 = 36π = 113.1 min.
The technician's estimate, using 0.159 m/min throughout, would be 6/0.159 = 37.7 min - less than a third of the true value. That rate applies only at h = 2 m.
First versus last metre. First metre (0 → 1 m): t = (π/6)(1³ − 0) = 0.52 min, about 31 s. Last metre (5 → 6 m): t = (π/6)(216 − 125) = (π/6)(91) = 47.6 min.
The last metre takes roughly 90 times as long as the first, which is exactly the kind of conclusion a constant-rate model conceals. A good answer reconciles this with the 1/h² dependence already found.