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(C) Given the curve $x = a(t + \sin t)$ and $y = a(1 - \cos t)$.First,find the derivative $\frac{dy}{dx}$:$\frac{dx}{dt} = a(1 + \cos t)$ and $\frac{d

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$2a \sin(t/2) 	an(t/2)$

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(C) Given the curve $x = a(t + \sin t)$ and $y = a(1 - \cos t)$.First,find the derivative $\frac{dy}{dx}$:$\frac{dx}{dt} = a(1 + \cos t)$ and $\frac{dy}{dt} = a \sin t$.$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin t}{a(1 + \cos t)} = \frac{2 \sin(t/2) \cos(t/2)}{2 \cos^2(t/2)} = 	an(t/2)$.The length of the normal is given by the formula $L = |y| \sqrt{1 + (\frac{dy}{dx})^2}$.Substituting the values:$L = |a(1 - \cos t)| \sqrt{1 + 	an^2(t/2)}$$L = a(2 \sin^2(t/2)) \sqrt{\sec^2(t/2)}$$L = 2a \sin^2(t/2) \sec(t/2)$Since $\sec(t/2) = \frac{1}{\cos(t/2)}$,we have:$L = 2a \sin(t/2) \cdot \frac{\sin(t/2)}{\cos(t/2)} = 2a \sin(t/2) 	an(t/2)$.

The length of the normal at point $t$ of the curve $x = a(t + \sin t)$,$y = a(1 - \cos t)$ is

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