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

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(D) Given the curve $x = a(t + \sin t)$ and $y = a(1 - \cos t)$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = a(1 + \cos t) = a(2

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$2a \sin \frac{t}{2} 	an \frac{t}{2}$

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(D) Given the curve $x = a(t + \sin t)$ and $y = a(1 - \cos t)$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = a(1 + \cos t) = a(2 \cos^2 \frac{t}{2}) = 2a \cos^2 \frac{t}{2}$$\frac{dy}{dt} = a(\sin t) = 2a \sin \frac{t}{2} \cos \frac{t}{2}$Now,the slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a \sin \frac{t}{2} \cos \frac{t}{2}}{2a \cos^2 \frac{t}{2}} = 	an \frac{t}{2}$.The slope of the normal is $-\frac{dx}{dy} = -\cot \frac{t}{2}$.The length of the normal is given by $|y \sqrt{1 + (\frac{dy}{dx})^2}|$.$|y \sqrt{1 + 	an^2 \frac{t}{2}}| = |y \sec \frac{t}{2}|$.Substituting $y = a(1 - \cos t) = 2a \sin^2 \frac{t}{2}$:Length of normal $= |2a \sin^2 \frac{t}{2} \cdot \frac{1}{\cos \frac{t}{2}}| = |2a \sin \frac{t}{2} 	an \frac{t}{2}|$.

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

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