If \(x = at^2\) and \(y = 2at\), then the value of \(\displaystyle\frac{d^2y}{dx^2}\) is
(where t is a parameter)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-\displaystyle\frac{1}{2at^3}\)
For a curve in parametric form, x and y are differentiated with respect to t:
\( \displaystyle \frac{dx}{dt} = 2at, \qquad \frac{dy}{dt} = 2a \)
\( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} \)
\( \displaystyle = \frac{1}{t} \)
The second derivative is the derivative of \(\frac{dy}{dx}\) with respect to x, so by the chain rule:
\( \displaystyle \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{1}{t}\right) \times \frac{dt}{dx} \)
\( \displaystyle = -\frac{1}{t^2} \times \frac{1}{2at} \)
\( \displaystyle = -\frac{1}{2at^3} \)
Tip: \(\frac{d^2y}{dx^2}\) is not \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\). Stopping at \(-\frac{1}{t^2}\) without multiplying by \(\frac{dt}{dx}\) gives Option (1), a very common mistake.