If $x=at^2$ and $y=2at$, then $\frac{d^2 y}{d x^2}$ is |
$\frac{1}{2 a t^3}$ $\frac{1}{a t^3}$ $\frac{-1}{2 a t^3}$ $-\frac{1}{t^2}$ |
$\frac{-1}{2 a t^3}$ |
The correct answer is Option (3) - $\frac{-1}{2 a t^3}$ $\frac{dx}{dt}=2at$, $\frac{dy}{dt}=2a$ $\frac{dy}{dx}=\frac{1}{t}$ so $\frac{d^2t}{dx^2}=\frac{-1}{t^2}\frac{dt}{dx}$ $=\frac{-1}{2 a t^3}$ |