The angle between the curves y^2=x and x^2=y | vedclass

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(A) Let the two curves be $C_1: y^2 = x$ and $C_2: x^2 = y$. First,we find the slopes of the tangents to these curves at the point $(1,1)$. For $C_1:

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$\operatorname{Tan}^{-1}\left(\frac{3}{4}\right)$

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(A) Let the two curves be $C_1: y^2 = x$ and $C_2: x^2 = y$. First,we find the slopes of the tangents to these curves at the point $(1,1)$. For $C_1: y^2 = x$,differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 1$,so $\frac{dy}{dx} = \frac{1}{2y}$. At $(1,1)$,the slope $m_1 = \frac{1}{2(1)} = \frac{1}{2}$. For $C_2: x^2 = y$,differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x$. At $(1,1)$,the slope $m_2 = 2(1) = 2$. The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_2 - m_1}{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}{1 + 1} ight| = \left| \frac{3/2}{2} ight| = \frac{3}{4}$. Therefore,$	heta = \operatorname{Tan}^{-1}\left(\frac{3}{4}ight)$.

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

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