The Leaking Tank and the Difference Quotient
A cylindrical tank is draining through a valve at its base. A student measures the volume V(t) (in liters) at several times t (in minutes) and records: V(0)=500, V(2)=392, V(4)=300, V(6)=224, V(8)=164. The student claims the tank drains at a constant rate and wants to model the process with a single number. A classmate argues the rate is not constant and points out that the average rate of change over [0,8] and the average rate of change over [6,8] tell different stories. The teacher asks the pair to decide what the data actually supports before choosing any model, and warns them that a limit-based argument is needed to talk about an instantaneous rate rather than merely an average one.
1. Compute the average rate of change of V over [0,8] and over [6,8]. What do these two numbers show about the student's constant-rate claim?
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Average rate over [0,8]: (V(8)-V(0))/(8-0) = (164-500)/8 = -336/8 = -42 L/min.
Average rate over [6,8]: (V(8)-V(6))/(8-6) = (164-224)/2 = -60/2 = -30 L/min.
Both are negative (draining), but they differ: -42 vs -30. A constant-rate model would force a single slope through all points, and these secant slopes disagree, so the data does not support a constant rate. The rate is changing, and the magnitude is shrinking over time, consistent with a draining tank whose outflow slows as head decreases.
2. The classmate proposes defining the instantaneous rate at t=6 as a limit of average rates. Write that limit in difference-quotient form and explain, using the table, why the value at t=6 should be closer to -30 than to -42.
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Instantaneous rate at t=6: V'(6) = lim_{h->0} (V(6+h) - V(6))/h, equivalently lim_{t->6} (V(t)-V(6))/(t-6).
The secant over [6,8] uses the nearby point t=8 and gives -30, while the secant over [0,8] spans the whole interval and gives -42. As h shrinks the secant approaches the tangent slope at t=6, so nearby-point estimates are the relevant ones. The [6,8] secant is a one-sided estimate from the right, and it is the best available approximation to V'(6) from this table; the [0,8] value averages over a long stretch where the rate is different, so it is not a good local estimate. A defensible answer notes V'(6) is near -30, perhaps slightly more negative if the rate continues decreasing in magnitude toward t=6.
3. Is V continuous on [0,8]? Does continuity of V guarantee that V'(6) exists? Justify both answers.
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Continuity on [0,8] is plausible from the data: volumes change smoothly with no jump, and a physical draining tank has continuous volume. But the table alone cannot prove continuity; it is consistent with it.
Continuity does NOT guarantee differentiability. Continuity is necessary but not sufficient for a derivative to exist. Differentiability requires the limit of the difference quotient to exist, which fails at corners, cusps, vertical tangents, or discontinuities. Here the physical situation (smooth draining) suggests V'(6) exists, but the logical point is that continuity alone is not enough to conclude it.