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Compound Interest

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

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Compound Interest

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

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

  3. Example – Compound Interest £1000 invested at 5% interest 1050.00 1102.50 1157.63 1215.51 1276.28

  4. Compound Interest Amount invested = £1000 Interest Rate = 5% Method 3 = 1.05nx£1000 Amount at end of Year n = 1.052x£1000 = £1102.50 Amount at end of Year2 = 1.0510x£1000 = £1628.89 Amount at end of Year10

  5. General Formulae Exponential Growth y = kamx k, a and m positive a > 1 Example – Compound Interest A = 1.05nx£1000 x is n y is A a = 1.05 k = 1000 m = 1 Can be written in other forms: A = 1.10250.5nx£1000 k = 1000 a = 1.1025 m = 0.5

  6. Example – Radioactive Decay Plutonium has a half-life of 24 thousand years 500 24 250 48 125 72 62.5 96 120 31.25

  7. A = 1000 x2-0.0416t where t = time in thousands of years Example – Radioactive Decay of Plutonium Decay functions A = 1000 x0.5n where n = no. of half lives A = 1000 x2-n where n = no. of half lives A = 1000 x2-t/24where t = time in thousands of years Exponential Decay a < 1 m positive k and a positive y = kamx a > 1 m negative

  8. General Shape of Graphs Exponential Growth Exponential Decay

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