# Ex 6.3, 14 (iii) - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Aug. 19, 2021 by

Last updated at Aug. 19, 2021 by

Transcript

Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (iii) π¦=π₯3 ππ‘ (1, 1)Given Curve is π¦=π₯3 Differentiating w.r.t.π₯ ππ¦/ππ₯=3π₯^2 We know that Slope of tangent is ππ¦/ππ₯ Given point is (1 , 3) Slope of tangent ππ¦/ππ₯ at (1 , 3) is γππ¦/ππ₯βγ_((1 , 3) )=3(1)^2 =3 We know that Slope of tangent Γ Slope of Normal =β1 3 Γ Slope of Normal =β1 Slope of Normal = (β1)/( 3) Hence Slope of tangent at (1 , 1)=3 Slope of Normal at ( 1 , 1)=(β1)/( 3) Finding equation of tangent & normal Now Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) Equation of tangent at (1, 1) & Slope 3 is (π¦ β1)=3(π₯β1) π¦ β1=3π₯β3 π¦=3π₯β3+1 π=ππβπ Equation of Normal at (1, 1) & Slope (β1)/3 is (π¦β1)=(β1)/( 3) (π₯β1) 3(π¦β1)=β(π₯β1) 3π¦β3=βπ₯+1 3π¦+π₯β3β1=0 π+ππβπ=π

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Ex 6.3, 14 (i)

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Ex 6.3, 14 (iii) You are here

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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.