Question

If \(x = \frac{1}{t^2}\) and \(y =\frac{1}{t^3}\), then \(\frac{d^2y}{dx^2}\) at \(t=1\) is:

• A. \(\frac{3}{2}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{3}{4}\)

Answer 1

The Correct Option is D

Given that \(x = \frac{1}{t^2}\) and \(y = \frac{1}{t^3}\), we need to find \(\frac{d^2y}{dx^2}\) at \(t=1\).

First, find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\).

\(\frac{dy}{dt} = \frac{d}{dt}\left(t^{-3}\right) = -3t^{-4}\)

\(\frac{dx}{dt} = \frac{d}{dt}\left(t^{-2}\right) = -2t^{-3}\)

Now, find \(\frac{dy}{dx}\) using the chain rule:

\(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{-3t^{-4}}{-2t^{-3}} = \frac{3}{2}t^{-1}\)

Differentiating \(\frac{dy}{dx}\) with respect to \(t\) gives:

\(\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{3}{2}t^{-1}\right) \div \frac{dx}{dt}\)

\(\frac{d}{dt}\left(\frac{3}{2}t^{-1}\right) = \frac{3}{2}(-1)t^{-2} = -\frac{3}{2}t^{-2}\)

Thus,

\(\frac{d^2y}{dx^2} = \frac{-\frac{3}{2}t^{-2}}{-2t^{-3}} = \frac{3}{4}t\)

Evaluating this at \(t=1\):

\(\frac{d^2y}{dx^2}\bigg|_{t=1} = \frac{3}{4}(1) = \frac{3}{4}\)

Therefore, the correct answer is \(\frac{3}{4}\).