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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

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Ex 6.3, 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

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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.