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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Ex 9.4, 19 The volume of spherical balloon being inflated changes at a constant rate. if initially its radius is 3 units and after 3 seconds it is 6 units . Find the radius of balloon after ๐‘ก seconds . Let volume of the spherical balloon = V V = 4/3 ๐œ‹๐‘Ÿ^3 Since volume changes at ๐‘Ž constant rate, โˆด ๐‘‘๐‘‰/๐‘‘๐‘ก=๐‘˜ ๐‘‘/๐‘‘๐‘ก (4/3 " " ๐œ‹๐‘Ÿ^3 ) = k 4/3 ๐œ‹ (๐‘‘๐‘Ÿ^3)/๐‘‘๐‘ก = k 4/3 ๐œ‹ 3r2 ๐‘‘๐‘Ÿ/๐‘‘๐‘ก = k 4๐œ‹r2 ๐‘‘๐‘Ÿ/๐‘‘๐‘ก = k 4๐œ‹r2 ๐‘‘๐‘Ÿ = k dt Integrating both sides 4๐œ‹โˆซ1โ–’ใ€–"r2 " ๐‘‘๐‘Ÿ" = k " ใ€— โˆซ1โ–’๐‘‘๐‘ก (4๐œ‹๐‘Ÿ^3)/3 = kt + C At T = 0, r = 3 units ใ€–4๐œ‹(3)ใ€—^3/3 = k(0) + C ใ€–"4" ๐œ‹(3)ใ€—^2 = C "4" ๐œ‹(9) = C 36๐œ‹ = C C = 36ฯ€ โ€ฆ(1) Also, At T = 3, r = 6 units Putting t = 3, r = 6 and C = 36ฯ€ in equation (1) (4๐œ‹๐‘Ÿ^3)/3 = kt + C ใ€–4๐œ‹(6)ใ€—^3/3 = 3k + 36๐œ‹ (4๐œ‹(216))/3 = 3k + 36๐œ‹ 288๐œ‹ = "3k + 36" ๐œ‹ 252๐œ‹ = "3k " 84๐œ‹ = "k" k = 84ฯ€ Putting value of k & C in equation (1) (4๐œ‹๐‘Ÿ^3)/3 = kt + C ใ€–4๐œ‹๐‘Ÿใ€—^3/3 = 84๐œ‹๐‘ก + 36๐œ‹ ใ€–"4" ๐œ‹๐‘Ÿใ€—^3 = 3[84๐œ‹๐‘ก+36๐œ‹] ใ€–"4" ๐œ‹๐‘Ÿใ€—^3 = 252๐œ‹t + 108๐œ‹ ๐‘Ÿ^3 = (252๐œ‹๐‘ก+ 108๐œ‹" " )/4๐œ‹ ๐‘Ÿ^3 = 63t + 27 r = ("63t + 27" )^(1/3) โˆด Radius of the balloon after t seconds is ("63t + 27" )^(๐Ÿ/๐Ÿ‘) units

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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.