Ex 6.1, 9 - A balloon has a variable radius. Find rate - Ex 6.1

Ex 6.1,9 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.1,9 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Ex 6.1, 9 A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.Since Balloon is spherical Let r be the radius of balloon . & V be the volume of balloon. We need to find rate at which balloon volume is increasing when radius is 10cm i.e. We need to find change of volume w.r.t radius when r = 10 i.e. we need to find š’…š‘½/š’…š’“ when r = 10 cm We know that Volume of sphere = V = 4/3 Ļ€r3 Now, š‘‘š‘‰/š‘‘š‘Ÿ = (š‘‘ (4/3 šœ‹š‘Ÿ3))/š‘‘š‘Ÿ š‘‘š‘‰/š‘‘š‘Ÿ = 4/3 Ļ€ š‘‘(š‘Ÿ3)/š‘‘š‘Ÿ š‘‘š‘‰/š‘‘š‘Ÿ = 4/3 Ļ€ 3š‘Ÿ^2 š‘‘š‘‰/š‘‘š‘Ÿ = 4šœ‹š‘Ÿ^2 When r = 10 š‘‘š‘‰/š‘‘š‘Ÿ = 4 Ɨ Ļ€ Ɨ (10)2 š‘‘š‘‰/š‘‘š‘Ÿ = 400Ļ€ Since volume is in cm3 & Radius is in cm So, š‘‘š‘‰/š‘‘š‘Ÿ = 400Ļ€ cm3/cm Hence, volume is increasing at the rate of 400 Ļ€ cm3/cm when r = 10 cm

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