Question

If \( x = a t^2, \; y = 2at \), then \( \frac{d^2 y}{dx^2} =\) _____

• A. \( \frac{a}{xy} \)
• B. \( \frac{ax}{y} \)
• C. \( -\frac{a}{xy} \)
• D. \( -\frac{ax}{y} \)

Hint:

Always convert parametric derivatives into \(x,y\) form at the end.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a parametric differentiation problem. We first find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. To find the second derivative, we use $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}$.
Step 2: Formula Application:
$dx/dt = 2at$ $dy/dt = 2a$ $\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$
Step 3: Explanation:
$\frac{d}{dt}(\frac{dy}{dx}) = \frac{d}{dt}(t^{-1}) = -t^{-2}$ Now, $\frac{d^2y}{dx^2} = -t^{-2} \times \frac{1}{2at} = -\frac{1}{2at^3}$ Since $y = 2at \implies t = y/2a$. $\frac{d^2y}{dx^2} = -\frac{1}{2a(y/2a)^3} = -\frac{4a^2}{y^3}$. (None of the original options match exactly; the calculated result is $-1/(2at^3)$ or equivalent).
Step 4: Final Answer:
The result is $-1/(2at^3)$.