If a neq 0,x=a(t+sin t) and y=a(1-cos t),then | vedclass

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(A) Given equations are:$x = a(t + \sin t) \quad \dots (i)$$y = a(1 - \cos t) \quad \dots (ii)$Differentiating $(i)$ with respect to $t$:$\frac{dx}{dt

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$\frac{4}{a}$

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(A) Given equations are:$x = a(t + \sin t) \quad \dots (i)$$y = a(1 - \cos t) \quad \dots (ii)$Differentiating $(i)$ with respect to $t$:$\frac{dx}{dt} = a(1 + \cos t)$Differentiating $(ii)$ with respect to $t$:$\frac{dy}{dt} = a(\sin t)$Now,find $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin t}{a(1 + \cos t)} = \frac{\sin t}{1 + \cos t} = 	an\left(\frac{t}{2}ight)$Now,find $\frac{d^2y}{dx^2}$:$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(	an\left(\frac{t}{2}ight)ight) = \frac{d}{dt}\left(	an\left(\frac{t}{2}ight)ight) \cdot \frac{dt}{dx}$$= \sec^2\left(\frac{t}{2}ight) \cdot \frac{1}{2} \cdot \frac{1}{a(1 + \cos t)}$$= \frac{1}{2 \cos^2(t/2)} \cdot \frac{1}{a(2 \cos^2(t/2))} = \frac{1}{4a \cos^4(t/2)}$At $t = \frac{2\pi}{3}$,$\frac{t}{2} = \frac{\pi}{3}$:$\frac{d^2y}{dx^2} = \frac{1}{4a \cos^4(\pi/3)} = \frac{1}{4a (1/2)^4} = \frac{1}{4a (1/16)} = \frac{16}{4a} = \frac{4}{a}$

If $a \neq 0$,$x=a(t+\sin t)$ and $y=a(1-\cos t)$,then $\frac{d^2 y}{d x^2}$ at $t=\frac{2 \pi}{3}$ is

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