Tangents and Normals (using Differentiation)

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Chapter 6 Class 12 Application of Derivatives
Serial order wise

Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 7

Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 8
Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 9


Transcript

Question 14 Find the equations of the tangent and normal to the given curves at the indicated points: (iii) 𝑦=π‘₯3 π‘Žπ‘‘ (1, 1)Given Curve is 𝑦=π‘₯3 Differentiating w.r.t.π‘₯ 𝑑𝑦/𝑑π‘₯=3π‘₯^2 We know that Slope of tangent is 𝑑𝑦/𝑑π‘₯ Given point is (1 , 3) Slope of tangent 𝑑𝑦/𝑑π‘₯ at (1 , 3) is 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_((1 , 3) )=3(1)^2 =3 We know that Slope of tangent Γ— Slope of Normal =βˆ’1 3 Γ— Slope of Normal =βˆ’1 Slope of Normal = (βˆ’1)/( 3) Hence Slope of tangent at (1 , 1)=3 Slope of Normal at ( 1 , 1)=(βˆ’1)/( 3) Finding equation of tangent & normal Now Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Equation of tangent at (1, 1) & Slope 3 is (𝑦 βˆ’1)=3(π‘₯βˆ’1) 𝑦 βˆ’1=3π‘₯βˆ’3 𝑦=3π‘₯βˆ’3+1 π’š=πŸ‘π’™βˆ’πŸ Equation of Normal at (1, 1) & Slope (βˆ’1)/3 is (π‘¦βˆ’1)=(βˆ’1)/( 3) (π‘₯βˆ’1) 3(π‘¦βˆ’1)=βˆ’(π‘₯βˆ’1) 3π‘¦βˆ’3=βˆ’π‘₯+1 3𝑦+π‘₯βˆ’3βˆ’1=0 𝒙+πŸ‘π’šβˆ’πŸ’=𝟎

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.