Exponential growth Any quantity that increases according to the exponential function y = b(ax), where a > 1, is said to be experiencing exponential growth.

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Presentation transcript:

Exponential growth Any quantity that increases according to the exponential function y = b(ax), where a > 1, is said to be experiencing exponential growth. Usually the growth is slow at first, then as time goes on the speed of growth increases. Some examples include growth in a population of people or bacteria or an investment using compound interest. For example the population amandadilis bacteria doubles every 2 hours. If originally there are 1000 bacteria present, after 2 hours there will be 2000, after 4 hours there will be 4000, after 6 hours there will be 8000 and after 8 hours there will be 16 000.

Example 1 The number of computers infected by a virus after t hours is modelled by the formula V = 30(1·7)t. What is the independent variable? What is the vertical intercept and what does it represent? How many computers are infected by the virus after: i) 2 hours ii) 5 hours Estimate when there will be 400 computers infected. a) t for time. b) 30, this represents the original number of computers infected. c) i) V = 30(1·7)t V = 30(1·7)2 V = 86·7 V = 87 d) V = 30(1·7)t 400 = 30(1·7)t ii) V = 30(1·7)t V = 30(1·7)5 V = 425·9 V = 426 13·3 = (1·7)t find t by trial and error t = 4hours 53minutes

Exponential decay Any quantity that decreases according to the exponential function y = b(ax), where a is between 0 and 1, is said to be experiencing exponential decay. Usually the decay is fast at first, then as time goes on the speed of growth decreases. Some examples include radioactive decay and cooling. For example the population chantillim bacteria halves every hours once treated with an antibiotic. If originally there are 10240 bacteria present, after an hour there will be 5120, after 2 hours there will be 2560, after 3 hours there will be 1280 and after 4 hours there will be 640, etc.

Example 2 a) Initially means at n = 0. V = 900(0·6)0 V = 900KL The volume of water in a local dam is given by V = 900(0·6)n where n is the number of years and V is in kilolitres. How much water is initially in the dam? How many water is in the tank after: i) 3 years ii) 5½ years Estimate when there will be 400 KL left. a) Initially means at n = 0. V = 900(0·6)0 V = 900KL b) i) V = 900(0·6)3 V = 194·4 V = 194·4KL c) 400 = 900(0·6)n 0·44 = 0·6n find n by trial and error n = 1·6 years ii) V = 900(0·6)5·5 V = 54·2 V = 54·2KL

Today’s work Exercise 12 D page 371 #6, 9, 10, 11