The document discusses exponential growth and decay models. It shows that the differential equation for population growth or decay (dP/dt) is proportional to the current population (P). This leads to an exponential model of the form P=Ae^kt, where A is the initial population, k is the growth or decay constant, and t is time. Examples are given to illustrate calculating population sizes at given times and time periods for populations growing or decaying exponentially.

1. Exponential Growth & Decay

2. Exponential Growth & Decay
Growth and decay is proportional to population.

3. Exponential Growth & Decay
Growth and decay is proportional to population.
dP
 kP
dt

4. Exponential Growth & Decay
Growth and decay is proportional to population.
dP
 kP
dt
dt 1

dP kP

5. Exponential Growth & Decay
Growth and decay is proportional to population.
dP
 kP
dt
dt 1

dP kP
1 dP
t 
k P

6. Exponential Growth & Decay
Growth 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 & Decay
Growth 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 & Decay
Growth 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 & Decay
Growth 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 & Decay
Growth 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 & Decay
Growth 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 & Decay
Growth 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 1000000
a) 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
1
b) 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
1
b) 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
1
b) 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
1
b) 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