Question

If \( x = at^4 \) and \( y = 2at^2 \), then \( \frac{d^2y}{dx^2} \) is equal to:

• A. \(-\frac{1}{4at^4}\)
• B. \(-\frac{2}{t^3}\)
• C. \(-\frac{1}{t}\)
• D. \(-\frac{1}{2at^6}\)

Answer 1

The Correct Option is D

Given \( x = at^4 \) and \( y = 2at^2 \), differentiate \( y \) with respect to \( t \) to find \( \frac{dy}{dx} \).

First, compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):

\[ \frac{dx}{dt} = 4at^3, \quad \frac{dy}{dt} = 4at. \]

Using the chain rule:

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4at}{4at^3} = \frac{1}{t^2}. \]

Next, differentiate \( \frac{dy}{dx} \) with respect to \( t \):

\[ \frac{d^2y}{dx^2} = \frac{d}{dt} \left( \frac{1}{t^2} \right)\times \frac{dt}{dx}. \]

\[ \frac{d}{dt} \left( \frac{1}{t^2} \right) = -\frac{2}{t^3}, \quad \frac{dt}{dx} = \frac{1}{4at^3}. \]

Substitute these values:

\[ \frac{d^2y}{dx^2} = -\frac{2}{t^3} \times \frac{1}{4at^3} = -\frac{1}{2at^6}. \]

Hence, the correct answer is Option (D).