Answer :
Sure, let's go through finding the number of infected individuals at various days step-by-step using the given function [tex]\( N(t) = \frac{2900}{1 + 21.8 e^{-0.5t}} \)[/tex].
### Given Data:
- Population of the town, [tex]\( P = 2900 \)[/tex]
- Infection function, [tex]\( N(t) = \frac{2900}{1 + 21.8 e^{-0.5t}} \)[/tex]
### Part b) Initial infection when [tex]\( t = 0 \)[/tex]:
To find the number of initially infected individuals, substitute [tex]\( t = 0 \)[/tex] into the infection function:
[tex]\[ N(0) = \frac{2900}{1 + 21.8 e^{-0.5 \cdot 0}} \][/tex]
Since [tex]\( e^0 = 1 \)[/tex]:
[tex]\[ N(0) = \frac{2900}{1 + 21.8 \cdot 1} = \frac{2900}{1 + 21.8} = \frac{2900}{22.8} \approx 127 \][/tex]
So, the number of initially infected individuals [tex]\( N(0) \)[/tex] is approximately 127.
### Part c) Number of infected individuals at specific days: [tex]\( t = 2 \)[/tex], [tex]\( t = 5 \)[/tex], [tex]\( t = 8 \)[/tex], [tex]\( t = 12 \)[/tex], [tex]\( t = 16 \)[/tex]:
#### For [tex]\( t = 2 \)[/tex]:
[tex]\[ N(2) = \frac{2900}{1 + 21.8 e^{-0.5 \cdot 2}} \][/tex]
Calculating the exponent:
[tex]\[ e^{-1} \approx 0.36788 \][/tex]
Thus,
[tex]\[ N(2) = \frac{2900}{1 + 21.8 \cdot 0.36788} \approx \frac{2900}{1 + 8.0178} \approx \frac{2900}{9.0178} \approx 322 \][/tex]
So, after 2 days, approximately 322 individuals are infected.
#### For [tex]\( t = 5 \)[/tex]:
[tex]\[ N(5) = \frac{2900}{1 + 21.8 e^{-0.5 \cdot 5}} \][/tex]
Calculating the exponent:
[tex]\[ e^{-2.5} \approx 0.08208 \][/tex]
Thus,
[tex]\[ N(5) = \frac{2900}{1 + 21.8 \cdot 0.08208} \approx \frac{2900}{1 + 1.788} \approx \frac{2900}{2.788} \approx 1040 \][/tex]
So, after 5 days, approximately 1040 individuals are infected.
#### For [tex]\( t = 8 \)[/tex]:
[tex]\[ N(8) = \frac{2900}{1 + 21.8 e^{-0.5 \cdot 8}} \][/tex]
Calculating the exponent:
[tex]\[ e^{-4} \approx 0.01832 \][/tex]
Thus,
[tex]\[ N(8) = \frac{2900}{1 + 21.8 \cdot 0.01832} \approx \frac{2900}{1 + 0.399376} \approx \frac{2900}{1.399376} \approx 2072 \][/tex]
So, after 8 days, approximately 2072 individuals are infected.
#### For [tex]\( t = 12 \)[/tex]:
[tex]\[ N(12) = \frac{2900}{1 + 21.8 e^{-0.5 \cdot 12}} \][/tex]
Calculating the exponent:
[tex]\[ e^{-6} \approx 0.00248 \][/tex]
Thus,
[tex]\[ N(12) = \frac{2900}{1 + 21.8 \cdot 0.00248} \approx \frac{2900}{1 + 0.054064} \approx \frac{2900}{1.054064} \approx 2751 \][/tex]
So, after 12 days, approximately 2751 individuals are infected.
#### For [tex]\( t = 16 \)[/tex]:
[tex]\[ N(16) = \frac{2900}{1 + 21.8 e^{-0.5 \cdot 16}} \][/tex]
Calculating the exponent:
[tex]\[ e^{-8} \approx 0.00034 \][/tex]
Thus,
[tex]\[ N(16) = \frac{2900}{1 + 21.8 \cdot 0.00034} \approx \frac{2900}{1 + 0.007412} \approx \frac{2900}{1.007412} \approx 2879 \][/tex]
So, after 16 days, approximately 2879 individuals are infected.
### In Summary:
- The number infected after 2 days is approximately 322.
- The number infected after 5 days is approximately 1040.
- The number infected after 8 days is approximately 2072.
