A Level Math Revision Free diagnostic
Themed maths · February

Women in maths: mathematician cards for A Level

A ready-to-teach lesson for February: three ‘mathematician cards’, each built on maths that carries a woman mathematician’s name or that she is known for. A 5-minute starter, a 35-minute main activity (one card per task), an extension and full worked answers.

Level
A Level Maths (Year 12 and Year 13)
Time
40 minutes, plus a 10-minute extension
Topics
Proof by deduction and by exhaustion; Differentiation: stationary points and points of inflection; Integration by substitution (Year 13); Radians, arcs and sectors
Equipment
The starter is non-calculator. A calculator is allowed in the main activity.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A11 minCard 1: Sophie Germain primes
Main: task B13 minCard 2: the witch of Agnesi
Main: task C11 minCard 3: Nightingale’s rose diagram
Extension10 minFast finishers or homework

Starter (5 minutes)

No calculator.

  1. Is 2 × 11 + 1 prime?
  2. Find the area of a sector of radius 4 and angle π/4.
  3. Find the gradient of y = (x2 + 4)−1 at x = 2.
  4. Write down the exact values of tan(π/4) and arctan 1.

Main activity (35 minutes)

Task A: Card 1: Sophie Germain primes (11 min)

A prime p is a Sophie Germain prime when 2p + 1 is also prime.

  1. How many Sophie Germain primes are less than 50?
  2. Prove that if p > 3 is a Sophie Germain prime then p ≡ 5 (mod 6).
  3. Disprove by counter-example: ‘if p is prime then 2p + 1 is prime’. Give the smallest counter-example.

Task B: Card 2: the witch of Agnesi (13 min)

The curve y = a3/(x2 + a2) is known as the witch of Agnesi, after Maria Gaetana Agnesi. Take a = 2: y = 8/(x2 + 4).

  1. Find the stationary point and show that it is a maximum.
  2. Show that d2y/dx2 = 16(3x2 − 4)/(x2 + 4)3, and find the points of inflection.
  3. (Year 13) Use the substitution x = 2tan θ to show that ∫02 8/(x2 + 4) dx = π.
  4. (Year 13) Find ∫02√3 8/(x2 + 4) dx exactly.

Task C: Card 3: Nightingale’s rose diagram (11 min)

Florence Nightingale is known for polar area diagrams (‘rose diagrams’): 12 months drawn as 12 equal sectors, where the area of each sector shows the number. Use 1 mm2 per case and these made-up numbers.

  1. Find the angle of each sector in radians.
  2. A month with 150 cases is drawn. Find the radius of its sector and its arc length, each to 3 significant figures.
  3. Another month has 4 times as many cases. By what factor is its radius larger? Why is drawing radius in proportion to the number misleading?

Extension (10 minutes)

For fast finishers.

  1. Start at 89 and keep applying p → 2p + 1. How many primes do you get before the chain breaks, and what is the first non-prime?
  2. Show that the area between the witch y = 8/(x2 + 4) and the x-axis from x = −N to N is 8 arctan(N/2). What does it approach as N → ∞?

For teachers

Teacher notes and full worked answers
  • Print the three tasks as cards and rotate groups between them.
  • The cards name each mathematician only through the maths named after her or that she is known for: invite students to research one of them for homework from a reliable source.
  • Card 1 practises three kinds of proof: exhaustion, deduction and counter-example.

Starter

  1. Yes, 23 is prime
    • 23 is not divisible by 2 or 3, and 52 > 23.
  2. 2π
    • ½ × 16 × π/4 = 2π
  3. −1/16
    • dy/dx = −2x(x2 + 4)−2 = −4/64 = −1/16
  4. 1 and π/4
    • tan(π/4) = 1, so arctan 1 = π/4.

Task A: Card 1: Sophie Germain primes

  1. 7: 2, 3, 5, 11, 23, 29, 41
    • Check 2p + 1 for each prime below 50: 5, 7, 11, 23, 47, 59, 83 are prime; the rest are not.
  2. Proof
    • Any integer is 6k, 6k + 1, …, 6k + 5. A prime above 3 is not divisible by 2 or 3, so it is 6k + 1 or 6k + 5.
    • If p = 6k + 1, then 2p + 1 = 3(4k + 1) is a multiple of 3 greater than 3: not prime.
    • So p = 6k + 5 (proof by exhaustion of the cases).
  3. p = 7: 2 × 7 + 1 = 15 = 3 × 5
    • 2, 3 and 5 give 5, 7 and 11, which are prime.
    • 7 gives 15, which is not.

Task B: Card 2: the witch of Agnesi

  1. Maximum at (0, 2)
    • dy/dx = −16x/(x2 + 4)2 = 0 only when x = 0, where y = 2.
    • dy/dx > 0 for x < 0 and < 0 for x > 0, so it is a maximum.
  2. (±2/√3, 3/2)
    • Quotient rule on −16x/(x2 + 4)2 and simplify.
    • Zero when x2 = 4/3, and the sign changes there, so x = ±2/√3.
    • y = 8/(16/3) = 3/2
  3. π
    • dx = 2sec2 θ dθ and x2 + 4 = 4sec2 θ.
    • The integral becomes ∫ 8 × 2sec2 θ/(4sec2 θ) dθ = ∫ 4 dθ.
    • x = 0 → θ = 0 and x = 2 → θ = π/4, so the value is 4 × π/4 = π.
  4. 4π/3
    • Same substitution: the upper limit becomes θ = arctan √3 = π/3.
    • 4 × π/3 = 4π/3

Task C: Card 3: Nightingale’s rose diagram

  1. π/6
    • 2π ÷ 12
  2. 23.9 mm; 12.5 mm
    • ½r2(π/6) = 150 gives r = √(1800/π) = 23.93…
    • Arc length = rθ = 23.93… × π/6 = 12.53…
  3. 2; doubling the radius would show 4 times the area
    • Area ∝ r2, so the radius scales by √4 = 2.
    • If the radius were drawn in proportion, a number twice as big would look 4 times as big.

Extension

  1. 6 primes; 5759 = 13 × 443
    • 89, 179, 359, 719, 1439, 2879 are prime.
    • 5759 = 13 × 443
  2. 4π
    • With x = 2tan θ the integral is 4θ between ±arctan(N/2): 8 arctan(N/2).
    • arctan(N/2) → π/2, so the area → 4π.

The same answers are in the answers PDF. Every answer was recomputed by computer before publishing.

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