If x=a(t+sin t) and y=a(1-cos t),then find fr | vedclass

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

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$\frac{1}{4 a \cos ^4(\frac{t}{2})}$

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(B) Given $x=a(t+\sin t)$ and $y=a(1-\cos t)$.First,find the derivatives with respect to $t$:$\frac{d x}{d t}=a(1+\cos t)$ and $\frac{d y}{d t}=a \sin t$.Now,find $\frac{d y}{d x}$:$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a \sin t}{a(1+\cos t)}=\frac{2 \sin(t/2) \cos(t/2)}{2 \cos^2(t/2)}=	an(t/2)$.Next,differentiate $\frac{d y}{d x}$ with respect to $x$:$\frac{d^2 y}{d x^2}=\frac{d}{d x}(	an(t/2))=\sec^2(t/2) \cdot \frac{1}{2} \cdot \frac{d t}{d x}$.Since $\frac{d x}{d t}=a(1+\cos t)=2a \cos^2(t/2)$,then $\frac{d t}{d x}=\frac{1}{2a \cos^2(t/2)}$.Substituting this back:$\frac{d^2 y}{d x^2}=\sec^2(t/2) \cdot \frac{1}{2} \cdot \frac{1}{2a \cos^2(t/2)}=\frac{1}{4a \cos^4(t/2)}$.

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

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