- The number infected after 12 days is approximately 2751.
- The number infected after 16 days is approximately 2879.
### Given Data:
- Population of the town, [tex]\( P = 2900 \)[/tex]
- Infection function, [tex]\( N(t) = \frac{2900}{1 + 21.8 e^{-0.5t}} \)[/tex]
### Part b) Initial infection when [tex]\( t = 0 \)[/tex]:
To find the number of initially infected individuals, substitute [tex]\( t = 0 \)[/tex] into the infection function:
[tex]\[ N(0) = \frac{2900}{1 + 21.8 e^{-0.5 \cdot 0}} \][/tex]
Since [tex]\( e^0 = 1 \)[/tex]:
[tex]\[ N(0) = \frac{2900}{1 + 21.8 \cdot 1} = \frac{2900}{1 + 21.8} = \frac{2900}{22.8} \approx 127 \][/tex]
So, the number of initially infected individuals [tex]\( N(0) \)[/tex] is approximately 127.
### Part c) Number of infected individuals at specific days: [tex]\( t = 2 \)[/tex], [tex]\( t = 5 \)[/tex], [tex]\( t = 8 \)[/tex], [tex]\( t = 12 \)[/tex], [tex]\( t = 16 \)[/tex]:
#### For [tex]\( t = 2 \)[/tex]:
[tex]\[ N(2) = \frac{2900}{1 + 21.8 e^{-0.5 \cdot 2}} \][/tex]
Calculating the exponent:
[tex]\[ e^{-1} \approx 0.36788 \][/tex]
Thus,
[tex]\[ N(2) = \frac{2900}{1 + 21.8 \cdot 0.36788} \approx \frac{2900}{1 + 8.0178} \approx \frac{2900}{9.0178} \approx 322 \][/tex]
So, after 2 days, approximately 322 individuals are infected.
#### For [tex]\( t = 5 \)[/tex]:
[tex]\[ N(5) = \frac{2900}{1 + 21.8 e^{-0.5 \cdot 5}} \][/tex]
Calculating the exponent:
[tex]\[ e^{-2.5} \approx 0.08208 \][/tex]
Thus,
[tex]\[ N(5) = \frac{2900}{1 + 21.8 \cdot 0.08208} \approx \frac{2900}{1 + 1.788} \approx \frac{2900}{2.788} \approx 1040 \][/tex]
So, after 5 days, approximately 1040 individuals are infected.
#### For [tex]\( t = 8 \)[/tex]:
[tex]\[ N(8) = \frac{2900}{1 + 21.8 e^{-0.5 \cdot 8}} \][/tex]
Calculating the exponent:
[tex]\[ e^{-4} \approx 0.01832 \][/tex]
Thus,
[tex]\[ N(8) = \frac{2900}{1 + 21.8 \cdot 0.01832} \approx \frac{2900}{1 + 0.399376} \approx \frac{2900}{1.399376} \approx 2072 \][/tex]
So, after 8 days, approximately 2072 individuals are infected.
#### For [tex]\( t = 12 \)[/tex]:
[tex]\[ N(12) = \frac{2900}{1 + 21.8 e^{-0.5 \cdot 12}} \][/tex]
Calculating the exponent:
[tex]\[ e^{-6} \approx 0.00248 \][/tex]
Thus,
[tex]\[ N(12) = \frac{2900}{1 + 21.8 \cdot 0.00248} \approx \frac{2900}{1 + 0.054064} \approx \frac{2900}{1.054064} \approx 2751 \][/tex]
So, after 12 days, approximately 2751 individuals are infected.
#### For [tex]\( t = 16 \)[/tex]:
[tex]\[ N(16) = \frac{2900}{1 + 21.8 e^{-0.5 \cdot 16}} \][/tex]
Calculating the exponent:
[tex]\[ e^{-8} \approx 0.00034 \][/tex]
Thus,
[tex]\[ N(16) = \frac{2900}{1 + 21.8 \cdot 0.00034} \approx \frac{2900}{1 + 0.007412} \approx \frac{2900}{1.007412} \approx 2879 \][/tex]
So, after 16 days, approximately 2879 individuals are infected.
### In Summary:
- The number infected after 2 days is approximately 322.
- The number infected after 5 days is approximately 1040.
- The number infected after 8 days is approximately 2072.
- The number infected after 12 days is approximately 2751.
- The number infected after 16 days is approximately 2879.
