Ex 5.6, 10 - Chapter 5 Class 12 Continuity Differentiability

Ex 5.6, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.6, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.6, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 4 Ex 5.6, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 5

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

Transcript

Ex 5.6, 10 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦/𝑑π‘₯, π‘₯ = π‘Ž (cosβ‘πœƒ + πœƒ sinβ‘πœƒ), 𝑦 = π‘Ž (sinβ‘πœƒ – πœƒ cosβ‘πœƒ)Here 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) Calculating π’…π’š/π’…πœ½ 𝑦 = π‘Ž (sinβ‘πœƒ – πœƒ cosβ‘πœƒ) 𝑑𝑦/π‘‘πœƒ = (𝑑(π‘Ž (sinβ‘πœƒ βˆ’ πœƒ cosβ‘πœƒ)) )/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ = a ((𝑑 (sinβ‘πœƒ βˆ’ πœƒ cosβ‘πœƒ ))/π‘‘πœƒ) 𝑑𝑦/π‘‘πœƒ = a ((𝑑 (sinβ‘πœƒ ) )/π‘‘πœƒ βˆ’ (𝑑 (πœƒ cosβ‘πœƒ ))/π‘‘πœƒ) 𝑑𝑦/π‘‘πœƒ = a (cosβ‘πœƒβˆ’ (𝑑 (πœƒ cosβ‘πœƒ ))/π‘‘πœƒ) 𝑑𝑦/π‘‘πœƒ = a (cosβ‘πœƒβˆ’((𝑑 (πœƒ) )/π‘‘πœƒ . cosβ‘πœƒ+(𝑑 (cosβ‘πœƒ ) )/π‘‘πœƒ . πœƒ)) 𝑑𝑦/π‘‘πœƒ = a (cosβ‘πœƒβˆ’(cosβ‘πœƒ+(βˆ’sinβ‘πœƒ )) πœƒ) Using Product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 𝑑𝑦/π‘‘πœƒ = a (π‘π‘œπ‘ β‘πœƒβˆ’cosβ‘πœƒ+πœƒ sinβ‘πœƒ ) 𝑑𝑦/π‘‘πœƒ = a (πœƒ sinβ‘πœƒ ) π’…π’š/π’…πœ½ = 𝒂 𝜽. π’”π’Šπ’β‘πœ½ Calculating 𝒅𝒙/π’…πœ½ π‘₯=π‘Ž (cosβ‘πœƒ+ πœƒ sinβ‘πœƒ ) 𝑑π‘₯/π‘‘πœƒ = 𝑑(π‘Ž (cosβ‘πœƒ+ πœƒ sinβ‘πœƒ ))/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = π‘Ž 𝑑(cosβ‘πœƒ+ πœƒ sinβ‘πœƒ )/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = π‘Ž (𝑑(cosβ‘πœƒ )/π‘‘πœƒ + 𝑑(πœƒ sinβ‘πœƒ )/π‘‘πœƒ) Using Product Rule As (𝑒𝑣)’ = 𝑒’𝑣 + 𝑣’𝑒 𝑑π‘₯/π‘‘πœƒ = π‘Ž (βˆ’sinβ‘πœƒ+ 𝑑(πœƒ sinβ‘πœƒ )/π‘‘πœƒ) 𝑑π‘₯/π‘‘πœƒ = π‘Ž (βˆ’sinβ‘πœƒ+(π‘‘πœƒ/π‘‘πœƒ . sinβ‘πœƒ+ 𝑑(sinβ‘πœƒ )/π‘‘πœƒ . πœƒ)) 𝑑π‘₯/π‘‘πœƒ = π‘Ž (βˆ’sinβ‘πœƒ+(sinβ‘πœƒ+cosβ‘πœƒ. πœƒ)) 𝑑π‘₯/π‘‘πœƒ = π‘Ž (βˆ’sinβ‘πœƒ+sinβ‘πœƒ+πœƒ.cosβ‘πœƒ ) 𝒅𝒙/π’…πœ½ = 𝒂 (𝜽 π’„π’π’”β‘πœ½ ) Therefore 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) 𝑑𝑦/𝑑π‘₯ = (π‘Ž (πœƒ. sinβ‘πœƒ ))/π‘Ž" " (πœƒ cosβ‘πœƒ ) 𝑑𝑦/𝑑π‘₯ = sinβ‘πœƒ/cosβ‘πœƒ π’…π’š/𝒅𝒙 = π’•π’‚π’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.