To solve this problem, we are tasked with finding the position of the particle at time $t = 18$ seconds. The particle is moving along the curve $y = \frac{2x^{3/2}}{3}$, and we are given the particle’s speed as 9 cm/sec. Here's a step-by-step breakdown of the solution:

1. Determine the distance traveled by the particle at $t = 18$ seconds.

Since the particle moves at a constant speed of 9 cm/sec, the distance traveled in 18 seconds is: $\text{Distance} = \text{Speed} \times \text{Time} = 9 \, \text{cm/sec} \times 18 \, \text{sec} = 162 \, \text{cm}$ So, the total distance traveled by the particle is 162 cm.

2. Set up the arc length integral for the curve.

The formula for the arc length $L$ of a curve $y = f(x)$ from $x = 0$ to $x = b$ is given by: $L = \int_0^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx$ In this case, the curve is $y = \frac{2x^{3/2}}{3}$.

First, compute $\frac{dy}{dx}$:

$y = \frac{2x^{3/2}}{3}$ Using the chain rule, the derivative of $y$ with respect to $x$ is: $\frac{dy}{dx} = \frac{d}{dx} \left( \frac{2x^{3/2}}{3} \right) = \frac{2}{3} \times \frac{3}{2} x^{1/2} = x^{1/2}$ Thus, $\frac{dy}{dx} = x^{1/2}$.

Substitute into the arc length formula:

$L = \int_0^b \sqrt{1 + \left( x^{1/2} \right)^2} \, dx = \int_0^b \sqrt{1 + x} \, dx$

3. Solve the arc length integral.

We need to solve the integral: $L = \int_0^b \sqrt{1 + x} \, dx$ To solve this, use a substitution: Let $u = 1 + x$, so $du = dx$ and when $x = 0$, $u = 1$, and when $x = b$, $u = 1 + b$. The integral becomes: $L = \int_1^{1 + b} \sqrt{u} \, du = \int_1^{1 + b} u^{1/2} \, du$ The integral of $u^{1/2}$ is: $\int u^{1/2} \, du = \frac{2}{3} u^{3/2}$ Evaluating this from $1$ to $1 + b$ gives: [ L = \frac{2}{3} \left[ (1 + b)^{3/2} - 1^{3/2} \right] = \frac{2}{3} \left[ (1 + b)^{3/2} - 1 \right] ]

4. Set the arc length equal to the distance traveled.

We know the total arc length traveled is 162 cm, so we set the arc length $L$ equal to 162: $\frac{2}{3} \left[ (1 + b)^{3/2} - 1 \right] = 162$ Multiply both sides by $\frac{3}{2}$ to simplify: $(1 + b)^{3/2} - 1 = 243$ Add 1 to both sides: $(1 + b)^{3/2} = 244$ Now, raise both sides to the power of $\frac{2}{3}$ to solve for $1 + b$: $1 + b = 244^{2/3}$ Using the fact that $244^{2/3} \approx 36.97$, we get: $1 + b = 36.97$ Thus, $b \approx 35.97$.

5. Final answer.

The value of $b$ is approximately 35.97 cm, which means the particle's position is approximately ( x