Implicit and parametric differentiation - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 17 exam-style questions on Implicit and parametric differentiation for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

17 questions
3–10 marks each
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17 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

A curve \(C\) has parametric equations
\[x = 3\sin^2\theta, \quad y = 2\cos^3\theta, \quad 0 < \theta < \dfrac{\pi}{2}\]
Show that
\[\dfrac{dy}{dx} = -\cos\theta\]

Model answer

\(\dfrac{dx}{d\theta} = 3 \times 2\sin\theta\cos\theta = 6\sin\theta\cos\theta\)
\(\dfrac{dy}{d\theta} = 2 \times 3\cos^2\theta \times (-\sin\theta) = -6\cos^2\theta\sin\theta\)
\(\dfrac{dy}{dx} = \dfrac{-6\cos^2\theta\sin\theta}{6\sin\theta\cos\theta} = -\cos\theta\) as required

Explanation

Differentiate \(x\) and \(y\) with respect to \(\theta\) using the chain rule: \(\sin^2\theta\) gives \(2\sin\theta\cos\theta\) and \(\cos^3\theta\) gives \(-3\cos^2\theta\sin\theta\). Dividing \(\dfrac{dy}{d\theta}\) by \(\dfrac{dx}{d\theta}\) and cancelling \(6\sin\theta\cos\theta\), which is not zero for \(0 < \theta < \dfrac{\pi}{2}\), gives \(-\cos\theta\). Common errors are losing the minus sign when differentiating \(\cos\theta\) and dividing the derivatives the wrong way round.

Mark scheme

1 mark
M1 Uses the chain rule to differentiate both: \(\dfrac{dx}{d\theta} = k\sin\theta\cos\theta\) and \(\dfrac{dy}{d\theta} = \pm m\cos^2\theta\sin\theta\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: this is a show-that question, and restating the given result earns nothing. \(\dfrac{dx}{d\theta} = 3\sin 2\theta\) is equivalent to \(6\sin\theta\cos\theta\).
1 mark
M1 (dM1) Divides \(\dfrac{dy}{d\theta}\) by \(\dfrac{dx}{d\theta}\) the right way round: \(\dfrac{dy}{dx} = \dfrac{-6\cos^2\theta\sin\theta}{6\sin\theta\cos\theta}\)
Depends on the first M mark. Dividing the wrong way round is dM0.
1 mark
A1* \(\dfrac{dy}{dx} = -\cos\theta\) cso
Given answer: needs both derivatives correct (\(6\sin\theta\cos\theta\) and \(-6\cos^2\theta\sin\theta\)), the cancelling of \(6\sin\theta\cos\theta\) shown or clearly implied, no sign errors, and "\(\dfrac{dy}{dx} =\)" seen. A common error is \(\dfrac{dy}{d\theta} = 6\cos^2\theta\sin\theta\) (sign lost), which cannot lead correctly to the result: A0*.

Question 2

6 marks

The curve \(C\) has equation
\[y + \sin(xy) = \dfrac{\pi}{6} + \dfrac{\sqrt{3}}{2}\]
The point \(P\left(2, \dfrac{\pi}{6}\right)\) lies on \(C\).
Find an equation of the tangent to \(C\) at \(P\), giving your answer in the form \(y = mx + c\), where \(m\) and \(c\) are exact constants.

Model answer

Differentiating with respect to \(x\):
\(\dfrac{dy}{dx} + \cos(xy)\left(y + x\dfrac{dy}{dx}\right) = 0\)
At \(P\): \(xy = 2 \times \dfrac{\pi}{6} = \dfrac{\pi}{3}\), so \(\cos(xy) = \dfrac{1}{2}\)
\(\dfrac{dy}{dx} + \dfrac{1}{2}\left(\dfrac{\pi}{6} + 2\dfrac{dy}{dx}\right) = 0\)
\(2\dfrac{dy}{dx} + \dfrac{\pi}{12} = 0\), so \(\dfrac{dy}{dx} = -\dfrac{\pi}{24}\)
Tangent: \(y - \dfrac{\pi}{6} = -\dfrac{\pi}{24}(x - 2)\)
\(y = -\dfrac{\pi}{24}x + \dfrac{\pi}{12} + \dfrac{\pi}{6}\)
\(y = -\dfrac{\pi}{24}x + \dfrac{\pi}{4}\)

Explanation

Differentiating \(\sin(xy)\) needs the chain rule and the product rule together: it becomes \(\cos(xy)\left(y + x\dfrac{dy}{dx}\right)\). At \(P\), \(\cos(xy) = \cos\dfrac{\pi}{3} = \dfrac{1}{2}\), which gives the gradient \(-\dfrac{\pi}{24}\) and the tangent \(y = -\dfrac{\pi}{24}x + \dfrac{\pi}{4}\). Common errors are differentiating \(\sin(xy)\) as \(\cos(xy)\) or as \(y\cos(xy)\) only, using the normal's gradient and working in degrees.

