Question

If \( x = a(\cos \theta + \theta \sin \theta), \, y = a(\sin \theta - \theta \cos \theta) \), then \( \frac{d^2 y}{dx^2} \) equals

• A. \( \frac{\sec^3 \theta}{a \theta} \)
• B. \( \frac{\sec^2 \theta}{a} \)
• C. \( a \theta \cos^3 \theta \)
• D. \( \frac{\sec^2 \theta}{a\theta} \)

Hint:

To compute higher-order derivatives in parametric equations, use the chain rule along with parametric differentiation.

Answer 1

The Correct Option is A

We are given the parametric equations: \[ x = a(\cos\theta + \theta\sin\theta), \quad y = a(\sin\theta - \theta\cos\theta). \] Step 1: Calculate \(\frac{dx}{d\theta}\) Differentiating \( x \) with respect to \(\theta\): \[ \frac{dx}{d\theta} = a\left(-\sin\theta + \theta\cos\theta + \sin\theta + \theta\cos\theta\right). \] Simplifying: \[ \frac{dx}{d\theta} = a\theta\cos\theta. \] Step 2: Calculate \(\frac{dy}{d\theta}\) Differentiating \( y \) with respect to \(\theta\): \[ \frac{dy}{d\theta} = a\left(\cos\theta + \theta(-\sin\theta) - \cos\theta - \theta(-\sin\theta)\right). \] Simplifying: \[ \frac{dy}{d\theta} = a\theta\sin\theta. \] Step 3: Calculate \(\frac{dy}{dx}\) Using the chain rule, we have: \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}. \] Substitute the values: \[ \frac{dy}{dx} = \frac{a\theta\sin\theta}{a\theta\cos\theta} = \tan\theta. \] Step 4: Calculate \(\frac{d^2y}{dx^2}\) Differentiate \(\frac{dy}{dx} = \tan\theta\) with respect to \( x \): \[ \frac{d^2y}{dx^2} = \frac{d}{dx}(\tan\theta). \] Using the chain rule: \[ \frac{d^2y}{dx^2} = \sec^2\theta \cdot \frac{d\theta}{dx}. \] From Step 1, we know: \[ \frac{dx}{d\theta} = a\theta\cos\theta \quad \Rightarrow \quad \frac{d\theta}{dx} = \frac{1}{a\theta\cos\theta}. \] Substitute: \[ \frac{d^2y}{dx^2} = \sec^2\theta \cdot \frac{1}{a\theta\cos\theta}. \] Simplifying: \[ \frac{d^2y}{dx^2} = \frac{\sec^3\theta}{a\theta}. \] Final Answer: \[ \boxed{\frac{\sec^3\theta}{a\theta} }. \]