The shape of the lesson
By the end of the lesson, learners can
- From chords to a tangent: In the lab, fix a point on the curve and drag the second point of the chord towards it.
- The derivative and its notation: Doing the chord calculation at a general point \(x\) gives a formula for the gradient at every point: the derivative.
- Using the gradient: Once you have \(\dfrac{dy}{dx}\), you can read off the gradient at any point, or run it backwards: set \(\dfrac{dy}{dx}\) equal to a value and solve for \(x\).
- Chord: gradient \(= \dfrac{f(a+h) - f(a)}{h}\); simplify, cancel \(h\), then let \(h \to 0\).
- Power rule: \(x^{n} \to nx^{n-1}\); constants vanish; work term by term.
- Gradient at a point: substitute \(x\) into \(\dfrac{dy}{dx}\).
- Point with a given gradient: solve \(\dfrac{dy}{dx} = m\), then use the curve for \(y\).
- Second derivative: differentiate \(\dfrac{dy}{dx}\) again.
One picture every idea returns to
The gradient of a curve at a point is the gradient of the tangent there, the limit of the gradients of chords as the second point slides onto the first. For \(y = x^{n}\), \(\dfrac{dy}{dx} = nx^{n-1}\): bring the power down, then reduce the power by one.
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, Chords that shrink to a tangent.. A curve has no single gradient, but at each point it has one: the gradient of the tangent there. 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: From chords to a tangent | In the lab, fix a point on the curve and drag the second point of the chord towards it. Project tab 4 of the lab (Functions and differentiation lab, gradient of a curve) and let a student drive it while the class predicts. Model the worked example Chords on a quadratic on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The derivative and its notation | Doing the chord calculation at a general point \(x\) gives a formula for the gradient at every point: the derivative. Model the worked example Term by term on the board, one line at a time. Your turn: Find the gradient of the curve \(y = x^{3} - 4x^{2} + 1\) at the point where \(x = 3\). | Pairs |
| 21 to 30 min | Teach and check: Using the gradient | Once you have \(\dfrac{dy}{dx}\), you can read off the gradient at any point, or run it backwards: set \(\dfrac{dy}{dx}\) equal to a value and solve for \(x\). Model the worked example Where is the gradient a given value? on the board, one line at a time. | Connect-Extend-Challenge |
| 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, 20 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 Using the gradient, 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.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
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 |
|---|---|
| Do not put \(h = 0\) into \(\dfrac{12h + 6h^{2} + h^{3}}{h}\) straight away: that gives \(\dfrac{0}{0}\), which means nothing. | Cancel the \(h\) first, THEN let \(h \to 0\). |
| \(\dfrac{d^{2}y}{dx^{2}}\) is NOT \(\left(\dfrac{dy}{dx}\right)^{2}\). | For \(y = x^{3}\), \(\dfrac{d^{2}y}{dx^{2}} = 6x\), but \(\left(\dfrac{dy}{dx}\right)^{2} = 9x^{4}\). |
| After solving \(\dfrac{dy}{dx} = 0\) for \(x\), substitute into the equation of the CURVE to get \(y\). | Putting \(x\) back into \(\dfrac{dy}{dx}\) just gives 0 again. |
The gradient of a curve: worksheet
6 questions, 20 marks. Written in the style of the papers; no past paper question is reproduced.
The points with \(x\)-coordinates \(2\) and \(2 + h\) lie on the curve \(y = x^{2} + x\).
Show that the gradient of the chord joining them is \(5 + h\), and state the gradient of the curve at \(x = 2\).
Find \(\dfrac{dy}{dx}\) and \(\dfrac{d^{2}y}{dx^{2}}\) for \(y = 3x^{4} - 2x^{3} + x - 7\).
\(f(x) = x^{3} - 4x^{2} + 2\). Find \(f'(3)\) and \(f''(3)\).
The curve \(y = x^{3} + kx^{2} - 4\), where \(k\) is a constant, has gradient \(4\) at the point where \(x = 2\). Find \(k\).
Find the coordinates of the points on the curve \(y = 2x^{3} - 3x^{2} - 12x + 1\) at which the gradient is zero.
A particle moves along a straight line. Its displacement after \(t\) seconds is \(s = t^{3} - 6t^{2} + 15t\) metres. Its velocity is \(\dfrac{ds}{dt}\) and its acceleration is \(\dfrac{d^{2}s}{dt^{2}}\).
