Implicit and parametric differentiation - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 17 exam-style questions on Implicit and parametric differentiation for Edexcel A-Level Maths (Paper 2). 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

4 marks

The curve \(C\) is defined by the parametric equations
\[x = 5t^2 - 2, \quad y = (t + 1)\mathrm{e}^{-t}, \quad t > 0\]
Express \(\dfrac{dy}{dx}\) in the form \(k\mathrm{e}^{-t}\), where \(k\) is a constant to be found.
Hence show that \(-\dfrac{1}{10} < \dfrac{dy}{dx} < 0\) at every point on \(C\).

Model answer

\(\dfrac{dx}{dt} = 10t\)
\(\dfrac{dy}{dt} = \mathrm{e}^{-t} - (t + 1)\mathrm{e}^{-t} = -t\mathrm{e}^{-t}\)
\(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{-t\mathrm{e}^{-t}}{10t}\)
\(\dfrac{dy}{dx} = -\dfrac{1}{10}\mathrm{e}^{-t}\)
For \(t > 0\), \(0 < \mathrm{e}^{-t} < 1\), so \(0 < \dfrac{1}{10}\mathrm{e}^{-t} < \dfrac{1}{10}\)
Hence \(-\dfrac{1}{10} < \dfrac{dy}{dx} < 0\) at every point on \(C\)

Explanation

Differentiate each equation with respect to \(t\), using the product rule on \((t + 1)\mathrm{e}^{-t}\), and then divide \(\dfrac{dy}{dt}\) by \(\dfrac{dx}{dt}\); the factor \(t\) cancels, leaving \(-\dfrac{1}{10}\mathrm{e}^{-t}\). Because \(t > 0\), \(\mathrm{e}^{-t}\) lies strictly between 0 and 1, which gives both bounds on the gradient. Common errors are dividing the wrong way round, losing the minus sign when differentiating \(\mathrm{e}^{-t}\) and not cancelling the \(t\).

Mark scheme

1 mark
M1 Attempts \(\dfrac{dx}{dt}\) and \(\dfrac{dy}{dt}\), using the product rule on \((t + 1)\mathrm{e}^{-t}\) (two terms, one with \(\mathrm{e}^{-t}\) and one with \((t + 1)\mathrm{e}^{-t}\)), and divides \(\dfrac{dy}{dt}\) by \(\dfrac{dx}{dt}\); or a correct final answer of \(\dfrac{dy}{dx} = -\dfrac{1}{10}\mathrm{e}^{-t}\) with no working
An answer of \(\dfrac{dy}{dx} = -\dfrac{1}{10}\mathrm{e}^{-t}\) (or \(-\dfrac{\mathrm{e}^{-t}}{10}\), \(-0.1\mathrm{e}^{-t}\), \(-\dfrac{1}{10\mathrm{e}^{t}}\)) with no working scores 3 marks; the final A1* needs a reason. Dividing the wrong way round, \(\dfrac{dx}{dt} \div \dfrac{dy}{dt}\), is M0. Condone a sign slip in \(\dfrac{dy}{dt}\) for this mark.
1 mark
A1 \(\dfrac{dx}{dt} = 10t\) and \(\dfrac{dy}{dt} = \mathrm{e}^{-t} - (t + 1)\mathrm{e}^{-t}\) (\(= -t\mathrm{e}^{-t}\)), giving \(\dfrac{dy}{dx} = \dfrac{-t\mathrm{e}^{-t}}{10t}\) oe
Unsimplified forms are fine for this mark, e.g. \(\dfrac{\mathrm{e}^{-t} - (t + 1)\mathrm{e}^{-t}}{10t}\). A common error is \(\dfrac{d}{dt}\left(\mathrm{e}^{-t}\right) = \mathrm{e}^{-t}\), losing the minus sign: A0.
1 mark
A1 \(\dfrac{dy}{dx} = -\dfrac{1}{10}\mathrm{e}^{-t}\) cao
Depends on the M mark. Accept \(-\dfrac{\mathrm{e}^{-t}}{10}\), \(-0.1\mathrm{e}^{-t}\) or \(-\dfrac{1}{10\mathrm{e}^{t}}\). Do not accept an answer in which \(t\) has not been cancelled, such as \(\dfrac{-t\mathrm{e}^{-t}}{10t}\): the question asks for the form \(k\mathrm{e}^{-t}\). Accept \(k = -\dfrac{1}{10}\) stated with the correct working.
1 mark
A1* Concludes: for \(t > 0\), \(0 < \mathrm{e}^{-t} < 1\), so \(0 < \dfrac{1}{10}\mathrm{e}^{-t} < \dfrac{1}{10}\) and therefore \(-\dfrac{1}{10} < \dfrac{dy}{dx} < 0\) cso
Depends on the M mark and needs the correct \(\dfrac{dy}{dx} = -\dfrac{1}{10}\mathrm{e}^{-t}\). Both bounds must be justified: \(\mathrm{e}^{-t} > 0\) for the upper bound and \(\mathrm{e}^{-t} < 1\) (because \(t > 0\)) for the lower bound. Substituting a few values of \(t\) is not a proof: A0*.

