Question

If \( x = \sin \theta, y = \sin^3 \theta \), then \( \frac{d^2y}{dx^2} \) at \( \theta = \frac{\pi}{6} \) is

• A. \( \frac{1}{2} \)
• B. \( \frac{\sqrt{3}}{2} \)
• C. 3
• D. 6

Hint:

If $y$ can be expressed directly in terms of $x$ (here $y=x^3$), use that instead of parametric differentiation to save time.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This is a parametric differentiation problem where we need the second derivative.
Step 2: Key Formula or Approach:
1. \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \).
2. \( \frac{d^2y}{dx^2} = \frac{d}{d\theta} \left( \frac{dy}{dx} \right) \cdot \frac{d\theta}{dx} \).
Step 3: Detailed Explanation:
Given \( x = \sin \theta \) and \( y = \sin^3 \theta \).
\( \frac{dx}{d\theta} = \cos \theta \).
\( \frac{dy}{d\theta} = 3 \sin^2 \theta \cos \theta \).
\( \frac{dy}{dx} = \frac{3 \sin^2 \theta \cos \theta}{\cos \theta} = 3 \sin^2 \theta \).
Now, find the second derivative:
\[ \frac{d^2y}{dx^2} = \frac{d}{d\theta}(3 \sin^2 \theta) \times \frac{1}{\cos \theta} \] \[ = (6 \sin \theta \cos \theta) \times \frac{1}{\cos \theta} = 6 \sin \theta \] At \( \theta = \frac{\pi}{6} \):
\[ \frac{d^2y}{dx^2} = 6 \sin\left(\frac{\pi}{6}\right) = 6 \times \frac{1}{2} = 3 \] Step 4: Final Answer:
The value is 3.