# Ex 6.3,21 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Aug. 19, 2021 by Teachoo

Ex 6.3

Ex 6.3, 1

Ex 6.3,2

Ex 6.3,3 Important

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 Important

Ex 6.3,10

Ex 6.3,11 Important

Ex 6.3,12

Ex 6.3,13

Ex 6.3, 14 (i)

Ex 6.3, 14 (ii) Important

Ex 6.3, 14 (iii)

Ex 6.3, 14 (iv) Important

Ex 6.3, 14 (v)

Ex 6.3,15 Important

Ex 6.3,16

Ex 6.3,17

Ex 6.3,18 Important

Ex 6.3,19

Ex 6.3,20

Ex 6.3,21 Important You are here

Ex 6.3,22

Ex 6.3,23 Important

Ex 6.3,24 Important

Ex 6.3,25

Ex 6.3,26 (MCQ) Important

Ex 6.3,27 (MCQ)

Last updated at Aug. 19, 2021 by Teachoo

Hello! We hope that the questions explained by Teachoo are helping you for your Board exams. If Teachoo has been of any help to you, would you consider making a donation to support us? Please click on this link to support us.

Hello! We hope that the questions explained by Teachoo are helping you for your Board exams. If Teachoo has been of any help to you, would you consider making a donation to support us? Please click on this link to support us.

Ex 6.3, 21 Find the equation of the normal to the curve 𝑦=𝑥^3+2𝑥+6 which are parallel to the line 𝑥+14𝑦+4=0.Let (ℎ , 𝑘) be the point on the Curve at which Normal is to be taken Given Curve is 𝑦=𝑥^3+2𝑥+6 Since point (ℎ , 𝑘) is on the Curve ∴ (ℎ , 𝑘) will satisfies the Equation of Curve Putting 𝑥=ℎ , 𝑦=𝑘 𝑘=ℎ^3+2ℎ+6 We know that Slope of a tangent to the Curve is 𝑑𝑦/𝑑𝑥 𝑦=𝑥^3+2𝑥+6 Differentiating w.r.t. 𝑥 𝑑𝑦/𝑑𝑥=3𝑥^2+2 Since tangent to be taken from (ℎ , 𝑘) Slope of tangent at (ℎ , 𝑘) is 〖𝑑𝑦/𝑑𝑥│〗_((ℎ, 𝑘) )=3ℎ^2+2 We know that Slope of tangent × Slope of Normal =−1 (3ℎ^2+2) × Slope of Normal =−1 Slope of Normal = (−1)/(3ℎ^2 + 2) Also, Given that Normal is parallel to the line 𝑥+14𝑦+4=0 If two lines are parallel then slopes are equal ⇒ Slopes of Normal = Slope of line 𝑥+14𝑦+4=0 Now, line is 𝑥+14𝑦+4=0 14𝑦=−𝑥−4 𝑦=(− 𝑥 − 4)/14 𝑦=((−1)/14)𝑥−(4/14) The above equation is of the form 𝑦=𝑚𝑥+𝑐 where m is slope ∴ Slope of line 𝑥+14𝑦+4=0 is (−1)/14 Now, Slope of Normal = Slope of line 𝑥+14𝑦+4=0 (−1)/(3ℎ^2 + 2)=(−1)/( 14) 1/(3ℎ^2 + 2)=1/( 14) 14=3ℎ^2+2 3ℎ^2+2=14 3ℎ^2=14−2 3ℎ^2=12 ℎ^2=12/3 ℎ^2=4 ℎ=±√4 ℎ=±2 When 𝒉=𝟐 𝑘=ℎ^3+2ℎ+6 𝑘=(2)^3+2(2)+6 𝑘=8+4+6 𝑘=18 ∴ Point is (𝟐 , 𝟏𝟖) When 𝒉=−𝟐 𝑘=ℎ^3+2ℎ+6 𝑘=(−2)^3+2(−2)+6 𝑘=−8−4+6 𝑘=− 6 ∴ Point is (−𝟐, −𝟔) Finding equation of normal We know that Equation of line at (𝑥1 , 𝑦1)& having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Equation of 1st Normal at (2 , 18) & having Slope (−1)/14 is (𝑦−18)=(−1)/14 (𝑥−2) 14(𝑦−18)=−(𝑥−2) 14𝑦−252=−𝑥+2 14𝑦+𝑥−252−2=0 𝒙+𝟏𝟒𝒚−𝟐𝟓𝟒=𝟎 Equation of 2nd Normal at (−2 , −6) & having Slope (−1)/14 is (𝑦−(−6))=(−1)/14 (𝑥−(−2)) 𝑦+6=(−1)/14 (𝑥+2) 14𝑦+84=−𝑥−2 14𝑦+𝑥+84+2=0 𝒙+𝟏𝟒𝒚+𝟖𝟔=𝟎