A single bacterium divides every 20 minutes. In 24 hours, unchecked, it would produce a colony with mass exceeding the Earth. This is the terrifying power of exponential growth. Conversely, radioactive carbon-14 decays so predictably that archaeologists use it to date artefacts 50,000 years old. Growth or decay, the mathematics is the same: a rate proportional to the current amount. This lesson brings together everything we have learned about exponentials, logarithms, and derivatives to solve real-world problems.
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A population doubles every 5 hours. Starting with 100 bacteria, how many will there be after 15 hours? Predict before calculating.
Core Content
Any quantity that changes at a rate proportional to its current value follows:
A = A_0 e^{kt}
where:
Why $e^{kt}$? If $\frac{dA}{dt} = kA$, then $\frac{dA}{A} = k \, dt$. Integrating (Y12): $\ln A = kt + C$, so $A = e^{kt+C} = e^C \cdot e^{kt} = A_0 e^{kt}$. At this level, we accept $e^{kt}$ as the solution to "rate proportional to amount."
The derivative confirms the model:
dA/dt = A_0 · k · e^{kt} = k · (A_0 e^{kt}) = kA
Half-life ($T_{1/2}$): the time for a decaying quantity to halve.
Set $A = \frac{A_0}{2}$:
A_0/2 = A_0 e^{kT} => 1/2 = e^{kT} => ln(1/2) = kT => -ln(2) = kT => T = ln(2)/|k|
Doubling time ($T_{double}$): the time for a growing quantity to double.
2A_0 = A_0 e^{kT} => 2 = e^{kT} => ln(2) = kT => T = ln(2)/k
Alternative form using half-life: $A = A_0 \cdot 2^{-t/T_{1/2}}$
A radioactive substance decays according to $A = A_0 e^{-0.05t}$ where $t$ is in years. Find the half-life.
The half-life in years.
T_{1/2} = ln(2)/|k| = ln(2)/0.05
ln(2) ≈ 0.693
T_{1/2} ≈ 0.693/0.05 = 13.86 years
Half-life $\approx 13.9$ years.
A bacterial population grows from 200 to 800 in 6 hours. Assuming exponential growth, find the growth constant $k$ and the doubling time.
Answer:
$800 = 200e^{6k} \Rightarrow 4 = e^{6k} \Rightarrow k = \frac{\ln 4}{6} = \frac{2\ln 2}{6} = \frac{\ln 2}{3} \approx 0.231$ per hour. Doubling time = $\frac{\ln 2}{k} = 3$ hours.
Given two data points $(t_1, A_1)$ and $(t_2, A_2)$:
A_2/A_1 = e^{k(t_2-t_1)} => k = ln(A_2/A_1)/(t_2 - t_1)
Example: A population is 500 at $t = 0$ and 1500 at $t = 4$.
k = ln(1500/500)/4 = ln(3)/4 ≈ 1.099/4 ≈ 0.275 per unit time
Once $k$ is known, the model $A = 500e^{0.275t}$ can predict future values.
$A = A_0 e^{kt}$
$T_{1/2} = \frac{\ln 2}{|k|}$
$T_{double} = \frac{\ln 2}{k}$
$k = \frac{\ln(A/A_0)}{t}$
After 15 hours (three 5-hour periods), the population doubles 3 times: $100 \times 2^3 = 800$ bacteria. Using the formula: $A = 100e^{kt}$ where $k = \frac{\ln 2}{5} \approx 0.139$. At $t = 15$: $A = 100e^{0.139 \times 15} = 100e^{2.079} \approx 100 \times 8 = 800$.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. A substance decays according to $A = 200e^{-0.04t}$ where $t$ is in years. Find the half-life to 1 decimal place. 3 MARKS
9. A bacterial population grows from 400 to 3200 in 9 hours. Assuming exponential growth, find the growth constant $k$ and the doubling time. 3 MARKS
10. An archaeological sample has 25% of the carbon-14 of a living organism. Carbon-14 has a half-life of 5730 years. Calculate the age of the sample. Explain why C-14 dating becomes unreliable for samples older than about 50,000 years, and describe what other dating methods might be used for older artefacts. 3 MARKS
1. $T_{1/2} = \frac{\ln 2}{0.03} \approx \frac{0.693}{0.03} \approx 23.1$ years.
2. $k = \frac{\ln 2}{8} \approx 0.0866$ per hour.
3. $k = -\frac{\ln 2}{5730} \approx -1.21 \times 10^{-4}$ per year.
4. $k = \frac{\ln(1200/300)}{5} = \frac{\ln 4}{5} \approx 0.277$ per hour. At $t = 10$: $A = 300e^{0.277 \times 10} = 300e^{2.77} \approx 300 \times 16 = 4800$.
1. $\frac{dA}{dt} = A_0 \cdot k \cdot e^{kt} = k \cdot (A_0 e^{kt}) = kA$. The growth rate is proportional to the current population.
2. $12.5\% = \frac{1}{8} = (\frac{1}{2})^3$, so 3 half-lives = $3 \times 5730 = 17{,}190$ years.
3. Real populations face resource limits, predators, and disease. The logistic model $A = \frac{L}{1 + Ce^{-kt}}$ accounts for carrying capacity $L$.
Q8 (3 marks): $T_{1/2} = \frac{\ln 2}{0.04}$ [1] $= \frac{0.693}{0.04}$ [0.5] $\approx 17.3$ years [1.5].
Q9 (3 marks): $3200 = 400e^{9k}$ [0.5]. $8 = e^{9k}$ [0.5]. $k = \frac{\ln 8}{9} = \frac{3\ln 2}{9} = \frac{\ln 2}{3} \approx 0.231$ per hour [1]. Doubling time = $\frac{\ln 2}{k} = \frac{\ln 2}{\ln 2 / 3} = 3$ hours [1].
Q10 (3 marks): $25\% = \frac{1}{4} = (\frac{1}{2})^2$ [0.5], so 2 half-lives have passed [0.5]. Age = $2 \times 5730 = 11{,}460$ years [0.5]. C-14 dating becomes unreliable beyond ~50,000 years because the remaining C-14 is too little to measure accurately — after about 9 half-lives ($9 \times 5730 \approx 51{,}570$ years), less than 0.2% remains, comparable to measurement error [0.5]. Background radiation and contamination make it impossible to distinguish the tiny remaining signal from noise [0.5]. For older artefacts, methods like potassium-argon dating (half-life 1.25 billion years), uranium-lead dating, or thermoluminescence are used [0.5].
Climb platforms using half-life, doubling time, and exponential models. Pool: lesson 13.
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