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Let's address the problem step-by-step:

The problem involves a bacteria culture that initially contains 100 cells and grows at a rate proportional to its size. After 1 hour, the population increases to 420 cells. We need to solve the following parts:

### Given Information:
- Initial number of cells (\( N_0 \)): 100
- Number of cells after 1 hour: 420
- The growth is exponential, so we can use the general formula for exponential growth:
\[
N(t) = N_0 e^{kt}
\]

where:
- \( N(t) \) is the number of cells at time \( t \),
- \( N_0 \) is the initial number of cells,
- \( k \) is the growth rate constant,
- \( t \) is the time in hours.

### Step-by-Step Solution:

#### (a) Find an expression for the number of bacteria after \( t \) hours.
1. **Start with the exponential growth formula:**
   \[
   N(t) = 100 e^{kt}
   \]
2. **Determine the value of \( k \) using the given information.**
   Since \( N(1) = 420 \), we have:
   \[
   420 = 100 e^{k \cdot 1}
   \]
   \[
   4.2 = e^{k}
   \]
   Taking the natural logarithm on both sides:
   \[
   k = \ln(4.2) \approx 1.4351
   \]
3. **Final expression:**
   \[
   N(t) = 100 e^{1.4351t}
   \]

#### (b) Find the number of bacteria after 3 hours.
1. Use the expression from part (a):
   \[
   N(3) = 100 e^{1.4351 \cdot 3}
   \]
2. Calculate:
   \[
   N(3) = 100 e^{4.3053} \approx 7,414
   \]

So, the number of bacteria after 3 hours is approximately **7,414**.

#### (c) Find the rate of growth after 3 hours.
1. The rate of growth is given by the derivative \( \frac{dN}{dt} \):
   \[
   \frac{dN}{dt} = 100 \cdot 1.4351 e^{1.4351t}
   \]
2. At \( t = 3 \):
   \[
   \frac{dN}{dt}\bigg|_{t=3} = 100 \cdot 1.4351 \cdot e^{4.3053} \approx 10,631
   \]

The rate of growth after 3 hours is approximately **10,631 cells per hour**.

#### (d) When will the population reach 10,000?
1. Set \( N(t) = 10,000 \) and solve for \( t \):
   \[
   10,000 = 100 e^{1.4351t}
   \]
   \[
   100 = e^{1.4351t}
   \]
   \[
   \ln(100) = 1.4351t
   \]
   \[
   t = \frac{\ln(100)}{1.4351} \approx 3.215
   \]

The population will reach 10,000 in approximately **3.22 hours**.

### Summary of Answers:
1. **(a)** \( N(t) = 100 e^{1.4351t} \)
2. **(b)** Approximately 7,414 bacteria after 3 hours.
3. **(c)** Growth rate of approximately 10,631 cells per hour after 3 hours.
4. **(d)** Population reaches 10,000 after about 3.22 hours.

Would you like more details or have any further questions?

### Related Questions:
1. What would the population be after 5 hours?
2. How does the growth rate change over time, and why?
3. If the initial population was 200 instead of 100, how would the equation change?
4. Can we model a scenario where the growth rate decreases over time?
5. What would happen if the growth rate was halved?

### Tip:
Exponential growth models are useful for understanding many real-world scenarios, including populations, finance (compound interest), and radioactive decay.

A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour, the population has increased to 420. (a) Find an expression for the number of bacteria after t hours. (b) Find the number of bacteria after 3 hours. (c) Find the rate of growth after 3 hours. (d) When will the population reach 10,000?

Explore how to solve an exponential growth problem involving a bacteria culture that starts with 100 cells and grows to 420 in 1 hour. Find the population after 3 hours, the rate of growth, and the time when the population reaches 10,000. Includes key concepts in exponential growth and calculus.

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