Write the correct alternative in the following:
If $x=a t^{2}, y=2 a t$, then $\frac{d^{2} y}{d x^{2}}=$
A. $-\frac{1}{t^{2}}$
B. $\frac{1}{2 \mathrm{at}^{3}}$
C. $-\frac{1}{t^{3}}$
D. $-\frac{1}{2 \mathrm{at}^{3}}$
Given:
$y=2 a t, x=a t^{2}$
$\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{at} ; \frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{a}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$
$=\frac{1}{t}$
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\frac{\mathrm{dx}}{\mathrm{dt}}}$
$=\frac{\frac{-1}{t^{2}}}{2 a t}$
$=\frac{-1}{2 a t^{3}}$
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