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Themed maths · 31 October

Halloween maths activities for IB Maths SL and HL

A ready-to-teach Halloween lesson for IB Maths classes: a 5-minute starter, a 35-minute main activity, an HL extension and full worked answers. It suits AA and AI, SL and HL.

Level
IB Maths SL and HL (AA and AI)
Time
40 minutes, plus a 10-minute extension
Topics
Exponential models and logarithms; Logistic models (HL, and a stretch for SL); Volume, surface area and rates of change; Conditional probability and the binomial distribution; 3D vectors (HL extension)
Equipment
The starter is non-calculator. A GDC is useful for the main activity.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A9 minZombie outbreak: an exponential model
Main: task B9 minA better model: the logistic limit
Main: task C8 minPumpkin geometry
Main: task D9 minTrick or treat
Extension10 minFast finishers or homework

Starter (5 minutes)

No GDC. Four quick questions to warm up the topics in the main activity.

  1. Solve 2x = 32.
  2. Find the value of log3 81.
  3. Events A and B are independent, with P(A) = 0.3 and P(B) = 0.5. Find P(A ∩ B).
  4. Find the distance between the points (1, 2) and (4, 6).

Main activity (35 minutes)

Task A: Zombie outbreak: an exponential model (9 min)

The number of zombies t hours after midnight on 31 October is modelled by Z(t) = 20e0.35t.

  1. Write down the number of zombies at midnight.
  2. Find the number of zombies after 5 hours, to the nearest whole number.
  3. Find the time when there are 1000 zombies. Give your answer to 3 significant figures.
  4. Find the time it takes for the number of zombies to double.

Task B: A better model: the logistic limit (9 min)

A village has 2000 people, so the zombie numbers cannot grow for ever. A better model is Z(t) = 2000 ÷ (1 + 99e−0.5t). HL; SL students can do every part with logarithms.

  1. Find Z(0).
  2. What happens to Z(t) as t gets very large? Explain what this means.
  3. Find the time when half the village are zombies, to 3 significant figures.
  4. Find Z(10) to 3 significant figures.

Task C: Pumpkin geometry (8 min)

A pumpkin is modelled as a sphere of radius 15 cm.

  1. Find the volume of the pumpkin, to 3 significant figures.
  2. Find its surface area, to 3 significant figures.
  3. Find dV/dr when r = 15. What do you notice?

Task D: Trick or treat (9 min)

60% of houses on a street have a pumpkin outside. At a house with a pumpkin, the probability of a treat is 0.9; at a house without, it is 0.4.

  1. Find the probability that a house chosen at random gives a treat.
  2. Given that a house gave a treat, find the probability that it had a pumpkin.
  3. You visit 10 houses chosen at random. Find the probability of at least 8 treats, to 3 significant figures.

Extension (10 minutes)

Extension 1 is for everyone; extension 2 is HL (3D vectors).

  1. Early in an outbreak, the logistic model Z(t) = 2000 ÷ (1 + 99e−0.5t) is close to the exponential model E(t) = 20e0.5t. Find the percentage error in using E instead of Z when t = 2, to 3 significant figures.
  2. (HL) A ghost floats along the line r = (1, 2, 0) + λ(2, −1, 2), where the floor of the haunted house is the plane z = 0. (a) Show that the ghost passes through the top of the chimney at (7, −1, 6). (b) Find the angle between the ghost’s path and the floor.

For teachers

Teacher notes and full worked answers
  • Ask the class to compare the two zombie models on one GDC screen: the curves agree early on and then separate. That is the whole point of a logistic model.
  • Part (c) of the pumpkin task links the derivative of volume to surface area: a nice discussion of why dV/dr = S for a sphere.
  • In the trick-or-treat task, students often divide by 0.6 instead of 0.7 for the conditional probability. A tree diagram on the board fixes it.

Starter

  1. x = 5
    • 32 = 25, so x = 5.
  2. 4
    • 34 = 81, so log3 81 = 4.
  3. 0.15
    • For independent events, P(A ∩ B) = P(A) × P(B) = 0.3 × 0.5 = 0.15.
  4. 5
    • √(32 + 42) = √25 = 5

Task A: Zombie outbreak: an exponential model

  1. 20
    • Z(0) = 20e0 = 20
  2. 115
    • 20e1.75 = 20 × 5.7546… = 115.09…
    • 115 zombies
  3. 11.2 hours
    • 20e0.35t = 1000, so e0.35t = 50.
    • 0.35t = ln 50, so t = ln 50 ÷ 0.35 = 11.177…
    • 11.2 hours
  4. 1.98 hours
    • e0.35t = 2, so t = ln 2 ÷ 0.35 = 1.980…
    • 1.98 hours (3 s.f.)

Task B: A better model: the logistic limit

  1. 20
    • 2000 ÷ (1 + 99) = 20
  2. It approaches 2000: the whole village
    • As t → ∞, e−0.5t → 0, so Z → 2000 ÷ 1 = 2000.
    • The model says the number of zombies levels off at the size of the village.
  3. 9.19 hours
    • 2000 ÷ (1 + 99e−0.5t) = 1000 gives 99e−0.5t = 1.
    • −0.5t = ln(1/99), so t = 2 ln 99 = 9.190…
    • 9.19 hours
  4. 1200
    • 2000 ÷ (1 + 99e−5) = 2000 ÷ 1.66706… = 1199.7…
    • 1200 (3 s.f.)

Task C: Pumpkin geometry

  1. 14 100 cm3
    • V = 4⁄3π × 153 = 4500π = 14 137.1…
    • 14 100 cm3
  2. 2830 cm2
    • S = 4π × 152 = 900π = 2827.4…
    • 2830 cm2
  3. 900π ≈ 2830 cm3 per cm: the same as the surface area
    • V = 4⁄3πr3, so dV/dr = 4πr2.
    • At r = 15 this is 900π, which is exactly the surface area: a thin extra skin of thickness δr adds about S × δr to the volume.

Task D: Trick or treat

  1. 0.7
    • 0.6 × 0.9 + 0.4 × 0.4 = 0.54 + 0.16 = 0.7
  2. 27/35 ≈ 0.771
    • P(pumpkin | treat) = 0.54 ÷ 0.7 = 27/35 = 0.771…
  3. 0.383
    • X ~ B(10, 0.7).
    • P(X ≥ 8) = P(8) + P(9) + P(10) = 0.2335 + 0.1211 + 0.0282 = 0.3828…
    • 0.383

Extension

  1. 1.72%
    • E(2) = 20e = 54.366…; Z(2) = 2000 ÷ (1 + 99e−1) = 53.445…
    • Percentage error = (54.366 − 53.445) ÷ 53.445 × 100 = 1.72%
  2. (a) λ = 3 (b) 41.8°
    • (a) λ = 3 gives (1 + 6, 2 − 3, 0 + 6) = (7, −1, 6).
    • (b) The floor has normal (0, 0, 1). sin θ = |2| ÷ (√(4 + 1 + 4) × 1) = 2/3.
    • θ = arcsin(2/3) = 41.8°

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

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