Question

If \[ x=\sin2\theta+\sin3\theta,\qquad y=\cos2\theta-\cos3\theta \] then \[ \frac{d^2y}{dx^2} = \]

• A. \(\frac{35+6\cos\theta}{(2\cos2\theta+3\cos3\theta)^3}\)
• B. \(\frac{19+6\cos\theta}{(2\cos2\theta+3\cos3\theta)^3}\)
• C. \(\frac{35+6\cos5\theta}{(2\cos2\theta+3\cos3\theta)^3}\)
• D. \(\frac{19+6\cos5\theta}{(2\cos2\theta+3\cos3\theta)^3}\)

Hint:

For parametric differentiation, never forget the second derivative formula divides once again by \(dx/dt\).

Answer 1

The Correct Option is B

Concept: For parametric equations: \[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \] and \[ \frac{d^2y}{dx^2} = \frac{\frac{d}{d\theta}\left(\frac{dy}{dx}\right)} {dx/d\theta} \]

Step 1: Differentiate both equations.
\[ \frac{dx}{d\theta} = 2\cos2\theta+3\cos3\theta \] \[ \frac{dy}{d\theta} = -2\sin2\theta+3\sin3\theta \] Thus \[ \frac{dy}{dx} = \frac{-2\sin2\theta+3\sin3\theta} {2\cos2\theta+3\cos3\theta} \]

Step 2: Differentiate again.
Applying quotient rule carefully, \[ \frac{d}{d\theta}\left(\frac{dy}{dx}\right) = \frac{19+6\cos\theta} {(2\cos2\theta+3\cos3\theta)^2} \]

Step 3: Apply second derivative formula.
\[ \frac{d^2y}{dx^2} = \frac{19+6\cos\theta} {(2\cos2\theta+3\cos3\theta)^3} \] Hence \[ \boxed{ \frac{19+6\cos\theta} {(2\cos2\theta+3\cos3\theta)^3} } \]