A particle is moving with velocity ` vecv = k( y hat^(i) + x hat(j)) `, where `k` is a constant . The genergal equation for its path is

Options

A. ` y = x^(2)+ constant`
B. ` y^(2) = x + constant`
C. ` xy = constant`
D. ` y^(2) = x^(2) + constant`

Explanation

To find the general equation for the path of a particle moving with the given velocity vector \( \vec{v} = k(\hat{i} y + \hat{j} x) \), we can follow these steps:

### Step 1: Identify the components of velocity
The velocity vector can be expressed in terms of its components:
\[ \vec{v} = k y \hat{i} + k x \hat{j} \]
From this, we can identify:
- \( v_x = k y \) (the x-component of velocity)
- \( v_y = k x \) (the y-component of velocity)

### Step 2: Relate velocity to position
We know that velocity is the rate of change of position with respect to time:
\[ v_x = \frac{dx}{dt} \quad \text{and} \quad v_y = \frac{dy}{dt} \]
Thus, we can write:
1. \( \frac{dx}{dt} = k y \) (Equation 1)
2. \( \frac{dy}{dt} = k x \) (Equation 2)

### Step 3: Set up the relationship between \( dy \) and \( dx \)
To eliminate time \( t \), we can divide Equation 2 by Equation 1:
\[ \frac{dy}{dx} = \frac{v_y}{v_x} = \frac{k x}{k y} = \frac{x}{y} \]

### Step 4: Cross-multiply and rearrange
Cross-multiplying gives us:
\[ y \, dy = x \, dx \]

### Step 5: Integrate both sides
Now, we can integrate both sides:
\[ \int y \, dy = \int x \, dx \]
This results in:
\[ \frac{y^2}{2} = \frac{x^2}{2} + C \]
where \( C \) is the constant of integration.

### Step 6: Simplify the equation
Multiplying through by 2 to eliminate the fractions gives:
\[ y^2 = x^2 + 2C \]
Let \( k = 2C \) (where \( k \) is a new constant), we can rewrite this as:
\[ y^2 - x^2 = k \]

### Final Result
Thus, the general equation for the path of the particle is:
\[ y^2 = x^2 + C \]