Question

The angle between the curves \( y^2 = x \) and \( x^2 = y \) at the point \( (1,1) \) is:

• A. \( \tan^{-1}\left(\frac{4}{3}\right) \)
• B. \( \tan^{-1}\left(\frac{3}{4}\right) \)
• C. \( 90^\circ \)
• D. \( 45^\circ \)

Hint:

To find the angle between two curves at a point, first compute the slopes of their tangents at that point, then use the formula: \[ \tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| \]

Answer 1

The Correct Option is B

We are asked to find the angle between two curves at their point of intersection \( (1,1) \). Step 1: Find the slopes of the tangents to both curves at \( (1,1) \) For \( y^2 = x \) \[ \frac{dy}{dx} = \frac{1}{2y} \] At \( (1,1) \) \[ m_1 = \frac{1}{2(1)} = \frac{1}{2} \] For \( x^2 = y \) \[ \frac{dy}{dx} = 2x \] At \( (1,1) \) \[ m_2 = 2(1) = 2 \] Step 2: Use the formula for angle between two curves \[ \tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| \] Substituting the values: \[ = \left|\frac{2 - \frac{1}{2}}{1 + \frac{1}{2} \times 2}\right| = \left|\frac{\frac{4}{2} - \frac{1}{2}}{1 + 1}\right| = \left|\frac{\frac{3}{2}}{2}\right| = \frac{3}{4} \] Step 3: Final Answer \[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \] % Final Answer Hence, the angle between the curves at \( (1,1) \) is \( \tan^{-1}\left(\frac{3}{4}\right) \).