B.4 - Applications of Mathematical Modelling

In the previous section, we completed the examples of mathematical modelling. We saw four examples altogether. Those examples helped us to learn the basics about mathematical modelling. In this section, we will see some interesting situations where mathematical modelling can be effectively used.

1. In human beings and all other animals, blood flows through blood vessels and reach all parts of the body. Blood carries the oxygen and nutrients which must reach all parts. Any obstruction to the flow can cause serious health issues. The characteristics of the blood vessels are such that there is no obstruction to the flow. But due to some illness, those characteristics can change. In such a situation, a mathematical model can be prepared. The model will relate the new characteristics and the resulting new pattern of blood flow. Based on the results given by the mathematical model, the physiologist can make better judgements.

2. In cricket, the third umpire has to take decisions about LBW. He analyses the trajectory of the ball assuming that the batsman is not present. A mathematical model of the trajectory will help the umpire to decide whether an LBW will occur if the batsman was actually present.

3. Meteorology department has to make whether predictions. The parameters that they have to consider are: temperature, air pressure, humidity, wind speed etc., A mathematical model is prepared by considering all the essential parameters. Precision instruments are available to measure these parameters. Using the values obtained from the instruments, the mathematical model will give the required results. Interpretation of the results enable the department to make predictions about the whether conditions.

4. Department of Agriculture can predict the quantity of rice that can be harvested in a year. The quantity of rice available from a unit area is an essential parameter. Fertility of the soil in different parts of the country, availability of water for proper irrigation etc., are also essential parameters. The mathematical model is prepared based on the principles of statistics. Interpretation of the results given by the model, enable the department to make predictions about the quantity of rice.

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Given below is an actual situation which occurred in the 18th century. It was solved by mathematical modelling. It can be written in 3 steps:

1. The Pregel river flows through Konigsberg town. So the town is on the two river banks. River bank A and River bank B. This is shown in the fig.B.4 below:

Fig.B.4

• There are two islands C and D in the river. Seven bridges connect the islands and the banks.
2. The problem is to go around the town, visiting both the islands.
• But three conditions must be satisfied:
(i) The person can start the journey from any river bank or any island. But the journey must end at the exact starting point.
(ii) The person must use all the bridges.
(iii) Each bridge must be used only once.
3. Both residents and authorities could not solve the problem. So they requested the famous mathematician Leonhard Euler to find a solution.  
• The solution put forward by Euler can be written in 5 steps:
(i) In a pictorial representation of the problem, bridges are denoted by arcs or lines. The end points of bridges are denoted by dots. Those dots are called vertices (plural of vertex).

(ii) Let us see some examples:
In fig.B.5(I) below,
• Three arcs start from A. This is because, there are three bridges starting from the river bank A. Two of them to island C and the third to island D.

Fig.B.5

• Four arcs and one line start from C. This is because, there are five bridges starting from the island C. Two of them to river bank A, two of them to river bank B and the fifth to island D.
(iii) We see that:
    ♦ A has 3 arcs/lines. 3 is an odd number.
    ♦ B has 3 arcs/lines. 3 is an odd number.
    ♦ C has 5 arcs/lines. 5 is an odd number.
    ♦ D has 3 arcs/lines. 3 is an odd number.
• If a vertex has an odd number of arcs/lines, that vertex is called an odd vertex.
• If a vertex has an even number of arcs/lines, that vertex is called an even vertex.
• So in this problem, all are odd vertices.
(iv) Euler said that, in a journey which has to satisfy the conditions in (2), there must be only two odd vertices.
• In the present problem, there are four odd vertices. So will never be able to satisfy the conditions.
(v) Later, a new bridge was added between A and B. This is shown in fig.B.5(II). Now C and D are the required two odd vertices while A and B are even vertices.

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We have completed a discussion on mathematical modelling. With that, we have completed all the chapters of class 11. In the next chapter, we will see the first chapter of class 12.

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B.3 - Mathematical Modelling Involving Exponential Functions

In the previous section, we saw the mathematical model involving inequalities. In this section, we will see another example.

