Related Rates: Filling a Conical Tank

Water poured into an inverted cone rises fast at first and slowly near the top, because the surface it has to fill keeps widening. The volume of water at depth h is V = ⅓πr²h, where r is the radius of the water's surface. Similar triangles tie r to h: r/h = R/H, the rim radius over the height. Substitute that before differentiating, so the volume depends on h alone, and dV/dt = π(Rh/H)²·dh/dt: the inflow equals the surface area times the rate the level rises. Worked example: a tank 4 m deep with a 2 m rim radius, filling at 2 m³/min. At h = 3 m the surface radius is 2 × 3/4 = 1.5 m and its area is π(1.5)² = 7.069 m², so dh/dt = 2 ÷ 7.069 = 0.283 m/min. At h = 1 m the surface is 0.785 m² and the level climbs at 2.546 m/min, nine times faster.

The classic mistake is differentiating V = ⅓πr²h with r held constant. r grows with h, and treating it as fixed gives dV/dt = ⅓πr²·dh/dt, which makes the level rise three times too fast. Eliminate r with similar triangles first, then differentiate. The underlying rule, that the level rises at the inflow divided by the area of the water's surface, holds for any container, which is why a conical hopper or clarifier cone fills quickly near its tip and slowly near the rim, and why a bottle seems to fill all at once when the water reaches its narrow neck.