Last updated at April 19, 2021 by Teachoo

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Ex 6.1, 12 The radius of an air bubble is increasing at the rate of 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?Since Air Bubble is spherical Let r be the radius of bubble & V be the volume of bubble Given that Radius of an air bubble is increasing at the rate of 1/2 cm/s i.e. ๐ ๐/๐ ๐ = ๐/๐ cm/sec We need to calculate the rate is the volume of the bubble increasing when the radius is 1 cm i.e. we need to calculate ๐ ๐ฝ/๐ ๐ when r = 1 cm We know that Volume of sphere = V = ๐/๐ ฯr3 Now, ๐๐/๐๐ก = ๐(4/3 ๐๐3)/๐๐ก ๐๐/๐๐ก = 4/3 ฯ (๐ (๐3))/๐๐ก ๐๐/๐๐ก = 4/3 ฯ (๐ (๐3))/๐๐ก ๐๐/๐๐ก = 4/3 ฯ . (๐(๐3))/๐๐ก ร ๐ ๐/๐ ๐ ๐๐/๐๐ก = 4/3 ฯ . (๐(๐3))/๐๐ก ร ๐๐/๐๐ก ๐๐/๐๐ก = 4/3 ฯ .3r2 . ๐๐/๐๐ก ๐๐/๐๐ก = 4/3 ฯ . 3r2 ร 1/2 ๐๐/๐๐ก = 2๐๐^2 We need to find ๐๐/๐๐ก at r = 1 cm ๐๐/๐๐ก = 2๐ใ(1)ใ^2 ("From (1): " ๐๐/๐๐ก=1/2 cm/s) ๐ ๐ฝ/๐ ๐ = ๐๐ Since Volume is in cm3 & time is in sec โด ๐๐/๐๐ก = ๐๐ cm3/sec Hence, Volume is increasing at rate of 2๐ cm3/sec

Ex 6.1

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Chapter 6 Class 12 Application of Derivatives

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About the Author

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.