If \( x=3t^2 \) and \( y=\sqrt{t} \), then the value of \( \displaystyle\frac{d^2y}{dx^2} \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( -\displaystyle\frac{1}{48t^3\sqrt{t}} \)
For a parametric curve, \( \frac{dy}{dx}=\frac{dy/dt}{dx/dt} \). Here \( \frac{dy}{dt}=\frac{1}{2}t^{-1/2} \) and \( \frac{dx}{dt}=6t \).
\( \displaystyle \frac{dy}{dx}=\frac{\frac{1}{2}t^{-1/2}}{6t} \)
\( \displaystyle =\frac{1}{12}t^{-3/2} \)
Differentiating again with respect to x, using \( \frac{dt}{dx}=\frac{1}{6t} \):
\( \displaystyle \frac{d^2y}{dx^2}=\frac{1}{12}\cdot\left(-\frac{3}{2}\right)t^{-5/2}\cdot\frac{1}{6t} \)
\( \displaystyle =-\frac{1}{48}t^{-7/2} \)
\( \displaystyle =-\frac{1}{48t^3\sqrt{t}} \)