Pearson Edexcel International Advanced Level Mathematics: Parametric and implicit differentiation (YMA01) – 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 cover Pearson Edexcel International Advanced Level Mathematics (Specification Issue 3, April 2019), unit P4 sections 3.1, 5.1 and 5.2. P4 is compulsory for IAL Mathematics and Pure Mathematics and has no IAS weighting. Teaching is in the study guide and the revision notes. See also the Edexcel A-level Mathematics hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. A curve has parametric equations x = 2 sec t, y = 3 tan t, where 0 ≤ t < π/2.

(a) Show that 9x² − 4y² = 36. [3] (b) State the range of values of x on the curve. [1]

2. A curve has parametric equations x = 2 sin t, y = cos 2t.

(a) Find y in terms of x. [3] (b) State the range of x. [1]

3. The curve C has equation 2x² − 3xy + y² = 12. Find the equation of the tangent to C at (4, 2), giving it as ax + by + c = 0 where the coefficients are integers. [5]

4. The curve C has equation x sin y + y = 1 + π/2 and passes through (1, π/2). Find the equation of the normal to C at this point. [5]

5. A curve has equation x² − 4xy + 5y² = 4.

(a) Show that dy/dx = (x − 2y)/(2x − 5y). [3] (b) Find the coordinates of the points where the tangent is parallel to the x-axis. [3]

6. A curve has parametric equations x = 3t − t³, y = 3t². Find the equation of the tangent and of the normal at the point where t = 2. [6]

7. A curve has parametric equations x = 2θ − sin 2θ, y = 1 − cos 2θ.

(a) Show that dy/dx = cot θ. [3] (b) Find, in exact form, the equation of the tangent at the point where θ = π/3. [3]

8. The curve C has parametric equations x = 2t, y = 3/t, t ≠ 0.

(a) Show that C has cartesian equation xy = 6. [1] (b) Find dy/dx in terms of t. [2] (c) The tangent at the point P where t = p (p > 0) meets the x-axis at A and the y-axis at B. Show that the area of triangle OAB, where O is the origin, does not depend on p. [6]

9. The mass M kg of a radioactive sample decreases at a rate proportional to M. When M = 60, the mass is decreasing at 1.5 kg per year. Form a differential equation for M in terms of time t years, and find the value of the constant. [4]

10. A tank has a rectangular base 4 m by 5 m. Water enters at 0.6 m³ per minute and leaves at k√h m³ per minute, where h metres is the depth.

(a) Show that dh/dt = (0.6 − k√h)/20. [2] (b) The depth stays constant at 9 m. Find k and write the differential equation for h. [2] (c) Find dh/dt when h = 4. [2]

11. Sand falls at 12 cm³ per second onto a flat surface and forms a cone whose height is always twice its base radius r.

(a) Find the rate at which r is increasing when the height is 10 cm. [4] (b) Find the rate at which the height is increasing at that instant. [2]

Answers

1. (a) sec t = x/2 and tan t = y/3 [1]. Use 1 + tan²t = sec²t: 1 + y²/9 = x²/4 [1]. Multiply by 36: 36 + 4y² = 9x², so 9x² − 4y² = 36 [1]. (b) sec t ≥ 1 for this range, so x ≥ 2 [1]. Examiner insight: in a “show that” question, write the identity you use and the line after multiplying out, so the route to the printed result is complete.

2. (a) cos 2t = 1 − 2 sin²t [1]. sin t = x/2, so y = 1 − 2(x/2)² [1], giving y = 1 − x²/2 [1]. (b) −1 ≤ sin t ≤ 1 gives −2 ≤ x ≤ 2 [1]. Examiner insight: the cartesian equation alone describes a whole parabola, so a stated range of x is part of a complete answer when the parameter is limited.

3. Differentiate: 4x − 3y − 3x dy/dx + 2y dy/dx = 0 [2], with the product rule applied to the xy term. Rearrange: dy/dx = (4x − 3y)/(3x − 2y) [1]. At (4, 2): (16 − 6)/(12 − 4) = 5/4 [1]. Tangent: y − 2 = (5/4)(x − 4), so 5x − 4y − 12 = 0 [1]. [5] Examiner insight: a coefficient error in the product rule term usually spoils the gradient, so write the xy derivative as its own line before combining.

4. Differentiate: sin y + x cos y dy/dx + dy/dx = 0 [2]. At (1, π/2): 1 + 0 + dy/dx = 0, so dy/dx = −1 [1]. Normal gradient = 1 [1]. Normal: y − π/2 = x − 1, so y = x + π/2 − 1 [1]. [5] Examiner insight: keep cos(π/2) = 0 and sin(π/2) = 1 exact; the question asks for an equation, so a decimal gradient or intercept loses clarity.