Mark scheme

1 mark
M1 Differentiates implicitly, using the chain rule on \(\sin(xy)\): \(y \to \dfrac{dy}{dx}\) and \(\sin(xy) \to \cos(xy) \times \dfrac{d}{dx}(xy)\); or a correct final answer of \(y = -\dfrac{\pi}{24}x + \dfrac{\pi}{4}\) with no working, which scores 2 marks (special case)
An answer of \(y = -\dfrac{\pi}{24}x + \dfrac{\pi}{4}\) with no working scores 2 marks (special case: the final M1 A1). Answers without working may not gain full credit: for full marks the implicit differentiation and the gradient \(-\dfrac{\pi}{24}\) at \(P\) must be seen. A common error is \(\sin(xy) \to \cos(xy)\) with no derivative of \(xy\): M0.
1 mark
M1 Uses the product rule on \(xy\): \(\dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx}\) within their derivative of \(\sin(xy)\)
Allow \(\cos(xy)\left(y \pm x\dfrac{dy}{dx}\right)\) for this mark. \(\cos(xy) \times x\dfrac{dy}{dx}\) or \(\cos(xy) \times y\) alone is M0.
1 mark
A1 \(\dfrac{dy}{dx} + \cos(xy)\left(y + x\dfrac{dy}{dx}\right) = 0\) oe, e.g. \(\dfrac{dy}{dx} = -\dfrac{y\cos(xy)}{1 + x\cos(xy)}\)
Needs both M marks. The right-hand side differentiates to 0; a non-zero constant left on the right is A0.
1 mark
A1 Gradient at \(P\): with \(xy = \dfrac{\pi}{3}\) and \(\cos\dfrac{\pi}{3} = \dfrac{1}{2}\), \(\dfrac{dy}{dx} + \dfrac{1}{2}\left(\dfrac{\pi}{6} + 2\dfrac{dy}{dx}\right) = 0\), so \(\dfrac{dy}{dx} = -\dfrac{\pi}{24}\)
Must be exact: \(-\dfrac{\pi}{24}\) (\(-0.1309\) alone is A0). Using \(\cos(xy)\) with \(xy\) in degrees is A0.
1 mark
M1 Uses \(y - \dfrac{\pi}{6} = m(x - 2)\) with their gradient from calculus, or \(y = mx + c\) through \(P\) and proceeds to \(c = \ldots\)
Depends on the first M mark. Using the gradient of the normal \(\left(\dfrac{24}{\pi}\right)\) is M0.
1 mark
A1 \(y = -\dfrac{\pi}{24}x + \dfrac{\pi}{4}\)
Accept exact equivalents in the form asked, e.g. \(y = \dfrac{\pi}{4} - \dfrac{\pi x}{24}\); typed y = -(pi/24)x + pi/4. Decimal constants (\(-0.131\), 0.785) are A0 unless the exact values are seen first (then isw). \(24y = -\pi x + 6\pi\) is not in the form asked: A0.

Question 3

6 marks

The curve \(C\) has parametric equations
\[x = \sin t + \cos t, \quad y = \sin t\cos t, \quad -\dfrac{\pi}{4} < t < \dfrac{\pi}{4}\]
Show that \(\dfrac{dy}{dx} = \sin t + \cos t\).
Hence find an equation of the tangent to \(C\) at the point where \(t = \dfrac{\pi}{6}\), giving your answer in the form \(y = mx + c\), where \(m\) and \(c\) are exact constants.

