Ex 5.6, 9 - Find dy/dx, x = a sec, y = b tan - Chapter 5 CBSE

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Ex 5.6, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Ex 5.6, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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

Transcript

Ex 5.6, 9 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦/𝑑π‘₯, π‘₯=π‘Ž sec β‘πœƒ, 𝑦=𝑏 tanβ‘πœƒHere 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) Calculating π’…π’š/π’…πœ½ 𝑑𝑦/π‘‘πœƒ = 𝑑(𝑏 tanβ‘πœƒ )/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ = 𝑏 𝑑(tanβ‘πœƒ )/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ = 𝒃 .γ€–π¬πžπœγ€—^𝟐⁑𝜽 Calculating 𝒅𝒙/π’…πœ½ 𝑑π‘₯/π‘‘πœƒ = 𝑑(π‘Ž sec β‘πœƒ)/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = π‘Ž 𝑑(sec β‘πœƒ)/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = 𝒂 (π’”π’†π’„β‘πœ½.π’•π’‚π’β‘πœ½ ) Therefore 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) 𝑑𝑦/𝑑π‘₯ = (𝑏 .γ€– secγ€—^2β‘πœƒ)/(π‘Ž (secβ‘πœƒ.tanβ‘πœƒ ) ) 𝑑𝑦/𝑑π‘₯ = (𝑏 secβ‘πœƒ)/(π‘Ž tanβ‘πœƒ ) 𝑑𝑦/𝑑π‘₯ = (𝑏 . 1/cosβ‘πœƒ )/(π‘Ž (sinβ‘πœƒ/cosβ‘πœƒ ) ) 𝑑𝑦/𝑑π‘₯ = 𝑏 Γ— 1/cosβ‘πœƒ Γ— cosβ‘πœƒ/γ€–a sinγ€—β‘πœƒ 𝑑𝑦/𝑑π‘₯ = 𝑏/π‘Ž (1/sinβ‘πœƒ ) π’…π’š/𝒅𝒙 = 𝒃/𝒂 𝒄𝒐𝒔𝒆𝒄 𝜽

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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.