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 (2š‘„ +1). Find the rate of change of its volume with respect to š‘„.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 = š‘‘/2 = šŸ‘/šŸ’ (2x + 1) The balloon is a spherical Volume of the balloon = 4/3 šœ‹š‘Ÿ^3 We need to find rate of change of volume with respect to x i.e. š‘‘š‘‰/š‘‘š‘„ Now, š‘‘š‘‰/š‘‘š‘„ = š‘‘/š‘‘š‘„ (4/3 šœ‹š‘Ÿ^3 ) = 4šœ‹/3 Ɨ (š‘‘š‘Ÿ^3)/š‘‘š‘„ = 4šœ‹/3 Ɨ š‘‘/š‘‘š‘„ (27/64 (2š‘„+1)^3 ) = 9šœ‹/16 Ɨ (š‘‘(2š‘„ + 1)^3)/š‘‘š‘„ = 9šœ‹/16 Ɨ 3(2x + 1)2 Ɨ 2 = šŸšŸ•š…/šŸ– (šŸš’™+šŸ)^šŸ

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