If x = at^2 and y = 2at,then frac{d^2y}{dx^2} | vedclass

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(D) Given $x = at^2$ and $y = 2at$.First,find $\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 2a$.Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at

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$-\frac{1}{2at^3}$

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(D) Given $x = at^2$ and $y = 2at$.First,find $\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 2a$.Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}$.Now,differentiate $\frac{dy}{dx}$ with respect to $x$:$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{1}{t}) = \frac{d}{dt}(\frac{1}{t}) \cdot \frac{dt}{dx}$.Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{2at}$,we have:$\frac{d^2y}{dx^2} = (-\frac{1}{t^2}) \cdot \frac{1}{2at} = -\frac{1}{2at^3}$.

If $x = at^2$ and $y = 2at$,then $\frac{d^2y}{dx^2} = $

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