1. A colony of bacteria increases according to the law of uninhibited growth.

a) If N is the number of cells and t is the time in hours, express N as a function of t.

b) If the number of bacteria doubles in 3 hours, find the function for the number of cells in the culture.

c) How long will it take for the size of the colony to triple?

d) How long will it take for the population to double a second time (that is, increase four times)?



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.