If $x=a(1-\cos \theta), \quad y=a(\theta-\sin \theta)$, then $\frac{d^{2} y}{d x^{2}}=$

If $x=a(1-\cos \theta), \quad y=a(\theta-\sin \theta)$, then $\frac{d^{2} y}{d x^{2}}=$
  1. $\frac{\cos ^{2}\left(\frac{\theta}{2}\right)}{2 \operatorname{acosec} \theta}$
  2. $\frac{\operatorname{cosec} \theta}{2 a \cos ^{2}\left(\frac{\theta}{2}\right)}$
  3. $\frac{\cos \left(\frac{\theta}{2}\right)}{2 \operatorname{asin} \theta}$
  4. $\frac{\sin \left(\frac{\theta}{2}\right)}{2 \mathrm{a} \cos \theta}$

Solution

$\frac{d x}{d \theta}=a \sin \theta, \frac{d y}{d \theta}=a(1-\cos \theta)$ $\frac{d y}{d x}=\frac{a(1-\cos \theta)}{a \sin \theta}=\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\tan \frac{\theta}{2}$ $\frac{d^{2} y}{d x^{2}}=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \cdot \frac{d \theta}{d x}$ $=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \times \frac{1}{a \sin \theta}=\frac{\operatorname{cosec} \theta}{2a \cos ^{2} \frac{\theta}{2}}$

Asked in: MHT CET 2020 (15 Oct Shift 1)

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