SACE Board SACE 2023-style practice: Calculator-assumed. A butterfly population B grows by dBdt = 0.1B(1 - BK), where K is a positive constant. You may assume KB(K-B) = 1B + 1K-B. Show that B = K1 + A e^-0.1t for some constant A.

Rewrite as dBdt = 0.1B(K-B)K and separate: KB(K-B)\,dB = 0.1\,dt. [1 mark] Use the given partial fractions: int (1B + 1K-B)dB = int 0.1\,dt, giving ln|B| - ln|K-B| = 0.1t + c, i.e. lnBK-B = 0.1t + c. [1 mark] Exponentiate: BK-B = Ce^0.1t. Take reciprocals: K-BB = 1Ce^-0.1t, so KB - 1 = A e^-0.1t with A = 1C. [1 mark] Then KB = 1 + A e^-0.1t, so B = K1 + A e^-0.1t, as required. [1 mark]

SACE Board SACE 2022-style practice: Calculator-assumed. The area covered by bacteria satisfies dAdt = 12A(50 - A)50 with A(0) = 1. Solve to show that A = 501 + 49 e^-0.5t.

Separate: 50A(50-A)\,dA = 12\,dt. Partial fractions: 50A(50-A) = 1A + 150-A. [1 mark] Integrate: ln|A| - ln|50-A| = 12t + c, so lnA50-A = 0.5t + c. [1 mark] Exponentiate: A50-A = Ce^0.5t. Apply A(0) = 1: 149 = C. [1 mark] So A50-A = 149e^0.5t, giving 49A = (50-A)e^0.5t, hence A = 50e^0.5t49 + e^0.5t. [1 mark] Divide through by e^0.5t: A = 501 + 49 e^-0.5t, as required. [1 mark]

SACE Board SACE 2021-style practice: Calculator-assumed. Given A = 501 + 49 e^-0.5t models the area covered by bacteria, state the maximum area available for growth.

As t → ∞, e^-0.5t → 0, so the denominator 1 + 49e^-0.5t → 1. Therefore A → 50. The maximum (limiting) area is 50, the carrying capacity of the logistic model. [1 mark]

§-Syllabus dot point

SASpecialist MathematicsSyllabus dot point

How do we translate a real situation into a differential equation and interpret its solution?

Formulate and solve differential equations modelling real situations such as growth, cooling and limited growth, and interpret the solutions.

Setting up differential equations from worded situations including exponential growth and decay, Newton's law of cooling and logistic limited growth, solving them, and interpreting the long-term behaviour.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Translating words into a differential equation
Exponential growth and decay
Newton's law of cooling
Limited (logistic) growth
Interpreting the logistic curve
Setting up from scratch

What this dot point is asking

You need to formulate differential equations from real situations, solve them, and interpret the resulting solutions including their long-term behaviour.

Translating words into a differential equation

The phrase describing the rate dictates the equation:

Exponential growth and decay

The simplest model, $\dfrac{dy}{dt} = ky$ , solves to $y = A e^{kt}$ (see the separable-equations method). Use two pieces of data: one to find $A$ , another to find $k$ . Interpretation: as $t \to \infty$ the quantity grows without bound if $k > 0$ or decays to $0$ if $k < 0$ .

Newton's law of cooling

A body at temperature $T$ in surroundings at $T_s$ cools at a rate proportional to the temperature difference:

\frac{dT}{dt} = -k(T - T_s).

Newton's law of cooling

A drink at $80^{\circ}$ C cools in a $20^{\circ}$ C room. After $5$ minutes it is $60^{\circ}$ C. Find $T(t)$

Step 1: Write the differential equation, separate variables and integrate

Newton's law of cooling says the rate of change of temperature is proportional to the gap between the object's temperature and the surrounding temperature. With room temperature $20^{\circ}$ C the model is $\dfrac{dT}{dt} = -k(T - 20)$ . Separating variables and integrating both sides gives:

\int \frac{1}{T - 20}\,dT = \int -k\,dt \implies \ln|T - 20| = -kt + C.

Step 2: Exponentiate and write the general solution

Exponentiating both sides removes the logarithm. Writing $A = e^{C}$ absorbs the integration constant into a single parameter.

$T - 20 = A e^{-kt}$ , so $T = 20 + A e^{-kt}$ .

Step 3: Apply the initial condition $T = 80$ at $t = 0$ to find $A$

Substituting $t = 0$ into the general solution immediately gives $A$ , because $e^{0} = 1$ .

$80 = 20 + A$ , so $A = 60$ . Thus $T = 20 + 60 e^{-kt}$ .

Step 4: Apply the second condition $T = 60$ at $t = 5$ to find $k$

The second piece of data gives an equation in $k$ alone, which we solve by taking a logarithm.

60 = 20 + 60 e^{-5k} \implies 40 = 60 e^{-5k} \implies e^{-5k} = \tfrac{2}{3}.

-5k = \ln\tfrac{2}{3}

, giving

k = -\tfrac{1}{5}\ln\tfrac{2}{3} = \tfrac{1}{5}\ln\tfrac{3}{2} \approx 0.0811

Step 5: Write the final model and interpret the long-term behaviour

Substituting the numerical value of $k$ gives the complete model. Examining what happens as time grows large confirms the physical meaning of the solution.

