If \(x=at^4\) and \(y=2at^2\), then \(\displaystyle\frac{d^2y}{dx^2}\) is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-\displaystyle\frac{1}{2at^6}\)
For parametric differentiation, differentiate both with respect to t:
\( \displaystyle \frac{dx}{dt} = 4at^3,\quad \frac{dy}{dt} = 4at \)
\( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4at}{4at^3} = \frac{1}{t^2} \)
dy/dx is a function of t, so differentiate it with respect to t and multiply by dt/dx (chain rule):
\( \displaystyle \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{1}{t^2}\right)\times\frac{dt}{dx} \)
\( \displaystyle = -\frac{2}{t^3}\times\frac{1}{4at^3} \)
\( \displaystyle = -\frac{1}{2at^6} \)
Tip: Stopping at \(\frac{d}{dt}\left(\frac{dy}{dx}\right) = -\frac{2}{t^3}\) gives Option (2). In the second derivative of a parametric curve, the factor dt/dx is easily forgotten.