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If $x=a t^2, y=2 a t$, then $\frac{d^2 y}{d x^2}=$
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The correct answer is:
$-\frac{1}{2 a t^3}$
$\begin{aligned} & \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{2 a}{2 a t} \Rightarrow \frac{d y}{d x}=\frac{1}{t}=\frac{2 a}{y} \\ & \Rightarrow y \frac{d y}{d x}=2 a \Rightarrow y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0 \\ & \Rightarrow \frac{d^2 y}{d x^2}=\frac{-(d y / d x)^2}{y}=\frac{-1}{2 a t^3} .\end{aligned}$
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