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In this section, we will: Investigate differential equations used to model population growth.

1.
9

DIFFERENTIAL EQUATIONS

DIFFERENTIAL EQUATIONS

2.
DIFFERENTIAL EQUATIONS

9.4

Models for

Population Growth

In this section, we will:

Investigate differential equations

used to model population growth.

9.4

Models for

Population Growth

In this section, we will:

Investigate differential equations

used to model population growth.

3.
NATURAL GROWTH

One of the models for population growth we

considered in Section 9.1 was based on the

assumption that the population grows at a rate

proportional to the size of the population:

dP

kP

dt

One of the models for population growth we

considered in Section 9.1 was based on the

assumption that the population grows at a rate

proportional to the size of the population:

dP

kP

dt

4.
NATURAL GROWTH

Is that a reasonable

Is that a reasonable

5.
NATURAL GROWTH

Suppose we have a population

(of bacteria, for instance) with size

P = 1000.

At a certain time, it is growing at

a rate of P’ = 300 bacteria per hour.

Suppose we have a population

(of bacteria, for instance) with size

P = 1000.

At a certain time, it is growing at

a rate of P’ = 300 bacteria per hour.

6.
NATURAL GROWTH

Now, let’s take another 1,000 bacteria

of the same type and put them with

the first population.

Each half of the new population was growing

at a rate of 300 bacteria per hour.

Now, let’s take another 1,000 bacteria

of the same type and put them with

the first population.

Each half of the new population was growing

at a rate of 300 bacteria per hour.

7.
NATURAL GROWTH

We would expect the total population

of 2,000 to increase at a rate of 600 bacteria

per hour initially—provided there’s enough

room and nutrition.

So, if we double the size, we double

the growth rate.

We would expect the total population

of 2,000 to increase at a rate of 600 bacteria

per hour initially—provided there’s enough

room and nutrition.

So, if we double the size, we double

the growth rate.

8.
NATURAL GROWTH

In general, it seems reasonable that

the growth rate should be proportional

to the size.

In general, it seems reasonable that

the growth rate should be proportional

to the size.

9.
LAW OF NATURAL GROWTH Equation 1

In general, if P(t) is the value of a quantity y

at time t and, if the rate of change of P with

respect to t is proportional to its size P(t) at

any time, then dP

kP

dt

where k is a constant.

This is sometimes called the law of natural growth.

In general, if P(t) is the value of a quantity y

at time t and, if the rate of change of P with

respect to t is proportional to its size P(t) at

any time, then dP

kP

dt

where k is a constant.

This is sometimes called the law of natural growth.

10.
LAW OF NATURAL GROWTH

If k is positive,

the population increases.

If k is negative, it decreases.

If k is positive,

the population increases.

If k is negative, it decreases.

11.
LAW OF NATURAL GROWTH

Equation 1 is a separable differential equation.

Hence, we can solve it by the methods of

dP

Section 9.3: P k dt

ln P kt C

kt C C kt

P e e e

P Ae kt

where A (= ±eC or 0) is an arbitrary constant.

Equation 1 is a separable differential equation.

Hence, we can solve it by the methods of

dP

Section 9.3: P k dt

ln P kt C

kt C C kt

P e e e

P Ae kt

where A (= ±eC or 0) is an arbitrary constant.

12.
LAW OF NATURAL GROWTH

To see the significance of the constant A,

we observe that:

P(0) = Aek·0 = A

Thus, A is the initial value of the function.

To see the significance of the constant A,

we observe that:

P(0) = Aek·0 = A

Thus, A is the initial value of the function.

13.
LAW OF NATURAL GROWTH Equation 2

The solution of the initial-value problem

dP

kP P (0) P0

dt

kt

is: P(t ) P0 e

The solution of the initial-value problem

dP

kP P (0) P0

dt

kt

is: P(t ) P0 e

14.
LAW OF NATURAL GROWTH

Another way of writing Equation 1 is:

1 dP

k

P dt

This says that the relative growth rate (the growth rate

divided by the population size) is constant.

Then, Equation 2 says that a population with

constant relative growth rate must grow exponentially.

Another way of writing Equation 1 is:

1 dP

k

P dt

This says that the relative growth rate (the growth rate

divided by the population size) is constant.

Then, Equation 2 says that a population with

constant relative growth rate must grow exponentially.

15.
LAW OF NATURAL GROWTH

We can account for emigration

(or “harvesting”) from a population

by modifying Equation 1—as follows.

We can account for emigration

(or “harvesting”) from a population

by modifying Equation 1—as follows.

