Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

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 ๐‘‘๐‘Ÿ/๐‘‘๐‘ก = 1/2 cm/s The Bubble is a sphere Volume of bubble = Volume of sphere = 4/3 ๐œ‹๐‘Ÿ^3 We need to find the rate at which volume of the bubble is increasing when radius is 1 cm โ€ฆ(1) i.e. ๐‘‘๐‘ฃ/๐‘‘๐‘ก at r = 1 cm ๐‘‘๐‘ฃ/๐‘‘๐‘ก = ๐‘‘(4/3 ๐œ‹๐‘Ÿ^3 )/๐‘‘๐‘ก = 4/3 ๐œ‹ (๐‘‘๐‘Ÿ^3)/๐‘‘๐‘ก = 4/3 ๐œ‹ (๐‘‘๐‘Ÿ^3)/๐‘‘๐‘Ÿ ๐‘‘๐‘Ÿ/๐‘‘๐‘ก = 4๐œ‹/3 ใ€–3๐‘Ÿใ€—^2 ๐‘‘๐‘Ÿ/๐‘‘๐‘ก = 4๐œ‹๐‘Ÿ^2 1/2 = 2๐œ‹๐‘Ÿ^2 (From (1): ๐‘‘๐‘Ÿ/๐‘‘๐‘ก=1/2) When radius is 1cm ๐‘‘๐‘ฃ/๐‘‘๐‘ก = 2๐œ‹(1)^2 ๐‘‘๐‘ฃ/๐‘‘๐‘ก = 2๐œ‹ Since volume is in cm3 & time is in sec So ๐‘‘๐‘ฃ/๐‘‘๐‘ก = ๐Ÿ๐… cm3/sec Hence, volume is increasing at rate of 2๐œ‹ cm3/sec when radius = 1cm

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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.