The equation of the normal to the curve y = s | vedclass

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(C) Given the curve $y = \sin \left(\frac{\pi x}{4}\right)$.First,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:$\frac{

Vedclass · Accepted Answer

$x = 2$

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(C) Given the curve $y = \sin \left(\frac{\pi x}{4}ight)$.First,we find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:$\frac{dy}{dx} = \frac{\pi}{4} \cos \left(\frac{\pi x}{4}ight)$.At the point $(2, 1)$,the slope of the tangent $m_t$ is:$m_t = \left(\frac{dy}{dx}ight)_{(2, 1)} = \frac{\pi}{4} \cos \left(\frac{2\pi}{4}ight) = \frac{\pi}{4} \cos \left(\frac{\pi}{2}ight) = \frac{\pi}{4} 	imes 0 = 0$.Since the slope of the tangent is $0$,the tangent is a horizontal line (parallel to the $X$-axis).Therefore,the normal,which is perpendicular to the tangent,must be a vertical line.The equation of a vertical line passing through $(2, 1)$ is $x = 2$.Thus,the correct option is $C$.

The equation of the normal to the curve $y = \sin \left(\frac{\pi x}{4}\right)$ at the point $(2, 1)$ is

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