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

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.