Average Rate versus Instantaneous Rate in a Cooling Tank
A tank of water cools so that its temperature T (°C) after t minutes is modelled by T(t) = 20 + 60e^(−0.05t) for 0 ≤ t ≤ 90. A technician measures the temperature at t = 0 and t = 40, computes an average rate of change, and reports that the water is cooling at that constant rate. A supervisor disputes this, noting the tank cools fastest at the start. You must determine the average rate over [0, 40], the instantaneous rate at t = 40, and decide which figure should be quoted in a report warning about the first 10 minutes.
1. Compute the average rate of change of T over [0, 40] and the instantaneous rate at t = 40.
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T(0) = 20 + 60 = 80. T(40) = 20 + 60e^(−2) ≈ 20 + 60(0.13534) = 20 + 8.120 = 28.12 °C. Average rate = (T(40) − T(0))/40 = (28.12 − 80)/40 = −51.88/40 ≈ −1.30 °C per minute. Derivative: T′(t) = 60(−0.05)e^(−0.05t) = −3e^(−0.05t). At t = 40: T′(40) = −3e^(−2) ≈ −3(0.13534) ≈ −0.406 °C per minute. The average rate is far steeper in magnitude than the instantaneous rate at t = 40 because early cooling dominates the average.
2. Which rate should the report quote for the first 10 minutes, and justify with a calculation.
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For a warning about the first 10 minutes, quote the average rate over [0, 10], since that interval is what matters, not [0, 40]. T(10) = 20 + 60e^(−0.5) ≈ 20 + 60(0.60653) = 20 + 36.39 = 56.39 °C. Average rate over [0,10] = (56.39 − 80)/10 = −2.36 °C per minute. Instantaneous at t = 0 is T′(0) = −3 °C/min, the steepest. The [0,40] figure (−1.30) understates early cooling; the [0,10] figure (−2.36) is more representative but still milder than the initial −3 °C/min. The judgement is matching the interval of the rate to the interval of concern; a single constant rate cannot describe exponential cooling.