Example 4:
• The situation is:
A population control unit wants to find out the population of a country after 10 years.
• This situation can be converted into a mathematical problem. For that, we engage in the process of mathematical modeling. The final result can be obtained in 4 steps:
 

Step 1:
First we study the situation. We will write it in order:

(i) The population changes with time.
For example, population next year will be different from the present population.
(ii) Population increases with births.
(iii) Population decreases with deaths.
(iv) Based on the above three information, we can write:
time, number of births and number of deaths are the parameters.

• We can write:
Step 1 involves studying the situation and identifying the parameters. Essential parameters should be carefully identified.

Step 2:
This can be written in order:
(i) Let us denote time by the letter ‘t’. It is time in years.
    ♦ So when t = 0, it indicates the present year 2023.
    ♦ When t = 1, it indicates the next year 2024
    ♦ When t = 2, it indicates the year 2025 so on . . .

(ii) Let us denote the population at any time by p(t).
• So when t = 0, the population will be p(0).
    ♦ It is the population as on first of January 2023
• When t = 1, the population will be p(1).
    ♦ It is the population as on first of January 2024 so on . . .

(iii) We can find p(1) as follows:
p(1) = population as on first of January 2024
= population as on first of January 2023 [this is p(0)]
+ Number of births in the year 2023
- Number of deaths in the year 2023

(iv) In general, we can write:
p(t+1) = Population as on the first of January of year t [this is p(t)]
+ Number of births in the year t
- Number of deaths in the year t
• Let us denote,
    ♦ Number of births in the year t as B(t)
    ♦ Number of deaths in the year t as D(t)
• Now the equation becomes:
p(t+1) = p(t) + B(t) - D(t)

(v) Now we consider the term birth rate.
• It is the number of births per 1000 of the population per year.
• For example, suppose that the birth rate is 7 and p(t) is 1500000.
• Number of “thousands” in p(t) = $\frac{1500000}{1000}$ = 1500.
• There will be 7 new births in each of those thousands.
• So the number of new births in the year t = 1500 × 7 = 10500.
• Let us denote birth rate by the letter ‘b’. Then for our present example, b = $\frac{7}{1000}$.
• We see that, number of births can be easily calculated. All we need to do is, multiply the population by b.
• That means, B(t) = p(t) × b

(vi) Similarly, if d is the death rate, then number of deaths can be calculated as:
D(t) = p(t) × d

(vii) Now the equation that we wrote in (iv) becomes:
p(t+1) = p(t) + b p(t) - d p(t)
⇒ p(t+1) = (1 + b - d) p(t)
• So if we put t = 9, we will get the population in the tenth year. But for that, we need to know the population in the ninth year.
• To know the population of the ninth year, we need to know the population of the eighth year. So on . . .
• This is a lengthy process. We do not have the populations in eighth or ninth years. All we have is p(0), which is the population of the present year.

(viii) Let us try to find an alternate method.
• Consider the equation: p(t+1) = (1 + b - d) p(t)
• Let us put t = 0. Then we get:
p(0+1) = (1 + b - d) p(0)
⇒ p(1) = (1 + b - d) p(0)
• Let us put t = 1. We get:
$\begin{array}{ll}{}    &{p(2)}    & {~=~}    &{(1+b-d) p(1)}    &{} \\
{}    &{}    & {~=~}    &{(1+b-d) (1+b-d)p(0)}    &{} \\
{}    &{}    & {~=~}    &{(1+b-d)^2 p(0)}    &{} \\
\end{array}$ 
• Let us put t = 2. We get:
$\begin{array}{ll}{}    &{p(3)}    & {~=~}    &{(1+b-d) p(2)}    &{} \\
{}    &{}    & {~=~}    &{(1+b-d) (1+b-d)^2p(0)}    &{} \\
{}    &{}    & {~=~}    &{(1+b-d)^3 p(0)}    &{} \\
\end{array}$

• We see a pattern. We can easily write:
p(t) = (1+b-d)^t p(0)

