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Tangents and Normals (using Differentiation)
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Tangents and Normals (using Differentiation)
Last updated at May 29, 2023 by Teachoo
Question 14 Find the equations of the tangent and normal to the given curves at the indicated points: (ii) 𝑦=𝑥4 −6𝑥3+13𝑥2 −10𝑥+5 𝑎𝑡 (1, 3) Given Curve is 𝑦=𝑥4 −6𝑥3+13𝑥2 −10𝑥+5 Differentiating w.r.t x 𝑑𝑦/𝑑𝑥=4𝑥^3−18𝑥^2+26𝑥−10 Now Point Given is (1 , 3) Finding slope of tangent at (1,3) 〖𝑑𝑦/𝑑𝑥│〗_((1 , 3) )=4(1)^3−18(1)^2+26(1)−10 =4−18+26−10 =2 ∴ Slope of tangent at (1, 3) =2 Also, We know that Slope of tangent × Slope of Normal =−1 2× Slope of Normal =−1 Slope of Normal = (−1)/( 2) Hence Slope of tangent at (1 , 3)=2 & Slope of Normal at (1 , 3)=(−1)/( 2) Finding equation of tangent & normal Now Equation of line at (𝑥1 , 𝑦1) & having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Equation of tangent at (1, 3) & Slope 2 is (𝑦−3)=2(𝑥−1) 𝑦−3=2𝑥−2 𝑦=2𝑥−2+3 𝒚=𝟐𝒙+𝟏 Equation of Normal at (1, 3) & Slope (−1)/2 is (𝑦−3)=(−1)/( 2) (𝑥−1) 2(𝑦−3)=−1(𝑥−1) 2𝑦−6=−𝑥+1 2𝑦+𝑥−7=0 𝒙+𝟐𝒚−𝟕=𝟎