The shape of the lesson
By the end of the lesson, learners can
- Parametric curves: When \(x\) and \(y\) both depend on a parameter \(t\), differentiate each with respect to \(t\) and divide.
- Implicit differentiation: For an equation such as \(x^2 + y^2 = 25\) (Fig. 1), differentiate both sides with respect to \(x\).
- Tangents and stationary points: Stationary points (horizontal tangents) occur where the numerator of \(\dfrac{dy}{dx}\) is zero.
- Parametric: \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\); find \(t\) at the point first.
- Implicit: differentiate every term with respect to \(x\); each \(y\) term gains \(\dfrac{dy}{dx}\).
- Products such as \(xy\) need the product rule.
- Collect the \(\dfrac{dy}{dx}\) terms on one side and factorise.
- Stationary points: numerator zero, then substitute back into the curve.
One picture every idea returns to
Parametric: \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\). Implicit: differentiate every term with respect to \(x\), treating \(y\) as a function of \(x\), so \(\dfrac{d}{dx}y^{n} = ny^{n-1}\dfrac{dy}{dx}\) and \(\dfrac{d}{dx}(xy) = x\dfrac{dy}{dx} + y\); then make \(\dfrac{dy}{dx}\) the subject.
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 without y = f(x).. Papers 2 and 3. 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: Parametric curves | When \(x\) and \(y\) both depend on a parameter \(t\), differentiate each with respect to \(t\) and divide. Model the worked example A tangent on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Implicit differentiation | For an equation such as \(x^2 + y^2 = 25\) (Fig. 1), differentiate both sides with respect to \(x\). Model the worked example A circle on the board, one line at a time. Your turn: Find the gradient of the curve \(x^3 + y^3 = 9\) at the point \((1, 2)\). | Pairs |
| 21 to 30 min | Teach and check: Tangents and stationary points | Stationary points (horizontal tangents) occur where the numerator of \(\dfrac{dy}{dx}\) is zero. Model the worked example Horizontal tangents on the board, one line at a time. | Round Table |
| 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, 25 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 Tangents and stationary points, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 formula is \(\dfrac{dy/dt}{dx/dt}\), y-derivative on TOP. | Dividing the other way gives \(\dfrac{dx}{dy}\), the reciprocal of the gradient. |
| \(\dfrac{d}{dx}(3xy)\) is NOT \(3\dfrac{dy}{dx}\) or \(3y\). | It is a product: \(3x\dfrac{dy}{dx} + 3y\). Missing one of the two terms is the most common error in implicit questions. |
| Do not stop at \(y = 2x\). | That is a line, not a point; you must substitute into the ORIGINAL curve to find where the line meets it. |
Parametric and implicit differentiation: worksheet
6 questions, 25 marks. Written in the style of the papers; no past paper question is reproduced.
A curve has parametric equations \(x = 4t\), \(y = \dfrac{2}{t}\). Find \(\dfrac{dy}{dx}\) in terms of \(t\).
A curve has parametric equations \(x = \theta - \sin\theta\), \(y = 1 - \cos\theta\). Find the equation of the tangent at the point where \(\theta = \dfrac{\pi}{2}\).
The curve \(y^2 + xy = 6\) passes through the point \((1, 2)\). Find the equation of the normal at this point, in the form \(ax + by = c\).
The curve \(xe^{y} + y = 2\) passes through \((2, 0)\). Find the gradient of the curve at this point.
Find the coordinates of the points on the curve \(x^2 + 2xy + 3y^2 = 18\) at which the tangent is parallel to the \(x\)-axis.
A curve has parametric equations \(x = t^2 + 2t\), \(y = t^3 - 12t\).
(a) Find \(\dfrac{dy}{dx}\) in terms of \(t\).
(b) Find the coordinates of the stationary points of the curve.
(c) Find the equation of the tangent at the point where \(t = 1\), in the form \(ax + by + c = 0\).
Answer key and marking notes
- \(\dfrac{dx}{dt} = 4\), \(\dfrac{dy}{dt} = -\dfrac{2}{t^2}\).
- \(\dfrac{dy}{dx} = -\dfrac{2}{t^2} \div 4 = -\dfrac{1}{2t^2}\).
- \(\dfrac{dx}{d\theta} = 1 - \cos\theta\), \(\dfrac{dy}{d\theta} = \sin\theta\), so \(\dfrac{dy}{dx} = \dfrac{\sin\theta}{1 - \cos\theta}\).
- At \(\theta = \dfrac{\pi}{2}\): gradient \(= \dfrac{1}{1 - 0} = 1\); point \(\left(\dfrac{\pi}{2} - 1, 1\right)\).
