Question 12 - Examples - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Question 12 If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its volume.Given Radius of sphere = r = 9 cm Error in radius = ∆r = 0.03 cm We need to find Error in calculating Volume Let Volume of sphere = V = 𝟒/𝟑 𝝅𝒓^𝟑 ∴ We need to find ∆V Now, ∆V = 𝑑𝑉/𝑑𝑟 ∆r = 𝑑(4/3 𝜋𝑟^3 )/𝑑𝑟 ∆r = 4/3 𝜋 (𝑑𝑟^3)/𝑑𝑟 ∆r = 4/3 𝜋 (3𝑟^2 )(0.03 ) = 4𝜋𝑟^2(0.03) = 4𝜋(9)2 (0.03) = 9.72𝝅 cm3 Hence, approximate error in calculating volume is 9.72𝜋 cm3
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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