a·bᵗ modelsA rumor spreads through a school of 1,200 students. On days 1–4, the counts are: 3, 6, 12, 24.
"It doubles daily — by day 11 that's 3,072 students." But the school only has 1,200 students. The exponential model is excellent early (constant ratio 2 — textbook exponential) and absurd later. Real modeling is exactly this two-step: pick the family the pattern demands, then say where the model must stop working. Both halves earn points; today drills both.
| Phrase | Model piece |
|---|---|
| "grows p% per unit time" | b = 1 + p/100 |
| "decays/loses p% per unit time" | b = 1 − p/100 |
| "doubles every T units" | b = 2^(1/T), i.e. a·2^(t/T) |
| "half-life T" | a·(1/2)^(t/T) |
| "tripled over k units" | b = 3^(1/k) |
All are the same skeleton: initial value × (factor)^(elapsed time / time per factor). Writing the exponent as t/T keeps the factor honest: at t = T the exponent is 1, so exactly one doubling (or halving) has occurred.
From two data points, divide to cancel a (Lesson 11, Example 4): the ratio over k steps is bᵏ; take the k-th root.
residual = actual value − predicted value
The residual plot is the verdict. If residuals show a clear curve or trend, the model family is wrong — even if each residual is smallish. Classic signature: fit a line to exponential data, and the residuals form a U (positive at both ends, negative in the middle): the data bend, the model doesn't.
[GRAPH: Two-panel figure. Left: scatterplot of gently curving data with a straight regression line cutting through; ends of data above the line, middle below. Right: the corresponding residual plot — a clear U-shaped pattern of points, labeled "pattern → linear model NOT appropriate."]
Given data and two candidate families:
The AP's grading language: "the residuals for model A are smaller in magnitude / show no pattern, so model A is more appropriate." Say it with the plot in evidence, not as an article of faith.
Every exponential model eventually collides with reality: populations hit capacity (the rumor can't pass 1,200), decaying quantities hit measurement floors, savings growth assumes rates never change. The standard limitation sentence: "The model assumes the growth rate stays constant; beyond [context boundary], that assumption fails because [reason]."
Problem: A 400 mg dose of a drug has a half-life of 8 hours. Write the model and find the amount after 24 hours.
Solution: M(t) = 400·(1/2)^(t/8). At t = 24: exponent 3 → 400·(1/8) = 50 mg.
Interpretation: 24 hours = 3 half-lives = three halvings. The t/T exponent counts halvings for you.
Problem: A city's population grows 6% per year. By what overall factor does it grow over a decade, and why isn't the answer 60%?
Solution: Factor per year: 1.06. Over 10 years: (1.06)¹⁰ ≈ 1.79 — about 79% total growth, not 60%. Growth compounds: each year's 6% applies to an already-grown population.
Interpretation: Percent rates multiply, never add, across time periods. "6% per year for 10 years = 60%" is the exam's favorite wrong answer.
Problem: A model predicts sales of S(5) = 78 units in week 5; actual sales were 82. Find and interpret the residual.
Solution: Residual = 82 − 78 = +4: the model underestimates week-5 sales by 4 units.
Interpretation: Order matters: actual − predicted. A student who computes predicted − actual gets the sign — and the interpretation — backwards.
Problem: Attendance data over 6 weeks: (1, 52), (2, 63), (3, 76), (4, 91), (5, 110), (6, 132). A student fits both a linear and an exponential model. Which is more appropriate?
Solution: Ratios: 63/52 ≈ 1.21, 76/63 ≈ 1.21, 91/76 ≈ 1.20, 110/91 ≈ 1.21, 132/110 = 1.20 — nearly constant ratios → exponential structure. (Differences: 11, 13, 15, 19, 22 — clearly not constant → not linear.) 📱 ExpReg: y ≈ 43.3·(1.204)ˣ; LinReg: y ≈ 15.9x + 31.7. The linear model's residuals form a U-pattern (overestimates the middle weeks, underestimates both ends); the exponential's are small and patternless. Conclusion: exponential, on two grounds — near-constant ratios over equal intervals, and a patternless residual plot.
