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JEE Mains · Maths · STD 12 - 6. Application of derivatives

वक्र \(x=2 \cos t+2 t \sin t, y=2 \sin t-2 t \cos t\) पर \(t=\frac{\pi}{4}\) पर खींचे गए अभिलंब की मूल बिंदु से दूरी है

  1. A \(2\)
  2. B \(4\)
  3. C \(\sqrt 2\)
  4. D \(2\sqrt 2\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2\)

Step-by-step Solution

Detailed explanation

Given that \(x = 2\cos t + 2t\sin t\) so, \(\frac{{dx}}{{dt}} = - 2\sin t + 2\left[ {t\cos t + \sin t} \right]\) \(\frac{{dy}}{{dt}} = 2\cos t - 2\left[ { - t\sin t + \cos t} \right]\) \(\frac{{dy}}{{dt}} = 2t\sin t\,\,\,\,\,\,\,\,....\left( {ii} \right)\)…
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