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