Year 12 Maths Extension 1 (2027) Further applications of calculus

Modelling with DEs (growth & decay)

Q: What is the exponential growth model?

dQ by dt equals k Q, solved by Q equals Q naught e to the k t.

Q: How do you find the growth constant k?

Use a second data point and solve for k.

Q: What is the doubling time?

ln 2 over k, the time for the quantity to double.

Q: What is the half-life?

Minus ln 2 over k for decay, the time to halve.

20 practice questions 2 video lessons Theory + worked examples

Apply calculus to the real world with modelling using differential equations in NSW Year 12 Mathematics Extension 1. Many natural processes change at a rate proportional to the current amount, giving exponential growth or decay.

You will learn to set up and solve growth and decay models, and interpret the limiting value they approach — modelling problems drawn from chemistry, biology and economics in the Extension 1 course.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

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Exponential Growth and Decay Calculus, Relative Growth Rate, Differential Equations, Word Problems Watch
Modelling exponential growth and decay | Introduction | ExamSolutions Watch

Theory

Exponential growth and decay follow $\frac{d Q}{d t} = k Q$ , solved by $Q = Q_{0} e^{k t}$ : $k > 0$ grows, $k < 0$ decays. This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-05) covers doubling time $\frac{\ln 2}{k}$ and half-life.

Exponential growth and decay arise when a quantity changes at a rate proportional to its current amount: $\frac{d Q}{d t} = k Q$ , with solution $Q = Q_{0} e^{k t}$ , where $Q_{0}$ is the initial value.

$k > 0$ gives growth; $k < 0$ gives decay. The doubling time is $\frac{\ln 2}{k}$ and the half-life is $- \frac{\ln 2}{k}$ .

Neither the doubling time nor the half-life depends on the starting amount.

NESA link. Part of the Year 12 Further applications of calculus focus area, outcome ME1-12-05 ("applies calculus to solve problems involving polynomials, further rates of change, areas and volumes and differential equations") with MAO-WM-01. The same modelling extends to $\frac{d Q}{d t} = k (Q - P)$ and logistic growth in chemistry, biology and economics.

\frac{d Q}{d t} = k Q

: growth for

k > 0

, decay for

k < 0

Half-life: the time for the quantity to fall to

\frac{M_{0}}{2}

\frac{d Q}{d t} = k Q ⟹ Q = Q_{0} e^{k t} .

dQ/dt = kQ gives Q = Q0 e^(kt)

doubling time = \frac{\ln 2}{k}, half-life = - \frac{\ln 2}{k} .

doubling time = ln2 / k; half-life = -ln2 / k

Keep logs exact. Carry values like $k = \frac{1}{2} \ln 3$ exactly until the final numerical step; the doubling time and half-life are independent of the starting amount.

How to model with $\frac{d Q}{d t} = k Q$

Write the solution $Q = Q_{0} e^{k t}$ with $Q_{0}$ the initial value.
Find $k$ from a second data point, keeping logs exact.
Answer the question: evaluate $Q$ at a time, or find a doubling time / half-life.
Interpret the sign of $k$ — growth or decay.

Example 1 — Growth model

\frac{d P}{d t} = 0.04 P

with

P (0) = 200

. Find

P (t)

and

P (10)

Solution

$P = 200 e^{0.04 t}$ .

P (10)

=

200 e^{0.4} \approx 298

P = 200 e^(0.04t), P(10) approx 298

$P = 200 e^{0.04 t}$ , $P (10) \approx 298$ .

Example 2 — Find k

A colony grows as

N = 1000 e^{k t}

, with

N = 3000

t = 2

. Find

k

Solution

$3000$	$=$	$1000 e^{2 k} \Rightarrow e^{2 k} = 3$
$k$	$=$	$\frac{1}{2} \ln 3$

k = (1/2) ln 3

$k = \frac{1}{2} \ln 3$ .

Example 3 — Half-life to k

A substance has a half-life of

20

years. Find

k

\frac{d M}{d t} = k M

Solution

$\frac{1}{2} = e^{20 k}$ .

$20 k$	$=$	$\ln \frac{1}{2}$
$k$	$=$	$- \frac{\ln 2}{20} \approx - 0.0347$

k = -ln2/20 approx -0.0347

$k \approx - 0.0347$ .

Example 4 — Doubling time

A population grows as

\frac{d P}{d t} = 0.02 P

. Find its doubling time.

Solution

$2 = e^{0.02 t}$ .

t

=

\frac{\ln 2}{0.02} \approx 34.7

doubling time approx 34.7 years

About $34.7$ years.

Common pitfalls

Sign of $k$ . For decay,

k < 0

(or write

Q = Q_{0} e^{- k t}

with

k > 0

Rounding early. Keep logs exact (e.g.

k = \frac{1}{2} \ln 3

) until the final numerical step.

Half-life depends on amount. It does not — doubling time and half-life are independent of the starting value.

Confusing the two formulas. Doubling time uses

2

; half-life uses

\frac{1}{2}

Frequently asked questions

What is the exponential growth model?

$\frac{d Q}{d t} = k Q$ , solved by $Q = Q_{0} e^{k t}$ .

How do you find the growth constant k?

Use a second data point, e.g. $N = 1000 e^{k t}$ with a known later value, and solve for $k$ .

What is the doubling time?

$\frac{\ln 2}{k}$ — the time for the quantity to double.

What is the half-life?

$- \frac{\ln 2}{k}$ for decay — the time to halve.

Does half-life depend on the starting amount?

No — it depends only on $k$ .

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Logistic equation dP/dt = kP(1 − P/C)

More in Further applications of calculus

Modelling with DEs (growth & decay)

📖 Theory

How to model with dQdt=kQ

Common pitfalls

Frequently asked questions

Theory

How to model with $\frac{d Q}{d t} = k Q$