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Misc 1 - Using differentials, find approximate value: 17/81 - Finding approximate value of numbers

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Misc 1 Using differentials, find the approximate value of each of the following: (a) 17﷮81﷯﷯﷮ 1﷮4﷯﷯ 17﷮81﷯﷯﷮ 1﷮4﷯﷯ = 17﷯﷮ 1﷮4﷯﷯﷮ 81﷯﷮ 1﷮4﷯﷯﷯ = 17﷯﷮ 1﷮4﷯﷯﷮3﷯ Let x = 16 and ∆𝑥=1 Since y = 𝑥﷮ 1﷮4﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 𝑥﷮ 1﷮4﷯﷯﷯﷮𝑑𝑥﷯ = 1﷮4﷯ 𝑥﷮ −3﷮4﷯﷯ = 1﷮4 𝑥﷮ 3﷮4﷯﷯﷯ Now, ∆𝑦 = 𝑑𝑦﷮𝑑𝑥﷯ ∆𝑥 ∆𝑦 = 1﷮4 𝑥﷮ 3﷮4﷯﷯﷯﷯∆𝑥 Putting values ∆𝑦 = 1﷮4 16﷯﷮ 3﷮4﷯﷯﷯ (1) = 1﷮4 16﷯﷮ 1﷮4﷯﷯﷯﷮3﷯﷯ = 1﷮4 2﷯﷮3﷯﷯ = 1﷮32﷯ Also, ∆𝑦 = f(x + ∆𝑥) − f(x) ∆𝑦 = x + ∆𝑥﷯﷮ 1﷮4﷯﷯ − 𝑥﷮ 1﷮4﷯﷯ ∆𝑦 = 16 + 1﷯﷮ 1﷮4﷯﷯ − (16)﷮ 1﷮4﷯﷯ ∆𝑦 = 17﷯﷮ 1﷮4﷯﷯ − (16)﷮ 1﷮4﷯﷯ 17﷯﷮ 1﷮4﷯﷯ = ∆𝑦+ 16﷯﷮ 1﷮4﷯﷯ 17﷯﷮ 1﷮4﷯﷯ = ∆𝑦+2 17﷯﷮ 1﷮4﷯﷯ = 1﷮32﷯+2 17﷯﷮ 1﷮4﷯﷯ = 0.03125 + 2 17﷯﷮ 1﷮4﷯﷯ = 2.03125 Now, Approximate value of 17﷮81﷯﷯﷮ 1﷮4﷯﷯ 17﷮81﷯﷯﷮ 1﷮4﷯﷯ = (17) 1﷮4﷯﷮3﷯ = 2.03125﷮3﷯ = 0.677 Hence approximate value of 17﷮81﷯﷯﷮ 1﷮4﷯﷯ is 0.677 Misc 1 Using differentials, find the approximate value of each of the following: (b) (33)﷮− 1﷮5﷯﷯ (33)﷮− 1﷮5﷯﷯ = 1﷮ 33﷯﷮ 1﷮5﷯﷯﷯ Let y = 𝑥﷮ 1﷮5﷯﷯ & also let 𝑥 = 32 & ∆ 𝑥 = 1 Now, 𝑦 = 𝑥﷮ 1﷮5﷯﷯ Diff w.r.t x 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 ( 𝑥﷮ 1﷮5﷯﷯)﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 1﷮5﷯ 𝑥﷮ 1 ﷮5﷯ −1﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮5﷯ 𝑥﷮ −4﷮5﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯= 1﷮ 5𝑥﷮ 4﷮5﷯﷯﷯ Using ∆y = 𝑑𝑦﷮𝑑𝑥﷯ ∆𝑥 ∆y = 1﷮ 5𝑥﷮ 4﷮5﷯﷯﷯ ∆𝑥 Putting values ∆y = 1﷮ 5(32)﷮ 4﷮5﷯﷯﷯ × (1) ∆y = 1﷮ 5(32)﷮ 4﷮5﷯﷯﷯ ∆y = 1﷮ 5(2)﷮5 × 4﷮5﷯﷯﷯ ∆y = 1﷮5 2﷮4﷯﷯﷯ ∆y = 1﷮5 × 16﷯ ∆y = 1﷮80﷯ We know that ∆y = f(x + ∆x) – f(x) So, ∆y = (x+ Δx)﷮ 1﷮5﷯﷯ −𝑥﷮ 1﷮5﷯﷯ Putting values 1﷮80﷯= (32+1)﷮ 1﷮5﷯﷯ − 32﷯﷮ 1﷮5﷯﷯ 1﷮80﷯= (33)﷮ 1﷮5﷯﷯ − 2﷯ ﷮5 × 1﷮5﷯﷯ 1﷮80﷯= (33)﷮ 1﷮5﷯﷯ − 2 1﷮80﷯+2= (33)﷮ 1﷮5﷯﷯ 1 + 160﷮80﷯= (33)﷮ 1﷮5﷯﷯ 161﷮80﷯= (33)﷮ 1﷮5﷯﷯ (33)﷮ 1﷮5﷯﷯ = 161﷮80﷯ But we need 1﷮ 33﷯﷮ 1﷮5﷯﷯﷯ So 1﷮ 33﷯﷮ 1﷮5﷯﷯﷯= 1﷮ 161﷮80﷯﷯ 1﷮ 33﷯﷮ 1﷮5﷯﷯﷯= 80﷮161﷯ 1﷮ 33﷯﷮ 1﷮5﷯﷯﷯= 0.497 33﷯﷮ −1﷮5﷯﷯= 0.497 Thus, the approximate value of 33﷯﷮ −1﷮5﷯﷯ 𝑖𝑠 𝟎.𝟒𝟗𝟕

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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