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

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(B) Given $x=a(	heta+\sin 	heta)$ and $y=a(1-\cos 	heta)$.First,find the derivatives with respect to $	heta$:$\frac{dx}{d	heta} = a(1+\cos 	heta

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$\frac{1}{a}$

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(B) Given $x=a(	heta+\sin 	heta)$ and $y=a(1-\cos 	heta)$.First,find the derivatives with respect to $	heta$:$\frac{dx}{d	heta} = a(1+\cos 	heta)$ and $\frac{dy}{d	heta} = a\sin 	heta$.Now,find $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a\sin 	heta}{a(1+\cos 	heta)} = \frac{\sin 	heta}{1+\cos 	heta}$.Using the identity $\sin 	heta = 2\sin(	heta/2)\cos(	heta/2)$ and $1+\cos 	heta = 2\cos^2(	heta/2)$,we get $\frac{dy}{dx} = 	an(	heta/2)$.Now,differentiate $\frac{dy}{dx}$ with respect to $x$:$\frac{d^2y}{dx^2} = \frac{d}{d	heta}\left(	an\frac{	heta}{2}ight) \cdot \frac{d	heta}{dx} = \frac{1}{2}\sec^2\frac{	heta}{2} \cdot \frac{1}{a(1+\cos 	heta)}$.Since $1+\cos 	heta = 2\cos^2(	heta/2)$,we have:$\frac{d^2y}{dx^2} = \frac{1}{2}\sec^2\frac{	heta}{2} \cdot \frac{1}{2a\cos^2(	heta/2)} = \frac{1}{4a\cos^4(	heta/2)}$.At $	heta = \pi/2$,$	heta/2 = \pi/4$ and $\cos(\pi/4) = 1/\sqrt{2}$.$\left(\frac{d^2y}{dx^2}ight)_{	heta=\pi/2} = \frac{1}{4a(1/\sqrt{2})^4} = \frac{1}{4a(1/4)} = \frac{1}{a}$.

If $x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$,then $\left(\frac{d^2 y}{dx^2}\right)_{\theta=\pi / 2}=$

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