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Ex 5.6, 3 - Find dy/dx, x = sin t, y = cos 2t - Class 12

Ex 5.6, 3 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Ex 5.6, 3 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦/𝑑π‘₯, π‘₯=sin⁑𝑑, 𝑦=cos⁑2𝑑Here, 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) Calculating π’…π’š/𝒅𝒕 𝑑𝑦/𝑑𝑑 " " = 𝑑(cos⁑2𝑑)/𝑑𝑑 𝑑𝑦/𝑑𝑑 = βˆ’sin⁑2𝑑 . 2 𝑑𝑦/𝑑𝑑 = βˆ’2 sin⁑2𝑑 Calculating 𝒅𝒙/𝒅𝒕 𝑑π‘₯/𝑑𝑑 " " = 𝑑(sin⁑𝑑 )/𝑑𝑑 𝑑π‘₯/𝑑𝑑 " " = cos⁑𝑑 Therefore 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) 𝑑𝑦/𝑑π‘₯ = (βˆ’2 sin⁑2𝑑" " )/cos⁑𝑑 𝑑𝑦/𝑑π‘₯ = (βˆ’2(2 sin⁑𝑑 .γ€– cos〗⁑𝑑 ))/cos⁑𝑑 π’…π’š/𝒅𝒙 = βˆ’πŸ’ π’”π’Šπ’β‘π’• (As sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.