Check sibling questions

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

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


Transcript

Ex 6.3, 17 Find the points on the curve 𝑦=π‘₯3 at which the slope of the tangent is equal to the y-coordinate of the pointLet the Point be (β„Ž , π‘˜) on the Curve 𝑦=π‘₯3 Where Slope of tangent at (β„Ž , π‘˜)=π‘¦βˆ’π‘π‘œπ‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’ π‘œπ‘“ (β„Ž, π‘˜) i.e. 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_((β„Ž, π‘˜) )=π‘˜ Given 𝑦=π‘₯^3 Differentiating w.r.t.π‘₯ 𝑑𝑦/𝑑π‘₯=3π‘₯^2 ∴ Slope of tangent at (β„Ž , π‘˜) is 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_((β„Ž, π‘˜) )=3β„Ž^2 From (1) 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_((β„Ž, π‘˜) )=π‘˜ 3β„Ž^2=π‘˜ Also Point (β„Ž , π‘˜) is on the Curve 𝑦=π‘₯^3 Point (β„Ž , π‘˜) must Satisfy the Equation of Curve i.e. π‘˜=β„Ž^3 Now our equations are 3β„Ž^2=π‘˜ …(1) & π‘˜=β„Ž^3 …(2) Putting Value of π‘˜=3β„Ž^2 in (3) 3β„Ž^2=β„Ž^3 β„Ž^3βˆ’3β„Ž^2=0 β„Ž^2 (β„Žβˆ’3)=0 β„Ž^2=0 β„Ž=0 β„Žβˆ’3=0 β„Ž=3 When 𝒉=𝟎 3β„Ž^2=π‘˜ 3(0)=π‘˜ π‘˜=0 Hence, point is (0, 0) When 𝒉=πŸ‘ 3β„Ž^2=π‘˜ 3(3)^2=π‘˜ π‘˜=27 Hence, point is (3 , 27)

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.