AS & A Level · Practice questions · Forming differential equations

Words in, equations out.

Six original Cambridge-style questions on forming differential equations: translating rates of change, finding constants, connected rates and interpreting models.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Which quantity, with respect to what, depends on what.

Derivative: rate of change of \(x\) means \(\dfrac{dx}{dt}\).
Proportional: \(= kf(x)\); inversely proportional: \(= \dfrac{k}{f(x)}\).
Decreasing: put in a minus sign and keep \(k > 0\).
Constant: substitute one value and its rate, with the correct sign.
Connected rates: \(\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}\).

Recall

[1 marks]

The number of bacteria \(N\) in a culture increases at a rate proportional to \(N\). Write down a differential equation relating \(N\) and the time \(t\).

\(\dfrac{dN}{dt} = kN\), where \(k\) is a positive constant. ✓

Recall

[2 marks]

A metal bar cools in a room at \(18^\circ\)C. Its temperature \(\theta\) °C decreases at a rate proportional to the amount by which \(\theta\) exceeds 18. When \(\theta = 98\), the temperature is falling at \(4^\circ\)C per minute. Form a differential equation.

\(\dfrac{d\theta}{dt} = -k(\theta - 18)\). ✓
\(-4 = -k(98 - 18) = -80k\), so \(k = 0.05\): \(\dfrac{d\theta}{dt} = -0.05(\theta - 18)\). ✓

Proof

[3 marks]

A spherical raindrop grows by condensation so that its volume increases at a rate proportional to its surface area. Show that its radius increases at a constant rate. [\(V = \tfrac{4}{3}\pi r^{3}\), \(A = 4\pi r^{2}\).]

\(\dfrac{dV}{dt} = kA = 4k\pi r^{2}\). ✓
Chain rule: \(\dfrac{dV}{dt} = 4\pi r^{2}\dfrac{dr}{dt}\). ✓
So \(4\pi r^{2}\dfrac{dr}{dt} = 4k\pi r^{2}\), giving \(\dfrac{dr}{dt} = k\), a constant. ✓

Calculation

[3 marks]

The depth \(h\) metres of water in a tank decreases at a rate proportional to \(\sqrt{h}\). When \(h = 1.44\), the depth is decreasing at \(0.03\) metres per hour. (a) Form a differential equation. (b) Find the rate of decrease when \(h = 0.49\).

(a) \(\dfrac{dh}{dt} = -k\sqrt{h}\); \(-0.03 = -k(1.2)\), so \(k = 0.025\): \(\dfrac{dh}{dt} = -0.025\sqrt{h}\). ✓
(b) \(\dfrac{dh}{dt} = -0.025 \times 0.7 = -0.0175\); decreasing at \(0.0175\) m per hour. ✓

Proof

[4 marks]

Show that \(N = \dfrac{1000}{1 + 9e^{-0.4t}}\) satisfies \(\dfrac{dN}{dt} = 0.4N\left(1 - \dfrac{N}{1000}\right)\), and state the value of \(N\) when \(t = 0\) and its limit as \(t \to \infty\).

Let \(u = 9e^{-0.4t}\), so \(N = 1000(1 + u)^{-1}\) and \(\dfrac{du}{dt} = -0.4u\). ✓
\(\dfrac{dN}{dt} = -1000(1 + u)^{-2} \times (-0.4u) = \dfrac{400u}{(1 + u)^{2}}\). ✓
\(1 - \dfrac{N}{1000} = 1 - \dfrac{1}{1 + u} = \dfrac{u}{1 + u}\), so \(0.4N\left(1 - \dfrac{N}{1000}\right) = 0.4 \times \dfrac{1000}{1 + u} \times \dfrac{u}{1 + u} = \dfrac{400u}{(1 + u)^{2}}\). The sides agree. ✓
\(t = 0\): \(N = \dfrac{1000}{10} = 100\). As \(t \to \infty\), \(u \to 0\) and \(N \to 1000\). ✓

Analysis

[6 marks]

Water is poured at \(30\ \text{cm}^{3}\,\text{s}^{-1}\) into an inverted cone whose radius always equals the depth of water \(h\) cm, so the volume is \(V = \tfrac{1}{3}\pi h^{3}\). Water leaks out of a hole at a rate of \(kh\ \text{cm}^{3}\,\text{s}^{-1}\). When \(h = 10\) the depth is not changing.

(a) Show that \(\pi h^{2}\dfrac{dh}{dt} = 30 - 3h\). (b) Find the rate at which the depth is rising when \(h = 5\). (c) Explain what happens to the depth in the long term if the cone starts empty.

(a) \(\dfrac{dV}{dt} = 30 - kh\) and \(\dfrac{dV}{dt} = \dfrac{dV}{dh}\dfrac{dh}{dt} = \pi h^{2}\dfrac{dh}{dt}\). ✓
At \(h = 10\), \(\dfrac{dh}{dt} = 0\), so \(30 - 10k = 0\) and \(k = 3\): \(\pi h^{2}\dfrac{dh}{dt} = 30 - 3h\). ✓
(b) \(h = 5\): \(25\pi\dfrac{dh}{dt} = 15\), so \(\dfrac{dh}{dt} = \dfrac{3}{5\pi} = 0.191\ \text{cm s}^{-1}\) (3 s.f.). ✓
(c) For \(h < 10\) the rate is positive, and it falls to zero as \(h \to 10\): the depth rises towards 10 cm. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.