A population of 480 bacteria is introduced into a culture and grows in number according to the equation below, where t is measured in hours. Find the rate at which the population is growing when t = 3. (Round your answer to two decimal places.) The 1 is outside the fraction but inside parenthesis
P(t) = 480	(	1 + 	5t	)
47 + t2


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.

A population of 480 bacteria is introduced into a culture and grows in number according to the equation below, where t is measured in hours. Find the rate at which the population is growing when t = 3. (Round your answer to two decimal places.) The 1 is outside the fraction but inside parenthesis P(t) = 480(1 + 5t / (47 + t^2)).

Find the rate of bacterial population growth using the derivative of the function P(t) = 480(1 + 5t / (47 + t^2)) at t = 3 hours. Learn how to apply calculus concepts, including the quotient and chain rules, to solve this population growth problem. Suitable for high school students in Grades 11-12.

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