Answer :
Let's solve the problem step-by-step.
### Given:
The number of infected bacteria after [tex]\( d \)[/tex] days is given by the equation:
[tex]\[ N(d) = (26d) e^{-\frac{d}{12}} \][/tex]
### a) How many bacteria are infected after 13 days?
To find the number of bacteria infected after 13 days, we substitute [tex]\( d = 13 \)[/tex] into the equation [tex]\( N(d) \)[/tex]:
[tex]\[ N(13) = (26 \cdot 13) e^{-\frac{13}{12}} \][/tex]
Calculating this gives:
[tex]\[ N(13) \approx 114.4013 \][/tex]
So, approximately 114.4013 bacteria are infected after 13 days.
### b) What is the rate of change of the infected population after 13 days? Is the infected population increasing or decreasing?
To find the rate of change of the infected population after 13 days, we need to differentiate [tex]\( N(d) \)[/tex] with respect to [tex]\( d \)[/tex]:
[tex]\[ N'(d) = \frac{d}{dd} \left[ (26d)e^{-\frac{d}{12}} \right] \][/tex]
Using the product rule:
[tex]\[ N'(d) = 26 e^{-\frac{d}{12}} + (26d)\left( -\frac{1}{12} \right)e^{-\frac{d}{12}} \][/tex]
Simplifying:
[tex]\[ N'(d) = 26 e^{-\frac{d}{12}} - \frac{26d}{12} e^{-\frac{d}{12}} \][/tex]
[tex]\[ N'(d) = 26 e^{-\frac{d}{12}} \left( 1 - \frac{d}{12} \right) \][/tex]
Substituting [tex]\( d = 13 \)[/tex] into [tex]\( N'(d) \)[/tex]:
[tex]\[ N'(13) = 26 e^{-\frac{13}{12}} \left( 1 - \frac{13}{12} \right) \][/tex]
Calculating this gives:
[tex]\[ N'(13) \approx -0.7333 \][/tex]
Since [tex]\( N'(13) \)[/tex] is negative, the infected population is decreasing after 13 days.
### c) When does the infected bacterial population peak? How many bacteria are infected at that time?
To find when the population peaks, we need to find the critical points by setting [tex]\( N'(d) = 0 \)[/tex]:
[tex]\[ 26 e^{-\frac{d}{12}} \left( 1 - \frac{d}{12} \right) = 0 \][/tex]
This equation holds when:
[tex]\[ 1 - \frac{d}{12} = 0 \][/tex]
[tex]\[ \frac{d}{12} = 1 \][/tex]
[tex]\[ d = 12 \][/tex]
To determine the number of infected bacteria at that peak, substitute [tex]\( d = 12 \)[/tex] into the original equation [tex]\( N(d) \)[/tex]:
[tex]\[ N(12) = (26 \cdot 12) e^{-\frac{12}{12}} \][/tex]
[tex]\[ N(12) = 312 e^{-1} \][/tex]
Calculating this gives:
[tex]\[ N(12) \approx 114.7784 \][/tex]
So, the population of infected bacteria peaks at 12 days with approximately 114.7784 bacteria infected.
### d) When is the number of infected bacteria in the population changing most rapidly?
To determine when the number of infected bacteria is changing most rapidly, we need to find the inflection points of [tex]\( N(d) \)[/tex].
This involves finding the second derivative:
[tex]\[ N''(d) \][/tex]
Taking the second derivative requires:
[tex]\[ N''(d) = \frac{d}{dd}[26 e^{-\frac{d}{12}} (1-\frac{d}{12})] \][/tex]
[tex]\[ N''(d) = \frac{d}{dd}[26 e^{-\frac{d}{12}}] (1-\frac{d}{12}) + 26 e^{-\frac{d}{12}} \frac{d}{dd}(1-\frac{d}{12}) \][/tex]
[tex]\[ N''(d) = -\frac{26}{12} e^{-\frac{d}{12}} (1-\frac{d}{12}) + 26 e^{-\frac{d}{12}} (-\frac{1}{12}) \][/tex]
Setting [tex]\( N''(d) = 0 \)[/tex] and solving for [tex]\( d \)[/tex] gives us the inflection point. Upon finding the roots, we find the most relevant inflection point within our context is:
[tex]\[ d \approx 11 \][/tex]
Therefore, the number of infected bacteria is changing most rapidly at approximately day 11, and the number of infected bacteria at that time is:
[tex]\[ N(11) \approx 114.3570 \][/tex]
So, the number of infected bacteria changes most rapidly around day 11 with approximately 114.3570 bacteria infected at that time.
