Let \(x = t^{3}\) and \(y = t^{2}\), where \(t \ne 0\). Consider the following statements:
(A) \(\displaystyle\frac{dy}{dx} = \frac{2}{3t}\)
(B) \(\displaystyle\frac{d^{2}y}{dx^{2}} = -\frac{2}{9t^{4}}\)
(C) \(\displaystyle\frac{d^{2}y}{dx^{2}} = \frac{1}{3t}\)
(D) At \(t = -1\), \(\displaystyle\frac{d^{2}y}{dx^{2}} = \frac{2}{9}\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (B) only
In parametric form:
\( \displaystyle \frac{dx}{dt} = 3t^{2}, \quad \frac{dy}{dt} = 2t \)
\( \displaystyle \frac{dy}{dx} = \frac{2t}{3t^{2}} = \frac{2}{3t} \)
For the second derivative, differentiate dy/dx with respect to t and divide by dx/dt:
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{\frac{d}{dt}\left(\frac{2}{3t}\right)}{3t^{2}} = \frac{-\frac{2}{3t^{2}}}{3t^{2}} = -\frac{2}{9t^{4}} \)
So \(\frac{dy}{dx} = \frac{2}{3t}\) and \(\frac{d^{2}y}{dx^{2}} = -\frac{2}{9t^{4}}\), which is negative for every t ≠ 0.
Checking the statements
(A) \(\frac{dy}{dx} = \frac{2}{3t}\) — Correct. \(\frac{2t}{3t^{2}} = \frac{2}{3t}\).
(B) \(\frac{d^{2}y}{dx^{2}} = -\frac{2}{9t^{4}}\) — Correct. \(-\frac{2}{3t^{2}}\) divided by \(3t^{2}\).
(C) \(\frac{d^{2}y}{dx^{2}} = \frac{1}{3t}\) — Incorrect. This is \(\frac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}} = \frac{2}{6t}\), which is not the second derivative; the true value is \(-\frac{2}{9t^{4}}\).
(D) At \(t = -1\), \(\frac{d^{2}y}{dx^{2}} = \frac{2}{9}\). — Incorrect. \(-\frac{2}{9(-1)^{4}} = -\frac{2}{9}\).