In a lab experiment, 4900 bacteria are placed in a petri dish. The conditions are such that the number of bacteria is able to double every 29 hours. Write a function showing the number of bacteria after tt hours, where the hourly growth rate can be found from a constant in the function. Round all coefficients in the function to four decimal places. Also, determine the percentage of growth per hour, to the nearest hundredth of a percent.

In this scenario, the number of bacteria doubles every 29 hours. This is an example of exponential growth. To model the number of bacteria after \( t \) hours, we use an exponential growth function of the form:

\[
N(t) = N_0 \cdot e^{kt}
\]

where:
- \( N(t) \) is the number of bacteria after \( t \) hours,
- \( N_0 \) is the initial number of bacteria,
- \( k \) is the growth constant,
- \( t \) is the time in hours, and
- \( e \) is the base of the natural logarithm.

### Step 1: Set up the growth function

We are given that \( N_0 = 4900 \) (the initial number of bacteria), and the number of bacteria doubles every 29 hours. Thus, after 29 hours, the population is \( 2N_0 \), or \( 2 \times 4900 = 9800 \).

We can use this information to find \( k \), the growth constant.

### Step 2: Solve for \( k \)

At \( t = 29 \) hours, \( N(29) = 2N_0 \), so:

\[
2N_0 = N_0 \cdot e^{29k}
\]

Dividing both sides by \( N_0 \):

\[
2 = e^{29k}
\]

Taking the natural logarithm of both sides:

\[
\ln(2) = 29k
\]

Solving for \( k \):

\[
k = \frac{\ln(2)}{29}
\]

Now, let's calculate \( k \).

\[
k \approx \frac{0.6931}{29} \approx 0.0239
\]

### Step 3: Write the function

Now that we know \( k \approx 0.0239 \), we can write the function for the number of bacteria after \( t \) hours:

\[
N(t) = 4900 \cdot e^{0.0239t}
\]

### Step 4: Calculate the percentage growth per hour

The growth rate per hour is given by the expression \( e^k - 1 \). This gives us the percentage growth per hour.

\[
\text{Growth rate per hour} = e^{0.0239} - 1
\]

Let's calculate this value.

\[
e^{0.0239} \approx 1.0242
\]

\[
\text{Growth rate per hour} = 1.0242 - 1 = 0.0242
\]

Thus, the growth rate per hour is approximately \( 2.42\% \).

### Final Answer:

1. The function showing the number of bacteria after \( t \) hours is:
   \[
   N(t) = 4900 \cdot e^{0.0239t}
   \]

2. The percentage growth per hour is approximately \( 2.42\%\).

---

Would you like more details on any of these steps or have further questions?

### Follow-up Questions:
1. How would the function change if the initial number of bacteria were different?
2. How would the hourly growth rate change if the bacteria doubled every 20 hours instead?
3. Can we find the time it would take for the bacteria to triple?
4. How does the natural logarithm help in solving for the growth constant?
5. What would the graph of the function \( N(t) = 4900 \cdot e^{0.0239t} \) look like?

**Tip:** When dealing with exponential growth, always look for key phrases like "doubles every X hours" to set up your equation.

In a lab experiment, 4900 bacteria are placed in a petri dish. The conditions are such that the number of bacteria is able to double every 29 hours. Write a function showing the number of bacteria after t hours, where the hourly growth rate can be found from a constant in the function. Round all coefficients in the function to four decimal places. Also, determine the percentage of growth per hour, to the nearest hundredth of a percent.

Learn how to calculate the exponential growth of bacteria in a petri dish, starting with 4900 bacteria. This step-by-step guide shows how to derive the growth function and calculate the hourly growth rate of 2.42%. Suitable for students in Grades 10-12.

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