Exponential Growth and Decay
Carbon dating, population booms, compound interest, radioactive decay — all share the same mathematical skeleton: $N = N_0 e^{kt}$. In this lesson you'll derive this model from the differential equation $\dfrac{dN}{dt} = kN$, master the key parameters $k$, $T_{1/2}$, and $T_2$, and solve applied problems the HSC loves to set.
A bacterial colony doubles every 3 hours. It starts with 100 bacteria. Without using a formula — estimate how many bacteria there will be after 12 hours. Write your reasoning below.
Any quantity whose rate of change is proportional to its current value satisfies the differential equation:
$$\frac{dN}{dt} = kN$$
Separating and integrating (from Lesson 13): $\dfrac{dN}{N} = k\,dt \Rightarrow \ln|N| = kt + C$
Exponentiating: $N = N_0 e^{kt}$ where $N_0 = N(0)$ is the initial value.
Key facts
- $\dfrac{dN}{dt} = kN \Rightarrow N = N_0 e^{kt}$
- Half-life: $T_{1/2} = \dfrac{\ln 2}{|k|}$ (when $k < 0$)
- Doubling time: $T_2 = \dfrac{\ln 2}{k}$ (when $k > 0$)
Concepts
- Why proportional growth/decay always produces an exponential function
- The geometric significance of $k$ as the continuous growth rate
- How two data points uniquely determine both $N_0$ and $k$
Skills
- Derive $N = N_0 e^{kt}$ from $\dfrac{dN}{dt} = kN$ using separation of variables
- Find $k$ and $N_0$ from given data, including non-zero start times
- Calculate half-life, doubling time, and unknown times/quantities
The derivation uses separation of variables exactly as in Lesson 13. Starting from $\dfrac{dN}{dt} = kN$:
- Separate: $\dfrac{dN}{N} = k\,dt$
- Integrate: $\ln|N| = kt + C$
- Exponentiate: $N = Ae^{kt}$ where $A = \pm e^C$
- Apply $N(0) = N_0$: $A = N_0$ (and $N_0 > 0$ for physical quantities, so drop $|\,|$)
Finding $k$ from two data points: If you know $N(t_1) = N_1$ and $N(t_2) = N_2$, divide the equations to eliminate $N_0$:
The derivation uses separation of variables exactly as in Lesson 13. Starting from $\dfrac{dN}{dt} = kN$:
Pause — copy the full derivation of $N=N_0 e^{kt}$ from $\frac{dN}{dt}=kN$ using separation of variables into your book.
Quick check: A radioactive substance has decay constant $k = -0.04$ per year. What is its half-life?
We just saw the derivation: $\frac{dN}{dt}=kN\Rightarrow\int\frac{dN}{N}=\int k\,dt\Rightarrow\ln|N|=kt+C\Rightarrow N=N_0 e^{kt}$. That raises a question: given only the formula $N=N_0 e^{kt}$, how does the sign and size of $k$ immediately tell you whether the quantity grows, decays, or is constant? This card answers it → $k>0$: exponential growth; $k<0$: exponential decay (write $k=-\lambda$); $k=0$: $N=N_0$ constant.
The sign and magnitude of $k$ tell you everything about how the quantity changes over time:
- Large positive $k$: rapid growth (steep curve). Example: unconstrained bacteria — $k \approx 1.4\,\text{hr}^{-1}$ for E. coli at 37°C.
- Small positive $k$: slow growth (gentle curve). Example: national population — $k \approx 0.01\,\text{yr}^{-1}$.
- Small negative $k$: slow decay. Example: carbon-14 — $k = -\dfrac{\ln 2}{5730} \approx -0.000121\,\text{yr}^{-1}$.
- Large negative $k$: fast decay. Example: a short-lived radioisotope.
Note: in many problems the decay constant is written as a positive number $\lambda$ with the equation $\dfrac{dN}{dt} = -\lambda N$, giving $N = N_0 e^{-\lambda t}$. This is the same model — just a sign convention.
The sign and magnitude of $k$ tell you everything about how the quantity changes over time:
Pause — copy the $k$-reading rule: $k>0\Rightarrow$ growth; $k<0\Rightarrow$ decay with half-life $t_{1/2}=\frac{\ln 2}{|k|}$; $k=0\Rightarrow$ constant into your book.
Did you get this? True or false: if a quantity satisfies $\dfrac{dN}{dt} = kN$ and the quantity is decreasing, then $k$ must be negative.
Worked examples · 3 in a row, reveal as you go
A town's population grows at 2% per year continuously. The population in 2020 is 50 000. Find the population in 2035 and the year it reaches 100 000.
A bacterial culture contains 500 bacteria at $t = 0$ and 2000 bacteria at $t = 4$ hours. Find $k$ and the number of bacteria at $t = 6$ hours.
