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: Differentiate the given equation for \( x \). We are given: \[ x = \int_0^y \frac{1}{\sqrt{1+9t^2}} \, dt \] Differentiating both sides with respect to \( y \): \[ \frac{dx}{dy} = \frac{1}{\sqrt{1+9y^2}} \]
Step 2: Differentiate again to find \( \frac{d^2x}{dy^2} \). Now 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: Relating \( \frac{d^2y}{dx^2} \) and \( a \). We are given: \[ \frac{d^2y}{dx^2} = ay \] From the chain rule, we know: \[ \frac{d^2y}{dx^2} = \frac{1}{\left( \frac{dx}{dy} \right)^3} \cdot \frac{d^2y}{dy^2} \] Substitute the values we have: \[ \frac{1}{\left( \frac{1}{\sqrt{1+9y^2}} \right)^3} \cdot \frac{9}{(1+9y^2)^{3/2}} = ay \] Simplify: \[ (1+9y^2)^{3/2} \cdot \frac{9}{(1+9y^2)^{3/2}} = ay \] Thus: \[ 9 = ay \]
Step 4: Conclude the value of \( a \). Thus: \[ a = \frac{9}{y} \] Since \( y \) is the variable of the original equation, the simplest scenario occurs when \( y = 1 \). This gives us: \[ a = 9 \]
Final Answer: \( a = 9 \).