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

Transcript

Ex 6.3, 15 Find the equation of the tangent line to the curve ๐‘ฆ=๐‘ฅ2 โˆ’2๐‘ฅ+7 which is : (a) parallel to the line 2๐‘ฅโˆ’๐‘ฆ+9=0 We know that Slope of tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘ฆ=๐‘ฅ2 โˆ’2๐‘ฅ+7 Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=2๐‘ฅโˆ’2 Finding Slope of line 2๐‘ฅโˆ’๐‘ฆ+9=0 2๐‘ฅโˆ’๐‘ฆ+9=0 ๐‘ฆ=2๐‘ฅ+9 ๐‘ฆ=2๐‘ฅ+9 The Above Equation is of form ๐‘ฆ=๐‘š๐‘ฅ+๐‘ where m is Slope of line Hence, Slope of line 2๐‘ฅโˆ’๐‘ฆ+9 is 2 Now, Given tangent is parallel to 2๐‘ฅโˆ’๐‘ฆ+9=0 Slope of tangent = Slope of line 2๐‘ฅโˆ’๐‘ฆ+9 = 0 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=2 2๐‘ฅโˆ’2=2 2(๐‘ฅโˆ’1)=2 ๐‘ฅ=2 Finding y when ๐‘ฅ=2 , ๐‘ฆ=๐‘ฅ^2โˆ’2๐‘ฅ+7= (2)^2โˆ’2(2)+7=4โˆ’4+7=7 We need to find Equation of tangent passes through (2, 7) & Slope is 2 Equation of tangent is (๐‘ฆโˆ’7)=2(๐‘ฅโˆ’2) ๐‘ฆโˆ’7=2๐‘ฅโˆ’4 ๐‘ฆโˆ’2๐‘ฅโˆ’7+4=0 ๐‘ฆโˆ’2๐‘ฅโˆ’3=0 Hence Required Equation of tangent is ๐’šโˆ’๐Ÿ๐’™โˆ’๐Ÿ‘=๐ŸŽ We know that Equation of line at (๐‘ฅ1 , ๐‘ฆ1) & having Slope m is ๐‘ฆโˆ’๐‘ฆ1=๐‘š(๐‘ฅโˆ’๐‘ฅ1) Ex 6.3,15 Find the equation of the tangent line to the curve ๐‘ฆ=๐‘ฅ2โˆ’2๐‘ฅ+7 which is (b) perpendicular to the line 5๐‘ฆโˆ’15๐‘ฅ=13 We know that Slope of tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘ฆ=๐‘ฅ2 โˆ’2๐‘ฅ+7 Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=2๐‘ฅโˆ’2 Finding Slope of line 5๐‘ฆโˆ’15๐‘ฅ=13 5๐‘ฆโˆ’15๐‘ฅ=13 5๐‘ฆ=15๐‘ฅ+13 ๐‘ฆ=1/5 (15๐‘ฅ+13) ๐‘ฆ=15/5 ๐‘ฅ+13/5 ๐‘ฆ=3๐‘ฅ+13/5 Above Equation is of form ๐‘ฆ=๐‘š๐‘ฅ+๐‘ , where m is Slope of a line โˆด Slope = 3 Now, Given tangent is perpendicular to 5๐‘ฆโˆ’15๐‘ฅ=13 Slope of tangent ร— Slope of line = โ€“1 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ร— 3=โˆ’1 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(โˆ’1)/( 3) 2๐‘ฅโˆ’2=(โˆ’1)/( 3) 2๐‘ฅ=(โˆ’1)/( 3)+2 2๐‘ฅ=(โˆ’1 + 6)/3 2๐‘ฅ=5/3 ๐‘ฅ=5/6 Finding y when ๐‘ฅ=5/6 ๐‘ฆ=๐‘ฅ^2โˆ’2๐‘ฅ+7=(5/6)^2โˆ’2(5/6)+7=25/36โˆ’10/6+7=217/36 โˆด Point is (5/6 ,217/36) Equation of tangent passing through (5/6 ,217/36) & having Slope (โˆ’1)/( 3) (๐‘ฆโˆ’217/36)=(โˆ’1)/( 3) (๐‘ฅโˆ’5/6) (36๐‘ฆ โˆ’217)/36=(โˆ’1)/( 3) (๐‘ฅโˆ’5/6) 36๐‘ฆ โˆ’217=(โˆ’36)/( 3) (๐‘ฅโˆ’5/( 6)) 36๐‘ฆ โˆ’217=โˆ’12(๐‘ฅโˆ’5/( 6)) 36๐‘ฆ โˆ’217=โˆ’12๐‘ฅ+(12 ร— 5)/6 36๐‘ฆ โˆ’217=โˆ’12๐‘ฅ+10 36๐‘ฆ โˆ’217=โˆ’12๐‘ฅ+10 36๐‘ฆ+12๐‘ฅโˆ’217โˆ’10=0 ๐Ÿ‘๐Ÿ”๐’š+๐Ÿ๐Ÿ๐’™โˆ’๐Ÿ๐Ÿ๐Ÿ•=๐ŸŽ is Required Equation of tangent We know that Equation of line at (๐‘ฅ1 , ๐‘ฆ1) & having Slope m is ๐‘ฆโˆ’๐‘ฆ1=๐‘š(๐‘ฅโˆ’๐‘ฅ1)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.