IB Math Revision Start free
Themed maths · 7 February

e Day maths for IB Maths SL and HL

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, the limit that defines e and continuous growth, an HL extension and full worked answers.

Level
IB Maths SL and HL (AA and AI)
Time
40 minutes, plus a 10-minute extension
Topics
Compound interest and financial mathematics; The limit definition of e; Exponential growth models and logarithms; Differentiating ex; Maclaurin series (HL extension)
Equipment
The starter is non-calculator. A GDC is needed for the main activity.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A12 minCompounding more often
Main: task B11 minThe limit that defines e
Main: task C12 minContinuous growth
Extension10 minFast finishers or homework

Starter (5 minutes)

No GDC.

  1. Write down the value of ln(e3).
  2. Solve ex = 1.
  3. Differentiate e4x.
  4. $100 is invested at 10% a year compound interest. Find its value after 2 years.

Main activity (35 minutes)

Task A: Compounding more often (12 min)

$1000 is invested for one year at a nominal annual interest rate of 6%, compounded n times a year. Give money to 2 decimal places.

  1. Find the value after one year when interest is compounded annually.
  2. Find the value after one year when interest is compounded monthly.
  3. Find the value after one year when interest is compounded daily (365 times).
  4. As n gets very large the value approaches 1000e0.06 (continuous compounding). Find it.

Task B: The limit that defines e (11 min)

Let f(n) = (1 + 1/n)n: the value of $1 after one year at 100% interest compounded n times.

  1. Find f(1), f(2) and f(10), giving f(10) to 4 decimal places.
  2. Use a table on your GDC to find the smallest whole number n for which f(n) is within 0.01 of e.
  3. Show that ln f(n) = n ln(1 + 1/n). Find this when n = 1000, to 4 decimal places. What does it suggest?

Task C: Continuous growth (12 min)

The number of bacteria in a dish t hours after the start is modelled by N = 500e0.2t.

  1. Find N when t = 5, to the nearest whole number.
  2. Find the time taken for the number of bacteria to double, to 3 significant figures.
  3. Find dN/dt when t = 5, to 3 significant figures, and show that dN/dt = 0.2N at every time.

Extension (10 minutes)

HL: e from a series.

  1. The Maclaurin series is ex = 1 + x + x2/2! + x3/3! + … Use the first six terms with x = 1 to estimate e, as a fraction and to 4 decimal places.
  2. Use the Maclaurin series for ln(1 + x) to explain why n ln(1 + 1/n) → 1 as n → ∞.

For teachers

Teacher notes and full worked answers
  • e Day is 7 February because 2/7, written month first, matches e = 2.7…
  • Task A works well as a class race: give each group a different n and collect the answers on the board in order.
  • In task C, AI students can find the doubling time from a graph; AA students should solve it with logs.

Starter

  1. 3
    • ln and ex undo each other.
  2. x = 0
    • e0 = 1
  3. 4e4x
    • Chain rule: the derivative of ekx is kekx.
  4. $121
    • 100 × 1.12 = 121

Task A: Compounding more often

  1. $1060.00
    • 1000 × 1.06 = 1060
  2. $1061.68
    • 1000 × (1 + 0.06/12)12 = 1000 × 1.00512 = 1061.677…
  3. $1061.83
    • 1000 × (1 + 0.06/365)365 = 1061.831…
  4. $1061.84
    • 1000e0.06 = 1061.836…
    • Compounding more often helps less and less: daily is only 1 cent short of continuous.

Task B: The limit that defines e

  1. 2, 2.25, 2.5937
    • f(1) = 21 = 2
    • f(2) = 1.52 = 2.25
    • f(10) = 1.110 = 2.59374…
  2. n = 135
    • f(134) = 2.70820…, which is 0.0100… below e = 2.71828…
    • f(135) = 2.70827…, which is 0.0099… below e. So n = 135.
  3. 0.9995: ln f(n) → 1, so f(n) → e
    • ln (1 + 1/n)n = n ln(1 + 1/n) by the power law of logs.
    • 1000 ln 1.001 = 0.99950…
    • ln f(n) approaches 1, so f(n) approaches e1 = e.

Task C: Continuous growth

  1. 1359
    • 500e1 = 1359.14…
  2. 3.47 hours
    • e0.2t = 2, so t = ln 2 ÷ 0.2 = 3.4657…
  3. 272 bacteria per hour
    • dN/dt = 500 × 0.2e0.2t = 100e0.2t = 0.2N.
    • At t = 5: 100e = 271.8…, which is 272 (3 s.f.).

Extension

  1. 163/60 = 2.7167
    • 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 = 163/60
    • 163/60 = 2.71666…, which is 2.7167 (4 d.p.); e = 2.71828…
  2. The limit is 1
    • ln(1 + x) = x − x2/2 + …, so n ln(1 + 1/n) = 1 − 1/(2n) + …
    • Every term after the 1 tends to 0, so the limit is 1, and (1 + 1/n)n → e.

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

Practise the topics

Valentine’s Day maths · Women in maths maths · Pi Day maths · All themed maths

More for lessons: Weekly starters · Free standard lessons · Worksheet builder · Competition maths