the angle of intersection of the curves y=x2 and x=y2 at (1,1) is
tan-1\(\frac{4}{3}\)
tan-1(1)
90\(^{\circ}\)
tan-1\(\frac{3}{4}\)
The Correct Option is D
Solution and Explanation
To find the angle of intersection of the curves \( y = x^2 \) and \( x = y^2 \) at the point \( (1, 1) \), we need to follow these steps:
- Find the derivatives:
- For the curve \( y = x^2 \), the derivative is \(\frac{dy}{dx} = 2x\).
- For the curve \( x = y^2 \), we can differentiate implicitly to find \(\frac{dy}{dx}\). Differentiating both sides with respect to \(x\) gives:
\[ 1 = 2y \cdot \frac{dy}{dx} \] Therefore, \[ \frac{dy}{dx} = \frac{1}{2y} \]
- Calculate slopes at the point (1,1):
- For \( y = x^2 \), substitute \( x = 1 \): \[ \frac{dy}{dx} = 2 \cdot 1 = 2 \]
- For \( x = y^2 \), substitute \( y = 1 \): \[ \frac{dy}{dx} = \frac{1}{2 \cdot 1} = \frac{1}{2} \]
- Find the angle of intersection:
The angle \(\theta\) between two curves with slopes \(m_1\) and \(m_2\) is given by the formula:
\[ \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right| \]- Substitute \( m_1 = 2 \) and \( m_2 = \frac{1}{2} \): \[ \tan\theta = \left|\frac{2 - \frac{1}{2}}{1 + 2 \cdot \frac{1}{2}}\right| = \left|\frac{\frac{4}{2} - \frac{1}{2}}{1 + 1}\right| = \left|\frac{\frac{3}{2}}{2}\right| = \left|\frac{3}{4}\right| \]
Therefore, the angle \(\theta\) can be expressed as \(\tan^{-1}\left(\frac{3}{4}\right)\).
Conclusion: The angle of intersection of the curves at the point (1,1) is \(\tan^{-1}\left(\frac{3}{4}\right)\). Thus, the correct answer is:
tan-1\(\frac{3}{4}\)
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