16.
LAW OF NATURAL GROWTH Equation 3

If the rate of emigration is a constant m,

then the rate of change of the population

is modeled by the differential equation

dP

kP m

dt

See Exercise 13 for the solution and consequences

of Equation 3.

If the rate of emigration is a constant m,

then the rate of change of the population

is modeled by the differential equation

dP

kP m

dt

See Exercise 13 for the solution and consequences

of Equation 3.

17.
LOGISTIC MODEL

As we discussed in Section 9.1, a population

often increases exponentially in its early

stages, but levels off eventually and

approaches its carrying capacity because

of limited resources.

As we discussed in Section 9.1, a population

often increases exponentially in its early

stages, but levels off eventually and

approaches its carrying capacity because

of limited resources.

18.
LOGISTIC MODEL

If P(t) is the size of the population at time t,

we assume that:

dP

kP if P is small

dt

This says that the growth rate is initially close to

being proportional to size.

In other words, the relative growth rate is almost

constant when the population is small.

If P(t) is the size of the population at time t,

we assume that:

dP

kP if P is small

dt

This says that the growth rate is initially close to

being proportional to size.

In other words, the relative growth rate is almost

constant when the population is small.

19.
LOGISTIC MODEL

However, we also want to reflect that

the relative growth rate:

Decreases as the population P increases.

Becomes negative if P ever exceeds its carrying

capacity K (the maximum population that the

environment is capable of sustaining in the long run).

However, we also want to reflect that

the relative growth rate:

Decreases as the population P increases.

Becomes negative if P ever exceeds its carrying

capacity K (the maximum population that the

environment is capable of sustaining in the long run).

20.
LOGISTIC MODEL

The simplest expression for the

relative growth rate that incorporates

these assumptions is:

1 dP P

k 1

P dt K

The simplest expression for the

relative growth rate that incorporates

these assumptions is:

1 dP P

k 1

P dt K

21.
LOGISTIC DIFFERENTIAL EQN. Equation 4

Multiplying by P, we obtain the model

for population growth known as the logistic

differential equation:

dP P

kP 1

dt K

Multiplying by P, we obtain the model

for population growth known as the logistic

differential equation:

dP P

kP 1

dt K

22.
LOGISTIC DIFFERENTIAL EQN.

Notice from Equation 4 that:

If P is small compared with K, then P/K is close to 0,

and so dP/dt ≈ kP.

If P → K (the population approaches its carrying

capacity), then P/K → 1, so dP/dt → 0.

Notice from Equation 4 that:

If P is small compared with K, then P/K is close to 0,

and so dP/dt ≈ kP.

If P → K (the population approaches its carrying

capacity), then P/K → 1, so dP/dt → 0.

23.
LOGISTIC DIFFERENTIAL EQN.

From Equation 4, we can deduce

information about whether solutions

increase or decrease directly.

From Equation 4, we can deduce

information about whether solutions

increase or decrease directly.

24.
LOGISTIC DIFFERENTIAL EQN.

If the population P lies between 0 and K,

the right side of the equation is positive.

So, dP/dt > 0 and the population increases.

If the population exceeds the carrying

capacity (P > K), 1 – P/K is negative.

So, dP/dt < 0 and the population decreases.

If the population P lies between 0 and K,

the right side of the equation is positive.

So, dP/dt > 0 and the population increases.

If the population exceeds the carrying

capacity (P > K), 1 – P/K is negative.

So, dP/dt < 0 and the population decreases.

25.
LOGISTIC DIFFERENTIAL EQN.

Let’s start our more detailed analysis

of the logistic differential equation by

looking at a direction field.

Let’s start our more detailed analysis

of the logistic differential equation by

looking at a direction field.

26.
LOGISTIC DIFFERENTIAL EQN. Example 1

Draw a direction field for the logistic equation

with k = 0.08 and carrying capacity K = 1000.

What can you deduce about the solutions?

Draw a direction field for the logistic equation

with k = 0.08 and carrying capacity K = 1000.

What can you deduce about the solutions?

27.
LOGISTIC DIFFERENTIAL EQN. Example 1

In this case, the logistic differential

equation is:

dP P

0.08P 1

dt 1000

In this case, the logistic differential

equation is:

dP P

0.08P 1

dt 1000

28.
LOGISTIC DIFFERENTIAL EQN. Example 1

A direction field for this equation is

shown here.

A direction field for this equation is

shown here.

29.
LOGISTIC DIFFERENTIAL EQN. Example 1

We show only the first quadrant because:

Negative populations aren’t meaningful.

We are interested only in what happens after t = 0.

