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# 12 x1 t04 02 growth & decay (2012)

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### 12 x1 t04 02 growth & decay (2012)

1. 1. Exponential Growth & Decay
2. 2. Exponential Growth & DecayGrowth and decay is proportional to population.
3. 3. Exponential Growth & DecayGrowth and decay is proportional to population. dP  kP dt
4. 4. Exponential Growth & DecayGrowth and decay is proportional to population. dP  kP dt dt 1  dP kP
5. 5. Exponential Growth & DecayGrowth and decay is proportional to population. dP  kP dt dt 1  dP kP 1 dP t  k P
6. 6. Exponential Growth & DecayGrowth and decay is proportional to population. dP  kP dt dt 1  dP kP 1 dP t  k P 1  log P  c k
7. 7. Exponential Growth & DecayGrowth and decay is proportional to population. dP  kP dt dt 1  dP kP 1 dP t  k P 1  log P  c k kt  log P  c
8. 8. Exponential Growth & DecayGrowth and decay is proportional to population. dP  kP dt dt 1  dP kP 1 dP t  k P 1  log P  c k kt  log P  c log P  kt  c
9. 9. Exponential Growth & DecayGrowth and decay is proportional to population. dP  kP dt dt 1  dP kP 1 dP t  k P 1  log P  c k kt  log P  c log P  kt  c P  e kt c
10. 10. Exponential Growth & DecayGrowth and decay is proportional to population. dP  kP dt dt 1  dP kP 1 dP t  k P 1  log P  c k kt  log P  c log P  kt  c P  e kt c P  e kt  e c
11. 11. Exponential Growth & DecayGrowth and decay is proportional to population. dP  kP dt dt 1  dP kP 1 dP t  k P 1  log P  c k P  Ae kt kt  log P  c log P  kt  c P  e kt c P  e kt  e c
12. 12. Exponential Growth & DecayGrowth and decay is proportional to population. dP  kP dt dt 1  dP kP 1 dP t  k P 1  log P  c k P  Ae kt kt  log P  c P = population at time t log P  kt  c P  e kt c A = initial population P  e kt  e c k = growth(or decay) constant t = time
13. 13. e.g.(i) The growth rate per hour of a population of bacteria is 10% of the population. The initial population is 1000000
14. 14. e.g.(i) The growth rate per hour of a population of bacteria is 10% of the population. The initial population is 1000000a) Show that P  Ae 0.1t is a solution t o the differential equation.
15. 15. e.g.(i) The growth rate per hour of a population of bacteria is 10% of the population. The initial population is 1000000 a) Show that P  Ae 0.1t is a solution t o the differential equation. P  Ae 0.1t
16. 16. e.g.(i) The growth rate per hour of a population of bacteria is 10% of the population. The initial population is 1000000 a) Show that P  Ae 0.1t is a solution t o the differential equation. P  Ae 0.1t dP  0.1Ae 0.1t dt
17. 17. e.g.(i) The growth rate per hour of a population of bacteria is 10% of the population. The initial population is 1000000 a) Show that P  Ae 0.1t is a solution t o the differential equation. P  Ae 0.1t dP  0.1Ae 0.1t dt  0.1P
18. 18. e.g.(i) The growth rate per hour of a population of bacteria is 10% of the population. The initial population is 1000000 a) Show that P  Ae 0.1t is a solution t o the differential equation. P  Ae 0.1t dP  0.1Ae 0.1t dt  0.1P 1b) Determine the population after 3 hours correct to 4 significant figures. 2
19. 19. e.g.(i) The growth rate per hour of a population of bacteria is 10% of the population. The initial population is 1000000 a) Show that P  Ae 0.1t is a solution t o the differential equation. P  Ae 0.1t dP  0.1Ae 0.1t dt  0.1P 1b) Determine the population after 3 hours correct to 4 significant figures. 2 when t  0, P  1000000  A  1000000 P  1000000 e 0.1t
20. 20. e.g.(i) The growth rate per hour of a population of bacteria is 10% of the population. The initial population is 1000000 a) Show that P  Ae 0.1t is a solution t o the differential equation. P  Ae 0.1t dP  0.1Ae 0.1t dt  0.1P 1b) Determine the population after 3 hours correct to 4 significant figures. 2 when t  0, P  1000000 when t  3 .5, P  1000000 e 0.13.5   A  1000000  1 4 19 000 P  1000000 e 0.1t
21. 21. e.g.(i) The growth rate per hour of a population of bacteria is 10% of the population. The initial population is 1000000 a) Show that P  Ae 0.1t is a solution t o the differential equation. P  Ae 0.1t dP  0.1Ae 0.1t dt  0.1P 1b) Determine the population after 3 hours correct to 4 significant figures. 2 when t  0, P  1000000 when t  3 .5, P  1000000 e 0.13.5   A  1 000000  1 4 19 000 P  1000000 e 0.1t 1  after 3 hours there is 1419000 bacteria 2
22. 22. (ii) On an island, the population in 1960 was 1732 and in 1970 it was 1260. a) Find the annual growth rate to the nearest %, assuming it is proportional to population.
