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

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

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$\frac{\sin t - 1}{a \cos^3 t}$

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(C) Given $x = a(1 - \sin t)$ and $y = a(t + \cos t)$.First,find $\frac{dx}{dt} = -a \cos t$ and $\frac{dy}{dt} = a(1 - \sin t)$.Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a(1 - \sin t)}{-a \cos t} = \frac{\sin t - 1}{\cos t} = 	an t - \sec t$.Now,differentiate with respect to $x$:$\frac{d^2y}{dx^2} = \frac{d}{dx}(	an t - \sec t) = \frac{d}{dt}(	an t - \sec t) \cdot \frac{dt}{dx}$.$\frac{d^2y}{dx^2} = (\sec^2 t - \sec t 	an t) \cdot \frac{1}{-a \cos t} = \frac{\sec t(\sec t - 	an t)}{-a \cos t} = \frac{\frac{1}{\cos t}(\frac{1 - \sin t}{\cos t})}{-a \cos t} = \frac{1 - \sin t}{-a \cos^3 t} = \frac{\sin t - 1}{a \cos^3 t}$.

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

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