A common inhabitant of human intestines is the bacterium Escherichia coli, named after the German pediatrician Theodor Escherich, who identified it in 1885. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 50 cells.
(a)
Find the relative growth rate.
k = 
  
Incorrect: Your answer is incorrect.
 hr−1
(b)
Find an expression for the number of cells after t hours.
P(t) = 
  
Incorrect: Your answer is incorrect.
(c)
Find the number of cells after 5 hours.
163,450,869
 
Incorrect: Your answer is incorrect.
 cells
(d)
Find the rate of growth after 5 hours. (Round your answer to the nearest integer.)
490,353,000
 
Incorrect: Your answer is incorrect.
 cells/h
(e)
When will the population reach a million cells? (Round your answer to two decimal places.)
3.30
 
Incorrect: Your answer is incorrect.
 h

To solve this problem involving the exponential growth of a population of *Escherichia coli* cells, we need to apply the standard exponential growth model:

\[
P(t) = P_0 e^{kt}
\]

where:
- \( P(t) \) is the population at time \( t \),
- \( P_0 \) is the initial population (in this case, 50 cells),
- \( k \) is the relative growth rate,
- \( t \) is time in hours,
- \( e \) is the base of the natural logarithm.

### Part (a): Find the relative growth rate \( k \)

We know that the bacterium doubles every 20 minutes, or \( \frac{1}{3} \) of an hour. This gives the condition:

\[
P\left(\frac{1}{3}\right) = 2P_0
\]

Using the exponential growth model, we substitute this into the equation:

\[
2P_0 = P_0 e^{k \cdot \frac{1}{3}}
\]

Dividing both sides by \( P_0 \):

\[
2 = e^{k \cdot \frac{1}{3}}
\]

Now, take the natural logarithm of both sides:

\[
\ln(2) = k \cdot \frac{1}{3}
\]

Solving for \( k \):

\[
k = 3 \ln(2) \approx 3 \times 0.6931 = 2.0794 \, \text{hr}^{-1}
\]

### Part (b): Find an expression for the number of cells after \( t \) hours

Now that we know the growth rate \( k \), we can plug it into the general exponential growth model. The initial population is \( P_0 = 50 \), so:

\[
P(t) = 50 e^{2.0794t}
\]

### Part (c): Find the number of cells after 5 hours

We can now use the formula from part (b) to calculate the population after 5 hours by substituting \( t = 5 \):

\[
P(5) = 50 e^{2.0794 \times 5}
\]

First, calculate \( 2.0794 \times 5 \):

\[
2.0794 \times 5 = 10.397
\]

Now, calculate \( e^{10.397} \):

\[
e^{10.397} \approx 32,489.94
\]

Now multiply by 50:

\[
P(5) = 50 \times 32,489.94 \approx 1,624,497
\]

Thus, the number of cells after 5 hours is approximately **1,624,497 cells**.

### Part (d): Find the rate of growth after 5 hours

The rate of growth at any time \( t \) is given by the derivative of \( P(t) \), which is:

\[
\frac{dP}{dt} = k P(t)
\]

At \( t = 5 \) hours, we already know \( P(5) \approx 1,624,497 \) and \( k \approx 2.0794 \). Therefore:

\[
\frac{dP}{dt} = 2.0794 \times 1,624,497 \approx 3,376,218 \, \text{cells/hr}
\]

Thus, the rate of growth after 5 hours is approximately **3,376,218 cells per hour**.

### Part (e): When will the population reach 1 million cells?

We use the formula \( P(t) = 50 e^{2.0794t} \) and set \( P(t) = 1,000,000 \):

\[
1,000,000 = 50 e^{2.0794t}
\]

Divide both sides by 50:

\[
20,000 = e^{2.0794t}
\]

Take the natural logarithm of both sides:

\[
\ln(20,000) = 2.0794t
\]

Solve for \( t \):

\[
t = \frac{\ln(20,000)}{2.0794} \approx \frac{9.9035}{2.0794} \approx 4.76 \, \text{hours}
\]

Thus, the population will reach 1 million cells after approximately **4.76 hours**.

---

### Summary of answers:
(a) \( k \approx 2.0794 \, \text{hr}^{-1} \)  
(b) \( P(t) = 50 e^{2.0794t} \)  
(c) \( P(5) \approx 1,624,497 \, \text{cells} \)  
(d) Rate of growth at \( t = 5 \) hours: \( \approx 3,376,218 \, \text{cells/hr} \)  
(e) Time to reach 1 million cells: \( \approx 4.76 \, \text{hours} \)

Do you want further details or clarifications?  

Here are 5 questions to deepen your understanding:
1. How is the relative growth rate of a population derived from a doubling time?
2. What are the assumptions behind the exponential growth model?
3. How does the growth rate \( k \) affect the steepness of the population curve over time?
4. What other real-world systems can be modeled by the same exponential growth formula?
5. How would the model change if there were limits to the nutrient supply?

**Tip**: When solving exponential growth problems, knowing the doubling time simplifies calculating the relative growth rate.

A common inhabitant of human intestines is the bacterium Escherichia coli, named after the German pediatrician Theodor Escherich, who identified it in 1885. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 50 cells. (a) Find the relative growth rate. k = (b) Find an expression for the number of cells after t hours. P(t) = (c) Find the number of cells after 5 hours. (d) Find the rate of growth after 5 hours. (e) When will the population reach a million cells?

This page explains how to model the exponential growth of Escherichia coli, calculating its relative growth rate, population after 5 hours, and rate of growth using the formula P(t) = P_0 e^{kt}. Learn how to determine when the population reaches 1 million cells with detailed steps.

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