A Level Math Revision Free diagnostic
Themed maths

Spring egg geometry for A Level

A ready-to-teach lesson for the spring term: an egg drawn with parametric equations, then a lopsided egg with exactly the same area, and an inflating egg. A 5-minute starter, a 35-minute main activity, an extension and full worked answers.

Level
A Level Maths (Year 12 and Year 13)
Time
40 minutes, plus a 10-minute extension
Topics
Parametric equations: tangents and area (Year 13); Connected rates of change (Year 13); Volume scale factors; Volumes of revolution (Further Maths taster)
Equipment
The starter is non-calculator. A calculator helps in the main activity.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A12 minAn ellipse egg
Main: task B11 minA lopsided egg
Main: task C12 minAn inflating egg
Extension10 minFast finishers or homework

Starter (5 minutes)

No calculator.

  1. A curve has x = 3cos t, y = 2sin t. Find its Cartesian equation.
  2. Work out ∫0π/2 sin2 t dt.
  3. Find the volume of a sphere of radius 3.
  4. For x = 3cos t, y = 2sin t, find dy/dx when t = π/4.

Main activity (35 minutes)

Task A: An ellipse egg (12 min)

An egg’s outline is x = 3cos t, y = 2sin t, 0 ≤ t < 2π (in cm).

  1. Find the equation of the tangent at t = π/4, in the form y = mx + c.
  2. Find the area of the egg using area = 2∫ y (dx/dt) dt over the top half, taking care with the limits.
  3. Find the point in the first quadrant where the gradient is −1. Give coordinates to 3 significant figures.

Task B: A lopsided egg (11 min)

A more egg-like outline is x = 3cos t, y = 2sin t(1 + 0.1cos t).

  1. Find the points where t = 0, π/2 and π.
  2. Find the height of the top of the egg above the x-axis at x = 1.5 and at x = −1.5, to 3 significant figures. Which end is fatter?
  3. Show that the area of the lopsided egg is still 6π.

Task C: An inflating egg (12 min)

A balloon egg keeps its shape as it is blown up: its half-length is always 1.5 times its half-width b, and its volume is 4⁄3π × 1.5b × b2 = 2πb3.

  1. Find the volume when b = 2 cm, exactly and to 3 significant figures.
  2. (Year 13) b increases at 0.1 cm per second. Find the rate of increase of the volume when b = 2, to 3 significant figures.
  3. When the volume is 128π cm3, find b and the length of the egg.

Extension (10 minutes)

A Further Maths taster: volumes of revolution.

  1. The ellipse x2/9 + y2/4 = 1 is rotated 2π about the x-axis. The volume is π∫−33 y2 dx. Find it.
  2. The same ellipse is rotated about the y-axis instead: volume π∫−22 x2 dy. Find it.

For teachers

Teacher notes and full worked answers
  • Plot both eggs on graphing software first: students enjoy predicting which has the bigger area before task B.
  • In task A (b), the limits matter: as t goes from 0 to π the point moves from right to left, so the integral comes out negative unless the limits are swapped.
  • Task C is a short connected-rates-of-change question in a new setting.

Starter

  1. x2/9 + y2/4 = 1
    • cos t = x/3, sin t = y/2, and cos2 t + sin2 t = 1.
  2. π/4
    • sin2 t = ½(1 − cos 2t), so the integral is [t/2 − sin 2t/4]0π/2 = π/4.
  3. 36π
    • 4⁄3π × 27
  4. −2/3
    • dy/dx = 2cos t/(−3sin t) = −(2/3)cot t = −2/3 at t = π/4.

Task A: An ellipse egg

  1. y = −(2/3)x + 2√2
    • The point is (3/√2, √2) and the gradient is −2/3.
    • y − √2 = −(2/3)(x − 3/√2), so y = −(2/3)x + √2 + √2 = −(2/3)x + 2√2.
  2. 6π cm2
    • Top half: t from π to 0 as x goes from −3 to 3.
    • 2∫π0 2sin t × (−3sin t) dt = 12∫0π sin2 t dt = 12 × π/2 = 6π
  3. (2.50, 1.11)
    • −(2/3)cot t = −1 gives tan t = 2/3.
    • Then cos t = 3/√13 and sin t = 2/√13, so the point is (9/√13, 4/√13) = (2.496…, 1.109…).

Task B: A lopsided egg

  1. (3, 0), (0, 2), (−3, 0)
    • t = 0: (3, 0). t = π/2: (0, 2). t = π: (−3, 0).
  2. 1.82 and 1.65: the right-hand end is fatter
    • x = 1.5 when t = π/3: y = 2 × (√3/2) × 1.05 = 1.05√3 = 1.818…
    • x = −1.5 when t = 2π/3: y = 2 × (√3/2) × 0.95 = 0.95√3 = 1.645…
  3. 6π
    • Area = 2∫0π 2sin t(1 + 0.1cos t) × 3sin t dt = 12∫0π sin2 t dt + 1.2∫0π sin2 t cos t dt.
    • The second integral is [sin3 t/3]0π = 0, so the area is 12 × π/2 = 6π.

Task C: An inflating egg

  1. 16π ≈ 50.3 cm3
    • 2π × 8 = 16π = 50.26…
  2. 7.54 cm3 per second
    • dV/db = 6πb2 = 24π at b = 2.
    • dV/dt = 24π × 0.1 = 2.4π = 7.539…
  3. b = 4 cm; length 12 cm
    • 2πb3 = 128π gives b3 = 64, so b = 4.
    • Length = 2 × 1.5 × 4 = 12 cm

Extension

  1. 16π
    • y2 = 4 − 4x2/9
    • π[4x − 4x3/27]−33 = π(8 + 8) = 16π
  2. 24π
    • x2 = 9 − 9y2/4
    • π[9y − 3y3/4]−22 = π(12 + 12) = 24π

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

Practise the topics

Big tournament statistics maths · Eclipse geometry maths · Maths about me maths · All themed maths

More for lessons: Weekly starters · Skill Builders · Worksheet builder · Competition maths