A particle moves along the parabola `y=x^(2)` in the first quadrant in such a way that its x-coordinate (measured in metres) increases at a rate of 10 m/sec. If the angle of inclination `theta` of the line joining the particle to the origin change, when `x = 3` m, at the rate of `k` rad/sec., then the value of `k` is

Options

A. `1`
B. `2`
C. `1//2`
D. `1//3`

Explanation

To solve the problem, we need to find the rate of change of the angle of inclination \( \theta \) of the line joining the particle moving along the parabola \( y = x^2 \) to the origin, when \( x = 3 \) m. The x-coordinate of the particle is increasing at a rate of \( \frac{dx}{dt} = 10 \) m/sec.

### Step-by-Step Solution:

1. **Identify the relationship between \( y \) and \( x \)**:
Since the particle moves along the parabola \( y = x^2 \), we have:
\[ y = x^2 \]

2. **Find the expression for \( \tan(\theta) \)**:
The angle \( \theta \) is defined as the angle of inclination of the line joining the particle to the origin. Thus,
\[ \tan(\theta) = \frac{y}{x} = \frac{x^2}{x} = x \]

3. **Differentiate \( \tan(\theta) \) with respect to time \( t \)**:
Using implicit differentiation:
\[ \sec^2(\theta) \frac{d\theta}{dt} = \frac{dx}{dt} \]
where \( \sec^2(\theta) = 1 + \tan^2(\theta) \).

4. **Substitute \( \tan(\theta) \)**:
Since \( \tan(\theta) = x \), we can write:
\[ \sec^2(\theta) = 1 + x^2 \]
Therefore, the equation becomes:
\[ (1 + x^2) \frac{d\theta}{dt} = \frac{dx}{dt} \]

5. **Substitute known values**:
We know \( \frac{dx}{dt} = 10 \) m/sec and we need to evaluate this when \( x = 3 \) m:
\[ (1 + 3^2) \frac{d\theta}{dt} = 10 \]
Simplifying this gives:
\[ (1 + 9) \frac{d\theta}{dt} = 10 \]
\[ 10 \frac{d\theta}{dt} = 10 \]

6. **Solve for \( \frac{d\theta}{dt} \)**:
Dividing both sides by 10:
\[ \frac{d\theta}{dt} = 1 \text{ rad/sec} \]

Thus, the value of \( k \) is:
\[ \boxed{1} \]