Population Growth Models

A **population model** is a function that predicts how many individuals (people, bacteria, animals) there will be after a time $t$. When a population grows by the same **percentage** every year, it follows an exponential model: $$P(t) = P_0 (1 + r)^t$$ - $P_0$ = initial population (at $t = 0$) - $r$ = yearly growth rate as a decimal (3% means $r = 0.03$) - $1 + r$ = growth factor - $t$ = time in years When we know the **doubling time** $d$ (how long it takes the population to double), we can write the same kind of model as: $$P(t) = P_0 \cdot 2^{t/d}$$ The exponent $\frac{t}{d}$ counts how many doublings have happened. A quick estimate links the two forms, the **rule of 70**: $$d \approx \frac{70}{\text{growth rate in percent}}$$ It is only an approximation, but it is very good for small rates. The bar chart shows a population of $10000$ that grows $10\%$ per year: $10000$, $11000$, $12100$, $13310$, $14641$. Each bar is $1.1$ times the one before it, so the jumps get bigger every year.

A Growing City

A city has $50000$ inhabitants and grows by $2\%$ per year. Estimate its population after 5 years.

Identify $P_0$, $r$ and $t$: $P_0 = 50000$, $r = 0.02$, $t = 5$ = Values identified

Find the growth factor: $1 + r = 1 + 0.02 = 1.02$ = Growth factor $1.02$

Write the model: $P(t) = 50000 \cdot 1.02^t$ = Model ready

Evaluate at $t = 5$: $P(5) = 50000 \cdot 1.02^5 \approx 50000 \cdot 1.10408 \approx 55204$ = $P(5) \approx 55204$

Answer: After 5 years the city will have about **55204** inhabitants.

Bacteria With a Doubling Time

A sample contains $300$ bacteria. The number doubles every $30$ minutes. How many bacteria are there after 3 hours?

Write the doubling model: $P(t) = 300 \cdot 2^{t/30}$, with $t$ in minutes = Doubling model

Convert the time to minutes: $3$ hours $= 180$ minutes = $t = 180$

Count the doublings: $\frac{t}{d} = \frac{180}{30} = 6$ = 6 doublings

Calculate: $P(180) = 300 \cdot 2^6 = 300 \cdot 64 = 19200$ = $19200$

Answer: After 3 hours there are **19200** bacteria.

Estimating Doubling Time With the Rule of 70

A country of $8$ million people grows by $3.5\%$ per year. About how long until it has $16$ million people? And $32$ million?

Apply the rule of 70: $d \approx \frac{70}{3.5} = 20$ years = Doubling time about 20 years

One doubling: $8$ million $\to 16$ million after about $20$ years = About 20 years

Two doublings: $16$ million $\to 32$ million after about $20 + 20 = 40$ years = About 40 years

Check with the exact model: $1.035^{20} \approx 1.99$, very close to $2$ = The estimate is good

Answer: The population reaches about **16 million in 20 years** and about **32 million in 40 years**. The rule of 70 is an estimate, and here the exact doubling time is about 20.15 years, so the estimate is off by less than a quarter of a year.

Building a Model From Two Counts

A deer population was counted at $1200$ in 2020 and $1320$ in 2021. Assuming the same yearly percentage growth, predict the population in 2025.

Find the growth factor: $1 + r = \frac{P_1}{P_0} = \frac{1320}{1200} = 1.1$ = Growth factor $1.1$

Find the growth rate: $r = 1.1 - 1 = 0.1 = 10\%$ per year = $r = 10\%$

Write the model (2020 is $t = 0$): $P(t) = 1200 \cdot 1.1^t$ = Model ready

Predict 2025 ($t = 5$): $P(5) = 1200 \cdot 1.1^5 = 1200 \cdot 1.61051 \approx 1933$ = About $1933$ deer

Answer: The model $P(t) = 1200 \cdot 1.1^t$ predicts about **1933 deer** in 2025.

Linear vs Exponential Forecast

A town had $20000$ people in 2020 and $22000$ in 2022. Forecast the population in 2030 with a linear model and with an exponential model.

Linear model: constant increase per year: $\frac{22000 - 20000}{2} = 1000$ people per year = $L(t) = 20000 + 1000t$

Linear forecast for 2030 ($t = 10$): $L(10) = 20000 + 1000 \cdot 10 = 30000$ = $30000$

Exponential model: factor per 2 years: $\frac{22000}{20000} = 1.1$ every 2 years, so $P(t) = 20000 \cdot 1.1^{t/2}$ = $\times 1.1$ every 2 years

Exponential forecast for 2030: $P(10) = 20000 \cdot 1.1^5 = 20000 \cdot 1.61051 \approx 32210$ = $32210$

Compare: $32210 - 30000 = 2210$ = Exponential is $2210$ higher

Answer: Linear forecast: **30000**. Exponential forecast: about **32210**. Both models fit the two counts, yet they disagree by about 2210 people after only 10 years.

