If $x = at^2, y = 2at$; then $\frac{d^2y}{dx^2}$ is equal to

Given $x = at^2$ and $y = 2at$, where both are expressed in terms of parameter $t$.

For parametric equations, the first derivative is:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

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Finding $\frac{dy}{dt}$:

$y = 2at$

$\frac{dy}{dt} = 2a$

Finding $\frac{dx}{dt}$:

$x = at^2$

$\frac{dx}{dt} = 2at$

Therefore:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

$\frac{dy}{dx} = \frac{2a}{2at}$

$\frac{dy}{dx} = \frac{1}{t}$

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For the second derivative in parametric form:

$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \times \frac{dt}{dx}$

Finding $\frac{d}{dt}\left(\frac{dy}{dx}\right)$:

$\frac{d}{dt}\left(\frac{1}{t}\right) = \frac{d}{dt}(t^{-1})$

$= -t^{-2}$

$= -\frac{1}{t^2}$

Finding $\frac{dt}{dx}$:

$\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}}$

$\frac{dt}{dx} = \frac{1}{2at}$

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Therefore:

$\frac{d^2y}{dx^2} = -\frac{1}{t^2} \times \frac{1}{2at}$

$\frac{d^2y}{dx^2} = -\frac{1}{2at^3}$