Question

If \[ x = \int_0^y \frac{1}{\sqrt{1+9t^2}} \, dt \quad \text{and} \quad \frac{d^2 y}{dx^2} = ay, \] then \( a \) is equal to:

• A. \( 3 \)
• B. \( 6 \)
• C. \( 9 \)
• D. \( 1 \)

Hint:

For integrals defining inverse functions: \begin{itemize} \item Use Fundamental Theorem of Calculus. \item Convert \( dx/dy \) to \( dy/dx \). \item Apply chain rule carefully. \end{itemize}

Answer 1

The Correct Option is C

Concept: Given inverse relation: \[ x = \int_0^y f(t)\,dt \Rightarrow \frac{dx}{dy} = f(y) \] Then: \[ \frac{dy}{dx} = \frac{1}{dx/dy} \] Step 1: {\color{red}Differentiate given integral.} \[ \frac{dx}{dy} = \frac{1}{\sqrt{1+9y^2}} \] So: \[ \frac{dy}{dx} = \sqrt{1+9y^2} \] Step 2: {\color{red}Find second derivative.} Differentiate w.r.t. \( x \): \[ \frac{d^2 y}{dx^2} = \frac{d}{dx}(\sqrt{1+9y^2}) \] Using chain rule: \[ = \frac{1}{2\sqrt{1+9y^2}} \cdot 18y \cdot \frac{dy}{dx} \] Substitute \( dy/dx = \sqrt{1+9y^2} \): \[ \frac{d^2 y}{dx^2} = \frac{18y}{2} = 9y \] Step 3: {\color{red}Compare with given form.} \[ \frac{d^2 y}{dx^2} = ay \Rightarrow a = 9 \]