Population models help people plan ahead: - **Cities** use growth forecasts to decide how many schools, homes and buses they will need in 10 or 20 years - **Biologists and doctors** predict how fast bacteria multiply in a sample or in an infection - **Wildlife managers** track endangered species and invasive animals to see whether a population is recovering or exploding - **Governments** compare linear and exponential forecasts to plan water, energy and food supplies Knowing the difference between a straight line and a curve that keeps doubling can change a plan completely: an exponential forecast can be thousands of people higher after only a few decades. Real populations cannot grow exponentially forever, so a good forecaster also knows when the model stops working.

Mistake: Using the percent itself in the growth factor: $50000 \cdot (1 + 2)^t$

Why: A rate of $2\%$ is $0.02$ as a decimal. Writing $1 + 2 = 3$ would mean the city triples every year.

Correct: Convert the percent first: $2\% = 0.02$, so the growth factor is $1.02$ and $P(t) = 50000 \cdot 1.02^t$.

Mistake: Forgetting to divide by the doubling time: $300 \cdot 2^{180}$

Why: The exponent must count doublings, not minutes. In 180 minutes there are only $\frac{180}{30} = 6$ doublings.

Correct: Use $P(t) = P_0 \cdot 2^{t/d}$: $300 \cdot 2^{180/30} = 300 \cdot 2^6 = 19200$.

Mistake: Using the decimal rate in the rule of 70: $\frac{70}{0.035} = 2000$ years

Why: The rule of 70 is built for the rate written as a percent number. With the decimal you get an answer 100 times too large.

Correct: Divide 70 by the percent number: $\frac{70}{3.5} = 20$ years, and remember it is an estimate.

Mistake: Dividing by the later count when finding the rate: $\frac{120}{1320} \approx 9.1\%$

Why: The growth rate is measured relative to where the population started, not where it ended.

Correct: Divide the new count by the old count: $\frac{1320}{1200} = 1.1$, so $r = 10\%$.

City Planning

City planners forecast how many residents a city will have so they can build enough schools, apartments and public transport in time.

A city of $250000$ that grows $1\%$ per year would have about $250000 \cdot 1.01^{10} \approx 276156$ residents after 10 years.

Bacteria in the Lab

Microbiologists use doubling times to predict when a culture will be large enough to study, or how fast an infection can spread before treatment.

Some bacteria double about every 20 minutes in ideal conditions, which is 3 doublings, or a factor of $2^3 = 8$, every hour.

Wildlife Recovery

Conservation teams count animals every year, build a growth model from the counts and check whether a protected species is recovering. They also know that a reserve can only feed a limited number of animals, its carrying capacity, so growth will slow down later.

If a herd grows from $500$ to $540$ animals in a year, the growth factor is $\frac{540}{500} = 1.08$, so the herd grows $8\%$ per year while food is plentiful.

A population that grows by the same percentage each year follows $P(t) = P_0 (1 + r)^t$, with $r$ as a decimal

With a doubling time $d$, the model is $P(t) = P_0 \cdot 2^{t/d}$, where $\frac{t}{d}$ counts the doublings

The rule of 70 estimates the doubling time: $d \approx \frac{70}{\text{percent rate}}$; it is an approximation, not an exact formula

From two yearly counts, the growth factor is $\frac{P_1}{P_0}$ and the rate is $r = \frac{P_1}{P_0} - 1$

Linear forecasts add the same number each year; exponential forecasts multiply, so they pull ahead over time

Real populations slow down as they approach the carrying capacity of their environment, so exponential models work best for early growth

Q: Why does the rule of 70 work?

A: The exact doubling time is $d = \frac{\ln 2}{\ln(1 + r)}$, and for small rates $\ln(1 + r)$ is very close to $r$. Since $\ln 2 \approx 0.693$, the doubling time is about $\frac{0.693}{r} = \frac{69.3}{\text{percent}}$, rounded to 70. You will meet logarithms in a later lesson; for now, use the rule as a quick estimate.

Q: How accurate is the rule of 70?

A: For rates up to about $10\%$ it is usually within a year. At $2\%$ it gives $35$ years and the exact value is about $35.0$ years. At $7\%$ it gives $10$ years and the exact value is about $10.2$ years. For large rates the error grows.

Q: What is carrying capacity?

A: The carrying capacity is the largest population an environment can support with its food, water and space. As a population gets close to it, growth slows down and the curve flattens into an S shape. This is called logistic growth; in this lesson we only describe it, without a formula.

Q: Should I use a linear or an exponential model?

A: Look at the data. If the population increases by about the same **number** each year, use a linear model. If it increases by about the same **percentage** each year (the ratios $\frac{P_1}{P_0}$, $\frac{P_2}{P_1}$ are about equal), use an exponential model.

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Population Growth Models

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Pre-Calculus / Exponential Functions

Population Growth Models

Model growing populations of bacteria, cities and animals with exponential functions, doubling time and the rule of 70.

Advanced30 min

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