If \(x = t^{2} + 1\) and \(y = t^{3} + t\), then the value of \(\displaystyle\frac{d^{2}y}{dx^{2}}\) at \(t = 1\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\frac{1}{2}\)
The parametric derivative comes from
\( \displaystyle \frac{dx}{dt} = 2t \)
\( \displaystyle \frac{dy}{dt} = 3t^{2} + 1 \)
\( \displaystyle \frac{dy}{dx} = \frac{3t^{2} + 1}{2t} = \frac{3t}{2} + \frac{1}{2t} \)
This is still a function of t, so to differentiate it with respect to x we differentiate with respect to t and divide by \(\frac{dx}{dt}\) again:
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt} \)
\( \displaystyle = \frac{\frac{3}{2} - \frac{1}{2t^{2}}}{2t} \)
At t = 1:
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{\frac{3}{2} - \frac{1}{2}}{2} \)
\( \displaystyle = \frac{1}{2} \)
Why the other options are incorrect
Option (1) 3 — Incorrect. This is \(\frac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}} = \frac{6t}{2}\) at t = 1, which is not the second derivative of y with respect to x.
Option (2) 2 — Incorrect. This is the first derivative \(\frac{dy}{dx} = \frac{3 + 1}{2}\) at t = 1.
Option (4) 1 — Incorrect. This is \(\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{3}{2} - \frac{1}{2}\) at t = 1, before dividing by \(\frac{dx}{dt} = 2\).