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To solve this problem, let's go through it step-by-step.

### Problem Breakdown
Given:
- The rate of growth of a population \( P \) is proportional to the square root of \( t \).
- We have a differential equation:
  \[
  \frac{dP}{dt} = 9 \sqrt{t}
  \]
- The initial population size at \( t = 0 \) is \( P(0) = 300 \).
- We need to find the population size after 7 days, rounded to the nearest integer.

### Step 1: Set Up the Differential Equation
We start with:
\[
\frac{dP}{dt} = 9 \sqrt{t}
\]
To find \( P(t) \), we integrate both sides with respect to \( t \).

### Step 2: Integrate Both Sides
Integrating \( \frac{dP}{dt} \) with respect to \( t \) gives us \( P(t) \):
\[
P(t) = \int 9 \sqrt{t} \, dt
\]
Rewrite \( \sqrt{t} \) as \( t^{1/2} \):
\[
P(t) = \int 9 t^{1/2} \, dt
\]
Now, integrate:
\[
P(t) = 9 \cdot \frac{t^{3/2}}{3/2} + C = 9 \cdot \frac{2}{3} t^{3/2} + C
\]
Simplify:
\[
P(t) = 6 t^{3/2} + C
\]

### Step 3: Solve for the Constant \( C \) Using the Initial Condition
Since \( P(0) = 300 \), substitute \( t = 0 \) and \( P = 300 \) into the equation:
\[
300 = 6 \cdot 0^{3/2} + C
\]
This gives:
\[
C = 300
\]
So, the solution to the differential equation is:
\[
P(t) = 6 t^{3/2} + 300
\]

### Step 4: Calculate \( P(7) \)
Now, substitute \( t = 7 \) to find \( P(7) \):
\[
P(7) = 6 \cdot 7^{3/2} + 300
\]
Calculate \( 7^{3/2} \):
\[
7^{3/2} = \sqrt{7^3} = \sqrt{343} \approx 18.52
\]
Now, plug this into the equation:
\[
P(7) \approx 6 \cdot 18.52 + 300
\]
\[
P(7) \approx 111.12 + 300
\]
\[
P(7) \approx 411
\]

### Answer
The population after 7 days is approximately **411**.

---

Would you like further clarification on any step? Here are some related questions to explore:

1. How would the solution change if the rate of growth constant was different?
2. Can we solve this problem using numerical methods instead of integration?
3. How does changing the initial condition \( P(0) = 300 \) affect the final answer?
4. What is the interpretation of \( t^{3/2} \) in population growth models?
5. How can we verify the solution by differentiating?

**Tip:** When solving differential equations, always check if initial conditions are provided to find the particular solution.

The rate of growth of a population of bacteria is proportional to the square root of t with a constant coefficient of 9, where P is the population size and t is the time in days (0 ≤ t ≤ 10). The initial size of the population is 300. Approximate the population after 7 days. (Round the answer to the nearest integer.)

Explore how to solve a differential equation to determine bacterial population growth over 7 days. This problem involves integrating the equation dP/dt = 9√t, with an initial population of 300. Suitable for Grades 11-12.

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