International Mathematics (0607) | Exponential growth and decay

Welcome to Exponential Growth and Decay!

Hello future Mathematician! This chapter is one of the most practical and useful topics you will study. Exponential functions describe processes that grow or shrink incredibly quickly, and they are everywhere—from the money in your bank account to the value of your phone.

Understanding exponential growth and decay is key to mastering the Number section, especially when dealing with financial problems like compound interest and depreciation. Don't worry if the formulas look complicated; we will break them down step-by-step!

Section 1: Linear vs. Exponential Change

Before jumping into the exponential formula, let's make sure we understand what makes exponential change so powerful, by comparing it to simple linear change.

1.1 Linear Change (Simple)

In linear change, you add or subtract the same fixed amount each time period.

• Example: If you earn $10 interest every year on $100.
• Year 1: $100 + $10 = $110
• Year 2: $110 + $10 = $120
• Year 3: $120 + $10 = $130

1.2 Exponential Change (Powerful)

In exponential change, you multiply by the same fixed factor (percentage increase or decrease) each time period.

• Example: If you earn 10% interest every year on $100.
• Year 1: $100 \(\times\) 1.10 = $110
• Year 2: $110 \(\times\) 1.10 = $121
• Year 3: $121 \(\times\) 1.10 = $133.10

See the difference? In exponential change, the amount added gets bigger each time because the percentage is applied to the ever-increasing total. This is why it's called growth or decay—it accelerates!

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Key Takeaway for Section 1

Linear change involves repeated addition/subtraction. Exponential change involves repeated multiplication by a constant growth or decay factor.

Section 2: The Universal Exponential Formula

All problems involving exponential growth (like compound interest) or decay (like depreciation) use a fundamental structure. You must know this formula well, even though it won't be provided in the exam.

2.1 The General Formula

The final amount (\(A\)) is calculated based on the initial amount (\(P\)), the rate (\(r\)), and the number of time periods (\(n\)).

$$A = P (1 \pm r)^n$$

2.2 Understanding the Variables

• \(A\) (Accumulated/Final Amount): The value after \(n\) periods of time.
• \(P\) (Principal/Initial Amount): The starting value or quantity.
• \(r\) (Rate): The percentage change per period, expressed as a decimal. (If the rate is 5%, then \(r = 0.05\)).
• \(n\) (Number of Periods): The number of times the growth or decay happens (usually time in years, months, etc.).
• \((1 \pm r)\): This is the Multiplier, or the Growth/Decay Factor.

2.3 The Importance of the Multiplier \((1 \pm r)\)

The term \((1 \pm r)\) is what makes the magic happen.

1. For Growth (Increase): Use the plus sign: \((1 + r)\).
Example: If something grows by 8% per year, the multiplier is \(1 + 0.08 = 1.08\). You are keeping 100% of the value (the '1') and adding 8%.

2. For Decay (Decrease/Depreciation): Use the minus sign: \((1 - r)\).
Example: If a car depreciates by 15% per year, the multiplier is \(1 - 0.15 = 0.85\). You are keeping 85% of the value, and 15% is lost.

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Key Takeaway for Section 2

Use \(A = P(1+r)^n\) for growth and \(A = P(1-r)^n\) for decay. The key is correctly identifying the decimal rate \(r\) and the number of periods \(n\).

Section 3: Exponential Growth (Compound Interest & Population)

Exponential growth occurs when a quantity increases by a fixed proportion over equal intervals of time. This is most commonly seen in finance (compound interest) and science (population increase).

3.1 Application 1: Compound Interest (The Bank Analogy)

Compound Interest is interest calculated on the initial principal *and* all the accumulated interest from previous periods.

Example: You invest $5000 at a rate of 4% compounded annually for 6 years.

Step 1: Identify variables.
Initial amount, \(P = 5000\)
Rate (as a decimal), \(r = 4\% = 0.04\)
Number of periods, \(n = 6\)
Multiplier, \(1 + r = 1.04\)

Step 2: Set up the formula.
$$A = 5000 (1 + 0.04)^6$$ $$A = 5000 (1.04)^6$$

Step 3: Calculate the final amount (using GDC).
\(A \approx 6326.60\) (Remember to round money to two decimal places.)
The total value after 6 years is $6326.60.

