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Ex 6.3

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Ex 6.3, 6 - Find slope of normal x = 1 - a cos, y = b cos2

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

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Transcript

Ex 6.3, 6 Find the slope of the normal to the curve π‘₯=1βˆ’π‘Ž sinβ‘πœƒ , 𝑦 =𝑏 cos^2 πœƒ at πœƒ= πœ‹/2Slope of tangent is 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=(𝑑𝑦/π‘‘πœƒ)/(𝑑π‘₯/π‘‘πœƒ) 𝒙=πŸβˆ’π’‚ π’”π’Šπ’β‘πœ½ Differentiating w.r.t. ΞΈ 𝑑π‘₯/π‘‘πœƒ=𝑑(1 βˆ’ π‘Ž sinβ‘πœƒ )/π‘‘πœƒ 𝑑π‘₯/π‘‘πœƒ=0βˆ’π‘Ž cosβ‘πœƒ 𝑑π‘₯/π‘‘πœƒ=βˆ’π‘Ž cosβ‘πœƒ π’š=𝒃 〖𝒄𝒐𝒔〗^𝟐⁑𝜽 Differentiating w.r.t. ΞΈ 𝑑𝑦/π‘‘πœƒ=𝑑(𝑏 cos^2β‘πœƒ )/π‘‘πœƒ 𝑑𝑦/π‘‘πœƒ=𝑏 . 2 cosβ‘γ€–πœƒ . 𝑑(cosβ‘πœƒ )/π‘‘πœƒγ€— 𝑑𝑦/π‘‘πœƒ=2𝑏 cosβ‘γ€–πœƒ . (βˆ’sinβ‘πœƒ )γ€— 𝑑𝑦/π‘‘πœƒ=βˆ’2𝑏 sinβ‘γ€–πœƒ cosβ‘πœƒ γ€— Now, 𝑑𝑦/𝑑π‘₯=(𝑑𝑦/π‘‘πœƒ)/(π‘‘πœƒ/π‘‘πœƒ) 𝑑𝑦/𝑑π‘₯=(βˆ’ 2𝑏 sinβ‘γ€–πœƒ cosβ‘πœƒ γ€—)/(βˆ’ π‘Ž cosβ‘πœƒ ) 𝑑𝑦/𝑑π‘₯=(2𝑏 sinβ‘πœƒ)/π‘Ž We need to find Slope of tangent at πœƒ=πœ‹/2 Putting πœƒ=πœ‹/2 β”œ 𝑑𝑦/𝑑π‘₯─|_(πœƒ = πœ‹/2)=2𝑏/π‘Ž 𝑠𝑖𝑛(πœ‹/2) =2𝑏/π‘Ž (1) We know that Tangent is Perpendicular to Normal Hence Slope of tangent Γ— Slope of Normal =βˆ’1 Slope of Normal =(βˆ’ 1)/(π‘†π‘™π‘œπ‘π‘’ π‘œπ‘“ π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘ ) Slope of Normal =(βˆ’ 1)/( 𝑑𝑦/𝑑π‘₯) Slope of Normal =βˆ’1 Γ— π‘Ž/2𝑏 Slope of Normal =(βˆ’π‘Ž)/2𝑏 Hence Slope of Normal at is (βˆ’π’‚)/πŸπ’ƒ

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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.