A particle moves in the plane `xy` with velocity `v=ai+bxj`, where i and j are the unit vectors of the x and y axes, and a and b are constants. At the initial moment of time the particle was located at the point `x=y=0`. Find:
(a) the equation of the particle's trajectory `y(x)`,
(b) the curvature radius of trajectory as a function of x.

Explanation

To solve the problem step by step, we will address both parts: finding the trajectory of the particle and then determining the radius of curvature of that trajectory.

### Part (a): Finding the Equation of the Particle's Trajectory \( y(x) \)

1. **Write the Velocity Components**:
The velocity of the particle is given as:
\[ \mathbf{v} = a \hat{i} + b \hat{j} \]
This can be expressed in terms of its components:
\[ \frac{dx}{dt} = a \quad \text{and} \quad \frac{dy}{dt} = b x \]

2. **Integrate the x-component**:
From the first equation, we can integrate to find \( x \):
\[ dx = a \, dt \implies x = a t + C \]
Since the particle starts at the origin \( (0,0) \) at \( t = 0 \), \( C = 0 \):
\[ x = a t \]

3. **Express \( t \) in terms of \( x \)**:
Rearranging gives:
\[ t = \frac{x}{a} \]

4. **Substitute \( t \) into the y-component**:
Now, substitute \( t \) into the equation for \( y \):
\[ \frac{dy}{dt} = b x \implies dy = b x \, dt \]
Substituting \( dt = \frac{dx}{a} \):
\[ dy = b x \left( \frac{dx}{a} \right) \implies dy = \frac{b}{a} x \, dx \]

5. **Integrate to find \( y \)**:
Integrating both sides:
\[ y = \frac{b}{a} \int x \, dx = \frac{b}{a} \cdot \frac{x^2}{2} + C \]
Again, since the initial position is \( (0,0) \), \( C = 0 \):
\[ y = \frac{b}{2a} x^2 \]

6. **Final Trajectory Equation**:
Thus, the equation of the trajectory is:
\[ y(x) = \frac{b}{2a} x^2 \]

### Part (b): Finding the Curvature Radius of the Trajectory as a Function of \( x \)

1. **Find the Velocity Magnitude**:
The velocity vector is:
\[ \mathbf{v} = a \hat{i} + b x \hat{j} \]
The magnitude of the velocity \( v \) is:
\[ v = \sqrt{a^2 + (b x)^2} \]

2. **Find the Acceleration**:
The acceleration \( \mathbf{a} \) is the derivative of the velocity:
\[ \mathbf{a} = \frac{d}{dt}(a \hat{i} + b x \hat{j}) = 0 \hat{i} + b \frac{dx}{dt} \hat{j} = b a \hat{j} \]

3. **Find the Radius of Curvature**:
The radius of curvature \( R \) is given by the formula:
\[ R = \frac{v^2}{|a_n|} \]
where \( a_n \) is the normal acceleration. The normal acceleration can be found from the centripetal acceleration:
\[ a_n = \frac{v^2}{R} \]
Thus:
\[ R = \frac{v^2}{\frac{d^2y}{dx^2} \cdot \left(1 + \left(\frac{dy}{dx}\right)^2\right)^{3/2}} \]

4. **Calculate \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \)**:
From the trajectory equation \( y = \frac{b}{2a} x^2 \):
\[ \frac{dy}{dx} = \frac{b}{a} x \]
\[ \frac{d^2y}{dx^2} = \frac{b}{a} \]

5. **Substituting into the Radius of Curvature Formula**:
Now substituting:
\[ R = \frac{(a^2 + (b x)^2)}{b \cdot \left(1 + \left(\frac{b}{a} x\right)^2\right)^{3/2}} \]
Simplifying gives:
\[ R = \frac{(a^2 + b^2 x^2)^{3/2}}{a^2 b} \]

### Final Results:
- **Trajectory Equation**:
\[ y(x) = \frac{b}{2a} x^2 \]
- **Radius of Curvature**:
\[ R = \frac{(a^2 + b^2 x^2)^{3/2}}{a^2 b} \]