Check sibling questions

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


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

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

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.