The acute angle between the curve `y^2=x` and `x^2=y` at (1,1) is

Options

A. `tan^-1(4/5)`
B. `tan^-1(3/4)`
C. `tan^-1(1)`
D. `tan^-1(4/3)`

Answer

To find the acute angle between the curves \( y^2 = x \) and \( x^2 = y \) at the point \( (1, 1) \), we will follow these steps:

### Step 1: Find the slopes of the tangents to the curves at the point (1, 1).

1. **For the curve \( y^2 = x \)**:
- Differentiate both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} = 1 \]
- Solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{2y} \]
- Substitute \( y = 1 \) (since we are evaluating at the point \( (1, 1) \)):
\[ \frac{dy}{dx} = \frac{1}{2 \cdot 1} = \frac{1}{2} \]
- Thus, the slope \( m_1 \) of the tangent to the curve \( y^2 = x \) at \( (1, 1) \) is \( \frac{1}{2} \).

2. **For the curve \( x^2 = y \)**:
- Differentiate both sides with respect to \( x \):
\[ 2x = \frac{dy}{dx} \]
- Substitute \( x = 1 \):
\[ \frac{dy}{dx} = 2 \cdot 1 = 2 \]
- Thus, the slope \( m_2 \) of the tangent to the curve \( x^2 = y \) at \( (1, 1) \) is \( 2 \).

### Step 2: Use the slopes to find the angle between the tangents.

The formula for the tangent of the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Substituting the values of \( m_1 \) and \( m_2 \):
\[ \tan \theta = \left| \frac{\frac{1}{2} - 2}{1 + \left(\frac{1}{2}\right)(2)} \right| \]
\[ = \left| \frac{\frac{1}{2} - \frac{4}{2}}{1 + 1} \right| \]
\[ = \left| \frac{-\frac{3}{2}}{2} \right| \]
\[ = \left| -\frac{3}{4} \right| = \frac{3}{4} \]

### Step 3: Find the angle \( \theta \).

To find \( \theta \), we take the inverse tangent:
\[ \theta = \tan^{-1} \left(\frac{3}{4}\right) \]

### Final Answer:
The acute angle between the curves \( y^2 = x \) and \( x^2 = y \) at the point \( (1, 1) \) is:
\[ \theta = \tan^{-1} \left(\frac{3}{4}\right) \]

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