What is translating a rate description into an equation?

The phrase "the rate of change of Q" is dQdt. Build the right-hand side from the description:

What is not interpreting the answer?

Modelling questions reward a sentence explaining the long-term behaviour or the meaning of a constant, not just the algebra.

What is units and context?

Keep track of units and check the answer is physically reasonable (a positive time, a mass between 0 and the initial value).

For dQdt = k(M - Q), what is the long-term value of Q? [1 mark]

A model gives Q = 100 - 80e^-0.5t. State the initial value and the limiting value. [2 marks]

SEAB Original-style practice: Water leaks from a tank so that the depth h falls at a rate proportional to sqrt(h). Initially h = 100\ cm and after 10 minutes h = 81\ cm. Form and solve a differential equation for h, and find the time for the tank to empty.

The depth falls at a rate proportional to sqrt(h), so dhdt = -ksqrt(h) for a positive constant k (negative because h decreases). Separate: h^-1/2\,dh = -k\,dt. Integrate: 2sqrt(h) = -kt + C. At t = 0, h = 100: 2(10) = C, so C = 20. At t = 10, h = 81: 2(9) = -10k + 20 18 = 20 - 10k k = 0.2. So 2sqrt(h) = 20 - 0.2t. The tank empties when h = 0: 0 = 20 - 0.2t t = 100 minutes. Markers reward the equation with the negative sign, separating and integrating to 2sqrt(h) = -kt + C, using both data points to find C and k, and the empty time t = 100 minutes.

SingaporeFurther MathsSyllabus dot point

How do we set up a differential equation to model a real situation and interpret its solution?

Formulate differential equations from descriptions of rates of change and interpret the solutions in context, including long-term behaviour

A focused answer to the H2 Further Mathematics outcome on modelling with differential equations. Translating a rate description into an equation, solving the resulting first- or second-order model, interpreting constants, and analysing long-term behaviour and limiting values.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

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

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What this dot point is asking

SEAB wants you to turn a verbal description of how a quantity changes into a differential equation, solve it with the methods you know, fit the constants to the given data, and then interpret the solution: what the constants mean, what happens in the long term, and any limiting value. The modelling step (writing the equation) and the interpretation step are as important as the solving.

The answer

Translating a rate description into an equation

The phrase "the rate of change of $Q$ " is $\dfrac{\mathrm{d}Q}{\mathrm{d}t}$ . Build the right-hand side from the description:

"proportional to $Q$ ": $\dfrac{\mathrm{d}Q}{\mathrm{d}t} = kQ$ (exponential growth or decay);
"proportional to $(M - Q)$ ": $\dfrac{\mathrm{d}Q}{\mathrm{d}t} = k(M - Q)$ (bounded approach to $M$ );
"proportional to $Q(M - Q)$ ": the logistic equation $\dfrac{\mathrm{d}Q}{\mathrm{d}t} = kQ(M - Q)$ .

Choose the sign of the constant so the equation matches whether the quantity rises or falls.

Setting the sign correctly

If the quantity decreases (cooling, leaking, decay), the rate is negative, so write $-k$ with $k > 0$ . Getting this sign right is essential for a sensible solution.

Solving and fitting constants

Solve the equation with the appropriate method (usually separation for these first-order models), giving a general solution with a constant. Use the initial value to find that constant, and any further data point to find the proportionality constant $k$ .

Interpreting long-term behaviour

Examine the solution (or the equation directly) as $t \to \infty$ :

exponential decay $\to 0$ ; exponential growth $\to \infty$ ;
a bounded model $\dfrac{\mathrm{d}Q}{\mathrm{d}t} = k(M - Q)$ approaches the limit $Q = M$ ;
the logistic model rises towards the carrying capacity $M$ and levels off.

A quick way to find a limiting (equilibrium) value is to set $\dfrac{\mathrm{d}Q}{\mathrm{d}t} = 0$ and solve.

Worked example

A chemical dissolves so that the mass $m$ still undissolved decreases at a rate proportional to $m$ . Initially $m = 50\ \text{g}$ , and $20\ \text{g}$ remain after $5$ minutes. Find $m$ as a function of $t$ and the time for the mass to fall to $5\ \text{g}$ .

Step 1: Form the equation

The mass decreases at a rate proportional to $m$ , so $\dfrac{\mathrm{d}m}{\mathrm{d}t} = -km$ with $k > 0$ .

Step 2: Solve by separation

$\dfrac{1}{m}\,\mathrm{d}m = -k\,\mathrm{d}t$ , so $\ln m = -kt + C$ , giving $m = m_0\mathrm{e}^{-kt}$ .

