CUET Preparation Today
If $x=at^2$ and $y=2at,$ then $\frac{d^2y}{dx^2}=$ |
0 $-\frac{1}{t^2}$ $-\frac{1}{2at^3}$ $-\frac{t^3}{2a}$ |
$-\frac{1}{2at^3}$ |
The correct answer is Option (3) → $-\frac{1}{2at^3}$ $x=at^2$ and $y=2at$ $\frac{dx}{dt}=2at$, $\frac{dy}{dt}=2a⇒\frac{dy}{dx}=\frac{1}{t}$ $\frac{d^2y}{dx^2}=-\frac{1}{t^2}×\frac{dt}{dx}$ $=-\frac{1}{2at^3}$ |
