The Leaking Tank Accumulation
A water tank drains through a valve. The flow rate at time t (minutes) is measured to be r(t) = 6 − 0.5t liters per minute for 0 ≤ t ≤ 12, after which the valve is closed and flow is zero. A technician wants to know the total volume that leaves the tank during the first 12 minutes. She also has a second, less reliable sensor that reports the cumulative drained volume as V(t) = 6t − 0.25t² and asks whether the two measurements are consistent. Finally, she notices that r(t) becomes negative if the formula is naively extended past t = 12 and worries the model predicts water flowing back in.
1. Compute the total volume drained during 0 ≤ t ≤ 12 using the rate function, and verify consistency with the cumulative sensor V(t).
Read the worked answer
Total drained = ∫₀¹² r(t) dt = ∫₀¹² (6 − 0.5t) dt = [6t − 0.25t²]₀¹².
At t = 12: 6(12) − 0.25(144) = 72 − 36 = 36 liters.
Check the cumulative sensor: V(12) = 6(12) − 0.25(144) = 72 − 36 = 36 liters, and V(0) = 0. Moreover V′(t) = 6 − 0.5t = r(t), so the two are consistent - the cumulative sensor is exactly the antiderivative of the rate, confirming both readings. Total drained = 36 L.
2. Address the technician's worry about r(t) going negative past t = 12. What does the integral ∫₁₂¹⁵ r(t) dt compute, and why is the negative value not a physical "backflow"?
Read the worked answer
∫₁₂¹⁵ (6 − 0.5t) dt = [6t − 0.25t²]₁₂¹⁵ = (90 − 56.25) − (72 − 36) = 33.75 − 36 = −2.25.
The negative sign is an artifact of extrapolating the linear formula beyond its valid domain. Physically the valve is closed at t = 12, so the true rate is r(t) = 0 for t > 12, and the true additional volume drained is 0, not −2.25 L. The linear model simply isn't valid past 12 minutes; a negative rate there does not mean water re-enters. The correct piecewise rate is r(t) = 6 − 0.5t on [0,12] and 0 afterward. The lesson: a definite integral only reports physical volume when the integrand is the true rate over that interval.