This tutorial shows a detailed example of modelling a population growth using an exponential mathematical model.

1. Exponential growth
Prepared by
Ismail Mohammad El-Badawy
ismailelbadawy@gmail.com

2. Linear Vs exponential growth
𝐏 = 𝐫𝐭 + 𝐏𝐨
𝐏 = 𝐏𝐨(𝟏 + 𝐫)𝐭
OR
𝐏 = 𝐏𝐨 𝐞 𝐫𝐭
e = 2.71828182846
✓ The growth is linear.
✓ The rate (gradient) is constant.
✓ The rate is not increasing by time.
✓ The growth is non-linear.
✓ The rate (gradient) is not constant.
✓ The rate is increasing by time.

3. Exponential growth
• Population growth is a common example of exponential
growth.
• Consider a population of bacteria, for instance:
• It seems plausible that the rate of population growth would be
proportional to the size of the population.
• After all, the more bacteria there are to reproduce, the faster
the population grows.
• The figure shows the growth of a population of bacteria with
an initial population of 200 bacteria and a rate of growth
(growth constant) of 0.02.
• Notice that after only 2 hours (120 minutes), the population is
10 times its original size!
200
✓ The growth is non-linear.
✓ The rate (gradient) is not constant.
✓ The rate is increasing by time.
e = 2.71828182846OR 𝐏 = 𝐏𝐨(𝟏 + 𝐫)𝐭

4. H1N1 flu outbreak 2009 in
Japan
Time (days) Recorded cases
0 98
25 183
50 326
75 534
100 768
(25, 183)
(50, 326)
(75, 534)
(100, 768)
(0, 98)
𝐏 = 𝐏𝐨 𝐞 𝐫𝐭
𝐏 = 𝐏𝐨(𝟏 + 𝐫)𝐭
OR
Let’s find the model
Rate of change (Δy/Δx) is not fixed …….
That’s why a straight line cannot be used to model the data

5. Make r subject of the formula
P = Poert P = Po(1 + r)t

6. Make r subject of the formula
P = Poert
P
Po
= ert
ln
P
Po
= ln ert
ln
P
Po
= rt ln e
ln
P
Po
= rt
∴ 𝐫 =
𝟏
𝐭
𝐥𝐧
𝐏
𝐏𝐨
P = Po(1 + r)t
P
Po
= (1 + r)t
Log
P
Po
= Log(1 + r)t
Log
P
Po
= t Log 1 + r
1
t
Log
P
Po
= Log 1 + r
10
1
t
log
P
Po = 1 + r
∴ 𝐫 = 𝟏𝟎
𝟏
𝐭
𝐥𝐨𝐠
𝐏
𝐏 𝐨 −𝟏

7. Let’s find the model
Time (days) Recorded cases Growth rate ‘ r ‘
0 98 -
25 183
50 326
75 534
100 768
𝐫 =
𝟏
𝐭
𝐥𝐧
𝐏
𝐏𝐨
𝐏 = 𝐏𝐨 𝐞 𝐫𝐭

8. Let’s find the model
Time (days) Recorded cases Growth rate ‘ r ‘
0 98 -
25 183 0.025
50 326 0.024
75 534 0.023
100 768 0.021
✓ Average growth rate = 0.023
𝐫 =
𝟏
𝐭
𝐥𝐧
𝐏
𝐏𝐨
𝐏 = 𝐏𝐨 𝐞 𝐫𝐭
𝐏𝐨

9. 98 𝑒0.023𝑡
98
𝐏 = 𝟗𝟖 𝐞 𝟎.𝟎𝟐𝟑𝐭
Time
(days)
Actual
cases
Predicted
cases
Error
0 98
25 183
50 326
75 534
100 768

10. 98 𝑒0.023𝑡
98
𝐏 = 𝟗𝟖 𝐞 𝟎.𝟎𝟐𝟑𝐭
Time
(days)
Actual
cases
Predicted
cases
Error
0 98 98 0
25 183 174.16 8.84
50 326 309.50 16.5
75 534 550.03 -16.03
100 768 977.47 -209.47
𝐀𝐯𝐞𝐫𝐚𝐠𝐞 𝐚𝐛𝐬𝐨𝐥𝐮𝐭𝐞 𝐞𝐫𝐫𝐨𝐫
=
𝟎 + 𝟖. 𝟖𝟒 + 𝟏𝟔. 𝟓 + 𝟏𝟔. 𝟎𝟑 + 𝟐𝟎𝟗. 𝟒𝟕
𝟓
= 𝟓𝟎. 𝟏𝟔

11. Let’s find the model
Time (days) Recorded cases Growth rate ‘ r ‘
0 98 -
25 183
50 326
75 534
100 768
𝐫 = 𝟏𝟎
𝟏
𝐭 𝐥𝐨𝐠
𝐏
𝐏 𝐨 − 𝟏
𝐏 = 𝐏𝐨(𝟏 + 𝐫)𝐭

12. Let’s find the model
Time (days) Recorded cases Growth rate ‘ r ‘
0 98 -
25 183 0.025
50 326 0.024
75 534 0.023
100 768 0.021
✓ Average growth rate = 0.023
𝐏𝐨
𝐏 = 𝐏𝐨(𝟏 + 𝐫)𝐭
𝐫 = 𝟏𝟎
𝟏
𝐭 𝐥𝐨𝐠
𝐏
𝐏 𝐨 − 𝟏

13. 98(1 + 0.023) 𝑡
98
𝐏 = 𝟗𝟖(𝟏 + 𝟎. 𝟎𝟐𝟑)𝐭
Time
(days)
Actual
cases
Predicted
cases
Error
0 98
25 183
50 326
75 534
100 768

14. 98(1 + 0.023) 𝑡
98
𝐏 = 𝟗𝟖(𝟏 + 𝟎. 𝟎𝟐𝟑)𝐭
Time
(days)
Actual
cases
Predicted
cases
Error
0 98 98 0
25 183 173.03 9.97
50 326 305.48 20.52
75 534 539.38 -5.38
100 768 952.33 -184.33
𝐀𝐯𝐞𝐫𝐚𝐠𝐞 𝐚𝐛𝐬𝐨𝐥𝐮𝐭𝐞 𝐞𝐫𝐫𝐨𝐫
=
𝟎 + 𝟗. 𝟗𝟕 + 𝟐𝟎. 𝟓𝟐 + 𝟓. 𝟑𝟖 + 𝟏𝟖𝟒. 𝟑𝟑
𝟓
= 𝟒𝟒. 𝟎𝟒

15. Which model shall you
choose ?
98
𝐏 = 𝟗𝟖𝐞 𝟎.𝟎𝟐𝟑𝐭
𝐏 = 𝟗𝟖(𝟏 + 𝟎. 𝟎𝟐𝟑)𝐭
OR
Average
error
50.16
44.04