Exponential Population Growth
Predicting the Unpredictable
In 1798, Thomas Malthus warned that human population grows exponentially while food production grows linearly, a recipe for disaster. While his dire predictions haven’t fully materialized, the mathematics of population growth remains crucial for ecologists, epidemiologists, and urban planners.
The key insight: when resources are unlimited, populations grow at a rate proportional to their size. A population of 1000 bacteria produces more offspring than a population of 100, simply because there are more parents. This proportional relationship leads directly to exponential growth.
Before You Start
Quick Self-Check
What is the general solution to $\frac{dy}{dt} = ky$? → Exponential Growth ODE
Solve for $k$: $e^{4k} = 3$ → Natural Logarithm
If you struggled with either question, review the linked prerequisite first.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Concept | Differential Equations |
| Chapter | 6, Section 5 |
| Difficulty | Intermediate |
| Time | ~15 minutes |
Key Concepts
The Population Model
If $P(t)$ is population at time $t$, and growth is proportional to size:
$$\frac{dP}{dt} = kP$$
The solution is:
$$\boxed{P(t) = P_0 e^{kt}}$$
where $P_0 = P(0)$ is the initial population.
Relative Growth Rate
The relative growth rate is:
$$\frac{dP/dt}{P} = k$$
This is the growth rate per capita, the rate of change divided by current population.
| $k$ value | Meaning |
|---|---|
| $k = 0.02$ | 2% growth per time unit |
| $k = 0.05$ | 5% growth per time unit |
| $k = -0.01$ | 1% decline per time unit |
Finding $k$ from Two Data Points
Given $P(t_1) = P_1$ and $P(t_2) = P_2$:
$$P_1 = P_0 e^{kt_1} \quad \text{and} \quad P_2 = P_0 e^{kt_2}$$
Dividing: $$\frac{P_2}{P_1} = e^{k(t_2 - t_1)}$$
Solving for $k$: $$\boxed{k = \frac{\ln(P_2/P_1)}{t_2 - t_1}}$$
Doubling Time
The doubling time $T_d$ is how long it takes the population to double:
$$P_0 e^{kT_d} = 2P_0$$ $$e^{kT_d} = 2$$ $$\boxed{T_d = \frac{\ln 2}{k} \approx \frac{0.693}{k}}$$
The Rule of 70: If growth rate is $r\%$ per year, doubling time $\approx \frac{70}{r}$ years.
Limitations of the Model
Exponential growth assumes:
- Unlimited resources
- No predators or disease
- Constant birth/death rates
- No environmental constraints
Real populations eventually face limits, leading to logistic growth models.
Common Pitfalls
| Mistake | Why It’s Wrong | Correct Approach |
|---|---|---|
| Confusing “2% growth rate” with “2% more each year” | If $k = 0.02$, after 1 year you have $e^{0.02} \approx 1.0202$, slightly more than 2% | Use the exponential formula; don’t round |
| Using the wrong formula for $k$ | $k \neq \frac{P_2 - P_1}{t_2 - t_1}$ (that’s linear growth) | Use $k = \frac{\ln(P_2/P_1)}{t_2 - t_1}$ |
| Expecting exponential growth forever | Real populations hit carrying capacity | Exponential models are short-term approximations |
| Doubling time confusion | Doubling time depends only on $k$, not on $P_0$ | $T_d = \frac{\ln 2}{k}$ always |
| Mixing time units | If $k$ is per year, $t$ must be in years | Convert all times to consistent units |
Worked Example
Problem: A bacterial culture has 500 cells initially. After 3 hours, the population is 4000 cells. Assuming exponential growth:
(a) Find the relative growth rate $k$. (b) Write the population function $P(t)$. (c) Find the population after 5 hours. (d) When will the population reach 100,000?
Solution:
(a) Using the formula for $k$: $$k = \frac{\ln(P_2/P_1)}{t_2 - t_1} = \frac{\ln(4000/500)}{3 - 0} = \frac{\ln 8}{3} = \frac{3\ln 2}{3} = \ln 2 \approx 0.693$$
The relative growth rate is about 69.3% per hour.