- \(y - 1 = x - \dfrac{\pi}{2} + 1\).
- \(y = x + 2 - \dfrac{\pi}{2}\).
- \(2y\dfrac{dy}{dx} + x\dfrac{dy}{dx} + y = 0\), so \(\dfrac{dy}{dx} = -\dfrac{y}{2y + x}\).
- At \((1, 2)\): \(\dfrac{dy}{dx} = -\dfrac{2}{5}\), so the normal gradient is \(\dfrac{5}{2}\).
- \(y - 2 = \dfrac{5}{2}(x - 1)\), so \(2y - 4 = 5x - 5\).
- \(5x - 2y = 1\).
- \(e^{y} + xe^{y}\dfrac{dy}{dx} + \dfrac{dy}{dx} = 0\) (product rule on \(xe^{y}\)).
- \(\dfrac{dy}{dx} = -\dfrac{e^{y}}{xe^{y} + 1}\).
- At \((2, 0)\): \(-\dfrac{1}{2 + 1} = -\dfrac{1}{3}\).
- \(2x + 2x\dfrac{dy}{dx} + 2y + 6y\dfrac{dy}{dx} = 0\), so \(\dfrac{dy}{dx} = -\dfrac{x + y}{x + 3y}\).
- Horizontal tangent: \(x + y = 0\), so \(y = -x\).
- Substitute: \(x^2 - 2x^2 + 3x^2 = 18\), so \(2x^2 = 18\) and \(x = \pm 3\).
- Points \((3, -3)\) and \((-3, 3)\) (the denominator \(x + 3y = \mp 6 \ne 0\)).
- (a) \(\dfrac{dy}{dx} = \dfrac{3t^2 - 12}{2t + 2}\).
- (b) \(3t^2 - 12 = 0\), so \(t = \pm 2\) (the denominator is not zero there).
- \(t = 2\): \((8, -16)\). \(t = -2\): \((0, 16)\).
- (c) At \(t = 1\): point \((3, -11)\), gradient \(\dfrac{3 - 12}{4} = -\dfrac{9}{4}\).
- \(y + 11 = -\dfrac{9}{4}(x - 3)\), so \(4y + 44 = -9x + 27\).
- \(9x + 4y + 17 = 0\).
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.
A curve has parametric equations \(x = t^3\), \(y = t^2 + 4t\). Find the gradient of the curve at the point where \(t = 1\).
- A\(2\)
- B\(\dfrac{1}{2}\)
- C\(6\)
- D\(3\)
Differentiate \(y^3\) with respect to \(x\).
- A\(3y^2\dfrac{dy}{dx}\)
- B\(3y^2\)
- C\(3\left(\dfrac{dy}{dx}\right)^2\)
- D\(0\)
Find the gradient of the curve \(x^2 + 3y^2 = 21\) at the point \((3, -2)\).
- A\(\dfrac{1}{2}\)
- B\(-\dfrac{1}{2}\)
- C\(2\)
- D\(-2\)
A curve has parametric equations \(x = 3\cos t\), \(y = 2\sin t\). Find \(\dfrac{dy}{dx}\) in terms of \(t\).
- A\(-\dfrac{2}{3}\cot t\)
- B\(\dfrac{2}{3}\cot t\)
- C\(-\dfrac{3}{2}\tan t\)
- D\(-\dfrac{2}{3}\tan t\)
Find the gradient of the curve \(x^2 + xy + y^2 = 7\) at the point \((1, 2)\).
- A\(-\dfrac{4}{5}\)
- B\(-\dfrac{2}{5}\)
- C\(-1\)
- D\(-\dfrac{5}{4}\)
A curve has parametric equations \(x = t^2 - 3\), \(y = t^3\). Find the equation of the normal to the curve at the point where \(t = 2\).
- A\(x + 3y = 25\)
- B\(y = 3x + 5\)
- C\(3x + y = 11\)
- D\(x - 3y + 23 = 0\)
Find the points on the curve \(x^2 - xy + y^2 = 12\) at which the tangent is parallel to the \(x\)-axis.
- A\((2, 4)\) and \((-2, -4)\)
- B\((4, 2)\) and \((-4, -2)\)
- C\((2, 4)\) only
- D\((0, 2\sqrt{3})\) and \((0, -2\sqrt{3})\)
A curve has parametric equations \(x = \ln t\), \(y = t^2 - 2t\), for \(t > 0\). Find the coordinates of the stationary point.
