Question

If $x=a{t}^2,y=2at$ then show that $\frac{d^{2}y}{d{x}^2}=\frac{-1}{2a{t}^2}$.

Answer 1

$x=a{t}^2$

$\Rightarrow \frac{dx}{dt}=2at$

$y=2at$

$\Rightarrow \frac{dy}{dt}=2a$

$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{2a}{2at}=\frac{1}{t}=\frac{1}{\frac{y}{2a}}$ since $y=2at$

$=\frac{2a}{y}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=2a\frac{d}{dx}\left(y^{-1}\right)$

$=\frac{2a\times -1}{y^2}=-\frac{2a}{y^2}$

$=\frac{-2a}{{\left(2at\right)}^2}$ since $y=2at$

$=\frac{-2a}{4a^2{t}^2}$

$=\frac{-1}{2a{t}^2}$

Hence proved.