(a) Find the velocity when \(t = 1\).
(b) Find the velocity at the instant when the acceleration is zero.
(c) Show that the particle is never at rest.
Answer key and marking notes
- At \(x = 2\): \(y = 6\). At \(x = 2 + h\): \(y = 4 + 4h + h^{2} + 2 + h = 6 + 5h + h^{2}\).
- Gradient \(= \dfrac{5h + h^{2}}{h} = 5 + h\).
- As \(h \to 0\), the gradient of the curve at \(x = 2\) is \(5\).
- \(\dfrac{dy}{dx} = 12x^{3} - 6x^{2} + 1\)
- \(\dfrac{d^{2}y}{dx^{2}} = 36x^{2} - 12x\)
- \(f'(x) = 3x^{2} - 8x\), so \(f'(3) = 27 - 24 = 3\).
- \(f''(x) = 6x - 8\), so \(f''(3) = 18 - 8 = 10\).
- \(\dfrac{dy}{dx} = 3x^{2} + 2kx\).
- At \(x = 2\): \(12 + 4k = 4\).
- \(k = -2\).
- \(\dfrac{dy}{dx} = 6x^{2} - 6x - 12 = 6(x^{2} - x - 2) = 6(x - 2)(x + 1)\).
- Zero when \(x = 2\) or \(x = -1\).
- \(x = 2\): \(y = 16 - 12 - 24 + 1 = -19\). \(\ x = -1\): \(y = -2 - 3 + 12 + 1 = 8\).
- Points \((2, -19)\) and \((-1, 8)\).
- (a) \(\dfrac{ds}{dt} = 3t^{2} - 12t + 15\); at \(t = 1\): \(3 - 12 + 15 = 6\ \text{m s}^{-1}\).
- (b) \(\dfrac{d^{2}s}{dt^{2}} = 6t - 12 = 0\) when \(t = 2\); velocity \(= 12 - 24 + 15 = 3\ \text{m s}^{-1}\).
- (c) At rest needs \(3t^{2} - 12t + 15 = 0\), i.e. \(t^{2} - 4t + 5 = 0\).
- Discriminant \(= 16 - 20 = -4 < 0\), so there are no real roots: the velocity is never zero.
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 points with \(x\)-coordinates \(1\) and \(1 + h\) lie on the curve \(y = x^{2} + 3x\). Find the gradient of the chord joining them, in terms of \(h\).
- A\(5 + h\)
- B\(5\)
- C\(5h + h^{2}\)
- D\(2 + h\)
Find \(\dfrac{dy}{dx}\) when \(y = 2x^{5} - 3x^{2} + 4\).
- A\(10x^{4} - 6x\)
- B\(10x^{4} - 6x + 4\)
- C\(10x^{6} - 6x^{3}\)
- D\(2x^{4} - 3x\)
\(f(x) = x^{3} - 5x^{2} + 2x\). Find \(f'(2)\).
- A\(-6\)
- B\(-8\)
- C\(2\)
- D\(4\)
For a curve \(y = f(x)\), what does \(f'(a)\) represent?
- AThe gradient of the tangent to the curve at the point where \(x = a\)
- BThe \(y\)-coordinate of the point on the curve where \(x = a\)
- CThe gradient of the chord from the origin to the point where \(x = a\)
- DThe square of the gradient of the curve at \(x = a\)
The curve \(y = ax^{3} + bx\), where \(a\) and \(b\) are constants, has gradient \(5\) at the point where \(x = 1\) and gradient \(23\) at the point where \(x = 2\). Find \(a\) and \(b\).
- A\(a = 2,\ b = -1\)
- B\(a = 6,\ b = -1\)
- C\(a = \dfrac{13}{6},\ b = \dfrac{17}{6}\)
- D\(a = 2,\ b = 1\)
Find the coordinates of the points on the curve \(y = x^{3} - 3x^{2} + 1\) at which the gradient is \(9\).
- A\((3, 1)\) and \((-1, -3)\)
- B\((3, 9)\) and \((-1, 9)\)
- C\((-3, -53)\) and \((1, -1)\)
- D\((0, 1)\) and \((2, -3)\)
The points \(P\) and \(Q\) on the curve \(y = x^{3} - 2x\) have \(x\)-coordinates \(2\) and \(2 + h\). Find the gradient of the chord \(PQ\) when \(h = 0.1\).
