stpm mathematics (t) assignment b: mathematical modelling (2013)

ASSIGNMENT B

MATHEMATICAL MODELLING

954/4 STPM

MATHEMATICS (T)

PAPER 4

MALAYSIAN EXAMINATIONS COUNCIL

MALAYSIA HIGHER SCHOOL CERTIFICATE

Second Term: Calculus (2013)

By

Stephen, P. Y. Bong

STPM MATHEMATICS (T) – ASSIGNMENT B: MATHEMATICAL MODELLING (2013)

STEPHEN, P. Y. BONG (APRIL 2013) of 11

ASSIGNMENT PROBLEMS

The population of Aedes mosquitoes which carry the Dengue virus can be modeled by a

differential equation which describes the rate of growth of the population. The population

growth rate d

d

P

tis given by

d1

d

P PrP

t k

= − , where r is a positive constant and k is the

carrying capacity.

1. When the population is small, the relative growth rate is almost constant. What do you

understand by the term relative growth rate?

2. (a) Show that if the population does not exceed its carrying capacity, then the population

is increasing.

(b) Show that if the population exceeds its carrying capacity, then the population is

decreasing.

3. (a) Suppose that the initial population is P0. Discuss, by considering the sign ofd

d

P

t, the

relationship between P and k if P0 is less than k and if P0 is greater than k.

(b)Determine the value of P for constant growth, increasing growth and declining growth.

4. Determine the maximum value of d

d

P

tand interpret the meaning of this maximum value.

5. Express P in terms of r, k and t. Using different initial population sizes P0, r, and k, plot,

on the same axes, a few graphs to show the behavior of P versus t.

6. The population sizes of the mosquitoes in a certain area at different times, in days, are

given in the table below.

Time Number of mosquitoes Time Number of mosquitoes

0 49 245 701

35 77 252 712

63 125 322 776

91 196 371 791

105 240 392 794

126 316 406 796

140 371 441 798

182 534 504 799

203 603 539 800

224 658 567 800

It is interesting to determine the carrying capacity and the growth rate based on the above data

to control the population of Aedes mosquitoes. Using different values of P and t, plot ∆P/∆t

against P and hence obtain the approximate values for r and k.



SOLUTION TO ASSIGNMENT PROBLEMS

Question 1

The term relative growth rate mentioned in Question 1 is the ratio of population growth rate

relative to the size of population which can be mathematically expressed by dividing P on

both sides of the differential equation:

d1

d

11 d1

d

1P PrP

t

P Pr

P t kkP P

⋅ = ⋅ − ⇒ ⋅ = −

When the size of population, P, is relatively small as compared to the carrying capacity, k,

then,

0P

k→

As P/k approaches 0, then

( )1 d1 0

1

d

d

d

P Pr

P tr

Pt⋅⋅ ≈ ⇒ ≈−

Hence, it can be verified that when the size of the population is small or relatively small as

compared to the carrying capacity, then the relative growth rate is approximately a constant.

Question 2(a) (If the population does not exceeds the carrying capacity, P < k)

If P < k, then

1 1 0P

k

PP k

k< ⇔ < ⇒ − >

Since the term 1 – (P/k) is greater than zero, thus, it will results in the upsurge of population

growth rate.

Question 2(b) (If the population exceeds the carrying capacity, P > k)

If P > k, then

1 1 0P

k

PP k

k> ⇔ > ⇒ − <

The inequality above has clearly portrayed that reduction in population growth rate will

occurred if the population exceeds the carrying capacity.



Question 3 (a)

In order to obtain the relationship between the initial population, P0, population, P, and the

carrying capacity, k; a general solution of the differential equation that governs the rate of

growth of the population is a necessary precursor. The general solution of the differential

equation can be derived by the employment of separable variable as follows:

dd1 d

d1

PP PrP r t

t P

kP

k

= − ⇔ = −

Integrating both sides with their corresponding variables

d d

1

r t

k

P

PP

= −

∫ ∫

In order to perform integration on the L. H. S. of the equation, partial fraction decomposition

is required. Thus, by letting

( )11

1 1 1

PA B P

A B k

P P P PP P

k k k

− + = + =

− − −

By comparing the numerators,

( )1 1P

A B Pk

= − +

By putting P = 0 and P = k gives A = 1 and B = 1/k respectively. Hence, the partial fraction

decomposition is given by

1 1 1

1 1P P P

P kk k

= + − −

Substituting the partial fraction into the L. H. S. of the equation and perform integration yields

