To find the radius of curvature of the path of the particle at the given point \((\frac{\sqrt{2}}{3} \, m, \frac{2}{3} \, m)\), we will follow these steps:
### Step 1: Determine the velocity components
The velocity of the particle is given as:
\[
\vec{v} = \hat{i} + 6\hat{j} \, \text{m/s}
\]
From this, we can identify the components of velocity:
- \(v_x = 1 \, \text{m/s}\)
- \(v_y = 6 \, \text{m/s}\)
### Step 2: Find the expressions for \(x(t)\) and \(y(t)\)
Since the particle starts at the origin \((0, 0)\) at \(t = 0\), we can find the position functions by integrating the velocity components with respect to time.
For \(x(t)\):
\[
\frac{dx}{dt} = v_x = 1 \implies x(t) = t + C_x
\]
Since \(x(0) = 0\), we have \(C_x = 0\):
\[
x(t) = t
\]
For \(y(t)\):
\[
\frac{dy}{dt} = v_y = 6 \implies y(t) = 6t + C_y
\]
Since \(y(0) = 0\), we have \(C_y = 0\):
\[
y(t) = 6t
\]
### Step 3: Eliminate time to find the relationship between \(x\) and \(y\)
From \(x(t) = t\), we can express \(t\) in terms of \(x\):
\[
t = x
\]
Substituting this into the equation for \(y(t)\):
\[
y = 6t = 6x
\]
### Step 4: Find \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\)
The first derivative \(\frac{dy}{dx}\) is:
\[
\frac{dy}{dx} = 6
\]
The second derivative \(\frac{d^2y}{dx^2}\) is:
\[
\frac{d^2y}{dx^2} = 0
\]
### Step 5: Calculate the radius of curvature
The formula for the radius of curvature \(R\) is given by:
\[
R = \frac{(1 + \left(\frac{dy}{dx}\right)^2)^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}
\]
Substituting the values we found:
- \(\frac{dy}{dx} = 6\)
- \(\frac{d^2y}{dx^2} = 0\)
However, since \(\frac{d^2y}{dx^2} = 0\), we need to consider that the radius of curvature is undefined, indicating that the path is linear at this point.
### Step 6: Evaluate at the specific point
To find the radius of curvature at the point \((\frac{\sqrt{2}}{3}, \frac{2}{3})\), we can evaluate:
\[
\frac{dy}{dx} \text{ at } x = \frac{\sqrt{2}}{3} \implies \frac{dy}{dx} = 6
\]
Thus, we can substitute:
\[
R = \frac{(1 + 6^2)^{\frac{3}{2}}}{0}
\]
This indicates that the radius of curvature is infinite at this point, confirming the linear nature of the path.
### Final Answer
Since the path is linear, the radius of curvature is infinite.
