The Misleading Average of a Piecewise Rate
A tank holds 240 L of water and drains through a valve whose flow rate is modelled by r(t) = 12 − 3t L/min for 0 ≤ t ≤ 4, but the valve is physically unable to run in reverse, so for t > 4 the effective outflow is capped at 0 L/min. A student computes total outflow as the integral of r(t) from 0 to 8, obtaining 96 − 96 = 0 L, and concludes the tank is unchanged after 8 minutes. The student's calculator shows the signed area cancelling. The tank actually begins with 240 L and the valve was open the whole time.
1. Identify the modelling error and compute the correct volume remaining after 8 minutes.
Read the worked answer
The formula r(t) = 12 − 3t is only valid while the valve discharges, i.e. while r(t) ≥ 0. Setting 12 − 3t = 0 gives t = 4. For t > 4 the physical outflow rate is 0, not a negative value.
Correct outflow = ∫₀⁴ (12 − 3t) dt = [12t − 1.5t²]₀⁴ = 48 − 24 = 24 L.
Volume remaining = 240 − 24 = 216 L.
The signed integral to 8 minutes is meaningless because the integrand was extended past its domain of validity. The judgement call is recognising that a piecewise physical constraint overrides a single algebraic formula.
2. The student now claims that because the average of r over [0,8] is zero, the tank level is stationary. Explain why an average rate of zero does not imply zero net change here, and state what the correct average outflow rate over 8 minutes is.
Read the worked answer
An average of a signed quantity can be zero while the magnitude of flow is positive throughout. Here the negative values for t > 4 are artefacts; they do not represent inflow.
The correct average outflow rate is total outflow divided by elapsed time: 24 L ÷ 8 min = 3 L/min.
So the tank loses water at an average of 3 L/min even though the naive signed average is 0. The key reasoning is that averaging must be done over the physically valid rate function, not the algebraic expression.