Question

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

• A. $\frac{-2}{a}$
• B. $\frac{4}{a}$
• C. $\frac{8}{a}$
• D. $\frac{-4}{a}$

Answer 1

The correct option is D $\frac{-4}{a}$

$\frac{dy}{dx}=\frac{1}{t}$
$\frac{d^{2}y}{d{x}^2}=\frac{1}{t^2}\frac{dy}{dx}=\frac{1}{t^{2}2at}=\frac{1}{2at3}$
Put $t=\frac{1}{2}$
$\frac{d^{2}y}{d{x}^2}=\frac{-4}{a}$