The magnitude of radius vector of a point varies with time as `r = beta t ( 1- alpha t) ` where `alpha` and `beta` are positive constant. The distance travelled by this body over a closed path must be

Options

A. `( beta)/( alpha)`
B. `(beta)/( 2 alpha)`
C. `alpha beta`
D. `( 2 alpha)/( beta)`

Answer

To solve the problem, we start with the given expression for the radius vector \( r \):

\[ r = \beta t (1 - \alpha t) \]

where \( \alpha \) and \( \beta \) are positive constants. We need to find the distance traveled by the body over a closed path.

### Step 1: Differentiate the radius vector to find velocity

The first step is to find the tangential component of the velocity by differentiating the radius vector \( r \) with respect to time \( t \):

\[ \frac{dr}{dt} = \frac{d}{dt}[\beta t (1 - \alpha t)] \]

Using the product rule, we differentiate:

\[ \frac{dr}{dt} = \beta (1 - \alpha t) + \beta t (-\alpha) = \beta (1 - \alpha t) - \alpha \beta t \]

This simplifies to:

\[ \frac{dr}{dt} = \beta - 2\alpha \beta t \]

### Step 2: Determine when the velocity is zero

Next, we find when the velocity becomes zero:

\[ \beta - 2\alpha \beta t = 0 \]

Solving for \( t \):

\[ 2\alpha \beta t = \beta \implies t = \frac{1}{2\alpha} \]

### Step 3: Find the time for one complete cycle

The radius vector returns to zero when \( r = 0 \):

\[ \beta t (1 - \alpha t) = 0 \]

This occurs at \( t = 0 \) and \( t = \frac{1}{\alpha} \). Thus, the time for one complete cycle is:

\[ T = \frac{1}{\alpha} \]

### Step 4: Calculate the distance traveled in two halves

We will calculate the distance traveled in two segments: from \( t = 0 \) to \( t = \frac{1}{2\alpha} \) and from \( t = \frac{1}{2\alpha} \) to \( t = \frac{1}{\alpha} \).

1. **From \( t = 0 \) to \( t = \frac{1}{2\alpha} \)**:

The distance \( s_1 \) is given by:

\[ s_1 = \int_0^{\frac{1}{2\alpha}} \left(\beta - 2\alpha \beta t\right) dt \]

Evaluating this integral:

\[ s_1 = \left[ \beta t - \alpha \beta t^2 \right]_0^{\frac{1}{2\alpha}} = \left(\beta \cdot \frac{1}{2\alpha} - \alpha \beta \left(\frac{1}{2\alpha}\right)^2\right) - 0 \]

\[ s_1 = \frac{\beta}{2\alpha} - \alpha \beta \cdot \frac{1}{4\alpha^2} = \frac{\beta}{2\alpha} - \frac{\beta}{4\alpha} = \frac{\beta}{4\alpha} \]

2. **From \( t = \frac{1}{2\alpha} \) to \( t = \frac{1}{\alpha} \)**:

The distance \( s_2 \) is given by:

\[ s_2 = \int_{\frac{1}{2\alpha}}^{\frac{1}{\alpha}} \left(\beta - 2\alpha \beta t\right) dt \]

Evaluating this integral:

\[ s_2 = \left[ \beta t - \alpha \beta t^2 \right]_{\frac{1}{2\alpha}}^{\frac{1}{\alpha}} = \left(\beta \cdot \frac{1}{\alpha} - \alpha \beta \left(\frac{1}{\alpha}\right)^2\right) - \left(\frac{\beta}{4\alpha}\right) \]

\[ s_2 = \left(\frac{\beta}{\alpha} - \frac{\beta}{\alpha}\right) + \frac{\beta}{4\alpha} = \frac{\beta}{4\alpha} \]

### Step 5: Total distance traveled

The total distance traveled \( S \) over one closed path is:

\[ S = s_1 + s_2 = \frac{\beta}{4\alpha} + \frac{\beta}{4\alpha} = \frac{\beta}{2\alpha} \]

### Conclusion

Thus, the total distance traveled by the body over a closed path is:

\[ \boxed{\frac{\beta}{2\alpha}} \]