MHT CET · Maths · Differentiation
If \(x=a(t+\sin t), y=a(1-\cos t)\), then \(\frac{d y}{d x}=\)
- A \(\tan \frac{t}{2}\)
- B \(-\frac{t}{2} \tan t\)
- C \(\frac{1}{2} \tan t\)
- D \(-\tan \frac{t}{2}\)
Answer & Solution
Correct Answer
(A) \(\tan \frac{t}{2}\)
Step-by-step Solution
Detailed explanation
\(x=a(t+\sin t) \text { and } y=a(1-\cos t) \)
\( \therefore \frac{d x}{d t}=a(1+\cos t) \text { and } \frac{d y}{d t}=a(\sin t) \)
\( \therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a \sin t}{a(1+\cos t)}=\frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \cos ^2 \frac{t}{2}}=\tan \frac{t}{2}\)
\( \therefore \frac{d x}{d t}=a(1+\cos t) \text { and } \frac{d y}{d t}=a(\sin t) \)
\( \therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a \sin t}{a(1+\cos t)}=\frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \cos ^2 \frac{t}{2}}=\tan \frac{t}{2}\)
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