If x=a(1-cos theta) and y=a(theta-sin theta), | vedclass

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(B) Given $x = a(1 - \cos 	heta)$ and $y = a(	heta - \sin 	heta)$.First,find $\frac{dx}{d	heta} = a \sin 	heta$ and $\frac{dy}{d	heta} = a(1 - \

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$\frac{\operatorname{cosec}^{4}(	heta/2)}{4a}$

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(B) Given $x = a(1 - \cos 	heta)$ and $y = a(	heta - \sin 	heta)$.First,find $\frac{dx}{d	heta} = a \sin 	heta$ and $\frac{dy}{d	heta} = a(1 - \cos 	heta)$.Then,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a(1 - \cos 	heta)}{a \sin 	heta} = \frac{2 \sin^{2}(	heta/2)}{2 \sin(	heta/2) \cos(	heta/2)} = 	an(	heta/2)$.Now,differentiate with respect to $x$: $\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(	an(	heta/2)) = \sec^{2}(	heta/2) \cdot \frac{1}{2} \cdot \frac{d	heta}{dx}$.Since $\frac{dx}{d	heta} = a \sin 	heta$,we have $\frac{d	heta}{dx} = \frac{1}{a \sin 	heta}$.Substituting this: $\frac{d^{2}y}{dx^{2}} = \frac{1}{2} \sec^{2}(	heta/2) \cdot \frac{1}{a(2 \sin(	heta/2) \cos(	heta/2))} = \frac{1}{4a \sin(	heta/2) \cos^{3}(	heta/2)} = \frac{1}{4a \sin(	heta/2) \cos(	heta/2) \cos^{2}(	heta/2)} = \frac{1}{2a \sin 	heta \cos^{2}(	heta/2)}$.Alternatively,$\frac{d^{2}y}{dx^{2}} = \frac{1}{2} \sec^{2}(	heta/2) \cdot \frac{1}{a(2 \sin(	heta/2) \cos(	heta/2))} = \frac{1}{4a \sin(	heta/2) \cos^{3}(	heta/2)} = \frac{\sec^{4}(	heta/2)}{4a}$.

If $x=a(1-\cos \theta)$ and $y=a(\theta-\sin \theta)$,then $\frac{d^{2} y}{d x^{2}}=$

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