Question 2

6 marks

The curve \(C\) has parametric equations
\[x = \ln(1 + \sin t), \quad y = \cos^2 t, \quad -\dfrac{\pi}{2} < t < \dfrac{\pi}{2}\]
Show that
\[\dfrac{dy}{dx} = -2\sin t(1 + \sin t)\]
Hence find the gradient of \(C\) at each of the two points on \(C\) where \(y = \dfrac{3}{4}\).

Model answer

\(\dfrac{dx}{dt} = \dfrac{\cos t}{1 + \sin t}, \quad \dfrac{dy}{dt} = -2\cos t\sin t\)
\(\dfrac{dy}{dx} = -2\cos t\sin t \times \dfrac{1 + \sin t}{\cos t}\)
\(\dfrac{dy}{dx} = -2\sin t(1 + \sin t)\) (\(\cos t ≠ 0\) for \(-\dfrac{\pi}{2} < t < \dfrac{\pi}{2}\))
When \(y = \dfrac{3}{4}\): \(\cos^2 t = \dfrac{3}{4}\); \(\cos t > 0\) in the range, so \(\cos t = \dfrac{\sqrt{3}}{2}\), \(t = \dfrac{\pi}{6}\) or \(t = -\dfrac{\pi}{6}\)
At \(t = \dfrac{\pi}{6}\): \(\dfrac{dy}{dx} = -2 \times \dfrac{1}{2} \times \dfrac{3}{2} = -\dfrac{3}{2}\)
At \(t = -\dfrac{\pi}{6}\): \(\dfrac{dy}{dx} = -2 \times \left(-\dfrac{1}{2}\right) \times \dfrac{1}{2} = \dfrac{1}{2}\)
The gradients are \(-\dfrac{3}{2}\) and \(\dfrac{1}{2}\)

Explanation

Differentiate \(\ln(1 + \sin t)\) with the chain rule to get \(\dfrac{\cos t}{1 + \sin t}\), and \(\cos^2 t\) to get \(-2\cos t\sin t\); dividing cancels the \(\cos t\). For \(y = \dfrac{3}{4}\) there are two parameters in the range, \(t = \pm\dfrac{\pi}{6}\), and they give different gradients because \(\sin t\) changes sign. Common errors are missing the chain rule on the logarithm and finding only the positive value of \(t\).

Mark scheme

1 mark
M1 Attempts \(\dfrac{dx}{dt} = \dfrac{\pm\cos t}{1 + \sin t}\) (chain rule on the logarithm) and \(\dfrac{dy}{dt} = \pm 2\cos t\sin t\) (or \(\pm\sin 2t\)), and divides \(\dfrac{dy}{dt}\) by \(\dfrac{dx}{dt}\). An answer with no working scores 0 marks for the show-that; gradients of \(-\dfrac{3}{2}\) and \(\dfrac{1}{2}\) with no working score 2 marks (special case).
An answer with no working scores 0 marks for the show-that: its three marks need the working, and restating the given result earns nothing. Special case: gradients of \(-\dfrac{3}{2}\) and \(\dfrac{1}{2}\) with no working score 2 marks (the final M1 A1). A common error is \(\dfrac{d}{dt}\ln(1 + \sin t) = \dfrac{1}{1 + \sin t}\), missing the chain rule: M0.
1 mark
A1 \(\dfrac{dy}{dx} = \dfrac{-2\cos t\sin t}{\cos t/(1 + \sin t)}\) oe
Both derivatives correct, unsimplified.
1 mark
A1* Cancels \(\cos t\) to reach \(\dfrac{dy}{dx} = -2\sin t(1 + \sin t)\) cso
Depends on the M mark. "\(\dfrac{dy}{dx} =\)" must be seen. Condone not stating that \(\cos t ≠ 0\) in the range.
1 mark
B1 \(\cos^2 t = \dfrac{3}{4}\) gives \(t = \dfrac{\pi}{6}\) and \(t = -\dfrac{\pi}{6}\) (both, and no others in the range)
Accept \(\sin t = \pm\dfrac{1}{2}\) used directly (from \(\sin^2 t = \dfrac{1}{4}\)). Values in degrees (\(\pm 30^\circ\)) may be condoned if the gradients are correct.
1 mark
M1 Substitutes at least one of their values of \(t\) (or of \(\sin t\)) into \(-2\sin t(1 + \sin t)\)
Uses the given result; must not substitute \(y = \dfrac{3}{4}\) or an \(x\)-value for \(t\).
1 mark
A1 \(-\dfrac{3}{2}\) (at \(t = \dfrac{\pi}{6}\)) and \(\dfrac{1}{2}\) (at \(t = -\dfrac{\pi}{6}\))
Both needed. Accept \(-1.5\) and \(0.5\).

