Let x = at^2 - 1, where a 0 and y = t^3 + 1 | vedclass

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(D) Given $x = at^2 - 1$ and $y = t^3 + 1$.First, find the derivatives with respect to $t$: $\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 3t^2$.Then, $\f

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$2$

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(D) Given $x = at^2 - 1$ and $y = t^3 + 1$.First, find the derivatives with respect to $t$: $\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 3t^2$.Then, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2at} = \frac{3t}{2a}$.Now, find the second derivative $\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{3t}{2a}) = \frac{d}{dt}(\frac{3t}{2a}) \cdot \frac{dt}{dx} = \frac{3}{2a} \cdot \frac{1}{2at} = \frac{3}{4a^2t}$.At $t = 1$, $\frac{d^2y}{dx^2} = \frac{3}{4a^2(1)} = \frac{3}{4a^2}$.Given $\frac{3}{4a^2} = \frac{3}{16}$, we get $4a^2 = 16$, so $a^2 = 4$.Since $a > 0$, $a = 2$.

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:

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