The given population growth equation is:

$P(t) = 480 \left( 1 + \frac{5t}{47 + t^2} \right)$

To find the rate at which the population is growing at $t = 3$, we need to calculate the derivative of $P(t)$ with respect to $t$, i.e., $\frac{dP(t)}{dt}$, and then evaluate it at $t = 3$.

Step 1: Differentiate $P(t)$

Let: $P(t) = 480 \left( 1 + \frac{5t}{47 + t^2} \right)$

The derivative is computed using the chain rule and the quotient rule. I'll first differentiate the inner function $1 + \frac{5t}{47 + t^2}$, and then multiply by the constant 480.

Step 2: Evaluate $\frac{dP}{dt}$ at $t = 3$

Let me compute that for you.The rate at which the population is growing at $t = 3$ hours is approximately $29.08$ bacteria per hour.

Would you like a more detailed explanation, or do you have any questions?

Here are 5 related questions to deepen your understanding:

1. How does the population growth rate change as $t$ increases?
2. What is the initial growth rate when $t = 0$?
3. Can you identify the maximum growth rate using calculus concepts?
4. How does the function $P(t)$ behave as $t$ approaches infinity?
5. What factors in the equation affect the growth rate most significantly?

Tip: To find the maximum or minimum rate of change, you can use the second derivative test.