OxfordAQA International AS and A-level Mathematics: Implicit and parametric differentiation (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the second part of content area P2.6 of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2: implicit and parametric differentiation, with tangents and normals. The specification places this content in International A-level (A2) Unit P2, so it is not examined at International AS. Second derivatives are not required.

Related pages: study guide, revision notes, course hub, printable checklist and free diagnostics.

Questions

1. Differentiate with respect to x:

(a) x³y² [2] (b) cos 2y [1]

2. The curve C has equation x³ − xy² + 2y = 8 and passes through (2, 1).

(a) Find dy/dx in terms of x and y. [3] (b) Find the equation of the tangent to C at (2, 1), in the form ax + by + c = 0 with integer coefficients. [2]

3. A curve has equation xeʸ + y = 1 and passes through (1, 0).

(a) Find the equations of the tangent and the normal at (1, 0). [6] (b) Explain why no point on the curve has a horizontal tangent. [2]

4. The curve D has equation 2x² + y³ = 10 and passes through P(1, 2). The tangent to D at P meets the x-axis at A and the y-axis at B. Find the area of triangle OAB, where O is the origin. [5]

5. A curve has parametric equations x = t² − 2t, y = t³ − 12t. Find the coordinates of the points where the tangent is horizontal. [4]

6. A curve has parametric equations x = t/(1 + t), y = t², for t > 0. Find the equations of the tangent and the normal at the point where t = 1. [6]

7. A curve has parametric equations x = sin t, y = t cos t, for 0 ≤ t < π/2.

(a) Show that dy/dx = 1 − t tan t. [3] (b) Find the exact gradient of the curve at t = π/4. [1]

8. A curve has parametric equations x = t³, y = 3t, for t ≠ 0.

(a) Show that the tangent at the point where t = p has equation x − p²y + 2p³ = 0. [4] (b) The tangent at one point passes through (8, 6). Find the coordinates of the two possible points of contact. [3]

9. The curve C has equation y² + 2xy − x² = 7 and passes through A(1, 2).

(a) Show that dy/dx = (x − y)/(x + y). [3] (b) Find the equation of the normal to C at A. [3] (c) The normal meets C again at B. Find the coordinates of B. [3]

10. A curve has parametric equations x = 6 cos θ, y = 2 sin θ, for 0 < θ < π/2.

(a) Show that the tangent at the point where θ = φ has equation x cos φ + 3y sin φ = 6. [5] (b) This tangent meets the x-axis at A and the y-axis at B. Show that the area of triangle OAB is 12/sin 2φ. [3] (c) Deduce the minimum area of triangle OAB and the value of φ that gives it. [2]

11. A curve has equation eˣy + y² = 5.

(a) Show that at the point where x = 0 and y > 0, y = (√21 − 1)/2. [2] (b) Find the exact gradient at this point, then give it to 3 significant figures. [3]

Answers

1. (a) Product rule: 3x²y² [1] + x³ × 2y dy/dx. Answer: 3x²y² + 2x³y dy/dx [1] (b) −2 sin 2y × dy/dx [1] [3] Examiner insight: write the product rule as two separate terms. The usual way to lose credit is to differentiate y² and forget its dy/dx.

2. (a) 3x² − y² − 2xy dy/dx + 2 dy/dx = 0 [2] (the product rule is needed for xy²). So dy/dx (2 − 2xy) = y² − 3x², giving dy/dx = (3x² − y²)/(2xy − 2) [1] (b) At (2, 1): (12 − 1)/(4 − 2) = 11/2 [1]. Tangent: y − 1 = (11/2)(x − 2), so 11x − 2y − 20 = 0 [1] [5] Examiner insight: the question names the form of the answer, so finish in that form. A correct line left as y = (11/2)x − 10 may not match what was asked.

3. (a) Differentiate: eʸ + xeʸ dy/dx + dy/dx = 0 [2]. So dy/dx = −eʸ/(xeʸ + 1) [1]. At (1, 0): −1/(1 + 1) = −1/2 [1]. Tangent: y = −(1/2)(x − 1), so x + 2y = 1 [1]. Normal gradient 2: y = 2x − 2 [1] (b) dy/dx is a fraction with numerator −eʸ [1]. Since eʸ > 0 it is never zero, so dy/dx is never zero, no horizontal tangent [1] [8] Examiner insight: an explanation needs the reason, not only the conclusion. Saying “eʸ is always positive” is the key step to state.

4. Differentiate: 4x + 3y² dy/dx = 0 [1]. So dy/dx = −4x/(3y²), which is −4/12 = −1/3 at P [1]. Tangent: y − 2 = −(1/3)(x − 1), so x + 3y = 7 [1]. Put y = 0: A = (7, 0). Put x = 0: B = (0, 7/3) [1]. Area = ½ × 7 × 7/3 = 49/6 [1] [5] Examiner insight: find the axis intercepts from the simplified tangent equation. Working from y − 2 = −(1/3)(x − 1) invites sign slips.

5. dy/dt = 3t² − 12 [1]. Horizontal tangent where dy/dt = 0, so t = ±2 [1]. dx/dt = 2t − 2 is 2 at t = 2 and −6 at t = −2, so it is non-zero at both [1]. t = 2 gives (0, −16) and t = −2 gives (8, 16). Points (0, −16) and (8, 16) [1] [4] Examiner insight: checking that dx/dt is non-zero is part of a complete answer. Quote the check, since it justifies calling the tangent horizontal.

