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Ex 5.6, 7 - Find dy/dx, x = sin3 t/ root cos 2t - CBSE - Ex 5.6

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
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Ex 5.6, 7 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦﷮𝑑𝑥﷯, 𝑥 = 𝑠𝑖𝑛﷮3﷯𝑡﷮ ﷮ cos﷮2𝑡﷯﷯﷯ , 𝑦 = 𝑐𝑜𝑠﷮3﷯𝑡﷮ ﷮ cos﷮2𝑡﷯﷯﷯ 𝑥 = 𝑠𝑖𝑛﷮3﷯𝑡﷮ ﷮ cos﷮2𝑡﷯﷯﷯ & 𝑦 = 𝑐𝑜𝑠﷮3﷯𝑡﷮ ﷮ cos﷮2𝑡﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑥﷯ × 𝑑𝑡﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑡﷯ × 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑡﷯﷮ 𝑑𝑥﷮𝑑𝑡﷯﷯ Calculating 𝒅𝒚﷮𝒅𝒕﷯ 𝑦 = 𝑐𝑜𝑠﷮3﷯𝑡﷮ ﷮ cos﷮2𝑡﷯﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑑﷮𝑑𝑡﷯ 𝑐𝑜𝑠﷮3﷯𝑡﷮ ﷮ cos﷮2𝑡﷯﷯﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑑 𝑐𝑜𝑠﷮3﷯𝑡﷯﷮𝑑𝑡﷯ . ﷮ cos﷮2﷯𝑡﷯ − 𝑑 ﷮ cos﷮2𝑡﷯﷯﷯﷮𝑑𝑡﷯ . cos﷮3﷯﷮𝑡﷯﷮ ﷮ cos﷮2﷯𝑡﷯﷯﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 3 cos﷮2﷯﷮𝑡 ﷯. 𝑑 cos﷮𝑡﷯﷯﷮𝑑𝑡﷯. ﷮ cos﷮2﷯𝑡﷯ − 1﷮2 ﷮ cos﷮2𝑡﷯﷯﷯ . 𝑑 cos﷮2𝑡﷯﷯﷮𝑑𝑡﷯ . cos﷮3﷯﷮𝑡﷯﷮ ﷮ cos﷮2﷯𝑡﷯﷯﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 3 cos﷮2﷯﷮𝑡 ﷯. − sin﷮𝑡﷯﷯ . ﷮ cos﷮2﷯𝑡﷯ − 1﷮2 ﷮ cos﷮2𝑡﷯﷯﷯ . (−2 sin﷮2𝑡﷯) . cos﷮3﷯﷮𝑡﷯﷮ ﷮ cos﷮2﷯𝑡﷯﷯﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = −3 cos﷮2﷯﷮𝑡 ﷯ sin﷮𝑡﷯ ﷮ cos﷮2﷯𝑡﷯ + 1﷮ ﷮ cos﷮2𝑡﷯﷯﷯ . sin﷮2𝑡﷯ . cos﷮3﷯﷮𝑡﷯﷮ ﷮ cos﷮2﷯ 𝑡﷯﷯﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = −3 cos﷮2﷯﷮𝑡 ﷯ sin﷮𝑡﷯ ﷮ cos﷮2﷯𝑡﷯ × ﷮ cos﷮2𝑡﷯﷯ + sin﷮2𝑡﷯ . cos﷮3﷯﷮𝑡﷯﷮ ﷮ cos﷮2𝑡﷯﷯﷯﷮ ﷮ cos﷮2﷯ 𝑡﷯﷯﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = −3 