Answer :
Certainly! Let's go through each part of the problem step-by-step.
### Part (a): Express N as a function of t
Given that the colony of bacteria increases according to the law of uninhibited growth, we use the formula:
[tex]\[ N(t) = N_0 \cdot e^{kt} \][/tex]
where:
- [tex]\( N(t) \)[/tex] is the number of cells at time [tex]\( t \)[/tex],
- [tex]\( N_0 \)[/tex] is the initial number of cells,
- [tex]\( k \)[/tex] is the growth rate constant,
- [tex]\( t \)[/tex] is the time in hours.
So the function is:
[tex]\[ N(t) = N_0 \cdot e^{kt} \][/tex]
### Part (b): If the number of bacteria doubles in 3 hours, find the function for the number of cells in the culture
We know that the population doubles in 3 hours, implying:
[tex]\[ N(3) = 2 \cdot N_0 \][/tex]
Using the uninhibited growth formula:
[tex]\[ 2 \cdot N_0 = N_0 \cdot e^{3k} \][/tex]
We can simplify this to:
[tex]\[ 2 = e^{3k} \][/tex]
To solve for [tex]\( k \)[/tex], take the natural logarithm of both sides:
[tex]\[ \ln(2) = 3k \][/tex]
So,
[tex]\[ k = \frac{\ln(2)}{3} \][/tex]
Thus, the function becomes:
[tex]\[ N(t) = N_0 \cdot e^{\frac{\ln(2)}{3} t} \][/tex]
### Part (c): How long will it take for the size of the colony to triple?
We need to find the time [tex]\( t \)[/tex] when the population triples, so:
[tex]\[ N(t) = 3 \cdot N_0 \][/tex]
Using the growth function:
[tex]\[ 3 \cdot N_0 = N_0 \cdot e^{\frac{\ln(2)}{3} t} \][/tex]
Simplify to:
[tex]\[ 3 = e^{\frac{\ln(2)}{3} t} \][/tex]
Taking the natural logarithm of both sides:
[tex]\[ \ln(3) = \frac{\ln(2)}{3} t \][/tex]
Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(3)}{\frac{\ln(2)}{3}} = \frac{3 \ln(3)}{\ln(2)} \][/tex]
Using the values we have determined, this calculation yields:
[tex]\[ t \approx 4.754887502163469 \][/tex]
### Part (d): How long will it take for the population to double a second time (that is, increase four times)?
We need to find the time [tex]\( t \)[/tex] when the population quadruples, so:
[tex]\[ N(t) = 4 \cdot N_0 \][/tex]
Using the growth function:
[tex]\[ 4 \cdot N_0 = N_0 \cdot e^{\frac{\ln(2)}{3} t} \][/tex]
Simplify to:
[tex]\[ 4 = e^{\frac{\ln(2)}{3} t} \][/tex]
Taking the natural logarithm of both sides:
[tex]\[ \ln(4) = \frac{\ln(2)}{3} t \][/tex]
Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(4)}{\frac{\ln(2)}{3}} = \frac{3 \ln(4)}{\ln(2)} \][/tex]
Using the values we have determined, this calculation yields:
[tex]\[ t \approx 6.0 \][/tex]
### Summary
- The growth rate constant [tex]\( k \approx 0.23104906018664842 \)[/tex]
- It takes approximately [tex]\( 4.754887502163469 \)[/tex] hours for the colony to triple in size.
- It takes exactly [tex]\( 6.0 \)[/tex] hours for the colony to quadruple in size.
### Part (a): Express N as a function of t
Given that the colony of bacteria increases according to the law of uninhibited growth, we use the formula:
[tex]\[ N(t) = N_0 \cdot e^{kt} \][/tex]
where:
- [tex]\( N(t) \)[/tex] is the number of cells at time [tex]\( t \)[/tex],
- [tex]\( N_0 \)[/tex] is the initial number of cells,
- [tex]\( k \)[/tex] is the growth rate constant,
- [tex]\( t \)[/tex] is the time in hours.
So the function is:
[tex]\[ N(t) = N_0 \cdot e^{kt} \][/tex]
### Part (b): If the number of bacteria doubles in 3 hours, find the function for the number of cells in the culture
We know that the population doubles in 3 hours, implying:
[tex]\[ N(3) = 2 \cdot N_0 \][/tex]
Using the uninhibited growth formula:
[tex]\[ 2 \cdot N_0 = N_0 \cdot e^{3k} \][/tex]
We can simplify this to:
[tex]\[ 2 = e^{3k} \][/tex]
To solve for [tex]\( k \)[/tex], take the natural logarithm of both sides:
[tex]\[ \ln(2) = 3k \][/tex]
So,
[tex]\[ k = \frac{\ln(2)}{3} \][/tex]
Thus, the function becomes:
[tex]\[ N(t) = N_0 \cdot e^{\frac{\ln(2)}{3} t} \][/tex]
### Part (c): How long will it take for the size of the colony to triple?
We need to find the time [tex]\( t \)[/tex] when the population triples, so:
[tex]\[ N(t) = 3 \cdot N_0 \][/tex]
Using the growth function:
[tex]\[ 3 \cdot N_0 = N_0 \cdot e^{\frac{\ln(2)}{3} t} \][/tex]
Simplify to:
[tex]\[ 3 = e^{\frac{\ln(2)}{3} t} \][/tex]
Taking the natural logarithm of both sides:
[tex]\[ \ln(3) = \frac{\ln(2)}{3} t \][/tex]
Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(3)}{\frac{\ln(2)}{3}} = \frac{3 \ln(3)}{\ln(2)} \][/tex]
Using the values we have determined, this calculation yields:
[tex]\[ t \approx 4.754887502163469 \][/tex]
### Part (d): How long will it take for the population to double a second time (that is, increase four times)?
We need to find the time [tex]\( t \)[/tex] when the population quadruples, so:
[tex]\[ N(t) = 4 \cdot N_0 \][/tex]
Using the growth function:
[tex]\[ 4 \cdot N_0 = N_0 \cdot e^{\frac{\ln(2)}{3} t} \][/tex]
Simplify to:
[tex]\[ 4 = e^{\frac{\ln(2)}{3} t} \][/tex]
Taking the natural logarithm of both sides:
[tex]\[ \ln(4) = \frac{\ln(2)}{3} t \][/tex]
Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(4)}{\frac{\ln(2)}{3}} = \frac{3 \ln(4)}{\ln(2)} \][/tex]
Using the values we have determined, this calculation yields:
[tex]\[ t \approx 6.0 \][/tex]
### Summary
- The growth rate constant [tex]\( k \approx 0.23104906018664842 \)[/tex]
- It takes approximately [tex]\( 4.754887502163469 \)[/tex] hours for the colony to triple in size.
- It takes exactly [tex]\( 6.0 \)[/tex] hours for the colony to quadruple in size.