Step 3: Apply the initial value

At $t = 0$ , $m = 50$ , so $m = 50\mathrm{e}^{-kt}$ .

Step 4: Use the second data point to find k

At $t = 5$ , $m = 20$ : $20 = 50\mathrm{e}^{-5k} \Rightarrow \mathrm{e}^{-5k} = 0.4 \Rightarrow k = -\tfrac{1}{5}\ln 0.4 \approx 0.1833$ .

Step 5: Find the required time

Set $m = 5$ : $5 = 50\mathrm{e}^{-kt} \Rightarrow \mathrm{e}^{-kt} = 0.1 \Rightarrow t = \dfrac{-\ln 0.1}{k} = \dfrac{\ln 10}{0.1833} \approx 12.6\ \text{minutes}$ .

Examples in context

Example 1. Radioactive decay. The number of undecayed nuclei satisfies $\dfrac{\mathrm{d}N}{\mathrm{d}t} = -\lambda N$ , giving $N = N_0\mathrm{e}^{-\lambda t}$ and the constant half-life $\dfrac{\ln 2}{\lambda}$ , the canonical exponential-decay model across physics and chemistry.

Example 2. Spread of a rumour or epidemic. The early spread of an infection through a fixed population follows the logistic equation: fast growth while susceptibles are plentiful, then a slowdown as the carrying capacity is approached, which is why the logistic curve is the standard first model of epidemics.

Try this

Q1. Write a differential equation for a quantity $Q$ decaying at a rate proportional to itself. [1 mark]

Cue. $\dfrac{\mathrm{d}Q}{\mathrm{d}t} = -kQ$ with $k > 0$ .

Q2. For $\dfrac{\mathrm{d}Q}{\mathrm{d}t} = k(M - Q)$ , what is the long-term value of $Q$ ? [1 mark]

Cue. $Q \to M$ (set the rate to zero to find the equilibrium $Q = M$ ).

Q3. A model gives $Q = 100 - 80\mathrm{e}^{-0.5t}$ . State the initial value and the limiting value. [2 marks]

Cue. At $t = 0$ , $Q = 20$ ; as $t \to \infty$ , $Q \to 100$ .

Exam-style practice questions

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

Original7 marksA population

P

grows so that its rate of increase is proportional to the product of

P

and

(M - P)

, where

M

is a constant carrying capacity. Write down a differential equation for

P

and describe, with reasons, the long-term value of

P

0 < P_0 < M

initially.

Show worked answer →

The rate of increase is proportional to $P(M - P)$ , so for a positive constant $k$ :

\frac{\mathrm{d}P}{\mathrm{d}t} = kP(M - P).

This is the logistic equation. For

0 < P < M

, both

P

and

(M - P)

are positive, so

\dfrac{\mathrm{d}P}{\mathrm{d}t} > 0

and the population increases. As

P \to M

, the factor

(M - P) \to 0

, so the growth rate falls to zero and the population levels off.

Therefore the long-term value is $P \to M$ : the population approaches the carrying capacity $M$ from below and stabilises there.

Markers reward the equation $\dfrac{\mathrm{d}P}{\mathrm{d}t} = kP(M - P)$ , the sign argument that growth is positive but slowing, and the conclusion $P \to M$ with reasoning.

Original6 marksWater leaks from a tank so that the depth

h

falls at a rate proportional to

\sqrt{h}

. Initially

h = 100\ \text{cm}

and after

10

minutes

h = 81\ \text{cm}

. Form and solve a differential equation for

h

, and find the time for the tank to empty.

Show worked answer →

The depth falls at a rate proportional to $\sqrt{h}$ , so $\dfrac{\mathrm{d}h}{\mathrm{d}t} = -k\sqrt{h}$ for a positive constant $k$ (negative because $h$ decreases).

Separate: $h^{-1/2}\,\mathrm{d}h = -k\,\mathrm{d}t$ . Integrate: $2\sqrt{h} = -kt + C$ .

At $t = 0$ , $h = 100$ : $2(10) = C$ , so $C = 20$ . At $t = 10$ , $h = 81$ : $2(9) = -10k + 20 \Rightarrow 18 = 20 - 10k \Rightarrow k = 0.2$ .

So $2\sqrt{h} = 20 - 0.2t$ . The tank empties when $h = 0$ : $0 = 20 - 0.2t \Rightarrow t = 100$ minutes.

Markers reward the equation with the negative sign, separating and integrating to $2\sqrt{h} = -kt + C$ , using both data points to find $C$ and $k$ , and the empty time $t = 100$ minutes.