Differential Equations · step 10 of 14

Modelling: growth, decay, cooling & mixing

Three model families cover most applications. Exponential growth or decay is dP/dt = kP, giving P = P₀e^(kt). Newton’s law of cooling is dT/dt = −k(T − T_env), whose solution approaches the ambient temperature. Mixing problems use the conservation statement rate of change = rate in − rate out, which is where the modelling marks lie.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Define the variable and its unitsLet A(t) be the amount of salt in kg at time t in minutes. Ambiguity here derails everything after.
Write the rate equationGrowth: dP/dt = kP. Cooling: dT/dt = −k(T − T_env). Mixing: dA/dt = (in concentration × in rate) − (A/V × out rate).
Solve by separation or integrating factorGrowth and cooling separate cleanly; mixing is usually linear.
Apply the conditions and interpretUse the given data to find k and the constant, then answer the actual question in context with units.

Worked example

A substance decays with half-life 10 years. How much of a 50 g sample remains after 25 years?

Model: dm/dt = km, so m = 50e^(kt).
Half-life: 25 = 50e^(10k), so e^(10k) = 1/2 and k = −ln2/10.
At t = 25: m = 50e^(−25ln2/10) = 50 × 2^(−2.5).
2^(2.5) = 4√2 ≈ 5.657.

Answer. About 8.84 g remains — 50/2^2.5, which is between the quarter (2 half-lives) and eighth (3 half-lives) marks as expected.

Where marks get dropped

These are the specific errors that cost credit on modelling: growth, decay, cooling & mixing questions — QED's rubric penalises each of them separately.

Getting the sign of k wrong. Decay needs k < 0, or equivalently write dm/dt = −km with k > 0 and stay consistent.
In mixing problems, assuming the volume is constant when inflow and outflow rates differ. Then V(t) = V₀ + (rate difference)t.
Using the ambient temperature as the initial temperature in cooling problems. They are different quantities and both appear in the solution.

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Modelling: growth, decay, cooling & mixing — frequently asked questions

How is the half-life related to k?

k = −ln2/T½ for the form m = m₀e^(kt). Equivalently m = m₀·2^(−t/T½), which is often easier to compute with.

Why does Newton’s cooling have a limit?

Because dT/dt → 0 as T → T_env, so the temperature approaches ambient asymptotically and never crosses it.

What makes mixing problems tricky?

The outflow concentration is A(t)/V(t), which changes as the tank contents change. Writing that ratio correctly is the whole difficulty.

The rest of Differential Equations

Solving and modelling with ODEs, from separable to systems. Each subtopic below has its own method, worked example and mark-losing traps.

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