Growth Decay and Iteration
This topic covers how quantities change when they are repeatedly multiplied by a fixed factor over time. It's essential for modelling real-world scenarios like population dynamics, radioactive decay, and financial investments without a calculator.
Part of the ESAT Mathematics 1 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- Growth and decay problems involve a quantity changing by a constant multiplicative factor in each time period.
- For growth, the multiplier is greater than 1 (e.g., a 5% increase corresponds to a multiplier of 1.05).
- For decay, the multiplier is between 0 and 1 (e.g., a 20% decrease corresponds to a multiplier of 0.8).
- Pay close attention to the number of time intervals. If a process occurs over 5 years starting from year 0, there are 5 multiplicative steps, so the power will be 5.
- Compound interest is a specific application of exponential growth where interest is earned on both the principal amount and the accumulated interest.
- An iterative process is any procedure that repeats a set of rules, using the output of one step as the input for the next.
Formulae
Final Value = Initial Value × (Multiplier)n For any problem involving repeated percentage change, growth, or decay. 'n' is the number of time periods.
A = P × (1 + r/100)n Specifically for compound interest calculations, where 'P' is the principal, 'r' is the annual interest rate percentage, 'n' is the number of years, and 'A' is the final amount.
Definitions
- Multiplier
- The constant factor by which a quantity is multiplied in each time period. Also known as the growth or decay factor.
- Compound Interest
- A method of calculating interest where it is added to the principal sum, so that from then on, interest is earned on the new, larger total.
- Iterative Process
- A sequence of operations where a rule is applied repeatedly to generate the next term from the previous term.
Worked example
A radioactive sample has a mass of 128mg. Its mass halves every 30 minutes. How many minutes will it take for the mass to reduce to 2mg?
- 1
Identify the initial mass (128mg), final mass (2mg), and the decay process (halving, so the multiplier is 1/2).
- 2
Set up the equation relating the values:
2 = 128 × (1/2)n, where 'n' is the number of 30-minute periods - 3
Isolate the term with the power:
(1/2)n = 2 / 128 - 4
Simplify the fraction:
2 / 128 = 1 / 64 - 5
Express the result as a power of the multiplier:
We know 64 = 26, so 1/64 = 1/(26) = (1/2)6 - 6
Equate the powers:
Since (1/2)n = (1/2)6, it follows that n = 6.
This means 6 half-life periods have passed.
- 7
Calculate the total time in minutes:
6 periods × 30 minutes/period = 180 minutes.
Answer: 180 minutes
Common mistakes
- ×Off-by-one error with the exponent 'n'. Be careful whether the process starts at time t=0 or t=1. Counting the number of multiplicative 'steps' is the safest way to find the correct power.
- ×Calculating the multiplier incorrectly. For a 15% decrease, the multiplier is (1 - 0.15) = 0.85, not 0.15. For a 3% increase, the multiplier is 1.03.
- ×Answering the wrong question. If asked for the 'interest earned', you must subtract the initial principal from the final amount. If asked when a value drops 'below' a threshold, you must find the first time period where this condition is met.
No-calculator tips
- ✓Break down calculations with powers and fractions. Instead of calculating (2/3)4 first, rewrite 81 × (2/3)4 as (81 × 24) / 34. Since 81 = 34, the calculation simplifies to just 24 = 16.
- ✓Memorise common powers, especially for 2, 3, 5, and 10. Knowing that 128 = 27 or 243 = 35 can save significant time.
- ✓When dealing with decimal multipliers like 1.1, it can be easier to work with them as fractions. For instance, 1.25 = 5/4. Calculating (5/4)3 is often simpler than 1.253.