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 need to determine the distance traveled by a point whose radius vector varies with time as \( r = \beta t (1 - \alpha t) \), where \( \alpha \) and \( \beta \) are positive constants.

### Step-by-Step Solution:

1. **Identify when the radius vector is zero:**
The radius vector \( r \) becomes zero when:
\[ r = \beta t (1 - \alpha t) = 0 \]
This occurs when either \( \beta t = 0 \) (which gives \( t = 0 \)) or \( 1 - \alpha t = 0 \) (which gives \( t = \frac{1}{\alpha} \)). Therefore, the radius vector is zero at \( t = \frac{1}{\alpha} \).

2. **Find the velocity vector:**
The velocity \( v \) is the derivative of the radius vector with respect to time:
\[ v = \frac{dr}{dt} = \frac{d}{dt} [\beta t (1 - \alpha t)] = \beta (1 - 2\alpha t) \]
The velocity becomes zero when:
\[ 1 - 2\alpha t = 0 \implies t = \frac{1}{2\alpha} \]

3. **Calculate the distance traveled:**
The distance traveled can be calculated by integrating the velocity over the appropriate intervals. The body travels from \( t = 0 \) to \( t = \frac{1}{2\alpha} \) and then from \( t = \frac{1}{2\alpha} \) to \( t = \frac{1}{\alpha} \).

- **First Interval (from \( t = 0 \) to \( t = \frac{1}{2\alpha} \)):**
\[ \text{Distance}_1 = \int_0^{\frac{1}{2\alpha}} v \, dt = \int_0^{\frac{1}{2\alpha}} \beta (1 - 2\alpha t) \, dt \]
Evaluating this integral:
\[ = \beta \left[ t - \alpha t^2 \right]_0^{\frac{1}{2\alpha}} = \beta \left( \frac{1}{2\alpha} - \alpha \left( \frac{1}{2\alpha} \right)^2 \right) \]
\[ = \beta \left( \frac{1}{2\alpha} - \frac{1}{8\alpha} \right) = \beta \left( \frac{4}{8\alpha} - \frac{1}{8\alpha} \right) = \frac{3\beta}{8\alpha} \]

- **Second Interval (from \( t = \frac{1}{2\alpha} \) to \( t = \frac{1}{\alpha} \)):**
\[ \text{Distance}_2 = \int_{\frac{1}{2\alpha}}^{\frac{1}{\alpha}} v \, dt = \int_{\frac{1}{2\alpha}}^{\frac{1}{\alpha}} \beta (1 - 2\alpha t) \, dt \]
Evaluating this integral:
\[ = \beta \left[ t - \alpha t^2 \right]_{\frac{1}{2\alpha}}^{\frac{1}{\alpha}} = \beta \left( \frac{1}{\alpha} - \alpha \left( \frac{1}{\alpha} \right)^2 - \left( \frac{1}{2\alpha} - \alpha \left( \frac{1}{2\alpha} \right)^2 \right) \right) \]
\[ = \beta \left( \frac{1}{\alpha} - \frac{1}{\alpha} - \left( \frac{1}{2\alpha} - \frac{1}{8\alpha} \right) \right) = \beta \left( 0 - \left( \frac{4}{8\alpha} - \frac{1}{8\alpha} \right) \right) = -\frac{3\beta}{8\alpha} \]

4. **Total Distance:**
The total distance traveled is the sum of the distances from both intervals:
\[ \text{Total Distance} = \text{Distance}_1 + |\text{Distance}_2| = \frac{3\beta}{8\alpha} + \frac{3\beta}{8\alpha} = \frac{6\beta}{8\alpha} = \frac{3\beta}{4\alpha} \]

### Final Answer:
The total distance traveled by the body over the closed path is \( \frac{3\beta}{4\alpha} \).