cos﷮2﷯﷮𝑡 ﷯ sin﷮𝑡﷯ cos﷮2﷯𝑡﷯ + sin﷮2𝑡﷯ . cos﷮𝑡﷯﷮ ﷮ cos﷮2﷯ 𝑡﷯﷯﷮2﷯ ﷮ cos﷮2﷯ 𝑡﷯﷯﷯ 𝑑𝑦﷮𝑑𝑡﷯ = cos﷮2﷯﷮𝑡﷯ −3 sin﷮𝑡﷯ . cos﷮2𝑡﷯ + cos﷮𝑡﷯ . sin﷮2𝑡﷯﷯﷮ cos﷮2﷯𝑡﷯﷮ 3﷮2﷯﷯ ﷯ Calculating 𝒅𝒙﷮𝒅𝒕﷯ 𝑥 = 𝑠𝑖𝑛﷮3﷯𝑡﷮ ﷮ cos﷮2𝑡﷯﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑑﷮𝑑𝑥﷯ 𝑠𝑖𝑛﷮3﷯𝑡﷮ ﷮ cos﷮2𝑡﷯﷯﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑑 𝑠𝑖𝑛﷮3﷯𝑡﷯﷮𝑑𝑡﷯ . ﷮ cos﷮2𝑡﷯﷯ − 𝑑 ﷮ cos﷮2𝑡﷯﷯﷯ ﷮𝑑𝑥﷯ . 𝑠𝑖𝑛﷮3﷯𝑡 ﷮ ﷮ cos﷮2𝑡﷯﷯﷯﷮2﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 3 𝑠𝑖𝑛﷮2﷯𝑡 . 𝑑 sin﷮𝑡﷯﷯ ﷮𝑑𝑡﷯ . ﷮ cos﷮2𝑡﷯﷯ − 1﷮2 ﷮ cos﷮2𝑡﷯﷯﷯ . 𝑑 cos﷮2𝑡﷯﷯ ﷮𝑑𝑥﷯ . 𝑠𝑖𝑛﷮3﷯𝑡 ﷮ ﷮ cos﷮2𝑡﷯﷯﷯﷮2﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 3 𝑠𝑖𝑛﷮2﷯𝑡 . cos﷮𝑡﷯ . ﷮ cos﷮2 𝑡﷯﷯ − 1﷮2 ﷮ cos﷮2 𝑡﷯﷯﷯ . − sin﷮2𝑡﷯﷯ . 2 . 𝑠𝑖𝑛﷮3﷯𝑡 ﷮ cos﷮2 𝑡﷯﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 3 𝑠𝑖𝑛﷮2﷯𝑡 . cos﷮𝑡﷯ . ﷮ cos﷮2𝑡﷯﷯﷯ . ﷮ cos﷮2𝑡﷯﷯﷯ + sin﷮2𝑡﷯ . 𝑠𝑖𝑛﷮3﷯𝑡 ﷮ ﷮ cos﷮2𝑡﷯﷯﷯ cos﷮2𝑡﷯﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 3 𝑠𝑖𝑛﷮2﷯𝑡 . cos﷮𝑡﷯ . cos﷮2𝑡﷯ + sin﷮2𝑡﷯ . 𝑠𝑖𝑛﷮3﷯𝑡 ﷮ cos﷮2𝑡﷯﷯﷮ 3﷮2﷯﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑠𝑖𝑛﷮2﷯𝑡 3 cos﷮𝑡﷯ . cos﷮2𝑡﷯ + sin﷮2𝑡﷯ . sin﷮𝑡﷯﷯ ﷮ cos﷮2𝑡﷯﷯﷮ 3﷮2﷯﷯﷯ Therefore 𝑑𝑦﷮𝑑𝑥﷯ = cos﷮2﷯﷮𝑡﷯ −3 sin﷮𝑡﷯ . cos﷮2𝑡﷯ + cos﷮𝑡﷯ . sin﷮2𝑡﷯﷯﷮ cos﷮2﷯𝑡﷯﷮ 3﷮2﷯﷯ ﷯﷮ 𝑠𝑖𝑛﷮2﷯𝑡 3 cos﷮𝑡﷯ . cos﷮2𝑡﷯ + sin﷮2𝑡﷯ . sin﷮𝑡﷯﷯ ﷮ cos﷮2𝑡﷯﷯﷮ 3﷮2﷯﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = cos﷮2﷯﷮𝑡﷯ −3 sin﷮𝑡﷯ . cos﷮2𝑡﷯ + cos﷮𝑡﷯ . sin﷮2𝑡﷯﷯﷮ 𝑠𝑖𝑛﷮2﷯𝑡 3 cos﷮𝑡﷯ . cos﷮2𝑡﷯ + sin﷮2𝑡﷯ . sin﷮𝑡﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = cot﷮2﷯𝑡 cos﷮2﷯﷮𝑡﷯ −3 sin﷮𝑡﷯ . cos﷮2𝑡﷯ + cos﷮𝑡﷯ . sin﷮2𝑡﷯﷯﷮ 𝑠𝑖𝑛﷮2﷯𝑡 3 cos﷮𝑡﷯ . cos﷮2𝑡﷯ + sin﷮2𝑡﷯ . sin﷮𝑡﷯﷯﷯﷯ Taking common cos⁡2𝑡 𝑑𝑦﷮𝑑𝑥﷯ = cot﷮2﷯𝑡 cos﷮2𝑡﷯ sin﷮2𝑡﷯﷮ cos﷮2𝑡﷯﷯ . 