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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


Ex 6.3, 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

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.