We show only the first quadrant because:

Negative populations aren’t meaningful.

We are interested only in what happens after t = 0.

30.
LOGISTIC DIFFERENTIAL EQN. Example 1

The logistic equation is autonomous

(dP/dt depends only on P, not on t).

So, the slopes

are the same

along any

horizontal line.

The logistic equation is autonomous

(dP/dt depends only on P, not on t).

So, the slopes

are the same

along any

horizontal line.

31.
LOGISTIC DIFFERENTIAL EQN. Example 1

As expected, the slopes are:

Positive for 0 < P < 1000

Negative for P > 1000

As expected, the slopes are:

Positive for 0 < P < 1000

Negative for P > 1000

32.
LOGISTIC DIFFERENTIAL EQN. Example 1

The slopes are small when:

P is close to 0 or 1000 (the carrying capacity).

The slopes are small when:

P is close to 0 or 1000 (the carrying capacity).

33.
LOGISTIC DIFFERENTIAL EQN. Example 1

Notice that the solutions move:

Away from the equilibrium solution P = 0

Toward the equilibrium solution P = 1000

Notice that the solutions move:

Away from the equilibrium solution P = 0

Toward the equilibrium solution P = 1000

34.
LOGISTIC DIFFERENTIAL EQN. Example 1

Here, we use the direction field to sketch

solution curves with initial populations

P(0) = 100, P(0) = 400, P(0) = 1300

Here, we use the direction field to sketch

solution curves with initial populations

P(0) = 100, P(0) = 400, P(0) = 1300

35.
LOGISTIC DIFFERENTIAL EQN. Example 1

Notice that solution curves that start:

Below P = 1000 are increasing.

Above P = 1000 are decreasing.

Notice that solution curves that start:

Below P = 1000 are increasing.

Above P = 1000 are decreasing.

36.
LOGISTIC DIFFERENTIAL EQN. Example 1

The slopes are greatest when P ≈ 500.

Thus, the solution curves that start below

P = 1000 have inflection points when P ≈ 500.

In fact, we can

prove that all

solution curves

that start below

P = 500 have

an inflection

point when P

is exactly 500.

The slopes are greatest when P ≈ 500.

Thus, the solution curves that start below

P = 1000 have inflection points when P ≈ 500.

In fact, we can

prove that all

solution curves

that start below

P = 500 have

an inflection

point when P

is exactly 500.

37.
LOGISTIC DIFFERENTIAL EQN.

The logistic equation 4 is separable.

So, we can solve it explicitly using

the method of Section 9.3

The logistic equation 4 is separable.

So, we can solve it explicitly using

the method of Section 9.3

38.
LOGISTIC DIFFERENTIAL EQN. Equation 5

dP P

kP 1

dt K

we have: dP

P(1 P / K ) k dt

dP P

kP 1

dt K

we have: dP

P(1 P / K ) k dt

39.
LOGISTIC DIFFERENTIAL EQN.

To evaluate the integral on the left side,

we write: 1 K

P (1 P / K ) P( K P )

Using partial fractions (Section 7.4),

we get:

K 1 1

P( K P) P K P

To evaluate the integral on the left side,

we write: 1 K

P (1 P / K ) P( K P )

Using partial fractions (Section 7.4),

we get:

K 1 1

P( K P) P K P

40.
LOGISTIC DIFFERENTIAL EQN. Equation 6

This enables us to rewrite Equation 5:

1 1

P K P dP k dt

ln | P | ln | K P |kt C

K P

ln kt C

P

K P kt C C kt

e e e

P

K P

Ae kt where A = ±e-C.

P

This enables us to rewrite Equation 5:

1 1

P K P dP k dt

ln | P | ln | K P |kt C

K P

ln kt C

P

K P kt C C kt

e e e

P

K P

Ae kt where A = ±e-C.

P

41.
LOGISTIC DIFFERENTIAL EQN.

Solving Equation 6 for P, we get:

K kt P 1

1 Ae kt

P K 1 Ae

Hence, K

P kt

1 Ae

Solving Equation 6 for P, we get:

K kt P 1

1 Ae kt

P K 1 Ae

Hence, K

P kt

1 Ae

42.
LOGISTIC DIFFERENTIAL EQN.

We find the value of A by putting t = 0

in Equation 6.

If t = 0, then P = P0 (the initial population),

K P0 0

Ae A

P0

We find the value of A by putting t = 0

in Equation 6.