6. Quotient rule: dx/dt = ((1 + t) − t)/(1 + t)² = 1/(1 + t)² [1]. dy/dt = 2t, so dy/dx = 2t(1 + t)² [1]. At t = 1: gradient 8 [1]; point (1/2, 1) [1]. Tangent: y − 1 = 8(x − 1/2), so y = 8x − 3 [1]. Normal: y − 1 = −(1/8)(x − 1/2), so 2x + 16y = 17 [1] [6] Examiner insight: find the point from the same value of t as the gradient. Mixing a gradient at t = 1 with a point from another value is a common slip.

7. (a) dx/dt = cos t [1]. dy/dt = cos t − t sin t, by the product rule [1]. dy/dx = (cos t − t sin t)/cos t = 1 − t tan t [1] (b) At t = π/4, tan t = 1, so gradient = 1 − π/4 [1] [4] Examiner insight: in a “show that” question, write each step down, including the division by cos t. Leave the answer exact when the question asks for exact.

8. (a) dx/dt = 3t², dy/dt = 3, so dy/dx = 1/t² and the gradient at t = p is 1/p² [2]. Tangent: y − 3p = (1/p²)(x − p³) [1]. Multiply by p²: p²y − 3p³ = x − p³, so x − p²y + 2p³ = 0 [1] (b) Substitute (8, 6): 8 − 6p² + 2p³ = 0 [1]. This factorises as 2(p − 2)²(p + 1) = 0, so p = 2 or p = −1 [1]. The points are (8, 6) and (−1, −3) [1] [7] Examiner insight: spot that p = 2 is a root because (8, 6) is itself on the curve. Using that as a trial root makes the cubic factorise quickly, and you can check the points by substitution.

9. (a) 2y dy/dx + 2y + 2x dy/dx − 2x = 0 [2]. Then dy/dx (y + x) = x − y, so dy/dx = (x − y)/(x + y) [1] (b) At A: (1 − 2)/(1 + 2) = −1/3 [1]. Normal gradient 3 [1]. Normal: y − 2 = 3(x − 1), so y = 3x − 1 [1] (c) Substitute: (3x − 1)² + 2x(3x − 1) − x² = 7 gives 14x² − 8x + 1 = 7 [1]. So 7x² − 4x − 3 = 0, which is (7x + 3)(x − 1) = 0 [1]. x = 1 is A, so x = −3/7 and y = −16/7. B = (−3/7, −16/7) [1] [9] Examiner insight: a quadratic for the second intersection always has the known point as a root. Use it as a check on your factorisation.

10. (a) dx/dθ = −6 sin θ, dy/dθ = 2 cos θ [1]. dy/dx = −cos θ/(3 sin θ) [1]. At θ = φ: y − 2 sin φ = −(cos φ/(3 sin φ))(x − 6 cos φ) [1]. Multiply by 3 sin φ: 3y sin φ − 6 sin²φ = −x cos φ + 6 cos²φ [1]. So x cos φ + 3y sin φ = 6(sin²φ + cos²φ) = 6 [1] (b) y = 0 gives x = 6/cos φ; x = 0 gives y = 2/sin φ [1]. Area = ½ × (6/cos φ) × (2/sin φ) = 6/(sin φ cos φ) [1]. Since sin 2φ = 2 sin φ cos φ, area = 12/sin 2φ [1] (c) sin 2φ ≤ 1, with equality when φ = π/4 [1]. Minimum area 12 at φ = π/4 [1] [10] Examiner insight: in part (a) the identity sin²φ + cos²φ = 1 is the step that reaches the printed result, so say you are using it.

11. (a) At x = 0: y + y² = 5, so y² + y − 5 = 0 [1]. The positive root is y = (√21 − 1)/2 [1] (b) Differentiate: eˣy + eˣ dy/dx + 2y dy/dx = 0 [1]. At x = 0: y + (1 + 2y) dy/dx = 0, and 1 + 2y = √21 [1]. So dy/dx = −y/√21 = (√21 − 21)/42 ≈ −0.391 [1] [5] Examiner insight: give the exact form first, then the rounded value if the question asks for both. Rounding too early, for example using y ≈ 1.79, can change the third significant figure.

Where marks are usually lost

• Omitting dy/dx on a y-term, or putting it on an x-term.
• Differentiating xy, x²y or xy² without the product rule.
• Rearranging for dy/dx with a sign error when moving terms across.
• Dividing dx/dt by dy/dt instead of dy/dt by dx/dt.
• Using the gradient from one value of the parameter and the point from another.
• Using the tangent gradient in the normal equation, or inverting without changing sign.
• Ending a “show that” question without the last step shown.
• Not giving the answer in the form asked.
• Ignoring the condition that dx/dt must be non-zero for a horizontal tangent.

Next steps

• Re-read the study guide for any method you found hard.
• Condense it with the revision notes.
• Go to the OxfordAQA A-level Mathematics hub and tick off the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, published by OxfordAQA. Content area P2.6: Differentiation.