Answer
$\frac{{{d^2}y}}{{d{x^2}}}{|_{t = 2}} = - \frac{{21}}{{512}}$.
Work Step by Step
We have
$x'\left( t \right) = 2t + 3{t^2}$, $x{\rm{''}}\left( t \right) = 2 + 6t$,
$y'\left( t \right) = 14t$, $y{\rm{''}}\left( t \right) = 14$.
Using Eq. (13) we get
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{x'\left( t \right)y{\rm{''}}\left( t \right) - y'\left( t \right)x{\rm{''}}\left( t \right)}}{{x'{{\left( t \right)}^3}}} = - \frac{{42}}{{t{{\left( {2 + 3t} \right)}^3}}}$.
$\frac{{{d^2}y}}{{d{x^2}}}{|_{t = 2}} = - \frac{{21}}{{512}}$.
