The shape of the lesson
By the end of the lesson, learners can
- Tangents and normals: Three steps every time: find the point (both coordinates), find the gradient by substituting into \(\dfrac{dy}{dx}\), then use \(y - y_{1} = m(x - x_{1})\).
- Increasing and decreasing: A function is increasing where its gradient is positive and decreasing where its gradient is negative.
- Connected rates of change: When two quantities both change with time, link their rates with the chain rule.
- Point: find both coordinates from the curve.
- Gradient: substitute \(x\) into \(\dfrac{dy}{dx}\); normal uses \(-\dfrac{1}{m}\).
- Line: \(y - y_{1} = m(x - x_{1})\); give it in the form asked.
- Increasing/decreasing: factorise \(f'(x)\), sketch it, read off where it is positive or negative.
- Connected rates: know one rate, want another, link with a formula, differentiate, then multiply or divide.
One picture every idea returns to
At a point where the gradient is \(m\), the tangent has gradient \(m\) and the normal has gradient \(-\dfrac{1}{m}\). \(f\) is increasing where \(f'(x) > 0\) and decreasing where \(f'(x) < 0\). Connected rates: \(\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt}\).
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, Gradients at work.. The derivative gives the gradient of the tangent, so it gives the tangent and the normal too. 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: Tangents and normals | Three steps every time: find the point (both coordinates), find the gradient by substituting into \(\dfrac{dy}{dx}\), then use \(y - y_{1} = m(x - x_{1})\). Model the worked example Tangent and normal to a cubic on the board, one line at a time. Your turn: Find the equation of the tangent to \(y = \dfrac{1}{x}\) at the point \(\left(2, \tfrac{1}{2}\right)\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: Increasing and decreasing | A function is increasing where its gradient is positive and decreasing where its gradient is negative. Model the worked example Where is a cubic decreasing? on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Connected rates of change | When two quantities both change with time, link their rates with the chain rule. Model the worked example A spreading circle on the board, one line at a time. | Numbered Heads Together |
| 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, 22 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 Connected rates of change, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| The normal gradient is \(-\dfrac{1}{m}\): flip AND change sign. | With \(m = 4\), \(-4\) and \(\tfrac{1}{4}\) are both wrong. And never use \(\dfrac{dy}{dx}\) as a formula in the line equation: substitute to get a number first. |
| For a positive quadratic \(f'(x)\), the negative part is between the roots, not outside them. | Sketch the parabola of \(f'(x)\) before writing the inequality. |
| Differentiate the formula BEFORE you substitute the particular value. | Putting \(r = 5\) into \(V = \tfrac{4}{3}\pi r^{3}\) first turns it into a constant, whose derivative is 0. |
Tangents, normals and rates of change: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
Find the equation of the tangent to the curve \(y = \dfrac{8}{x^{2}}\) at the point where \(x = 2\).
Find the equation of the normal to the curve \(y = x^{2} - 5x + 6\) at the point where \(x = 4\). Give your answer in the form \(ax + by = c\).
\(f(x) = 2x^{3} + 3x^{2} - 12x + 5\). Find the set of values of \(x\) for which \(f\) is increasing.
Show that \(f(x) = x^{3} + 3x^{2} + 5x - 2\) is an increasing function.
The edges of a metal cube expand at \(0.2\ \text{cm s}^{-1}\) when it is heated. Find, at the instant when the edge length is \(5\) cm, the rate of increase of (a) the volume, (b) the total surface area.
Water is poured into an inverted conical tank at a constant rate of \(30\ \text{cm}^{3}\,\text{s}^{-1}\). When the depth of water is \(h\) cm, the volume of water is \(V = \tfrac{1}{3}\pi h^{3}\ \text{cm}^{3}\).
(a) Find the rate at which the depth is increasing when \(h = 6\).
(b) Find the depth of water at the instant when the depth is increasing at \(0.15\ \text{cm s}^{-1}\).
Answer key and marking notes
- Point: \(y = \dfrac{8}{4} = 2\), so \((2, 2)\).
- \(\dfrac{dy}{dx} = -16x^{-3} = -\dfrac{16}{8} = -2\) at \(x = 2\).
