The length of the subtangent at t on the curv | vedclass

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

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$a \sin t$

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(A) Given,$x=a(t+\sin t)$ and $y=a(1-\cos t)$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = a(1+\cos t) = 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 given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a \sin \frac{t}{2} \cos \frac{t}{2}}{2a \cos^2 \frac{t}{2}} = 	an \frac{t}{2}$.The length of the subtangent is defined as $\left| \frac{y}{dy/dx} ight|$.Substituting the values,we get:Length of subtangent $= \frac{a(1-\cos t)}{	an \frac{t}{2}} = \frac{2a \sin^2 \frac{t}{2}}{	an \frac{t}{2}} = \frac{2a \sin^2 \frac{t}{2}}{\sin \frac{t}{2} / \cos \frac{t}{2}} = 2a \sin \frac{t}{2} \cos \frac{t}{2} = a \sin t$.

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

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