If t is a parameter and x=t+frac{1}{t}, y=t-f | vedclass

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

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$\frac{-4 t^3}{\left(t^2-1\right)^3}$

Vedclass · Answer

(D) Given $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$.First, find the derivatives with respect to $t$:$\frac{dx}{dt} = 1 - \frac{1}{t^2} = \frac{t^2-1}{t^2}$$\frac{dy}{dt} = 1 + \frac{1}{t^2} = \frac{t^2+1}{t^2}$Now, find $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{(t^2+1)/t^2}{(t^2-1)/t^2} = \frac{t^2+1}{t^2-1}$Next, find $\frac{d^2y}{dx^2}$ by differentiating with respect to $x$:$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{t^2+1}{t^2-1}\right) \cdot \frac{dt}{dx}$Using the quotient rule for $\frac{d}{dt}\left(\frac{t^2+1}{t^2-1}\right)$:$= \frac{(t^2-1)(2t) - (t^2+1)(2t)}{(t^2-1)^2} = \frac{2t^3 - 2t - 2t^3 - 2t}{(t^2-1)^2} = \frac{-4t}{(t^2-1)^2}$Since $\frac{dx}{dt} = \frac{t^2-1}{t^2}$, then $\frac{dt}{dx} = \frac{t^2}{t^2-1}$.Therefore, $\frac{d^2y}{dx^2} = \frac{-4t}{(t^2-1)^2} \cdot \frac{t^2}{t^2-1} = \frac{-4t^3}{(t^2-1)^3}$.

If $t$ is a parameter and $x=t+\frac{1}{t}, y=t-\frac{1}{t}$, then $\frac{d^2 y}{d x^2}=$

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