Check sibling questions

Ex 9.4, 19 - Volume of spherical balloon being inflated changes

Ex 9.4, 19 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.4, 19 - Chapter 9 Class 12 Differential Equations - Part 3
Ex 9.4, 19 - Chapter 9 Class 12 Differential Equations - Part 4

Solve all your doubts with Teachoo Black (new monthly pack available now!)


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 = 4/3 𝜋𝑟^3 Since volume changes at 𝑎 constant rate, ∴ 𝑑𝑉/𝑑𝑡=𝑘 𝑑/𝑑𝑡 (4/3 " " 𝜋𝑟^3 ) = k 4/3 𝜋 (𝑑𝑟^3)/𝑑𝑡 = k 4/3 𝜋 3r2 𝑑𝑟/𝑑𝑡 = k 4𝜋r2 𝑑𝑟/𝑑𝑡 = k 4𝜋r2 𝑑𝑟 = k dt Integrating both sides 4𝜋∫1▒〖"r2 " 𝑑𝑟" = k " 〗 ∫1▒𝑑𝑡 (4𝜋𝑟^3)/3 = kt + C At T = 0, r = 3 units 〖4𝜋(3)〗^3/3 = k(0) + C 〖"4" 𝜋(3)〗^2 = C "4" 𝜋(9) = C 36𝜋 = C C = 36π …(1) Also, At T = 3, r = 6 units Putting t = 3, r = 6 and C = 36π in equation (1) (4𝜋𝑟^3)/3 = kt + C 〖4𝜋(6)〗^3/3 = 3k + 36𝜋 (4𝜋(216))/3 = 3k + 36𝜋 288𝜋 = "3k + 36" 𝜋 252𝜋 = "3k " 84𝜋 = "k" k = 84π Putting value of k & C in equation (1) (4𝜋𝑟^3)/3 = kt + C 〖4𝜋𝑟〗^3/3 = 84𝜋𝑡 + 36𝜋 〖"4" 𝜋𝑟〗^3 = 3[84𝜋𝑡+36𝜋] 〖"4" 𝜋𝑟〗^3 = 252𝜋t + 108𝜋 𝑟^3 = (252𝜋𝑡+ 108𝜋" " )/4𝜋 𝑟^3 = 63t + 27 r = ("63t + 27" )^(1/3) ∴ Radius of the balloon after t seconds is ("63t + 27" )^(𝟏/𝟑) units

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.