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. \( -\frac{1}{2at^6} \)

Hint:

Remember: \( \frac{d^2y}{dx^2} \neq \frac{y''(t)}{x''(t)} \).
A helpful mnemonic for the parametric second derivative is:
\( \text{Derivative of Slope} \div \text{Derivative of x} \).
Always express your final answer in the simplest power form to avoid confusion with the options.

Answer 1

The Correct Option is D

Step 1: Differentiate each part with respect to $t$.
From $x = at^4$ we get $\dfrac{dx}{dt} = 4at^3$. From $y = 2at^2$ we get $\dfrac{dy}{dt} = 4at$.

Step 2: Get the first slope.
Divide one by the other.
\[ \frac{dy}{dx} = \frac{4at}{4at^3} = \frac{1}{t^2} \]

Step 3: Differentiate the slope, then divide again.
The second derivative needs us to differentiate $\dfrac{dy}{dx}$ with respect to $t$, and then divide by $\dfrac{dx}{dt}$ once more. First, $\dfrac{d}{dt}\left(t^{-2}\right) = -2t^{-3}$.
\[ \frac{d^2y}{dx^2} = \frac{-2t^{-3}}{4at^3} \]

Step 4: Tidy up.
\[ \frac{d^2y}{dx^2} = \frac{-2}{4at^6} = -\frac{1}{2at^6} \]
That is option 4.
\[ \boxed{-\dfrac{1}{2at^6}} \]