# Ex 6.3,15 - Chapter 6 Class 12 Application of Derivatives

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 6.3,15 Find the equation of the tangent line to the curve = 2 2 +7 which is : (a) parallel to the line 2 +9=0 We know that Slope of tangent is = 2 2 +7 Differentiating w.r.t. =2 2 Finding Slope of line 2 +9=0 2 +9=0 =2 +9 =2 +9 The Above Equation is of the form = + where m is Slope of line Hence Slope of line 2 +9 is 2 Now, Given tangent is parallel to 2 +9=0 Slope of tangent = Slope of line 2 +9 = 0 =2 2 2=2 2 1 =2 =2 Finding y when =2 , = 2 2 +7= 2 2 2 2 +7=4 4+7=7 We need to find Equation of tangent passes through 2 , 7 & Slope is 2 Equation of tangent is 7 =2 2 7=2 4 2 7+4=0 2 3=0 Hence Required Equation of tangent parallel to = Ex 6.3,15 Find the equation of the tangent line to the curve = 2 2 +7 which is (b) perpendicular to the line 5 15 =13 We know that Slope of tangent is = 2 2 +7 Differentiating w.r.t. =2 2 Finding Slope of line 5 15 =13 5 15 =13 5 =15 +13 = 1 5 15 +13 = 15 5 + 13 5 =3 + 13 5 Above Equation is of the form = + , where m is Slope of a line Slope = 3 Now, Given tangent is perpendicular to 5 15 =13 Slope of tangent Slope of line = 1 3= 1 = 1 3 2 2= 1 3 2 = 1 3 +2 2 = 1 + 6 3 2 = 5 3 = 5 6 Finding y when = 5 6 = 2 2 +7= 5 6 2 2 5 6 +7= 25 36 10 6 +7= 217 36 Point is 5 6 , 217 36 Equation of tangent passing through 5 6 , 217 36 & having Slope 1 3 217 36 = 1 3 5 6 36 217 36 = 1 3 5 6 36 217= 36 3 5 6 36 217= 12 5 6 36 217= 12 + 12 5 6 36 217= 12 +10 36 217= 12 +10 36 +12 217 10=0 + = is Required Equation of tangent

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Chapter 6 Class 12 Application of Derivatives

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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.