If x=at^2 and y=4at , then the value of (d^2y)/(dx^2) at t = 1/a, is

The correct answer is Option (2) → -a²

For these parametric equations,

    \( \displaystyle \frac{dx}{dt}=2at \)

    \( \displaystyle \frac{dy}{dt}=4a \)

    \( \displaystyle \frac{dy}{dx}=\frac{4a}{2at} \)

    \( \displaystyle =\frac{2}{t} \)

The second derivative is the t-derivative of \( \frac{dy}{dx} \), multiplied by \( \frac{dt}{dx} \):

    \( \displaystyle \frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{2}{t}\right)\cdot\frac{dt}{dx} \)

    \( \displaystyle =-\frac{2}{t^2}\cdot\frac{1}{2at} \)

    \( \displaystyle =-\frac{1}{at^3} \)

At \( t=\frac{1}{a} \):

    \( \displaystyle \frac{d^2y}{dx^2}=-\frac{1}{a\cdot\frac{1}{a^3}} \)

    \( \displaystyle =-a^2 \)

Tip: \( \frac{d^2y}{dx^2} \) is not \( \frac{d^2y/dt^2}{d^2x/dt^2} \); here that ratio would give 0.