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.

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.