Bean Plant Models
Two students grow bean plants for a science fair. Plant A is 4 cm tall on day 0 and grows at a steady 2 cm per day. Plant B is 2 cm tall on day 0; its height doubles every 5 days. The students record heights only through day 10, when Plant A is 24 cm and Plant B is 8 cm. Their teacher asks them to build models, compare the plants at day 12, and predict when Plant B will catch Plant A. She warns that any prediction beyond the 10-day data is an extrapolation and must be justified. The students also need to state a reasonable domain for each model in this experiment.
1. Write a function for each plant's height in terms of day d. Use the models to find each height on day 12 and state which plant is taller and by how much.
Read the worked answer
Plant A: A(d) = 4 + 2d. Plant B: B(d) = 2 · 2^(d/5).
At d = 12: A(12) = 4 + 24 = 28 cm. B(12) = 2 · 2^(12/5) = 2 · 2^2.4 ≈ 2 · 5.278 = 10.56 cm.
Plant A is taller by about 28 − 10.56 = 17.44 cm. The equations fit the recorded points: A(0) = 4, A(10) = 24; B(0) = 2, B(5) = 4, B(10) = 8.
2. Solve for when Plant B catches Plant A. Discuss whether the answer is reliable and state a reasonable domain for the models.
Read the worked answer
Set 4 + 2d = 2 · 2^(d/5). Testing integer days: d = 20: A = 44, B = 32; d = 23: A = 50, B ≈ 48.5; d = 24: A = 52, B ≈ 55.7. So B catches A between day 23 and day 24, about day 23.3 (A ≈ 50.6 cm).
Justification/caution: this is an extrapolation more than 13 days beyond the data. The exponential model assumes B keeps doubling every 5 days indefinitely, which may fail because of light, nutrients, or pot size. A reasonable experimental domain is d ≥ 0, but predictions should be treated as tentative beyond d = 10; the observed domain is 0 ≤ d ≤ 10.
3. A classmate says, “By day 10 Plant B must be growing faster because it is exponential.” Use average rates of change and the daily multiplier to evaluate this claim.
Read the worked answer
Average rate over 0–10 days: A = (24 − 4)/10 = 2 cm/day; B = (8 − 2)/10 = 0.6 cm/day. So over the observed period B is not growing faster on average.
B's 5-day multiplier is 2, so its daily multiplier is 2^(1/5) ≈ 1.1487, a 14.87% daily increase. On day 10 its one-day increase is about 8(0.1487) ≈ 1.19 cm/day, still below A's constant 2 cm/day. B is accelerating, but its current growth rate is lower. The classmate is confusing “increasing rate” with “greater rate.”