Slide25.JPG

Slide26.JPG
Slide27.JPG Slide28.JPG Slide29.JPG Slide30.JPG Slide31.JPG

Subscribe to our Youtube Channel - https://you.tube/teachoo

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

Transcript

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 π‘₯=π‘Ž cosβ‘πœƒ+π‘Ž πœƒ sinβ‘πœƒ , 𝑦=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‘₯ 𝑑𝑦/𝑑π‘₯= (𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) 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) 𝑑𝑦/π‘‘πœƒ=π‘Ž cosβ‘πœƒβˆ’π‘Ž(cosβ‘πœƒ+(βˆ’sinβ‘πœƒ )πœƒ) 𝑑𝑦/π‘‘πœƒ=π‘Ž cosβ‘πœƒβˆ’(π‘Ž cosβ‘πœƒβˆ’π‘Ž πœƒ sinβ‘πœƒ ) 𝑑𝑦/π‘‘πœƒ=π‘Ž cos πœƒβˆ’π‘Ž cosβ‘πœƒ + π‘Ž . πœƒ sinβ‘πœƒ. 𝑑𝑦/π‘‘πœƒ= π‘Ž. πœƒ .sin. πœƒ Now, 𝑑𝑦/𝑑π‘₯= (π‘‘π‘¦βˆ•π‘‘πœƒ)/(𝑑π‘₯βˆ•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 𝑑𝑦/𝑑π‘₯= tanβ‘πœƒ 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’cotβ‘πœƒ Equation of normal which passes through the curve π‘₯ = a cos ΞΈ + a ΞΈ sin ΞΈ & 𝑦 = a sin ΞΈ - a ΞΈ cos ΞΈ & has slope βˆ’cotβ‘πœƒ is (π‘¦βˆ’(π‘Ž sinβ‘πœƒβˆ’π‘Ž cosβ‘πœƒ ))=βˆ’cotβ‘πœƒ(π‘₯βˆ’(π‘Ž cosβ‘πœƒ+π‘Ž πœƒ sinβ‘πœƒ )) 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’cotβ‘πœƒ Equation of normal which passes through the curve π‘₯ = a cos ΞΈ + a ΞΈ sin ΞΈ & 𝑦 = a sin ΞΈ - a ΞΈ cos ΞΈ & has slope βˆ’cotβ‘πœƒ is (π‘¦βˆ’(π‘Ž sinβ‘πœƒβˆ’π‘Ž cosβ‘πœƒ ))=βˆ’cotβ‘πœƒ(π‘₯βˆ’(π‘Ž cosβ‘πœƒ+π‘Ž πœƒ sinβ‘πœƒ )) 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œƒ+π‘Ž πœƒ cosβ‘πœƒ )=(βˆ’cosβ‘πœƒ)/sinβ‘πœƒ (π‘₯βˆ’π‘Ž cosβ‘πœƒ+π‘Ž πœƒ sinβ‘πœƒ ) sinβ‘πœƒ(π‘¦βˆ’π‘Ž sinβ‘πœƒ+π‘Ž πœƒ cosβ‘πœƒ )=βˆ’cosβ‘πœƒ(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 (sin2 πœƒ+cos2 πœƒ)=0 𝑦 sinβ‘πœƒ+π‘₯ cosβ‘πœƒβˆ’π‘Ž (1)=0 π‘₯ cosβ‘πœƒ+𝑦 sinβ‘πœƒ βˆ’π‘Ž = 0 Thus, Equation of normal is π‘₯ cosβ‘πœƒ+𝑦 sinβ‘πœƒ βˆ’π‘Ž = 0 Finding Distance of Normal from Origin Equation of normal is π‘₯ cosβ‘πœƒ+𝑦 sinβ‘πœƒ βˆ’π‘Ž = 0 Comparing with aπ‘₯ + b𝑦 + c = 0 where a = cos ΞΈ, b = sin ΞΈ & c = βˆ’ a We need to find distance of normal from origin i.e. π‘₯1 = 0 & 𝑦1 = 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 We know that Distance of line ax + by + c = 0 from point (x1, y1) is d = |π‘Žπ‘₯1 + 𝑏𝑦1 +𝑐|/√(𝐴^2 + 𝐡^2 ) d = |β‰Ίπ‘Ž|/1 d = π‘Ž/1 d = a (constant) Hence, distance of normal from origin is a constant. Hence proved.

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.