The Leaking Tank and the Ambiguous Average
A cylindrical water tank of radius 1.5 m is being drained. Engineers model the volume with V(t) = 18 - 3t + 0.25t^2 for 0 <= t <= 8, where V is in cubic metres and t is in hours. A supervisor claims the tank drains at a constant rate of 3 cubic metres per hour because the coefficient of t is 3. Meanwhile a technician measures the actual water level at t = 0 and t = 4 and computes an average rate of change over that interval. The tank's height is 2 m. A second technician argues that the average rate over the whole 8 hours is the meaningful figure for the report.
1. Evaluate the supervisor's claim that the tank drains at a constant rate of 3 cubic metres per hour. What is the actual instantaneous rate of change at t = 2, and what does the sign tell you?
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The supervisor confuses a coefficient with a rate. The rate of change is dV/dt = -3 + 0.5t, which depends on t, so the rate is not constant. At t = 2: dV/dt = -3 + 0.5(2) = -2 cubic metres per hour. The negative sign means volume is decreasing, i.e. water is leaving at 2 m^3/h at that instant. The constant -3 is only the initial rate at t = 0. This is a judgement call: reading a quadratic model requires differentiating, not quoting a coefficient.
2. Compute the average rate of change over [0,4] and over [0,8]. Which figure should the report use, and why might the two technicians disagree?
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Average rate over [0,4]: (V(4) - V(0))/4. V(0) = 18, V(4) = 18 - 12 + 4 = 10, so average = (10 - 18)/4 = -2 m^3/h. Over [0,8]: V(8) = 18 - 24 + 16 = 10, so average = (10 - 18)/8 = -1 m^3/h. Both are correct averages but answer different questions. The [0,4] figure describes the early draining; the [0,8] figure is diluted because the quadratic term slows net loss later. Neither is 'the' rate. The report should state the interval explicitly, since an average without an interval is meaningless. The disagreement is about scope, not arithmetic.
3. The tank has height 2 m. Determine whether the tank ever empties during 0 <= t <= 8, and interpret the result physically.
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Set V(t) = 0: 0.25t^2 - 3t + 18 = 0, i.e. t^2 - 12t + 72 = 0. Discriminant = 144 - 288 = -144 < 0, so no real roots; V never reaches zero. The minimum volume occurs where dV/dt = 0, t = 6, giving V(6) = 18 - 18 + 9 = 9 m^3. Physically the model predicts the tank stops draining (rate reaches zero at t = 6) and then volume increases, which is unphysical for a draining tank. A good answer flags that the model is only valid while dV/dt <= 0, i.e. 0 <= t <= 6, and that beyond t = 6 the model breaks down. Recognising domain-of-validity limits is the judgement being tested.