Mathematics Advanced • Year 11 • Module 4 • Lesson 3

The Number e and Natural Exponentials

Build fluency in evaluating e^x, sketching the natural exponential, and using continuous-compounding A = Pe^rt.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Approximate value of e to 5 decimal places: e ≈ ____________ .

Q1.2 Fill in the special property of the natural exponential:

d/dx ( e^x ) = ____________ . ∫ e^x dx = ____________ .

Q1.3 Write the continuous-compounding formula for an initial deposit P at continuous rate r for time t:

A = ____________ .

Stuck? Revisit lesson § Concept and § Key Terms.

2. Worked example — evaluate e², e⁻¹, e^0.5

Problem. Evaluate e², e⁻¹, and e^0.5, each to 3 decimal places.

Step 1 — Use the e^x button on the calculator.

e² ≈ 7.389

Reason: the calculator's e^x key gives more precision than the approximation 2.718^x.

Step 2 — Negative exponent gives the reciprocal.

e⁻¹ = 1/e ≈ 0.368

Reason: e⁻ⁿ = 1/eⁿ.

Step 3 — Fractional exponent is e to a half.

e^0.5 = √e ≈ 1.649

Reason: e^1/2 is the principal square root of e.

Conclusion. e² ≈ 7.389, e⁻¹ ≈ 0.368, e^0.5 ≈ 1.649.

3. Faded example — continuous compounding

$10 000 is invested at 5% per annum compounded continuously. Find the balance after 10 years. 4 marks

Step 1 — Identify P, r, t.

P = ____________ , r = ____________ (as a decimal), t = ____________ years.

Step 2 — Write the formula.

A = ____________ · e^{____ × ____} = ____________ · e^____.

Step 3 — Evaluate e^0.5.

e^0.5 ≈ ____________ (to 4 dp).

Step 4 — Compute and round.

A = 10 000 × ____________ ≈ $____________ (nearest dollar).

Stuck? Revisit lesson § Worked Example 3.

4. Graduated practice

Use a calculator. Express decimals to 3 dp unless stated.

Foundation — evaluate e to a power (4 questions)

Q	Compute	Answer
4.1 1	e¹
4.2 1	e³
4.3 1	e⁻²
4.4 1	e⁰

Standard — apply A = Pe^rt (6 questions)

4.5 $5000 at 4% p.a. continuously compounded for 5 years. Find A. 2 marks

4.6 $2500 at 3% p.a. continuously compounded for 8 years. Find A. 2 marks

4.7 $1200 at 6% p.a. continuously compounded for 12 years. Find A. 2 marks

4.8 Sketch y = e^x and y = e^−x on the same axes. State which is growth and which is decay. 2 marks

4.9 Compute (e²) × (e³) and (e²) / (e⁵) using index laws (no calculator), then approximate. 2 marks

4.10 The temperature of a hot drink obeys T = 60 e^{−0.05 t} + 20 (t in min). Find T(0) and T(10). 2 marks

Extension — reason about e (2 questions)

4.11 Compute (1 + 1/n)ⁿ for n = 10, n = 100, n = 1000 (to 4 dp). What value do they appear to approach? 3 marks

4.12 A student writes d/dx (e^2x) = e^2x. Identify the error and write the correct derivative, with a one-line reason (refer to the chain rule). 2 marks

Stuck on 4.12? Revisit lesson § Trap 03 — derivative is only e^x when the exponent is exactly x.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 (e¹) I used the e^x button (not 2.718) and got 2.718...

For 4.3 (e⁻²) I wrote 1/e² ≈ 0.135, not −e².

For 4.4 (e⁰) I wrote 1, not e or 0.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Value of e

e ≈ 2.71828 (to 5 dp).

Q1.2 — Calculus properties

d/dx (e^x) = e^x. ∫ e^x dx = e^x + C.

Q1.3 — Continuous compounding

A = P e^rt.

Q3 — Faded example, $10 000 at 5% for 10 yr

P = 10 000; r = 0.05; t = 10.
A = 10 000 · e^{0.05 × 10} = 10 000 · e^0.5.
e^0.5 ≈ 1.6487.
A = 10 000 × 1.6487 ≈ $16 487.

Q4.1–4.4 — Powers of e

4.1: e¹ ≈ 2.718. 4.2: e³ ≈ 20.086. 4.3: e⁻² = 1/e² ≈ 0.135. 4.4: e⁰ = 1.

Q4.5 — $5000 at 4% for 5 yr

A = 5000 · e^{0.04 × 5} = 5000 · e^0.2 ≈ 5000 × 1.2214 ≈ $6 107.

Q4.6 — $2500 at 3% for 8 yr

A = 2500 · e^0.24 ≈ 2500 × 1.2712 ≈ $3 178.

Q4.7 — $1200 at 6% for 12 yr

A = 1200 · e^0.72 ≈ 1200 × 2.0544 ≈ $2 465.

Q4.8 — Sketch y = e^x and y = e^−x

Both pass through (0, 1); both have asymptote y = 0. y = e^x is growth (rising right). y = e^−x is decay (falling right). The graphs are reflections of each other in the y-axis.

Q4.9 — Index laws with e

e² · e³ = e²⁺³ = e⁵ ≈ 148.413. e² / e⁵ = e²⁻⁵ = e⁻³ ≈ 0.050.

Q4.10 — T = 60 e^{−0.05 t} + 20

T(0) = 60 · e⁰ + 20 = 60 + 20 = 80°C. T(10) = 60 · e^−0.5 + 20 ≈ 60 × 0.6065 + 20 ≈ 36.4 + 20 = 56.4°C.

Q4.11 — (1 + 1/n)ⁿ → e

n = 10: (1.1)¹⁰ ≈ 2.5937. n = 100: (1.01)¹⁰⁰ ≈ 2.7048. n = 1000: (1.001)¹⁰⁰⁰ ≈ 2.7169. The values approach e ≈ 2.71828 from below — exactly the limit definition of e.

Q4.12 — Derivative of e^2x

The student forgot to apply the chain rule. Correct: d/dx (e^2x) = 2 e^2x. Reason: let u = 2x, then d/dx(e^u) = e^u · du/dx = e^2x · 2.