If x = at2, y = 2 at, then d^2 y/d x^2 =

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(d) - $\frac{1}{2}$ \ a t^3 x = a t^2 \text { and y } = 2\text { at } \text { Differentiating w . r . t . t, we get } d x/d t = 2\text { at and } d y/d t = 2a \therefore d y/d x = 2a/2at = 1/t \text{ Differentiating again w . r . t . t, we get } d^2 y/d x^2 = - 1/t^2d t/d x = - $\frac{1}{2}$ \ a t^3

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