Given the parametric equations \( x = at^4 \) and \( y = 2at^2 \), the objective is to determine the second derivative of \( y \) with respect to \( x \), denoted as \( \frac{d^2y}{dx^2} \).
First, compute the derivatives of \( y \) and \( x \) with respect to \( t \):
Employ the chain rule to find \( \frac{dy}{dx} \):
\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4at}{4at^3} = \frac{1}{t^2}\)
Next, compute the derivative of \( \frac{dy}{dx} \) with respect to \( t \):
\(\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(t^{-2}\right) = -2t^{-3} = -\frac{2}{t^3}\)
Finally, apply the chain rule again to obtain \( \frac{d^2y}{dx^2} \):
\(\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt} = \left(-\frac{2}{t^3}\right) \div (4at^3) = -\frac{2}{t^3} \cdot \frac{1}{4at^3} = -\frac{1}{2at^6}\)
The computed result is: \(-\frac{1}{2at^6}\)