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Let's solve the problem step by step.

We are given the bacterial population \( N(t) = 3t(t - 10)^2 + 40 \), where \( t \) is in hours and \( N(t) \) is the total number of bacteria in millions.

### Part a: Find \( N'(t) \)
We need to differentiate the function \( N(t) = 3t(t - 10)^2 + 40 \).

#### Step 1: Differentiate the function
We'll apply the **product rule** to differentiate \( 3t(t - 10)^2 \) and the constant \( +40 \) will vanish upon differentiation.

Let \( u(t) = 3t \) and \( v(t) = (t - 10)^2 \). We use the product rule:
\[
\frac{d}{dt} [u(t) v(t)] = u'(t) v(t) + u(t) v'(t)
\]

Now:
- \( u(t) = 3t \), so \( u'(t) = 3 \)
- \( v(t) = (t - 10)^2 \), so \( v'(t) = 2(t - 10) \cdot 1 = 2(t - 10) \)

Using the product rule:
\[
N'(t) = u'(t)v(t) + u(t)v'(t)
\]
Substitute the expressions:
\[
N'(t) = 3(t - 10)^2 + 3t \cdot 2(t - 10)
\]
Simplify:
\[
N'(t) = 3(t - 10)^2 + 6t(t - 10)
\]

#### Step 2: Simplify \( N'(t) \)
Expand both terms:
\[
N'(t) = 3(t^2 - 20t + 100) + 6t(t - 10)
\]
\[
N'(t) = 3(t^2 - 20t + 100) + 6t^2 - 60t
\]
\[
N'(t) = 3t^2 - 60t + 300 + 6t^2 - 60t
\]
\[
N'(t) = 9t^2 - 120t + 300
\]

So, \( N'(t) = 9t^2 - 120t + 300 \).

### Part b: Find \( N'(t) \) at \( t = 8 \)
Substitute \( t = 8 \) into \( N'(t) \):
\[
N'(8) = 9(8)^2 - 120(8) + 300
\]
\[
N'(8) = 9(64) - 960 + 300
\]
\[
N'(8) = 576 - 960 + 300
\]
\[
N'(8) = -84
\]

So, \( N'(8) = -84 \).

### Part c: Find \( N'(t) \) at \( t = 11 \)
Substitute \( t = 11 \) into \( N'(t) \):
\[
N'(11) = 9(11)^2 - 120(11) + 300
\]
\[
N'(11) = 9(121) - 1320 + 300
\]
\[
N'(11) = 1089 - 1320 + 300
\]
\[
N'(11) = 69
\]

So, \( N'(11) = 69 \).

### Part d: Interpretation
- **At \( t = 8 \)**, the rate of change is negative, meaning the bacterial population is decreasing.
- **At \( t = 11 \)**, the rate of change is positive, meaning the bacterial population is increasing.

This indicates that at \( t = 8 \), the bacteria population is shrinking, but by \( t = 11 \), the bacteria are growing again.

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Would you like further details on any part of this solution, or have any questions?

### Related Questions:
1. How would you solve for when the population is neither increasing nor decreasing?
2. What is the population size at \( t = 8 \)?
3. What does the second derivative \( N''(t) \) tell us about the population's growth?
4. How can you determine the time when the population reaches its minimum?
5. What would happen if the growth formula had a different constant instead of \( +40 \)?

### Tip:
When interpreting the derivative, remember: a negative value means a decrease in the population, and a positive value indicates an increase.

Find the rate of change of bacterial population given by N(t) = 3t(t - 10)^2 + 40 at specific times.

This problem explores the rate of change of a bacterial population using the function N(t) = 3t(t - 10)^2 + 40. It applies the product rule for differentiation to find the rate at 8 and 11 hours. Suitable for high school students.

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