- \(y - 2 = -2(x - 2)\), so \(y = -2x + 6\).
- Point: \(y = 16 - 20 + 6 = 2\), so \((4, 2)\).
- \(\dfrac{dy}{dx} = 2x - 5 = 3\); normal gradient \(-\tfrac{1}{3}\).
- \(y - 2 = -\tfrac{1}{3}(x - 4)\), so \(3y - 6 = -x + 4\).
- \(x + 3y = 10\)
- \(f'(x) = 6x^{2} + 6x - 12 = 6(x + 2)(x - 1)\).
- Positive quadratic, so \(f'(x) > 0\) outside the roots.
- \(x < -2\) or \(x > 1\)
- \(f'(x) = 3x^{2} + 6x + 5\).
- \(= 3(x + 1)^{2} + 2\).
- Since \(3(x + 1)^{2} \ge 0\), \(f'(x) \ge 2 > 0\) for all \(x\), so \(f\) is increasing.
- (a) \(V = x^{3}\), \(\dfrac{dV}{dx} = 3x^{2} = 75\). \(\dfrac{dV}{dt} = 75 \times 0.2 = 15\ \text{cm}^{3}\,\text{s}^{-1}\).
- (b) \(S = 6x^{2}\), \(\dfrac{dS}{dx} = 12x = 60\). \(\dfrac{dS}{dt} = 60 \times 0.2 = 12\ \text{cm}^{2}\,\text{s}^{-1}\).
- (a) \(\dfrac{dV}{dh} = \pi h^{2} = 36\pi\) when \(h = 6\).
- \(\dfrac{dh}{dt} = \dfrac{30}{36\pi} = \dfrac{5}{6\pi} = 0.265\ \text{cm s}^{-1}\) (3 s.f.).
- (b) \(\dfrac{dh}{dt} = \dfrac{30}{\pi h^{2}} = 0.15\), so \(\pi h^{2} = 200\).
- \(h = \sqrt{\dfrac{200}{\pi}} = 7.98\) cm (3 s.f.).
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 tangent to a curve at the point \(P\) has gradient \(-\dfrac{2}{3}\). What is the gradient of the normal at \(P\)?
- A\(\dfrac{3}{2}\)
- B\(-\dfrac{3}{2}\)
- C\(\dfrac{2}{3}\)
- D\(-\dfrac{2}{3}\)
Find the equation of the tangent to the curve \(y = x^{3} - 4x\) at the point where \(x = 1\).
- A\(y = -x - 2\)
- B\(y = x - 4\)
- C\(y = -x + 4\)
- D\(y = -x - 4\)
\(f(x) = x^{3} - 12x + 2\). Find the set of values of \(x\) for which \(f\) is decreasing.
- A\(-2 < x < 2\)
- B\(x < -2\) or \(x > 2\)
- C\(x < 2\)
- D\(-4 < x < 4\)
The side of a square is increasing at \(0.2\ \text{cm s}^{-1}\). Find the rate of increase of the area of the square, in \(\text{cm}^{2}\,\text{s}^{-1}\), at the instant when the side is \(5\) cm.
- A\(2\)
- B\(10\)
- C\(50\)
- D\(5\)
The normal to the curve \(y = \dfrac{8}{x}\) at the point \((2, 4)\) meets the \(x\)-axis at \(A\). Find the \(x\)-coordinate of \(A\).
- A\(-6\)
- B\(4\)
- C\(10\)
- D\(0\)
The tangent to the curve \(y = x^{2} - 6x + 10\) at the point \(P\) is parallel to the line \(y = 2x + 1\). Find the coordinates of \(P\).
- A\((4, 2)\)
- B\((4, 9)\)
- C\((3, 1)\)
- D\((2, 2)\)
The area of a circle is increasing at \(12\ \text{cm}^{2}\,\text{s}^{-1}\). Find the exact rate of increase of the radius, in \(\text{cm s}^{-1}\), when the radius is \(3\) cm.
- A\(\dfrac{2}{\pi}\)
- B\(72\pi\)
- C\(\dfrac{4}{3\pi}\)
- D\(2\)
The volume of a cube is increasing at \(6\ \text{cm}^{3}\,\text{s}^{-1}\). Find the rate of increase of its total surface area, in \(\text{cm}^{2}\,\text{s}^{-1}\), at the instant when the edge is \(3\) cm.
