CUET Preparation Today
If $x=4 t^2, y=8 t$, then $\frac{d^2 y}{d x^2}$ is: |
$\frac{1}{t}$ $-\frac{1}{8 t}$ $-\frac{1}{8 t^3}$ $-\frac{1}{t^2}$ |
$-\frac{1}{8 t^3}$ |
The correct answer is Option (3) → $-\frac{1}{8 t^3}$ $x=4 t^2, y=8 t$ $\frac{dx}{dt}=8t$ $\frac{dy}{dt}=8$ $⇒\frac{dy}{dx}=\frac{1}{t}⇒\frac{d^2y}{dx^2}=-\frac{1}{t}\frac{dt}{dx}=-\frac{1}{8 t^3}$ $\frac{d^2y}{dx^2}=-\frac{1}{8 t^3}$ |
