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

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

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 𝑑𝑦𝑑𝑥=4𝑥3−18𝑥2+26𝑥−10 Now Point Given is 0 ,5 Hence 𝑥=0 , 𝑦=5 Putting 𝑥=0 in (1) Slope of tangent at 0 , 5 𝑑𝑦𝑑𝑥│0 , 5=403−1802+260−10 𝑑𝑦𝑑𝑥│0 , 5=0−0+0−10 𝑑𝑦𝑑𝑥=−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=110 Hence Slope of tangent at 0 , 5=−10 & Slope of Normal at 0 , 5=110 Finding equation of tangent & normal Ex 6.3,14 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 𝑑𝑦𝑑𝑥=4𝑥3−18𝑥2+26𝑥−10 Now Point Given is 1 , 3 Finding slope of tangent at (1,3) 𝑑𝑦𝑑𝑥│1 , 3=413−1812+261−10 =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 Ex 6.3,14 Find the equations of the tangent and normal to the given curves at the indicated points: (iii) 𝑦=𝑥3 𝑎𝑡 (1, 1) Given Curve is 𝑦=𝑥3 Differentiating w.r.t.𝑥 𝑑𝑦𝑑𝑥=3𝑥2 We know that Slope of tangent is 𝑑𝑦𝑑𝑥 Given point is 1 , 3 Slope of tangent 𝑑𝑦𝑑𝑥 at 1 , 3 is 𝑑𝑦𝑑𝑥│1 , 3=312 =3 We know that Slope of tangent × Slope of Normal =−1 3 × Slope of Normal =−1 Slope of Normal = −1 3 Hence Slope of tangent at 1 , 1=3 Slope of Normal at 1 , 1=−1 3 Finding equation of tangent & normal Ex 6.3,14 Find the equations of the tangent and normal to the given curves at the indicated points: (iv) 𝑦=𝑥2 𝑎𝑡 (0, 0) Given Curve is 𝑦=𝑥2 Differentiating w.r.t.𝑥 𝑑𝑦𝑑𝑥=2𝑥 We know that Slope of tangent is 𝑑𝑦𝑑𝑥 Given point is 0 , 0 Slope of tangent at 0 , 0 𝑑𝑦𝑑𝑥│0 , 0=20 =0 We know that Slope of tangent × Slope if Normal =−1 0 × Slope if Normal =−1 Slope if Normal =−1 0 Equation of tangent at 0 , 0 & having Slope zero is 𝑦−0=0𝑥−0 Finding equation of tangent & normal Ex 6.3,14 Find the equations of the tangent and normal to the given curves at the indicated points: (v) 𝑥=cos𝑡, 𝑦=sin𝑡 𝑎𝑡 𝑡= 𝜋4 At 𝒕= 𝝅𝟒 x = cos 𝜋4 = 12 y = sin 𝜋4 = 12 ∴ At 𝑡=𝜋4 , the point is 12 ,12 Finding 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥=𝑑𝑦𝑑𝑡𝑑𝑥𝑑𝑡 Now, Hence, 𝑑𝑦𝑑𝑥=𝑑𝑦𝑑𝑡𝑑𝑥𝑑𝑡 𝑑𝑦𝑑𝑥=cos𝑡−sin𝑡 𝑑𝑦𝑑𝑥=−cot𝑡 At 𝒕=𝝅𝟒 𝑑𝑦𝑑𝑥│𝑡=𝜋4=−𝑐𝑜𝑡𝜋4 =−1 ∴ Slope of tangent at 12 ,12 is – 1 We know that Slope of tangent × Slope of Normal =−1 −1 × Slope of Normal =−1 Slope of Normal =−1−1 Slope of Normal =1 Hence Slope of tangent is – 1 & Slope of Normal is 1 Finding equation of tangent & normal

Ex 6.3

Ex 6.3,1

Ex 6.3,2

Ex 6.3,3

Ex 6.3,4

Ex 6.3,5 Important

Ex 6.3,6

Ex 6.3,7 Important

Ex 6.3,8

Ex 6.3,9

Ex 6.3,10

Ex 6.3,11

Ex 6.3,12 Important

Ex 6.3,13

Ex 6.3,14 You are here

Ex 6.3,15 Important

Ex 6.3,16

Ex 6.3,17

Ex 6.3,18

Ex 6.3,19

Ex 6.3,20

Ex 6.3,21

Ex 6.3,22

Ex 6.3,23

Ex 6.3,24

Ex 6.3,25

Ex 6.3,26 Important

Ex 6.3,27

Chapter 6 Class 12 Application of Derivatives

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.