Paper 3 only. Cooling coffee, leaking tanks and growing populations are all described by how fast something changes. Writing that rate as a derivative, with a constant of proportionality, is the first mark in every modelling question.
'The rate of change of \(x\)' means \(\dfrac{dx}{dt}\). 'Proportional to \(f(x)\)' means \(= k f(x)\) for a constant \(k\). If the quantity is DECREASING, write \(\dfrac{dx}{dt} = -k f(x)\) with \(k > 0\), then use the given data to find \(k\).
Fig. 1: tea at 80°C cooling in a 20°C room; the rate of cooling is proportional to the excess temperature, so the curve starts with gradient −0.05 × 60 = −3 °C per minute and flattens towards 20°C
Section 01
Rates of change as derivatives.
Every modelling statement has the same three parts: WHICH quantity changes (the top of the derivative), with respect to WHAT (usually time \(t\)), and what the rate depends on. Translate each phrase in turn.
Statement
Differential equation
The population \(P\) increases at a rate proportional to \(P\).
\(\dfrac{dP}{dt} = kP\), \(k > 0\)
The mass \(m\) decreases at a rate proportional to \(\sqrt{m}\).
\(\dfrac{dm}{dt} = -k\sqrt{m}\), \(k > 0\)
The rate of change of \(v\) is inversely proportional to \(v^{2}\).
\(\dfrac{dv}{dt} = \dfrac{k}{v^{2}}\)
The temperature \(\theta\) falls at a rate proportional to the excess over 20.
A cup of tea is at \(\theta\) °C at time \(t\) minutes, in a room at \(20^\circ\)C. The temperature decreases at a rate proportional to the difference between the temperature of the tea and the room. Write down a differential equation.
The quantity changing is \(\theta\), with respect to \(t\): \(\dfrac{d\theta}{dt}\).
The difference between the tea and the room is \(\theta - 20\) (positive while the tea is hotter).
The temperature is decreasing, so the derivative is negative: use \(-k\) with \(k > 0\).
Answer: \(\dfrac{d\theta}{dt} = -k(\theta - 20)\), where \(k\) is a positive constant.
Worked example A gradient condition
At every point \((x, y)\) of a curve, the gradient is proportional to \(\dfrac{x}{y}\). The gradient at \((2, 3)\) is 4. Form a differential equation for the curve.
Gradient is \(\dfrac{dy}{dx}\), so \(\dfrac{dy}{dx} = \dfrac{kx}{y}\).
'Rate of change' with no other words means with respect to TIME. Writing \(\dfrac{d\theta}{dx}\) or forgetting the minus sign on a decreasing quantity both lose the first mark, and the error carries into the solution.
Section 02
Finding the constant.
The constant of proportionality is found from one pair of simultaneous values: a value of the quantity and its rate of change at that moment. Watch the sign: a quantity 'decreasing at 2 units per second' has rate \(-2\).
Worked example A leaking tank
Water leaks from a tank so that the depth \(h\) metres decreases at a rate proportional to \(\sqrt{h}\). When \(h = 0.81\), the depth is decreasing at \(0.018\) metres per minute. Form a differential equation and find the rate when \(h = 0.25\).
Answer: \(\dfrac{dh}{dt} = -0.02\sqrt{h}\); the depth is falling at \(0.01\) m per minute when \(h = 0.25\).
Worked example Connected rates: an inflating balloon
Air is pumped into a spherical balloon at a constant rate of \(50\ \text{cm}^{3}\,\text{s}^{-1}\). Form a differential equation for the radius \(r\) cm.
The information is about \(V = \tfrac{4}{3}\pi r^{3}\): \(\dfrac{dV}{dt} = 50\).
If you write \(\dfrac{dh}{dt} = -k\sqrt{h}\), then \(k\) must come out POSITIVE. Substituting a rate of \(+0.018\) gives \(k = -0.02\) and the double negative makes the water rise.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Which equation says 'the rate of change of \(x\) is proportional to \(x^{2}\)'?
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\(y\) decreases at a rate inversely proportional to \(\sqrt{y}\). Which is correct (\(k > 0\))?
Quick check+10 XP
\(\dfrac{dP}{dt} = 0.03P\). How fast is \(P\) growing when \(P = 2000\)?
Quick check+10 XP
\(\dfrac{d\theta}{dt} = -k(\theta - 15)\) with \(k > 0\) and \(\theta\) starting at 60. What happens to \(\theta\) in the long term?
Section 03
Reading a model.
Before solving anything, a differential equation tells you a lot: where the quantity grows or shrinks, where it stops changing, and how fast it changes at any moment. You can also check a proposed solution by differentiating it and substituting.
Worked example Checking a solution
Show that \(\theta = 20 + 60e^{-0.05t}\) satisfies \(\dfrac{d\theta}{dt} = -0.05(\theta - 20)\) with \(\theta = 80\) when \(t = 0\), and describe what happens to \(\theta\) in the long term.
Right-hand side: \(\theta - 20 = 60e^{-0.05t}\), so \(-0.05(\theta - 20) = -3e^{-0.05t}\). The two sides agree.
At \(t = 0\): \(\theta = 20 + 60 = 80\) ✓.
As \(t \to \infty\), \(e^{-0.05t} \to 0\), so \(\theta \to 20\).
Answer: It satisfies both; the tea cools towards room temperature, \(20^\circ\)C, without ever reaching it (Fig. 1).
Worked example Logistic growth
A population of fish is modelled by \(\dfrac{dN}{dt} = 0.2N\left(1 - \dfrac{N}{500}\right)\). For which values of \(N\) is the population growing, and at what population is it growing fastest?
For \(0 < N < 500\) both factors are positive, so \(\dfrac{dN}{dt} > 0\): growing. At \(N = 500\) the rate is zero.
The rate is \(f(N) = 0.2N - 0.0004N^{2}\), a quadratic in \(N\).
\(f'(N) = 0.2 - 0.0008N = 0\) gives \(N = 250\).
Maximum rate: \(f(250) = 0.2 \times 250 \times 0.5 = 25\) fish per unit time.
Answer: Growing for \(0 < N < 500\), levelling off at 500; fastest growth (25 per unit time) when \(N = 250\).
Examiner trap
'Growing fastest' means the RATE \(\dfrac{dN}{dt}\) is a maximum. Differentiate the rate with respect to \(N\) (or complete the square); do not just set \(\dfrac{dN}{dt} = 0\), which finds where growth STOPS.
Stage 2 · Exam
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The volume \(V\) of liquid in a container decreases at a rate proportional to \(V\). When \(V = 400\), the volume is decreasing at 8 units per second. Find the differential equation.
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A spherical balloon is inflated at a constant \(20\ \text{cm}^{3}\,\text{s}^{-1}\). Find the rate of increase of the radius when \(r = 5\) cm.
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At any point \((x, y)\) on a curve, the gradient is proportional to the product \(xy\). At \((2, 3)\) the gradient is 12. Find the differential equation.
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A population is modelled by \(\dfrac{dN}{dt} = 0.4N\left(1 - \dfrac{N}{1000}\right)\). At what population is the growth rate greatest, and what is that rate?
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