Ex 5.6, 2 - Find dy/dx, x = a cos, y = b cos - Chapter 5 - Ex 5.6

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Ex 5.6, 2 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦/𝑑π‘₯, π‘₯=π‘Ž cosβ‘πœƒ, 𝑦=𝑏 cosβ‘πœƒ π‘₯=π‘Ž cosβ‘πœƒ, 𝑦=𝑏 cosβ‘πœƒ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑦/𝑑π‘₯ Γ— π‘‘πœƒ/π‘‘πœƒ (β–ˆ("Multiplying Numerator" @"& Denominator by d" πœƒ)) 𝑑𝑦/𝑑π‘₯ = 𝑑𝑦/π‘‘πœƒ Γ— π‘‘πœƒ/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) Calculating π’…π’š/π’…πœ½ 𝑦 = 𝑏 cosβ‘πœƒ 𝑑𝑦/π‘‘πœƒ = 𝑑(𝑏 π‘π‘œπ‘ β‘πœƒ)/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ = b . (βˆ’sinβ‘πœƒ ) 𝑑𝑦/π‘‘πœƒ = βˆ’ b sin πœƒ Calculating 𝒅𝒙/π’…πœ½ π‘₯=π‘Ž cosβ‘πœƒ 𝑑π‘₯/π‘‘πœƒ = 𝑑(π‘Ž cosβ‘πœƒ )/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = π‘Ž . (βˆ’sinβ‘πœƒ ) 𝑑π‘₯/π‘‘πœƒ = βˆ’π‘Ž sinβ‘πœƒ Therefore 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) 𝑑𝑦/𝑑π‘₯ = (βˆ’ b sin πœƒ)/(βˆ’ π‘Ž sinβ‘πœƒ ) π’…π’š/𝒅𝒙 = 𝒃/𝒂

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.