- A\((0, -1)\)
- B\((1, -1)\)
- C\((e, e^2 - 2e)\)
- D\((\ln 2, 0)\)
Quiz key
\(\dfrac{dy}{dt} = 2t + 4 = 6\) and \(\dfrac{dx}{dt} = 3t^2 = 3\) at \(t = 1\), so \(\dfrac{dy}{dx} = \dfrac{6}{3} = 2\). The tempting \(\dfrac{1}{2}\) divides the other way, giving \(\dfrac{dx}{dy}\). 6 is only \(\dfrac{dy}{dt}\), and 3 is only \(\dfrac{dx}{dt}\).
\(\dfrac{d}{dx}y^3 = \dfrac{d}{dy}(y^3) \times \dfrac{dy}{dx} = 3y^2\dfrac{dy}{dx}\). The tempting \(3y^2\) differentiates with respect to \(y\) and forgets the chain rule factor. \(3\left(\dfrac{dy}{dx}\right)^2\) puts the derivative in place of \(y\), and 0 treats \(y\) as a constant, which it is not.
\(2x + 6y\dfrac{dy}{dx} = 0\), so \(\dfrac{dy}{dx} = -\dfrac{x}{3y} = -\dfrac{3}{3(-2)} = \dfrac{1}{2}\). The tempting \(-\dfrac{1}{2}\) loses the sign of \(y = -2\) when substituting. 2 and \(-2\) are the reciprocals, from making \(\dfrac{dx}{dy}\) the subject or writing \(-\dfrac{3y}{x}\).
\(\dfrac{dx}{dt} = -3\sin t\) and \(\dfrac{dy}{dt} = 2\cos t\), so \(\dfrac{dy}{dx} = \dfrac{2\cos t}{-3\sin t} = -\dfrac{2}{3}\cot t\). The tempting \(\dfrac{2}{3}\cot t\) forgets that \(\dfrac{d}{dt}\cos t = -\sin t\). \(-\dfrac{3}{2}\tan t\) is \(\dfrac{dx}{dy}\), the reciprocal, and \(-\dfrac{2}{3}\tan t\) turns \(\dfrac{\cos t}{\sin t}\) into \(\tan t\) instead of \(\cot t\).
\(2x + x\dfrac{dy}{dx} + y + 2y\dfrac{dy}{dx} = 0\), so \(\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}\). At \((1, 2)\): \(-\dfrac{4}{5}\). The tempting \(-\dfrac{2}{5}\) loses the \(y\) from the product rule (\(\dfrac{d}{dx}(xy) = x\dfrac{dy}{dx}\) only). \(-1\) loses the other half (\(\dfrac{d}{dx}(xy) = y\) only), and \(-\dfrac{5}{4}\) is the reciprocal of the gradient.
At \(t = 2\): \(x = 1\), \(y = 8\). \(\dfrac{dy}{dx} = \dfrac{3t^2}{2t} = \dfrac{3t}{2} = 3\), so the normal has gradient \(-\dfrac{1}{3}\): \(y - 8 = -\dfrac{1}{3}(x - 1)\), that is \(x + 3y = 25\). The tempting \(y = 3x + 5\) is the tangent. \(3x + y = 11\) starts from the upside-down gradient \(\dfrac{1}{3}\) and so uses normal gradient \(-3\), and \(x - 3y + 23 = 0\) uses gradient \(+\dfrac{1}{3}\), forgetting the minus sign.
\(2x - x\dfrac{dy}{dx} - y + 2y\dfrac{dy}{dx} = 0\), so \(\dfrac{dy}{dx} = \dfrac{y - 2x}{2y - x}\). It is zero when \(y = 2x\). Substituting: \(x^2 - 2x^2 + 4x^2 = 3x^2 = 12\), so \(x = \pm 2\), giving \((2, 4)\) and \((-2, -4)\). The tempting \((4, 2)\) and \((-4, -2)\) set the DENOMINATOR to zero (\(x = 2y\)), which gives tangents parallel to the \(y\)-axis. \((2, 4)\) only takes just the positive root, and \((0, \pm 2\sqrt{3})\) are where the curve crosses the \(y\)-axis, not where \(\dfrac{dy}{dx} = 0\).
\(\dfrac{dy}{dx} = \dfrac{2t - 2}{1/t} = 2t(t - 1)\). This is zero when \(t = 1\) (\(t = 0\) is not allowed). At \(t = 1\): \(x = \ln 1 = 0\), \(y = 1 - 2 = -1\), so the point is \((0, -1)\). The tempting \((1, -1)\) writes the parameter value \(t = 1\) as the \(x\)-coordinate. \((e, e^2 - 2e)\) solves \(x = 1\) for \(t\), mixing up \(t\) and \(x\), and \((\ln 2, 0)\) is where the curve crosses the \(x\)-axis (\(y = 0\) at \(t = 2\)).
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.