Question

If \( x = \int_0^y \frac{1}{\sqrt{1+9t^2}} \, dt \) and \( \frac{d^2y}{dx^2} = ay \), then the value of \( a \) is:

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

Hint:

For problems involving integrals with square roots and second derivatives, use the chain rule carefully and always substitute intermediate results into your final expressions to avoid errors.

Answer 1

The Correct Option is C

Step 1: Find \( \frac{dx}{dy} \). Given: \[ x = \int_0^y \frac{1}{\sqrt{1+9t^2}} \, dt \] Differentiate with respect to \( y \): \[ \frac{dx}{dy} = \frac{1}{\sqrt{1+9y^2}} \]
Step 2: Find \( \frac{d^2x}{dy^2} \). Differentiate \( \frac{dx}{dy} \) with respect to \( y \): \[ \frac{d^2x}{dy^2} = \frac{d}{dy} \left( \frac{1}{\sqrt{1+9y^2}} \right) \] Using the chain rule: \[ \frac{d^2x}{dy^2} = -\frac{9y}{(1+9y^2)^{3/2}} \]
Step 3: Relate \( \frac{d^2y}{dx^2} \) and \( a \). Given: \[ \frac{d^2y}{dx^2} = ay \] From the chain rule: \[ \frac{d^2y}{dx^2} = \frac{1}{\left( \frac{dx}{dy} \right)^3} \cdot \frac{d^2y}{dy^2} \] Substitute: \[ \frac{1}{\left( \frac{1}{\sqrt{1+9y^2}} \right)^3} \cdot \frac{9}{(1+9y^2)^{3/2}} = ay \] Simplify: \[ (1+9y^2)^{3/2} \cdot \frac{9}{(1+9y^2)^{3/2}} = ay \] Therefore: \[ 9 = ay \]
Step 4: Determine the value of \( a \). Thus: \[ a = \frac{9}{y} \] Assuming \( y = 1 \): \[ a = 9 \]
Final Answer: \( a = 9 \).