Question

If \(x = a \cos^3 \theta, y = a \sin^3 \theta\), then \(\frac{d^2y}{dx^2}\) at \(\theta = \pi/4\) is

• A. \(\frac{4\sqrt{2}}{3a}\)
• B. \(\frac{2}{3a}\)
• C. \(\frac{2\sqrt{2}}{3a}\)
• D. \(\frac{7\sqrt{2}}{3a}\)

Hint:

For parametric curves, use \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\) and \(\frac{d^2y}{dx^2} = \frac{d(dy/dx)/d\theta}{dx/d\theta}\).

Answer 1

The Correct Option is A

Step 1: Find first derivative.
Parametric form: \(x = a \cos^3 \theta, y = a \sin^3 \theta\)
\(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} = -\tan \theta\)

Step 2: Find derivative of dy/dx w.r.t x.
\(\frac{d^2y}{dx^2} = \frac{d(dy/dx)/d\theta}{dx/d\theta}\)

Step 3: Compute derivative w.r.t \(\theta\).
\(\frac{d}{d\theta}(-\tan \theta) = -\sec^2 \theta\)

Step 4: Divide by dx/d\(\theta\).
\(dx/d\theta = -3a \cos^2 \theta \sin \theta\)
\(\frac{d^2y}{dx^2} = \frac{-\sec^2 \theta}{-3a \cos^2 \theta \sin \theta} = \frac{1}{3a \sin \theta}\)

Step 5: Evaluate at \(\theta = \pi/4\).
\(\frac{d^2y}{dx^2} = \frac{1}{3a \cdot 1/\sqrt{2}} = \frac{4\sqrt{2}}{3a}\)

Step 6: Final conclusion.
Hence, \[ \boxed{\frac{4\sqrt{2}}{3a}} \]