If x=t-sin t, y=1-cos t and frac{d^2 y}{d x^2 | vedclass

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(C) Given $x = t - \sin t$ and $y = 1 - \cos t$. First, find the derivatives with respect to $t$: $\frac{dx}{dt} = 1 - \cos t$ and $\frac{dy}{dt} = \s

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$\frac{2}{\pi-2}$

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(C) Given $x = t - \sin t$ and $y = 1 - \cos t$. First, find the derivatives with respect to $t$: $\frac{dx}{dt} = 1 - \cos t$ and $\frac{dy}{dt} = \sin t$. Then, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sin t}{1 - \cos t} = \frac{2 \sin(t/2) \cos(t/2)}{2 \sin^2(t/2)} = \cot(t/2)$. Now, find the second derivative: $\frac{d^2y}{dx^2} = \frac{d}{dx}(\cot(t/2)) = \frac{d}{dt}(\cot(t/2)) \cdot \frac{dt}{dx} = -\frac{1}{2} \csc^2(t/2) \cdot \frac{1}{1 - \cos t} = -\frac{1}{2} \csc^2(t/2) \cdot \frac{1}{2 \sin^2(t/2)} = -\frac{1}{4} \csc^4(t/2)$. Given $\frac{d^2y}{dx^2} = -1$ at $t=K$: $-\frac{1}{4} \csc^4(K/2) = -1 \implies \csc^4(K/2) = 4 \implies \csc^2(K/2) = 2 \implies \sin^2(K/2) = 1/2$. Since $K>0$, $\sin(K/2) = 1/\sqrt{2}$, so $K/2 = \pi/4$, which means $K = \pi/2$. Now, evaluate the limit: $\lim_{t \rightarrow K} \frac{y}{x} = \lim_{t \rightarrow \pi/2} \frac{1 - \cos t}{t - \sin t} = \frac{1 - \cos(\pi/2)}{\pi/2 - \sin(\pi/2)} = \frac{1 - 0}{\pi/2 - 1} = \frac{1}{(\pi - 2)/2} = \frac{2}{\pi - 2}$.

If $x=t-\sin t, y=1-\cos t$ and $\frac{d^2 y}{d x^2}=-1$ at $t=K, K>0$, then $\lim_{t \rightarrow K} \frac{y}{x}=$

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