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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise


Misc 5 Show that the normal at any point ΞΈ to the curve x = a cos πœƒ + a πœƒ sin πœƒ, y = a sin πœƒ – a πœƒ cos πœƒ is at a constant distance from the origin.Given curve 𝒙=π‘Ž cosβ‘πœƒ+π‘Ž πœƒ sinβ‘πœƒ , π’š=π‘Ž sinβ‘πœƒβ€“ π‘Ž πœƒ cosβ‘πœƒ We need to show distance of a normal from (0, 0) is constant First , calculating Equation of Normal We know that Slope of tangent is 𝑑𝑦/𝑑π‘₯ π’…π’š/𝒅𝒙= (π’…π’š/π’…πœ½)/(𝒅𝒙/π’…πœ½) Finding 𝒅𝒙/π’…πœ½ Given π‘₯=π‘Ž cosβ‘πœƒ+π‘Ž πœƒ sinβ‘πœƒ Diff. w.r.t ΞΈ 𝑑π‘₯/π‘‘πœƒ= 𝑑(π‘Ž cosβ‘πœƒ + π‘Ž πœƒ sinβ‘πœƒ )/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ= (𝑑(π‘Ž cosβ‘πœƒ))/π‘‘πœƒ + (𝑑(π‘Žπœƒ sinβ‘πœƒ))/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ = π‘Ž (βˆ’sinβ‘πœƒ )+π‘Ž (𝑑(πœƒ sinβ‘πœƒ))/π‘‘πœƒ Using product rule (u v)’ = u’ v + v’ u 𝑑π‘₯/π‘‘πœƒ=βˆ’ a sin ΞΈ+a (π‘‘πœƒ/π‘‘πœƒ sin ΞΈ+ (𝑑(𝑠𝑖𝑛 πœƒ))/π‘‘πœƒ ΞΈ) 𝑑π‘₯/π‘‘πœƒ=βˆ’ a sin ΞΈ+a ( sin ΞΈ+ΞΈ cos⁑〖θ γ€— ) 𝑑π‘₯/π‘‘πœƒ=βˆ’ a sin ΞΈ+a sin ΞΈ+π‘Ž ΞΈ cos⁑〖θ γ€— 𝒅𝒙/π’…πœ½=𝒂 𝜽 π’„π’π’”β‘γ€–πœ½ γ€— Finding π’…π’š/π’…πœ½ Given 𝑦=π‘Ž π‘ π‘–π‘›β‘πœƒβˆ’π‘Ž πœƒ π‘π‘œπ‘ β‘πœƒ Diff. w.r.t ΞΈ 𝑑𝑦/π‘‘πœƒ= 𝑑(π‘Ž π‘ π‘–π‘›β‘πœƒβˆ’π‘Ž πœƒ π‘π‘œπ‘ β‘πœƒ)/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ=a cos ΞΈβˆ’a (π‘‘πœƒ/π‘‘πœƒ cos ΞΈ+ (𝑑(π‘π‘œπ‘  πœƒ))/π‘‘πœƒ ΞΈ) 𝑑𝑦/π‘‘πœƒ=a cos ΞΈβˆ’a ( cos ΞΈβˆ’ΞΈ sin⁑〖θ γ€— ) 𝑑𝑦/π‘‘πœƒ=a cos ΞΈβˆ’a cos ΞΈ+π‘Ž ΞΈ sin⁑〖θ γ€— π’…π’š/π’…πœ½=𝒂 𝜽 π’”π’Šπ’β‘γ€–πœ½ γ€— Now, π’…π’š/𝒅𝒙= (π’…π’šβˆ•π’…πœ½)/(π’…π’™βˆ•π’…πœ½) 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 We know that Slope of tangent of Γ— Slope of normal = βˆ’1 tan ΞΈ Γ— Slope of normal = βˆ’1 