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: Use the parametric second-derivative formula.
For \(x=a\cos^3\theta, y=a\sin^3\theta\), use \(\frac{d^2y}{dx^2}=\frac{x'y''-y'x''}{(x')^3}\), primes meaning derivatives with respect to \(\theta\).

Step 2: Compute the first and second derivatives.
\(x' = -3a\cos^2\theta\sin\theta\), \(y' = 3a\sin^2\theta\cos\theta\)
\(x'' = 3a\cos\theta(2\sin^2\theta-\cos^2\theta)\), \(y'' = 3a\sin\theta(2\cos^2\theta-\sin^2\theta)\)

Step 3: Evaluate all four at \(\theta=\pi/4\).
Here \(\sin\theta=\cos\theta=\frac{1}{\sqrt2}\), so \(x'=-\frac{3a}{2\sqrt2}\), \(y'=\frac{3a}{2\sqrt2}\), and \(x''=y''=\frac{3a}{2\sqrt2}\) (since \(2\sin^2\theta-\cos^2\theta=2\cos^2\theta-\sin^2\theta=\frac12\) here).

Step 4: Substitute into the formula.
\[ x'y''-y'x'' = -2\left(\frac{3a}{2\sqrt2}\right)^2 = -\frac{9a^2}{4}, \qquad (x')^3=-\frac{27a^3}{16\sqrt2} \]
\[ \frac{d^2y}{dx^2}=\frac{-9a^2/4}{-27a^3/(16\sqrt2)}=\frac{4\sqrt2}{3a} \]

Step 5: Final conclusion.
\[ \boxed{\frac{4\sqrt2}{3a}} \]