# Ex 9.4, 19 - Chapter 9 Class 12 Differential Equations

Last updated at Nov. 14, 2019 by Teachoo

Last updated at Nov. 14, 2019 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

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Chapter 9 Class 12 Differential Equations

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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.