a particle starts the origin and moves along the curve y=(2x^3/2)/3 in the positive x-direction at a speed of 9 cm/sec, where x and y are in cm. Find the position of the particle at t=18. First determine the distance that the particle has traveled at t=18. the arc length from the origin to this point, say x=b is equal to the distance traveled. set the arc length integral equal to the distance traveled and solve for b

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

A particle starts at the origin and moves along the curve y = (2x^3/2)/3 in the positive x-direction at a speed of 9 cm/sec, where x and y are in cm. Find the position of the particle at t = 18 seconds. First, determine the distance the particle has traveled at t = 18. The arc length from the origin to this point, say x = b, is equal to the distance traveled. Set the arc length integral equal to the distance traveled and solve for b.

Learn how to calculate the position of a particle moving along the curve y = (2x^3/2)/3 after 18 seconds using calculus. This problem involves the arc length formula and integration techniques, suitable for advanced high school or early college students.

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