If \(x = at^2, y = 2at\); then \(\displaystyle\frac{d^2y}{dx^2}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-\displaystyle\frac{1}{2at^3}\)
Since x and y are given in terms of the parameter t, the derivative comes from the parametric form:
\( \displaystyle \frac{dx}{dt} = 2at,\quad \frac{dy}{dt} = 2a \)
\( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t} \)
Differentiating again with respect to x, by the chain rule through t:
\( \displaystyle \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{1}{t}\right)\cdot\frac{dt}{dx} \)
\( \displaystyle = -\frac{1}{t^2}\cdot\frac{1}{2at} \)
\( \displaystyle = -\frac{1}{2at^3} \)
Why the other options are incorrect
Option (1) 0 — Incorrect. \(\frac{dy}{dx} = \frac{1}{t}\) is not constant, so its derivative is not 0.
Option (2) \(\displaystyle\frac{1}{t}\) — Incorrect. This is the first derivative \(\frac{dy}{dx}\).
Option (3) \(-\displaystyle\frac{1}{t^2}\) — Incorrect. This is only \(\frac{d}{dt}\left(\frac{1}{t}\right)\); it must still be multiplied by \(\frac{dt}{dx} = \frac{1}{2at}\).