Question 3

7 marks

The curve \(C\) has equation
\[y^3 - 3y + x^2 = 4\]
Find the exact coordinates of each point on \(C\) at which the tangent to \(C\) is parallel to the \(y\)-axis.

Model answer

Differentiating with respect to \(x\):
\(3y^2\dfrac{dy}{dx} - 3\dfrac{dy}{dx} + 2x = 0\)
\(\dfrac{dy}{dx} = \dfrac{-2x}{3y^2 - 3}\)
The tangent is parallel to the \(y\)-axis where the denominator is zero (and the numerator is not): \(3y^2 - 3 = 0\), so \(y = 1\) or \(y = -1\)
\(y = 1\): \(1 - 3 + x^2 = 4\), \(x^2 = 6\), \(x = \pm\sqrt{6}\)
\(y = -1\): \(-1 + 3 + x^2 = 4\), \(x^2 = 2\), \(x = \pm\sqrt{2}\)
At each of these points \(x ≠ 0\), so the numerator is not zero.
The points are \((\sqrt{6}, 1)\), \((-\sqrt{6}, 1)\), \((\sqrt{2}, -1)\) and \((-\sqrt{2}, -1)\)

Explanation

A tangent parallel to the \(y\)-axis has an undefined gradient, so the denominator of \(\dfrac{dy}{dx}\) must be zero (equivalently \(\dfrac{dx}{dy} = 0\)). Here that means \(3y^2 - 3 = 0\), so \(y = \pm 1\), and substituting each value into the curve gives two points. Examiners report candidates setting the numerator to zero instead, which finds the tangents parallel to the \(x\)-axis.

Mark scheme

1 mark
M1 Differentiates implicitly: \(y^3 \to ky^2\dfrac{dy}{dx}\), \(3y \to 3\dfrac{dy}{dx}\), \(x^2 \to 2x\), \(4 \to 0\); or correct final answers of \((\sqrt{6}, 1)\), \((-\sqrt{6}, 1)\), \((\sqrt{2}, -1)\) and \((-\sqrt{2}, -1)\) with no working, which score 2 marks (special case)
Correct answers \((\pm\sqrt{6}, 1)\) and \((\pm\sqrt{2}, -1)\) with no working score 2 marks (special case: the final two A marks). Answers without working may not gain full credit: for full marks the derivative and the condition \(3y^2 - 3 = 0\) must be seen. Alternative: differentiating with respect to \(y\), \(3y^2 - 3 + 2x\dfrac{dx}{dy} = 0\), and setting \(\dfrac{dx}{dy} = 0\), earns the same marks.
1 mark
A1 \(\dfrac{dy}{dx} = \dfrac{-2x}{3y^2 - 3}\) oe (or \(\dfrac{dx}{dy} = \dfrac{3 - 3y^2}{2x}\))
1 mark
M1 Uses the condition for a tangent parallel to the \(y\)-axis: sets the denominator of their \(\dfrac{dy}{dx}\) to zero, \(3y^2 - 3 = 0\) (or \(\dfrac{dx}{dy} = 0\))
A common error is setting the numerator to zero (\(x = 0\)), which gives tangents parallel to the \(x\)-axis: M0 and no further marks.
1 mark
A1 \(y = 1\) and \(y = -1\)
Both values needed.
1 mark
M1 (dM1) Substitutes at least one of their values of \(y\) into the equation of \(C\) to find \(x\)
Depends on the previous M mark (setting the denominator to zero).
1 mark
A1 \((\sqrt{6}, 1)\) and \((-\sqrt{6}, 1)\)
Exact values needed; \(\pm 2.45\) only is A0. Both points needed.
1 mark
A1 \((\sqrt{2}, -1)\) and \((-\sqrt{2}, -1)\)
Depends on the third M mark. Exact values needed; both points needed. Withhold this mark if extra points are given (e.g. points with \(x = 0\)).

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
\[x = t^2, \quad y = \dfrac{2}{t}, \quad t \in \mathbb{R},\ t ≠ 0\]
Show that the tangent to \(C\) at the point with parameter \(p\) has equation
\[x + p^3y = 3p^2\]
The point \(A\) has coordinates \((2, -1)\).
Hence show that there are exactly three points on \(C\) at which the tangent to \(C\) passes through \(A\), and find the exact value of the parameter at each of these points.