Slope of normal = (βˆ’1)/tanβ‘πœƒ Slope of normal =βˆ’π’„π’π’•β‘πœ½ Equation of normal which passes through the curve 𝒙 = a cos ΞΈ + a ΞΈ sin ΞΈ & π’š = a sin ΞΈ βˆ’ a ΞΈ cos ΞΈ & has slope βˆ’π’„π’π’•β‘πœ½ is We know that Equation of line passing through (π‘₯1 , 𝑦1) & having slope m is (π‘¦βˆ’π‘¦1) = m(π‘₯βˆ’π‘₯1) (π‘¦βˆ’(π‘Ž sinβ‘πœƒβˆ’π‘Ž cosβ‘πœƒ ))=βˆ’πœπ¨π­β‘πœ½(π‘₯βˆ’(π‘Ž cosβ‘πœƒ+π‘Ž πœƒ sinβ‘πœƒ )) (π‘¦βˆ’π‘Ž sinβ‘πœƒ+π‘Ž πœƒ cosβ‘πœƒ )=(βˆ’π’„π’π’”β‘πœ½)/π’”π’Šπ’β‘πœ½ (π‘₯βˆ’π‘Ž cosβ‘πœƒ+π‘Ž πœƒ sinβ‘πœƒ ) 𝐬𝐒𝐧⁑𝜽(π‘¦βˆ’π‘Ž sinβ‘πœƒ+π‘Ž πœƒ cosβ‘πœƒ )=βˆ’πœπ¨π¬β‘πœ½(π‘₯βˆ’π‘Ž cosβ‘πœƒβˆ’π‘Ž πœƒ sinβ‘πœƒ ) 𝑦 sinβ‘πœƒβˆ’π‘Ž sin2 πœƒ+π‘Ž πœƒ .cosβ‘πœƒ sinβ‘πœƒ=βˆ’π‘₯ cosβ‘πœƒ+π‘Ž cos2 πœƒ+π‘Ž πœƒsinβ‘πœƒ cosβ‘πœƒ 𝑦 sinβ‘πœƒβˆ’π‘Ž sin2 πœƒ+π‘₯ cosβ‘πœƒβˆ’π‘Ž cos2 πœƒ=π‘Ž πœƒ sinβ‘πœƒ cosβ‘πœƒβˆ’π‘Ž πœƒ sinβ‘πœƒ cosβ‘πœƒ 𝑦 sinβ‘πœƒ+π‘₯ cosβ‘πœƒβˆ’π‘Ž sin2 πœƒβˆ’π‘Ž cos2 πœƒ=0 𝑦 sinβ‘πœƒ+ π‘₯ cosβ‘πœƒβˆ’π‘Ž (π’”π’Šπ’πŸ 𝜽+π’„π’π’”πŸ 𝜽)=0 𝑦 sinβ‘πœƒ+π‘₯ cosβ‘πœƒβˆ’π‘Ž (𝟏)=0 𝒙 π’„π’π’”β‘πœ½+π’š π’”π’Šπ’β‘πœ½ βˆ’π’‚ = 𝟎 We know that Distance of line ax + by + c = 0 from point (x1, y1) is d = |π’‚π’™πŸ + π’ƒπ’šπŸ +𝒄|/√(𝑨^𝟐 + 𝑩^𝟐 ) Finding Distance of Normal from Origin Equation of normal is 𝒙 π’„π’π’”β‘πœ½+π’š π’”π’Šπ’β‘πœ½ βˆ’π’‚ = 𝟎 Comparing with aπ‘₯ + b𝑦 + c = 0 ∴ a = cos ΞΈ, b = sin ΞΈ & c = βˆ’ a We need to find distance of normal from origin i.e. π’™πŸ = 0 & π’šπŸ = 0 𝑑= |cosβ‘γ€–πœƒ(0) + sinβ‘γ€–πœƒ(0) βˆ’ π‘Žγ€— γ€— |/√(cos^2β‘πœƒ + sin^2β‘πœƒ ) d = |0 + 0 βˆ’ π‘Ž|/√1 d = |βˆ’π‘Ž|/1 d = π‘Ž/1 d = a d = Constant Hence, distance of normal from origin is a constant. Hence proved.

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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.