The Leaking Storage Tank
A cylindrical storage tank of radius 2 m and height 5 m is being drained. The volume of water remaining, in cubic metres, is modelled by V(t) = 20π − 0.4πt for 0 ≤ t ≤ 50 minutes, where t is time after draining begins. A technician claims the tank can be modelled instead by a rational function W(t) = 20π/(1 + 0.02t) because "drainage slows as pressure drops." The tank has a sight glass marked in 0.5 m increments. At t = 25 minutes the technician reads a depth of 2.5 m and declares the linear model wrong because "half the time should leave half the water, but the depth is exactly half the height, which is suspicious." A safety officer wants to know when the depth falls below 1 m and whether the rational model would give a materially different answer for planning purposes.
1. Convert the linear volume model to a depth model, and evaluate the technician's claim that a depth of 2.5 m at t = 25 min is evidence against it.
Read the worked answer
For a cylinder, V = πr²d with r = 2, so V = 4πd. Thus d(t) = V(t)/(4π) = (20π − 0.4πt)/(4π) = 5 − 0.1t metres.
At t = 25: d = 5 − 2.5 = 2.5 m. This is exactly what the linear model predicts, so the reading confirms rather than refutes it. The technician's reasoning conflates "half the time leaves half the water" with "half the time leaves half the depth." Because the tank is a cylinder, depth is proportional to volume, so halving volume does halve depth. The confusion arises from imagining a cone or sphere where depth and volume are not proportional. The judgement call is to check the geometry before rejecting a model: the linear-in-time volume model does imply linear-in-time depth for a uniform cross-section.
2. Using the linear model, find when the depth first falls below 1 m, and compare with the rational model's prediction.
Read the worked answer
Linear model: solve 5 − 0.1t = 1, giving t = 40 minutes. The depth is below 1 m for t > 40 min, within the stated domain (draining ends at t = 50 when d = 0).
Rational model: depth is W(t)/(4π) = 20π/(4π(1 + 0.02t)) = 5/(1 + 0.02t). Set 5/(1 + 0.02t) = 1, so 1 + 0.02t = 5, giving t = 200 minutes.
The rational model predicts the 1 m threshold is reached at 200 min, five times later, and never reaches zero depth in finite time (it asymptotes to 0). For planning, the difference is material: the linear model empties the tank in 50 min, the rational model predicts a long tail. Since the tank physically empties, the rational model's non-zero asymptote is physically implausible for a finite tank; the linear model is preferable for emptying-time planning, while the rational model might better describe early flow if pressure-driven. The judgement is to match model behaviour to physical constraints (finite emptying) rather than fit alone.