Solveeit Logo
Login

Question

Question: If \(x = a{t^2}\) and \(y = 2at\), then \(\dfrac{{{d^2}y}}{{d{x^2}}}\) at \(t = \dfrac{1}{2}\) is ...

If x=at2x = a{t^2} and y=2aty = 2at, then d2ydx2\dfrac{{{d^2}y}}{{d{x^2}}} at t=12t = \dfrac{1}{2} is
A.−2a\dfrac{{ - 2}}{a}
B.4a\dfrac{4}{a}
C.8a\dfrac{8}{a}
D.−4a\dfrac{{ - 4}}{a}

Explanation

Solution

Hint : Use differentiation of parametric form method
The given equations are x=at2x = a{t^2}and y=2aty = 2at
On differentiating xx with respect to tt we get,
dxdt=d(at2)dt=2at\dfrac{{dx}}{{dt}} = \dfrac{{d(a{t^2})}}{{dt}} = 2at ……(i)
dtdx=12at\dfrac{{dt}}{{dx}} = \dfrac{1}{{2at}} ……(ii)
Similarly differentiating yy with respect to tt we get,
dydt=d(2at)dt=2a\dfrac{{dy}}{{dt}} = \dfrac{{d(2at)}}{{dt}} = 2a ……(iii)
To get dydx\dfrac{{dy}}{{dx}} divide (iii) by (i)
So, dydx=1t\dfrac{{dy}}{{dx}} = \dfrac{1}{t}
Double differentiating the above equation we get
d2ydx2=−1t2dtdx\dfrac{{{d^2}y}}{{d{x^2}}} = - \dfrac{1}{{{t^2}}}\dfrac{{dt}}{{dx}}
d2ydx2=−1t2(12at)=−12at3\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
d2ydx2\dfrac{{{d^2}y}}{{d{x^2}}} at t=12t = \frac{1}{2}
On putting t=12t = \dfrac{1}{2} in equation (iv) we get,
d2ydx2=−12a(12)3=−12a(18)=−4a\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 dydx\dfrac{{dy}}{{dx}} then double differentiate it as done above, at last put the value of the variable provided to get the answer.