Model answer

\(\dfrac{dx}{dt} = \cos t - \sin t, \quad \dfrac{dy}{dt} = \cos t\cos t + \sin t(-\sin t) = \cos^2 t - \sin^2 t\)
\(\dfrac{dy}{dx} = \dfrac{\cos^2 t - \sin^2 t}{\cos t - \sin t} = \dfrac{(\cos t - \sin t)(\cos t + \sin t)}{\cos t - \sin t} = \sin t + \cos t\) as required
At \(t = \dfrac{\pi}{6}\): \(x = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} = \dfrac{1 + \sqrt{3}}{2}\), \(y = \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{4}\), gradient \(= \dfrac{1 + \sqrt{3}}{2}\)
\(y - \dfrac{\sqrt{3}}{4} = \dfrac{1 + \sqrt{3}}{2}\left(x - \dfrac{1 + \sqrt{3}}{2}\right)\)
\(y = \dfrac{1 + \sqrt{3}}{2}x - \dfrac{4 + 2\sqrt{3}}{4} + \dfrac{\sqrt{3}}{4}\)
\(y = \dfrac{1 + \sqrt{3}}{2}x - \dfrac{4 + \sqrt{3}}{4}\)

Explanation

Differentiate \(x\) and \(y\) with respect to \(t\) (the product rule, or \(y = \dfrac{1}{2}\sin 2t\), gives \(\cos^2 t - \sin^2 t\)) and divide; factorising \(\cos^2 t - \sin^2 t\) as a difference of two squares cancels \(\cos t - \sin t\). At \(t = \dfrac{\pi}{6}\) the point is \(\left(\dfrac{1 + \sqrt{3}}{2}, \dfrac{\sqrt{3}}{4}\right)\) and the gradient is \(\dfrac{1 + \sqrt{3}}{2}\). Common errors are dividing the derivatives the wrong way round, a sign error in differentiating \(\cos t\) and giving \(c\) as a decimal when exact constants are asked for.

Mark scheme

1 mark
M1 Finds \(\dfrac{dx}{dt} = \cos t - \sin t\) and \(\dfrac{dy}{dt} = \cos^2 t - \sin^2 t\) (or \(\cos 2t\)), allowing sign slips, and divides the right way round. An answer with no working scores 0 marks for the show-that; a correct tangent \(y = \dfrac{1 + \sqrt{3}}{2}x - \dfrac{4 + \sqrt{3}}{4}\) with no working scores 1 mark (special case: the final A1).
An answer with no working scores 0 marks for the show-that: its three marks need the working. Special case: a correct tangent \(y = \dfrac{1 + \sqrt{3}}{2}x - \dfrac{4 + \sqrt{3}}{4}\) with no working scores 1 mark (the final A1); the B1 and M1 need the point and the gradient to be seen, as the question says "Hence". \(\dfrac{dy}{dt}\) may come from the product rule or from \(y = \dfrac{1}{2}\sin 2t\).
1 mark
M1 (dM1) Uses \(\cos^2 t - \sin^2 t = (\cos t - \sin t)(\cos t + \sin t)\) (difference of two squares, or \(\cos 2t = \cos^2 t - \sin^2 t\)) to simplify
Depends on the first M mark. Alternative: shows that \((\sin t + \cos t)(\cos t - \sin t) = \cos^2 t - \sin^2 t = \cos 2t\), working from the right-hand side.
1 mark
A1* \(\dfrac{dy}{dx} = \dfrac{(\cos t - \sin t)(\cos t + \sin t)}{\cos t - \sin t} = \sin t + \cos t\) cso
Given answer: needs a correct line before the result, no errors and "\(\dfrac{dy}{dx} =\)" seen. The cancelling is valid because \(\cos t - \sin t ≠ 0\) for \(-\dfrac{\pi}{4} < t < \dfrac{\pi}{4}\); this need not be stated.
1 mark
B1 The point where \(t = \dfrac{\pi}{6}\) is \(\left(\dfrac{1 + \sqrt{3}}{2}, \dfrac{\sqrt{3}}{4}\right)\)
Accept exact equivalents, e.g. \(x = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\). Decimals (1.37, 0.433) with no exact values seen are B0.
1 mark
M1 Uses gradient \(\sin\dfrac{\pi}{6} + \cos\dfrac{\pi}{6} = \dfrac{1 + \sqrt{3}}{2}\) with their point in \(y - y_1 = m(x - x_1)\) or \(y = mx + c\)
May use the given result even if the show-that was not completed. Using the gradient of the normal is M0.
1 mark
A1 \(y = \dfrac{1 + \sqrt{3}}{2}x - \dfrac{4 + \sqrt{3}}{4}\)
Accept exact equivalents in the form asked, e.g. \(m = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\) and \(c = -1 - \dfrac{\sqrt{3}}{4}\); typed y = (1 + sqrt3)/2 x - (4 + sqrt3)/4. Decimal values of \(m\) and \(c\) only (1.37, \(-1.43\)) are A0.

