Question

# If $x = a{t^2}$ and $y = 2at$, then $\dfrac{{{d^2}y}}{{d{x^2}}}$ at $t = \dfrac{1}{2}$ isA.$\dfrac{{ - 2}}{a}$B.$\dfrac{4}{a}$C.$\dfrac{8}{a}$D.$\dfrac{{ - 4}}{a}$

Hint : Use differentiation of parametric form method
The given equations are $x = a{t^2}$and $y = 2at$
On differentiating $x$ with respect to $t$ we get,
$\dfrac{{dx}}{{dt}} = \dfrac{{d(a{t^2})}}{{dt}} = 2at$ ……(i)
$\dfrac{{dt}}{{dx}} = \dfrac{1}{{2at}}$ ……(ii)
Similarly differentiating $y$ with respect to $t$ we get,
$\dfrac{{dy}}{{dt}} = \dfrac{{d(2at)}}{{dt}} = 2a$ ……(iii)
To get $\dfrac{{dy}}{{dx}}$ divide (iii) by (i)
So, $\dfrac{{dy}}{{dx}} = \dfrac{1}{t}$
Double differentiating the above equation we get
$\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{1}{{{t^2}}}\dfrac{{dt}}{{dx}}$
$\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{1}{{{t^2}}}\left( {\dfrac{1}{{2at}}} \right) = - \dfrac{1}{{2a{t^3}}}$ (From (ii)) ……(iv)
We have been asked to find the value of
$\dfrac{{{d^2}y}}{{d{x^2}}}$ at $t = \frac{1}{2}$
On putting $t = \dfrac{1}{2}$ in equation (iv) we get,
$\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{1}{{2a{{\left( {\dfrac{1}{2}} \right)}^3}}} = - \dfrac{1}{{2a\left( {\dfrac{1}{8}} \right)}} = - \dfrac{4}{a}$
Hence the correct option is D.

Note :-In these type of question of finding double differentiation at a particular point where the equations are in parametric form, we first have to differentiate it with respect to the variable assigned then divide them to get $\dfrac{{dy}}{{dx}}$ then double differentiate it as done above, at last put the value of the variable provided to get the answer.