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.2 - Mathematical Modelling Involving Inequalities

In the previous section, we saw the mathematical model for the motion of a simple pendulum. In this section, we will see another example.

Example 3:
• The situation is:
A farm house uses atleast 800 kg of special food daily. The special food is a mixture of corn and soyabean with the following compositions:
$\begin{array}{cc}{}    &{\textbf{Material}}    &{\textbf{Protein present per Kg}}    &{\textbf{Fibre present per Kg}}    &{\textbf{Cost per Kg}}\\
{}    &{\textbf{Corn}}    &{0.09}    &{0.02}    &{\text{Rs 10}}\\
{}    &{\textbf{Soyabean}}    &{0.60}    &{0.06}    &{\text{Rs 20}}\\
\end{array}$
The dietary requirements of the special food stipulate atleast 30% protein and at most 5% fibre. How can this dietary requirement be satisfied at the least possible cost?
• 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) “Atleast 800 kg” means:
    ♦ The total quantity should not be less than 800 kg.
    ♦ The total quantity can be greater than 800 kg.
(ii) The available nutrients and costs can be listed:
• One kg corn will
    ♦ give 0.09 kg protein.
    ♦ give 0.02 kg fibre.
    ♦ cost Rs 10
• One kg soya bean will give
    ♦ give 0.60 kg protein.
    ♦ give 0.06 kg fibre.
    ♦ cost Rs 20
(iii) Our aim is:
• The 800 kg (or more) food must
    ♦ contain atleast 30% protein.
    ♦ contain at most 5% fibre.
    ♦ cost the least possible amount.

• 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 the food be made up of x kg corn and y kg of soya bean.
• Then from (i) of step 1, we can write: x + y ≥ 800

(ii) Available protein and fibre:
• From the x kg corn, we will get some protein and fibre.
• From the y kg soya bean also, we will get some protein and fibre.
• Then from (ii) of step 1, we can write:
    ♦ Total protein than can be obtained from x kg corn and y kg soya bean is:
0.09x + 0.60y   
    ♦ Total fibre than can be obtained from x kg corn and y kg soya bean is:
0.02x + 0.06y

(iii) Checking whether the dietary requirements are fulfilled:
• We are taking (x+y) kg food. This will contain (0.09x + 0.60y) kg of protein.
Then from (iii) of step 1, we can write:
$\begin{array}{ll}{}    &{\frac{0.09x + 0.60y}{x+y} \times 100}    & {~\ge~}    &{30}    &{} \\
{\Rightarrow}    &{\frac{0.09x + 0.60y}{x+y}}    & {~\ge~}    &{0.30}    &{} \\
{\Rightarrow}    &{0.09x + 0.60y}    & {~\ge~}    &{0.30 x + 0.30y}    &{} \\
{\Rightarrow}    &{0.30y}    & {~\ge~}    &{0.21 x}    &{} \\
{\Rightarrow}    &{0}    & {~\ge~}    &{0.21 x – 0.30 y}    &{} \\
{\Rightarrow}    &{0.21 x – 0.30 y}    & {~\le~}    &{0}    &{} \\
\end{array}$

• We are taking (x+y) kg food. This will contain (0.02x + 0.06y) kg of fibre.
Then from (iii) of step 1, we can write:
$\begin{array}{ll}{}    &{\frac{0.02x + 0.06y}{x+y} \times 100}    & {~\le~}    &{5}    &{} \\
{\Rightarrow}    &{\frac{0.02x + 0.06y}{x+y}}    & {~\le~}    &{0.05}    &{} \\
{\Rightarrow}    &{0.02x + 0.06y}    & {~\le~}    &{0.05 x + 0.05y}    &{} \\
{\Rightarrow}    &{0.01y}    & {~\le~}    &{0.03 x}    &{} \\
{\Rightarrow}    &{0}    & {~\le~}    &{0.03 x – 0.01 y}    &{} \\
{\Rightarrow}    &{0.03 x – 0.01 y}    & {~\ge~}    &{0}    &{} \\
\end{array}$

(iv) Checking cost:
We are taking (x+y) kg food. That food will cost (10x + 20y)
Then from (iii) of step 1, we can write: (10x + 20y) should be as small as possible.

• 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 equations and inequalities into actual use. This can be written in order:

(i) From (i) of step 2, we have:
x + y ≥ 800
• The red line in fig.B.2 below represents
x + y = 800

Fig.B.2

• So any (x,y) in the upper half plane II of the red line will satisfy the inequality. See section 6.3.

(ii) From (iii) of step 2, we have:
0.21x - 0.30y ≤ 0
• The magenta line in fig.B.2 above represents
0.21x - 0.30y = 0
• So any (x,y) in the upper half plane II of the magenta line will satisfy the inequality.

(iii) Again from (iii) of step 2, we have:
0.03x - 0.01y ≥ 0
• The green line in fig.B.2 above represents
0.03x - 0.01y = 0
• So any (x,y) in the lower half plane I of the green line will satisfy the inequality.

(iv) So we have three regions:
    ♦ upper half plane II of the red line.
    ♦ upper half plane II of the magenta line.
    ♦ lower half plane I of the green line.
• The three regions will over lap in the yellow region shown in fig.B.2.
• So any (x,y) in the yellow region will satisfy all the three inequalities.

(v) From (iv) of step 2, we have:
(10x + 20y) must be as small as possible.
• For that, x and y must be as small as possible. At the same time, x and y must satisfy all three inequalities.
• This point is marked by a cyan dot in fig.B.2 above.
• If we draw a vertical line through the cyan dot, then that line will meet the x-axis at 470.59
• If we draw a horizontal line through the cyan dot, then that line will meet the y-axis at 329.41
• So the required point is (470.6,329.4)
• 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 are going to buy 470.6 kg corn and 329.4 kg soya bean.
• So total quantity = (470.6 + 329.4) = 800.
Thus “atleast 800” is satisfied.
• In this way all the dietary requirements can be checked. This will complete the validation process.
(ii) Next we do the interpretation.
• We state that:
If we buy 470.6 kg corn and 329.4 kg soya bean, then all the dietary requirements will be satisfied and at the same time, the cost will be the least possible value. The least possible cost is (470.6 × 10 + 329.4 × 20) = Rs 11294.00
• We can write:
In step 4, we do validation and interpretation.

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In the next section, we will see one more example.

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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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