1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
  3. Ex 6.3

Ex 6.3,17 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by

Ex 6.3, 17 - Find points on y = x3 at which slope of tangent - Finding point when tangent is parallel/ perpendicular

Slide41.JPG
Slide42.JPG

  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
    3. Ex 6.3
    Ex 6.4 →
Ask Download

Transcript

Ex 6.3,17 Find the points on the curve = 3 at which the slope of the tangent is equal to the y-coordinate of the point Let the Point be , on the Curve = 3 Where Slope of tangent at , = , i.e. , = Given = 3 Differentiating w.r.t. =3 2 Slope of tangent at , is , =3 2 From (1) , = 3 2 = Also Point , is on the Curve = 3 Point , must Satisfy the Equation of Curve i.e. = 3 Now our equations are 3 2 = (1) & = 3 (2) Putting Value of =3 2 in (3) 3 2 = 3 3 3 2 =0 2 3 =0 Hence the Required Point on the Curve are , & ,

Chapter 6 Class 12 Application of Derivatives
Serial order wise

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.