Ex 6.1, 13 - A balloon has a variable diameter 3/2 (2x + 1)

Ex 6.1,13 - Chapter 6 Class 12 Application of Derivatives - Part 2

  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Ex 6.1, 13 A balloon, which always remains spherical, has a variable diameter 3/2 (2๐‘ฅ +1). Find the rate of change of its volume with respect to ๐‘ฅ.Let d be the diameter of the balloon Given that Diameter = d = 3/2 (2x + 1) Let r be the radius of the balloon r = ๐‘‘/2 = ๐Ÿ‘/๐Ÿ’ (2x + 1) The balloon is a spherical Volume of the balloon = 4/3 ๐œ‹๐‘Ÿ^3 We need to find rate of change of volume with respect to x i.e. ๐‘‘๐‘‰/๐‘‘๐‘ฅ Now, ๐‘‘๐‘‰/๐‘‘๐‘ฅ = ๐‘‘/๐‘‘๐‘ฅ (4/3 ๐œ‹๐‘Ÿ^3 ) = 4๐œ‹/3 ร— (๐‘‘๐‘Ÿ^3)/๐‘‘๐‘ฅ = 4๐œ‹/3 ร— ๐‘‘/๐‘‘๐‘ฅ (27/64 (2๐‘ฅ+1)^3 ) = 9๐œ‹/16 ร— (๐‘‘(2๐‘ฅ + 1)^3)/๐‘‘๐‘ฅ = 9๐œ‹/16 ร— 3(2x + 1)2 ร— 2 = ๐Ÿ๐Ÿ•๐…/๐Ÿ– (๐Ÿ๐’™+๐Ÿ)^๐Ÿ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.