A Level Math Revision Free diagnostic
Themed maths · 14 February

Valentine’s Day maths for A Level

A ready-to-teach lesson for 14 February: a 5-minute starter, a 35-minute main activity on a cardioid and a heart curve given parametrically and the probability of a perfect match, an extension and full worked answers. Tasks A and B are Year 13 parametric work; task C suits Year 12.

Level
A Level Maths (Year 12 and Year 13)
Time
40 minutes, plus a 10-minute extension
Topics
Parametric equations: points, gradients and area (Year 13); Trigonometric identities; Counting and probability; Arc length of a parametric curve (extension)
Equipment
The starter is non-calculator. A calculator helps in tasks A and C.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A13 minThe cardioid
Main: task B10 minA heart curve
Main: task C12 minA perfect match?
Extension10 minFast finishers or homework

Starter (5 minutes)

No calculator.

  1. A curve has x = cos t, y = sin t. Find its Cartesian equation.
  2. Write cos 2t in terms of cos t.
  3. x = t2, y = 2t. Find dy/dx when t = 1.
  4. Work out 5C2.

Main activity (35 minutes)

Task A: The cardioid (13 min)

A cardioid (heart-shaped curve) has parametric equations x = 2cos t − cos 2t, y = 2sin t − sin 2t, for 0 ≤ t < 2π.

  1. Find the points where t = 0, t = π/2 and t = π.
  2. Find dy/dx in terms of t, and its value when t = π/2.
  3. Show that dx/dt and dy/dt are both 0 when t = 0. What does the curve look like there?
  4. The area enclosed by the cardioid is |∫02π y (dx/dt) dt|. Use your calculator to evaluate it, and show it equals 6π.

Task B: A heart curve (10 min)

Another heart curve is x = 16sin3 t, y = 13cos t − 5cos 2t − 2cos 3t − cos 4t, for 0 ≤ t < 2π.

  1. Find the points where t = 0 and t = π.
  2. Find the point where t = π/2. Why is this the furthest point to the right?
  3. Explain why the curve is symmetrical in the y-axis.

Task C: A perfect match? (12 min)

Five friends each write their name on a card. The cards are shuffled and each friend takes one at random.

  1. In how many ways can the cards be handed out?
  2. There are 44 ways in which nobody gets their own card. Find the probability that nobody does.
  3. There are 9 ways for four people to all miss their own cards. Find the probability that exactly one of the five friends gets their own card.
  4. Find the probability that at least one friend gets their own card, and compare it with 1 − 1/e.

Extension (10 minutes)

Year 13 stretch.

  1. For the cardioid in task A, show that (dx/dt)2 + (dy/dt)2 = 8 − 8cos t = 16sin2(t/2). The length of the curve is ∫02π 4sin(t/2) dt: find it.

For teachers

Teacher notes and full worked answers
  • Plot both curves on graphing software first: the reveal is part of the fun.
  • In task A (d), the integral comes out negative because the curve is traced anticlockwise from t = 0 with x decreasing on top: discuss why we take the modulus.
  • Task C works as a class experiment: shuffle five name cards ten times and compare the results with 11/30.

Starter

  1. x2 + y2 = 1
    • cos2 t + sin2 t = 1
  2. 2cos2 t − 1
    • Double-angle formula.
  3. 1
    • dy/dx = (dy/dt) ÷ (dx/dt) = 2 ÷ 2t = 1 when t = 1.
  4. 10
    • 5 × 4 ÷ 2 = 10

Task A: The cardioid

  1. (1, 0), (1, 2), (−3, 0)
    • t = 0: (2 − 1, 0 − 0) = (1, 0)
    • t = π/2: (0 − (−1), 2 − 0) = (1, 2)
    • t = π: (−2 − 1, 0 − 0) = (−3, 0)
  2. dy/dx = (cos t − cos 2t)/(sin 2t − sin t); −1 at t = π/2
    • dx/dt = −2sin t + 2sin 2t and dy/dt = 2cos t − 2cos 2t.
    • At t = π/2: dy/dt = 0 + 2 = 2 and dx/dt = −2 + 0 = −2, so dy/dx = −1.
  3. Both are 0: a sharp point (cusp) at (1, 0)
    • −2sin 0 + 2sin 0 = 0 and 2cos 0 − 2cos 0 = 0.
    • The point stops moving for an instant, which makes the dip of the heart.
  4. 6π ≈ 18.8
    • ∫02π (2sin t − sin 2t)(−2sin t + 2sin 2t) dt = −18.849…
    • |−18.849…| = 18.849… = 6π

Task B: A heart curve

  1. (0, 5) and (0, −17)
    • t = 0: x = 0 and y = 13 − 5 − 2 − 1 = 5.
    • t = π: x = 0 and y = −13 − 5 + 2 − 1 = −17.
  2. (16, 4)
    • x = 16 × 1 = 16, y = 0 − 5(−1) − 2(0) − 1 = 4.
    • sin3 t ≤ 1, so x ≤ 16, with equality only at t = π/2.
  3. Replacing t by −t changes the sign of x but not y
    • sin3(−t) = −sin3 t, and every cosine term is unchanged, so the point (x, y) goes to (−x, y).

Task C: A perfect match?

  1. 120
    • 5! = 120
  2. 11/30
    • 44/120 = 11/30
  3. 3/8
    • Choose the lucky friend: 5 ways; the other four all miss: 9 ways.
    • 5 × 9 = 45, and 45/120 = 3/8.
  4. 19/30 ≈ 0.633; 1 − 1/e ≈ 0.632
    • 1 − 11/30 = 19/30 = 0.6333…
    • 1 − 1/e = 0.6321…: very close already for five people.

Extension

  1. 16
    • (−2sin t + 2sin 2t)2 + (2cos t − 2cos 2t)2 = 4 + 4 − 8(cos t cos 2t + sin t sin 2t) = 8 − 8cos t.
    • 8 − 8cos t = 16sin2(t/2), and sin(t/2) ≥ 0 for 0 ≤ t ≤ 2π.
    • ∫02π 4sin(t/2) dt = [−8cos(t/2)]02π = 8 + 8 = 16

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

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