Growth and Decay

Growth and decay extend the compound interest and depreciation concepts into broader contexts. This topic is particularly important for Higher tier students in the Edexcel GCSE Mathematics (1MA1) specification, though Foundation students also encounter simpler versions. Questions involve exponential models, multipliers and interpreting growth/decay in real-world settings.

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Key Vocabulary

Term	Meaning
Exponential growth	A quantity increases by a fixed percentage each time period
Exponential decay	A quantity decreases by a fixed percentage each time period
Multiplier	The factor applied each time period (>1 for growth, between 0 and 1 for decay)
Half-life	The time taken for a quantity to reduce to half its original value
Model	A mathematical equation that represents a real-world situation

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Exponential Growth

When a quantity grows by a fixed percentage each period, we use the formula:

$A = P \times k^n$A=P×kn

where:

• A = final amount
• P = initial amount
• k = multiplier (k > 1 for growth)
• n = number of time periods

Worked Example 1

A colony of bacteria starts with 500 organisms and doubles every hour. How many bacteria are there after 6 hours?

• Multiplier k = 2
• A = 500 x 2^6 = 500 x 64 = 32,000

Worked Example 2

A town's population is 25,000 and grows by 3% per year. Find the population after 8 years.

• k = 1.03
• A = 25,000 x 1.03^8
• A = 25,000 x 1.26677...
• A = 31,669 (nearest whole number)

Worked Example 3

An investment of 2,000 pounds grows at 4.5% compound interest per annum. After how many complete years will it first exceed 3,000 pounds?

Year	Value (pounds)
0	2,000.00
5	2,492.36
8	2,868.49
9	2,997.57
10	3,132.46

It first exceeds 3,000 after 10 complete years.

Alternatively using trial: 2,000 x 1.045^n > 3,000, so 1.045^n > 1.5. Testing: 1.045^9 = 1.4889 (not enough), 1.045^10 = 1.5530 (exceeds). Answer: 10 years.

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Exponential Decay

Decay uses a multiplier between 0 and 1.

$A = P \times k^n$A=P×kn where 0 < k < 1

Worked Example 4

A radioactive substance has a mass of 400 g and decays at 8% per hour. Find the mass after 5 hours.

• k = 1 - 0.08 = 0.92
• A = 400 x 0.92^5
• A = 400 x 0.65908...
• A = 263.63 g (to 2 d.p.)

Worked Example 5

The value of a car is modelled by V = 20,000 x 0.82^t, where t is the number of years.

(a) What was the initial value? When t = 0: V = 20,000 x 0.82^0 = 20,000 x 1 = 20,000 pounds

(b) What annual percentage rate of depreciation does this represent? Multiplier = 0.82 = 1 - 0.18, so the rate of depreciation is 18% per year

(c) Find the value after 4 years. V = 20,000 x 0.82^4 = 20,000 x 0.45212... = 9,042.40 pounds (to nearest penny)

Edexcel Exam Tip: When interpreting a growth/decay formula, the number in front of k^t is always the initial value, and the multiplier k tells you the percentage change. If k = 0.82, the decrease is 100% - 82% = 18%.

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Half-Life

The half-life is the time taken for a quantity to decay to half its initial value.

Worked Example 6

A substance has a half-life of 3 hours. Starting with 240 g, find the mass after 12 hours.

• Number of half-lives = 12 / 3 = 4
• Mass = 240 x (1/2)^4 = 240 x 1/16 = 15 g

Worked Example 7

A radioactive isotope decays from 500 g to 62.5 g. How many half-lives have passed?

• 500 -> 250 -> 125 -> 62.5
• That is 3 half-lives

Check: 500 x (1/2)^3 = 500 x 1/8 = 62.5

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Interpreting Growth and Decay Graphs

Exponential growth curves rise steeply. Exponential decay curves fall steeply at first, then level off but never reach zero.

Key Features of Growth Curves

• Start slowly, then increase rapidly
• The y-intercept gives the initial value
• The curve gets steeper over time

Key Features of Decay Curves

• Start steeply, then flatten
• The curve approaches the x-axis but never touches it (asymptote)
• The y-intercept gives the initial value

Worked Example 8

The graph of P = 100 x 1.05^t shows population growth. Describe the trend.

• The initial population is 100.
• The population grows by 5% each time period.
• The curve gets steeper as t increases.
• This is exponential growth.

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Modelling with Growth and Decay

On Higher papers, you may need to set up or interpret models.

Worked Example 9

The number of fish in a lake is modelled by N = 800 x 1.12^t, where t is the number of years. How many years until the population first exceeds 2,000?

• 800 x 1.12^t > 2,000
• 1.12^t > 2.5

Testing values:

t	1.12^t
7	2.2107
8	2.4760
9	2.7731

1.12^8 = 2.476 (not enough), 1.12^9 = 2.773 (exceeds 2.5).

The population first exceeds 2,000 after 9 years.

Worked Example 10

A cup of coffee cools. Its temperature is modelled by T = 20 + 60 x 0.9^t, where T is in degrees Celsius and t is in minutes.

(a) What is the initial temperature? When t = 0: T = 20 + 60 x 1 = 80 degrees C

(b) What temperature does the coffee approach as t gets very large? As t increases, 0.9^t approaches 0, so T approaches 20 + 0 = 20 degrees C (room temperature).

(c) Find the temperature after 10 minutes. T = 20 + 60 x 0.9^10 = 20 + 60 x 0.34868... = 20 + 20.92 = 40.92 degrees C (to 2 d.p.)

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Practice Problems