Example 36 - Find dy/dx, if  x = a (theta + sin theta), y = a - Derivatives in parametric form

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Example 36 Find 𝑑𝑦/𝑑π‘₯ , if π‘₯ = π‘Ž (πœƒ+sinβ‘πœƒ), 𝑦 = π‘Ž (1 – cosβ‘πœƒ) 𝑑𝑦/𝑑π‘₯ = 𝑑𝑦/𝑑π‘₯ Γ— π‘‘πœƒ/π‘‘πœƒ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑦/π‘‘πœƒ Γ— π‘‘πœƒ/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) Therefore 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) 𝑑𝑦/𝑑π‘₯ = π‘Ž" " (sinβ‘πœƒ )/π‘Ž" " (1 +γ€– cosγ€—β‘πœƒ ) 𝑑𝑦/𝑑π‘₯ = sinβ‘πœƒ/(1 +γ€– cosγ€—β‘πœƒ ) 𝑑𝑦/𝑑π‘₯ = (γ€–πŸ π’”π’Šπ’γ€—β‘γ€– 𝜽/πŸγ€— .γ€– 𝒄𝒐𝒔〗⁑〖 𝜽/πŸγ€—)/(1 + γ€–πŸ 〖𝒄𝒐𝒔〗^𝟐 γ€—β‘γ€–πœ½/πŸγ€— βˆ’ 𝟏) 𝑑𝑦/𝑑π‘₯ = (γ€–2 sin〗⁑〖 πœƒ/2γ€— .γ€– cos〗⁑〖 πœƒ/2γ€—)/γ€–2 cos^2 γ€—β‘γ€–πœƒ/2γ€— 𝑑𝑦/𝑑π‘₯ = (sin⁑〖 πœƒ/2γ€— )/γ€–cos γ€—β‘γ€–πœƒ/2γ€— π’…π’š/𝒅𝒙 = π­πšπ§β‘γ€–πœ½/πŸγ€—

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