Why do epidemiologists speak of 'flattening the curve'? Why do banks advertise compound interest as 'exponential growth'? Why do archaeologists trust carbon-14 dating? All these questions revolve around a single differential equation: the rate of change is proportional to the current amount. This equation, $\frac{dN}{dt} = kN$, is the simplest and most powerful model in applied mathematics. Its solution, $N(t) = N_0 e^{kt}$, describes everything from bacterial colonies to Bitcoin prices, from radioactive decay to viral pandemics. Understanding this model is not just HSC preparation — it is scientific literacy.
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A bank account earns 5% interest compounded continuously. If $A(t)$ is the amount after $t$ years, the differential equation is $\frac{dA}{dt} = 0.05A$. If you start with \$1000, how much will you have after 10 years? Can you solve this without a calculator?
Core Content
The fundamental assumption: the rate of change is proportional to the current amount.
dN/dt = kN
Solving: Separate variables: $\frac{dN}{N} = k\,dt$
Integrate: $\ln|N| = kt + C$
Exponentiate: $N = e^{kt+C} = e^C \cdot e^{kt} = N_0 e^{kt}$
where $N_0 = N(0) = e^C$ is the initial amount.
Growth ($k > 0$):
Decay ($k < 0$):
Half-life: Time for $N$ to halve: $\frac{N_0}{2} = N_0 e^{kt_{1/2}} \Rightarrow t_{1/2} = \frac{\ln 2}{|k|}$
Doubling time: Time for $N$ to double: $2N_0 = N_0 e^{kt_{double}} \Rightarrow t_{double} = \frac{\ln 2}{k}$
Example 1 — Compound Interest: \$5000 is invested at 6% per annum compounded continuously. Find the value after 8 years.
dA/dt = 0.06A, A(0) = 5000
A(t) = 5000e^{0.06t}
A(8) = 5000e^{0.48} ≈ $8080.37
Example 2 — Radioactive Decay: A substance has half-life 12 years. Find the decay constant and the amount remaining from 200g after 30 years.
k = -ln(2)/12 ≈ -0.0578 per year
N(30) = 200e^{-0.0578 × 30} = 200e^{-1.733} ≈ 35.4g
Example 3 — Bacterial Growth: A colony doubles every 3 hours. Starting with 100 bacteria, how many after 10 hours?
k = ln(2)/3 ≈ 0.231 per hour
N(10) = 100e^{0.231 × 10} = 100e^{2.31} ≈ 1006 bacteria
The population of a town was 10,000 in 2010 and 12,000 in 2015. Assuming exponential growth, when will it reach 20,000?
The year when population reaches 20,000.
P(t) = 10000e^{kt} where t = years since 2010
P(5) = 12000: 10000e^{5k} = 12000
e^{5k} = 1.2
5k = ln(1.2)
k = ln(1.2)/5 ≈ 0.0365 per year
Now solve 10000e^{kt} = 20000
e^{kt} = 2
t = ln(2)/k = ln(2) × 5/ln(1.2) ≈ 19.0 years
Approximately year 2029.
A drug decays in the bloodstream with half-life 4 hours. If a patient is given 100mg, how much remains after 12 hours?
Answer:
$k = -\frac{\ln 2}{4} \approx -0.173$ per hour. $N(12) = 100e^{-0.173 \times 12} = 100e^{-2.08} \approx 12.5$ mg.
Alternatively: 12 hours = 3 half-lives, so $100 \times (\frac{1}{2})^3 = 12.5$ mg.
$\frac{dN}{dt} = kN$
$N(t) = N_0 e^{kt}$
$t_{1/2} = \frac{\ln 2}{|k|}$
$t_{double} = \frac{\ln 2}{k}$
For continuous compounding at 5%, $A(t) = 1000e^{0.05t}$. After 10 years: $A(10) = 1000e^{0.5} \approx 1000 \times 1.6487 = \$1648.72$. This is slightly more than annual compounding ($1000 \times 1.05^{10} \approx \$1628.89$) because interest earns interest more frequently. The limit as compounding frequency approaches infinity gives the continuous formula — and this limit is exactly what the differential equation $\frac{dA}{dt} = rA$ describes. The exponential function $e^{rt}$ is therefore the natural language of continuous growth.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. A radioactive isotope has half-life 8 days. Starting with 50g, how much remains after 20 days? Show all working. 3 MARKS
9. A city's population grows from 50,000 to 65,000 in 10 years. Assuming exponential growth: (a) Find the growth constant $k$. (b) Predict the population in 25 years. (c) When will the population reach 100,000? 4 MARKS
10. The simple exponential model $\frac{dP}{dt} = kP$ assumes unlimited resources. In reality, populations face carrying capacity. Explain how this limitation changes the model, what the modified differential equation looks like (logistic growth), and why the exponential model is still useful in the early stages of growth. 4 MARKS
1. $t_{1/2} = \frac{\ln 2}{0.05} \approx 13.9$ years.
2. $A(15) = 2000e^{0.045 \times 15} = 2000e^{0.675} \approx \$3927.80$.
3. $3 = e^{5k}$, so $k = \frac{\ln 3}{5}$. $10 = e^{kt}$, so $t = \frac{\ln 10}{k} = \frac{5\ln 10}{\ln 3} \approx 10.5$ hours.
1. Unlimited resources don't exist. The logistic model $\frac{dP}{dt} = kP(1 - \frac{P}{K})$ includes carrying capacity $K$.
2. Annual: $1000 \times 1.05^{10} = \$1628.89$. Continuous: $1000e^{0.5} = \$1648.72$. Difference: \$19.83.
3. All rely on $\frac{dN}{dt} = kN$ with $k < 0$. The half-life is an intrinsic property independent of initial amount.
Q8 (3 marks): $k = -\frac{\ln 2}{8} \approx -0.0866$ per day [1]. $N(t) = 50e^{kt}$ [0.5]. $N(20) = 50e^{-0.0866 \times 20} = 50e^{-1.733} \approx 8.84$ g [1.5].
Q9 (4 marks): (a) $65000 = 50000e^{10k}$ [0.5], $e^{10k} = 1.3$, $k = \frac{\ln 1.3}{10} \approx 0.0263$ per year [1]. (b) $P(25) = 50000e^{0.0263 \times 25} = 50000e^{0.657} \approx 92950$ [1.5]. (c) $100000 = 50000e^{kt}$, $e^{kt} = 2$, $t = \frac{\ln 2}{k} \approx 26.4$ years [1].
Q10 (4 marks): The simple model predicts unbounded growth, which is unrealistic as resources are finite [1]. The logistic equation is $\frac{dP}{dt} = kP(1 - \frac{P}{K})$ where $K$ is carrying capacity [1]. When $P \ll K$, the term $(1 - P/K) \approx 1$, so the logistic equation approximates exponential growth [1]. This is why exponential models work well in early pandemic stages or when a species first colonises a new habitat [1].
Climb platforms by solving exponential growth and decay problems, calculating half-lives and doubling times. Pool: lesson 13.
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