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

Answer: (D)

$\frac{-4}{a}$

$\frac{dy}{dt}=2a,\frac{dx}{dt}=2at$

$\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{1}{t}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-1}{t^2}\frac{dx}{dt}=\frac{1}{t^{2}2at}=\frac{-1}{2a(t{)}^3}$
Put $t=\frac{1}{2}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-4}{a}$