Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 4

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Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 5

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

  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: (ii) ๐‘ฆ=๐‘ฅ4 โˆ’6๐‘ฅ3+13๐‘ฅ2 โˆ’10๐‘ฅ+5 ๐‘Ž๐‘ก (1, 3) Given Curve is ๐‘ฆ=๐‘ฅ4 โˆ’6๐‘ฅ3+13๐‘ฅ2 โˆ’10๐‘ฅ+5 Differentiating w.r.t x ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4๐‘ฅ^3โˆ’18๐‘ฅ^2+26๐‘ฅโˆ’10 Now Point Given is (1 , 3) Finding slope of tangent at (1,3) ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_((1 , 3) )=4(1)^3โˆ’18(1)^2+26(1)โˆ’10 =4โˆ’18+26โˆ’10 =2 โˆด Slope of tangent at (1, 3) =2 Also, We know that Slope of tangent ร— Slope of Normal =โˆ’1 2ร— Slope of Normal =โˆ’1 Slope of Normal = (โˆ’1)/( 2) Hence Slope of tangent at (1 , 3)=2 & Slope of Normal at (1 , 3)=(โˆ’1)/( 2) Finding equation of tangent & normal Now Equation of line at (๐‘ฅ1 , ๐‘ฆ1) & having Slope m is ๐‘ฆโˆ’๐‘ฆ1=๐‘š(๐‘ฅโˆ’๐‘ฅ1) Equation of tangent at (1, 3) & Slope 2 is (๐‘ฆโˆ’3)=2(๐‘ฅโˆ’1) ๐‘ฆโˆ’3=2๐‘ฅโˆ’2 ๐‘ฆ=2๐‘ฅโˆ’2+3 ๐’š=๐Ÿ๐’™+๐Ÿ Equation of Normal at (1, 3) & Slope (โˆ’1)/2 is (๐‘ฆโˆ’3)=(โˆ’1)/( 2) (๐‘ฅโˆ’1) 2(๐‘ฆโˆ’3)=โˆ’1(๐‘ฅโˆ’1) 2๐‘ฆโˆ’6=โˆ’๐‘ฅ+1 2๐‘ฆ+๐‘ฅโˆ’7=0 ๐’™+๐Ÿ๐’šโˆ’๐Ÿ•=๐ŸŽ

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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.