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

Ex 5.6, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.6, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.