Zero Average Rate of Change and a Model Past Its Domain
Mr. Duarte's precalculus class models a ball tossed from a balcony with h(t) = −16t² + 48t + 5, where h is height in feet and t is seconds after release. Ana computes the average rate of change on [0, 3] as (h(3) − h(0))/3 = (5 − 5)/3 = 0 ft/s and concludes, 'The ball didn't move between 0 and 3 seconds, so the model must be wrong.' Ben evaluates h(4) = −16(16) + 192 + 5 = −59 feet and says, 'At 4 seconds the ball is 59 feet below the balcony.' Mr. Duarte must address both claims while keeping attention on what an average rate of change does and does not tell you; the upcoming quiz asks students to interpret an average rate of change over an interval in context and to connect the vertex to the physical maximum. He also knows the model stops being valid at some point and wants students to state the domain themselves rather than be told.
1. Check Ana's computation, explain what the zero actually means, and give the ball's maximum height and when it occurs.
Read the worked answer
The computation is correct. h(0) = 5 and h(3) = −16(9) + 144 + 5 = 5, so the average rate of change on [0, 3] is 0/3 = 0 ft/s.
The interpretation is wrong. Average rate of change is net change divided by elapsed time; it summarizes only the endpoints. A value of 0 means the ball was at the same height at t = 0 and t = 3 - it returned to 5 ft - not that it was motionless. The ball rose and then fell, and averaging cancels the rise against the fall. A useful probe: the instantaneous velocity is +48 ft/s at t = 0 and −48 ft/s at t = 3, both nonzero, yet the interval average is 0.
Maximum. The vertex is at t = −b/(2a) = −48/(2·(−16)) = 1.5 s, and h(1.5) = −16(2.25) + 72 + 5 = −36 + 72 + 5 = 41 ft. So the ball rose 36 ft above the balcony and came back down, all hidden by the single number 0.
This is also a chance to name the symmetry: h is a parabola symmetric about t = 1.5, so h(0) = h(3). A zero average rate of change on an interval centered at the vertex is expected, not suspicious.
2. Evaluate Ben's claim, determine when the ball reaches the ground, and state the domain on which the model is valid.
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h(4) = −59 is arithmetically correct but physically meaningless. The model stops describing the ball once it lands. The balcony is 5 ft above the ground, so landing means h = 0:
−16t² + 48t + 5 = 0 → 16t² − 48t − 5 = 0
t = [48 ± √(2304 + 320)]/32 = [48 ± √2624]/32.
√2624 ≈ 51.22, so t ≈ 99.22/32 ≈ 3.10 s (the root ≈ −0.10 s is rejected, since t ≥ 0).
At t = 4 s the ball has already been on the ground for about 0.9 s; the value −59 ft is an extrapolation about 59 ft below ground level and has no referent. Ben's error is treating a function rule as valid for every real input instead of checking whether the input lies in the domain of the situation.
Domain: 0 ≤ t ≤ 3.10 s (exactly 0 ≤ t ≤ [48 + √2624]/32). Restricting the domain is a modeling decision, not an algebraic one: the quadratic continues forever in both directions, but the ball does not. A good follow-up is to ask what −16t² + 48t + 5 would describe if the ball were thrown from the top of a 59-ft cliff instead, so students see the formula is not 'wrong' - it is simply being applied outside its situation.
3. Distinguish average and instantaneous rate of change using [1, 2] and t = 1.5, and write one quiz item that tests interpretation rather than computation.
Read the worked answer
On [1, 2]: h(1) = −16 + 48 + 5 = 37 and h(2) = −64 + 96 + 5 = 37, so the average rate of change is (37 − 37)/1 = 0 ft/s. The ball is moving the whole time - rising at t = 1, falling at t = 2 - but its net displacement is zero because the interval is symmetric about the vertex.
At t = 1.5: h′(t) = −32t + 48, so h′(1.5) = 0. Here the instantaneous rate is genuinely zero: the ball is momentarily at rest at its peak. The two zeros mean different things. Compare with h′(1) = 16 ft/s and h′(2) = −16 ft/s.
The interval average of 0 is consistent with the instantaneous values because velocity is linear in t here: the average of 16 and −16 is 0. That is the Mean Value Theorem visible in a concrete model - there is a time in (1, 2), namely t = 1.5, where the instantaneous rate equals the average rate.
Quiz item. 'A ball's height is h(t) = −16t² + 48t + 5 feet after t seconds. (a) Find the average rate of change on [1, 2] and explain in one sentence what it means physically. (b) Find all times in (1, 2) when the instantaneous rate of change equals that average, and explain why your answer is consistent with the graph. (c) Explain how the average rate on [0, 3] could be 0 while the ball is never at rest for an interval.' Part (c) is the interpretation test; a student who only knows the quotient formula cannot answer it.