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Question: If $x=2\sin\theta - \sin2\theta$ and $y=2\cos\theta - \cos2\theta, \theta \in [0,2\pi]$, then $\frac...

If x=2sin⁡θ−sin⁡2θx=2\sin\theta - \sin2\theta and y=2cos⁡θ−cos⁡2θ,θ∈[0,2π]y=2\cos\theta - \cos2\theta, \theta \in [0,2\pi], then d2ydx2\frac{d^2y}{dx^2} at θ=π\theta = \pi is:

Answer

38\frac{3}{8}

Explanation

Solution

  1. Calculate dxdθ=2cos⁡θ−2cos⁡2θ\frac{dx}{d\theta} = 2\cos\theta - 2\cos2\theta and dydθ=2sin⁡2θ−2sin⁡θ\frac{dy}{d\theta} = 2\sin2\theta - 2\sin\theta.

  2. Evaluate at θ=π\theta = \pi: dxdθ∣θ=π=2(−1)−2(1)=−4\frac{dx}{d\theta}\bigg|_{\theta=\pi} = 2(-1) - 2(1) = -4 dydθ∣θ=π=2(0)−2(0)=0\frac{dy}{d\theta}\bigg|_{\theta=\pi} = 2(0) - 2(0) = 0

  3. Find dydx=dy/dθdx/dθ=2sin⁡2θ−2sin⁡θ2cos⁡θ−2cos⁡2θ=sin⁡2θ−sin⁡θcos⁡θ−cos⁡2θ\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2\sin2\theta - 2\sin\theta}{2\cos\theta - 2\cos2\theta} = \frac{\sin2\theta - \sin\theta}{\cos\theta - \cos2\theta}. Using sum-to-product identities, this simplifies to cot⁡(3θ2)\cot\left(\frac{3\theta}{2}\right).

  4. Differentiate dydx\frac{dy}{dx} with respect to θ\theta: ddθ(dydx)=ddθ(cot⁡(3θ2))=−csc⁡2(3θ2)⋅32\frac{d}{d\theta}\left(\frac{dy}{dx}\right) = \frac{d}{d\theta}\left(\cot\left(\frac{3\theta}{2}\right)\right) = -\csc^2\left(\frac{3\theta}{2}\right) \cdot \frac{3}{2}.

  5. Evaluate this derivative at θ=π\theta = \pi: ddθ(dydx)∣θ=π=−csc⁡2(3π2)⋅32=−(−1)2⋅32=−32\frac{d}{d\theta}\left(\frac{dy}{dx}\right)\bigg|_{\theta=\pi} = -\csc^2\left(\frac{3\pi}{2}\right) \cdot \frac{3}{2} = -(-1)^2 \cdot \frac{3}{2} = -\frac{3}{2}.

  6. Calculate d2ydx2=ddθ(dydx)dxdθ∣θ=π=−3/2−4=38\frac{d^2y}{dx^2} = \frac{\frac{d}{d\theta}\left(\frac{dy}{dx}\right)}{\frac{dx}{d\theta}}\bigg|_{\theta=\pi} = \frac{-3/2}{-4} = \frac{3}{8}.