The model is $T = 20 + 60 e^{-0.0811 t}$ . As $t \to \infty$ , $e^{-0.0811t} \to 0$ , so $T \to 20^{\circ}$ C. The drink approaches room temperature, as expected.

Limited (logistic) growth

When growth slows as a population nears a carrying capacity $M$ , the logistic equation $\dfrac{dP}{dt} = kP(M - P)$ applies. Its solution is an S-shaped curve that rises from a small initial value and levels off at $M$ . The key interpretation: $\dfrac{dP}{dt} \to 0$ as $P \to M$ , so the population stabilises at the carrying capacity, and the steepest growth occurs at the halfway point $P = \tfrac{M}{2}$ .

Interpreting the logistic curve

The logistic solution $P = \dfrac{M}{1 + A e^{-kt}}$ has features SACE frequently asks you to read off. The carrying capacity $M$ is the horizontal asymptote the curve approaches as $t \to \infty$ . The constant $A$ is fixed by the initial population: substituting $t = 0$ gives $P(0) = \dfrac{M}{1 + A}$ , so $A = \dfrac{M - P(0)}{P(0)}$ . The growth rate is slow at first when $P$ is small, fastest at the inflection point $P = \tfrac{M}{2}$ , then slows again as $P$ nears $M$ . Being able to identify the capacity, the initial value, and the point of fastest growth from the formula is exactly what the interpretation marks reward.

Setting up from scratch

A worked setup: a tank gains salt at a constant rate and loses it proportionally to the amount present, giving $\dfrac{dS}{dt} = a - bS$ . This is separable and solves to an exponential approaching the equilibrium $S = \tfrac{a}{b}$ , where the rate of change is zero. Recognising the equilibrium (where $\dfrac{dS}{dt} = 0$ ) often answers the long-term question without fully solving.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20234 marksCalculator-assumed. A butterfly population

B

grows by

\dfrac{dB}{dt} = 0.1B\left(1 - \dfrac{B}{K}\right)

, where

K

is a positive constant. You may assume

\dfrac{K}{B(K-B)} = \dfrac{1}{B} + \dfrac{1}{K-B}

. Show that

B = \dfrac{K}{1 + A e^{-0.1t}}

for some constant

A

Show worked answer →

Rewrite as $\dfrac{dB}{dt} = \dfrac{0.1B(K-B)}{K}$ and separate: $\dfrac{K}{B(K-B)}\,dB = 0.1\,dt$ . [1 mark]

Use the given partial fractions: $\int \left(\dfrac{1}{B} + \dfrac{1}{K-B}\right)dB = \int 0.1\,dt$ , giving $\ln|B| - \ln|K-B| = 0.1t + c$ , i.e. $\ln\dfrac{B}{K-B} = 0.1t + c$ . [1 mark]

Exponentiate: $\dfrac{B}{K-B} = Ce^{0.1t}$ . Take reciprocals: $\dfrac{K-B}{B} = \dfrac{1}{C}e^{-0.1t}$ , so $\dfrac{K}{B} - 1 = A e^{-0.1t}$ with $A = \tfrac{1}{C}$ . [1 mark]

Then $\dfrac{K}{B} = 1 + A e^{-0.1t}$ , so $B = \dfrac{K}{1 + A e^{-0.1t}}$ , as required. [1 mark]

SACE 20225 marksCalculator-assumed. The area covered by bacteria satisfies

\dfrac{dA}{dt} = \tfrac{1}{2}A\dfrac{(50 - A)}{50}

with

A(0) = 1

. Solve to show that

A = \dfrac{50}{1 + 49 e^{-0.5t}}

Show worked answer →

Separate: $\dfrac{50}{A(50-A)}\,dA = \tfrac{1}{2}\,dt$ . Partial fractions: $\dfrac{50}{A(50-A)} = \dfrac{1}{A} + \dfrac{1}{50-A}$ . [1 mark]

Integrate: $\ln|A| - \ln|50-A| = \tfrac{1}{2}t + c$ , so $\ln\dfrac{A}{50-A} = 0.5t + c$ . [1 mark]

Exponentiate: $\dfrac{A}{50-A} = Ce^{0.5t}$ . Apply $A(0) = 1$ : $\dfrac{1}{49} = C$ . [1 mark]

So $\dfrac{A}{50-A} = \tfrac{1}{49}e^{0.5t}$ , giving $49A = (50-A)e^{0.5t}$ , hence $A = \dfrac{50e^{0.5t}}{49 + e^{0.5t}}$ . [1 mark]

Divide through by $e^{0.5t}$ : $A = \dfrac{50}{1 + 49 e^{-0.5t}}$ , as required. [1 mark]

SACE 20211 marksCalculator-assumed. Given

A = \dfrac{50}{1 + 49 e^{-0.5t}}

models the area covered by bacteria, state the maximum area available for growth.

Show worked answer →

As $t \to \infty$ , $e^{-0.5t} \to 0$ , so the denominator $1 + 49e^{-0.5t} \to 1$ .

Therefore $A \to 50$ . The maximum (limiting) area is $50$ , the carrying capacity of the logistic model. [1 mark]