### Given:
The number of infected bacteria after [tex]\( d \)[/tex] days is given by the equation:
[tex]\[ N(d) = (26d) e^{-\frac{d}{12}} \][/tex]
### a) How many bacteria are infected after 13 days?
To find the number of bacteria infected after 13 days, we substitute [tex]\( d = 13 \)[/tex] into the equation [tex]\( N(d) \)[/tex]:
[tex]\[ N(13) = (26 \cdot 13) e^{-\frac{13}{12}} \][/tex]
Calculating this gives:
[tex]\[ N(13) \approx 114.4013 \][/tex]
So, approximately 114.4013 bacteria are infected after 13 days.
### b) What is the rate of change of the infected population after 13 days? Is the infected population increasing or decreasing?
To find the rate of change of the infected population after 13 days, we need to differentiate [tex]\( N(d) \)[/tex] with respect to [tex]\( d \)[/tex]:
[tex]\[ N'(d) = \frac{d}{dd} \left[ (26d)e^{-\frac{d}{12}} \right] \][/tex]
Using the product rule:
[tex]\[ N'(d) = 26 e^{-\frac{d}{12}} + (26d)\left( -\frac{1}{12} \right)e^{-\frac{d}{12}} \][/tex]
Simplifying:
[tex]\[ N'(d) = 26 e^{-\frac{d}{12}} - \frac{26d}{12} e^{-\frac{d}{12}} \][/tex]
[tex]\[ N'(d) = 26 e^{-\frac{d}{12}} \left( 1 - \frac{d}{12} \right) \][/tex]
Substituting [tex]\( d = 13 \)[/tex] into [tex]\( N'(d) \)[/tex]:
[tex]\[ N'(13) = 26 e^{-\frac{13}{12}} \left( 1 - \frac{13}{12} \right) \][/tex]
Calculating this gives:
[tex]\[ N'(13) \approx -0.7333 \][/tex]
Since [tex]\( N'(13) \)[/tex] is negative, the infected population is decreasing after 13 days.
### c) When does the infected bacterial population peak? How many bacteria are infected at that time?
To find when the population peaks, we need to find the critical points by setting [tex]\( N'(d) = 0 \)[/tex]:
[tex]\[ 26 e^{-\frac{d}{12}} \left( 1 - \frac{d}{12} \right) = 0 \][/tex]
This equation holds when:
[tex]\[ 1 - \frac{d}{12} = 0 \][/tex]
[tex]\[ \frac{d}{12} = 1 \][/tex]
[tex]\[ d = 12 \][/tex]
To determine the number of infected bacteria at that peak, substitute [tex]\( d = 12 \)[/tex] into the original equation [tex]\( N(d) \)[/tex]:
[tex]\[ N(12) = (26 \cdot 12) e^{-\frac{12}{12}} \][/tex]
[tex]\[ N(12) = 312 e^{-1} \][/tex]
Calculating this gives:
[tex]\[ N(12) \approx 114.7784 \][/tex]
So, the population of infected bacteria peaks at 12 days with approximately 114.7784 bacteria infected.
### d) When is the number of infected bacteria in the population changing most rapidly?
To determine when the number of infected bacteria is changing most rapidly, we need to find the inflection points of [tex]\( N(d) \)[/tex].
This involves finding the second derivative:
[tex]\[ N''(d) \][/tex]
Taking the second derivative requires:
[tex]\[ N''(d) = \frac{d}{dd}[26 e^{-\frac{d}{12}} (1-\frac{d}{12})] \][/tex]
[tex]\[ N''(d) = \frac{d}{dd}[26 e^{-\frac{d}{12}}] (1-\frac{d}{12}) + 26 e^{-\frac{d}{12}} \frac{d}{dd}(1-\frac{d}{12}) \][/tex]
[tex]\[ N''(d) = -\frac{26}{12} e^{-\frac{d}{12}} (1-\frac{d}{12}) + 26 e^{-\frac{d}{12}} (-\frac{1}{12}) \][/tex]
Setting [tex]\( N''(d) = 0 \)[/tex] and solving for [tex]\( d \)[/tex] gives us the inflection point. Upon finding the roots, we find the most relevant inflection point within our context is:
[tex]\[ d \approx 11 \][/tex]
Therefore, the number of infected bacteria is changing most rapidly at approximately day 11, and the number of infected bacteria at that time is:
[tex]\[ N(11) \approx 114.3570 \][/tex]
So, the number of infected bacteria changes most rapidly around day 11 with approximately 114.3570 bacteria infected at that time.
