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

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(B) Given the curve $y = \sin \left( \frac{\pi x}{2} \right)$.First,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent at $(1,

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$x = 1$

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(B) Given the curve $y = \sin \left( \frac{\pi x}{2} \right)$.First,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent at $(1, 1)$:$\frac{dy}{dx} = \frac{d}{dx} \left( \sin \left( \frac{\pi x}{2} \right) \right) = \cos \left( \frac{\pi x}{2} \right) \cdot \frac{\pi}{2}$.Now,evaluate the slope of the tangent at $(1, 1)$:$m_{tangent} = \left. \frac{dy}{dx} \right|_{(1, 1)} = \frac{\pi}{2} \cos \left( \frac{\pi(1)}{2} \right) = \frac{\pi}{2} \cos \left( \frac{\pi}{2} \right) = \frac{\pi}{2} \cdot 0 = 0$.Since the slope of the tangent is $0$,the tangent is a horizontal line.The slope of the normal is the negative reciprocal of the slope of the tangent,which is $-\frac{1}{0}$,representing a vertical line.The equation of a vertical line passing through $(1, 1)$ is $x = 1$.

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

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