3. The number N of bacteria in a culture after ‘t’ days is modeled by
N=100 (1−(4/(t3+1)^2)) , find the rate of change of N with respect to ‘t’ when t = 0 and 1.

We are tasked with finding the rate of change of \( N \) with respect to \( t \) for the function \( N(t) = 100 \left( 1 - \frac{4}{(t^3 + 1)^2} \right) \).

### Step 1: Differentiate \( N(t) \) with respect to \( t \)
We need to compute the derivative \( \frac{dN}{dt} \).

Given:
\[
N(t) = 100 \left( 1 - \frac{4}{(t^3 + 1)^2} \right)
\]

To differentiate this, we apply the chain rule. Let's first find the derivative of the function inside the parentheses:

\[
N(t) = 100 \left( 1 - u(t) \right), \quad \text{where} \ u(t) = \frac{4}{(t^3 + 1)^2}
\]

Now, let's compute \( \frac{du}{dt} \).

### Step 2: Differentiate \( u(t) = \frac{4}{(t^3 + 1)^2} \)
Using the quotient rule or chain rule, we have:

\[
u(t) = 4(t^3 + 1)^{-2}
\]

Differentiating \( u(t) \) with respect to \( t \) using the chain rule:

\[
\frac{du}{dt} = -8(t^3 + 1)^{-3} \cdot \frac{d}{dt}(t^3 + 1)
\]

The derivative of \( t^3 + 1 \) is \( 3t^2 \), so:

\[
\frac{du}{dt} = -8(t^3 + 1)^{-3} \cdot 3t^2 = -24t^2(t^3 + 1)^{-3}
\]

### Step 3: Differentiate \( N(t) \)
Now, using \( \frac{dN}{dt} = 100 \left( 0 - \frac{du}{dt} \right) \):

\[
\frac{dN}{dt} = -100 \cdot \left( -24t^2(t^3 + 1)^{-3} \right)
\]

Thus:

\[
\frac{dN}{dt} = 2400t^2(t^3 + 1)^{-3}
\]

### Step 4: Evaluate at \( t = 0 \) and \( t = 1 \)

1. **At \( t = 0 \)**:
\[
\frac{dN}{dt}\bigg|_{t=0} = 2400 \cdot 0^2 \cdot (0^3 + 1)^{-3} = 0
\]

So, the rate of change at \( t = 0 \) is \( 0 \).

2. **At \( t = 1 \)**:
\[
\frac{dN}{dt}\bigg|_{t=1} = 2400 \cdot 1^2 \cdot (1^3 + 1)^{-3} = 2400 \cdot 1 \cdot (1 + 1)^{-3} = 2400 \cdot \frac{1}{8} = 300
\]

So, the rate of change at \( t = 1 \) is \( 300 \).

### Final Answer:
- The rate of change of \( N \) with respect to \( t \) at \( t = 0 \) is \( 0 \).
- The rate of change of \( N \) with respect to \( t \) at \( t = 1 \) is \( 300 \).

Would you like further details or have any questions?

---

1. How does the chain rule apply when differentiating complex functions?
2. What is the quotient rule and how can it be used in this context?
3. How does the value of \( t \) affect the rate of change in the bacteria population?
4. What happens to the rate of change as \( t \) becomes very large?
5. How does the shape of the curve for \( N(t) \) relate to the bacteria's growth?

**Tip:** Always apply the chain rule carefully when differentiating nested functions to avoid errors in the differentiation process.

The number N of bacteria in a culture after ‘t’ days is modeled by N=100 (1−(4/(t^3+1)^2)), find the rate of change of N with respect to ‘t’ when t = 0 and 1.

Learn how to find the rate of change of bacteria population modeled by N=100(1−(4/(t^3+1)^2)) using differentiation. This problem covers advanced calculus concepts such as the chain rule and quotient rule, suitable for undergraduate students.

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