Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 10

Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 11
Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 12

Go Ad-free

Transcript

Question 14 Find the equations of the tangent and normal to the given curves at the indicated points: (iv) 𝑦=π‘₯2 π‘Žπ‘‘ (0, 0) Given Curve is 𝑦=π‘₯^2 Differentiating w.r.t.π‘₯ 𝑑𝑦/𝑑π‘₯=2π‘₯ We know that Slope of tangent is 𝑑𝑦/𝑑π‘₯ Given point is (0 , 0) Slope of tangent at (0 , 0) 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_((0 , 0) )=2(0) =0 We know that Slope of tangent Γ— Slope if Normal =βˆ’1 0 Γ— Slope if Normal =βˆ’1 Slope if Normal =(βˆ’1)/( 0) Equation of tangent at (0 , 0) & having Slope zero is (π‘¦βˆ’0)=0(π‘₯βˆ’0) We know that Equation of line at (π‘₯1 , 𝑦1)& having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Finding equation of tangent & normal Now Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Equation of tangent at (0, 0) & Slope 0 is (π‘¦βˆ’0)=0(π‘₯βˆ’0) π‘¦βˆ’0=0 π’š=𝟎 Equation of Normal at (0, 0) & Slope (βˆ’1)/0 is (π‘¦βˆ’0)=1/0 (π‘₯βˆ’0) 0 Γ— (π‘¦βˆ’0)=1(π‘₯βˆ’0) 0=π‘₯βˆ’0 𝒙=𝟎

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo