Accumulation, Average Value, and a Modeling Assumption
A factory's pollutant discharge rate is modeled by R(t) = 50 + 30 sin(t/4) gallons per hour, for 0 ≤ t ≤ 24 hours. The plant manager reports 'total discharge = rate × time = 50 × 24 = 1200 gallons' and claims the sinusoidal term 'averages out to zero anyway.' An environmental analyst objects that the average of sin(t/4) over a partial period is not zero, and that the correct total is an accumulation integral. The analyst also wants the average discharge rate over the 24 hours and the time(s) when the instantaneous rate equals that average, to interpret the report.
1. Compute the total discharge over 0 ≤ t ≤ 24 using an accumulation integral. Show the antiderivative and evaluate exactly.
Read the worked answer
Total discharge = ∫₀²⁴ (50 + 30 sin(t/4)) dt.
Antiderivative: 50t − 120 cos(t/4), since d/dt[−120 cos(t/4)] = −120·(−sin(t/4))·(1/4) = 30 sin(t/4).
Evaluate: At t = 24: 50(24) − 120 cos(6) = 1200 − 120 cos 6. At t = 0: 0 − 120 cos 0 = −120.
Total = (1200 − 120 cos 6) − (−120) = 1320 − 120 cos 6 gallons.
Numerically, cos 6 ≈ 0.9603, so total ≈ 1320 − 115.2 ≈ 1204.8 gallons. The manager's 1200 is close but not exact: the sinusoidal term does not integrate to zero over [0, 24] because that interval is not a whole number of periods (period = 8π ≈ 25.13 h).
2. Find the average discharge rate over the 24 hours, and determine the time(s) in (0, 24) when the instantaneous rate equals this average. Interpret the result.
Read the worked answer
Average value: (1/24)∫₀²⁴ R(t) dt = (1320 − 120 cos 6)/24 = 55 − 5 cos 6 ≈ 50.20 gal/h.
Set R(t) = average: 50 + 30 sin(t/4) = 55 − 5 cos 6 30 sin(t/4) = 5 − 5 cos 6 sin(t/4) = (1 − cos 6)/6.
Numerically, (1 − 0.9603)/6 ≈ 0.00662, so t/4 = arcsin(0.00662) ≈ 0.00662 (small angle). Then t ≈ 0.0265 h, and the second solution in the period is t/4 = π − 0.00662 → t ≈ 4(3.1350) ≈ 12.54 h.
So the instantaneous rate equals the 24-hour average at roughly t ≈ 0.03 h and t ≈ 12.5 h. Interpretation: the average rate (~50.2 gal/h) is only slightly above the baseline 50, and the rate matches the average near the start and near the midpoint - a consequence of the partial-period offset, not a coincidence. The 'averages to zero' claim is wrong because the interval is not a full period.