Ex 5.6, 6 - Find dy/dx, x = a (theta - sin), y = a (1 + cos)

Ex 5.6, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.6, 6 - 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, 6 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦/𝑑π‘₯, π‘₯ = π‘Ž (πœƒ – sinβ‘πœƒ), 𝑦 = π‘Ž (1 + cosβ‘πœƒ)Here 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) Calculating π’…π’š/π’…πœ½ 𝑦 = π‘Ž (1+cosβ‘πœƒ) 𝑑𝑦/π‘‘πœƒ = 𝑑(π‘Ž (1+cosβ‘πœƒ))/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ = π‘Ž (𝑑(1 + cosβ‘πœƒ )/π‘‘πœƒ) 𝑑𝑦/π‘‘πœƒ = π‘Ž (0βˆ’sinβ‘πœƒ ) 𝑑𝑦/π‘‘πœƒ = π‘Ž (βˆ’sinβ‘πœƒ ) π’…π’š/π’…πœ½ = βˆ’π’‚ π’”π’Šπ’β‘πœ½ Calculating 𝒅𝒙/π’…πœ½ π‘₯=π‘Ž (πœƒ –sinβ‘πœƒ ) 𝑑π‘₯/π‘‘πœƒ = 𝑑(π‘Ž πœƒ βˆ’π‘Ž sinβ‘πœƒ )/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = 𝑑(π‘Ž πœƒ)/π‘‘πœƒ βˆ’ 𝑑(π‘Ž sinβ‘πœƒ )/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = π‘Žβˆ’γ€–π‘Ž cosγ€—β‘πœƒ 𝒅𝒙/π’…πœ½ = 𝒂(γ€–πŸβˆ’π’„π’π’”γ€—β‘πœ½ ) Therefore, 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) 𝑑𝑦/𝑑π‘₯ = (βˆ’π‘Ž sinβ‘πœƒ)/π‘Ž(γ€–1 βˆ’ cosγ€—β‘πœƒ ) 𝑑𝑦/𝑑π‘₯ = (βˆ’sinβ‘πœƒ)/γ€–1 βˆ’ cosγ€—β‘πœƒ 𝑑𝑦/𝑑π‘₯ = (βˆ’2 γ€–sin γ€—β‘γ€–πœƒ/2γ€— γ€–cos γ€—β‘γ€–πœƒ/2γ€—)/(2 γ€–sin^2 γ€—β‘γ€–πœƒ/2γ€— ) 𝑑𝑦/𝑑π‘₯ = (βˆ’γ€–cos γ€—β‘γ€–πœƒ/2γ€—)/(sin πœƒ/2) π’…π’š/𝒅𝒙 = βˆ’π’„π’π’•β‘γ€–πœ½/πŸγ€— Rough We know that sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ Replacing ΞΈ by πœƒ/2 sin ΞΈ = 2 π’”π’Šπ’β‘γ€–πœ½/πŸγ€— π’„π’π’”β‘γ€–πœ½/πŸγ€— and cos 2ΞΈ = 2cos2 ΞΈ – 1 Replacing ΞΈ by πœƒ/2 cos ΞΈ = 2cos2 𝜽/𝟐 – 1

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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.