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, we start with the given velocity vector:

\[ \vec{v} = 3 \hat{i} + 6x \hat{j} \]

### Step 1: Relate velocity to position
The velocity vector can be expressed in terms of the derivatives of the position coordinates \(x\) and \(y\):

\[ \vec{v} = \frac{dx}{dt} \hat{i} + \frac{dy}{dt} \hat{j} \]

From this, we can equate the components:

\[ \frac{dx}{dt} = 3 \quad \text{(1)} \]
\[ \frac{dy}{dt} = 6x \quad \text{(2)} \]

### Step 2: Solve for \(x\)
From equation (1), we can integrate to find \(x\):

\[ \frac{dx}{dt} = 3 \]

Integrating both sides with respect to \(t\):

\[ \int dx = \int 3 \, dt \]

This gives us:

\[ x = 3t + C_1 \]

Since the particle is projected from the origin, we can set \(C_1 = 0\):

\[ x = 3t \quad \text{(3)} \]

### Step 3: Substitute \(x\) into \(y\) equation
Now we substitute \(x\) from equation (3) into equation (2):

\[ \frac{dy}{dt} = 6x = 6(3t) = 18t \]

### Step 4: Solve for \(y\)
Now we integrate to find \(y\):

\[ \frac{dy}{dt} = 18t \]

Integrating both sides with respect to \(t\):

\[ \int dy = \int 18t \, dt \]

This gives us:

\[ y = 9t^2 + C_2 \]

Again, since the particle starts from the origin, we set \(C_2 = 0\):

\[ y = 9t^2 \quad \text{(4)} \]

### Step 5: Eliminate \(t\)
Now we have expressions for \(x\) and \(y\):

From equation (3):
\[ t = \frac{x}{3} \]

Substituting \(t\) into equation (4):

\[ y = 9\left(\frac{x}{3}\right)^2 \]

This simplifies to:

\[ y = 9 \cdot \frac{x^2}{9} = x^2 \]

### Final Equation of the Path
Thus, the equation of the path followed by the particle is:

\[ y = x^2 \]

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