5. (a) 2x − 4y − 4x dy/dx + 10y dy/dx = 0 [2]. So (10y − 4x) dy/dx = 4y − 2x, and dividing by −2 gives dy/dx = (x − 2y)/(2x − 5y) [1]. (b) Tangent parallel to the x-axis: x − 2y = 0, so x = 2y [1]. Substitute: 4y² − 8y² + 5y² = 4, so y² = 4 [1]. The points are (4, 2) and (−4, −2); the denominator 2x − 5y is −2 and 2, not zero [1]. Examiner insight: set only the numerator to zero, then use the curve equation to get y. Using the gradient formula alone gives x in terms of y, which is not yet a point.

6. dx/dt = 3 − 3t² and dy/dt = 6t [2]. dy/dx = 6t/(3 − 3t²). At t = 2: 12/(−9) = −4/3 [1]. Point: x = 6 − 8 = −2, y = 12 [1]. Tangent: y − 12 = −(4/3)(x + 2), so 4x + 3y = 28 [1]. Normal gradient 3/4: y − 12 = (3/4)(x + 2), so 3x − 4y + 54 = 0 [1]. [6] Examiner insight: work out both coordinates of the point from t as well as the gradient, since a correct gradient with a wrong point still gives the wrong line.

7. (a) dx/dθ = 2 − 2 cos 2θ and dy/dθ = 2 sin 2θ [1]. dy/dx = sin 2θ/(1 − cos 2θ) [1]. Use sin 2θ = 2 sin θ cos θ and 1 − cos 2θ = 2 sin²θ: dy/dx = 2 sin θ cos θ/(2 sin²θ) = cot θ [1]. (b) At θ = π/3: gradient = cot(π/3) = 1/√3 [1]. Point: x = 2π/3 − √3/2, y = 1 − (−1/2) = 3/2 [1]. Tangent: y − 3/2 = (1/√3)(x − 2π/3 + √3/2) [1]. Examiner insight: a “show that” on a trig derivative nearly always needs a double-angle step; name each identity so the simplification can be followed.

8. (a) xy = 2t × 3/t = 6 [1]. (b) dx/dt = 2 and dy/dt = −3/t² [1], so dy/dx = −3/(2t²) [1]. (c) At t = p the point is (2p, 3/p) and the gradient is −3/(2p²). Tangent: y − 3/p = −(3/(2p²))(x − 2p) [1]. Put y = 0: −3/p = −(3/(2p²))(x − 2p), so x − 2p = 2p and A = (4p, 0) [2]. Put x = 0: y = 3/p + 3/p, so B = (0, 6/p) [1]. Area = ½ × 4p × 6/p [1] = 12, which does not depend on p [1]. [6] Examiner insight: in a “show that” with a general parameter p, keep p symbolic throughout and cancel it visibly at the end.

9. Decreasing and proportional to M: dM/dt = −kM, k > 0 [2]. Substitute: −1.5 = −k × 60 [1], so k = 0.025 (per year) [1]. Examiner insight: state the sign of k as you define it, since the minus sign for a decrease is separate from the proportionality.

10. (a) V = 4 × 5 × h = 20h, so dV/dt = 20 dh/dt [1]. Net rate dV/dt = 0.6 − k√h, so dh/dt = (0.6 − k√h)/20 [1]. (b) Constant depth means dh/dt = 0: 0.6 − 3k = 0 [1], so k = 0.2 and dh/dt = (3 − √h)/100 [1]. (c) At h = 4: (3 − 2)/100 [1] = 0.01 m per minute (depth rising) [1]. Examiner insight: explain the link dV/dt = 20 dh/dt in words or symbols, because connecting volume and depth is the key idea, not just the algebra.

11. (a) Height 10 means r = 5. V = (1/3)πr²(2r) = 2πr³/3 [1], so dV/dr = 2πr² [1]. Then 12 = 2π(25) dr/dt [1], so dr/dt = 6/(25π) ≈ 0.0764 cm s⁻¹ [1]. (b) h = 2r, so dh/dt = 2 dr/dt [1] = 12/(25π) ≈ 0.153 cm s⁻¹ [1]. Examiner insight: eliminate h before differentiating the volume. Differentiating a formula with both r and h in it makes the chain rule harder to apply.

Where marks are usually lost

• Writing dy/dx for a y term but forgetting it on a product such as xy.
• Dividing by dx/dt instead of by dy/dt, or the reverse.
• Using a tangent gradient as the normal gradient, or forgetting to change the sign.
• Finding a gradient but then using the wrong point on the line.
• Giving a numerical answer for an exact-form question.
• Stopping at the cartesian equation without the range of the parameter.
• Forgetting the minus sign for a decreasing quantity.
• Differentiating a volume with two variables before eliminating one.
• Leaving out units on a rate of change.

Next steps

• Re-read the study guide for any question you found hard.
• Use the revision notes for the self-test.
• Tick progress on the printable checklist and see the Edexcel A-level Mathematics hub.
• Try the free 10-minute diagnostics.
• Book a free trial class for help with this unit.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, unit P4 sections 3.1, 5.1 and 5.2.