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If x=a(θ+sin⁡θ)x=a(\theta+\sin\theta), y=a(1−cos⁡θ)y=a(1-\cos\theta), then d2ydx2=\dfrac{d^2y}{dx^2}=

Solution

✅ Correct Option: 1

dydx=asin⁡θa(1+cos⁡θ)=tan⁡θ2\dfrac{dy}{dx}=\dfrac{a\sin\theta}{a(1+\cos\theta)}=\tan\dfrac{\theta}{2}. Then d2ydx2=12sec⁡2θ2a(1+cos⁡θ)=12sec⁡2θ22acos⁡2θ2=14asec⁡4θ2\dfrac{d^2y}{dx^2}=\dfrac{\frac12\sec^2\frac{\theta}{2}}{a(1+\cos\theta)}=\dfrac{\frac12\sec^2\frac{\theta}{2}}{2a\cos^2\frac{\theta}{2}}=\dfrac{1}{4a}\sec^4\dfrac{\theta}{2}.

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