What is the slope of the normal to the curve | vedclass

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(C) Given the parametric equations of the curve: $x = a(	heta - \sin 	heta)$ and $y = a(1 - \cos 	heta)$. First,we find the derivative $\frac{dy}{d

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$-1$

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(C) Given the parametric equations of the curve: $x = a(	heta - \sin 	heta)$ and $y = a(1 - \cos 	heta)$. First,we find the derivative $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta}$. $\frac{dx}{d	heta} = a(1 - \cos 	heta)$ and $\frac{dy}{d	heta} = a(\sin 	heta)$. Thus,the slope of the tangent is $\frac{dy}{dx} = \frac{a \sin 	heta}{a(1 - \cos 	heta)} = \frac{\sin 	heta}{1 - \cos 	heta}$. At $	heta = \pi / 2$: $\frac{dy}{dx} = \frac{\sin(\pi / 2)}{1 - \cos(\pi / 2)} = \frac{1}{1 - 0} = 1$. The slope of the normal is given by $-\frac{1}{	ext{slope of tangent}}$. Therefore,the slope of the normal is $-1 / 1 = -1$.

What is the slope of the normal to the curve $x = a(\theta - \sin \theta)$,$y = a(1 - \cos \theta)$ at the point $\theta = \pi / 2$?

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