A particle projected from origin moves in x-y plane with a velocity `vecv=3hati+6xhatj`, where `hati" and "hatj` are the unit vectors along x and y axis. Find the equation of path followed by the particle :-

Options

A. y = `x^(2)`
B. y = `(1)/(x^(2))`
C. y = `2x^(2) `
D. `y = (1)/(x)`

Answer

To find the equation of the path followed by the particle projected from the origin with a given velocity vector \(\vec{v} = 3 \hat{i} + 6x \hat{j}\), we can follow these steps:

### Step 1: Write the equations for the components of velocity
The velocity vector can be expressed in terms of its components:
\[ \vec{v} = \frac{dx}{dt} \hat{i} + \frac{dy}{dt} \hat{j} \]
From the given velocity, we identify:
1. \(\frac{dx}{dt} = 3\) (Equation 1)
2. \(\frac{dy}{dt} = 6x\) (Equation 2)

### Step 2: Solve for \(x\)
From Equation 1, we can express \(dx\) in terms of \(dt\):
\[ dx = 3 dt \]
Integrating both sides:
\[ \int dx = \int 3 dt \]
This gives:
\[ x = 3t + C_1 \]
Since the particle is projected from the origin, we can take \(C_1 = 0\):
\[ x = 3t \]

### Step 3: Substitute \(x\) into the equation for \(dy\)
Now, we substitute \(x = 3t\) into Equation 2:
\[ dy = 6x dt = 6(3t) dt = 18t dt \]
Integrating both sides:
\[ \int dy = \int 18t dt \]
This gives:
\[ y = 9t^2 + C_2 \]
Again, since the particle starts from the origin, we take \(C_2 = 0\):
\[ y = 9t^2 \]

### Step 4: Eliminate \(t\) to find the relationship between \(x\) and \(y\)
From the equation \(x = 3t\), we can express \(t\) in terms of \(x\):
\[ t = \frac{x}{3} \]
Substituting this into the equation for \(y\):
\[ y = 9\left(\frac{x}{3}\right)^2 = 9 \cdot \frac{x^2}{9} = x^2 \]

### Final Equation
Thus, the equation of the path followed by the particle is:
\[ y = x^2 \]