If \( x=4t^3 \) and \( y=6t^2 \) then \( \displaystyle\frac{d^2y}{dx^2} \) at \( t=1 \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -\displaystyle\frac{1}{12} \)
By parametric differentiation:
\( \displaystyle \frac{dx}{dt}=12t^2 \)
\( \displaystyle \frac{dy}{dt}=12t \)
\( \displaystyle \frac{dy}{dx}=\frac{12t}{12t^2}=\frac{1}{t} \)
Differentiating again with respect to x means multiplying by \( \frac{dt}{dx} \):
\( \displaystyle \frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{1}{t}\right)\cdot\frac{dt}{dx} \)
\( \displaystyle =-\frac{1}{t^2}\cdot\frac{1}{12t^2} \)
\( \displaystyle =-\frac{1}{12t^4} \)
At t = 1:
\( \displaystyle \frac{d^2y}{dx^2}=-\frac{1}{12} \)
Why the other options are incorrect
Option (1) 12 — Incorrect. 12 is \( \frac{d^2y}{dt^2}=12 \), the second derivative with respect to t, not x.
Option (3) \( \displaystyle\frac{1}{3} \) — Incorrect. \( \frac{1}{3} \) is positive, but \( -\frac{1}{12t^4} \) is negative for every t.
Option (4) 1 — Incorrect. 1 is the value of \( \frac{dy}{dx}=\frac{1}{t} \) at t = 1, the first derivative.
Tip: \( \frac{d^2y}{dx^2} \) is not \( \frac{d^2y/dt^2}{d^2x/dt^2} \): differentiate \( \frac{dy}{dx} \) with respect to t, then divide by \( \frac{dx}{dt} \).