P = 500(4 + 5t/(40 + t^2)) Find the rate of change of the population at 

*t* = 3.

 (Round your answer in bacteria/hour to two decimal places.)

We are given the population function:

\[
P(t) = 500\left(4 + \frac{5t}{40 + t^2}\right)
\]

We need to find the rate of change of the population at \( t = 3 \), which means we need to compute \( \frac{dP}{dt} \) and evaluate it at \( t = 3 \).

### Step 1: Differentiate \( P(t) \)

To find the rate of change, we need to differentiate \( P(t) \) with respect to \( t \). Let's differentiate each part.

\[
P(t) = 500\left(4 + \frac{5t}{40 + t^2}\right)
\]

The derivative of \( P(t) \) is:

\[
\frac{dP}{dt} = 500 \cdot \frac{d}{dt}\left(4 + \frac{5t}{40 + t^2}\right)
\]

Since the derivative of a constant (4) is 0, we just need to focus on differentiating \( \frac{5t}{40 + t^2} \) using the quotient rule.

### Step 2: Apply the quotient rule

The quotient rule is given by:

\[
\frac{d}{dt}\left(\frac{u(t)}{v(t)}\right) = \frac{u'(t)v(t) - u(t)v'(t)}{v(t)^2}
\]

Here, \( u(t) = 5t \) and \( v(t) = 40 + t^2 \).

- The derivative of \( u(t) = 5t \) is \( u'(t) = 5 \).
- The derivative of \( v(t) = 40 + t^2 \) is \( v'(t) = 2t \).

Now, applying the quotient rule:

\[
\frac{d}{dt}\left(\frac{5t}{40 + t^2}\right) = \frac{(5)(40 + t^2) - (5t)(2t)}{(40 + t^2)^2}
\]

Simplifying:

\[
\frac{d}{dt}\left(\frac{5t}{40 + t^2}\right) = \frac{200 + 5t^2 - 10t^2}{(40 + t^2)^2}
\]

\[
= \frac{200 - 5t^2}{(40 + t^2)^2}
\]

### Step 3: Multiply by 500

Now, multiplying the derivative by 500:

\[
\frac{dP}{dt} = 500 \cdot \frac{200 - 5t^2}{(40 + t^2)^2}
\]

### Step 4: Evaluate at \( t = 3 \)

Substitute \( t = 3 \) into the derivative:

\[
\frac{dP}{dt} = 500 \cdot \frac{200 - 5(3^2)}{(40 + 3^2)^2}
\]

\[
= 500 \cdot \frac{200 - 45}{(40 + 9)^2}
\]

\[
= 500 \cdot \frac{155}{49^2}
\]

\[
= 500 \cdot \frac{155}{2401}
\]

\[
\approx 500 \cdot 0.0646
\]

\[
\approx 32.3
\]

### Final Answer:

The rate of change of the population at \( t = 3 \) is approximately **32.30 bacteria/hour**.

Would you like more details on any part of this solution? 

Here are 5 related questions:
1. How do we use the quotient rule in other contexts?
2. What does the rate of change represent in population models?
3. How would the derivative change if the formula had a different exponent?
4. Can we estimate the population increase over a longer period from this rate?
5. How do higher-order derivatives help in understanding population growth?

**Tip:** When dealing with complex functions, breaking down the function using rules like product, quotient, or chain rule step-by-step ensures accuracy.

P = 500(4 + 5t/(40 + t^2)) Find the rate of change of the population at t = 3. (Round your answer in bacteria/hour to two decimal places.)

Find the rate of change of population using calculus at t = 3. This problem involves differentiation using the quotient rule and is suitable for college-level calculus students. Key concepts include finding derivatives and applying the quotient rule.

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