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Example 25 - If radius of a sphere is 9 cm with error 0.03 cm

Example 25 - Chapter 6 Class 12 Application of Derivatives - Part 2


Transcript

Example 25 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 = βˆ†r = 0.03 cm We need to find Error in calculating Volume Let Volume of sphere = V = πŸ’/πŸ‘ 𝝅𝒓^πŸ‘ ∴ We need to find βˆ†V Now, βˆ†V = 𝑑𝑉/π‘‘π‘Ÿ βˆ†r = 𝑑(4/3 πœ‹π‘Ÿ^3 )/π‘‘π‘Ÿ βˆ†r = 4/3 πœ‹ (π‘‘π‘Ÿ^3)/π‘‘π‘Ÿ βˆ†r = 4/3 πœ‹ (3π‘Ÿ^2 )(0.03 ) = 4πœ‹π‘Ÿ^2(0.03) = 4πœ‹(9)2 (0.03) = 9.72𝝅 cm3 Hence, approximate error in calculating volume is 9.72πœ‹ cm3

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.