The angle of intersection of the curves y^2 = | vedclass

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(B) The curves $y^2 = \frac{2x}{\pi}$ and $y = \sin x$ intersect at $(0, 0)$ and $(\pi/2, 1)$.Let the gradients of the tangents to the curves be $m_1$

Vedclass · Accepted Answer

$\cot^{-1}(\pi)$

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(B) The curves $y^2 = \frac{2x}{\pi}$ and $y = \sin x$ intersect at $(0, 0)$ and $(\pi/2, 1)$.Let the gradients of the tangents to the curves be $m_1$ and $m_2$ respectively.For $y^2 = \frac{2x}{\pi}$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = \frac{2}{\pi}$,so $m_1 = \frac{dy}{dx} = \frac{1}{\pi y}$.For $y = \sin x$,differentiating with respect to $x$ gives $m_2 = \frac{dy}{dx} = \cos x$.At the point of intersection $(\pi/2, 1)$:$m_1 = \frac{1}{\pi(1)} = \frac{1}{\pi}$.$m_2 = \cos(\pi/2) = 0$.The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.$	an 	heta = \left| \frac{1/\pi - 0}{1 + (1/\pi)(0)} ight| = \frac{1}{\pi}$.Therefore,$	heta = 	an^{-1}(1/\pi) = \cot^{-1}(\pi)$.

The angle of intersection of the curves $y^2 = \frac{2x}{\pi}$ and $y = \sin x$ is

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