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If the solve the problem

Question:

If $x=2 a t, y=a t^{2}$, where $a$ is a constant, then find $\frac{d^{2} y}{d x^{2}}$ at $x=\frac{1}{2}$.

Solution:

Given:

$x=2 a t, y=a t^{2}$

$\frac{d x}{d t}=2 a ; \frac{d y}{d t}=2 a t$

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

$=\mathrm{t}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

$=\frac{1}{2 a}$

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