If t = 0, then P = P0 (the initial population),

K P0 0

Ae A

P0

43.
LOGISTIC DIFFERENTIAL EQN. Equation 7

Therefore, the solution to the logistic

equation is:

K K P0

P(t ) kt

where A

1 Ae P0

Therefore, the solution to the logistic

equation is:

K K P0

P(t ) kt

where A

1 Ae P0

44.
LOGISTIC DIFFERENTIAL EQN.

Using the expression for P(t) in Equation 7,

we see that

lim P(t ) K

t

which is to be expected.

Using the expression for P(t) in Equation 7,

we see that

lim P(t ) K

t

which is to be expected.

45.
LOGISTIC DIFFERENTIAL EQN. Example 2

Write the solution of the initial-value problem

dP P

0.08 P 1 P(0) 100

dt 1000

and use it to find the population sizes P(40)

and P(80).

At what time does the population reach 900?

Write the solution of the initial-value problem

dP P

0.08 P 1 P(0) 100

dt 1000

and use it to find the population sizes P(40)

and P(80).

At what time does the population reach 900?

46.
LOGISTIC DIFFERENTIAL EQN. Example 2

The differential equation is a logistic

equation with:

k = 0.08

Carrying capacity K = 1000

Initial population P0 = 100

The differential equation is a logistic

equation with:

k = 0.08

Carrying capacity K = 1000

Initial population P0 = 100

47.
LOGISTIC DIFFERENTIAL EQN. Example 2

Thus, Equation 7 gives the population at

time t as:

1000 1000 100

P(t ) where A 9

1 Ae 0.08t 100

Therefore, 1000

P(t ) 0.08t

1 9e

Thus, Equation 7 gives the population at

time t as:

1000 1000 100

P(t ) where A 9

1 Ae 0.08t 100

Therefore, 1000

P(t ) 0.08t

1 9e

48.
LOGISTIC DIFFERENTIAL EQN. Example 2

Hence, the population sizes when t = 40

and 80 are:

1000

P (40) 3.2

731.6

1 9e

1000

P (80) 6.4

985.3

1 9e

Hence, the population sizes when t = 40

and 80 are:

1000

P (40) 3.2

731.6

1 9e

1000

P (80) 6.4

985.3

1 9e

49.
LOGISTIC DIFFERENTIAL EQN. Example 2

The population reaches 900

1000

0.08t

900

1 9e

The population reaches 900

1000

0.08t

900

1 9e

50.
LOGISTIC DIFFERENTIAL EQN. Example 2

Solving this equation for t, we get:

0.08t

1 9e 109

0.08t

e 811

0.08t ln 811 ln 81

ln 81

t 54.9

0.08

So, the population reaches 900 when t is

approximately 55.

Solving this equation for t, we get:

0.08t

1 9e 109

0.08t

e 811

0.08t ln 811 ln 81

ln 81

t 54.9

0.08

So, the population reaches 900 when t is

approximately 55.

51.
LOGISTIC DIFFERENTIAL EQN. Example 2

As a check on our work, we graph

the population curve and observe where

it intersects the line P = 900.

The cursor

indicates that

t ≈ 55.

As a check on our work, we graph

the population curve and observe where

it intersects the line P = 900.

The cursor

indicates that

t ≈ 55.

52.
COMPARING THE MODELS

In the 1930s, the biologist G. F. Gause

conducted an experiment with the protozoan

Paramecium and used a logistic equation to

model his data.

In the 1930s, the biologist G. F. Gause

conducted an experiment with the protozoan

Paramecium and used a logistic equation to

model his data.

53.
COMPARING THE MODELS

The table gives his daily count of

the population of protozoa.

He estimated the initial relative growth rate

to be 0.7944 and the carrying capacity to be 64.

The table gives his daily count of

the population of protozoa.

He estimated the initial relative growth rate

to be 0.7944 and the carrying capacity to be 64.

54.
COMPARING THE MODELS Example 3

Find the exponential and logistic models

for Gause’s data.

Compare the predicted values with the

observed values and comment on the fit.

Find the exponential and logistic models

for Gause’s data.

Compare the predicted values with the

observed values and comment on the fit.

55.
COMPARING THE MODELS Example 3

Given the relative growth rate k = 0.7944

and the initial population P0 = 2,

the exponential model is:

kt 0.7944 t

P (t ) P0 e 2e

Given the relative growth rate k = 0.7944

and the initial population P0 = 2,

the exponential model is:

kt 0.7944 t

P (t ) P0 e 2e

56.
COMPARING THE MODELS Example 3

Gause used the same value of k for

his logistic model.

This is reasonable as P0 = 2 is small compared

with the carrying capacity (K = 64).