(ix) Consider the term (1+b-d). All items inside the brackets are constants. So the term as a whole will be a constant. We denote it by the letter ‘r’. It is called the growth rate. It is also known as Malthusian parameter, in honor of Robert Malthus who first brought this model to popular attention.
• So the equation in (viii) can be modified as:
p(t) = r^t p(0)

(x) Consider the equation p(t) = r^t p(0).
• With the derivation of this equation, we have completed step 2.
• We see that population is a function of t. This is because, t is the only variable. r and p(0) are constants.
• p(t) is an exponential function. Any function of the form c r^t, where c and r are constants is called an exponential function.

• We can write:
Step 2 involves drawing the necessary diagrams and writing the relevant mathematical equations/inequalities. In short, we write a mathematical problem in this step. Any mathematical problem will have a definite solution. So in this step, it is necessary to recheck all the works done thus far.

Step 3:
In this step, we put the equation into actual use. This can be written in order:

(i) Suppose that, the present population p(0) is 250000000 and the rates b and d are 0.02 and 0.01 respectively.

(ii) We get r = (1+b-d) = (1+0.02-0.01) = 1.01

(iii) So the population in the tenth year will be:
p(10) = 1.01¹⁰ × 250000000 = 276,155,531.25

• We can write:
Step 3 involves the actual application of the result obtained in step 2. Calculators or digital computers can be used for lengthy problems.

Step 4:
This can be written in order:
(i) First we do the validation.
We have obtained the population in step 3. But it has decimal values. Population cannot be in the decimal form. It must be a whole number. We can say that, the equation derived in step 2 is not a very accurate equation.
• However, we can make an approximation. We can state that, the population in the tenth year will be 276,155,531 approximately.

(ii) Next we do the interpretation.
We state that the equation derived in step 2 is not completely dependable. This is because, we assumed b and d to be constant for all the years. In actual practice, this is not possible. Improvements or decline in health management systems in the society can alter b and d. Also, if there is migration into or out of the society, population will change considerably. To get an accurate mathematical model, we must take such factors also into consideration.

• We can write:
In step 4, we do validation and interpretation.

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It is clear that, after validation and interpretation, we may need to go back to step 1 and look for additional factors and parameters that may be playing crucial roles. This is shown in the flow chart in fig.B.3 below:

Fig.B.3

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In the next section, we will see some interesting situations where mathematical modelling can be used effectively.

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B.1 Mathematical Model of a Simple Pendulum

In the previous section, we saw an example which demonstrated the basics of mathematical modeling. In this section, we will see another example.

Example 2:
• The situation is:
A physicist (scientist who specializes in the field of physics) wants to understand the motion of simple pendulum. A simple pendulum consists of a mass (called bob) attached to one end of a string. The other end of the string is fixed at a point.
• This situation can be converted into a mathematical problem. For that, we engage in the process of mathematical modeling. The final result can be obtained in 4 steps:
 

Step 1:
The physicist studies the situation. He realizes that, the factors that may be affecting the motion of a pendulum are:
(i) Period of oscillation (T)
(ii) Mass of the bob (m)
(iii) Effective length of the pendulum (l). This is the distance between the point of suspension to the center of mass of the bob.
(iv) Acceleration due to gravity (g)

• Now, the physicist tries to confirm that, the above four are indeed the parameters. For that, he performs simple experiments.
• Time period of the following two pendulums are determined experimentally:
(i) Pendulum with effective length l and mass of bob m₁.
(ii) Pendulum with same effective length l but mass of bob m₂.
• He finds that, there is no appreciable change in T.
• Again, T of following two pendulums are determined experimentally:
(i) Pendulum with effective length l₁ and mass of bob m.
(ii)Pendulum with a different effective length l₂ but same mass m.
• He finds that, there is appreciable change in T.
• So he concludes that:
    ♦ m is not an essential parameter.
    ♦ l is an essential parameter.
• We can write:
Step 1 involves studying the situation and identifying the parameters. Essential parameters should be carefully identified.

