If x=at2 and y=2at then d2ydx2 at t=12 is

x=at2, y=2at dxdt=a(2t), dydt=2a dydx=2a2at=1t d2ydx2=-1t2·dtdx=-1t2 12at d2ydx2t=12=-12a123=-4a

If x=at2 and y=2at then d2ydx2 at t=12 is

x=at2, y=2at dxdt=a(2t), dydt=2a dydx=2a2at=1t d2ydx2=-1t2·dtdx=-1t2 12at d2ydx2t=12=-12a123=-4a

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

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$\frac{2}{a}$

$\frac{4}{a}$

$\frac{8}{a}$

$\frac{- 4}{a}$

answer is D.

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Detailed Solution

$x = {at}^{2}, y = 2 at \frac{dx}{dt} = a (2 t), \frac{dy}{dt} = 2 a \frac{dy}{dx} = \frac{2 a}{2 at} = \frac{1}{t} \frac{d^{2} y}{{dx}^{2}} = \frac{- 1}{t^{2}} \cdot \frac{dt}{dx} = \frac{- 1}{t^{2}} \frac{1}{2 at} {(\frac{d^{2} y}{{dx}^{2}})}_{t = \frac{1}{2}} = \frac{- 1}{2 a {(\frac{1}{2})}^{3}} = \frac{- 4}{a}$