Bellwood Reservoir's Drawdown and Mussel Colony: Fitting Growth Models

Hook (5 min): Present two logged reservoir readings (Day 5: 62 feet; Day 20: 32 feet) and two logged mussel-colony readings (Week 1: 4 square feet; Week 4: 13.5 square feet) on the board, with no other information. Ask students whether these two numbers are enough to predict the water level on Day 12, or the colony's area at Week 3, and what they would need to figure out first.

Direct instruction (17 min): Model fitting the linear function first. From the points (5, 62) and (20, 32), calculate the slope: (32 − 62) / (20 − 5) = −30 / 15 = −2 feet per day. Use point-slope form with (5, 62) to find L(d) = 72 − 2d, and confirm the fit against the other point: L(20) = 72 − 40 = 32, which matches. Then model fitting the exponential function from (1, 4) and (4, 13.5). Explain that with a constant weekly growth factor r, the area three weeks later is the starting area times r cubed, so r³ = 13.5 / 4 = 3.375, and r is the cube root of 3.375, which is 1.5, a 50 percent weekly increase. Write A(w) = 4(1.5)^(w−1) and confirm: A(4) = 4(1.5)³ = 4(3.375) = 13.5, which matches. Pose the operational question: engineers must close the intake gate and begin manual removal once the colony's area reaches 50 square feet, the manufacturer's fouling limit for that screen material. Demonstrate solving 4(1.5)^(w−1) = 50 for w: divide both sides by 4 to get (1.5)^(w−1) = 12.5, then take the logarithm of both sides, (w − 1) × log(1.5) = log(12.5), so w − 1 = log(12.5) / log(1.5) ≈ 1.0969 / 0.1761 ≈ 6.23, giving w ≈ 7.23 weeks.

Guided practice (15 min): In pairs, students re-derive both functions step by step from the same two data points for each model, checking each computed value against a partner's. They then use the linear function to solve, on their own, for the day the reservoir reaches its 20-foot minimum inspection level: 72 − 2d = 20, so −2d = −52, so d = 26. Pairs also check a third, previously unused reading against each fitted model, a Day 8 sensor log of 56 feet and a Week 2 colony reading of 6 square feet, confirming both models predict these values exactly (L(8) = 72 − 16 = 56; A(2) = 4(1.5)¹ = 6), which shows the two-point fit was not a coincidence.

Independent practice (8 min): Each student solves two new threshold problems individually.

1. A second, older section of intake screen has a stricter fouling limit of 30 square feet. Using A(w) = 4(1.5)^(w−1), find the week, to the nearest hundredth, when the colony first reaches this limit.
2. Using L(d) = 72 − 2d, find the day the reservoir would reach 10 feet, the depth at which the pump intake itself would be exposed.

Closure / exit ticket (5 min): Students answer one question on an index card, continuing the numbered set:

3. A third mussel colony, on a different intake screen and tracked with the same weekly growth factor of 1.5, is modeled by B(w) = 6(1.5)^(w−1). Find the week, to the nearest hundredth, when this colony first reaches 50 square feet, and state in one sentence why this answer is exactly one week smaller than the 7.23 weeks found for the original colony even though both colonies share the same growth factor.