1 1 d d

1

ln ln 1 C

P r tP P

kk

PP rt

k

+ = −

− − = +

∫ ∫



Question 3(a) (Continued)

By using the property of logarithms: ln ln ln AB

A B− =

Cln C e1 1

rt

P P

k k

P Prt += + ⇔ =

− −

Since ert + C

= ert · e

C, therefore,

C Ce e e ; e1 1

rt rt

P P

k k

P PA A= ⋅ ⇒ = =

− −

Rearranging terms to make P the subject of the equation gives

e

e

rt

rt

kAP

k A=

+

Multiplying both the numerator and denominator by e rt− result in

e rt

kAP

A k −=

+

Given that the initial population is P0, i.e. P(t = 0) = P0, thus,

00

0

kPkA

P AA k k P

= ⇒ =+ −

Substituting 0

0

kPA

k P=− into

e

e

rt

rt

kAP

k A=+

and simplifying gives

( )0

0

0

0 0

0

0 =e

ert

rt

kP

k P

kP

k

kk

PP

P

P

P kk

−−

⋅

=+ −

−

−+

Therefore, the general solution of the differential equation that governs the population of

Aedes mosquitoes is given by

( )( )

0

0 0e rt

kPP t

P k P −=

+ −



For r > 0 and t ≥ 0, the range of the exponential function e–rt

is 0 < e–rt

≤ 1. Multiply each term

in the inequality by (k – P0) gives

( ) ( )0 0

0 e 1

0 e

rt

rt k P k P

−

− −< ≤ −

< ≤

Adding P0 on all the terms in the inequality above yields

( ) ( )( )0 0

0

0

0

0 0

0

0 e

e

rt

rt

k P k P

P P

P P P

k P k

−

−

+ < + − ≤ + −

< + − ≤

Inverting all the terms in the inequality gives

( )0 0 0

1 1 1

e rtP kP k P− −≥

+>

Multiplying kP0 to each term in the inequality:

( )0 00

0 0 0e rt

P kP k

P kP k

k

P

P

− −≥

+>

Simplifying gives

0k P P> ≥

Thus, for the case if P0 < k or k – P0 > 0, then

k > P ≥ P0

Likewise, if P0 > k or k – P0 < 0, then

k < P ≤ P0

Question 3(b)

For constant growth, dP/dt = 0. If dP/dt = 0, then

1 0

0 or 1 0

or 0

PrP

k

P

p P k

rpk

− =

= =

= =

−

Based on the table below, the range of values of P that result in increasing growth is 0 < P < k.

0P < 0 P k< < P k>

rP − + +

1 P

k− + + −

Resulting sign ( )− ( )+ ( )−



Question 3(b) (Continued)

Since it is impossible to have population that is less than zero, thus, declining growth will be

consequence if and only if the population exceeds the carrying capacity, P > k.

Question 4

In order to determine the maximum value of dP/dt, second derivative test is employed.

Differentiating dP/dt implicitly with respect to time gives

( ) ( )

( )

( )2

2

2 2

d d

d d

d d d

d d d

d d d d

d d d d

d2

d

rrP Pt t k

P rrP P

t t t k

r rr P P r P Pt t k t t k

P rr P

t k

= − = ⋅ + ⋅ − ⋅ + ⋅

= ⋅ − ⋅

��

22

2

d 2

d

d2 d

1d

1 1d

P P Pr

P

tP P

r

Pk kt

k t

= − −

⋅

= −

⋅

Likewise, differentiate d2P/dt

2 implicitly with respect to time yields

( )