(b) The population function: $$P(t) = 500e^{(\ln 2)t} = 500 \cdot 2^t$$
(c) After 5 hours: $$P(5) = 500 \cdot 2^5 = 500 \cdot 32 = 16,000 \text{ cells}$$
(d) Solve $P(t) = 100,000$: $$500 \cdot 2^t = 100000$$ $$2^t = 200$$ $$t = \log_2 200 = \frac{\ln 200}{\ln 2} \approx 7.64 \text{ hours}$$
Practice Problems
A population grows exponentially with $k = 0.04$ per year. If the initial population is 5000, what is the population after 10 years?
A city’s population was 120,000 in 2000 and 156,000 in 2020. Assuming exponential growth, find the relative growth rate as a percentage.
A deer population in a forest was 80 in 2015 and 125 in 2022.
- Find the relative growth rate.
- Predict the population in 2030.
- Find the doubling time.
A population triples every 8 years.
- Find the relative growth rate $k$.
- How long does it take for the population to increase by 50%?
- If the population is 10,000 now, what was it 5 years ago?
Country A has population 50 million with 1.5% annual growth. Country B has population 200 million with 0.5% annual growth.
- Write population functions for both countries.
- When will the populations be equal?
- Show that if two populations have exponential growth with rates $k_1$ and $k_2$ (where $k_1 > k_2$), they will eventually be equal regardless of initial populations (assuming $P_{01} < P_{02}$).
CCI-Style Conceptual Questions
Question 1: A population has a relative growth rate of 3% per year. Does this mean the population increases by exactly 3% of its initial value each year?
Answer
No. A 3% relative growth rate means the population grows continuously at 3% of its current value. After one year, the population is $P_0 e^{0.03} \approx 1.0305P_0$, about 3.05% more than the initial value. The difference grows larger over longer periods.
Question 2: If population A doubles every 10 years and population B doubles every 20 years, is A’s growth rate exactly twice B’s?
Answer
Yes! Doubling time $T_d = \frac{\ln 2}{k}$, so $k = \frac{\ln 2}{T_d}$.
- $k_A = \frac{\ln 2}{10}$
- $k_B = \frac{\ln 2}{20}$
Therefore $k_A = 2k_B$.
Question 3: A population grows from 1000 to 2000 in 5 years. Without any calculation, estimate how long it will take to grow from 2000 to 4000.
Answer
5 years. In exponential growth, doubling time is constant. If it took 5 years to double from 1000 to 2000, it will take another 5 years to double from 2000 to 4000.
Common Misconceptions
doubling time depends on the initial population size, so a larger population takes longer to double.
This is the rate-as-fixed-number error. For the exponential model $P(t) = P_0 e^{kt}$, the doubling time $t_d = \frac{\ln 2}{k}$ depends only on the growth rate $k$, not on $P_0$. A colony of 1000 bacteria with $k = \ln 2$ per hour doubles in exactly 1 hour; a colony of 1{,}000{,}000 bacteria with the same $k$ also doubles in exactly 1 hour. This is a hallmark of exponential growth: the proportional rate is constant regardless of scale.
Mastery Checklist
Mental Model
The “Copying Machine” Analogy:
Imagine every organism can make one copy of itself per time unit. With 100 organisms, you get 100 copies. With 1000 organisms, you get 1000 copies. The more you have, the more you produce, leading to explosive growth. The growth rate $k$ is like the “copying efficiency” of the population.
Still Confused?
| If you’re struggling with... | Review this |
|---|---|
| The general solution $y = Ce^{kt}$ | Exponential Growth ODE |
| Solving $e^{kt} = c$ for $t$ | Natural Logarithm |
| When exponential models break down | Logistic Growth |
Connections
Looking back:
- Exponential Growth ODE provides the mathematical foundation
- The derivative of $e^x$ is what makes this model work
Looking ahead:
- Logistic growth (Chapter 9) adds carrying capacity limits
- Predator-prey models consider interacting populations
- Epidemiological models (SIR) track disease spread
Real-world connections:
- World population projections
- Bacterial growth in biology labs
- Invasive species spread
- Viral content propagation online
| Previous | Up | Next |
|---|---|---|
| Exponential Growth ODE | Chapter 6 Skills | Half-Life & Decay |
Last updated: 2026-01-22