Ex 5.6, 8 - Find dy/dx, x = a (cos t + log tan t/2) - Ex 5.6

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Ex 5.6, 8 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦﷮𝑑𝑥﷯, 𝑥 = 𝑎 cos﷮𝑡+ log﷮ tan﷮ 𝑡﷮2﷯﷯﷯﷯﷯ , 𝑦=𝑎 sin⁡𝑡 𝑥 = 𝑎 cos﷮𝑡+ log﷮ tan﷮ 𝑡﷮2﷯﷯﷯﷯﷯ , 𝑦=𝑎 sin⁡𝑡 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 𝑑𝑦﷮𝑑𝑡﷯ = 𝑑 𝑎 sin⁡𝑡﷯﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎 𝑑 sin⁡𝑡﷯﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑡﷯ = 𝑎 cos﷮𝑡﷯ Calculating 𝒅𝒙﷮𝒅𝒕﷯ 𝑥 = 𝑎 cos﷮𝑡+ log﷮ tan﷮ 𝑡﷮2﷯﷯﷯﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = 𝑑 𝑎 cos﷮𝑡+ log﷮ tan ﷮ 𝑡﷮2﷯﷯﷯﷯﷯﷯﷮𝑑𝑡﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a 𝑑 cos﷮𝑡﷯﷯ ﷮𝑑𝑡﷯ + 𝑑 log﷮ tan ﷮ 𝑡﷮2﷯﷯ ﷯﷯﷯﷮𝑑𝑡﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a − sin﷮𝑡﷯ + 𝑑 log﷮ tan﷮ 𝑡﷮2﷯﷯ ﷯﷯﷯﷮𝑑𝑡﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a − sin﷮𝑡﷯ + 1﷮ tan﷮ 𝑡﷮2﷯﷯﷯ . 𝑑 tan ﷮ 𝑡﷮2﷯﷯﷯﷮𝑑𝑡﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a − sin﷮𝑡﷯ + 1﷮ tan﷮ 𝑡﷮2﷯﷯﷯ . sec﷮2﷯﷮ 𝑡﷮2﷯﷯﷯. 𝑑 𝑡﷮2﷯﷯﷮𝑑𝑡﷯ ﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a − sin﷮𝑡﷯ + 1﷮ tan﷮ 𝑡﷮2﷯﷯﷯ . sec﷮2﷯ 𝑡﷮2﷯﷯. 1﷮2﷯ ﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a − sin﷮𝑡﷯ + cot﷮ 𝑡﷮2﷯﷯ . 1﷮ cos﷮2﷯ 𝑡﷮2﷯﷯ . 1﷮2﷯ ﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a − sin﷮𝑡﷯ + cos ﷮ 𝑡﷮2﷯﷯﷮ sin﷮ 𝑡﷮2﷯﷯﷯ . 1﷮ cos﷮2﷯ 𝑡﷮2﷯﷯ . 1﷮2﷯ ﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a − sin﷮𝑡﷯ + 1﷮ sin﷮ 𝑡﷮2﷯﷯ cos﷮ 𝑡﷮2﷯﷯﷯ . 1﷮2﷯ ﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a − sin﷮𝑡﷯ + 1﷮2 sin﷮ 𝑡﷮2﷯﷯ cos﷮ 𝑡﷮2﷯﷯﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a − sin﷮𝑡﷯ + 1﷮ sin﷮𝑡﷯﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a − sin﷮2﷯﷮𝑡﷯ + 1﷮ sin﷮𝑡﷯﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a 1 − sin﷮2﷯﷮𝑡﷯﷮ sin﷮𝑡﷯﷯﷯ 𝑑𝑥﷮𝑑𝑡﷯ = a cos﷮2﷯﷮𝑡﷯﷮ sin﷮𝑡﷯﷯﷯ Therefore, 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑦﷮𝑑𝑡﷯﷮ 𝑑𝑥﷮𝑑𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑎 cos﷮𝑡﷯﷮ 𝑎 cos﷮2﷯﷮𝑡﷯﷮ sin﷮𝑡﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑎 cos﷮𝑡﷯ × sin﷮𝑡﷯﷮𝑎 cos﷮2﷯𝑡﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑎 cos﷮𝑡﷯ . sin﷮𝑡﷯﷮𝑎 cos﷮𝑡﷯ . cos﷮𝑡﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = sin﷮𝑡﷯﷮ cos﷮𝑡﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝐭𝐚𝐧﷮𝒕﷯

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