1. Compound Interest Amount invested = £1000 Interest Rate = 5% Interest at end of Year 1= 5% of £1000 = 0.05 x  £1000 = £50 Amount at end of Year 1= £1050 Interest at end of Year 2= 5% of £1050 = 0.05 x  £1050 = £52.50 Amount at end of Year 2= £1050 + £52.50 = £1102.50 and so on Method 1

2. Compound Interest Amount invested = £1000 Interest Rate = 5% Amount at end of Year 1= 105% of £1000 = 1.05 x  £1000= £1050 and so on Method 2 Amount at end of Year 2 = 1.05 x  £1050 = £1102.50

3. Example – Compound Interest £1000 invested at 5% interest End of Year nAmount A(£) 01000.00 1 2 3 4 5 1050.00 1102.50 1157.63 1215.51 1276.28

4. Compound Interest Amount invested = £1000 Interest Rate = 5% Method 3 Amount at end of Year n = 1.05 n x  £1000 Amount at end of Year 2 Amount at end of Year 10 = 1.05 2 x  £1000 = 1.05 10 x  £1000 = £1102.50 = £1628.89

5. General Formulae k, a and m positivea > 1 Exponential Growth y = ka mx A = 1.05 n x  £1000 Example – Compound Interest y is A x is n m = 1 k = 1000 a = 1.05 Can be written in other forms: A = 1.1025 0.5n x  £1000 m = 0.5 k = 1000 a = 1.1025

6. Example – Radioactive Decay Plutonium has a half-life of 24 thousand years Number of half-livesTime (000s years)Amount (g) 001000 1 2 3 4 5 24 500 48 250 72 125 96 62.5 120 31.25

7. Example – Radioactive Decay of Plutonium Decay functions A = 1000 x 0.5 n where n = no. of half lives A = 1000 x 2 -t/24 where t = time in thousands of years A = 1000 x 2 -n where n = no. of half lives k and a positive a < 1 Exponential Decay m positive y = ka mx a > 1m negative A = 1000 x 2 -0.0416t where t = time in thousands of years

8. General Shape of Graphs Exponential Growth Exponential Decay