Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 10

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Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 11

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

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (iv) 𝑦=π‘₯2 π‘Žπ‘‘ (0, 0) Given Curve is 𝑦=π‘₯^2 Differentiating w.r.t.π‘₯ 𝑑𝑦/𝑑π‘₯=2π‘₯ We know that Slope of tangent is 𝑑𝑦/𝑑π‘₯ Given point is (0 , 0) Slope of tangent at (0 , 0) 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_((0 , 0) )=2(0) =0 We know that Slope of tangent Γ— Slope if Normal =βˆ’1 0 Γ— Slope if Normal =βˆ’1 Slope if Normal =(βˆ’1)/( 0) Equation of tangent at (0 , 0) & having Slope zero is (π‘¦βˆ’0)=0(π‘₯βˆ’0) We know that Equation of line at (π‘₯1 , 𝑦1)& having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) 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 (0, 0) & Slope 0 is (π‘¦βˆ’0)=0(π‘₯βˆ’0) π‘¦βˆ’0=0 π’š=𝟎 Equation of Normal at (0, 0) & Slope (βˆ’1)/0 is (π‘¦βˆ’0)=1/0 (π‘₯βˆ’0) 0 Γ— (π‘¦βˆ’0)=1(π‘₯βˆ’0) 0=π‘₯βˆ’0 𝒙=𝟎

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.