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

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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 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.