The equation 1 dP 2

k 1 k

P0 dt t 0 64

shows that the value of k for the logistic model is

very close to the value for the exponential model.

Gause used the same value of k for

his logistic model.

This is reasonable as P0 = 2 is small compared

with the carrying capacity (K = 64).

The equation 1 dP 2

k 1 k

P0 dt t 0 64

shows that the value of k for the logistic model is

very close to the value for the exponential model.

57.
COMPARING THE MODELS Example 3

Then, the solution of the logistic equation

in Equation 7 gives

K 64

P (t ) kt

0.7944 t

1 Ae 1 Ae

where K P0 64 2

A 31

P0 2

64

So, P (t ) 0.7944 t

1 31e

Then, the solution of the logistic equation

in Equation 7 gives

K 64

P (t ) kt

0.7944 t

1 Ae 1 Ae

where K P0 64 2

A 31

P0 2

64

So, P (t ) 0.7944 t

1 31e

58.
COMPARING THE MODELS Example 3

We use these equations to calculate

the predicted values and compare them

We use these equations to calculate

the predicted values and compare them

59.
COMPARING THE MODELS Example 3

Now, let’s compare

the table with this

Now, let’s compare

the table with this

60.
COMPARING THE MODELS Example 3

For the first three

or four days:

The exponential model

gives results

comparable to those of

the more sophisticated

logistic model.

For the first three

or four days:

The exponential model

gives results

comparable to those of

the more sophisticated

logistic model.

61.
COMPARING THE MODELS Example 3

However, for t ≥ 5:

The exponential model

is hopelessly

inaccurate.

The logistic model fits

the observations

reasonably well.

However, for t ≥ 5:

The exponential model

is hopelessly

inaccurate.

The logistic model fits

the observations

reasonably well.

62.
COMPARING THE MODELS

Many countries that formerly experienced

exponential growth are now finding that their

rates of population growth are declining and

the logistic model provides a better model.

Many countries that formerly experienced

exponential growth are now finding that their

rates of population growth are declining and

the logistic model provides a better model.

63.
COMPARING THE MODELS

The table shows midyear values of B(t),

the population of Belgium, in thousands,

at time t, from 1980 to 2000.

The table shows midyear values of B(t),

the population of Belgium, in thousands,

at time t, from 1980 to 2000.

64.
COMPARING THE MODELS

This figure shows the data points of the table

together with a shifted logistic function

obtained from a calculator with the ability to fit

a logistic function to these points by

This figure shows the data points of the table

together with a shifted logistic function

obtained from a calculator with the ability to fit

a logistic function to these points by

65.
COMPARING THE MODELS

We see that the logistic model

provides a very good fit.

We see that the logistic model

provides a very good fit.

66.
MODELS FOR POPULATION GROWTH

The Law of Natural Growth and the logistic

differential equation are not the only equations

that have been proposed to model population

The Law of Natural Growth and the logistic

differential equation are not the only equations

that have been proposed to model population

67.
OTHER MODELS FOR POPULATION GROWTH

In Exercise 18, we look at the Gompertz

growth function.

In Exercises 19 and 20, we investigate

seasonal-growth models.

In Exercise 18, we look at the Gompertz

growth function.

In Exercises 19 and 20, we investigate

seasonal-growth models.

68.
OTHER MODELS FOR POPULATION GROWTH

Two of the other models

are modifications of the logistic

Two of the other models

are modifications of the logistic

69.
OTHER MODELS FOR POPULATION GROWTH

The differential equation dP P

kP 1 c

dt K

has been used to model populations that are

subject to harvesting of one sort or another.

Think of a population of fish being caught

at a constant rate.

This equation is explored in Exercises 15 and 16.

The differential equation dP P

kP 1 c

dt K

has been used to model populations that are

subject to harvesting of one sort or another.

Think of a population of fish being caught

at a constant rate.

This equation is explored in Exercises 15 and 16.

70.
OTHER MODELS FOR POPULATION GROWTH

For some species, there is a minimum

population level m below which the species

tends to become extinct.

Adults may not be able to find suitable mates.

For some species, there is a minimum

population level m below which the species

tends to become extinct.

Adults may not be able to find suitable mates.

71.
OTHER MODELS FOR POPULATION GROWTH

Such populations have been modeled by

the differential equation

dP P m

kP 1 1

dt K P

where the extra factor, 1 – m/P,

takes into account the consequences

of a sparse population.

Such populations have been modeled by

the differential equation

dP P m

kP 1 1

dt K P

where the extra factor, 1 – m/P,

takes into account the consequences

of a sparse population.