For `0 lt r lt 1,` let `n_(r)` dennotes the line that is normal to the curve `y = x ^r` at the point `(1,1)` Let `S _(r)` denotes the region in the first quadrant bounded by the curve `y = x ^(r )` ,the x-axis and the line `n _(r )'` Then the value of r the minimizes the area of ` S_(r )` is :

Options

A. `(1)/(sqrt2)`
B. `sqrt2-1`
C. `(sqrt2-1)/(2)`
D. `sqrt2-1/2`

Answer

To solve the problem, we need to find the value of \( r \) that minimizes the area \( S_r \) in the first quadrant bounded by the curve \( y = x^r \), the x-axis, and the normal line to the curve at the point \( (1, 1) \).

### Step-by-Step Solution:

1. **Differentiate the Curve**:
The curve is given by \( y = x^r \). We differentiate it to find the slope of the tangent line:
\[ \frac{dy}{dx} = r x^{r-1} \]

2. **Find the Slope of the Normal Line**:
The slope of the normal line \( m_r \) is the negative reciprocal of the slope of the tangent line at the point \( (1, 1) \):
\[ m_r = -\frac{1}{\frac{dy}{dx}} = -\frac{1}{r} \]

3. **Equation of the Normal Line**:
The equation of the normal line in point-slope form is:
\[ y - 1 = m_r (x - 1) \]
Substituting \( m_r \):
\[ y - 1 = -\frac{1}{r} (x - 1) \]
Rearranging gives:
\[ y = -\frac{1}{r}x + 1 + \frac{1}{r} \]
Thus, the equation of the normal line is:
\[ y = -\frac{1}{r}x + \left(1 + \frac{1}{r}\right) \]

4. **Find the x-intercept of the Normal Line**:
To find the x-intercept, set \( y = 0 \):
\[ 0 = -\frac{1}{r}x + 1 + \frac{1}{r} \]
Solving for \( x \):
\[ \frac{1}{r}x = 1 + \frac{1}{r} \implies x = r + 1 \]

5. **Calculate the Area \( S_r \)**:
The area \( S_r \) consists of two parts: the area under the curve from \( x = 0 \) to \( x = 1 \) and the area of the triangle formed by the x-axis, the normal line, and the vertical line at \( x = 1 \).

- **Area under the curve**:
\[ A_{\text{curve}} = \int_0^1 x^r \, dx = \left[ \frac{x^{r+1}}{r+1} \right]_0^1 = \frac{1}{r+1} \]

- **Area of the triangle**:
The height of the triangle is \( 1 + \frac{1}{r} \) and the base is \( r \):
\[ A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times r \times \left(1 + \frac{1}{r}\right) = \frac{1}{2} \times (r + 1) \]

Therefore, the total area \( S_r \) is:
\[ S_r = \frac{1}{r+1} + \frac{1}{2}(r + 1) \]

6. **Simplifying the Area Expression**:
\[ S_r = \frac{1}{r+1} + \frac{1}{2}r + \frac{1}{2} \]

7. **Finding the Minimum Area**:
To minimize \( S_r \), we differentiate with respect to \( r \) and set the derivative to zero:
\[ \frac{dS_r}{dr} = -\frac{1}{(r+1)^2} + \frac{1}{2} = 0 \]
Solving gives:
\[ \frac{1}{(r+1)^2} = \frac{1}{2} \implies (r+1)^2 = 2 \implies r + 1 = \sqrt{2} \implies r = \sqrt{2} - 1 \]

8. **Conclusion**:
The value of \( r \) that minimizes the area \( S_r \) is:
\[ \boxed{\sqrt{2} - 1} \]