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

Answer: (D)

$-\frac{1}{2at^6}$

Answer 2

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).