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If x=2sin⁡θx = 2\sin\theta and y=2cos⁡θy = 2\cos\theta, then the value of d2ydx2\frac{d^2y}{dx^2} at θ=0\theta = 0 is

Solution

✅ Correct Option: 1

x=2sin⁡θ, y=2cos⁡θx = 2\sin\theta,\ y = 2\cos\theta.

dydx=−2sin⁡θ2cos⁡θ=−tan⁡θ\frac{dy}{dx} = \frac{-2\sin\theta}{2\cos\theta} = -\tan\theta.

d2ydx2=−sec⁡2θ⋅dθdx=−sec⁡2θ⋅12cos⁡θ=−sec⁡3θ2\frac{d^2y}{dx^2} = -\sec^2\theta \cdot \frac{d\theta}{dx} = -\sec^2\theta \cdot \frac{1}{2\cos\theta} = -\frac{\sec^3\theta}{2}.

At θ=0\theta=0: −12-\frac{1}{2}.

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