Interpretation: Lead with the ratio evidence (it's checkable by hand), close with the residual plot. Two independent justifications make the FRQ point bulletproof.
1. (A). Three half-lives (24/8 = 3): 400 → 200 → 100 → 50 mg. (C) stops at two halvings; (D) does four.
2. (D). Doubling every 12 → 2^(t/12): at t = 12 the exponent is 1 ✓. (A) doubles 12 times per year; (B) is a half-life model.
3. (B). Compounding: (1.06)¹⁰ ≈ 1.79. (A) adds 6% ten times — the classic error.
4. (C). Actual − predicted = 82 − 78 = +4 (model underestimates). (A) is the flipped order.
5. (A). A patterned residual plot means the functional form misses the data's curvature — try another family. (B) misreads "balanced"; residual plots aren't about symmetry.
6. (D). Constant ratios = multiplicative growth = exponential. (A) is linear's signature; (B) quadratic's.
7. (B). Keep 100% − 3.5% = 96.5% → b = 0.965. (D)/(A) use the loss rate as the base.
8. (B). Solve 300(1.08)ᵗ = 600 → (1.08)ᵗ = 2 📱 (intersection or logs): t ≈ 9.0 years (1.08⁹ ≈ 1.999). (D) is the simple-interest answer (300·0.08·12.5 = 300); (A) is 50/8.
9. (C). Keep 96% per washing: a·(0.96)ⁿ. (B) keeps 4% instead; (A) is linear.
f(t) = a·bᵗ passes through (0, 12) and (3, 96). Then f(t) =10. (A). a = 12; b³ = 96/12 = 8 → b = 2: 12·2ᵗ. (B) forgets the cube root; (D) uses b² = 16.
P(t) = a·2^(t/7), the quantity11. (C). At t = 7 the exponent is 1: one doubling per 7 time units. (A) reads the fraction upside down.
12. (FRQ-style, 6 points) (i) [2 pts] a = M(0) = 20. b³ = 67.5/20 = 27/8 → b = 3/2. M(t) = 20·(3/2)ᵗ. (ii) [1 pt] b = 1.5 → 50% growth per day. (iii) [3 pts] The exponential model's residuals would be zero/patternless (it matches the true growth) [1 pt]. The linear model matches at the two fitted points but a straight line through two points of a concave-up exponential runs above the curve between them and below it outside them: residuals negative between t = 0 and 3, then increasingly positive beyond — a clear pattern → linear inappropriate [1 pt]. At t = 6: true mass 20(1.5)⁶ = 20·11.390625 ≈ 227.8 g vs. linear 20 + 15.83(6) ≈ 115 g — the exponential model remains accurate; the linear model underestimates by half [1 pt].
12. (FRQ-style) 🚫 Data: a colony's mass is 20 g at t = 0 and 67.5 g at t = 3 days, and mass grows exponentially: M(t) = a·bᵗ. (i) Find a and b exactly. (67.5/20 = 3.375 = 27/8) (ii) State the daily percent growth rate. (iii) A student proposes the linear model L(t) = 20 + 15.83t, which also passes through both data points. Describe what the residual plots would show for each model if the colony truly grows exponentially, and which model would remain accurate at t = 6.
A lab tracks a bacteria population, in thousands:
| hour t | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| N (thousands) | 150 | 180 | 216 | 259 | 311 |
(a) (i) Find the average rate of change of N over [1, 3]. (ii) Interpret it in context with units.
(b) (i) Demonstrate that an exponential model is appropriate, using ratios over equal-length intervals. (ii) Find an exponential regression model N(t) = a·bᵗ.
(c) (i) Use the model to predict the population at t = 7 hours. (ii) The lab dish supports at most 1.2 million bacteria. Explain the implication for using the model at t = 12.
(a) [2 pts] (i) [1 pt] (259 − 180)/(3 − 1) = 79/2 = 39.5 thousand bacteria per hour. (ii) [1 pt] Between hours 1 and 3, the population grew at an average rate of about 39,500 bacteria per hour.
(b) [2 pts] (i) [1 pt] Consecutive ratios: 180/150 = 1.20, 216/180 = 1.20, 259/216 ≈ 1.199, 311/259 ≈ 1.201 — approximately constant ratio ≈ 1.2 over equal 1-hour intervals, the defining property of exponential growth. (Differences 30, 36, 43, 52 rule out linear.) (ii) [1 pt] ExpReg: N(t) ≈ 150.021·(1.200)ᵗ (three decimals; data are within rounding of N = 150·1.2ᵗ; store full precision).
(c) [2 pts] (i) [1 pt] N(7) ≈ 150·(1.2)⁷ ≈ 150·3.583 ≈ 537.5 → about 537,000 bacteria (values from 535–540 thousand accepted per rounding). (ii) [1 pt] At t = 12 the model gives N(12) ≈ 150·(1.2)¹² ≈ 150·8.916 ≈ 1,337 thousand ≈ 1.34 million — above the dish's 1.2 million capacity, so the model's assumption of unconstrained growth fails before t = 12; the prediction is not reliable there (growth must slow as the population approaches capacity).
1. (A). Three half-lives (24/8 = 3): 400 → 200 → 100 → 50 mg. (C) stops at two halvings; (D) does four.
2. (D). Doubling every 12 → 2^(t/12): at t = 12 the exponent is 1 ✓. (A) doubles 12 times per year; (B) is a half-life model.
3. (B). Compounding: (1.06)¹⁰ ≈ 1.79. (A) adds 6% ten times — the classic error.
4. (C). Actual − predicted = 82 − 78 = +4 (model underestimates). (A) is the flipped order.
5. (A). A patterned residual plot means the functional form misses the data's curvature — try another family. (B) misreads "balanced"; residual plots aren't about symmetry.
6. (D). Constant ratios = multiplicative growth = exponential. (A) is linear's signature; (B) quadratic's.
7. (B). Keep 100% − 3.5% = 96.5% → b = 0.965. (D)/(A) use the loss rate as the base.
8. (B). Solve 300(1.08)ᵗ = 600 → (1.08)ᵗ = 2 📱 (intersection or logs): t ≈ 9.0 years (1.08⁹ ≈ 1.999). (D) is the simple-interest answer (300·0.08·12.5 = 300); (A) is 50/8.
9. (C). Keep 96% per washing: a·(0.96)ⁿ. (B) keeps 4% instead; (A) is linear.
10. (A). a = 12; b³ = 96/12 = 8 → b = 2: 12·2ᵗ. (B) forgets the cube root; (D) uses b² = 16.
11. (C). At t = 7 the exponent is 1: one doubling per 7 time units. (A) reads the fraction upside down.
12. (FRQ-style, 6 points) (i) [2 pts] a = M(0) = 20. b³ = 67.5/20 = 27/8 → b = 3/2. M(t) = 20·(3/2)ᵗ. (ii) [1 pt] b = 1.5 → 50% growth per day. (iii) [3 pts] The exponential model's residuals would be zero/patternless (it matches the true growth) [1 pt]. The linear model matches at the two fitted points but a straight line through two points of a concave-up exponential runs above the curve between them and below it outside them: residuals negative between t = 0 and 3, then increasingly positive beyond — a clear pattern → linear inappropriate [1 pt]. At t = 6: true mass 20(1.5)⁶ = 20·11.390625 ≈ 227.8 g vs. linear 20 + 15.83(6) ≈ 115 g — the exponential model remains accurate; the linear model underestimates by half [1 pt].
🎯 Exam tip: For any "which model is better?" part, write two sentences: one about structure (differences vs. ratios over equal intervals) and one about residuals (pattern or none). Either alone can earn the point; together they always do.