Answer key: Population Growth Activity
For teachers. This page is hidden from search engines and not linked from student pages, but it is not password-protected, so please don’t share the link with students. Student activity: Population Growth Activity.
Goal: Use compound, exponential and logistic growth models to predict population size and doubling time, solve percentage-growth word problems, and explain the effect of carrying capacity.
Completed results table
| What to find | Your result |
|---|---|
| Years for 20,000 to reach 210,000 at 12% per year | 20.75 years |
| The same problem if 0.12 is a continuous rate | 19.59 years |
| Population of 1,100 after 180 years, doubling every 60 years | 8,800 |
| Doubling time if r = 0.013 is a continuous rate | 53.32 years |
| Rule of 70 estimate for 1.3% per year | 53.85 years |
| Rabbits on an island: N after 15 years | 2,000 |
| Rabbits on an island: N after 5 years | 1,286 |
Answers
Write the equation for the town problem using N = N₀(1 + r)^t, then show how taking logarithms gives the time. Round to the nearest year.
210,000 = 20,000 × 1.12^t, so 1.12^t = 10.5. Taking natural logs: t = ln 10.5 ÷ ln 1.12 = 2.3514 ÷ 0.11333 = 20.75 years, about 21 years.
Why is the continuous-rate answer (N = N₀e^(rt)) shorter than the compound answer for the same 12%?
Continuous growth compounds every instant, so e^0.12 = 1.1275: an effective 12.75% a year instead of 12%. Faster effective growth reaches 210,000 sooner, in 19.59 years instead of 20.75.
A town of 1,100 doubles every 60 years. Without the calculator, explain why it has 8,800 people after 180 years.
180 ÷ 60 = 3 doublings, and each doubling multiplies the population by 2, so N = 1,100 × 2³ = 1,100 × 8 = 8,800.
The rule of 70 says doubling time ≈ 70 ÷ (percent growth rate). How close is it to the exact answer for a continuous rate r = 0.013, and why does it work?
70 ÷ 1.3 = 53.85 years, compared with ln 2 ÷ 0.013 = 53.32 years: about half a year apart. It works because ln 2 ≈ 0.693, which is close to 0.70.
Compare the rabbit population after 5 and after 15 years. Why does it grow so much in the first 5 years but barely change after 15? Use the term carrying capacity.
After 5 years N ≈ 1,286 (64.3% of K) and after 15 years N ≈ 2,000, essentially the carrying capacity K = 2,000. Early on resources are plentiful and growth is nearly exponential; as N approaches K, competition for food and space slows per-capita growth to almost zero.
Challenge: logistic growth is fastest at N = K/2. For the rabbits, about when is that, and what is the fastest growth rate rK/4?
t = ln((K − N₀) ÷ N₀) ÷ r = ln(1,976 ÷ 24) ÷ 1 = 4.41 years. The fastest growth is rK/4 = 1 × 2,000 ÷ 4 = 500 rabbits per year. The calculator shows both as Time to reach K/2 and Fastest growth.