Carbon-14 decays with half-life $T_{1/2} = 5730$ years. A fossil contains 25% of its original carbon-14. How old is the fossil?
Fill the gap: If a radioactive substance has half-life 20 years, then after 60 years, $\dfrac{1}{8}$ of the original amount remains because half-lives have elapsed.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the half-life $T_{1/2}$ of a radioactive substance depends on the initial amount $N_0$.
Activities · practice with the ideas
A colony of 200 bacteria doubles every 5 hours. Write the model $N(t) = N_0 e^{kt}$ with an exact value for $k$.
A sample of iodine-131 has half-life 8.02 days. What fraction remains after 24 days? Give an exact answer.
An investment of $1000 grows to $1500 in 10 years (continuous compounding). Find $k$ exactly and determine when it reaches $3000.
A radioactive substance has $N(0) = 600$ and $N(3) = 300$. Find $k$, write the model, and find when $N = 75$.
Show that the doubling time of $N = N_0 e^{kt}$ (with $k>0$) is $T_2 = \dfrac{\ln 2}{k}$ and is independent of $N_0$.
Odd one out: Three of these statements about $N = N_0 e^{kt}$ are correct. Which one is NOT?
Earlier you estimated the number of bacteria after 12 hours for a colony that doubles every 3 hours, starting at 100.
Four doublings occur in 12 hours (12 ÷ 3 = 4), so $N = 100 \times 2^4 = 1600$ bacteria. In the exponential model: $k = \dfrac{\ln 2}{3}$, and $N(12) = 100\,e^{(\ln 2/3) \times 12} = 100 \cdot 2^4 = 1600$. Did your estimate capture the idea that each doubling multiplies by 2?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A radioactive element decays so that $\dfrac{dM}{dt} = -0.1M$. If $M(0) = 400$ g, find $M(t)$ and calculate the half-life. (2 marks)
Q2. A population is 2000 at $t=0$ and 5000 at $t=10$ years. Find $k$ (exact) and determine when the population reaches 20 000. (3 marks)
Q3. An archaeological site yields wood with 12.5% of its original carbon-14 content. Given $T_{1/2} = 5730$ years for carbon-14, find the exact age of the wood and explain why the result is independent of the original amount of carbon-14 present. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $T_2 = 5 \Rightarrow k = \dfrac{\ln 2}{5}$; $N = 200\,e^{(\ln 2/5)t}$.
2. $N/N_0 = e^{kt}$; $k = -\dfrac{\ln 2}{8.02}$; after 24 days: $e^{-3\ln 2} = (1/2)^3 = 1/8$. Exact fraction: $\dfrac{1}{8}$.
3. $1500 = 1000\,e^{10k} \Rightarrow k = \dfrac{\ln(3/2)}{10}$; reaches 3000 when $3 = e^{kt} \Rightarrow t = \dfrac{\ln 3}{k} = \dfrac{10\ln 3}{\ln(3/2)}$.
4. $k = \dfrac{\ln(300/600)}{3} = \dfrac{-\ln 2}{3}$; $N = 600\,e^{-(\ln 2/3)t}$; $N=75$ when $e^{-(\ln 2/3)t} = \dfrac{1}{8} = 2^{-3}$, so $t = 9$ hours.
5. $2N_0 = N_0\,e^{kT_2} \Rightarrow 2 = e^{kT_2} \Rightarrow kT_2 = \ln 2 \Rightarrow T_2 = \dfrac{\ln 2}{k}$. $N_0$ cancels in step 1, confirming independence.
Q1 (2 marks): $M = 400\,e^{-0.1t}$ [1]. $T_{1/2} = \dfrac{\ln 2}{0.1} = 10\ln 2 \approx 6.93$ years [1].
Q2 (3 marks): $k = \dfrac{\ln(5000/2000)}{10} = \dfrac{\ln(5/2)}{10}$ [1]. $N = 2000\,e^{kt}$. $20\,000 = 2000\,e^{kt} \Rightarrow e^{kt} = 10$ [1]. $t = \dfrac{\ln 10}{k} = \dfrac{10\ln 10}{\ln(5/2)}$ years $\approx 28.3$ years [1].
Q3 (3 marks): $k = -\dfrac{\ln 2}{5730}$; $0.125\,N_0 = N_0\,e^{kt}$ → $N_0$ cancels [1]; $0.125 = 2^{-3}$, so $3\ln 2 = |k|t$ → $t = 3 \times 5730 = 17\,190$ years [1]. Independence: $N_0$ cancels in $N/N_0 = e^{kt}$, so the age depends only on the fraction remaining, not the original quantity [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering exponential growth and decay questions. Lighter alternative to the boss.
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