# Misc 1 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 1 Using differentials, find the approximate value of each of the following: (a) 1781 14 1781 14 = 17 14 81 14 = 17 143 Let x = 16 and ∆𝑥=1 Since y = 𝑥 14 𝑑𝑦𝑑𝑥 = 𝑑 𝑥 14𝑑𝑥 = 14 𝑥 −34 = 14 𝑥 34 Now, ∆𝑦 = 𝑑𝑦𝑑𝑥 ∆𝑥 ∆𝑦 = 14 𝑥 34∆𝑥 Putting values ∆𝑦 = 14 16 34 (1) = 14 16 143 = 14 23 = 132 Also, ∆𝑦 = f(x + ∆𝑥) − f(x) ∆𝑦 = x + ∆𝑥 14 − 𝑥 14 ∆𝑦 = 16 + 1 14 − (16) 14 ∆𝑦 = 17 14 − (16) 14 17 14 = ∆𝑦+ 16 14 17 14 = ∆𝑦+2 17 14 = 132+2 17 14 = 0.03125 + 2 17 14 = 2.03125 Now, Approximate value of 1781 14 1781 14 = (17) 143 = 2.031253 = 0.677 Hence approximate value of 1781 14 is 0.677 Misc 1 Using differentials, find the approximate value of each of the following: (b) (33)− 15 (33)− 15 = 1 33 15 Let y = 𝑥 15 & also let 𝑥 = 32 & ∆ 𝑥 = 1 Now, 𝑦 = 𝑥 15 Diff w.r.t x 𝑑𝑦𝑑𝑥= 𝑑 ( 𝑥 15)𝑑𝑥 𝑑𝑦𝑑𝑥= 15 𝑥 1 5 −1 𝑑𝑦𝑑𝑥 = 15 𝑥 −45 𝑑𝑦𝑑𝑥= 1 5𝑥 45 Using ∆y = 𝑑𝑦𝑑𝑥 ∆𝑥 ∆y = 1 5𝑥 45 ∆𝑥 Putting values ∆y = 1 5(32) 45 × (1) ∆y = 1 5(32) 45 ∆y = 1 5(2)5 × 45 ∆y = 15 24 ∆y = 15 × 16 ∆y = 180 We know that ∆y = f(x + ∆x) – f(x) So, ∆y = (x+ Δx) 15 −𝑥 15 Putting values 180= (32+1) 15 − 32 15 180= (33) 15 − 2 5 × 15 180= (33) 15 − 2 180+2= (33) 15 1 + 16080= (33) 15 16180= (33) 15 (33) 15 = 16180 But we need 1 33 15 So 1 33 15= 1 16180 1 33 15= 80161 1 33 15= 0.497 33 −15= 0.497 Thus, the approximate value of 33 −15 𝑖𝑠 𝟎.𝟒𝟗𝟕

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.