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Ex 6.3, 14 - Find equations of tangent and normal to - Ex 6.3

Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 3

 

 


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Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (i) 𝑦=𝑥4 −6𝑥3+13𝑥2 −10𝑥+5 𝑎𝑡 (0, 5) 𝑦=𝑥4 −6𝑥3+13𝑥2 −10𝑥+5 Differentiating w.r.t. 𝑥 𝑑𝑦/𝑑𝑥=4𝑥^3−18𝑥^2+26𝑥−10 Now Point Given is (0 ,5) Hence 𝑥=0 , 𝑦=5 Putting 𝑥=0 in (1) Slope of tangent at (0 , 5) 〖𝑑𝑦/𝑑𝑥│〗_((0, 5) )=4(0)^3−18(0)^2+26(0)−10 〖𝑑𝑦/𝑑𝑥│〗_((0, 5) )=0−0+0−10 𝑑𝑦/𝑑𝑥=−10 Hence, Slope of tangent =−10 We know that Slope of tangent × Slope of Normal =−1 −10 ×"Slope of Normal "=−1 "Slope of Normal" =(−1)/(−10)=1/10 Hence Slope of tangent at (0, 5)=−10 & Slope of Normal at (0, 5)=1/10 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 (0, 5) & Slope –10 is (𝑦−5)=−10(𝑥−0) 𝑦−5=−10𝑥 10𝑥+𝑦−5=0 𝟏𝟎𝒙+𝒚=𝟓 Equation of Normal at (0, 5) & Slope 1/10 is (𝑦−5)=1/10 (𝑥−0) 𝑦−5=1/10 𝑥 10(𝑦−5)=𝑥 10𝑦−50=𝑥 𝒙−𝟏𝟎𝒚+𝟓𝟎=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.