Ex 5.6, 5 - Find dy/dx, x = cos - cos 2, y = sin - sin 2

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

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

Transcript

Ex 5.6, 5 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦/𝑑π‘₯, π‘₯ = cosβ‘πœƒ – cos⁑2πœƒ, 𝑦 = sinβ‘πœƒ – sin⁑2πœƒ Here 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) Calculating π’…π’š/π’…πœ½ 𝑑𝑦/π‘‘πœƒ = 𝑑(sinβ‘πœƒ – sin⁑2πœƒ)/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ = 𝑑(sin πœƒ)/π‘‘πœƒ βˆ’ 𝑑(sin⁑2πœƒ)/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ = cosβ‘πœƒβˆ’cos⁑2πœƒ. 2 𝑑𝑦/π‘‘πœƒ = π’„π’π’”β‘πœ½βˆ’πŸ π’„π’π’”β‘πŸπœ½ Calculating 𝒅𝒙/π’…πœ½ 𝑑π‘₯/π‘‘πœƒ = 𝑑(cosβ‘πœƒ – cos⁑2πœƒ)/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ " " = 𝑑(cosβ‘πœƒ)/π‘‘πœƒ βˆ’ 𝑑(cos⁑2πœƒ)/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ " " = βˆ’sinβ‘πœƒβˆ’(βˆ’sin⁑2πœƒ . 2) 𝑑π‘₯/π‘‘πœƒ = βˆ’π’”π’Šπ’β‘πœ½ + 𝟐 π’”π’Šπ’β‘πŸπœ½ Therefore 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/(π‘‘πœƒ ))/(𝑑π‘₯/π‘‘πœƒ) 𝑑𝑦/𝑑π‘₯ = (cosβ‘πœƒ βˆ’ 2 cos⁑2πœƒ)/(βˆ’sinβ‘πœƒ " +" 2 sin⁑2πœƒ ) π’…π’š/𝒅𝒙 = (π’„π’π’”β‘πœ½ βˆ’ 𝟐 π’„π’π’”β‘πŸπœ½)/(𝟐 π’”π’Šπ’β‘πŸπœ½ " βˆ’" π’”π’Šπ’β‘πœ½ )

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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.