Ex 6.3, 17 - Find points on y = x3 at which slope of tangent

Ex 6.3,17 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,17 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.3,17 - Chapter 6 Class 12 Application of Derivatives - Part 4

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

Transcript

Ex 6.3, 17 Find the points on the curve ๐‘ฆ=๐‘ฅ3 at which the slope of the tangent is equal to the y-coordinate of the pointLet the Point be (โ„Ž , ๐‘˜) on the Curve ๐‘ฆ=๐‘ฅ3 Where Slope of tangent at (โ„Ž , ๐‘˜)=๐‘ฆโˆ’๐‘๐‘œ๐‘œ๐‘Ÿ๐‘‘๐‘–๐‘›๐‘Ž๐‘ก๐‘’ ๐‘œ๐‘“ (โ„Ž, ๐‘˜) i.e. ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_((โ„Ž, ๐‘˜) )=๐‘˜ Given ๐‘ฆ=๐‘ฅ^3 Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=3๐‘ฅ^2 โˆด Slope of tangent at (โ„Ž , ๐‘˜) is ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_((โ„Ž, ๐‘˜) )=3โ„Ž^2 From (1) ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_((โ„Ž, ๐‘˜) )=๐‘˜ 3โ„Ž^2=๐‘˜ Also Point (โ„Ž , ๐‘˜) is on the Curve ๐‘ฆ=๐‘ฅ^3 Point (โ„Ž , ๐‘˜) must Satisfy the Equation of Curve i.e. ๐‘˜=โ„Ž^3 Now our equations are 3โ„Ž^2=๐‘˜ โ€ฆ(1) & ๐‘˜=โ„Ž^3 โ€ฆ(2) Putting Value of ๐‘˜=3โ„Ž^2 in (3) 3โ„Ž^2=โ„Ž^3 โ„Ž^3โˆ’3โ„Ž^2=0 โ„Ž^2 (โ„Žโˆ’3)=0 โ„Ž^2=0 โ„Ž=0 โ„Žโˆ’3=0 โ„Ž=3 When ๐’‰=๐ŸŽ 3โ„Ž^2=๐‘˜ 3(0)=๐‘˜ ๐‘˜=0 Hence, point is (0, 0) When ๐’‰=๐Ÿ‘ 3โ„Ž^2=๐‘˜ 3(3)^2=๐‘˜ ๐‘˜=27 Hence, point is (3 , 27)

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
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.