# Ex 5.6, 8 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 5.6, 8 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦𝑑𝑥, 𝑥 = 𝑎 cos𝑡+ log tan 𝑡2 , 𝑦=𝑎 sin𝑡 𝑥 = 𝑎 cos𝑡+ log tan 𝑡2 , 𝑦=𝑎 sin𝑡 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑𝑥 × 𝑑𝑡𝑑𝑡 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑𝑡 × 𝑑𝑡𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑𝑡 𝑑𝑥𝑑𝑡 Calculating 𝒅𝒚𝒅𝒕 𝑦 = 𝑎 sin𝑡 𝑑𝑦𝑑𝑡 = 𝑑 𝑎 sin𝑡𝑑𝑡 𝑑𝑦𝑑𝑡 = 𝑎 𝑑 sin𝑡𝑑𝑡 𝑑𝑦𝑑𝑡 = 𝑎 cos𝑡 Calculating 𝒅𝒙𝒅𝒕 𝑥 = 𝑎 cos𝑡+ log tan 𝑡2 𝑑𝑥𝑑𝑡 = 𝑑 𝑎 cos𝑡+ log tan 𝑡2𝑑𝑡 𝑑𝑥𝑑𝑡 = a 𝑑 cos𝑡 𝑑𝑡 + 𝑑 log tan 𝑡2 𝑑𝑡 𝑑𝑥𝑑𝑡 = a − sin𝑡 + 𝑑 log tan 𝑡2 𝑑𝑡 𝑑𝑥𝑑𝑡 = a − sin𝑡 + 1 tan 𝑡2 . 𝑑 tan 𝑡2𝑑𝑡 𝑑𝑥𝑑𝑡 = a − sin𝑡 + 1 tan 𝑡2 . sec2 𝑡2. 𝑑 𝑡2𝑑𝑡 𝑑𝑥𝑑𝑡 = a − sin𝑡 + 1 tan 𝑡2 . sec2 𝑡2. 12 𝑑𝑥𝑑𝑡 = a − sin𝑡 + cot 𝑡2 . 1 cos2 𝑡2 . 12 𝑑𝑥𝑑𝑡 = a − sin𝑡 + cos 𝑡2 sin 𝑡2 . 1 cos2 𝑡2 . 12 𝑑𝑥𝑑𝑡 = a − sin𝑡 + 1 sin 𝑡2 cos 𝑡2 . 12 𝑑𝑥𝑑𝑡 = a − sin𝑡 + 12 sin 𝑡2 cos 𝑡2 𝑑𝑥𝑑𝑡 = a − sin𝑡 + 1 sin𝑡 𝑑𝑥𝑑𝑡 = a − sin2𝑡 + 1 sin𝑡 𝑑𝑥𝑑𝑡 = a 1 − sin2𝑡 sin𝑡 𝑑𝑥𝑑𝑡 = a cos2𝑡 sin𝑡 Therefore, 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑𝑡 𝑑𝑥𝑑𝑡 𝑑𝑦𝑑𝑥 = 𝑎 cos𝑡 𝑎 cos2𝑡 sin𝑡 𝑑𝑦𝑑𝑥 = 𝑎 cos𝑡 × sin𝑡𝑎 cos2𝑡 𝑑𝑦𝑑𝑥 = 𝑎 cos𝑡 . sin𝑡𝑎 cos𝑡 . cos𝑡 𝑑𝑦𝑑𝑥 = sin𝑡 cos𝑡 𝒅𝒚𝒅𝒙 = 𝐭𝐚𝐧𝒕

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.