The angle of intersection of the curves y = x | vedclass

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(D) Given curves are $y = x^2$ and $x = y^2$.For $y = x^2$,differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x$.At point $(1, 1)$,the slope

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$	an^{-1}\left(\frac{3}{4}ight)$

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(D) Given curves are $y = x^2$ and $x = y^2$.For $y = x^2$,differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x$.At point $(1, 1)$,the slope $m_1 = 2(1) = 2$.For $x = y^2$,differentiating with respect to $x$,we get $1 = 2y \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{2y}$.At point $(1, 1)$,the slope $m_2 = \frac{1}{2(1)} = \frac{1}{2}$.The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values,$	an 	heta = \left| \frac{2 - 1/2}{1 + 2(1/2)} ight| = \left| \frac{3/2}{2} ight| = \frac{3}{4}$.Therefore,$	heta = 	an^{-1}\left(\frac{3}{4}ight)$.

The angle of intersection of the curves $y = x^2$ and $x = y^2$ at $(1, 1)$ is

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