If x = int_0^y frac{1}{sqrt{1 + 9t^2}} dt and | vedclass

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(C) Given $x = \int_0^y \frac{1}{\sqrt{1 + 9t^2}} dt$. By the Fundamental Theorem of Calculus, differentiating both sides with respect to $y$, we get

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$9$

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(C) Given $x = \int_0^y \frac{1}{\sqrt{1 + 9t^2}} dt$. By the Fundamental Theorem of Calculus, differentiating both sides with respect to $y$, we get $\frac{dx}{dy} = \frac{1}{\sqrt{1 + 9y^2}}$. Taking the reciprocal, we have $\frac{dy}{dx} = \sqrt{1 + 9y^2}$. Differentiating both sides with respect to $x$ using the chain rule, we get $\frac{d^2y}{dx^2} = \frac{d}{dx}(\sqrt{1 + 9y^2}) = \frac{1}{2\sqrt{1 + 9y^2}} \cdot (18y) \cdot \frac{dy}{dx}$. Substituting $\frac{dy}{dx} = \sqrt{1 + 9y^2}$, we get $\frac{d^2y}{dx^2} = \frac{9y}{\sqrt{1 + 9y^2}} \cdot \sqrt{1 + 9y^2} = 9y$. Comparing $\frac{d^2y}{dx^2} = 9y$ with $\frac{d^2y}{dx^2} = ay$, we find $a = 9$.

If $x = \int_0^y \frac{1}{\sqrt{1 + 9t^2}} dt$ and $\frac{d^2y}{dx^2} = ay$, then $a$ is equal to

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