What is the angle between the curves y = x^2 | vedclass

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(D) For the curve $y = x^2$,the slope $m_1$ is given by $\frac{dy}{dx} = 2x$.At the point $(1, 1)$,$m_1 = 2(1) = 2$.For the curve $x = y^2$,differenti

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

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(D) For the curve $y = x^2$,the slope $m_1$ is given by $\frac{dy}{dx} = 2x$.At the point $(1, 1)$,$m_1 = 2(1) = 2$.For the curve $x = y^2$,differentiating with respect to $x$ gives $1 = 2y \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{1}{2y}$.At the point $(1, 1)$,$m_2 = \frac{1}{2(1)} = \frac{1}{2}$.The angle $	heta$ between the two 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)$.

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

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