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CIE IGCSE Mathematics (0580) Exponential growth and decay Study Notes

CIE IGCSE Mathematics (0580) Exponential growth and decay Study Notes - New Syllabus

CIE IGCSE Mathematics (0580) Exponential growth and decay Study Notes

LEARNING OBJECTIVE

  • Exponential Growth and Decay

Key Concepts: 

  • Exponential Growth and Decay

CIE iGCSE Maths (0580)  Study Notes – All topics

Exponential Growth and Decay

Exponential Growth and Decay

Exponential growth and decay describe situations where the rate of change of a quantity is proportional to the current value. This means the amount increases or decreases by the same percentage in each time period.

  • Growth: When a quantity increases over time (e.g. population, investments).
  • Decay: When a quantity decreases over time (e.g. depreciation, radioactive decay).

General Formula

$\text{Final Amount} = \text{Initial Amount} \times (1 \pm r)^n $

  • \( r \): rate of growth or decay as a decimal (e.g. $5\% → 0.05$)
  • \( n \): number of time periods (years, months, etc.)
  • Use \( +r \) for growth and \( -r \) for decay

Why Exponential?

The word “exponential” comes from the fact that the quantity is multiplied by a fixed number repeatedly — the variable is in the exponent. This leads to rapid changes compared to linear growth or decay.

Important Characteristics:

  • Growth curves: become steeper over time.
  • Decay curves: fall rapidly at first, then level off.
  • Useful for modeling real-life processes like compound interest, inflation, population growth, depreciation, and cooling.

Graphical Interpretation:

  • Exponential Growth: Curve slopes upwards — it gets steeper.
  • Exponential Decay: Curve slopes downwards — it flattens out over time.

Example:

A town has a population of 12,000. It increases by 4% per year. What will the population be after 3 years?

▶️ Answer/Explanation

\( r = 0.04,\ n = 3 \)

\( \text{Population} = 12000 \times (1 + 0.04)^3 \)

\( = 12000 \times (1.04)^3 = 12000 \times 1.124864 \)

\( \approx 13,498.37 \)

Answer: Approximately 13,498 people

Example:

A car is bought for $25,000. It depreciates by 15% per year. What will be its value after 4 years?

▶️ Answer/Explanation

\( r = 0.15,\ n = 4 \)

\( \text{Value} = 25000 \times (1 – 0.15)^4 \)

\( = 25000 \times (0.85)^4 = 25000 \times 0.52200625 \)

\( \approx 13,050.16 \)

Answer: Approximately $\$13,050.16$

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