- A\(10.61\)
- B\(10\)
- C\(1.061\)
- D\(10.6\)
A particle moves in a straight line so that its displacement after \(t\) seconds is \(s = 2t^{3} - 9t^{2} + 12t\) metres. The velocity is \(\dfrac{ds}{dt}\) and the acceleration is \(\dfrac{d^{2}s}{dt^{2}}\). The particle is at rest at two instants. Find its acceleration at the later of these instants.
- A\(6\ \text{m s}^{-2}\)
- B\(-6\ \text{m s}^{-2}\)
- C\(0\ \text{m s}^{-2}\)
- D\(4\ \text{m s}^{-2}\)
Quiz key
\(f(1) = 4\) and \(f(1 + h) = 1 + 2h + h^{2} + 3 + 3h = 4 + 5h + h^{2}\). Rise \(= 5h + h^{2}\), run \(= h\), so the gradient is \(5 + h\). \(5h + h^{2}\) is the rise only (no division by \(h\)); \(5\) is the limit as \(h \to 0\), the gradient of the tangent, not of the chord; \(2 + h\) ignores the \(3x\) term.
\(2x^{5} \to 10x^{4}\), \(-3x^{2} \to -6x\), and the constant \(4 \to 0\). So \(\dfrac{dy}{dx} = 10x^{4} - 6x\). Keeping the \(+4\) forgets that a constant has zero gradient; \(10x^{6} - 6x^{3}\) raises the powers instead of reducing them; \(2x^{4} - 3x\) reduces the powers but never brings them down as multipliers.
\(f'(x) = 3x^{2} - 10x + 2\), so \(f'(2) = 12 - 20 + 2 = -6\). \(-8\) is \(f(2)\), the \(y\)-value rather than the gradient; \(2\) is \(f''(2) = 6(2) - 10\); \(4\) comes from differentiating \(5x^{2}\) as \(5x\) instead of \(10x\).
\(f'(a)\) is the limit of the gradients of chords as the second point slides onto the point where \(x = a\): the gradient of the tangent there. The \(y\)-coordinate is \(f(a)\), not \(f'(a)\). A chord from the origin has gradient \(\dfrac{f(a)}{a}\), which is generally different. The dash means "differentiated", not "squared".
\(\dfrac{dy}{dx} = 3ax^{2} + b\). At \(x = 1\): \(3a + b = 5\). At \(x = 2\): \(12a + b = 23\). Subtract: \(9a = 18\), so \(a = 2\) and \(b = -1\). \(a = \dfrac{13}{6}, b = \dfrac{17}{6}\) comes from substituting into \(y = ax^{3} + bx\) instead of the derivative; \(a = 6\) comes from differentiating \(ax^{3}\) as \(ax^{2}\) without bringing the 3 down.
\(\dfrac{dy}{dx} = 3x^{2} - 6x = 9\), so \(x^{2} - 2x - 3 = 0\), \((x - 3)(x + 1) = 0\), \(x = 3\) or \(x = -1\). From the curve: \(y(3) = 27 - 27 + 1 = 1\), \(y(-1) = -1 - 3 + 1 = -3\). \((3, 9)\) and \((-1, 9)\) put the gradient in as the \(y\)-coordinate; \((-3, -53)\) and \((1, -1)\) come from a sign slip in the factors; \((0, 1)\) and \((2, -3)\) are where the gradient is zero.
\(f(2 + h) - f(2) = (8 + 12h + 6h^{2} + h^{3} - 4 - 2h) - 4 = 10h + 6h^{2} + h^{3}\). Divide by \(h\): \(10 + 6h + h^{2}\). At \(h = 0.1\): \(10 + 0.6 + 0.01 = 10.61\). \(10\) is the limit, the gradient of the tangent at \(P\); \(1.061\) is the rise, not divided by \(h\); \(10.6\) drops the \(h^{2}\) term from the expansion.
\(\dfrac{ds}{dt} = 6t^{2} - 18t + 12 = 6(t - 1)(t - 2)\), so the particle is at rest at \(t = 1\) and \(t = 2\). \(\dfrac{d^{2}s}{dt^{2}} = 12t - 18\); at \(t = 2\) this is \(24 - 18 = 6\ \text{m s}^{-2}\). \(-6\) is the acceleration at the earlier instant \(t = 1\); \(4\) is the displacement \(s\) at \(t = 2\); \(0\) assumes that zero velocity means zero acceleration.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
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.