Ex 6.3, 5 - Find slope of normal to the curve x = a cos^3 , y=sin^3

Ex 6.3, 5 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3, 5 - Chapter 6 Class 12 Application of Derivatives - Part 3


Question 5 Find the slope of the normal to the curve π‘₯=π‘Ž cos^3β‘πœƒ, 𝑦=π‘Ž sin3 πœƒ at πœƒ=πœ‹/4Given π‘₯=π‘Ž cos^3β‘πœƒ Differentiating w.r.t. ΞΈ 𝑑π‘₯/π‘‘πœƒ=𝑑(γ€–a cosγ€—^3β‘πœƒ )/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ=π‘Ž .𝑑(cos^3β‘πœƒ )/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ=π‘Ž . 3 cos^2β‘πœƒ. (βˆ’sinβ‘πœƒ ) 𝑑π‘₯/π‘‘πœƒ=βˆ’ 3π‘Ž sinβ‘γ€–πœƒ cos^2β‘πœƒ γ€— Similarly 𝑦=π‘Ž sin3 πœƒ Differentiating w.r.t. ΞΈ 𝑑𝑦/π‘‘πœƒ=𝑑(π‘Ž sin3 πœƒ" " )/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ=π‘Ž .𝑑(sin3 πœƒ)/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ=π‘Ž . 3 sin^2β‘πœƒ. (cosβ‘πœƒ ) 𝑑𝑦/π‘‘πœƒ= 3π‘Ž sin^2β‘γ€–πœƒ .π‘π‘œπ‘ β‘πœƒ γ€— We know that Slope of tangent = 𝑑𝑦/𝑑π‘₯ =𝑑𝑦/π‘‘πœƒΓ·π‘‘π‘₯/π‘‘πœƒ =(3π‘Ž sin^2β‘γ€–πœƒ cosβ‘πœƒ γ€—)/(βˆ’ 3π‘Ž sinβ‘γ€–πœƒ cos^2β‘πœƒ γ€— ) =(βˆ’sinβ‘πœƒ)/cosβ‘πœƒ =βˆ’tanβ‘πœƒ Putting πœƒ=πœ‹/4 β”œ 𝑑𝑦/𝑑π‘₯─|_(πœƒ = πœ‹/4)=βˆ’π‘‘π‘Žπ‘›(πœ‹/4) =βˆ’1 Now we know that Tangent is perpendicular to Normal Hence, Slope of tangent Γ— Slope of Normal = βˆ’1 βˆ’1 Γ— Slope of Normal = βˆ’1 Slope of Normal =(βˆ’1)/(βˆ’1) Slope of Normal = 1 Hence, Slope of Normal is 1

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.