Step 2:
This can be written in order:
(i) The physicist repeats the experiment with same m and different l. He notes T in each trial.
(ii) Using the observations, he plots a graph with T along the y-axis and l along the x-axis.
• The graph thus obtained is a parabola.
(iii) Equation of a parabola is of the form y² = kx.
• So he can conclude that, the relation between T and L will be of the form T² = kl
(iv) From the values of T and l of the various trials, k can be obtained. He gets:
$k = \frac{4 \pi^2}{g}$
• Substituting in (iii), we get:
$\begin{array}{ll}{}    &{T^2}    & {~=~}    &{\frac{4 \pi^2}{g} l}    &{} \\
{\Rightarrow}    &{T}    & {~=~}    &{\sqrt{\frac{4 \pi^2}{g} l}}    &{} \\
{\Rightarrow}    &{T}    & {~=~}    &{2 \pi \sqrt{\frac{l}{g}}}    &{} \\
\end{array}$

• We can write:
Step 2 involves drawing the necessary diagrams and writing the relevant mathematical equations. In short, we write a mathematical problem in this step. Any mathematical problem will have a definite solution. So in this step, it is necessary to recheck all the works done thus far.

Step 3:
This can be written in order:
(i) The physicist puts the equation into actual use.
(ii) He assumes two values for l. They are 225 cm and 275 cm.
(iii) When l = 225 cm, he gets:
$\begin{array}{ll}{}    &{T}    & {~=~}    &{2 \pi \sqrt{\frac{l}{g}}}    &{} \\
{}    &{}    & {~=~}    &{2 \pi \sqrt{\frac{2.25}{9.81}}}    &{} \\
{}    &{}    & {~=~}    &{3.04 ~\text{sec}}    &{} \\
\end{array}$ 
(iv) When l = 275 cm, he gets:
$\begin{array}{ll}{}    &{T}    & {~=~}    &{2 \pi \sqrt{\frac{l}{g}}}    &{} \\
{}    &{}    & {~=~}    &{2 \pi \sqrt{\frac{2.75}{9.81}}}    &{} \\
{}    &{}    & {~=~}    &{3.36 ~\text{sec}}    &{} \\
\end{array}$
• We can write:
Step 3 involves the actual application of the result obtained in step 2. Calculators or digital computers can be used for lengthy problems.

Step 4:
This can be written in order:
(i) The physicist tries to confirm the results obtained in step 3. For that, he performs a number of trials of the actual experiment.

(ii) When mass (m) of the bob is 385 gms and l is 275 cm, T is 3.371 sec
• This T obtained experimentally, is comparable with the 3.36 sec that he got by the mathematical calculations in step 3.  

(iii) When m is 385 gms and l is 225 cm, T is 3.056 sec
• This T obtained experimentally, is comparable with the 3.04 sec that he got by the mathematical calculations in step 3.  

(iv) When m is 230 gms and l is 275 cm, T is 3.351 sec
• This T obtained experimentally, is comparable with the 3.36 sec that he got by the mathematical calculations in step 3.  

(v) When m is 230 gms and l is 225 cm, T is 3.042 sec
• This T obtained experimentally, is comparable with the 3.04 sec that he got by the mathematical calculations in step 3.

(vi) However, there are some small errors. For example, in (ii) above, we can find an error of (3.371 - 3.36) = 0.011.
• But such errors are small. So the mathematical model in step 3 is acceptable.
• We can conclude that:
    ♦ T is directly proportional to l.
    ♦ T is inversely proportional to g.
(vii) We see that, the mathematical model is in good agreement with the practical values. But small errors are also seen.
• These errors are due to:
    ♦ mass of the string, which we did not consider.
    ♦ resistance of the air, which also we did not consider.
• When the physicist sees the errors, he investigates further. Such an investigation will help to find the reasons for the errors.
• Once the reasons are found out, the mathematical model can be improved. The improved model may contain many complex equations.
• We can surely say that, a simple mathematical model is a good starting point.

