MCQ
If $x = a{t^2},y = 2at$, then ${{{d^2}y} \over {d{x^2}}} = $
- A$ - {1 \over {{t^2}}}$
- B${1 \over {2a{t^3}}}$
- C$ - {1 \over {{t^3}}}$
- ✓$ - {1 \over {2a{t^3}}}$
==> $\frac{{dy}}{{dx}} = \frac{1}{t} = \frac{{2a}}{y}$
==> $y\frac{{dy}}{{dx}} = 2a$
==> $y\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^2} = 0$
$ \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - {{(dy/dx)}^2}}}{y} = - \frac{1}{{2a{t^3}}}$.
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