The length of the normal to the curve x = a(t | vedclass

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(C) The length of the normal is given by the formula: $L = |y| \sqrt{1 + (dy/dx)^2}$.First,we find the derivatives with respect to $	heta$:$dx/d	het

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$\sqrt{2}a$

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(C) The length of the normal is given by the formula: $L = |y| \sqrt{1 + (dy/dx)^2}$.First,we find the derivatives with respect to $	heta$:$dx/d	heta = a(1 + \cos 	heta)$$dy/d	heta = a \sin 	heta$Thus,$dy/dx = (dy/d	heta) / (dx/d	heta) = (a \sin 	heta) / (a(1 + \cos 	heta)) = \sin 	heta / (1 + \cos 	heta)$.Using trigonometric identities,$\sin 	heta = 2 \sin(	heta/2) \cos(	heta/2)$ and $1 + \cos 	heta = 2 \cos^2(	heta/2)$.So,$dy/dx = 	an(	heta/2)$.At $	heta = \pi/2$,$dy/dx = 	an(\pi/4) = 1$.Also,at $	heta = \pi/2$,$y = a(1 - \cos(\pi/2)) = a(1 - 0) = a$.Substituting these values into the formula:$L = |a| \sqrt{1 + (1)^2} = a \sqrt{2}$.

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

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