Model answer

\(\dfrac{dx}{dt} = 2t, \quad \dfrac{dy}{dt} = -\dfrac{2}{t^2}\)
\(\dfrac{dy}{dx} = -\dfrac{2}{t^2} \div 2t = -\dfrac{1}{t^3}\)
At the point with parameter \(p\), \(\left(p^2, \dfrac{2}{p}\right)\), the gradient is \(-\dfrac{1}{p^3}\)
Tangent: \(y - \dfrac{2}{p} = -\dfrac{1}{p^3}\left(x - p^2\right)\)
\(p^3y - 2p^2 = -x + p^2\)
\(x + p^3y = 3p^2\)
Through \(A(2, -1)\): \(2 - p^3 = 3p^2\), so \(p^3 + 3p^2 - 2 = 0\)
\(p = -1\): \(-1 + 3 - 2 = 0\), so \((p + 1)\) is a factor
\(p^3 + 3p^2 - 2 = (p + 1)\left(p^2 + 2p - 2\right)\)
\(p^2 + 2p - 2 = 0\): \((p + 1)^2 = 3\), so \(p = -1 \pm \sqrt{3}\)
The cubic has three distinct real roots, \(p = -1\), \(p = -1 + \sqrt{3}\) and \(p = -1 - \sqrt{3}\), and none is zero. Different values of \(p\) give different points of \(C\) (different values of \(y = \dfrac{2}{p}\)), so there are exactly three points on \(C\) at which the tangent passes through \(A\).

Explanation

The general tangent comes from the gradient \(-\dfrac{1}{p^3}\) at the point \(\left(p^2, \dfrac{2}{p}\right)\). Requiring it to pass through \(A\) gives the cubic \(p^3 + 3p^2 - 2 = 0\); the factor theorem with \(p = -1\) and the quadratic formula give three distinct non-zero roots, so there are exactly three tangents through \(A\). Because the question rules out calculator-only solutions, the factorisation and the solution of the quadratic must be shown, and a conclusion is needed.

Mark scheme

1 mark
M1 Attempts \(\dfrac{dx}{dt} = 2t\) and \(\dfrac{dy}{dt} = \pm 2t^{-2}\) (at least one correct) and divides \(\dfrac{dy}{dt}\) by \(\dfrac{dx}{dt}\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the question demands all stages of working, and the tangent is a show-that. Alternative: the Cartesian equation \(xy^2 = 4\) differentiated implicitly, \(y^2 + 2xy\dfrac{dy}{dx} = 0\), earns the M1 A1.
1 mark
A1 Gradient at the point with parameter \(p\) is \(-\dfrac{1}{p^3}\)
1 mark
M1 (dM1) Forms the tangent at \(\left(p^2, \dfrac{2}{p}\right)\) with their gradient: \(y - \dfrac{2}{p} = -\dfrac{1}{p^3}\left(x - p^2\right)\)
Depends on the first M mark. Must use the point in terms of \(p\).
1 mark
A1* Correct algebra to \(x + p^3y = 3p^2\), with an intermediate line such as \(p^3y - 2p^2 = -x + p^2\) cso
No errors in the working.
1 mark
M1 Substitutes \(x = 2\), \(y = -1\) into the given tangent and rearranges to a cubic: \(p^3 + 3p^2 - 2 = 0\)
Uses the given equation; may be scored even if the show-that was not completed.
1 mark
B1 Shows that \(p = -1\) is a root (\((-1)^3 + 3(-1)^2 - 2 = 0\)), so \((p + 1)\) is a factor
The substitution must be seen; stating the roots from a calculator scores nothing here.
1 mark
M1 Divides by \((p + 1)\) (or compares coefficients) to obtain the quadratic factor \(p^2 + 2p - 2\) and solves it by the formula or completing the square
Depends on the previous M mark. \(p^3 + 3p^2 - 2 = (p + 1)(p^2 + 2p - 2)\). Roots of the quadratic stated from a calculator without the formula or completing the square are M0.
1 mark
A1 \(p = -1\), \(p = -1 + \sqrt{3}\) and \(p = -1 - \sqrt{3}\)
All three, exact. Decimals (0.732, \(-2.73\)) only are A0.
1 mark
A1 Concludes: the cubic has three distinct real roots, none of them zero, and different values of \(p\) give different points of \(C\) (different \(y = \dfrac{2}{p}\)), so exactly three points cso
Depends on all the M marks. A minimal conclusion such as "three different non-zero values of \(p\), so exactly three points" is acceptable; the cubic having at most three roots may be implied.

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