A Level Math Revision Free diagnostic
Themed maths · 7 February

e Day maths for A Level

7 February is e Day: written month first, 2/7 matches e = 2.718… This ready-to-teach lesson has a 5-minute starter, a 35-minute main activity on compounding, where e comes from and continuous growth, an extension and full worked answers.

Level
A Level Maths (Year 12 and Year 13)
Time
40 minutes, plus a 10-minute extension
Topics
Exponentials and logarithms; Compound interest; The binomial expansion and the limit for e; Differentiating ex from first principles; Exponential models
Equipment
The starter is non-calculator. A calculator is needed for the main activity.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A11 minCompound interest
Main: task B12 minWhere does e come from?
Main: task C12 minContinuous growth
Extension10 minFast finishers or homework

Starter (5 minutes)

No calculator. Exact answers.

  1. Write down ln e5.
  2. Solve e2x = 9.
  3. Differentiate 3ex/2.
  4. Simplify eln 7 + ln 2.

Main activity (35 minutes)

Task A: Compound interest (11 min)

£2000 is invested for 3 years at a nominal rate of 5% a year. Give money to the nearest penny.

  1. Find the value if interest is compounded annually.
  2. Find the value if interest is compounded monthly.
  3. With continuous compounding the value is 2000e0.15. Find it, and the extra gained over monthly compounding.

Task B: Where does e come from? (12 min)

e is the limit of (1 + 1/n)n as n → ∞.

  1. Show that the first four terms of the binomial expansion of (1 + 1/n)n are 1 + 1 + (n − 1)/(2n) + (n − 1)(n − 2)/(6n2). Evaluate them for n = 10.
  2. As n → ∞, (n − 1)/(2n) → 1/2! and (n − 1)(n − 2)/(6n2) → 1/3!, and so on. Add 1 + 1 + 1/2! + … + 1/6! to estimate e to 4 decimal places.
  3. The gradient of y = ax at x = 0 is close to (ah − 1)/h for small h. Using h = 0.001, estimate it for a = 2 and for a = 3, to 3 decimal places. What does this tell you about e?

Task C: Continuous growth (12 min)

A population of rare birds on an island is modelled by P = 40e0.3t, where t is the time in years.

  1. Write down the initial population.
  2. Find when the population reaches 400, to 3 significant figures.
  3. Show that dP/dt = 0.3P. Find the rate of growth when the population is 400.

Extension (10 minutes)

For fast finishers.

  1. Find the smallest whole number n for which (1 + 1/n)n > 2.7.
  2. Money grows continuously at 4% a year, so its value is Ae0.04t. How long does it take to double? Give 3 significant figures, and compare with ‘70 ÷ 4’.

For teachers

Teacher notes and full worked answers
  • e Day is 7 February because 2/7, written month first, matches e = 2.7…
  • Task B (c) is the key idea: e is the base for which the gradient of ax at x = 0 is 1, which is why d/dx ex = ex.
  • In task A, students often divide the rate by 12 but forget to multiply the number of years by 12.

Starter

  1. 5
    • ln ek = k
  2. x = ln 3
    • 2x = ln 9 = 2 ln 3, so x = ln 3.
  3. (3/2)ex/2
    • 3 × ½ ex/2
  4. 14
    • eln 7 + ln 2 = eln 14 = 14

Task A: Compound interest

  1. £2315.25
    • 2000 × 1.053 = 2315.25
  2. £2322.94
    • 2000 × (1 + 0.05/12)36 = 2322.944…
  3. £2323.67; 73p more
    • 2000e0.15 = 2323.668…, so £2323.67.
    • 2323.67 − 2322.94 = 0.73

Task B: Where does e come from?

  1. 2.57
    • nC2/n2 = n(n − 1)/(2n2) = (n − 1)/(2n), and nC3/n3 = (n − 1)(n − 2)/(6n2).
    • n = 10: 1 + 1 + 9/20 + 72/600 = 1 + 1 + 0.45 + 0.12 = 2.57
  2. 2.7181
    • 1 + 1 + 0.5 + 0.16667 + 0.04167 + 0.00833 + 0.00139 = 2.71806
    • 2.7181 (4 d.p.); e = 2.71828…
  3. 0.693 and 1.099: e lies between 2 and 3
    • (20.001 − 1)/0.001 = 0.6934…
    • (30.001 − 1)/0.001 = 1.0992…
    • e is the base whose gradient at x = 0 is exactly 1, so 2 < e < 3.

Task C: Continuous growth

  1. 40
    • P(0) = 40e0 = 40
  2. 7.68 years
    • e0.3t = 10, so t = ln 10 ÷ 0.3 = 7.675…
  3. 120 birds per year
    • dP/dt = 40 × 0.3e0.3t = 0.3P.
    • When P = 400: 0.3 × 400 = 120.

Extension

  1. n = 74
    • (1 + 1/73)73 = 2.69989…, which is less than 2.7.
    • (1 + 1/74)74 = 2.70013…, which is more than 2.7.
  2. 17.3 years; 70 ÷ 4 = 17.5
    • e0.04t = 2, so t = ln 2 ÷ 0.04 = 17.33…
    • ln 2 = 0.693…, which is why ‘70 ÷ rate’ is a good rule of thumb.

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

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