Check sibling questions

Ex 6.3

Ex 6.3, 1 Deleted for CBSE Board 2023 Exams

Ex 6.3,2 Deleted for CBSE Board 2023 Exams

Ex 6.3,3 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,4 Deleted for CBSE Board 2023 Exams

Ex 6.3, 5 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,6 Deleted for CBSE Board 2023 Exams

Ex 6.3,7 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,8 Deleted for CBSE Board 2023 Exams

Ex 6.3,9 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,10 Deleted for CBSE Board 2023 Exams

Ex 6.3,11 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,12 Deleted for CBSE Board 2023 Exams

Ex 6.3,13 Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (i) Deleted for CBSE Board 2023 Exams You are here

Ex 6.3, 14 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (iii) Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (iv) Important Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (v) Deleted for CBSE Board 2023 Exams

Ex 6.3,15 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,16 Deleted for CBSE Board 2023 Exams

Ex 6.3,17 Deleted for CBSE Board 2023 Exams

Ex 6.3,18 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,19 Deleted for CBSE Board 2023 Exams

Ex 6.3,20 Deleted for CBSE Board 2023 Exams

Ex 6.3,21 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,22 Deleted for CBSE Board 2023 Exams

Ex 6.3,23 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,24 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,25 Deleted for CBSE Board 2023 Exams

Ex 6.3,26 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 6.3,27 (MCQ) Deleted for CBSE Board 2023 Exams

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

 

 


Transcript

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

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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.