MHT CET · Maths · Differentiation
If \(x=a(1-\cos \theta), \quad y=a(\theta-\sin \theta)\), then \(\frac{d^{2} y}{d x^{2}}=\)
- A \(\frac{\cos ^{2}\left(\frac{\theta}{2}\right)}{2 \operatorname{acosec} \theta}\)
- B \(\frac{\operatorname{cosec} \theta}{2 a \cos ^{2}\left(\frac{\theta}{2}\right)}\)
- C \(\frac{\cos \left(\frac{\theta}{2}\right)}{2 \operatorname{asin} \theta}\)
- D \(\frac{\sin \left(\frac{\theta}{2}\right)}{2 \mathrm{a} \cos \theta}\)
Answer & Solution
Correct Answer
(B) \(\frac{\operatorname{cosec} \theta}{2 a \cos ^{2}\left(\frac{\theta}{2}\right)}\)
Step-by-step Solution
Detailed explanation
\(\frac{d x}{d \theta}=a \sin \theta, \frac{d y}{d \theta}=a(1-\cos \theta)\)
\(\frac{d y}{d x}=\frac{a(1-\cos \theta)}{a \sin \theta}=\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\tan \frac{\theta}{2}\)
\(\frac{d^{2} y}{d x^{2}}=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \cdot \frac{d \theta}{d x}\)
\(=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \times \frac{1}{a \sin \theta}=\frac{\operatorname{cosec} \theta}{2 \cos ^{2} \frac{\theta}{2}}\)
\(\frac{d y}{d x}=\frac{a(1-\cos \theta)}{a \sin \theta}=\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\tan \frac{\theta}{2}\)
\(\frac{d^{2} y}{d x^{2}}=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \cdot \frac{d \theta}{d x}\)
\(=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \times \frac{1}{a \sin \theta}=\frac{\operatorname{cosec} \theta}{2 \cos ^{2} \frac{\theta}{2}}\)
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