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: First Derivative:
\(x=at^2-1\Rightarrow\dfrac{dx}{dt}=2at\). \(y=t^3+1\Rightarrow\dfrac{dy}{dt}=3t^2\). So
\[ \frac{dy}{dx}=\frac{3t^2}{2at}=\frac{3t}{2a} \]

Step 2: Second Derivative:
\[ \frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{3t}{2a}\right)\Big/\frac{dx}{dt}=\frac{3/(2a)}{2at}=\frac{3}{4a^2t} \]

Step 3: Use t = 1:
At \(t=1\): \(\dfrac{3}{4a^2}=\dfrac3{16}\Rightarrow a^2=4\). Since \(a>0\), \(a=2\).

Step 4: Check the Options:
\(a=-2\) is excluded by \(a>0\). For \(a=3\) and \(a=1\) the value would be \(1/12\) and \(3/4\). Only \(a=2\) works.

Final Answer:
The value is \(a=2\), option (D). \[ \boxed{\text{(D) } 2} \]