If x=a t^2 and y=2 a t,then frac{d^2 y}{d x^2 | vedclass

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

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$-\frac{4}{a}$

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(D) Given $x=a t^2$ and $y=2 a t$. 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 $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{1}{t}) = -\frac{1}{t^2} \cdot \frac{dt}{dx}$. Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{2at}$,we have: $\frac{d^2y}{dx^2} = -\frac{1}{t^2} \cdot \frac{1}{2at} = -\frac{1}{2at^3}$. At $t=\frac{1}{2}$,$\frac{d^2y}{dx^2} = -\frac{1}{2a(1/2)^3} = -\frac{1}{2a(1/8)} = -\frac{1}{a/4} = -\frac{4}{a}$.

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

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