Question

Let \(x = at^2-1\), where \(a > 0\) and \(y = t^3+1\). If at \(t = 1\), \(\frac{d^2y}{dx^2} = \frac{3}{16}\), then the value of \(a\) is...

• A. \(3\)
• B. \(-2\)
• C. \(1\)
• D. \(2\)

Hint:

Differentiate in t twice, using dy/dx = (dy/dt)/(dx/dt).

Answer 1

The Correct Option is D

Step 1: Eliminate t Instead:
From $x=at^2-1$, $t^2=\dfrac{x+1}{a}$. For $t>0$, $y=t^3+1=\left(\dfrac{x+1}{a}\right)^{3/2}+1$.

Step 2: Differentiate:
$\dfrac{dy}{dx}=\dfrac{3}{2a}\left(\dfrac{x+1}{a}\right)^{1/2}$. $\dfrac{d^2y}{dx^2}=\dfrac{3}{4a^2}\left(\dfrac{x+1}{a}\right)^{-1/2}=\dfrac{3}{4a^2t}$.

Step 3: Solve:
At $t=1$, $\dfrac3{4a^2}=\dfrac3{16}$, so $a=2$. Option (D).

Final Answer:
Option (D). \[ \boxed{\text{(D) } 2} \]