1. Exponential Growth & Decay Functions Recall from unit 1 that the graph of f(x) = a x (a>1) looks like y = a x As x   then y   but as x  -  then y  0 GROWTH FUNCTION

2. While the graph of f(x) = a -x ie 1 (a>1) a x looks like y = a -x As x   then y  0 but as x  -  then y   DECAY FUNCTION From function transformations in Unit 1 If f(x) = a x then f(-x) = a -x and f(-x) is obtained by reflecting f(x) in the Y-axis.

3. Example1 (a)It is estimated that the value of a house increases by 3.5% per annum. If a house is bought for £70000 then how much will it be worth in 5 years time? (b)At this rate of increase how long does it take for the value to double? (answer to nearest year!) ************** (a) An increase of 3.5 % gives us103.5% or 1.035 X original. Let V n be the value in year n then … V 0 = £70000 V 1 = 1.035V 0 V 2 = 1.035V 1 = (1.035) 2 V 0 etc

4. So V 5 = (1.035) 5 V 0 = (1.035) 5 X £70000 = £83100 to nearest £100 (b) If the value doubles then V n = 2V 0 And since V n = (1.035) n V 0 Then we need(1.035) n V 0 = 2V 0 Or simply (1.035) n = 2 Using trial & error Take n =10then(1.035) 10 = 1.41..too small Take n =20then(1.035) 20 = 1.99..too small Take n =21then(1.035) 21 = 2.06..too large ** To the nearest year time required to double value = 20 years

5. Example 2 In chemistry if it takes 10 years for half of a sample of radioactive material to decay then we say that it has a half-life of 10 years. If we started with 80g of the material then … Time in YearsMass Remaining 080g 10 40g 20 20g 30 10g 40 5g 50 2.5g etc

6. Example 3 A radioactive substance has a half-life of 7 years. Find how long it takes for any given sample to decay to only 10% of its initial mass. ******** Suppose the An is the amount remaining after n half-lives. If A 0 is the original amount then we have A 1 = 0.5A 0 A 2 = 0.5A 1 = (0.5) 2 A 0 A 3 = 0.5A 2 = (0.5) 3 A 0 etc In general A n = (0.5) n A 0

7. We also need A n = 10% of A 0 or 0.1A 0 This means that (0.5) n A 0 = 0.1A 0 Or (0.5) n = 0.1 Again by trial & error n = 5 (0.5) 5 = 0.03….too small n = 3 (0.5) 3 = 0.125too large n = 4 (0.5) 4 = 0.0625too small n = 3.2 (0.5) 3.2 = 0.11…too large n = 3.1 (0.5) 3.1 = 0.12…too large n = 3.3 (0.5) 3.3 = 0.10…closest

8. So required time = 3.3 half-lives = 3.3 X 7 years = 23.1 years The long period required for the radioactive substance to decay significantly should give us an indication of the dangers of radioactive materials and the potential problems they can create for the environment.