A particle moves in xy-plane. The position vector of particle at any time is `vecr={(2t)hati+(2t^(2))hatj}m.` The rate of change of 0 at time `t = 2` second. ( where 0 is the angle which its velocity vector makes with positive x-axis) is :

Options

A. `(2)/(17)" rad s"^(-1)`
B. `(1)/(14)" rad s"^(-1)`
C. `(4)/(7)" rad s"^(-1)`
D. `(6)/(5)" rad s"^(-1)`

Answer

To find the rate of change of the angle \( \theta \) (which the velocity vector makes with the positive x-axis) at time \( t = 2 \) seconds, we can follow these steps:

### Step 1: Determine the Position Vector
The position vector of the particle is given by:
\[ \vec{r} = (2t) \hat{i} + (2t^2) \hat{j} \]

### Step 2: Find the Velocity Vector
To find the velocity vector \( \vec{v} \), we differentiate the position vector with respect to time \( t \):
\[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}[(2t) \hat{i} + (2t^2) \hat{j}] = 2 \hat{i} + 4t \hat{j} \]

### Step 3: Calculate the Components of the Velocity
From the velocity vector, we can identify the components:
- \( v_x = 2 \)
- \( v_y = 4t \)

### Step 4: Find the Angle \( \theta \)
The angle \( \theta \) that the velocity vector makes with the positive x-axis can be found using the tangent function:
\[ \tan \theta = \frac{v_y}{v_x} = \frac{4t}{2} = 2t \]

### Step 5: Differentiate \( \tan \theta \)
To find the rate of change of \( \theta \) with respect to time, we differentiate both sides with respect to \( t \):
\[ \frac{d}{dt}(\tan \theta) = \sec^2 \theta \frac{d\theta}{dt} \]
Using the chain rule, we have:
\[ \frac{d}{dt}(\tan \theta) = \frac{d}{dt}(2t) = 2 \]
Thus, we can write:
\[ \sec^2 \theta \frac{d\theta}{dt} = 2 \]

### Step 6: Express \( \sec^2 \theta \)
We know that:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
Substituting \( \tan \theta = 2t \):
\[ \sec^2 \theta = 1 + (2t)^2 = 1 + 4t^2 \]

### Step 7: Substitute and Solve for \( \frac{d\theta}{dt} \)
Substituting \( \sec^2 \theta \) into the equation:
\[ (1 + 4t^2) \frac{d\theta}{dt} = 2 \]
Thus, we have:
\[ \frac{d\theta}{dt} = \frac{2}{1 + 4t^2} \]

### Step 8: Evaluate at \( t = 2 \)
Now, we substitute \( t = 2 \) into the equation:
\[ \frac{d\theta}{dt} = \frac{2}{1 + 4(2^2)} = \frac{2}{1 + 16} = \frac{2}{17} \]

### Final Answer
The rate of change of the angle \( \theta \) at \( t = 2 \) seconds is:
\[ \frac{d\theta}{dt} = \frac{2}{17} \text{ radians per second} \]

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