# Ex 6.1,12 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 by Teachoo

Transcript

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? Let r be the radius of air bubble Given ๐๐/๐๐ก = 1/2 cm/s The Bubble is a sphere Volume of bubble = Volume of sphere = 4/3 ๐๐^3 We need to find the rate at which volume of the bubble is increasing when radius is 1 cm โฆ(1) i.e. ๐๐ฃ/๐๐ก at r = 1 cm ๐๐ฃ/๐๐ก = ๐(4/3 ๐๐^3 )/๐๐ก = 4/3 ๐ (๐๐^3)/๐๐ก = 4/3 ๐ (๐๐^3)/๐๐ ๐๐/๐๐ก = 4๐/3 ใ3๐ใ^2 ๐๐/๐๐ก = 4๐๐^2 1/2 = 2๐๐^2 (From (1): ๐๐/๐๐ก=1/2) When radius is 1cm ๐๐ฃ/๐๐ก = 2๐(1)^2 ๐๐ฃ/๐๐ก = 2๐ Since volume is in cm3 & time is in sec So ๐๐ฃ/๐๐ก = ๐๐ cm3/sec Hence, volume is increasing at rate of 2๐ cm3/sec when radius = 1cm

Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.