Question

If x = 4t², \(y=\frac{3}{t^3}\), then \(\frac{d^2y}{dx^2}\) at t = 1 is :

• A. \(\frac{15}{8}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{15}{16}\)
• D. \(\frac{45}{64}\)

Answer 1

The Correct Option is D

To find \(\frac{d^2y}{dx^2}\) at t = 1, we need to go through the following steps:

Given: \(x = 4t^2\), \(y = \frac{3}{t^3}\).

1. Find \(\frac{dx}{dt}\):
   \(x = 4t^2 \Rightarrow \frac{dx}{dt} = 8t\).
2. Find \(\frac{dy}{dt}\):
   \(y = \frac{3}{t^3} \Rightarrow \frac{dy}{dt} = -\frac{9}{t^4}\) (using power rule).
3. Apply the chain rule to find \(\frac{dy}{dx}\):
   \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-\frac{9}{t^4}}{8t} = -\frac{9}{8t^5}\).
4. Differentiate \(\frac{dy}{dx}\) with respect to t to find \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\):
   \(\frac{d}{dt}\left(-\frac{9}{8t^5}\right) = -\frac{d}{dt}\left(\frac{9}{8t^5}\right)\).
   Applying the power rule: \(-\frac{9}{8}\cdot(-5)t^{-6} = \frac{45}{8t^6}\).
5. Find \(\frac{d^2y}{dx^2}\):
   \(\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}\).
   Substituting the derivative values: \(\frac{\frac{45}{8t^6}}{8t} = \frac{45}{64t^7}\).
6. Evaluate at t = 1:
   \(\frac{45}{64 \times 1^7} = \frac{45}{64}\).

Thus, \(\frac{d^2y}{dx^2}\) at t=1 is \(\frac{45}{64}\).