Last updated at Dec. 16, 2024 by Teachoo
Question 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (i) √25.3Let y = √𝒙 Thus, √(𝟐𝟓.𝟑) = y + ∆𝒚 Here, ∆𝒚 = 𝒅𝒚/𝒅𝒙 △x where x = 25 & △x = 0.3 Since y = √𝑥 𝒅𝒚/𝒅𝒙 = (𝑑(√𝑥))/𝑑𝑥 = 𝟏/(𝟐√𝒙) Now, ∆𝒚 = 𝒅𝒚/𝒅𝒙 △x = 1/(2√𝑥) △x Putting x = 25 & △x = 0.3 = 1/(2√25) (0.3) = 1/(2 × 5) × 0.3 = 0.3/10 = 0.03 Therefore, √25.3 = y + ∆𝑦 Putting values √25.3 =√25+0.03 √(𝟐𝟓. 𝟑)=𝟓. 𝟎𝟑 Hence, approximate value of √25.3 is 5.03
Approximations (using Differentiation)
Question 1 (ii)
Question 1 (iii)
Question 1 (iv)
Question 1 (v) Important
Question 1 (vi)
Question 1 (vii)
Question 1 (viii)
Question 1 (ix)
Question 1 (x)
Question 1 (xi) Important
Question 1 (xii)
Question 1 (xiii)
Question 1 (xiv) Important
Question 1 (xv)
Question 2
Question 3 Important
Question 4
Question 5 Important
Question 6
Question 7
Question 8 (MCQ) Important
Question 9 (MCQ)
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo