# Ex 9.4, 19

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

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 𝑡 seconds . Let volume of the spherical balloon = V V = 43 𝜋 𝑟3 Since volume changes at 𝑎 constant rate, ∴ 𝑑𝑉𝑑𝑡=𝑘 𝑑𝑑𝑡 43 𝜋 𝑟3 = k 43 𝜋 𝑑 𝑟3𝑑𝑡 = k 43 𝜋 3r2 𝑑𝑟𝑑𝑡 = k 4𝜋r2 𝑑𝑟𝑑𝑡 = k 4𝜋r2 𝑑𝑟 = k dt Integrating both sides 4𝜋 r2 𝑑𝑟 = k 𝑑𝑡 4𝜋 𝑟33 = kt + C At T = 0, r = 3 units 4𝜋(3)33 = k(0) + C 4𝜋(3)2 = C 4𝜋(9) = C 36𝜋 = C C = 36π Also, At T = 3, r = 6 units Putting t = 3, r = 6 and C = 36π in equation (1) 4𝜋(6)33 = 3k + 36𝜋 4𝜋(216)3 = 3k + 36𝜋 288𝜋 = 3k + 36𝜋 252𝜋 = 3k 84𝜋 = k k = 84π Putting value of k & C in equation (1) 4𝜋𝑟33 = 84𝜋𝑡 + 36𝜋 4𝜋𝑟3 = 3 84𝜋𝑡+36𝜋 4𝜋𝑟3 = 252𝜋t + 108𝜋 𝑟3 = 252𝜋𝑡+ 108𝜋 4𝜋 𝑟3 = 63t + 27 r = 63t + 27 13 ∴ Radius of the balloon after 3 seconds is 63t + 27 𝟏𝟑 units

Chapter 9 Class 12 Differential Equations

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .