Population Growth Models

Show your work for each problem.

1.
A town of $2000$ people grows by $4\%$ per year. Which function models its population after $t$ years?
1. a) $P(t) = 2000 \cdot 0.04^t$
2. b) $P(t) = 2000 \cdot 1.04^t$
3. c) $P(t) = 2000 + 1.04t$
4. d) $P(t) = 2000 \cdot 4^t$
Answer: $P(t) = 2000 \cdot 1.04^t$
$4\% = 0.04$ , so the growth factor is $1.04$ and $P(t) = 2000 \cdot 1.04^t$ . The base $0.04$ would shrink the town, $4$ would quadruple it every year, and $2000 + 1.04t$ is linear.
2.
A population of $6000$ doubles every $15$ years. Which model gives the population after $t$ years?
1. a) $P(t) = 6000 \cdot 2^{15t}$
2. b) $P(t) = 6000 \cdot 2^t$
3. c) $P(t) = 6000 \cdot 2^{t/15}$
4. d) $P(t) = 6000 \cdot 15^{t/2}$
Answer: $P(t) = 6000 \cdot 2^{t/15}$
The doubling model is $P(t) = P_0 \cdot 2^{t/d}$ with $d = 15$ , so $P(t) = 6000 \cdot 2^{t/15}$ . After $15$ years the exponent is $1$ and the population is $12000$ .
3.
A country's population grows by $2\%$ per year. Using the rule of 70, about how long does it take to double?
1. a)14 years
2. b)35 years
3. c)140 years
4. d)3500 years
Answer: 35 years
$d \approx \frac{70}{2} = 35$ years. Dividing by $0.02$ would give $3500$ , which is far too large. The exact doubling time is also about $35$ years.
4.
A rabbit population on an island grew exponentially for a few years, but now its growth is slowing down. What is the most likely reason?
1. a)The growth factor of an exponential model drops below 1 after a few years
2. b)Populations always switch to linear growth after exactly 5 years
3. c)Food and space are limited, so the population is approaching the carrying capacity
4. d)Exponential functions always reach a maximum value
Answer: Food and space are limited, so the population is approaching the carrying capacity
Food and space on the island are limited. As the population approaches the carrying capacity, growth slows and the curve levels off (logistic growth). Exponential functions with $b > 1$ have no maximum, so the slowdown comes from the environment, not from the math.
5.
Town A and town B both have $10000$ inhabitants today. Town A grows by $500$ people per year; town B grows by $5\%$ per year. Which statement about the populations in 20 years is true? (Use $1.05^{20} \approx 2.6533$ .)
1. a)Town B is larger by about 6533
2. b)Both towns have 20000 inhabitants
3. c)Town B is larger by exactly 500
4. d)Town A is larger by about 6533
Answer: Town B is larger by about 6533
Town A: $10000 + 500 \cdot 20 = 20000$ . Town B: $10000 \cdot 1.05^{20} \approx 26533$ . Both start by adding $500$ in the first year, but the exponential town pulls ahead by about $26533 - 20000 = 6533$ people.
6.
A colony of $1500$ rabbits grows by $10\%$ per year. How many rabbits are there after 1 year?
Answer: 1650
$P(1) = 1500 \cdot 1.1 = 1650$ rabbits.
7.
A city of $4000$ people doubles every $8$ years. What is its population after $24$ years?
Answer: 32000
$\frac{24}{8} = 3$ doublings, so $P(24) = 4000 \cdot 2^3 = 4000 \cdot 8 = 32000$ .
8.
A fish population grows by $7\%$ per year. Use the rule of 70 to estimate its doubling time in years.
Answer: 10
$d \approx \frac{70}{7} = 10$ years. This is an estimate: the exact doubling time is about $10.2$ years.
9.
A village had $2500$ inhabitants in 2023 and $2600$ in 2024. What is its yearly growth rate in percent? (Write only the number, for example 5.)
Answer: 4
$\frac{2600}{2500} = 1.04$ , so $r = 1.04 - 1 = 0.04 = 4\%$ per year.
10.
A culture of $200$ bacteria doubles every $25$ minutes. After how many minutes will there be $12800$ bacteria?
Answer: 150
$\frac{12800}{200} = 64 = 2^6$ , so 6 doublings are needed: $6 \cdot 25 = 150$ minutes.
11.
A town of $30000$ people grows by $2\%$ per year. What is its population after 2 years?
Answer: 31212
1. What is the growth factor $1 + r$ ? 1.02
2. What is $1.02^2$ ? 1.0404
3. Multiply by $30000$ . What is the population after 2 years? 31212
12.
A sample of $50$ bacteria doubles every $10$ minutes. How many bacteria are there after 1 hour?
Answer: 3200
1. How many doublings happen in 60 minutes? 6
2. What is $2^6$ ? 64
3. How many bacteria are there after 1 hour? 3200
13.
A lake had $800$ fish last year and has $880$ fish this year. Assuming the same yearly percentage growth, how many fish will there be 2 years after the count of 800?
Answer: 968
1. What is the growth factor $\frac{880}{800}$ ? 1.1
2. What is the yearly growth rate in percent? (Write only the number.) 10
3. Compute $P(2) = 800 \cdot 1.1^2$ . 968
14.
A country of $5$ million people grows by $1.4\%$ per year. Use the rule of 70 to estimate its population in 100 years (in millions).
Answer: 20
1. Estimate the doubling time in years with the rule of 70. 50
2. How many doublings happen in 100 years? 2
3. Estimate the population in 100 years, in millions. 20
15.
A deer population of $1000$ doubles every $10$ years. Estimate the population after $5$ years. (Round $2^{0.5}$ to 3 decimal places.)
Answer: 1414
1. What is the exponent $\frac{t}{d}$ ? 0.5
2. What is $2^{0.5}$ , rounded to 3 decimal places? 1.414
3. Estimate the population after 5 years. 1414
16.
A city had $40000$ inhabitants in 2020 and $44000$ in 2021. Forecast 2023 with a linear model and with an exponential model. How much larger is the exponential forecast?
Answer: 1240
1. Linear model: how many inhabitants are added per year? 4000
2. What is the linear forecast for 2023? 52000
3. Exponential model: what is the growth factor? 1.1
4. What is the exponential forecast for 2023? 53240
5. How much larger is the exponential forecast? 1240
17.
A city had $5000$ inhabitants in 2018 and $7200$ in 2020. Assuming the same yearly percentage growth, predict the population in 2022.
Answer: 10368
1. What is the growth factor over 2 years, $\frac{7200}{5000}$ ? 1.44
2. What is the yearly growth factor $b$ ? 1.2
3. What is the yearly growth rate in percent? (Write only the number.) 20
4. Predict the population in 2022. 10368
18.
A population of $10000$ grows by $5\%$ per year. Estimate the doubling time with the rule of 70, then check the estimate: how large is the population after that many years?
Answer: 19800
1. Estimate the doubling time in years with the rule of 70. 14
2. Compute $1.05^{14}$ with a calculator, rounded to 2 decimal places. 1.98
3. Using $1.98$ , what is the population after 14 years? 19800