The shape of the lesson
By the end of the lesson, learners can
- 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.
- 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.
- 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.
- 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: \(\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}\).
One picture every idea returns to
'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 \(k\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Turn a sentence into a derivative.. Paper 3 only. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: 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. Model the worked example Newton's law of cooling on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: 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. Model the worked example A leaking tank on the board, one line at a time. Your turn: A population \(P\) grows at a rate proportional to \(P\). When \(P = 2500\), it is growing at 75 per year. Form the differential equation. | Pairs |
| 21 to 30 min | Teach and check: 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. Model the worked example Checking a solution on the board, one line at a time. | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 19 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Reading a model, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| '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. |
| 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. |
| '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. |
Forming differential equations: worksheet
6 questions, 19 marks. Written in the style of the papers; no past paper question is reproduced.
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 \(t\).
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.
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. [\(V = \tfrac{4}{3}\pi r^{3}\), \(A = 4\pi r^{2}\).]
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 \(h = 0.49\).
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 \(t \to \infty\).
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.
Answer key and marking notes
\(\dfrac{dN}{dt} = kN\), where \(k\) is a positive constant.
- \(\dfrac{d\theta}{dt} = -k(\theta - 18)\).
- \(-4 = -k(98 - 18) = -80k\), so \(k = 0.05\): \(\dfrac{d\theta}{dt} = -0.05(\theta - 18)\).
- \(\dfrac{dV}{dt} = kA = 4k\pi r^{2}\).
- 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.
- (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.
- 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 \(N \to 1000\).
- (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.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The rate of increase of a quantity \(x\) with respect to time \(t\) is proportional to \(x^{2}\). Which differential equation models this?
- A\(\dfrac{dx}{dt} = kx^{2}\)
- B\(\dfrac{dx}{dt} = kt^{2}\)
- C\(\dfrac{dx}{dt} = x^{2} + k\)
- D\(\dfrac{dx}{dt} = kx\)
The volume \(V\) of liquid in a container DECREASES at a rate proportional to \(\sqrt{V}\). With \(k\) a positive constant, which equation is correct?
- A\(\dfrac{dV}{dt} = -k\sqrt{V}\)
- B\(\dfrac{dV}{dt} = k\sqrt{V}\)
- C\(\dfrac{dV}{dt} = -\dfrac{k}{\sqrt{V}}\)
- D\(\dfrac{dV}{dt} = -kt\)
A cup of tea at temperature \(\theta^{\circ}\text{C}\) is in a room at \(20^{\circ}\text{C}\). The rate of decrease of \(\theta\) is proportional to the difference between \(\theta\) and the room temperature. With \(k > 0\), which equation models this?
- A\(\dfrac{d\theta}{dt} = -k(\theta - 20)\)
- B\(\dfrac{d\theta}{dt} = k(\theta - 20)\)
- C\(\dfrac{d\theta}{dt} = -k\theta + 20\)
- D\(\dfrac{d\theta}{dt} = -k(\theta + 20)\)
The gradient of a curve at any point \((x, y)\) is inversely proportional to the square of the \(y\)-coordinate. Which equation is correct?
- A\(\dfrac{dy}{dx} = \dfrac{k}{y^{2}}\)
- B\(\dfrac{dy}{dx} = ky^{2}\)
- C\(\dfrac{dy}{dx} = \dfrac{k}{x^{2}}\)
- D\(\dfrac{dy}{dx} = \dfrac{k}{y}\)
The number \(N\) of bacteria increases at a rate proportional to \(N\). When \(N = 200\), the rate of increase is 30 per hour. Find the differential equation, with \(t\) in hours.
- A\(\dfrac{dN}{dt} = 0.15N\)
- B\(\dfrac{dN}{dt} = 6000N\)
- C\(\dfrac{dN}{dt} = \dfrac{20}{3}N\)
- D\(\dfrac{dN}{dt} = 0.15\)
A spherical balloon is inflated so that its volume increases at a constant rate of \(50\ \text{cm}^{3}\) per second. Given \(V = \dfrac{4}{3}\pi r^{3}\), find a differential equation for the radius \(r\) cm at time \(t\) seconds.
