Question
Find $\frac{d^2 y}{d x^2}$, if $y =2 at , x = at t ^2$.

Answer

$
x=a t^2, y=2 a t
$
Differentiating $x$ and $y$ w.r.t. $t$, we get
$
\begin{aligned}
& \frac{d x}{d t}=\frac{d}{d t}\left(a t^2\right)=a \frac{d}{d t}\left(t^2\right) \\
& =a \times 2 t=2 a t \\
& \text { and } \frac{d y}{d t}=\frac{d}{d t}(2 a t)=2 a \frac{d}{d t}(t) \\
& =2 a \times 1=2 a \\
& \therefore \frac{d y}{d x}=\frac{(d y / d t)}{(d x / d t)}=\frac{2 a}{2 a t}=\frac{1}{t} \\
& \therefore \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{1}{t}\right)=\frac{d}{d t}\left(\frac{1}{t}\right) \cdot \frac{d t}{d x} \\
& =-\frac{1}{t^2} \times \frac{1}{\left(\frac{d x}{d t}\right)}=-\frac{1}{t^2} \times \frac{1}{2 a t} \\
& =-\frac{1}{2 a t^3} \text {. } \\
\end{aligned}
$

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