3.2 Application 2: Population Change

Population growth often follows an exponential model if resources are not limited.

Example: A town has 15,000 residents. The population increases by 2.5% every year. What will the population be in 10 years?

Variables: \(P = 15000\), \(r = 0.025\), \(n = 10\).
$$A = 15000 (1 + 0.025)^{10}$$ $$A = 15000 (1.025)^{10}$$

Calculation: \(A \approx 19201.27\)
Since population must be a whole number, we round: 19,201 residents.

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Key Takeaway for Section 3

Growth problems use \((1+r)\). Ensure you handle the time period correctly (e.g., if interest is compounded half-yearly, you double \(n\) and halve \(r\), but IGCSE 0607 usually focuses on annual compounding in this context).

Section 4: Exponential Decay (Depreciation)

Exponential decay describes a quantity decreasing by a fixed percentage over time. The most common real-world example is depreciation, where the value of an asset (like a car or machinery) falls over time.

4.1 Depreciation Formula

The process is identical to growth, but we use the minus sign in the multiplier:

$$A = P (1 - r)^n$$

4.2 Example: Car Depreciation

A new car is bought for $35,000. It depreciates in value by 12% per year. What is its value after 5 years?

Step 1: Identify variables.
Initial value, \(P = 35000\)
Rate (as a decimal), \(r = 12\% = 0.12\)
Number of periods, \(n = 5\)
Decay Multiplier, \(1 - r = 1 - 0.12 = 0.88\)

Step 2: Set up the formula.
$$A = 35000 (1 - 0.12)^5$$ $$A = 35000 (0.88)^5$$

Step 3: Calculate the final amount (using GDC).
\(A \approx 18471.19\)
The car's value after 5 years is $18,471.19.

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Key Takeaway for Section 4

Decay problems use \((1-r)\). Ensure you calculate the value remaining, not just the loss percentage.

Section 5: Solving for Time (\(n\)) or Rate (\(r\))

Sometimes, you are given the initial and final amounts and asked to find the time it took (\(n\)) or the rate of change (\(r\)). Since the variable is in the exponent (or inside the power), algebraic methods (like logarithms) are often required.

Don't worry! The IGCSE International Mathematics (0607) syllabus usually expects you to use your Graphic Display Calculator (GDC) to solve these equations.

5.1 Finding the Time Period (\(n\))

How long does it take for an investment of $2000 to double if the interest rate is 7% per year?

Equation: \(4000 = 2000 (1.07)^n\)

Step 1: Simplify the equation algebraically.
Divide both sides by 2000: $$2 = (1.07)^n$$

Step 2: Use the GDC (Graphing Method).

1. Define \(Y1 = 2\) (The target amount, or ratio).
2. Define \(Y2 = (1.07)^x\) (Where \(x\) represents \(n\)).
3. Graph both functions and use the GDC's "Intersect" function to find the value of \(x\).

The intersection will be at \(x \approx 10.24\). Therefore, it takes about 10.24 years for the investment to double.

5.2 Finding the Rate (\(r\))

A machine initially worth $10,000 is worth $6,500 after 4 years. Find the constant annual rate of depreciation (\(r\)).

Equation: \(6500 = 10000 (1 - r)^4\)

Step 1: Isolate the multiplier.
Divide by 10000: $$0.65 = (1 - r)^4$$

Step 2: Solve for \((1 - r)\).
Take the fourth root of both sides: $$1 - r = \sqrt[4]{0.65}$$ $$1 - r \approx 0.90048$$

Step 3: Solve for \(r\).
$$r = 1 - 0.90048$$ $$r \approx 0.09952$$

Step 4: Convert back to a percentage.
The rate of depreciation is approximately \(0.09952 \times 100 = 9.95\%\) (to 3 significant figures).

5.3 What if the compounding is frequent?

While standard IGCSE problems usually use annual periods, sometimes you might see problems where the change happens monthly, quarterly, or half-yearly.

If a bank offers 6% interest compounded monthly for 5 years:

• Rate \(r\): Divide the annual rate by 12. \(r = 0.06 / 12 = 0.005\)
• Time \(n\): Multiply the years by 12. \(n = 5 \times 12 = 60\) periods.
• Formula: \(A = P (1 + 0.005)^{60}\)

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