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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

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Ex 6.1, 13 A balloon, which always remains spherical, has a variable diameter 3/2 (2π‘₯ +1). Find the rate of change of its volume with respect to π‘₯.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 = 𝑑/2 = πŸ‘/πŸ’ (2x + 1) The balloon is a spherical Volume of the balloon = 4/3 πœ‹π‘Ÿ^3 We need to find rate of change of volume with respect to x i.e. 𝑑𝑉/𝑑π‘₯ Now, 𝑑𝑉/𝑑π‘₯ = 𝑑/𝑑π‘₯ (4/3 πœ‹π‘Ÿ^3 ) = 4πœ‹/3 Γ— (π‘‘π‘Ÿ^3)/𝑑π‘₯ = 4πœ‹/3 Γ— 𝑑/𝑑π‘₯ (27/64 (2π‘₯+1)^3 ) = 9πœ‹/16 Γ— (𝑑(2π‘₯ + 1)^3)/𝑑π‘₯ = 9πœ‹/16 Γ— 3(2x + 1)2 Γ— 2 = πŸπŸ•π…/πŸ– (πŸπ’™+𝟏)^𝟐

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.