(viii) We can write:
• In step 4, we do validation and interpretation.
◼ What is validation?
The answer can be written in 4 steps:
(a) We have two items:
    ♦ Results obtained from the mathematical model.
    ♦ Known facts about the real problem.
(b) We compare the two items
(c) If there is no appreciable difference, the mathematical model can be considered to be valid.     
(d) If there is appreciable difference, it cannot be considered to be valid.

◼ Suppose that a mathematical model is found to be valid. Once found valid, we have to do the interpretation. What is interpretation?
The answer can be written in 2 steps:
(a) We have the results obtained from the mathematical model. We state, why and how those results are obtained. In our present case, the physicist state that, T is directly proportional to l and inversely proportional to g.
(b) We also state why and how some errors (if any) occur. In our present case, the physicist state that, mass of the string and air resistance are the cause of the errors.

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So we have seen the mathematical model related to the oscillation of a simple pendulum. In the next section, we will see a mathematical model related to inequalities.

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Appendix B - Mathematical Modelling

In the previous section, we completed a discussion on infinite series. In this appendix B, we will see mathematical modelling.

• Many difficult situations that we face in science, engineering, economics, finance, management, etc., can be written in the form of mathematical problems.
• When we write such mathematical problems, there are two advantages:
   ♦ A better understanding of the situation can be obtained.
   ♦ Better solutions can be obtained to overcome the situation.
• Nowadays it has become a trend to translate difficult situations into mathematical problems. This is partly due to the fact that, advanced digital computers are now available to solve lengthy mathematical problems.

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• The process of translating a situation into a mathematical problem is called mathematical modeling. We can become experts in mathematical modeling by practicing different types of problems. Let us see a few examples:

Example 1:
• The situation is:
A surveyor wants to measure the height of a tower. Since the tower is very tall, a measuring tape cannot be used.
• This situation can be converted into a mathematical problem. For that, we engage in the process of mathematical modeling. The final result can be obtained in 4 steps:
 

Step 1:
The surveyor studies the situation. He realizes that, a triangle can be drawn and trigonometry can be applied. For applying trigonometry, he will need three parameters:
(i) Height of the tower (h)
(ii) Angle of elevation (𝜃)
(iii) Distance from the foot of the tower to the point at which angle of elevation is measured (d).
• We can write:
Step 1 involves studying the situation and identifying the parameters.
Once identified, the surveyor proceeds to obtain those parameters which can be measured.
   ♦ He measures d using a tape measure. It is 450 m
   ♦ He measures 𝜃 using a theodolite. It is 40^o.

Step 2:
The surveyor draws the triangle and writes the relevant mathematical equations. The triangle is shown in fig.B.1 below:

Fig.B.1

   ♦ Length AB is h
   ♦ Length OA is d
   ♦ Angle AOB is 𝜃.
• The equation connecting the parameters is:
$\begin{array}{ll}{}    &{\tan \theta}    & {~=~}    &{\frac{AB}{OA}}    &{} \\
{\Rightarrow}    &{\tan \theta}    & {~=~}    &{\frac{h}{d}}    &{} \\
{\Rightarrow}    &{h}    & {~=~}    &{d \tan \theta}    &{} \\
{\Rightarrow}    &{h}    & {~=~}    &{450 \times \tan 40}    &{} \\
\end{array}$
• We can write:
Step 2 involves drawing the necessary diagrams and writing the relevant mathematical equations. In short, we write a mathematical problem in this step. Any mathematical problem will have a definite solution. So in this step, it is necessary to recheck all the works done thus far.

Step 3:
The surveyor solves the equation. He gets h = 377.6 m
• We can write:
Step 3 involves solving the mathematical problem obtained in step 2. Calculators or digital computers can be used for lengthy problems.

Step 4:
The surveyor writes the height of the tower:
The height of the tower is 378 m.
• We can write:
In step 4, we write the result of the mathematical calculations. With this final step, the situation is overcome.

• This final step is called:
Interpreting the mathematical solution to the real solution.

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In the next section, we will see the case of simple pendulum.

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