Question 4

9 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

The curve \(C\) has parametric equations
\[x = t^2 - 2t, \quad y = t^3 - 2t^2 - 3t, \quad t > 0\]
The point \(P\) on \(C\) has parameter \(t = 2\).
The normal to \(C\) at \(P\) meets \(C\) again at the point \(Q\).
Find the coordinates of \(Q\).

Model answer

\(\dfrac{dx}{dt} = 2t - 2, \quad \dfrac{dy}{dt} = 3t^2 - 4t - 3\)
\(\dfrac{dy}{dx} = \dfrac{3t^2 - 4t - 3}{2t - 2}\)
At \(t = 2\): \(x = 4 - 4 = 0\), \(y = 8 - 8 - 6 = -6\), so \(P(0, -6)\); \(\dfrac{dy}{dx} = \dfrac{12 - 8 - 3}{4 - 2} = \dfrac{1}{2}\)
Gradient of the normal \(= -2\): \(y + 6 = -2(x - 0)\), so \(y = -2x - 6\)
Substituting: \(t^3 - 2t^2 - 3t = -2(t^2 - 2t) - 6\)
\(t^3 - 2t^2 - 3t + 2t^2 - 4t + 6 = 0\), so \(t^3 - 7t + 6 = 0\)
\(t = 2\) is a root (the point \(P\)): \((t - 2)(t^2 + 2t - 3) = 0\), so \((t - 2)(t - 1)(t + 3) = 0\)
\(t = -3\) is not allowed because \(t > 0\), so at \(Q\), \(t = 1\)
\(x = 1 - 2 = -1\), \(y = 1 - 2 - 3 = -4\)
\(Q\) is \((-1, -4)\)

Explanation

Parametric differentiation gives \(\dfrac{dy}{dx} = \dfrac{3t^2 - 4t - 3}{2t - 2}\), so the gradient at \(P(0, -6)\) is \(\dfrac{1}{2}\) and the normal is \(y = -2x - 6\). Substituting the parametric equations into the normal gives \(t^3 - 7t + 6 = 0\); one root is \(t = 2\) (the point \(P\)), and of the other two only \(t = 1\) satisfies \(t > 0\), giving \(Q(-1, -4)\). Common errors are using the tangent gradient, solving the cubic only on a calculator and not rejecting \(t = -3\).

Mark scheme

1 mark
M1 Attempts \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\) with \(\dfrac{dx}{dt} = 2t - 2\) or \(\dfrac{dy}{dt} = 3t^2 - 4t - 3\) correct. An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the question demands all stages of working. The derivatives may be evaluated at \(t = 2\) straight away (\(\dfrac{dx}{dt} = 2\), \(\dfrac{dy}{dt} = 1\)).
1 mark
A1 \(\dfrac{dy}{dx} = \dfrac{3t^2 - 4t - 3}{2t - 2}\) oe
1 mark
B1 \(P(0, -6)\)
From \(x = 4 - 4\), \(y = 8 - 8 - 6\).
1 mark
M1 Finds the gradient at \(t = 2\) \(\left(\dfrac{1}{2}\right)\) and uses its negative reciprocal with their \(P\): \(y + 6 = -2(x - 0)\)
Depends on the first M mark. Using the tangent gradient is M0.
1 mark
A1 \(y = -2x - 6\) oe
1 mark
M1 Substitutes \(x = t^2 - 2t\) and \(y = t^3 - 2t^2 - 3t\) into their normal to form a cubic equation in \(t\)
\(t^3 - 2t^2 - 3t = -2(t^2 - 2t) - 6\). A Cartesian route is not practical here.
1 mark
A1 \(t^3 - 7t + 6 = 0\) oe
1 mark
M1 (dM1) Uses the known root \(t = 2\) to factorise their cubic, e.g. by division by \((t - 2)\): \((t - 2)(t^2 + 2t - 3) = 0\), and solves their quadratic
Depends on the previous M mark (forming the cubic), not only on the point just before. Roots stated directly from a calculator are M0.
1 mark
A1 \(Q(-1, -4)\) cso, from \(t = 1\), with \(t = -3\) rejected because \(t > 0\)
Both coordinates needed; accept \(x = -1, y = -4\). The root \(t = -3\) must be rejected (or clearly not used) with a reason; giving \((15, -36)\) as a second point \(Q\) is A0.

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