Misc 17 - If x = a (cos t + t sin t), y = a (sin t - t cos t) - Derivatives in parametric form

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
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Misc 17 If 𝑥=𝑎 (cos⁡𝑡 + 𝑡 sin⁡𝑡) and y=𝑎 (sin⁡𝑡 – 𝑡 cos⁡𝑡), Find 𝑑﷮2﷯𝑦﷮ 𝑑𝑥﷮2﷯﷯ If 𝑥=𝑎 (cos⁡𝑡 + 𝑡 sin⁡𝑡) & 𝑦=𝑎 (sin⁡𝑡 – 𝑡 cos⁡𝑡) We need to find 𝑑﷮2﷯𝑦﷮ 𝑑𝑥﷮2﷯﷯ First we find 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑥﷯ . 𝑑𝑡﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑡﷯ . 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑡﷯﷮ 𝑑𝑥﷮𝑑𝑡﷯﷯ Calculating 𝒅𝒚/𝒅𝒕 𝑦=𝑎 sin﷮𝑡﷯– 𝑡 cos﷮𝑡﷯﷯ Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦﷮𝑑𝑡﷯ = 𝑑 𝑎 sin﷮𝑡﷯– 𝑡 cos﷮𝑡﷯﷯﷯﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎 𝑑 sin﷮𝑡﷯– 𝑡 cos﷮𝑡﷯﷯﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎 𝑑 sin﷮𝑡﷯﷯﷮𝑑𝑡﷯ − 𝑑 𝑡 cos﷮𝑡﷯﷯﷮𝑑𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎 cos﷮𝑡﷯ − 𝑑 𝑡 cos﷮𝑡﷯﷯﷮𝑑𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎 cos﷮𝑡﷯ − 𝑑𝑡﷮𝑑𝑡﷯ . cos﷮𝑡﷯+ 𝑑 cos﷮𝑡﷯﷮𝑑𝑡﷯ . 𝑡 ﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎 cos﷮𝑡﷯ − cos﷮𝑡﷯+ −sin﷮𝑡﷯﷯ . 𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎 cos﷮𝑡﷯ − cos﷮𝑡﷯− sin﷮𝑡﷯﷯ . 𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎 cos﷮𝑡﷯ − cos﷮𝑡﷯+𝑡 . sin﷮𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎 0+𝑡 sin﷮𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎 .𝑡. sin﷮𝑡﷯ Hence 𝑑𝑦/𝑑𝑡 = 𝑎 .𝑡. sin﷮𝑡﷯ Calculating 𝒅𝒙/𝒅𝒕 𝑥=𝑎 (cos⁡𝑡 + 𝑡 sin⁡𝑡) Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑥﷮𝑑𝑡﷯ = 𝑑 𝑎 (cos⁡𝑡 + 𝑡 sin⁡𝑡) ﷯﷮𝑑𝑡﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑎 𝑑 cos⁡𝑡 + 𝑡 sin⁡𝑡﷯﷮𝑑𝑡﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑎 𝑑 cos⁡𝑡﷯﷮𝑑𝑡﷯ + 𝑑 𝑡 sin⁡𝑡﷯﷮𝑑𝑡﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑎 −sin﷮𝑡﷯ + 𝑑 𝑡 sin﷮𝑡﷯﷯﷮𝑑𝑡﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑎 −sin﷮𝑡﷯+ 𝑑𝑡﷮𝑑𝑡﷯ . sin﷮𝑡﷯+ 𝑑 sin﷮𝑡﷯﷯﷮𝑑𝑡﷯ . 𝑡 ﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑎 −sin﷮𝑡﷯+ sin﷮𝑡﷯+ cos﷮𝑡﷯ . 𝑡﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯= 𝑎 − sin﷮𝑡﷯+ sin﷮𝑡﷯+𝑡 . c𝑜𝑠﷮𝑡﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑎 .𝑡. cos﷮𝑡﷯ Now , 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦/𝑑𝑡﷮𝑑𝑥/𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑎 .𝑡. sin﷮𝑡﷯﷮𝑎 .𝑡. cos﷮𝑡﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝒕𝒂𝒏⁡𝒕 Again Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯ = 𝑑 tan⁡𝑡﷯﷮𝑑𝑥﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = 𝑑 tan⁡𝑡﷯﷮𝑑𝑥﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = 𝑑 tan⁡𝑡﷯﷮𝑑𝑥﷯ . 𝑑𝑡﷮𝑑𝑡﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = sec﷮2﷯﷮𝑡﷯ . 𝑑𝑡﷮𝑑𝑥﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = sec﷮2﷯﷮𝑡﷯ ÷ 𝒅𝒙﷮𝒅𝒕﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = sec﷮2﷯﷮𝑡﷯ ÷ 𝒂.𝒕. 𝐜𝐨𝐬﷮𝒕﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = sec﷮2﷯﷮𝑡﷯ ﷮𝑎 . 𝑡. cos﷮𝑡﷯﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = sec﷮2﷯﷮𝑡﷯ ﷮𝑎 . 𝑡 × 1﷮ sec﷮𝑡﷯﷯﷯ 𝑑﷮2﷯𝑦﷮𝑑 𝑥﷮2﷯﷯ = sec﷮3﷯﷮𝑡﷯ ﷮𝑎 . 𝑡﷯ Hence 𝒅﷮𝟐﷯𝒚﷮𝒅 𝒙﷮𝟐﷯﷯ = 𝒔𝒆𝒄﷮𝟑﷯﷮𝒕﷯ ﷮𝒂 . 𝒕﷯

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