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

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  • 1. Exponential Growth & Decay
  • 2. Exponential Growth & DecayGrowth and decay is proportional to population.
  • 3. Exponential Growth & DecayGrowth and decay is proportional to population. dP  kP dt
  • 4. Exponential Growth & DecayGrowth and decay is proportional to population. dP  kP dt dt 1  dP kP
  • 5. Exponential Growth & DecayGrowth and decay is proportional to population. dP  kP dt dt 1  dP kP 1 dP t  k P
  • 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. 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. 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. 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. 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. 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. 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. e.g.(i) The growth rate per hour of a population of bacteria is 10% of the population. The initial population is 1000000
  • 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. 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. 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. 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. 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. 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. 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. 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. (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. (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. (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. (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. (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. (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. (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. (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. (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. b) In how many years will the population be half that in 1960?
  • 32. b) In how many years will the population be half that in 1960? when P  866, 866  1732e kt
  • 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. 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. 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. 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

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