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Module 9 · L14 of 20 ~35 min ⚡ +95 XP available

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.

Today's hook, Carbon-14 has a half-life of 5730 years. A fossil contains 25% of its original carbon-14. Before working through this lesson, estimate how old the fossil is. You'll calculate the exact answer in card 09.

0/5QUESTS

Recall, your gut answer first

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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.

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The exponential growth/decay model

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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.

$N = N_0 e^{kt}$

$k > 0$ means growth

When $k > 0$, $e^{kt}$ increases without bound, exponential growth (population, compound interest).

$k < 0$ means decay

When $k < 0$, $e^{kt} \to 0$ as $t \to \infty$, exponential decay (radioactive decay, drug concentration).

$N_0$ is always the value at $t=0$

Even if the problem gives $N$ at a time other than $t=0$, you can still find $N_0$ by substituting that data point into $N = N_0 e^{kt}$.

What you'll master

Know

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 $k > 0$)

Understand

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 $k$

Can do

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

Key terms

Exponential growthGrowth where $\dfrac{dN}{dt} = kN$ with $k > 0$. The quantity grows by a fixed percentage per unit time.

Exponential decayDecay where $\dfrac{dN}{dt} = kN$ with $k < 0$. The quantity decreases by a fixed percentage per unit time.

$N_0$ (initial value)The value of $N$ at time $t = 0$. Determined by the initial condition $N(0) = N_0$.

Growth/decay constant $k$The constant of proportionality in $\dfrac{dN}{dt} = kN$. Units are $\text{time}^{-1}$. $k>0$ growth; $k<0$ decay.

Half-life $T_{1/2}$Time for a decaying quantity to halve: $T_{1/2} = \dfrac{\ln 2}{|k|}$. Used in radioactive decay, drug pharmacokinetics.

Doubling time $T_2$Time for a growing quantity to double: $T_2 = \dfrac{\ln 2}{k}$. Used in population and investment models.

Deriving $N = N_0 e^{kt}$

core concept

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 $|\,|$)

$$N(t) = N_0\,e^{kt}$$

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 $N_0$:

$$\frac{N_2}{N_1} = e^{k(t_2-t_1)} \;\Rightarrow\; k = \frac{\ln(N_2/N_1)}{t_2 - t_1}$$

Half-life and doubling time. Set $N = \dfrac{N_0}{2}$ (or $N = 2N_0$) and solve for $t$: in both cases $e^{kt} = 2$, giving $t = \dfrac{\ln 2}{|k|}$. The result is independent of $N_0$ every piece of a radioactive sample has the same half-life.

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?

Growth vs decay, reading $k$

core concept

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.

NESA exam tip. When asked to "find $k$", always start from the differential equation $\dfrac{dN}{dt} = kN$, derive $N = N_0 e^{kt}$, then substitute the given data. Do not assume the form without showing the derivation, it earns a method mark.

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

PROBLEM 1 · POPULATION GROWTH

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.

Model: $P = P_0 e^{kt}$ with $k = 0.02$. Let $t = 0$ correspond to 2020, so $P_0 = 50\,000$.

2% per year continuous growth means $k = 0.02$. Choose a convenient origin for $t$.

PROBLEM 2 · FINDING $k$ FROM TWO DATA POINTS

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.

$N(0)=500$ gives $N_0=500$. Use $N(4)=2000$: $2000 = 500\,e^{4k} \Rightarrow e^{4k} = 4$

Substitute both data points. The ratio eliminates $N_0$ or we can directly find $k$.

PROBLEM 3 · CARBON DATING (HOOK ANSWER)

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?

Find $k$: from $T_{1/2} = \dfrac{\ln 2}{|k|} = 5730$, we get $|k| = \dfrac{\ln 2}{5730}$, so $k = -\dfrac{\ln 2}{5730}$ (decay).

Rearrange the half-life formula for $k$. The negative sign confirms decay.

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

Trap 01

Confusing percentage rate with $k$

"Grows at 5% per year" means $k = 0.05$ (continuous rate), not $k = 5$. Common mistake: writing $k = 5$ or using $N = N_0 \times 1.05^t$ (which is the discrete model). For differential equation models, $k$ is the continuous rate per unit time.

Trap 02

Forgetting that half-life is independent of $N_0$

Students sometimes write half-life as $T_{1/2} = \dfrac{N_0}{2k}$ or include $N_0$ in the formula. Half-life is $T_{1/2} = \dfrac{\ln 2}{|k|}$, it depends only on $k$, not on how much substance you started with. This is why carbon dating works.

Trap 03

Using $\ln$ laws incorrectly

When solving $N_0 e^{kt} = N_1$, take $\ln$ of both sides: $\ln N_0 + kt = \ln N_1$. A common error is writing $\ln(N_0 e^{kt}) = \ln(N_0) \cdot kt$, incorrect. The correct law is $\ln(AB) = \ln A + \ln B$, giving $\ln N_0 + kt = \ln N_1$.

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?

Revisit your thinking

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?

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Multiple choice

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Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

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)

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ApplyBand 43 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)

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AnalyseBand 53 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)

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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].

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Module overview · Extension 1 · Checkpoint 3