3 tan﷮1﷯ ﷯﷮ cos﷮2𝑡﷯ 3 + tan﷮𝑡﷯ . sin﷮2𝑡﷯﷮ cos﷮2𝑡﷯﷯ ﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = cot﷮2﷯𝑡 tan﷮2𝑡﷯﷮3 + tan﷮𝑡﷯ . tan﷮2𝑡﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = cot﷮2﷯𝑡 2 tan﷮𝑡﷯﷮1 − tan﷮2﷯﷮𝑡﷯﷯ − 3 tan﷮𝑡﷯﷮3 + tan﷮𝑡﷯﷯ 2 tan﷮𝑡﷯﷮1 − tan﷮2﷯﷮𝑡﷯﷯﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = cot﷮2﷯𝑡 2 tan﷮𝑡﷯ −3 tan﷮𝑡﷯ 1 − tan﷮2﷯﷮𝑡﷯﷯﷮3 + 1− tan﷮𝑡﷯﷯ + tan﷮𝑡﷯ 2 tan﷮𝑡﷯﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = cot﷮2﷯𝑡 2 tan﷮𝑡﷯ −3 tan﷮𝑡﷯ + 3 tan﷮3﷯﷮𝑡﷯﷮3 − 3 tan﷮2﷯﷮𝑡﷯ + 2 tan﷮2﷯﷮𝑡﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = cot﷮2﷯𝑡 tan﷮𝑡﷯ −3 tan﷮3﷯﷮𝑡﷯﷯ ﷮ 3 − tan﷮2﷯﷮𝑡﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = cot﷮2﷯﷮𝑡﷯ × tan﷮𝑡﷯ − 3 cot﷮2﷯﷮𝑡﷯ tan﷮3﷯﷮𝑡﷯﷮3 − tan﷮2﷯﷮𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮ tan﷮𝑡﷯﷯ . 3 tan﷮𝑡﷯ ﷮3 − tan﷮2﷯﷮𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1 − 3 tan﷮2﷯﷮𝑡﷯ ﷮ tan﷮𝑡﷯ 3 − tan﷮2﷯﷮𝑡﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1 − 3 tan﷮2﷯﷮𝑡﷯ ﷮3 tan﷮𝑡﷯ − tan﷮3﷯﷮𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮ 3 tan﷮𝑡﷯ − tan﷮3﷯﷮𝑡﷯﷮1 −3 tan﷮2﷯﷮𝑡﷯﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮ tan﷮3𝑡﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝒄𝒐𝒕﷮𝟑𝒕﷯

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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