22

2

3 42

2

3 4 3 42 2

2 2

2 3

2

d d d 21 1

d dd

d 3 2

d

d 3 2 3 2 d

d d

9 82

P P PrP

t t k kt

P Pr P

t k k

P P P Pr P P rt k k tk k

P Pr P

k k

= − − = − + = ⋅ − + + − + ⋅

= ⋅ − +

3 22 2

2

2

3

2

3

d

d

9 82 1

d 9 81 2

d

P P P Pr P

k kt k

P

t

P P Pr P rP

k kk

⋅ = ⋅ − + ⋅ −

= − − +



Question 4 (Continued)

For maximum value of dP/dt, d2P/dt

2 = 0 (i.e. stationary points)

2

2

2

2

d0

d2

1 1 0

20 or 1 0 or 1 0

or or 0 2

P

tP P

rPk k

P PrP

kk

P P P k

k

=

− − =

=

= = =

− = − =

Substituting P = 0, P = k/2, and P = k into d3P/dt

3 for second derivative test gives

For P = 0,

( )( ) ( )23

22

3 20

0 000

80

9d1 2

d P

Pr

k kt k=

= − − + =

For P = k/2,

2 232

3 2

22

2

2

2

d 1 9 81 2

d

1 91

2 2 2 2

16

2 24 2 2

kP

k kPr

k kt k

k

r

r

k

k

k

=

= − − + = − − +

−=

For P = k,

( ) ( ) ( ) ( )

( )( )

32 22

3 2

2 2

d 1 9 81 2

d

1 1 2

0

9 8P k

Pr

k kk k

t

r

k kk

k=

= − − + = − − +=

Based on the second derivative test conducted above, it can be concluded that when the

population of Aedes mosquitoes is half the carrying capacity, a maximum population

growth rate can be obtained.




Therefore, the maximum value of dP/dt is

Max.2

d d 11

d d 2 2

1

4kP

P P k kr rk

t t k=

= = − =

Question 5

In order to clearly visualize growth behavior of Aedes mosquitoes, both the effects of (k & P0)

and r are taken into considerations.

(a) Effects of carrying capacity, k, and initial population, P0 (r = 1)

Case I: P0 < k (Assuming P0 = 2 and k = 10)

Thus, the population is given by

( )( )( )( ) ( )110 2

2 10 2 e

20

2 8et tP t

− −= =+ +−

Case II: P0 > k (Assuming P0 = 20 and k = 10)

Likewise, the population of the Aedes mosquitoes is

( )( )( )( ) ( )110 20

20 10 2

20

2 e0 et t

P t−−

= =+ − −

Based on the two functions derived above, two graphs are plotted by the aid of Microsoft

Excel with time interval of 100 days as shown in Figure 1 and 2 below. Both the plots of

population versus time have led to a good agreement with the calculations in Question 2

above. As illustrated in Figure 1 below, if the population is less than the carrying capacity,

the population increases which also implies that the environment has the capability to

support a denser population. On the contrary, a great reduction or declination in

population growth rate will existed if the population exceeds the carrying capacity.



Figure 1: Plot of Population vs. Time (P < k)

Figure 2: Plot of Population vs. Time (P > k)

2

3

4

5

6

7

8

9

10

0 20 40 60 80 100

Popula

tion, P

Time, t (Days)

Population, P vs. Time, t (P < k)

10

11

12

13

14

15

16

17

18

19

20

0 20 40 60 80 100

Popula

tion, P

Time, t (Days)

Population, P vs. Time, t (P > k)




(b) Effect of the positive constant, r (P0 = 2 and k = 10)

By assuming r = 2, 3 and 4, the populations are:

( )

( )

( )

2

3

4

20

2 8e20

2;

3;2 8

4e

;

e20

2 8

t

t

t

r P t

r P t

r P t

−

−

−

= =

= =

=

+

+

+=

Three graphs with different values of r are plotted by using Microsoft Excel with time

interval of 10 days. As shown in Figure 3 below, r = 4 results in highest rate of

population growth as compared to the others.

Figure 3: Plot of Population, P vs. Time, t with Variable r

2

3

4

5

6

7

8

9

10

0 2 4 6 8 10

Popula

tion, P

Time, t (Days)

Plot of Population, P vs. Time, t

r = 2

r = 3

r = 4

stpm mathematics (t) assignment b: mathematical modelling (2013)

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