- A\(\dfrac{dr}{dt} = \dfrac{25}{2\pi r^{2}}\)
- B\(\dfrac{dr}{dt} = 200\pi r^{2}\)
- C\(\dfrac{dr}{dt} = \dfrac{75}{2\pi r^{3}}\)
- D\(\dfrac{dr}{dt} = 50\)
Water leaks from a tank so that the depth \(h\) m decreases at a rate proportional to \(\sqrt{h}\). When \(h = 4\), the depth is decreasing at \(0.1\) m per hour. Find the differential equation, with \(t\) in hours.
- A\(\dfrac{dh}{dt} = -0.05\sqrt{h}\)
- B\(\dfrac{dh}{dt} = -0.025\sqrt{h}\)
- C\(\dfrac{dh}{dt} = -0.2\sqrt{h}\)
- D\(\dfrac{dh}{dt} = 0.05\sqrt{h}\)
In a fish farm the population \(P\) would grow at a rate proportional to \(P\), but 100 fish are removed each year at a steady rate. With \(k > 0\) and \(t\) in years, which model fits?
- A\(\dfrac{dP}{dt} = kP - 100\)
- B\(\dfrac{dP}{dt} = k(P - 100)\)
- C\(\dfrac{dP}{dt} = kP + 100\)
- D\(\dfrac{dP}{dt} = -kP - 100\)
Quiz key
\(\dfrac{dx}{dt} = kx^{2}\), with \(k > 0\) since \(x\) increases. The tempting \(kt^{2}\) makes the rate depend on time, but the statement says it depends on the quantity \(x\) itself. \(x^{2} + k\) adds the constant instead of multiplying, and \(kx\) is proportional to \(x\), not \(x^{2}\).
\(\dfrac{dV}{dt} = -k\sqrt{V}\): the minus sign makes the rate negative while \(k > 0\). The tempting \(k\sqrt{V}\) describes an INCREASING volume when \(k > 0\). \(-\dfrac{k}{\sqrt{V}}\) is inversely proportional to \(\sqrt{V}\), and \(-kt\) makes the rate depend on time instead of on \(V\).
\(\dfrac{d\theta}{dt} = -k(\theta - 20)\): while \(\theta > 20\), the right-hand side is negative, so the tea cools, and the rate drops to 0 as \(\theta\) approaches 20. The tempting \(k(\theta - 20)\) makes a hot cup get hotter. \(-k\theta + 20\) applies \(k\) to \(\theta\) only, and \(\theta + 20\) is a sum, not the difference.
\(\dfrac{dy}{dx} = \dfrac{k}{y^{2}}\). The tempting \(ky^{2}\) is DIRECTLY proportional to \(y^{2}\). \(\dfrac{k}{x^{2}}\) uses the wrong coordinate, and \(\dfrac{k}{y}\) forgets the square.
\(30 = k \times 200\) gives \(k = 0.15\), so \(\dfrac{dN}{dt} = 0.15N\). The tempting \(6000N\) multiplies 30 by 200 instead of dividing. \(\dfrac{20}{3}N\) divides the wrong way round (\(\dfrac{200}{30}\)), and \(\dfrac{dN}{dt} = 0.15\) loses the \(N\), turning the model into a constant rate.
\(\dfrac{dV}{dr} = 4\pi r^{2}\), so \(50 = 4\pi r^{2}\dfrac{dr}{dt}\) and \(\dfrac{dr}{dt} = \dfrac{50}{4\pi r^{2}} = \dfrac{25}{2\pi r^{2}}\). The tempting \(200\pi r^{2}\) multiplies by \(\dfrac{dV}{dr}\) instead of dividing. \(\dfrac{75}{2\pi r^{3}}\) divides 50 by \(V\) itself rather than by \(\dfrac{dV}{dr}\), and \(\dfrac{dr}{dt} = 50\) confuses the rate for the volume with the rate for the radius.
\(-0.1 = -k\sqrt{4} = -2k\), so \(k = 0.05\) and \(\dfrac{dh}{dt} = -0.05\sqrt{h}\). The tempting \(-0.025\sqrt{h}\) divides by 4 instead of by \(\sqrt{4}\). \(-0.2\sqrt{h}\) multiplies by 2 instead of dividing, and \(+0.05\sqrt{h}\) loses the sign that shows the depth is decreasing.
Net rate = growth \(-\) removal \(= kP - 100\). The tempting \(k(P - 100)\) multiplies the removal by \(k\) too, so the fish removed per year would be \(100k\), not 100. \(kP + 100\) adds fish instead of removing them, and \(-kP - 100\) turns natural growth into decline.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.