23. 23. (ii) On an island, the population in 1960 was 1732 and in 1970 it was 1260. a) Find the annual growth rate to the nearest %, assuming it is proportional to population. dP  kP dt
24. 24. (ii) On an island, the population in 1960 was 1732 and in 1970 it was 1260. a) Find the annual growth rate to the nearest %, assuming it is proportional to population. dP  kP dt P  Ae kt
25. 25. (ii) On an island, the population in 1960 was 1732 and in 1970 it was 1260. a) Find the annual growth rate to the nearest %, assuming it is proportional to population. dP  kP dt P  Ae kt when t  0, P  1732  A  1732 P  1732e kt
26. 26. (ii) On an island, the population in 1960 was 1732 and in 1970 it was 1260. a) Find the annual growth rate to the nearest %, assuming it is proportional to population. dP when t  10, P  1260  kP dt i.e. 1260  1732e10 k P  Ae kt when t  0, P  1732  A  1732 P  1732e kt
27. 27. (ii) On an island, the population in 1960 was 1732 and in 1970 it was 1260. a) Find the annual growth rate to the nearest %, assuming it is proportional to population. dP when t  10, P  1260  kP dt i.e. 1260  1732e10 k P  Ae kt 1260 e10 k  when t  0, P  1732 1732  A  1732 P  1732e kt
28. 28. (ii) On an island, the population in 1960 was 1732 and in 1970 it was 1260. a) Find the annual growth rate to the nearest %, assuming it is proportional to population. dP when t  10, P  1260  kP dt i.e. 1260  1732e10 k P  Ae kt 1260 e10 k  when t  0, P  1732 1732  A  1732  1260  10k  log  P  1732e kt  1732  1 k  log  1260   10  1732 
29. 29. (ii) On an island, the population in 1960 was 1732 and in 1970 it was 1260. a) Find the annual growth rate to the nearest %, assuming it is proportional to population. dP when t  10, P  1260  kP dt i.e. 1260  1732e10 k P  Ae kt 1260 e10 k  when t  0, P  1732 1732  A  1732  1260  10k  log  P  1732e kt  1732  1 k  log  1260   10  1732  k  0.0318165
30. 30. (ii) On an island, the population in 1960 was 1732 and in 1970 it was 1260. a) Find the annual growth rate to the nearest %, assuming it is proportional to population. dP when t  10, P  1260  kP dt i.e. 1260  1732e10 k P  Ae kt 1260 e10 k  when t  0, P  1732 1732  A  1732  1260  10k  log  P  1732e kt  1732  1 k  log  1260   10  1732  k  0.0318165  growth rate is - 3%
31. 31. b) In how many years will the population be half that in 1960?
32. 32. b) In how many years will the population be half that in 1960? when P  866, 866  1732e kt
33. 33. b) In how many years will the population be half that in 1960? when P  866, 866  1732e kt 1 e  kt 2 1 kt  log 2 1 1 t  log k 2
34. 34. b) In how many years will the population be half that in 1960? when P  866, 866  1732e kt 1 e  kt 2 1 kt  log 2 1 1 t  log k 2 t  21.786
35. 35. b) In how many years will the population be half that in 1960? when P  866, 866  1732e kt 1 e  kt 2 1 kt  log 2 1 1 t  log k 2 t  21.786  In 22 years the population has halved
36. 36. b) In how many years will the population be half that in 1960? when P  866, 866  1732e kt 1 e  kt 2 1 kt  log 2 1 1 t  log k 2 t  21.786  In 22 years the population has halved Exercise 7G; 2, 3, 7, 9, 11, 12