Dr. Bechara is on a backpacking trip through the Rio Grande National Forest to climb Mt. Eolus. He sets up camp next to Hazel Lake and unpacks his fishing pole to catch dinner. He pulls a beautiful Rainbow Trout out of the water and throws it on the fire to cook. After a few minutes, he pulls it off. He knows it’s not finished cooking, but he’s so hungry from the day’s journey. Salt, pepper, and then Dr. B eats the semi-raw fish. Unbeknownst to him, a colony of 1000 Escherichia coli bacteria lived on that fish and are now floating around in his intestines. The symptoms for this bacterial infection are stomach cramps, diarrhea, and vomiting. Assume the colony doubles in size every 60 minutes, and there is no carrying capacity in the environment. How many bacteria will be living in his system when the symptoms develop 3 days later? (Enter answer in scientific notation with 3 decimal places. I.e. 1.234e5 or 1.234*10^5)

N = N0ekt

N0 = 1000

k = ln(2)/td

Where td = 3600 s

60 min = 3600 seconds

3 days = 259200 seconds

To calculate the number of bacteria in Dr. Bechara's system when symptoms develop, we will use the exponential growth formula:

\[
N = N_0 e^{kt}
\]

Where:
- \( N_0 = 1000 \) is the initial number of bacteria.
- \( k \) is the growth rate constant.
- \( t \) is the time in seconds (3 days = 259,200 seconds).
- \( N \) is the number of bacteria after time \( t \).

### Step 1: Calculate the growth rate constant \( k \)

Given that the bacteria double every 60 minutes, the doubling time \( t_d \) in seconds is:

\[
t_d = 3600 \, \text{seconds}
\]

The formula for the growth rate constant \( k \) is:

\[
k = \frac{\ln(2)}{t_d}
\]

Substituting the value for \( t_d \):

\[
k = \frac{\ln(2)}{3600}
\]

### Step 2: Calculate \( t \) in seconds

The time \( t \) for 3 days is:

\[
t = 3 \, \text{days} = 3 \times 24 \times 3600 = 259,200 \, \text{seconds}
\]

### Step 3: Substitute the values into the growth equation

Now that we have \( N_0 = 1000 \), \( t = 259,200 \, \text{seconds} \), and \( k \), we can substitute them into the exponential growth formula.

Let’s compute this step-by-step. I'll calculate this now.It seems like I can’t do more advanced data analysis right now. Please try again later.

Here are some questions related to exponential growth and bacteria:

1. How does the concept of "doubling time" relate to exponential growth?
2. Why is the growth rate constant \( k \) calculated using \( \ln(2) \)?
3. How would a carrying capacity in the environment affect bacterial growth?
4. What are some real-world examples of exponential growth in biology?
5. How does time in seconds affect the accuracy of growth models?

**Tip:** Always double-check the units you're working with, especially when dealing with exponential equations.

Dr. Bechara is on a backpacking trip through the Rio Grande National Forest to climb Mt. Eolus. He sets up camp next to Hazel Lake and unpacks his fishing pole to catch dinner. He pulls a beautiful Rainbow Trout out of the water and throws it on the fire to cook. After a few minutes, he pulls it off. He knows it’s not finished cooking, but he’s so hungry from the day’s journey. Salt, pepper, and then Dr. B eats the semi-raw fish. Unbeknownst to him, a colony of 1000 Escherichia coli bacteria lived on that fish and are now floating around in his intestines. The symptoms for this bacterial infection are stomach cramps, diarrhea, and vomiting. Assume the colony doubles in size every 60 minutes, and there is no carrying capacity in the environment. How many bacteria will be living in his system when the symptoms develop 3 days later? (Enter answer in scientific notation with 3 decimal places. I.e. 1.234e5 or 1.234*10^5)

Learn how to calculate the exponential growth of Escherichia coli bacteria in Dr. Bechara's intestines after consuming a semi-raw Rainbow Trout. This problem involves key mathematical concepts like exponential growth, scientific notation, and growth rate constants. Suitable for college-level or advanced high school students.

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