Growth, decay, cooling and logistic models in QCE Specialist Mathematics Unit 4

Syllabus dot point

“Model and solve practical situations with first-order differential equations, including exponential growth and decay, Newton's law of cooling and the logistic equation, and interpret the long-term behaviour of solutions”

QCESpecialist MathematicsUnit 4: Further calculus, and…7 min readUpdated 29 May 2026

Quick answer

Three standard first-order models: exponential growth/decay $\frac{dN}{dt} = kN$ with solution $N = N_0 e^{kt}$ ; Newton's cooling $\frac{dT}{dt} = -k(T - T_s)$ with solution $T = T_s + (T_0 - T_s)e^{-kt} \to T_s$ ; and the logistic equation $\frac{dP}{dt} = kP(1 - P/M)$ with S-shaped solution tending to the carrying capacity $M$ . Solve by separation of variables, fit constants with initial and later data, and interpret limits: decay to $0$ , cooling to ambient, logistic to $M$ .

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What this dot point is asking
The answer
Exam-style questions

What this dot point is asking

QCAA wants you to recognise the standard first-order models, set up the differential equation from a worded scenario, solve it (usually by separation of variables), apply initial conditions, and interpret long-term behaviour. The three core models are exponential growth and decay, Newton's law of cooling, and the logistic equation, each with a characteristic solution and limiting value.

The answer

Exponential growth and decay

When a quantity changes at a rate proportional to its current size,

\frac{dN}{dt} = kN \ \Rightarrow\ N = N_0 e^{kt}.

If $k > 0$ the quantity grows without bound; if $k < 0$ it decays toward zero. This models populations, radioactive decay and continuously compounded interest. The half-life or doubling time follows from $e^{kt}$ .

Newton's law of cooling

The temperature $T$ of an object changes at a rate proportional to the difference from the ambient temperature $T_s$ :

\frac{dT}{dt} = -k(T - T_s), \qquad k > 0.

Separating variables gives the solution

T = T_s + (T_0 - T_s)e^{-kt},

where $T_0$ is the initial temperature. As $t \to \infty$ , $e^{-kt} \to 0$ , so $T \to T_s$ : the object approaches the surrounding temperature. The ambient temperature $T_s$ is the equilibrium.

The logistic equation

Population growth limited by resources is modelled by

\frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right),

where $M$ is the carrying capacity. When $P$ is small the bracket is near $1$ and growth is nearly exponential; as $P$ approaches $M$ the bracket approaches $0$ and growth slows. The solution is the S-shaped logistic curve

P = \frac{M}{1 + A e^{-kt}}, \qquad A = \frac{M - P_0}{P_0}.

Long-term behaviour and equilibria

Each model has a limiting value. Exponential growth has none (it diverges); decay tends to $0$ ; cooling tends to $T_s$ ; the logistic model tends to its carrying capacity $M$ . Equilibria are the constant solutions where the rate is zero: for the logistic equation, $\frac{dP}{dt} = 0$ at $P = 0$ (unstable) and $P = M$ (stable).

Setting up from a scenario

Translate phrases into the equation. "Rate proportional to amount" gives $\frac{dN}{dt} = kN$ . "Rate proportional to the temperature difference" gives cooling. "Growth limited by a maximum sustainable level" signals the logistic model with $M$ that maximum. Then separate variables and integrate, applying the initial condition to find the constant.

Interpreting the constant $k$

The constant $k$ sets the timescale: a larger $|k|$ means faster change. Solve for $k$ using a second data point (for example a known value at a later time) when the scenario provides one.

Worked example

A cup of coffee at $90^\circ\text{C}$ cools in a $20^\circ\text{C}$ room. After 5 minutes it is $60^\circ\text{C}$ . Find its temperature after 10 minutes.

This is a Newton's law of cooling problem. We write down the model, use the 5-minute reading to find $k$ , then substitute $t = 10$ .

Step 1: Write the cooling model

Newton's law gives $T = T_s + (T_0 - T_s)e^{-kt}$ with ambient temperature $T_s = 20$ and initial temperature $T_0 = 90$ :

T = 20 + 70 e^{-kt}.

Step 2: Find the cooling constant $k$ using the 5-minute data

At $t = 5$ , $T = 60$ . Substituting and solving for $e^{-5k}$ :

60 = 20 + 70 e^{-5k} \;\Rightarrow\; 40 = 70 e^{-5k} \;\Rightarrow\; e^{-5k} = \frac{4}{7}.

Taking logarithms: $-5k = \ln\dfrac{4}{7}$ , so $k = \dfrac{1}{5}\ln\dfrac{7}{4} \approx 0.1119$ per minute.

Step 3: Find the temperature at 10 minutes

Rather than recomputing $k$ numerically, use the exact relation $e^{-10k} = (e^{-5k})^2$ to keep the calculation clean:

e^{-10k} = \left(\frac{4}{7}\right)^2 = \frac{16}{49}.

T(10) = 20 + 70\cdot\frac{16}{49} = 20 + \frac{1120}{49} \approx 20 + 22.86 = 42.86^\circ\text{C}.

Step 4: Check the rate of cooling is realistic

The coffee dropped $30^\circ$ in the first 5 minutes and about $17^\circ$ in the next 5. The rate slows as the coffee approaches room temperature, exactly as Newton's law predicts.

Final answer: The temperature after 10 minutes is approximately $42.86^\circ\text{C}$ .

Common trap

A common error is misreading the equilibrium: the logistic solution tends to the carrying capacity $M$ , not to zero, and cooling tends to the ambient temperature $T_s$ , not to zero. State the correct limit. Another slip is forgetting the sign in cooling: the rate is $-k(T - T_s)$ with $k > 0$ , so a hot object cools. Finally, use a second data point to find $k$ rather than guessing it, and keep $k$ exact (in logarithmic form) until the final numerical step.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice7 marks

A research organisation plans to use a drone to drop a scientific instrument vertically from a stationary position above the ocean surface. The acceleration (m s^-2) of the falling instrument can be modelled by 9.8 - 0.1v, where v is its velocity (m s^-1). In order for the instrument sensors to activate, its speed as it hits the ocean surface must reach at least 20 m s^-1. However, if it hits with a speed above 50 m s^-1, the sensors will be damaged. Determine the range of the drone's flying height above the ocean surface to ensure that the sensors are activated but not damaged.

Show worked answer →

To link speed to height, write acceleration as v dv/dx = 9.8 - 0.1v (x is distance fallen). Separate variables:
x = integral of v / (9.8 - 0.1v) dv, taken from v = 0 (released at rest) up to the impact speed V.

Evaluate the integral [working]: v/(9.8 - 0.1v) = -10 - 980/(v - 98), so the antiderivative is -10v - 980 ln|v - 98|.
Between 0 and V this gives x(V) = -10V + 980 ln(98 / (98 - V)).

Lower height (impact speed 20): x = -10(20) + 980 ln(98/78) = -200 + 223.7 = 23.7 m.
Upper height (impact speed 50): x = -10(50) + 980 ln(98/48) = -500 + 699.5 = 199.5 m.

So the drone's flying height must lie between about 23.7 m and 199.5 m for the sensors to activate without being damaged. (The model has a terminal velocity of 98 m s^-1, found by setting the acceleration 9.8 - 0.1v = 0, so an impact speed of 50 m s^-1 is physically attainable.)

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Sources & how we know this

Specialist Mathematics General senior syllabus 2025 — QCAA