- A\(8\)
- B\(216\)
- C\(4.5\)
- D\(\dfrac{2}{9}\)
Quiz key
Normal gradient \(= -\dfrac{1}{m} = -\dfrac{1}{-\frac{2}{3}} = \dfrac{3}{2}\). Check: \(-\dfrac{2}{3} \times \dfrac{3}{2} = -1\). \(-\dfrac{3}{2}\) flips the fraction but forgets to change the sign; \(\dfrac{2}{3}\) changes the sign but forgets to flip; \(-\dfrac{2}{3}\) is the tangent's own gradient.
At \(x = 1\): \(y = 1 - 4 = -3\), and \(\dfrac{dy}{dx} = 3x^{2} - 4 = -1\). Tangent: \(y + 3 = -1(x - 1)\), so \(y = -x - 2\). \(y = x - 4\) is the normal (gradient \(+1\)); \(y = -x + 4\) uses \(y_{1} = +3\); \(y = -x - 4\) uses \(x_{1} = -1\).
\(f'(x) = 3x^{2} - 12 = 3(x - 2)(x + 2)\). This U-shaped quadratic is negative between its roots, so \(f\) is decreasing for \(-2 < x < 2\). "\(x < -2\) or \(x > 2\)" is where \(f\) is increasing; \(x < 2\) takes only one root; \(\pm 4\) solves \(x^{2} = 4\) as \(x = \pm 4\).
\(\dfrac{dA}{dx} = 2x = 10\) when \(x = 5\). \(\dfrac{dA}{dt} = 10 \times 0.2 = 2\ \text{cm}^{2}\,\text{s}^{-1}\). \(10\) is \(\dfrac{dA}{dx}\) only; \(50\) divides by \(0.2\) instead of multiplying; \(5\) multiplies the area \(25\) by \(0.2\) instead of the derivative.
\(\dfrac{dy}{dx} = -\dfrac{8}{x^{2}} = -2\) at \(x = 2\), so the normal has gradient \(\tfrac{1}{2}\). Normal: \(y - 4 = \tfrac{1}{2}(x - 2)\). At \(y = 0\): \(-8 = x - 2\), \(x = -6\). \(4\) comes from using the tangent (gradient \(-2\)); \(10\) from a normal gradient of \(-\tfrac{1}{2}\) (sign not changed); \(0\) from a normal gradient of \(2\) (reciprocal not taken).
\(\dfrac{dy}{dx} = 2x - 6 = 2\) gives \(x = 4\); on the curve \(y = 16 - 24 + 10 = 2\), so \(P(4, 2)\). \((4, 9)\) takes \(y\) from the line, not the curve; \((3, 1)\) is where the gradient is zero; \((2, 2)\) solves \(2x - 6 = -2\), a sign slip.
\(\dfrac{dA}{dr} = 2\pi r = 6\pi\). So \(12 = 6\pi \times \dfrac{dr}{dt}\), giving \(\dfrac{dr}{dt} = \dfrac{2}{\pi} \approx 0.637\ \text{cm s}^{-1}\). \(72\pi\) multiplies \(12\) by \(6\pi\) instead of dividing; \(\dfrac{4}{3\pi}\) divides by the area \(9\pi\) instead of by \(\dfrac{dA}{dr}\); \(2\) loses the \(\pi\).
\(\dfrac{dV}{dx} = 3x^{2} = 27\), so \(\dfrac{dx}{dt} = \dfrac{6}{27} = \dfrac{2}{9}\). \(\dfrac{dS}{dx} = 12x = 36\), so \(\dfrac{dS}{dt} = 36 \times \dfrac{2}{9} = 8\ \text{cm}^{2}\,\text{s}^{-1}\). \(\dfrac{2}{9}\) is \(\dfrac{dx}{dt}\), only half way; \(216 = 6 \times 36\) multiplies \(\dfrac{dV}{dt}\) by \(\dfrac{dS}{dx}\) with no link through \(x\); \(4.5\) inverts the ratio, \(\dfrac{6 \times 27}{36}\).
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.