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Misc 5 - Show that normal at any point is at constant distance

Misc 5 - Chapter 6 Class 12 Application of Derivatives - Part 2
Misc 5 - Chapter 6 Class 12 Application of Derivatives - Part 3 Misc 5 - Chapter 6 Class 12 Application of Derivatives - Part 4 Misc 5 - Chapter 6 Class 12 Application of Derivatives - Part 5 Misc 5 - Chapter 6 Class 12 Application of Derivatives - Part 6 Misc 5 - Chapter 6 Class 12 Application of Derivatives - Part 7 Misc 5 - Chapter 6 Class 12 Application of Derivatives - Part 8

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.